ASM Exam LC Study Manual (90 Days)

Study Manual for

CAS Exam LC Models for Life Contingencies First Edition Third Printing

by Abraham Weishaus, Ph.D., F.S.A., CFA, M.A.A.A. Note: NO RETURN IF OPENED

Study Manual for

CAS Exam LC Models for Life Contingencies First Edition Third Printing

by Abraham Weishaus, Ph.D., F.S.A., CFA, M.A.A.A. Note: NO RETURN IF OPENED

TO OUR READERS: Please check A.S.M.’s web site at www.studymanuals.com for errata and updates. If you have any comments or reports of errata, please e-mail us at [email protected].

©Copyright 2014 by Actuarial Study Materials (A.S.M.), PO Box 69, Greenland, NH 03840. All rights reserved. Reproduction in whole or in part without express written permission from the publisher is strictly prohibited.

Contents 1

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1 1 2 3 3 4 4 5 7 9 10 14

Survival Distributions: Probability Functions 2.1 Probability notation . . . . . . . . . . . . . 2.2 Actuarial notation . . . . . . . . . . . . . . 2.3 Life tables . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . .

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19 19 22 23 25 31

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Survival Distributions: Force of Mortality Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

37 41 51

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Survival Distributions: Mortality Laws 4.1 Mortality laws that may be used for human mortality . . . . . 4.1.1 Gompertz’s law . . . . . . . . . . . . . . . . . . . . . . . 4.1.2 Makeham’s law . . . . . . . . . . . . . . . . . . . . . . . 4.1.3 Weibull Distribution . . . . . . . . . . . . . . . . . . . . 4.2 Mortality laws for exam questions . . . . . . . . . . . . . . . . 4.2.1 Exponential distribution, or constant force of mortality 4.2.2 Uniform distribution . . . . . . . . . . . . . . . . . . . . 4.2.3 Beta distribution . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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61 61 63 64 64 65 65 65 67 69 72

Survival Distributions: Moments 5.1 Complete . . . . . . . . . . . . 5.1.1 General . . . . . . . . . 5.1.2 Special mortality laws 5.2 Curtate . . . . . . . . . . . . . Exercises . . . . . . . . . . . . Solutions . . . . . . . . . . . .

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77 77 77 79 82 86 95

Survival Distributions: Percentiles and Recursions 6.1 Percentiles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

107 107

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Probability Review 1.1 Functions and moments . . . . . . . . . 1.2 Probability distributions . . . . . . . . . 1.2.1 Bernoulli distribution . . . . . . 1.2.2 Uniform distribution . . . . . . . 1.2.3 Exponential distribution . . . . . 1.3 Variance . . . . . . . . . . . . . . . . . . 1.4 Normal approximation . . . . . . . . . . 1.5 Conditional probability and expectation 1.6 Conditional variance . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . .

LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

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CONTENTS

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6.2

Recursive formulas for life expectancy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

108 109 114

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Survival Distributions: Fractional Ages Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

121 126 130

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Insurance: Annual—Moments 8.1 Review of Financial Mathematics . . . 8.2 Moments of annual insurances . . . . 8.3 Standard insurances and notation . . . 8.4 Illustrative Life Table . . . . . . . . . . 8.5 Constant force and uniform mortality 8.6 Normal approximation . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . .

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139 139 140 141 143 145 147 148 161

Insurance: Continuous—Moments—Part 1 9.1 Definitions and general formulas . . . 9.2 Constant force of mortality . . . . . . . Exercises . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . .

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173 173 174 180 188

10 Insurance: Continuous—Moments—Part 2 10.1 Uniform survival function . . . . . . . . . . . . . 10.2 Other mortality functions . . . . . . . . . . . . . 10.2.1 Integrating at n e −ct (Gamma Integrands) 10.3 Variance of endowment insurance . . . . . . . . 10.4 Normal approximation . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . .

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197 197 199 199 201 202 203 209

11 Insurance: Probabilities and Percentiles 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . 11.2 Probabilities for Continuous Insurance Variables 11.3 Probabilities for Discrete Variables . . . . . . . . 11.4 Percentiles . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . .

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219 219 220 222 223 226 230

12 Insurance: Recursive Formulas Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

237 239 242

13 Annuities: Continuous, Expectation 13.1 Whole life annuity . . . . . . . . . . . 13.2 Temporary and deferred life annuities 13.3 n-year certain-and-life annuity . . . . Exercises . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . .

247 248 250 253 255 260

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LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

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CONTENTS

14 Annuities: Discrete, Expectation 14.1 Annuities-due . . . . . . . . . 14.2 Annuities-immediate . . . . . 14.3 Actuarial Accumulated Value Exercises . . . . . . . . . . . . Solutions . . . . . . . . . . . .

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267 267 271 273 274 286

15 Annuities: Variance 15.1 Whole Life and Temporary Life Annuities . . . . . . . . . . . 15.2 Other Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . 15.3 Typical Exam Questions . . . . . . . . . . . . . . . . . . . . . 15.4 Combinations of Annuities and Insurances with No Variance Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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297 297 299 300 302 303 313

16 Annuities: Probabilities and Percentiles 16.1 Probabilities for continuous annuities . . . . . . 16.2 Distribution functions of annuity present values 16.3 Probabilities for discrete annuities . . . . . . . . 16.4 Percentiles . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . .

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329 329 332 333 334 336 341

17 Annuities: Recursive Formulas Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

349 351 353

18 Premiums: Net Premiums for Fully Continuous Insurances 18.1 Future loss . . . . . . . . . . . . . . . . . . . . . . . . . . 18.2 Net premium . . . . . . . . . . . . . . . . . . . . . . . . 18.3 Expected value of future loss . . . . . . . . . . . . . . . 18.4 International Actuarial Premium Notation . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . .

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357 357 358 360 362 362 369

19 Premiums: Net Premiums from Life Tables Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

377 378 382

20 Premiums: Net Premiums from Formulas Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

387 389 397

21 Premiums: Variance of Future Loss, Continuous Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

407 410 415

22 Premiums: Variance of Future Loss, Discrete Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

423 425 430

LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

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CONTENTS

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23 Premiums: Probabilities and Percentiles of Future Loss 23.1 Probabilities . . . . . . . . . . . . . . . . . . . . . . . 23.1.1 Fully continuous insurances . . . . . . . . . . 23.1.2 Discrete insurances . . . . . . . . . . . . . . . 23.1.3 Annuities . . . . . . . . . . . . . . . . . . . . 23.2 Percentiles . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . .

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437 437 437 441 441 444 445 448

24 Multiple Lives: Joint Life Probabilities 24.1 Introduction . . . . . . . . . . . . . 24.2 Independence . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . .

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455 455 456 458 463

25 Multiple Lives: Last Survivor Probabilities Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

467 471 476

26 Multiple Lives: Moments Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

481 486 490

27 Multiple Decrement Models: Probabilities 27.1 Introduction . . . . . . . . . . . . . . . . . . . . 27.2 Life Tables . . . . . . . . . . . . . . . . . . . . . 27.3 Examples of Multiple Decrement Probabilities 27.4 Discrete Insurances . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . .

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497 497 499 501 502 505 515

28 Multiple Decrement Models: Forces of Decrement Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

523 525 532

29 Multi-State Models (Markov Chains): Probabilities Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

541 549 557

30 Multi-State Models (Markov Chains): Premiums and Reserves 30.1 Prototype Example . . . . . . . . . . . . . . . . . . . . . . . 30.2 Insurances . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.3 Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.4 Premiums and Reserves . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

565 565 565 568 570 572 581

Practice Exams 1

Practice Exam 1

LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

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2

Practice Exam 2

597

3

Practice Exam 3

603

4

Practice Exam 4

609

5

Practice Exam 5

615

6

Practice Exam 6

621

7

Practice Exam 7

627

8

Practice Exam 8

633

9

Practice Exam 9

639

10 Practice Exam 10

645

Appendices A Solutions to the Practice Exams Solutions for Practice Exam 1 . . Solutions for Practice Exam 2 . . Solutions for Practice Exam 3 . . Solutions for Practice Exam 4 . . Solutions for Practice Exam 5 . . Solutions for Practice Exam 6 . . Solutions for Practice Exam 7 . . Solutions for Practice Exam 8 . . Solutions for Practice Exam 9 . . Solutions for Practice Exam 10 . .

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653 653 658 663 668 673 679 685 690 696 702

B Solutions to Old Exams B.1 Solutions to CAS Exam 3, Spring 2005 . . B.2 Solutions to CAS Exam 3, Fall 2005 . . . . B.3 Solutions to CAS Exam 3, Spring 2006 . . B.4 Solutions to CAS Exam 3, Fall 2006 . . . . B.5 Solutions to CAS Exam 3, Spring 2007 . . B.6 Solutions to CAS Exam 3, Fall 2007 . . . . B.7 Solutions to CAS Exam 3L, Spring 2008 . B.8 Solutions to CAS Exam 3L, Fall 2008 . . . B.9 Solutions to CAS Exam 3L, Spring 2009 . B.10 Solutions to CAS Exam 3L, Fall 2009 . . . B.11 Solutions to CAS Exam 3L, Spring 2010 . B.12 Solutions to CAS Exam 3L, Fall 2010 . . . B.13 Solutions to CAS Exam 3L, Spring 2011 . B.14 Solutions to CAS Exam 3L, Fall 2011 . . . B.15 Solutions to CAS Exam 3L, Spring 2012 . B.16 Solutions to SOA Exam MLC, Spring 2012 B.17 Solutions to CAS Exam 3L, Fall 2012 . . . B.18 Solutions to SOA Exam MLC, Fall 2012 . .

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709 709 713 717 720 723 727 730 733 735 738 741 744 747 749 752 755 757 760

LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

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CONTENTS

viii

B.19 B.20 B.21 B.22 B.23 B.24 B.25 B.26

Solutions to CAS Exam 3L, Spring 2013 . Solutions to SOA Exam MLC, Spring 2013 Solutions to CAS Exam 3L, Fall 2013 . . . Solutions to SOA Exam MLC, Fall 2013 . . Solutions to CAS Exam LC, Spring 2014 . Solutions to SOA Exam MLC, Spring 2014 Solutions to CAS Exam LC, Fall 2014 . . . Solutions to SOA Exam MLC, Fall 2014 . .

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C Lessons Corresponding to Questions on Released and Practice Exams

LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

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762 766 771 775 778 784 786 790 793

Preface Welcome to the new CAS life contingencies exam! You will be studying probability of death and amount of time until death, and the evaluation of payments made only while one is alive or only at death. Such payment streams come up not only in the life insurance/annuity industry, but in the property/casualty industry as well when damages may involve payment to someone for life. The concepts are also useful in a more general context, when evaluating time until a device fails. Thus they can be used to estimate the cost of a warranty. The exam has 15 questions and is 1.5 hours long. The material in this manual has been largely taken from the corresponding manual for the SOA’s MLC exam. However, the weights put on the material by the CAS differ from the SOA weights. Accordingly, you should concentrate your efforts on a small part of this manual and speed through the rest. Here is the estimated distribution of questions on the exam, and the actual distribution of the 2014 exams:

Topic

Lessons

Estimated Questions

Spring 2014

Fall 2014

Single-life survival models Fractional ages Single-life insurances, annuities, and premiums Multiple-life survival models Multiple-decrement survival models Multi-state survival models Multi-state insurances, annuities, and premiums

2–6 7 8–23 24–26 27–28 29 30

2–3 1 3–4 2–3 2–3 1–2 1–2

3 1 2 3 3 2 1

3 1 2 2 3 2 2

As you can see, very little weight is put on single-life insurances, annuities, and premiums, even though more than half of this manual deals with those topics. You should study those topics rapidly and do just enough exercises to understand the basic concepts.

Comparison of Exam LC to the former Exam 3L Exam LC covers the life contingency topics from the former Exam 3L. Statistics and Poisson processes are not covered. In addition, some life contingencies topics were dropped. The exam no longer covers reserves. Also, it does not cover insurances with annually varying benefits and annuities with annually varying payments. They can still ask questions about insurances with two benefit levels, such as an insurance with a benefit of 1000 for the first 20 years and 500 thereafter.

Tables Download the tables you will be given on the exam. They will often be needed for the exercises. Go to the LC syllabus page at www.casact.org/admissions/exams/ and click on Tables for Exam LC under Syllabus Material. The direct address at this writing is http://www.casact.org/admissions/syllabus/LC-Tables.pdf, but this may change. These tables include tables of the normal distribution and life tables. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

ix

x

PREFACE

Features of this manual Old exam questions There are many original exercises in the manual, but most of the questions are from old exams. They come from old Part 4, Part 4A, Course 150, and Course 151 exams, as well as 2000-syllabus Exam 3. SOA Part 4 in 1986 had morning and afternoon sessions. I indicate afternoon session questions with “A”. The morning session had the more basic topics (through reserves), while the afternoon session had advanced topics (multiple lives, multiple decrements, etc.) Both sessions were multiple choice questions. SOA Course 150 from 1987 through 1991 had multiple choice questions in the morning and written answer questions in the afternoon. No written answer questions are included in this manual. The CAS Part 4A exams awarded varying numbers of points to questions; some are 1 point and some are 2 points. The 1 point questions are probably too easy for a modern exam, but they’ll give you a little practice. The pre-1987 exams probably were still based on Jordan (an old textbook that did not use a stochastic approach to life contingencies), but the questions I provided, while ancient, still have value. Similarly, the cluster questions on SOA Course 150 in the 1990’s generally were awarded 1 point per question. Course 151 is the least relevant to this subject. I’ve only included a small number of questions from 151 in the background lesson on probability. Back in 1999, the CAS and SOA created a sample exam for the then-new 2000 syllabus. This exam had some questions from previous exams but also some new questions, some of them not multiple choice. This sample exam was never a real exam, and some of its questions were defective. This sample exam is no longer available on the web. I have included appropriate questions from it. Whenever an exercise is labeled C3 Sample, it refers to the 1999 sample. Questions from exams Spring 2005 and later are not included in this manual. You may use those as a final test of your knowledge. Check Appendix C for a listing of questions that are not on the syllabus. For the period since Spring 2005, the SOA released Spring 2005, Fall 2005, Fall 2006, and Spring 2007, and all exams since Spring 2012. The CAS releases all exams, but does not provide worked-out solutions. However, I have provided worked-out solutions to all relevant multiple-choice questions from the released SOA and CAS exams in Appendix B. Questions from old exams are marked xxx:yy, where xxx is the time the exam was given, with S for spring and F for fall followed by a 2-digit year, and yy is the question number. Sometimes xxx is preceded with SOA or CAS to indicate the sponsoring organization. From about 1986 to 2000, SOA exams had 3digit numbers (like 150) and CAS exams were a number and a letter (like 4A). From 2000 to Spring 2003, the exams were jointly sponsored. There was a period in the 1990’s when the SOA, while it allowed use of its old exam questions, did not want people to reveal which exam they came from. As a result, I sometimes had study notes for old exams in this period and could not identify the exam they came from. In such a case, I mark the question aaa-bb-cc:yy, where aaa-bb-cc is the study note number and yy is the question number. Generally aaa is the exam number (like 150), and cc is the 2-digit year the study note was published.

Practice Exams Ten practice exams are provided. Each one has 15 questions, just like the real exam.

Other Useful Features of This Manual Almost every lesson has a summary. I omitted the summary for short lessons with no formulas or only one set of formulas. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

PREFACE

xi

This manual has an index. Whenever you remember some topic in this manual but can’t remember where you saw it, check the index. If it isn’t in the index but you’re sure it’s in the manual and an index listing would be appropriate, contact the author. This may be the first CAS exam you are taking. CAS exams have a somewhat different style from SOA or the former jointly sponsored exams. CAS exams have a guessing penalty, so omit questions unless you can at least eliminate some choices. CAS exams usually provide ranges rather then specific answer choices. The ranges are usually equal in size, and your answer should usually not be more than the size of a range lower than the first choice or higher than the last choice. For example, if the choices offered are A. Less than 5 B. At least 5, but less than 7 C. At least 7, but less than 9 D. At least 9, but less than 11 E. At least 11 then your answer should not be less than 3 or more than 13. However, this rule is not hard-and-fast, so if you get an answer far out of range, it is suspect but not necessarily wrong. Every CAS exam is released. The CAS will not release your score if you pass. CAS exams frequently have defective questions. If you feel sure of your answer and it is far out of the ranges given, or not one of the five answer choices, move on.

Acknowledgements I would like to thank the SOA and CAS for allowing me to use their old exam questions. I’d also like to thank Harold Cherry for providing three of the pre-2000 SOA exams and all of the pre-2000 CAS exams I used. The creators of TEX, LATEX, and its multitude of packages all deserve thanks for making possible the professional typesetting of this mathematical material. I would like to thank all readers who submitted errata. Some of the readers who submitted errata are: Jonathan Bell, Ira Blassberger, James Bruce, Mark Doering, Griffin Winton-LaVieri, Christina Malleo, Russell Mawk, Alan May, Sarah Nimphie, Mou Jian Teo.

Errata Please report all errors you find in these notes to the author. You may send them to the publisher at [email protected] or directly to me at [email protected]. Please identify the manual and edition the error is in. This is the 1st edition 3rd printing of the Exam LC manual. An errata list will be posted at errata.aceyourexams.net. Check this errata list frequently.

LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

xii

LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

PREFACE

Lesson 1

Probability Review This lesson is a brief summary of probability concepts you will need in the course. You will not be directly tested on these topics, but they are essential background. If you find this review too brief, you should review your favorite probability textbook for more details. Conversely, you may skip this lesson if you are familiar with the concepts.

1.1

Functions and moments

The cumulative distribution function F ( x ) of a random variable X, usually just called the distribution function, is the probability that X is less than or equal to x: F ( x )  Pr ( X ≤ x ) It defines X, and is right-continuous, meaning limh→0 F ( x + h )  F ( x ) for h positive. Some random variables are discrete (there are isolated points x at which Pr ( X  x ) is nonzero) and some are continuous (meaning F ( x ) is continuous, and differentiable except at a countable number of points). Some are mixed—they are continuous except at a countable number of points. Here are some important functions that are related to F ( x ) : • S ( x ) is the survival function, the complement of F ( x ) , the probability that X is strictly greater than x. S ( x )  Pr ( X > x ) It is called the survival function since if X represents survival time, it is the probability of surviving longer than x. d F ( x ) is the probability density function. For a discrete • For a continuous random variable, f ( x )  dx random variable, the probability mass function f ( x )  Pr ( X  x ) serves a similar purpose.

f (x ) d ln S ( x ) − is the hazard rate function.1 SomeS (x ) dx times the hazard rate function is denoted by h ( x ) instead of λ ( x ) . The hazard rate function is like a conditional density function, the conditional density to time x. We can reverse the  R x given survival  operations to go from λ ( x ) to S ( x ) : S ( x )  exp − −∞ λ ( u ) du .

• For a continuous random variable, λ ( x ) 

Why do we bother differentiating F to obtain f ? Because the density is needed for calculating moments. Moments of a random variable measure its center and dispersion. The expected value of X is defined by

Z

∞

E[X] 

x f ( x ) dx −∞

and more generally the expected value of a function of a random variable is defined by

Z

∞

E[g ( X ) ] 

g ( x ) f ( x ) dx −∞

1As we’ll learn in Lesson 3, in International Actuarial Notation (IAN), µ x is used for the hazard rate function. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

1

1. PROBABILITY REVIEW

2

For discrete variables, the integrals are replaced with sums. The n th raw moment of X is defined as µ0n  E[X n ]. µ  µ01 is the mean. The n th central moment of X (n , 1) is defined as2 µ n  E[ ( X − µ ) n ]. Usually n is a positive integer, but it need not be. When we mention moments in this manual and don’t state otherwise, we mean raw moments. Expectation is linear, so the central moments can be calculated from the raw moments by binomial expansion. In the binomial expansion, the last two terms always merge, so we have µ2  µ02 − µ2 µ3  µ4 

µ03 µ04

− −

3µ02 µ 4µ03 µ

instead of µ02 − 2µ01 µ + µ2 + 2µ +

3

instead of

6µ02 µ2

− 3µ

4

instead of

µ03 µ04

− −

3µ02 µ 4µ03 µ

+ +

3µ01 µ2 6µ02 µ2

(1.1) −µ −

3

4µ01 µ3

(1.2) +µ

4

Special functions of moments are: • The variance is Var ( X )  µ2 , and is denoted by σ 2 . • The standard deviation σ is the positive square root of the variance. • The coefficient of variation is σ/µ. This concept does not appear in this course. However, it plays a big role in Exam C. We will discuss important things you should know about variance in Section 1.3. For the meantime, I will repeat formula (1.1) using different notation, since it’s so important: Var ( X )  E[X 2 ] − E[X]2

(1.3)

Many times this is the best way to calculate variance. For two random variables X and Y: • The covariance is defined by Cov ( X, Y )  E[ ( X − µ X )( Y − µY ) ]. • The correlation coefficient is defined by ρ XY 

Cov ( X,Y ) σ X σY .

As with the variance, another formula for covariance is Cov ( X, Y )  E[XY] − E[X] E[Y] Note that E[XY] , E[X] E[Y] in general. In fact, E[XY]  E[X] E[Y] if and only if X and Y are uncorrelated, in other words if their correlation is 0. For independent random variables, Cov ( X, Y )  0. A 100p th percentile is a number π p such that F ( π p ) ≥ p and F ( π−p ) ≤ p. If F is continuous and strictly increasing, it is the unique point at which F ( π p )  p. In this course, we will only discuss percentiles for strictly increasing distribution functions, and that will simplify matters. A median is a 50th percentile. A mode is x such that f ( x ) (or Pr ( X  x ) for a discrete distribution) is maximized.

1.2

Probability distributions

A couple of probability distributions will be used frequently during this course.

2This µ n has no connection to µ x , the force of mortality. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

1.2. PROBABILITY DISTRIBUTIONS

1.2.1

3

Bernoulli distribution

A random variable has a Bernoulli distribution if it only assumes the values of 0 and 1. The value of 1 is assumed with probability q. If X is Bernoulli, then its mean is q—the same as the probability of 1. Its variance is q (1 − q ) . We can generalize the variable. Consider a random variable Y which can assume only two values, but the two values are x 1 and x2 instead of 0 or 1; the probability of x 2 is q. Then Y  x1 + ( x 2 − x1 ) X, where X is Bernoulli. It follows that the mean is x1 + ( x2 − x1 ) q. More importantly, the variance is ( x2 − x 1 ) 2 q (1 − q ) . This is a fast way to calculate variance, faster than calculating E[Y] and E[Y 2 ], so remember it. To repeat: To compute the variance of a Bernoulli-type variable assuming only two values, multiply the product of the probabilities of the two values by the square of the difference between the two values. I call this trick for calculating the variance the Bernoulli shortcut. Example 1A For a one-year term life insurance policy of 1000: • • • •

The premium is 30. The probability of death during the year is 0.02. The company has expenses of 2. If the insured survives to the end of the year, the company pays a dividend of 3.

Ignore interest. Calculate the variance in the amount of profit the company makes on this policy. Answer: There are only two possibilities—either the insured dies or he doesn’t—so we have a Bernoulli here. We can ignore premium and expenses, since they don’t vary, so they generate no variance. Either the company pays 1000 (probability 0.02) or it pays 3 (probability 0.98). The variance is therefore

(1000 − 3) 2 (0.02)(0.98)  19,482.5764 .

?



Quiz 1-1 3 A random variable X has the following distribution: x

Pr ( X  x )

20 120

0.7 0.3

Calculate Var ( X ) . A sum of m Bernoulli random variables each with the same mean q is a binomial random variable. Its mean is mq and its variance is mq (1 − q ) .

1.2.2

Uniform distribution

The uniform distribution on [a, b] is a continuous distribution with constant density 1/ ( b − a ) on the interval [a, b] and 0 elsewhere. Its mean is its midpoint, ( a + b ) /2, and its variance is ( b − a ) 2 /12. It is a simple distribution, and will be used heavily in examples throughout the course.

3Quiz solutions are at the end of the lesson, after exercise solutions. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

1. PROBABILITY REVIEW

4

1.2.3

Exponential distribution

The exponential distribution is defined by cumulative distribution function F ( x )  1 − e −x/θ , where θ is the mean. The density function is f ( x )  e −x/θ /θ. This density function—an exponentiated variable— is very convenient to use in conjunction with other exponentiated items, such as those that arise from compound interest. Therefore, this distribution will be used heavily in examples throughout the course. The sum of n independent exponential random variables all having the same mean is a gamma random variable. If Y is gamma and is the sum of n exponential random variables with mean θ, then its density function is x n−1 e −x/θ fY ( x )  Γ(n ) θn where Γ ( n ) , the gamma function, is a continuous extension of the factorial function; for n an integer, Γ ( n )  ( n − 1) ! By using Γ ( n ) instead of ( n − 1) !, the gamma function can be defined for non-integral n.

1.3

Variance

Expected value is linear, meaning that E[aX + bY]  a E[X] + b E[Y], regardless of whether X and Y are independent or not. Thus E[ ( X + Y ) 2 ]  E[X 2 ] + 2 E[XY] + E[Y 2 ], for example. This means that E[ ( X + Y ) 2 ] is not equal to E[X 2 ] + E[Y 2 ] (unless E[XY]  0).

f

g







2

Also, it is not true in general that E g ( X )  g E[X] . So E[X 2 ] , E[X] . Since variance is defined in terms of expected value, Var ( X )  E[X 2 ] − E[X]2 , this allows us to develop a formula for Var ( aX + bY ) . If you work it out, you get Var ( aX + bY )  a 2 Var ( X ) + 2ab Cov ( X, Y ) + b 2 Var ( Y )

(1.4)

In particular, if Cov ( X, Y )  0 (which is true if X and Y are independent), then Var ( X + Y )  Var ( X ) + Var ( Y ) and generalizing to n independent variables, Var *

n X

Xi + 

, i1

n X

Var ( X i )

i1

-

If all the X i ’s are independent and have identical distributions, and we set X  X i for all i, then Var *

n X

, i1

X i +  n Var ( X )

(1.5)

-

However, Var ( nX )  n 2 Var ( X ) , not n Var ( X ) . You must distinguish between these two situations, which are quite different. Think of the following example. The stock market goes up or down randomly each day. We will assume that each day’s change is independent of the previous day’s, and has the same distribution. Compare the variance of the following possibilities: 1. You put $1 in the market, and leave it there for 10 days. 2. You put $10 in the market, and leave it there for 1 day. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

1.4. NORMAL APPROXIMATION

5

In the first case, there are going to be potential ups and downs each day, and the variance of the change of your investment will be 10 times the variance of one day’s change because of this averaging. In the second case, however, you are multiplying the variation of a single day’s change by 10—there’s no dampening of the change by 10 different independent random events, the change depends on a single random event. As a result, you are multiplying the variance of a single day’s change by 100. In the more general case where the variables are not independent, you need to know the covariance. This can be provided in a covariance matrix. If you have n random variables X1 , . . . , X n , this n × n matrix A has a i j  Cov ( X i , X j ) for i , j. For i  j, a ii  Var ( X i ) . This matrix is symmetric and non-negative definite. However, the covariance of two random variables may be negative. Example 1B For a loss X on an insurance policy, let X1 be the loss amount and X2 the loss adjustment expenses, so that X  X1 + X2 . The covariance matrix for these random variables is 25 5 5 2

!

Calculate the variance in total cost of a loss including loss adjustment expenses. Answer: In formula (1.4), a  b  1. From the matrix, Var ( X1 )  25, Cov ( X1 , X2 )  5, and Var ( X2 )  2. So Var ( X1 + X2 )  Var ( X1 ) + 2 Cov ( X1 , X2 ) + Var ( X2 )  25 + 2 (5) + 2  37



A sample is a set of observations from n independent identically distributed random variables. The sample mean X¯ is the sum of the observations divided by n. The variance of the sample mean of X1 , . . . , X n , which are observations from the random variable X, is

Pn Var ( X¯ )  Var

1.4

i1

n

Xi

! 

n Var ( X ) Var ( X )  n n2

(1.6)

Normal approximation

The Central Limit Theorem says that for any distribution with finite variance, the sample mean of a set of independent identically distributed random variables approaches a normal distribution. By the previous section, the mean of the sample mean of observations of X is E[X] and the variance is σ2 /n. These parameters uniquely determine the normal distribution that the sample mean converges to. A random variable Y with normal distribution with mean µ and variance σ2 can be expressed in terms of a standard normal random variable Z in the following way: Y  µ + σZ and you can look up the distribution of Z in a table of the standard normal distribution function that you get at the exam. The normal approximation of a percentile of a random variable is performed by finding the corresponding percentile of a normal distribution with the same mean and variance. Let Φ ( x ) be the cumulative distribution function of the standard normal distribution. (The standard normal distribution has µ  0, σ  1. Φ is the symbol generally used for this distribution function.) Suppose we are given that X is a normal random variable with mean µ, variance σ2 ; we will write X ∼ n ( µ, σ2 ) to describe X. And suppose we want to calculate the 95th percentile of X; in other words, we want a number x such that LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

1. PROBABILITY REVIEW

6

Pr ( X ≤ x )  0.95. We would reason as follows: Pr ( X ≤ x )  0.95 X−µ x−µ  0.95 Pr ≤ σ σ

!

x−µ  0.95 σ x−µ  Φ−1 (0.95) σ x  µ + σΦ−1 (0.95)

!

Φ

Note that Φ−1 (0.95)  1.645 is a commonly used percentile of the normal distribution, and is listed at the bottom of the table you get at the exam. You should internalize the above reasoning so you don’t have to write it out each time. Namely, to calculate a percentile of a random variable being approximated normally, find the value of x such that Φ ( x ) is that percentile. Then scale x: multiply by the standard deviation, and then translate x: add the mean. This approximation will be used repeatedly throughout the course. Example 1C A big fire destroyed a building in which 100 of your insureds live. Each insured has a fire insurance policy. The losses on this policy follow a distribution with mean 1000 and variance 3,000,000. Even though all the insureds live in the same building, the losses are independent. You are now setting up a reserve for the cost of these losses. Using the normal approximation, calculate the size of the reserve you should put up if you want to have a 95% probability of having enough money in the reserve to pay all the claims. Answer: The expected total loss is the sum of the means, or (100)(1000)  100,000. The variance √ of the 8 total loss is the sum of the variances, or 100 (3,000,000)  3 × 10 . The standard deviation σ  3 × 108  17,320.51. For a standard normal distribution, the 95th percentile is 1.645. We scale this by 17,320.51 and translate it by 100,000: 100,000 + 17,320.51 (1.645)  128,492.24 .  The normal approximation is also used for probabilities. To approximate the probability that a random variable is less than x, calculate the probability that  a normal random variable with the same mean and variance is less than x. In other words, calculate Φ ( x − µ ) /σ . In this course, however, it will be rare that we approximate probabilities. Example 1D A big fire destroyed a building in which 100 of your insureds live. Each insured has a fire insurance policy. The losses on this policy follow a distribution with mean 1000 and variance 3,000,000. Even though all the insureds live in the same building, the losses are independent. You are now setting up a reserve for the cost of these losses. Using the normal approximation, calculate the probability that the average loss is less than 1100. Answer: The mean of the average is 1000 and the variance of the average is 3,000,000/100  30,000, as we just mentioned in formula (1.6). Therefore

!

1100 − 1000 Pr ( X¯ < 1100) ≈ Φ √  Φ (0.577)  0.7190 30,000 where we’ve evaluated Φ (0.577) as Φ (0.58) from the printed normal distribution tables.

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1.5. CONDITIONAL PROBABILITY AND EXPECTATION

1.5

7

Conditional probability and expectation

The probability of event A given B, assuming Pr ( B ) , 0, is Pr ( A | B ) 

Pr ( A ∩ B ) Pr ( B )

where Pr ( A ∩ B ) is the probability of both A and B occurring. A corresponding definition for continuous distributions uses the density function f instead of Pr: fX ( x | y ) 

f ( x, y ) f ( y)

R

where f ( y )  f ( x, y ) dx , 0. Two important theorems are Bayes Theorem and the Law of Total Probability: Theorem 1.1 (Bayes Theorem) Pr ( A | B ) 

Pr ( B | A ) Pr ( A ) Pr ( B )

(1.7)

fY ( y | x ) fX ( x ) fY ( y )

(1.8)

Correspondingly for continuous distributions fX ( x | y ) 

Theorem 1.2 (Law of Total Probability) If B i is a set of exhaustive (in other words, i Pr ( B i )  1) and mutually exclusive (in other words Pr ( B i ∩ B j )  0 for i , j) events, then for any event A,

P

Pr ( A ) 

X

Pr ( A ∩ B i ) 

i

X

Pr ( B i ) Pr ( A | B i )

(1.9)

i

Correspondingly for continuous distributions,

Z Pr ( A ) 

Pr ( A | x ) f ( x ) dx

(1.10)

Expected values can be factored through conditions too. In other words, the mean of the means is the mean, or: Conditional Mean Formula

f

EX [X]  EY EX [X | Y]

g

(1.11)

This formula is one of the double expectation formulas. More generally for any function g EX [g ( X ) ]  EY [EX [g ( X ) | Y]]

(1.12)

Here are examples of this important theorem. Versions of the first example appear very frequently on this exam. Example 1E There are 2 types of actuarial students, bright and not-so-bright. The bright ones pass 80% of the exams they take and the not-so-bright ones pass 40% of the exams they take. All students start with Exam 1 and take the exams in sequence, and drop out as soon as they fail one exam. An equal number of bright and not-so-bright students take Exam 1. Determine the probability that a randomly selected student taking Exam 3 will pass.

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1. PROBABILITY REVIEW

8

Answer: A common wrong answer to this question is 0.5 (0.8) + 0.5 (0.4)  0.6. This is an incorrect application of the Law of Total Probability. The probability that a student taking Exam 3 is bright is more than 0.5, because of the elimination of the earlier exams. A correct way to calculate the probability is to first calculate the probability that a student is taking Exam 3 given the two types of students. Let I1 be the event of being bright initially (before taking Exam 1) and I2 the event of not being bright initially. Let E be the event of taking Exam 3. Then by Bayes Theorem and the Law of Total Probability, Pr (E | I1 ) Pr ( I1 ) Pr (E ) Pr (E )  Pr (E | I1 ) Pr ( I1 ) + Pr (E | I2 ) Pr ( I2 )

Pr ( I1 | E ) 

Now, the probability that one takes Exam 3 if bright is the probability of passing the first two exams, or 0.82  0.64. If not-so-bright, the probability is 0.42  0.16. So we have Pr (E )  0.64 (0.5) + 0.16 (0.5)  0.4 (0.64)(0.5)  0.8 Pr ( I1 | E )  0.4 and Pr ( I2 | E )  1 − 0.8  0.2 (or you could go through the above derivation with I2 instead of I1 ). Now we’re ready to apply the Law of Total Probability to the conditional distributions given E to answer the question. Let P be the event of passing Exam 3. Then Pr ( P | E )  Pr ( P | I1 &E ) Pr ( I1 | E ) + Pr ( P | I2 &E ) Pr ( I2 | E )  (0.8)(0.8) + (0.4)(0.2)  0.72



Now let’s do a continuous example. Example 1F Claim sizes follow an exponential distribution with mean θ. θ varies by insured. Over all insureds, θ has a distribution with the following density function: f (θ) 

1 θ2

1≤θ 0.5) 

!

e −0.5/θ 1

∞ 1 

1 dθ θ2

 2e −0.5/θ



 2 1 − e −0.5  2 (1 − 0.606531)  0.786939



If f ( x | y )  f ( x ) for all x and y, we say that X and Y are independent random variables. Independent random variables are uncorrelated (but not conversely), so for X, Y independent, E[XY]  E[X] E[Y]. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

1.6. CONDITIONAL VARIANCE

1.6

9

Conditional variance

Suppose we wish to calculate the variance of a random variable X. Rather than calculating it directly, it may be more convenient to condition X on Y, and then calculate moments of the conditional variable X | Y. Consider the following example: Example 1G A life insurance agent may be happy or sad. The probability of happiness is 0.8. On a day when the agent is happy, the number of policies sold is exponentially distributed with mean 0.3. When the agent is sad, the number of policies sold is exponentially distributed with mean 0.1. Calculate the variance of the number of policies sold per day. Answer: One way to attack this problem is to calculate first and second moments and then variance. We condition X, the number of policies sold, on happiness, or Y. From equation (1.11) and the more general (1.12) with g ( X )  X 2 ,

f

g

E[X]  E E[X | Y]  E[0.3, 0.1]  0.8 (0.3) + 0.2 (0.1)  0.26 The second moment of an exponential is the variance plus the mean squared, and the variance equals the mean squared, so the second moment of an exponential is twice the mean squared.

f

g

E[X 2 ]  E E[X 2 | Y]  E[2 (0.32 ) , 2 (0.1) 2 ]  0.8 (0.18) + 0.2 (0.02)  0.148 So the variance is Var ( X )  0.148 − 0.262  0.0804 . It is tempting, for those not in the know, to try to calculate the variance by weighting the two variances of happiness and sadness. In each state, the variance is the square of the mean, so the calculation would go 0.8 (0.32 ) + 0.2 (0.12 )  0.074. But that is the wrong answer. It is too low. Do you see what is missing? What is missing is the variance of the states. To capture the full variance, you must add the expected value of the variance of the states and the variance of the expected values of the states. The correct formula is Conditional Variance Formula VarX ( X )  EY [VarX ( X | Y ) ] + VarY (EX [X | Y])

(1.13)

In our example, we’ve computed the expected value of the variances as 0.074. The state is a Bernoulli variable, and the expected values of the states are 0.3 and 0.1. So by the Bernoulli shortcut, the variance of the expected values is (0.8)(0.2)(0.3 − 0.1) 2  0.0064. The variance of the number of policies sold is 0.074 + 0.0064  0.0804 . 

?

Quiz 1-2 Given Y, the variable X has a normal distribution with mean Y and variance Y 2 . Y is uniformly distributed on [−10, 2]. Calculate the variance of X.

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1. PROBABILITY REVIEW

10

Exercises Functions and moments 1.1. [CAS3-F04:24] A pharmaceutical company must decide how many experiments to run in order to maximize its profits. •

The company will receive a grant of $1 million if one or more of its experiments is successful.

•

Each experiment costs $2,900.

•

Each experiment has a 2% probability of success, independent of the other experiments.

•

All experiments run simultaneously.

•

Fixed expenses are $500,000.

•

Ignore investment income. The company performs the number of experiments that maximizes its expected profit. Determine the company’s expected profit before it starts the experiments.

A. 77,818

B. 77,829

C. 77,840

D. 77,851

E. 77,862

Variance 1.2. [4B-S93:9] (1 point) If X and Y are independent random variables, which of the following statements are true? 1.

Var ( X + Y )  Var ( X ) + Var ( Y )

2.

Var ( X − Y )  Var ( X ) + Var ( Y )

3.

Var ( aX + bY )  a 2 E[X 2 ] − a (E[X]) 2 + b 2 E[Y 2 ] − b (E[Y]) 2

A. 1

B. 1,2

C. 1,3

D. 2,3

E. 1,2,3

1.3. [4B-F95:28] (2 points) Two numbers are drawn independently from a uniform distribution on [0,1]. What is the variance of their product? A. 1/144

B. 3/144

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C. 4/144

D. 7/144

E. 9/144

Exercises continue on the next page . . .

EXERCISES FOR LESSON 1

11

Table 1.1: Important formulas from this lesson

Var ( X )  E[X 2 ] − E[X]2

(1.3)

Var ( aX + bY )  a Var ( X ) + 2ab Cov ( X, Y ) + b Var ( Y ) Var ( X ) Var ( X¯ )  n Pr ( B | A ) Pr ( A ) Pr ( A | B )  Pr ( B ) fY ( y | x ) fX ( x ) fX ( x | y )  fY ( y ) 2

Pr ( A ) 

2

X

Pr ( A ∩ B i ) 

i

X

Pr ( B i ) Pr ( A | B i )

(1.4) (1.6) (Bayes Theorem—discrete)

(1.7)

(Bayes Theorem—continuous)

(1.8)

Law of Total Probability—discrete

(1.9)

i

Z Pr ( A ) 

Pr ( A | x ) f ( x ) dx

Law of Total Probability—continuousS (1.10)

f

EX [X]  EY EX [X | Y]

g

VarX ( X )  EY [VarX ( X | Y ) ] + VarY (EX [X | Y]) Distribution

Mean

(Double expectation)

(1.11)

(Conditional variance)

(1.13)

Variance

Bernoulli

q

q (1 − q )

Binomial

mq

mq (1 − q )

a+b 2

(b − a )2

θ

θ2

Uniform on [a, b] Exponential

12

Bernoulli shortcut: If a random variable can only assume two values a and b with probabilities q and 1 − q respectively, then its variance is q (1 − q )( b − a ) 2 .

1.4. [151-82-92:4] A company sells group travel-accident life insurance with b payable in the event of a covered individual’s death in a travel accident. The gross premium for a group is set equal to the expected value plus the standard deviation of the group’s aggregate claims. The standard premium is based on the following assumptions: • All individual claims within the group are mutually independent; and • b 2 q (1 − q )  2500, where q is the probability of death by travel accident for an individual. In a certain group of 100 lives, the independence assumption fails because three specific individuals always travel together. If one dies in an accident, all three are assumed to die. Determine the difference between this group’s premium and the standard premium. A. 0

B. 15

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C. 30

D. 45

E. 60

Exercises continue on the next page . . .

1. PROBABILITY REVIEW

12

1.5. You are given the following information about the random variables X and Y: • Var ( X )  9 • Var ( Y )  4 • Var (2X − Y )  22 Determine the correlation coefficient of X and Y. A. 0

B. 0.25

C. 0.50

D. 0.75

E. 1

1.6. [151-82-93:9] (1 point) For a health insurance policy, trended claims will be equal to the product of the claims random variable X and a trend random variable Y. You are given: • • • • •

E[X]  10 Var ( X )  100 E[Y]  1.20 Var ( Y )  0.01 X and Y are independent

Determine the variance of trended claims. A. 144

B. 145

C. 146

D. 147

E. 148

1.7. X and Y are two independent exponentially distributed random variables. You are given that Var ( X )  25 and Var ( XY )  7500. Determine Var ( Y ) . A. 25

B. 50

C. 100

D. 200

E. 300

Normal approximation 1.8. The number of policies a life insurance agent sells in one day is 1 with probability 1/5 and 0 with probability 4/5. Assume the agent works 252 days a year. Using the normal approximation, determine the 95th percentile of the number of policies sold in one year. 1.9. A life insurance company has determined that the present value of profit on selling one contract is uniformly distributed on [−50, 70]. Using the normal approximation, calculate the probability of making a profit on a portfolio of 50 policies.

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Exercises continue on the next page . . .

EXERCISES FOR LESSON 1

13

Bernoulli shortcut 1.10. [4B-F99:7] (2 points) A player in a game may select one of two fair, six-sided dice. Die A has faces marked with 1, 2, 3, 4, 5 and 6. Die B has faces marked with 1, 1, 1, 6, 6, and 6. If the player selects Die A, the payoff is equal to the result of one roll of Die A. If the player selects Die B, the payoff is equal to the mean of the results of n rolls of Die B. The player would like the variance of the payoff to be as small as possible. Determine the smallest value of n for which the player should select Die B. A. 1

B. 2

C. 3

D. 4

E. 5

Conditional probability 1.11. [M-F05:17] The length of time, in years, that a person will remember an actuarial statistic is modeled by an exponential distribution with mean 1/Y. In a certain population, the probability density function of Y is ye −y/2 y≥0 f ( y)  4 Calculate the probability that a person drawn at random from this population will remember an actuarial statistic less than 1/2 year. A. 0.125

B. 0.250

C. 0.500

D. 0.750

E. 0.875

Conditional variance 1.12.

A population consists of smokers and non-smokers. 80% of the population is non-smokers.

Survival time is normally distributed. For smokers, mean survival time is 40 with variance 800. For non-smokers, mean survival time is 45 with variance 600. Calculate the variance of survival time for an individual randomly selected from the population. 1.13. [C3 Sample:10] An insurance company is negotiating to settle a liability claim. If a settlement is not reached, the claim will be decided in the courts 3 years from now. You are given: •

There is a 50% probability that the courts will require the insurance company to make a payment. The amount of the payment, if there is one, has a lognormal distribution with mean 10 and standard deviation 20.

•

In either case, if the claim is not settled now, the insurance company will have to pay 5 in legal expenses, which will be paid when the claim is decided, 3 years from now.

•

The most that the insurance company is willing to pay to settle the claim is the expected present value of the claim and legal expenses plus 0.02 times the variance of the present value.

•

Present values are calculated using i  0.04. Calculate the insurance company’s maximum settlement value for this claim.

A. 8.89

B. 9.93

C. 12.45

Additional old CAS Exam 3/3L questions: S06:25,30

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D. 12.89

E. 13.53

1. PROBABILITY REVIEW

14

Solutions 1.1. The probability of success for n experiments is 1 − 0.98n , so profit, ignoring fixed expenses, is 1,000,000 (1 − 0.98n ) − 2900n Differentiating this and setting it equal to 0: −106 (0.98n )(ln 0.98) − 2900  0 0.98n  n

−2900 106 ln 0.98 ln 10−2900 6 ln 0.98

 96.0815 ln 0.98 Thus either 96 or 97 experiments are needed. Plugging those numbers into the original expression g ( n )  1,000,000 (1 − 0.98n ) − 2900n gets g (96)  577,818.4 and g (97)  577,794.0, so 96 is best, and the expected profit is 577,818.4 − 500,000  77,818.4 . (A) An alternative to calculus which is more appropriate for this discrete exercise is to note that as n increases, at first expected profit goes up and then it goes down. Let X n be the expected profit with n experiments. Then X n  106 (1 − 0.98n ) − 2900n − 500,000 and the incremental profit generated by experiment #n is





X n − X n−1  106 0.98n−1 − 0.98n − 2900. We want this difference to be greater than 0, which occurs when





106 0.98n−1 − 0.98n > 2900 0.98n−1 (0.02) > 0.0029 0.0029  0.145 0.98n−1 > 0.02 ( n − 1) ln 0.98 > ln 0.145 ln 0.145 −1.93102   95.582 n−1 < ln 0.98 −0.02020 On the last line, the inequality got reversed because we divided by ln 0.98, a negative number. We conclude that the n th experiment increases profit only when n < 96.582, or n ≤ 96, the same conclusion as above. 1.2. The first and second are true by formula (1.4). The third should have squares on the second a and second b, since Var ( aX )  E[ ( aX ) 2 ] − E[aX]2  a 2 E[X 2 ] − a 2 E[X]2 for example. (B) 1.3. The mean of the uniform distribution is

1 2

and the second moment is 13 . So

Var ( XY )  E[X 2 Y 2 ] − E[X]2 E[Y]2 1  3 

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!

!

1 1 − 3 4

!

1 4

1 1 7 −  9 16 144

!

(D)

EXERCISE SOLUTIONS FOR LESSON 1

15

1.4. The number of fatal accidents for each life, N, has a Bernoulli distribution with mean q and variance q (1−q ) , so the variance in one life’s aggregate claims is the variance of bN. Var ( bN )  b 2 Var ( N )  b 2 q (1− q )  2500. For 100 independent lives, aggregate claims are 100bN, with variance 100 Var ( bN )  100 (2500) . For three lives always traveling together, aggregate claims are 3bN with variance 32 Var ( bN )  9 (2500) . If we add this to the variance of aggregate claims for the other 97 independent lives, the variance is 9 (2500) + 97 (2500)  106 (2500) . The expected value of aggregate claims, however, is no different from the expected value of the totally independent group’s aggregate claims. The difference in premiums is therefore

p

p

106 (2500) − 100 (2500)  14.7815

(B)

1.5. From formula (1.4), 22  Var (2X − Y )  4 (9) + 4 − 2 (2) Cov ( X, Y ) Cov ( X, Y )  4.5 4.5 ρ XY  √ √  0.75 9 4

(D)

1.6. E[XY]  (10)(1.20)  12













E[ ( XY ) 2 ]  E[X 2 ] E[Y 2 ]  102 + 100 1.202 + 0.01  290 Var ( XY )  290 − 122  146

(C)

1.7. For an exponential variable, the variance is the square of the mean. Let θ be the parameter for Y Var ( XY )  E[X 2 ] E[Y 2 ] − E[X]2 E[Y]2 7500  (25 + 25)(2θ 2 ) − 25θ 2  75θ 2 θ  10 Var ( Y )  θ 2  100

(C)

1.8. The mean number of policies sold in one year is 252 (0.2)  50.4. The variance of the Bernoulli number sold per day is (0.2)(0.8)  0.16, so the variance of the number of policies sold in one year is √ 252 (0.16)  40.32. The 95th percentile of the number of policies sold is 50.4 + 1.645 40.32  60.85 . Rounding this to 61, since an integral number of policies is sold, is appropriate. 1.9. The mean of the uniform is (70−50) /2  10, and the variance is 1202 /12  1200. Multiply these moments by 50 for 50 policies. The probability that profit is greater than 0, using the normal approximation, is ! 0 − 500 1−Φ √  Φ (2.0412)  0.9794 60,000

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1. PROBABILITY REVIEW

16

1.10.

The variance of Die A is 7 1* . 1− 6 2



2

 + 2−

7 2

2

 + 3−

7 2

2

 + 4−

7 2

2

 + 5−

7 2

2

 + 6−

,

7 2

2

+/  35 . 12 -

Die B is Bernoulli, only two possibilities with probabilities 1/2 and values 1 and 6, so the variance of one toss is 52 (1/2) 2  25/4. The variance of the mean is the variance of one toss over n (equation (1.6)). So 35 25 < 4n 12 140n > 300 n>2 The answer is 3 . (C) 1.11. Use the Law of Total Probability. Let X be the length of time. It’s a little easier to calculate the probability that X > 1/2. Pr ( X > 1/2 | Y )  e −y/2 ∞

Z Pr ( X > 1/2) 

0.25ye −y/2 e −y/2 dy 0 ∞

Z  0.25

ye −y dy 0

∞ −ye −y 0

 0.25

∞

Z +

! e

−y

dy

0

 ∞  0.25 −e −y 

0

 0.25 Then Pr ( X < 1/2  1 − Pr ( X > 1/2)  0.75 . (D) (since X is continuous, making Pr ( X  1/2)  0). 1.12.

Let I be the indicator variable for whether the individual is a smoker. If survival time is T, then Var (T )  Var (E[T | I]) + E[Var (T | I ) ]

The expected value of T | I is 40 with probability 0.2 and 45 with probability 0.8. Since it has only two values, it is a Bernoulli variable, and its variance is Var (E[T | I])  (0.2)(0.8)(45 − 40) 2  4. The variance of T | I is 800 with probability 0.2 and 600 with probability 0.8. The mean of these two values is E[Var (T | I ) ]  0.2 (800) + 0.8 (600)  640. Thus Var (T )  4 + 640  644 . As a check, you may calculate the second moment and subtract the first moment squared.

f

g

E[T]  E E[T | I]  0.2 (40) + 0.8 (45)  44

f

g

E[T 2 ]  E E[T 2 | I]  0.2 (402 + 800) + 0.8 (452 + 600)  2580 Var (T )  2580 − 442  644

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QUIZ SOLUTIONS FOR LESSON 1

17

1.13. The expected value of the present value of the claim is 0.5 (10/1.043 ) , and the present value of legal fees is 5/1.043 , for a total of 10/1.043  8.89. We will compute the variance using the conditional variance formula. The legal expenses are not random and have no variance, so we’ll ignore them. Let I be the indicator variable for whether a payment is required, and X the settlement value.





f

Var ( X )  Var E[X | I] + E Var ( X | I )

g

The expected value of the claim is 0 with probability 50% and 10/1.043 with probability 50%. Thus the expected value can only have one of two values. It is a Bernoulli random variable. The Bernoulli shortcut says that its variance is !2   10  19.7579 Var E[X | I]  (0.5)(0.5) 1.043 The variance of the claim is 0 with probability 50% and (20/1.043 ) 2 with probability 50%. The expected value of the variance is therefore

!2 20 +  158.0629 * E Var ( X | I )  (0.5) 0 + 1.043 , f

g

Therefore, Var ( X )  19.7579 + 158.0629  177.8208. The answer is 8.89 + 0.02 (177.8208)  12.4463

(C)

Quiz Solutions 1-1.





Var ( X )  (0.7)(0.3) 1002  2100

1-2. Var ( X )  E[Var ( X | Y ) ] + Var (E[X | Y])  E[Y 2 ] + Var ( Y ) Since Y is uniform on [−10, 2], its variance is the range squared over 12, or Var ( Y ) 

(−10 − 2) 2 12

 12

The second moment of Y is the sum of its variance and the square of its mean. The mean is the midpoint of [−10, 2], or −4. So E[Y 2 ]  (−4) 2 + 12  28 Var ( X )  28 + 12  40

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18

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1. PROBABILITY REVIEW

Lesson 2

Survival Distributions: Probability Functions Reading: Models for Quantifying Risk (4th or 5th edition) 5.1.1, 5.1.2, 5.3.1, 5.3.2, 6.1, 6.2 We will study the probability distribution of future lifetime. Once we have specified the probability distribution, we will be able to answer questions like What is the probability that someone age 30 will survive to age 80? What is the probability that someone age 40 will die between ages 75 and 85? With regard to the second question, in this course, whenever we say “between ages x and y”, we mean between the x th birthday and the y th birthday. To say someone dies between ages 75 and 85 means that the person dies after reaching the 75th birthday, but before reaching the 85th birthday. If the person dies one month after his 85th birthday, he has not died between ages 75 and 85. For our survival models, we will use two styles of notation: probability (the type you use in probability courses, which writes arguments of functions with parentheses after the function symbol, like f ( x ) ) and actuarial (which, as you will see, writes arguments as subscripts). We will use actuarial notation most of the time, but since you are probably already familiar with probability notation, we will start by discussing that, and then we’ll define actuarial notation in terms of probability notation.

2.1

Probability notation

We first define Tx as the random variable for time to death for someone age x. Thus, for someone age 50, T50 is the amount of time until he dies, and to say T50  32.4 means that the person who was originally age 50 died when he was age 82.4, so that he lived exactly 32.4 years. We will use the symbol ( x ) to mean “someone age x”, so (50) means “someone age 50”. It is very common in this course to use the letter x to mean age. In probability notation, FT ( t ) is the cumulative distribution function of T, or Pr (T ≤ t ) . Usually, the cumulative distribution function is called the distribution function, dropping the word “cumulative”. As an example of a cumulative distribution function, FT50 (30) is the probability that (50) does not survive 30 years. Rather than using a double subscript, we will abbreviate the notation for the cumulative distribution function of Tx as Fx ( t ) , or F50 (30) in our example. The complement of the distribution function is called the survival function and is denoted by ST ( t ) . In other words, ST ( t )  Pr (T > t ) . Thus ST50 (30) is the probability that (50) survives 30 years. In general, ST ( t )  1 − FT ( t ) . Once again, we’ll abbreviate the notation as S x ( t ) , or S50 (30) in our example. If we wanted to express the probability that (40) will die between ages 75 and 85 in terms of distribution functions, we would write it as Pr (35 < T40 ≤ 45)  F40 (45) − F40 (35) and if we wanted to express it in terms of survival functions, we’d write Pr (35 < T40 ≤ 45)  S40 (35) − S40 (45) LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

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2. SURVIVAL DISTRIBUTIONS: PROBABILITY FUNCTIONS

20

In the probability expressions, you may wonder why I made the inequality strict on one side and not on the other. I did this to be consistent with the definitions of F and S. However, we will always assume that Tx is a continuous random variable, so it doesn’t matter whether the inequalities are strict or not. Since we mentioned continuity as a desirable property of a survival function, let’s discuss required and desirable characteristics of a survival function. A survival function must have the following properties: 1. S x (0)  1. Negative survival times are impossible. 2. S x ( t ) ≥ S x ( u ) for u > t. The function is monotonically nonincreasing. The probability of surviving a longer amount of time is never greater than the probability of surviving a shorter amount of time. 3. limt→∞ S x ( t )  0. Eventually everyone dies; Tx is never infinite. Those are the required properties of a survival function. Examples of valid survival functions (although they may not represent human mortality) are • S x ( t )  e −0.01t x+1 • Sx ( t )  x+1+t   1 − 0.01t • Sx ( t )   0

t ≤ 100 t > 100

 Examples of invalid survival functions are

  50 − t • Sx ( t )   0 

• S x ( t )  | cos t |.

t ≤ 50 . t > 50

Violates first property.

Violates second and third properties.

We will also assume the following properties for our survival functions: • S x ( t ) is differentiable for t ≥ 0, with at most only a finite number of exceptions. Differentiability will allow us to define the probability density function (except at a finite number of points). • limt→∞ tS x ( t )  0. This will assure that mean survival time exists. • limt→∞ t 2 S x ( t )  0. This will assure that the variance of survival time exists. In the three examples of valid survival functions given above, you may verify that the first and third ones satisfy all of these properties but the second one does not satisfy the second and third properties. We would now like to relate the various Tx variables (one variable for each x) to each other. To do this, note that each Tx is a conditional random variable: it is the distribution of survival time, given that someone survived to age x. We can relate them using conditional probability: Probability ( x ) survives t + u years  Probability ( x ) survives t years × Probability ( x + t ) survives u years or Pr (Tx > t + u )  Pr (Tx > t ) Pr (Tx+t > u ) S x ( t + u )  S x ( t ) S x+t ( u ) Sx ( t + u ) S x+t ( u )  Sx ( t )

(2.1)

In English: If you’re given the survival function for ( x ) , and you want to know the probability that someone t years older than x survives another u years, calculate the probability of ( x ) surviving t + u years, and divide by the probability that ( x ) survives t years. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

2.1. PROBABILITY NOTATION

21

A special case is x  0, for which (changing variables: the new x is the old t, the new t is the old u) Sx ( t ) 

S0 ( x + t ) S0 ( x )

(2.2)

The corresponding relationship for distribution functions is Pr (T0 ≤ x + t ) − Pr (T0 ≤ x ) Pr (T0 > x ) F0 ( x + t ) − F0 ( x ) Fx ( t )  1 − F0 ( x )

Pr (Tx ≤ t ) 

(2.3)

Example 2A The survival function for newborns is

q 100−t   100 S0 ( t )    0

t ≤ 100 t > 100

Calculate 1. The probability that a newborn survives to age 75 but does not survive to age 84. 2. The probability that (20) survives to age 75 but not to age 84. 3. F60 (20) . Answer: Write each of the items we want to calculate in terms of survival functions. 1. We want S0 (75) − S0 (84) .

r

25  0.5 100

r

16  0.4 100

S0 (75)  S0 (84) 

Pr (75 < T0 ≤ 84)  0.5 − 0.4  0.1 2. We want S20 (55) − S20 (64) . We’ll use equation (2.1) to calculate the needed survival functions. S0 (75) 0.5 √  0.559017 S0 (20) 80/100 S0 (84) 0.4 S20 (64)  √  0.447214 S0 (20) 80/100 S20 (55) 

Pr (55 < T20 ≤ 64)  0.559017 − 0.447214  0.111803 3. F60 (20)  1 − S60 (20) , and √ S0 (80) 0.2 √ S60 (20)  √  0.5 S0 (60) 0.4 √ F60 (20)  1 − 0.5  0.292893

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2. SURVIVAL DISTRIBUTIONS: PROBABILITY FUNCTIONS

22

?

Quiz 2-1 (40) is subject to the survival function

  1 − 0.005t S40 ( t )    1.3 − 0.02t

t < 20 20 ≤ t ≤ 65

 Calculate the probability that (50) survives at least 30 years.

2.2

Actuarial notation

Actuarial notation puts arguments of functions in subscripts and sometimes superscripts before and after the base function symbol instead of using parenthesized arguments. The first function we work with is S x ( t )  Pr (Tx > t ) . The actuarial symbol for this is t p x . The letter p denotes the concept of probability of survival. The x subscript is the age; the t presubscript is the duration. The complement of the survival function is Fx ( t )  Pr (Tx ≤ t ) . The actuarial symbol for this is t q x . The letter q denotes the concept of probability of death. A further refinement to this symbol is the probability of delayed death: the probability that Tx is between u and u + t, which is denoted by u|t q x .1 For all three symbols, the t (but not the u) is usually omitted if it is 1. To summarize the notation: tpx

 Sx ( t )

tqx

 Fx ( t )

u|t q x  Fx ( t + u ) − Fx ( u )  S x ( u ) − S x ( t + u )

The following relationships are clear: px  1 − qx tpx t p x u p x+t t|u q x

 1 − tqx  t+u p x  t p x u q x+t

There are two additional useful formulas for t|u q x . The probability of ( x ) dying in the period from t to t + u is the probability that ( x ) survives t years and does not survive t + u years, or t|u q x

 t p x − t+u p x

(2.4)

The probability of ( x ) dying in the period from t to t + u is the probability that ( x ) dies within t + u years minus the probability that ( x ) dies within t years, or t|u q x

 t+u q x − t q x

(2.5)

Example 2B You are given the following mortality table:

1Both textbooks use a large line on the baseline for this symbol, like u |t q x . I think this is ugly, and none of the older textbooks write it this way. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

2.3. LIFE TABLES

23

x

qx

60 61 62 63 64

0.001 0.002 0.003 0.004 0.005

Calculate the probability that a person age 60 will die sometime between 2 and 5 years from now. Answer: The actuarial notation for what we are calculating is 2|3 q60 . One way to calculate this is as the probability of living 2 years minus the probability of living 5 years, or 2 p60 − 5 p60 . We calculate: 2 p 60

 p60 p61  (1 − q 60 )(1 − q 61 )  (1 − 0.001)(1 − 0.002)  0.997002

5 p 60

 2 p60 p 62 p63 p64  0.997002 (1 − q62 )(1 − q 63 )(1 − q64 )  0.997002 (1 − 0.003)(1 − 0.004)(1 − 0.005)  0.985085

2|3 q 60

 0.997002 − 0.985085  0.011917



In the following example, we’ll relate actuarial and probability notation. Example 2C You are given that 100 S0 ( t )  100 + t

!2

Calculate 5| q 40 Answer: First express the desired probability in terms of survival functions, using equation (2.4). 5| q 40

 5p 40 − 6p 40  S40 (5) − S40 (6)

Then express these in terms of S0 , using equation (2.2). S0 (45) (100/145) 2   0.932224 S0 (40) (100/140) 2 S0 (46) (100/146) 2 S40 (6)    0.919497 S0 (40) (100/140) 2

S40 (5) 

The answer2 is 5| q 40  0.932224 − 0.919497  0.012727

2.3



Life tables

A life table is a concrete way to look at the survivorship random variable. A life table specifies a certain number of lives at a starting integer age x0 . Usually x0  0. This number of lives at age x0 is called the radix. Then for each integer x > x0 , the expected number of survivors is listed. The notation for the number of lives listed in the table for age x is l x . 2Without intermediate rounding, the answer would be 0.012726. We will often show rounded values but use unrounded values in our calculations. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

2. SURVIVAL DISTRIBUTIONS: PROBABILITY FUNCTIONS

24

Assuming for simplicity that x0  0, the random variable for the number of lives at each age, L ( x ) , is a binomial random variable with parameters l0 and x p 0 , so the expected number of lives is l x  l 0 x p 0 . Similarly, l x+t  l x t p x . More importantly, t p x can be calculated from the table using t p x  l x+t /l x . A life table also lists the expected number of deaths at each age; d x is the notation for this concept. Thus d x  l x − l x+1  l x q x . Therefore, q x  d x /l x . The life table also makes it easy to compute u|t q x  ( l x+u − l x+t+u ) /l x . Here’s an example of a life table: x

lx

dx

70 71 72 73 74

100 90 75 60 40

10 15 15 20 18

Notice that on each line, d x  l x − l x+1 . We can deduce that l75  l 74 − d74  40 − 18  22. Example 2D Using the life table above, calculate 3p 71 . Answer: 3p 71



l 74 40 4   l 71 90 9



The notation n d x is the number of deaths occurring within n years after age x; d x  1 d x , and n d x  j0 d x+ j . On exams, they base questions on the Illustrative Life Table, an abridged version of a table in the Bowers textbook from the previous syllabus. In this table, the column for d x is omitted, and must be deduced from l x if needed. However, a column of mortality rates, 1000q x , is given, even though mortality rates could also be computed from the l x ’s. Since life tables are so convenient, it is sometimes easier to build a life table to solve a probability question than to work the question out directly.

Pn−1

Continuation of Example 2B. Redo Example 2B using life tables. Answer: We will arbitrarily use a radix of 1,000,000 at age 60. Then we recursively calculate l x , x  61, . . . , 65 using l x+1  l x (1 − q x ) . x

qx

lx

60 61 62 63 64 65

0.001 0.002 0.003 0.004 0.005

1,000,000 999,000 997,002 994,011 990,035 985,085

The answer is (997,002 − 985,085) /1,000,000  0.011917 . An exam question may ask you to fill in the blanks in a life table using probabilities. Example 2E You are given the following life table: LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM



EXERCISES FOR LESSON 2

25

x 0 1 2

lx

dx

px

50 0.98 890

Calculate 2p 0 . Answer: We back out l0 : 890  908.16 0.98 l0  908.16 + 50  958.16

l1 

Hence 2p 0  890/958.2  0.9288 .

?



Quiz 2-2 You are given: • d48  80 • l50  450 • 3|2 q45  1/6 • 3 p45  2/3 Determine d49 .

Note Although we’ve spoken about human mortality throughout this lesson, everything applies equally well to any situation in which you want to study the time of failure random variable. Failure doesn’t even have to be a bad thing. Thus, we can study random variables such as time until becoming an FSA, time until first marriage, etc. Define a random variable measuring time until the event of interest, and then you can define the actuarial functions and build a life table.

Exercises Probability Notation 2.1. [CAS4-S87:16] (1 point) You are given the following survival function:

  (10000 − x 2 ) /10000 S0 ( x )   0

0 ≤ x ≤ 100 x > 100

 Calculate q32 . A. B. C. D. E.

Less than 0.005 At least 0.005, but less than 0.006 At least 0.006, but less than 0.007 At least 0.007, but less than 0.008 At least 0.008

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2. SURVIVAL DISTRIBUTIONS: PROBABILITY FUNCTIONS

26

Table 2.1: Summary for this lesson

Probability notation Fx ( t )  Pr (Tx ≤ t ) S x ( t )  1 − Fx ( t )  Pr (Tx > t ) Sx ( t + u ) S x+u ( t )  Sx ( u ) S0 ( x + t ) Sx ( t )  S0 ( x ) F0 ( x + t ) − F0 ( x ) Fx ( t )  1 − F0 ( x )

(2.1) (2.2) (2.3)

Actuarial notation  S x ( t )  probability that ( x ) survives t years q  Fx ( t )  probability that ( x ) dies within t years t x q  probability that ( x ) survives t years and then dies in the next u years. t|u x tpx

t+u p x

 t p x u p x+t

t|u q x

 t p x u q x+t  t p x − t+u p x

(2.4)

 t+u q x − t q x

(2.5)

Life table functions l x+t lx l x − l x+t t dx  tqx  lx lx l d x+t − l x+t+u u x+t  t|u q x  lx lx tpx



2.2. You are given • •

S10 (25)  0.9 F20 (15)  0.05.

Determine S10 (10) . 2.3.

You are given the survival function S0 ( t ) 

t 2 − 190t + 9000 9000

t ≤ 90

Determine the probability that a life currently age 36 dies between ages 72 and 81.

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EXERCISES FOR LESSON 2

27

Actuarial Notation 2.4. [CAS4-S88:16] (1 point) Which of the following are equivalent to t p x ? A. B. C. D. E.

t|u q x

− t+u p x t+u q x − t q x + t+u p x t q x − t+u q x + t p x+u t q x − t+u q x − t p x+u The correct answer is not given by (A), (B), (C), or (D).

2.5. You are given: • The probability that a person age 50 is alive at age 55 is 0.9. • The probability that a person age 55 is not alive at age 60 is 0.15. • The probability that a person age 50 is alive at age 65 is 0.54. Calculate the probability that a person age 55 dies between ages 60 and 65. 2.6. [4-S86:13] You are given that t| q x  0.10 for t  0, 1, . . . , 9. Calculate 2 p x+5 . A. 0.40

B. 0.60

C. 0.72

D. 0.80

E. 0.81

2.7. [150-82-94:10] You are given the following: • The probability that a person age 20 will survive 30 years is 0.7. • The probability that a person age 45 will die within 5 years and that another person age 40 will survive 5 years is 0.0475. • The probability that a person age 20 will survive 20 years and that another person age 40 will die within 5 years is 0.04. Calculate the probability that a person age 20 will survive 25 years. A. 0.74

B. 0.75

C. 0.76

D. 0.77

E. 0.78

2.8. [CAS4A-S98:13] (2 points) You are given the following information: 1. 2. 3. 4.

The probability that two 70-year-olds are both alive in 20 years is 16%. The probability that two 80-year-olds are both alive in 20 years is 1%. There is an 8% chance of a 70-year-old living 30 years. All lives are independent and have the same expected mortality.

Determine the probability of an 80-year-old living 10 years. A. B. C. D. E.

Less than 0.35 At least 0.35, but less than 0.45 At least 0.45, but less than 0.55 At least 0.55, but less than 0.65 At least 0.65

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Life Tables 2.9. You are given the following mortality table: x

lx

60 61 62 63

1000

dx

x−60| q 60

100 0.07 780

Calculate q60 . 2.10.

[CAS4A-S93:2] (1 point) You are given the following information: l 1  9700 q 1  q 2  0.020 q 4  0.026 d3  232

Determine the expected number of survivors to age 5. A. B. C. D. E. 2.11.

Less than 8,845 At least 8,845, but less than 8,850 At least 8,850, but less than 8,855 At least 8,855, but less than 8,860 At least 8,860 [CAS4A-F93:1] (1 point) You are given the following mortality table: x 20 21 22 23 24

qx

lx

dx

30,000

1,200

27,350 0.0700 0.0790

23,900

Determine the probability that a life age 21 will die within two years. A. B. C. D. E.

Less than 0.0960 At least 0.0960, but less than 0.1010 At least 0.1010, but less than 0.1060 At least 0.1060, but less than 0.1110 At least 0.1110

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EXERCISES FOR LESSON 2

2.12.

29

Jack enters a mortality study at age 25. He dies between ages 65 and 67.

Which of the following does not express the likelihood of this event? A. B. C. D. E. 2.13.

40p 25

· 2q 65 S0 (65) − S0 (67) S0 (25) p 40 25 − 42p 25 d66 + d67 l 25 40|2q 25 [CAS4A-S92:4] (2 points) You are given the following mortality table: x

lx

qx

50 51 52 53 54

1,000

0.020

dx 32 30 28

0.028

In a group of 800 people age 50, determine the expected number who will die while age 54. A. B. C. D. E. 2.14.

Less than 21 At least 21, but less than 24 At least 24, but less than 27 At least 27, but less than 30 At least 30 [CAS4A-S99:12] (2 points) Given the following portion of a life table: x

lx

dx

px

qx

0 1 2 3 4 5 6 7

1,000 — 750 — — 200 — —

— — — — — 120 — 20

0.875 — — — — — — —

— — 0.25 — — — — 1.00

Determine the value of p1 · p2 · p3 · p 4 · p5 · q6 . A. B. C. D. E.

Less than 0.055 At least 0.055, but less than 0.065 At least 0.065, but less than 0.075 At least 0.075 The answer cannot be determined from the given information.

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[3-S00:28] For a mortality study on college students:

2.15.

Students entered the study on their birthdays in 1963. You have no information about mortality before birthdays in 1963. Dick, who turned 20 in 1963, died between his 32nd and 33rd birthdays. Jane, who turned 21 in 1963, was alive on her birthday in 1998, at which time she left the study. All lifetimes are independent. Likelihoods are based upon the Illustrative Life Table.

• • • • • •

Calculate the likelihood for these two students. A. 0.00138

B. 0.00146

C. 0.00149

D. 0.00156

E. 0.00169

2.16. Mortality follows the Illustrative Life Table. Jack and Jill are two independent lives of ages 25 and 30 respectively. Calculate the probability of Jack and Jill both living to at least age 65 but not to age 90. 2.17.

[CAS4A-F98:15] (2 points) Light bulbs burn out according to the following life table: l0 l1 l2 l3 l4

1,000,000 800,000 600,000 300,000 0

A new plant has 2,500 light bulbs. Burned out light bulbs are replaced with new light bulbs at the end of each year. What is the expected number of new light bulbs that will be needed at the end of year 3? A. B. C. D. E. 2.18.

Less than 800 At least 800, but less than 860 At least 860, but less than 920 At least 920, but less than 980 At least 980 You are given the following life table: x

lx

dx

80 81 82 83

5000 4941 4864 4790

59 77 74 80

Let X be the number of survivors to age 83 from a cohort of 5000 lives at age 80. Calculate Var ( X ) . 2.19. [M-F05:31] The graph of a piecewise linear survival function, S0 ( t ) , consists of 3 line segments with endpoints (0,1), (25,0.50), (75,0.40), (100,0). 20|55q 15 Calculate . 55q 35 A. 0.69

B. 0.71

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C. 0.73

D. 0.75

E. 0.77

EXERCISES FOR LESSON 2

31

Additional old CAS Exam 3/3L questions: S05:29, S07:5, S08:14, F08:14, S09:2, F09:1, S10:2, F10:2, S11:2, S12:4, F13:1 Additional old CAS Exam LC questions: S14:3, F14:1

Solutions 2.1. Translate q 32 into survival functions and use using equation (2.2). q32  F32 (1)  1 − S32 (1) S0 (33) S32 (1)  S0 (32) 10000 − 332   0.992758 10000 − 322 F32 (1)  1 − 0.992758  0.007242

(D)

2.2. The probability of (10) surviving 25 years is the probability of (10) surviving 10 years and then another 15 years. S10 (25)  S10 (10) S20 (15) 0.9  S10 (10)(1 − 0.05) 0.9 S10 (10)   0.9474 0.95 2.3. We will need S0 (36) , S0 (72) , and S0 (81) . 362 − 190 (36) + 9000  0.384 9000 2 72 − 190 (72) + 9000 S0 (72)   0.056 9000 812 − 190 (81) + 9000 S0 (81)   0.019 9000 36|9q 36  36p 36 − 45p 36 S0 (72) S0 (81)  − S0 (36) S0 (36) 0.056 0.019  −  0.096354 0.384 0.384 S0 (36) 

2.4. Using t|u q x  t p x − t+u p x , (A) becomes t p x − 2 t+u p x , so it doesn’t work. In (B), we note that t+u q x + t+u p x  1, so it becomes 1 − t q x  t p x , the correct answer. (C) and (D) each have t p x+u which is a function of survival to age x + u, and no other term to cancel it, so those expressions cannot possibly equal t p x which only depends on survival to age x. (B)

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2.5. We need 5|5 q 55  5 p55 − 10 p55 . 5 p 55

 1 − 5 q55  1 − 0.15  0.85

15 p 50

 0.54

5 p 50 10 p 55

 0.54

0.9 10 p55  0.54 10 p 55

 0.6

The answer is 0.85 − 0.6  0.25 . 2.6. It’s probably easiest to go from t| q x to 2 p x+5 , which is what we need, by using a life table. If we start with a radix of l x  10, then since 1/10 of the population dies each year, l x+t  10 − t. Then 2 p x+5



l x+7 10 − 7   0.6 l x+5 10 − 5

(B)

2.7. In a question like this, your first step should be to determine what the variables are and what the equations are. There should be the same number of equations and variables. In this question, if we analyze the three statements and the probability we’re looking for, we notice that all probabilities start or end at ages 20, 40, 45, or 50. The variables are therefore 20 p20 , 5 p40 , and 5 p45 , and everything can be expressed in terms of these. Let u  20 p20 , v  5 p40 , and w  5 p45 . Then the 3 statements give us these 3 equations: uvw  0.7 v (1 − w )  0.0475 u (1 − v )  0.04 We must find uv. It’s therefore reasonable to substitute w 

0.7 uv

!

v−v

0.7  0.0475 uv

or

v−

into the second equation:

0.7  0.0475 u

Then we substitute u  0.04/ (1 − v ) into this equation. 0.7 (1 − v )  0.0475 0.04 v − 17.5 + 17.5v  0.0475 v−

18.5v  17.5475 v  0.948514 0.04 u  0.776903 1−v The answer is uv  (0.948514)(0.776903)  0.736903 . (A) 2.8. There are three variables: x  10 p70 , y  10 p80 , and z  10 p90 . We are given 1. ( x y ) 2  0.16 ⇒ x y  0.4 2. ( yz ) 2  0.01 ⇒ yz  0.1 3. x yz  0.08 and we want y. From the first and third statement, z  0.08/0.4  0.2. Then from the second statement, y  0.1/0.2  0.5 . (C) LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

EXERCISE SOLUTIONS FOR LESSON 2

33

2.9. Did you notice that you are given x−60| q 60 rather than q x ? Since 2| q 60  0.07, then d62  0.07l 60  70 and l62  l 63 + d62  780 + 70  850. Then l61  850 + d61  950 and d60  l60 − l61  1000 − 950  50, so q60  d60 /l60  50/1000  0.05 . 2.10.

We recursively compute l x through x  5. l2  l1 (1 − q 1 )  9700 (1 − 0.020)  9506 l3  l2 (1 − q 2 )  9506 (1 − 0.020)  9315.88 l4  l3 − d3  9315.88 − 232  9083.88 l5  l4 (1 − q 4 )  9083.88 (1 − 0.026)  8847.70

2.11.

(B)

We need l 21 and l 23 . l 21  30,000 − 1,200  28,800 l 23  23,900/ (1 − 0.0700)  25,698.92 28,800 − 25,698.92  0.1077 2 q 21  28,800

2.12.

(D)

All of these expressions are fine except for (D), which should have d65 + d66 in the numerator.

2.13. l54  1000 (1 − 0.020) − 32 − 30 − 28  890. Then 4| q50  (890/1000)(0.028)  0.02492. For 800 people, 800 (0.02492)  19.936 . (A) 2.14. They gave you superfluous information for year 2. It’s the usual CAS type of humor—it’s not that the answer cannot be determined from the given information (almost never the right answer choice), rather there is too much information provided. We back out l 7  20, since everyone dies that year. We calculate l 6  l 5 − d5  80. l 1  1000 (0.875)  875. Then what the question is asking for is 5| q 1



l 6 − l7 80 − 20  0.06857  l1 875

(C)

2.15. For independent lifetimes, we multiply the likelihood for each life together to get the likelihood of the joint event. d32 l32 − l33 For Dick, the condition is age 20, and death occurs at age 32, so we need  . l20 l 20 l56 For Jane, the condition is age 21 and she survived to age 56, so we need . l21 Looking up the Illustrative Life Table, we find

The answer is

x

lx

20 21 32 33 56

9,617,802 9,607,896 9,471,591 9,455,522 8,563,435

9,471,591 − 9,455,522 9,617,802

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!

!

8,563,435  0.001489 9,607,896

(C)

2. SURVIVAL DISTRIBUTIONS: PROBABILITY FUNCTIONS

34

2.16. For Jack, we need ( l65 − l 90 ) /l 25 , and for Jill we need ( l65 − l90 ) /l30 . From the Illustrative Life Table, we have x

lx

25 30 65 90

9,565,017 9,501,381 7,533,964 1,058,491

7,533,964 − 1,058,491 9,565,017

!

!

7,533,964 − 1,058,491  0.4614 9,501,381

2.17. We have to keep track of three cohorts of light bulbs, the ones installed at times 0, 1, and 2. From the life table, the unconditional probabilities of failure are q0  0.2 in year 0, 1| q 0  0.2 in year 1, and 2| q 0  0.3 in year 3. Of the 2500 original bulbs, 500 apiece fail in years 1 and 2 and 750 in year 3. 500 new ones are installed in year 1, and 100 apiece fail in years 2 and 3. Finally 500 + 100  600 are installed in year 2, of which 120 fail in year 3. The answer is 750 + 100 + 120  970. (D). Here’s a table with these results: Number installed

Failures year 1

Failures year 2

Failures year 3

2500 500 600

500

500 100

500

600

750 100 120 970

Installed in year 0 Installed in year 1 Installed in year 2 Total failures 2.18.

X is binomial with parameters 5000 and 3p 80  4790/5000, so its variance is 5000

4790 5000

!

4790  201.18 5000



1−

The variance is l 80 3p 80 3q 80 . Notice that the variance of the number who die in the three years is the same. 2.19.

The given fraction is 20|55q 15 55q 35



20p 15 55q 35 55q 35

 20p 15 

S0 (35) S0 (15)

By linear interpolation, S0 (15)  0.7 and S0 (35)  0.48. So the quotient is 0.48/0.7  0.685714 . (A)

Quiz Solutions 2-1. Use equation (2.1) to relate S50 to S40 . S40 (40) S40 (10) S40 (40)  1.3 − 0.02 (40)  0.5 S50 (30) 

S40 (10)  1 − 0.005 (10)  0.95 0.5 S50 (30)   0.5263 0.95

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QUIZ SOLUTIONS FOR LESSON 2

35

2-2. Since 3|2 q 45  3 p45 2 q 48 , we deduce 2 q 48  (1/6) / (2/3)  1/4, and therefore l 50 /l 48  1 − 2 q 48  3/4 and l 48  450 (4/3)  600. Then l 50  l 48 − d48 − d49 , or 450  600 − 80 − d49 , implying d49  70 .

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36

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2. SURVIVAL DISTRIBUTIONS: PROBABILITY FUNCTIONS

Lesson 3

Survival Distributions: Force of Mortality Reading: Models for Quantifying Risk (4th or 5th edition) 5.1.3, 5.1.4, 5.3.3, 5.3.4, 6.3.1, 6.3.2 As discussed in the previous lesson, we are assuming that the distribution function of the time-to-death random variable Tx , or Fx ( t ) , is differentiable, and therefore the probability density function of the timeto-death random variable, or f x ( t ) , is fx (t ) 

dS x ( t ) dFx ( t ) − dt dt

We can derive F or S from f by integrating: t

Z Fx ( t ) 

f x ( t ) dt

Z0 ∞ Sx ( t ) 

f x ( t ) dt t

A function related to density is the force of mortality, denoted by µ x . For the time-to-death random variable Tx , this is the hazard rate function discussed in Lesson 1, in the third bullet on page 1. It measures the rate of mortality at age x, given survival to age x. It can also be defined as the probability of death in a small amount of time after age x, given survival to age x, or µ x  lim

dx→0+

1 Pr (T0 ≤ x + dx | T0 > x ) dx

Now, Pr (T0 ≤ x + h | T0 > x ) 

S0 ( x ) − S0 ( x + h ) S0 ( x )

so the force of mortality can be written as 1 S0 ( x ) − S0 ( x + h ) lim S0 ( x ) h→0 h dS0 ( x ) /dx − S0 ( x ) f0 ( x )  S0 ( x )

µx 

Since S x ( t )  S0 ( x + t ) /S0 ( x ) , differentiating this expression with respect to t, we get f x ( t )  f0 ( x + t ) /S0 ( x ) It follows that

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fx (t ) f0 ( x + t ) /S0 ( x ) f0 ( x + t )   S x ( t ) S0 ( x + t ) /S0 ( x ) S0 ( x + t ) 37

(*) (**)

3. SURVIVAL DISTRIBUTIONS: FORCE OF MORTALITY

38

and by (**) µ x+t 

f0 ( x + t ) fx (t )  S0 ( x + t ) S x ( t )

Therefore µ x+t 

fx (t ) Sx ( t )

(3.1)

regardless of which x or t you pick. As long as you hold the sum x + t fixed, you’re free to pick any x and t you wish in order to calculate µ x+t . For example, if you want to calculate µ50 and you are given probability functions f30 ( t ) and S30 ( t ) , you may calculate µ50  f30 (20) /S30 (20) . In the Bowers textbook formerly on the syllabus, despite the fact that µ x+t is only a function of x + t and not a function of x and t separately, they used the notation µ x ( t ) to mean the same as what we mean by µ x+t . This emphasizes that x is generally held fixed and only t varies. For example, µ50  µ30 (20)  µ15 (35) . The “default” value of the argument t is 0, not 1. µ20 is not µ20 (1) . Sometimes, the notation µ ( x ) is used to mean µ x . In this manual, we will almost always use the notation µ x and not µ x ( t ) or µ ( x ) . Since S x ( t )  t p x and Fx ( t )  t q x so that f x ( t )  dt q x /dt, formula (3.1) can be written as µ x+t 

dt q x /dt tpx

(3.2)

The derivative of the log of a function equals the derivative of the function divided by the function: d ln h ( x ) dh ( x ) /dx  dx h (x )

for any function h

Therefore from (*), we have µx  −

d ln S0 ( x ) dx

or d ln S x ( t ) dt d ln t p x − dt

µ x+t  −

(3.3) (3.4)

where we’re free to choose any x and t that add up to the subscript of µ. Going in the other direction t

Z S x ( t )  exp −

! µ x+s ds

(3.5)

0 t

Z t p x  exp −

! µ x+s ds

(3.6)

0 x+t

Z  exp −

! µ s ds

(3.7)

x

The only difference between the last two expressions is the variable of integration; otherwise they are equivalent. What you should remember about equations (3.6) and (3.7) is if you integrate µ x from a lower bound to an upper bound, and then exponentiate the negative of the integral, you get the probability of survival to the age represented by the upper bound, conditional on survival to the age represented by the lower bound. You can calculate t q x as the complement of an exponentiated integral, but you cannot calculate t|u q x directly in this fashion. Instead, you express it as t p x − t+u p x and calculate the two survival probabilities in terms of µs . LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

3. SURVIVAL DISTRIBUTIONS: FORCE OF MORTALITY

39

Example 3A You are given1 that µ35+t  1/ (100 + t ) . 1. Calculate 10p 35 . 2. Calculate 20q 45 . 3. Calculate 10|20q 40 . Answer: 1. We want the probability of survival to 45 given survival to 35, so we’ll integrate µ from age 35 to age 45. µ35 (0) represents age 35 and µ35 (10) represents age 45. 10

Z 10 p 35

 exp −

! µ35+s ds

0 10

Z  exp − 0

ds 100 + s

!



 exp − (ln 110 − ln 100)  exp ln



100 10  110 11

2. We calculate the probability of survival to 65 given survival to 45 and then take the complement.

Z 20 p 45

30

 exp −

! µ35+s ds

10 Z 30

 exp − 10

ds 100 + s



!

 exp − (ln 130 − ln 110)  exp ln 20 q 45





110 11  130 13

2 13

3. This can be evaluated as 10p 40 − 30p 40 or as 10p 40 20q 50 ; either way, we need two integrals to evaluate this. We’ll use the former expression. We already saw in the previous two solutions that for this force of mortality, t p x  (65 + x ) / (65 + x + t ) . 10|20q 40

?

 10p 40 − 30p 40 65 + 40 65 + 40  − 65 + 50 65 + 70 105 105  −  0.135266 115 135



√ Quiz 3-1 The force of mortality is µ x  0.001 x. Calculate the probability of someone age 20 surviving 30 years and then dying in the next 10 years.

1This is a Pareto distribution, a poor model for human mortality, and is used only to illustrate the concept of calculating probabilities from the force of mortality. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

3. SURVIVAL DISTRIBUTIONS: FORCE OF MORTALITY

40

By equation (3.1), the density function for Tx is f x ( t )  S x ( t ) µ x+t  t p x µ x+t

(3.8)

Thus, probabilities of Tx being in a given range can be computed by integrating t p x µ x+t over that range:



t+u

Z



Pr t < Tx ≤ t + u  t|u q x 

spx

µ x+s ds

(3.9)

t

and in particular, t

Z tqx



spx

µ x+s ds

(3.10)

0

In order for µ x to be a legitimate force ofR mortality, the survival function must go to zero as x goes to x infinity, which by equation (3.5) means that 0 µ t dt must go to infinity as x goes to infinity. µ x itself does not have to go to infinity, although for a realistic mortality function it would keep increasing as x increases for x greater than 30 or so. You should understand how a linear transformation of µ affects p. Look at equation (3.6). Since µ is exponentiated to get p, adding something to µ corresponds to multiplying t p x by e to negative t times that something. Multiplying µ by a constant corresponds to raising p to that power. In other words,

• Suppose one person has force of mortality µ x+s and survival probability t p x and another one has force of mortality µ0x+s and survival probability t p 0x . Then a person who has force of mortality µ x+s + µ0x+s for all s has survival probability t p x t p 0x . In particular, if µ0x+s is constant k, then the survival probability for the third person is e −kt t p x .

• If a force of mortality µ x+s corresponds to survival probability t p x , then kµ x+s for all s corresponds to ( t p x ) k .

Example 3B Sarah’s force of mortality is µ x , and her probability of dying at age 70, q 70 , is 0.01. Toby’s force of mortality is µ0x  0.5µ x + 0.1. Calculate Toby’s probability of dying at age 70. Answer: As discussed above, multiplying the force of mortality by 0.5 results in raising the probability of survival to the 0.5 power. Adding 0.1 to the force of mortality for one year multiplies the survival probability by e −0.1 . So the answer will be 1 − (1 − 0.01) 0.5 e −0.1 . Once you get used to this, you will not have to carry out the math. But this time around, let’s work it out step by step. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

EXERCISES FOR LESSON 3

41

Denoting Toby’s functions with primes, we have 1

Z 0 p70

! µ070+t

 exp − 0

by equation (3.6) applied to primed functions

dt

1

Z  exp −

!

using the information given in the example to expand µ0

(0.5µ70+t + 0.1) dt 0 1

Z  . exp −

!

1

Z

0.5µ70+t dt / . exp −

*

0

! 0.1 dt /

+*

+

0

,

-, !  Z 1  * +  . exp −0.5 µ70+t dt / e −0.1 0 , ! 0.5  Z 1  * +  . exp − e −0.1 µ70+t dt / 0 ,    0.5  p70

-

by equation (3.6) applied to unprimed functions

e −0.1



 (1 − q 70 ) 0.5



e −0.1



 (0.990.5 )( e −0.1 )  0.90030 0 Hence, q 70  0.09970 .



Exam questions will expect you to go between µ and S, p, or q, in either direction. Table 3.1 reviews the relationships between these functions. Figure 3.1 diagrams the relationships, with the actuarial notation on top of each circle and the mathematical notation on the bottom.

Exercises 3.1. A person age 70 is subject to the following force of mortality:

  0.01 µ70+t    0.02

t≤5 t>5

 Calculate 20p 70 for this person.

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3. SURVIVAL DISTRIBUTIONS: FORCE OF MORTALITY

42

t qx

 1 − t px

t qx

sT (x ) ( t )

t px

 1 − t qx

FT ( x ) ( t )

0

s px 0

t

Z

px

s

d t qx dt

t 0

t qx 

! µ x ( s ) ds t

Z

px dt t − d

−

Z

1

µx (t ) 

)



(s

µx

)d

t p x  exp −

(t

µx

px

t

d ln p d p x /dt t x t or − dt t px

t px

px

t

µx (t )  −

µ x ( s ) ds

t px

s

µx (t )

t px

hT (x ) ( t )

µx (t )

fT ( x ) ( t )

Figure 3.1: Relationships between t p x , t q x , and µ x+t

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EXERCISES FOR LESSON 3

43

Table 3.1: Formula Summary for this lesson

µ x+t 

fx (t ) Sx ( t )

(3.1)

 µ x+t 

dt q x /dt tpx

(3.2)

d ln S x ( t ) dt d ln t p x − dt −

(3.3) (3.4)

t

Z S x ( t )  exp −

! µ x+s ds

(3.5)

0 t

Z tpx

 exp −

! µ x+s ds

(3.6)

0 x+t

Z  exp −

! µ s ds

(3.7)

x

f x ( t )  t p x µ x+t t|u q x

(3.8)

t+u

Z 

spx

µ x+s ds

(3.9)

t t

Z tqx 

spx

µ x+s ds

(3.10)

0

If µ0x+t  µ x+t + k for all t, then t p 0x  t p x e −kt . More generally, if µ x+t  µˆ x+t + µ˜ x+t for all t, then t p x  t pˆ x t p˜ x If µ0x+t  kµ x+t for all t, then t p 0x 

tpx


k

.

3.2. [CAS4-S88:15] (1 point) The force of mortality is µx 

1 100 − x

Calculate 10 p50 . A. B. C. D. E.

Less than 0.82 At least 0.82, but less than 0.84 At least 0.84, but less than 0.86 At least 0.86, but less than 0.88 At least 0.88

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3. SURVIVAL DISTRIBUTIONS: FORCE OF MORTALITY

44

3.3. [CAS4A-F97:8] (1 point) Given that the force of mortality µ x  2x, determine the cumulative distribution function for the random variable time until death, F0 ( x ) , the density function for that random variable, f0 ( x ) , and the survival function S0 ( x ) . (A) (B) (C) (D) (E)

F0 ( x ) F0 ( x ) F0 ( x ) F0 ( x ) F0 ( x )

2

 1 − ex 2  1 − e −x  1 − 2x 2  1 − ex 2  1 − e −x

f0 ( x ) f0 ( x ) f0 ( x ) f0 ( x ) f0 ( x )

2

 2xe −x 2  2e −x 2  2e −x 2  xe −x 2  2xe −x

S0 ( x ) S0 ( x ) S0 ( x ) S0 ( x ) S0 ( x )

    

2

e −x 2 ex 2 ex e 2x 2 e −x

3.4. [CAS4A-F99:12] (1 point) If l x  100 ( k − 0.5x ) 2/3 and µ50  A. B. C. D. E.

1 48 ,

what is the value of k?

Less than 40 At least 40, but less than 42 At least 42, but less than 44 At least 44, but less than 46 At least 46

3.5. The force of mortality for Kevin is µ x  kx 2 . The force of mortality for Kira is µˆ x  2. Determine the k for which 5p 10 is the same for Kevin and Kira. 3.6. You are given that the force of mortality is µ x  1.5 (1.1x ) , x > 0. Calculate 2 p 1 . 3.7. You are given that xp0



−x 2 − 30x + 18000 30000

0 ≤ x ≤ 120

Develop an expression for µ x valid for 0 < x < 120. 3.8. For a standard life, 5p 45  0.98. Since Boris, age 45, is recovering from surgery, he is subject to extra mortality. Therefore, the µ x applying to Boris is increased for x between 45 and 50. The increase over the µ x for a standard life is 0.002 at x  45, decreasing in a straight line to 0 at age 50. Calculate 5p 45 for Boris. 3.9. [4-S86:26] You are given µ x  2x/ (10,000 − x 2 ) for 0 ≤ x ≤ 100. Determine q x . A. B. C. D. E.

2x + 1 10,000 − x 2 4x + 2 10,000 − x 2 6x + 3 10,000 − x 2 2x + 1 29,999 − 3x 2 − 3x 6x + 1 29,999 − 3x 2 − 3x

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EXERCISES FOR LESSON 3

45

3.10. [4-S86:31] A mortality table has a force of mortality µ x+t and mortality rate q x . A second mortality table has a force of mortality µ∗x+t and mortality rate q ∗x . You are given µ∗x+t  0.5µ x+t for 0 ≤ t ≤ 1. Calculate q ∗x .

p

A. 1 − 1 − q x

B.

p

C. 0.5q x

qx

D. ( q x ) 2

E. q x − ( q x ) 2

3.11. [CAS4A-S93:12] (2 points) A mortality table for a subset of the population with better than average health is constructed by dividing the force of mortality in the standard table by 2. The probability of an 80-year-old dying within the next year is defined in the standard table as q80 and in the revised table it is 0 defined as q 80 . In the standard table q80  0.30. 0 Determine the value of q80 in the revised table.

A. B. C. D. E.

Less than 0.150 At least 0.150, but less than 0.155 At least 0.155, but less than 0.160 At least 0.160, but less than 0.165 At least 0.165 [150-S88:1] You are given:

3.12.

µˆ x+t  µ x+t − k, 0 ≤ t ≤ 1 qˆ x  0, where qˆ x is based on the force of mortality µˆ x+t .

• •

Determine k. A. − ln p x

B. ln p x

C. − ln q x

D. ln q x

E. q x

[150-F88:6] Which of the following functions can serve as a force of mortality?

3.13.

B > 0, 0 < c < 1, x ≥ 0 B > 0, x ≥ 0 k > 0, n > 0, x ≥ 0

Bc x B ( x + 1) −0.5 k ( x + 1) n

I. II. III.

A. I and II only B. I and III only C. II and III only E. The correct answer is not given by A. , B. , C. , or D.

D. I, II and III

[CAS4A-F99:13] (2 points) Which of the following equations define valid mortality functions?

3.14. 1.

µ x  (1 + x ) −3

2.

µx 

3.

f0 ( x ) 

A. 1

x≥0

0.05 (1.01) x e −x/2

x≥0 x≥0 B. 2

C. 1,2

D. 1,3

E. 2,3

[CAS3-F04:7] Which of the following formulas could serve as a force of mortality?

3.15. 1. 2. 3.

µ x  BC x , µ x  a ( b + x ) −1 , µ x  (1 + x ) −3 ,

A. 1 only

B > 0, C > 1 a > 0, b > 0 x≥0

B. 2 only

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C. 3 only

D. 1 and 2 only

E. 1 and 3 only

Exercises continue on the next page . . .

3. SURVIVAL DISTRIBUTIONS: FORCE OF MORTALITY

46

You are given:

3.16.

f40 (15)  0.0010 10p 40  0.96

• •

Calculate f50 (5) . The probability density function for survival of a newborn is

3.17.

f0 ( t ) 

30t 4 (100 − t ) , 1006

0 < t ≤ 100

Calculate µ80 . [150-F88:15] You are given F0 ( x )  1 −

3.18. I.

x p0

1 x+1

for x ≥ 0. Which of the following are true?

 1/ ( x + 1)

µ49  0.02

II. III.

10 p 39

 0.80

A. I and II only B. I and III only C. II and III only E. The correct answer is not given by A. , B. , C. , or D.

D. I, II and III

[150-S90:12] You are given

3.19. • •

µ x  A + e x for x ≥ 0 0.5 p 0  0.50

Calculate A. A. −0.26 3.20.

B. −0.09

C. 0.00

D. 0.09

E. 0.26

[150-81-94:48] You are given: S0 ( x ) 

(10 − x ) 2 100

0 ≤ x ≤ 10.

Calculate the difference between the force of mortality at age 1, and the probability that (1) dies before age 2. A. 0.007

B. 0.010

C. 0.012

D. 0.016

E. 0.024

3.21. [CAS4-S86:17] (1 point) A subgroup of lives is subject to twice the normal force of mortality. In other words, µ0x  2µ x where a prime indicates the rate for the subgroup. Express q 0x in terms of q x . A. B. C. D. E.

(qx )2 ( q x ) 2 − 2q x 2q x − ( q x ) 2 2(qx )2 − qx 2q x + ( q x ) 2

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EXERCISES FOR LESSON 3

47

3.22. [CAS4A-S92:16] (2 points) You are given that p30  0.95 for a standard insured with force of mortality µ30+t , 0 ≤ t ≤ 1. For a preferred insured, the force of mortality is µ30+t − c for 0 ≤ t ≤ 1. Determine c such that the probability that (30) will die within one year is 25% lower for a preferred insured than for a standard. A. B. C. D. E.

Less than 0.014 At least 0.014, but less than 0.015 At least 0.015, but less than 0.016 At least 0.016, but less than 0.017 At least 0.017

3.23. [CAS4A-S96:16] (2 points) A life table for severely disabled lives is created by modifying an existing life table by doubling the force of mortality at all ages. In the original table, q 75  0.12. Calculate q 75 in the modified table. A. B. C. D. E.

Less than 0.21 At least 0.21, but less than 0.23 At least 0.23, but less than 0.25 At least 0.25, but less than 0.27 At least 0.27 [CAS4A-F92:2] (1 point) You are given S0 ( x )  e −x

3.24.

3 /12

for x ≥ 0.

Determine µ x . A. −x 2 /4

B. 1 − x 2 /4

C. x 2 /4

D. ( x 2 /4) e −x

2 /12

E. −x 3 /12

You are given that t p x  1 − t 2 /100 for 0 < t ≤ 10.

3.25.

Calculate µ x+5 . [150-83-96:15] You are given:

3.26.

• µ35+t  µ, 0 ≤ t ≤ 1. • p35  0.985 • µ035+t is the force of mortality for (35) subject to an additional hazard, 0 ≤ t ≤ 1. • µ035+t  µ + c, 0 ≤ t ≤ 0.5 • The additional force of mortality decreases uniformly from c to 0 between age 35.5 and 36. Determine the probability that (35) subject to the additional hazard will not survive to age 36. A. B. C. D. E.

0.015e −0.25c 0.015e 0.25c 1 − 0.985e −c 1 − 0.985e −0.5c 1 − 0.985e −0.75c

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3. SURVIVAL DISTRIBUTIONS: FORCE OF MORTALITY

48

[3-F00:36] Given:

3.27.

µ ( x )  F + e 2x , x ≥ 0 0.4 p 0  0.50

• •

Calculate F. A. −0.20

B. −0.09

C. 0.00

D. 0.09

E. 0.20

You are given the force of failure for a battery is 0.1x, with x measured in hours of use.

3.28.

Calculate the probability that the battery will last 10 hours. [3-S01:28] For a population of individuals, you are given:

3.29.

• Each individual has a constant force of mortality. • The forces of mortality are uniformly distributed over the interval (0, 2) . Calculate the probability that an individual drawn at random from this population dies within one year. A. 0.37

B. 0.43

C. 0.50

D. 0.57

E. 0.63

3.30. [CAS4A-F96:8] (2 points) At birth, infants are subject to a decreasing force of mortality during the early months of life. Assume a newborn infant is subject to a force of mortality given by µx 

1 10 + x

for x ≥ 0, where x is expressed in months.

Calculate the probability that a newborn infant will survive 5 months and die in the ensuing 15 months. A. B. C. D. E. 3.31.

Less than 0.15 At least 0.15, but less than 0.20 At least 0.20, but less than 0.25 At least 0.25, but less than 0.30 At least 0.30 [CAS4A-F97:11] (2 points) You are given a life, age 30, subject to a force of mortality of µ x  0.02x 0.5

for 20 ≤ x ≤ 50

Determine the probability that this life will survive 5 years and die during the following year. A. B. C. D. E.

Less than 0.044 At least 0.044, but less than 0.052 At least 0.052, but less than 0.060 At least 0.060, but less than 0.068 At least 0.068

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Exercises continue on the next page . . .

EXERCISES FOR LESSON 3

49

[150-F97:17] You are given:

3.32.

• The force of mortality is a constant, µ. • µ ≤ 1. • 3|3 q33  0.0030 Calculate 1000µ. A. 0.8

B. 0.9

C. 1.0

D. 1.1

E. 1.2

3.33. You are given that p x  0.85 for a person whose future lifetime has force of mortality µ x+s for s ≤ 1. For another person, future lifetime has force of mortality µ0x ( s )  1.1µ x+s − 0.05 for s ≤ 1. Calculate p x for this other person. 3.34. For a certain individual, mortality follows the Illustrative Life Table, except that the force of mortality is double the force of mortality underlying the Illustrative Life Table between ages 45 and 50. Calculate 5|5 q 42 for this individual. [3-S00:17] The future lifetimes of a certain population can be modeled as follows:

3.35.

• Each individual’s future lifetime is exponentially distributed with constant hazard rate θ. • Over the population, θ is uniformly distributed over (1, 11) . Calculate the probability of surviving to time 0.5, for an individual randomly selected at time 0. A. 0.05

B. 0.06

C. 0.09

D. 0.11

E. 0.12

[3-F02:1] Given: The survival function S0 ( x ) , where

3.36.

 1,     0≤x≤1  x  S0 ( x )   1 − e /100 , 1 ≤ x < 4.5   0, 4.5 ≤ x  Calculate µ4 . A. 0.45

B. 0.55

C. 0.80

D. 1.00

E. 1.20

[150-82-94:15] You are given:

3.37.

R

• • •

1

R  1 − e −R 0 −

µ x+t dt

.

1 ( µ x+t −k ) dt 0

S 1−e . k is a positive constant

Determine an expression for k such that S  23 R.





















A.

ln (1 − p x ) / (1 − 23 q x )

B.

ln (1 − 32 q x ) / (1 − p x )

C.

ln (1 − 23 p x ) / (1 − p x )

D.

ln (1 − q x ) / (1 − 23 q x )

E.

ln (1 − 32 q x ) / (1 − q x )

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Exercises continue on the next page . . .

3. SURVIVAL DISTRIBUTIONS: FORCE OF MORTALITY

50

[3-F02:35] You are given:

3.38. •

µ x+t is the force of mortality.

•

R  1 − e −R 0

R

• •

−

1

µ x+t dt

1 ( µ x+t +k ) dt 0

S 1−e k is a constant such that S  0.75R

Determine an expression for k.





















A.

ln (1 − q x ) / (1 − 0.75q x )

B.

ln (1 − 0.75q x ) / (1 − p x )

C.

ln (1 − 0.75p x ) / (1 − p x )

D.

ln (1 − p x ) / (1 − 0.75q x )

E.

ln (1 − 0.75q x ) / (1 − q x )



3.39. [CAS3-F04:8] Given S0 ( x )  1 − ( x/100) age 36 will die between ages 51 and 64. A. B. C. D. E.

 1/2

, for 0 ≤ x ≤ 100, calculate the probability that a life

Less than 0.15 At least 0.15, but less than 0.20 At least 0.20, but less than 0.25 At least 0.25, but less than 0.30 At least 0.30

3.40. [SOA3-F04:4] For a population which contains equal numbers of males and females at birth: • For males, µ m x  0.10, x ≥ 0 f

• For females, µ x  0.08, x ≥ 0 Calculate q 60 for this population. A. 0.076

B. 0.081

C. 0.086

D. 0.091

E. 0.096

3.41. [M-S05:33] You are given:

  0.05 50 ≤ x < 60 µx    0.04 60 ≤ x < 70  Calculate 4|14q 50 . A. 0.38

B. 0.39

C. 0.41

D. 0.43

E. 0.44

3.42. [M-F05:32] For a group of lives aged 30, containing an equal number of smokers and non-smokers, you are given: • For non-smokers, µ nx  0.08, x ≥ 30 • For smokers, µ sx  0.16, x ≥ 30 Calculate q 80 for a life randomly selected from those surviving to age 80. A. 0.078

B. 0.086

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C. 0.095

D. 0.104

E. 0.112 Exercises continue on the next page . . .

EXERCISES FOR LESSON 3

51

[MLC-S07:1] You are given:

3.43. • • •

3p 70

 0.95  0.96

2p 71 R 75 µx 71

dx  0.107

Calculate 5p 70 . A. 0.85 B. 0.86 C. 0.87 D. 0.88 E. 0.89 Additional old CAS Exam 3/3L questions: S05:30, F05:11,12, S06:10,11, S08:15, F08:12, F10:1, F11:1, S12:1, F12:2 Additional old CAS Exam LC questions: S14:1

Solutions 3.1. We can use equation (3.6) to calculate 20p 70 . Namely, 20

Z 20p 70

 exp −

! µ70+s ds

0

Break the integral up into two parts:

Z

5

5

Z µ70+t dt 

0 20

Z

0.01 dt  0.05 0 20

Z µ70+t dt 

0.02 dt  0.3

5

It follows that 0.70469 .

20 p 70

5

is the product of the negative exponentiated integrals,

20 p 70

 e −0.05−0.3 

3.2. This will be easier after the next lesson. For now, we do it from first principles, using equation (3.6): 10

Z 10 p 50  exp −

! µ50+s ds

0 10

Z  exp − 0

ds 50 − s



!

 exp ln (50 − 10) − ln (50) 

40  0.8 50



(A)

3.3. The choices are so poor that only (E) has S0 ( x )  1−F0 ( x ) , so it is the only choice that could possibly be right, regardless of the mortality assumption. 2 Anyhow, the integral of 2t from 0 to x is x 2 . We then negate and exponentiate to obtain S0 ( x )  e −x , which after complementing and differentiating verifies all three parts of (E).

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3. SURVIVAL DISTRIBUTIONS: FORCE OF MORTALITY

52

3.4.

l x doesn’t work for x > 2k (it’s negative), so the question is not totally accurate. We calculate S0 ( x )  x p0  l x /l 0 and then use equation (3.3). 100 ( k − 0.5x ) 2/3 100k 2/3  2 ln S0 ( x )  ln ( k − 0.5x ) − ln k 3 d 2 0.5 µx  − ln S0 ( x )  dx 3 k − 0.5x 1 1 µ50   3 ( k − 25) 48 S0 ( x ) 

k  41

(B)

3.5. Since for both Kevin and Kira 5p 10 will be the exponential of negative an integral, it suffices to compare the integrals. For Kira, the integral of 2 from 10 to 15 is 10. For Kevin,

Z

15

kx 2 dx  10

k (153 − 103 )  791.6667k 3

So k  10/791.6667  0.01263 . 3.6. This is a Gompertz force of mortality, as we’ll learn in the next lesson. 3

Z 2 p1

 exp −

! µ x dx

1 3

Z  exp −

! x

1.5 (1.1 ) dx 1

1.5 (1.13 − 1.1)  exp − ln 1.1

!

 e −3.63550  0.02637

3.7. Notice that the numerator of x p 0 factors as xp0



−1 ( x − 120)( x + 150) 30000

Thus ln x p 0  − ln 30000 + ln (120 − x ) + ln ( x + 150) . Notice that 120 − x is positive in the range 0 < x < 120. Differentiating and negating, µx 

1 1 1 1 − − − 120 − x x + 150 x − 120 x + 150

If you desire to put this over one denominator, you get µx  −

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2x + 30 x 2 + 30x − 18000

EXERCISE SOLUTIONS FOR LESSON 3

53

3.8. As indicated at the bottom of Table 3.1, if µ x+t  µˆ x+t + µ˜ x+t , then t p x  t pˆ x t p˜ x . So we need to multiply 0.98 by the t p x based on the increase in force. The increase in force, which we’ll call µ˜ x , is a linear function with slope −0.0004 and equal to 0 at 50, so it can be written as (The constant 0.02 is selected as 50 (0.0004) so that µ˜ 50  0.) µ˜ x  0.02 − 0.0004x Then

Z

50

50

Z µ˜ x dx 

45

45

502 − 452  0.005 (0.02 − 0.0004x ) dx  0.02 (5) − 0.0004 2

!

and 5 p˜ 45  e −0.005 . So 5 p 45  0.98e −0.005  0.9751 . One way to do the above integral quickly is to note that since the function is linear, its average value is its median, or 0.001 (half way between 0.002 and 0), and multiplying this average by the length of the interval from 45 to 50, we get 0.005. 3.9. Using equation (3.7), x+1

Z q x  1 − exp −

! µ t dt

x x+1

Z  1 − exp − x

2t dt 10,000 − t 2

!

10,000 − ( x + 1) 2 1− 10,000 − x 2 ( x + 1) 2 − x 2  10,000 − x 2 

3.10.

2x + 1 (A) 10,000 − x 2

As we discussed, halving µ means taking the square root of p.

p √ q ∗x  1 − p ∗x  1 − p x  1 − 1 − q x 3.11.

(A)

√ 0 This is the same situation as the previous exercise. q80  1 − 1 − 0.30  0.1633 . (D)

3.12. 1

Z 1  pˆ x  exp −

! ( µ t − k ) dt

0 1

Z  p x exp −

! (−k ) dt

0

 px e k So e k  1/p x , and k  − ln p x . (A) Note: The only way q x can be 0 is if p x is 1, which means the integral of µ is 0 for an interval of length 1. Barring negative µ, which would be an invalid force of mortality, this can only happen if µ is 0 almost everywhere (to use the language of measure theory). So this exercise represents an artificial situation. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

3. SURVIVAL DISTRIBUTIONS: FORCE OF MORTALITY

54

All of these functions are nonnegative. We need

3.13. I.

R

II.

R R

III.

∞ 0

µ x dx  ∞.

Bc x ∞

 − B , ∞, since 0 < c < 1. #

∞ 0

Bc x dx 

∞ 0

B ( x + 1) −0.5 dx  2B ( x + 1) 0.5  ∞. !

∞ 0

R

ln c 0

ln c

∞ 0

k ( x + 1) n dx 

∞ 1) n+1

k (x + n+1

 ∞. ! 0

(C) 3.14. 1. ∞

Z

∞

∞

Z µ x dx 

(1 + x ) −3 dx  −

0

0

1 1  2 2 2 (1 + x ) 0

The integral does not go to infinity. # 2. ∞

Z

0.05 (1.01) x dx  0

0.05 (1.01x ) →∞ ! ln 1.01

This is a Gompertz law, which we’ll learn more about in subsection 4.1.1, page 63. 3. ∞

Z

∞

Z f ( x ) dx 

0

e −x/2 dx  2 0

A proper density function integrated from 0 to ∞ should integrate to 1. # If 3 had an extra factor 1/2, then it would be OK. (B) 3.15.

1 and 2 are non-negative and integrate to infinity. 3, however, has a finite integral. (D)

3.16.

Since f x ( t )  t p x µ x+t , we have 15p 40

µ55  0.0010

f50 (5)  5p 50 µ55 However, 15p 40  10p 40 5p 50 , so f50 (5) 

15p 40 10p 40

µ55 

0.0010  0.001042 0.96

You can also do this without using force of mortality, as follows: f50 (5)  −

f40 (15) 0.0010 d S40 ( t ) d   S50 ( t )  −  0.001042 dt dt S ( 10 ) 0.96 40 10p 40 t5 t15

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EXERCISE SOLUTIONS FOR LESSON 3

55

3.17. We need f0 (80) /S0 (80) . Let’s integrate f0 to obtain S0 . Since 100 is the upper limit, the integral goes up to 100 rather than ∞.

Z

100

S0 (80) 

f0 ( t ) dt 80 Z 100

 80

30 (100t 4 − t 5 ) dt 1006

30 100 (1005 − 805 ) 1006 − 806  −  0.344640 5 6 1006

!

30 (804 )(20)  0.024576 1006 0.024576   0.071309 0.344640

f0 (80)  µ80

3.18. x p0  S0 ( x )  1 − F0 ( x ) , so I is true. ln S0 ( x )  − ln ( x + 1) , and then µ x  49) 40 II true. 10 p39  SS00 ((39 )  50  0.80 making III true. (D) 3.19.

1 x+1 ,

We use equation (3.7). 0.5

Z 0.50  0.5 p0  exp −

! x

(A + e ) dx 0

0.5

Z ln 2 

(A + e x ) dx 0

0.693147  0.5A + ( e 0.5 − 1) 0.693147  0.5A + 0.648721 0.693147 − 0.648721  0.08885 A 0.5 3.20.

(D)

The probability that (1) dies before age 2 is q1 

S0 (1) − S0 (2) 0.81 − 0.64   0.2099 S0 ( 1 ) 0.81

The force of mortality at age 1 is d (10 − x ) 2 µ1  − ln dx 100

! 2 2    0.2222 10 − x 1 9 1

So the difference is 0.2222 − 0.2099  0.0123 (C) 3.21.

p 0x  ( p x ) 2 since doubling the force of mortality squares the survival rate. 1 − q 0x  (1 − q x ) 2  1 − 2q x + q 2x q 0x  2q x − ( q x ) 2

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(C)

so µ49  0.02, making

3. SURVIVAL DISTRIBUTIONS: FORCE OF MORTALITY

56

3.22. Let primes be used for preferred insureds. Subtracting c from the force of mortality multiplies the survival probability by e c . 0 1 − 0.75 (0.05)  p30  p 30 e c  0.95e c 0.9625  ec 0.95 ! 0.9625  0.013072 c  ln 0.95

(A)

3.23. Let primes be used for substandard insureds. Doubling the force of mortality squares the survival probability. 0 p75  (1 − q 75 ) 2  0.7744 0 q 75  1 − 0.7744  0.2256

3.24.

Using equation (3.3), we log e −x

entiate getting 3.25.

x 2 /4

3 /12

getting −x 3 /12, negate the expression getting x 3 /12, and differ-

. (C)

Using equation (3.4), µ x+t  −

so µ x+5  3.26.

(B)

10 75

d (1 − t 2 /100) /dt 2t  1 − t 2 /100 100 − t 2

 0.13333 .

0 Let p35 be the modified probability of survival. Then, by equation (3.7), 0 p35  e−

R

1 0

µ035+t dt

 e−

R

1 0

µ35+t dt −

e

R

1 0 ( µ35+t −µ35+t ) dt 0

 p35 e −

So p 0x  p x times e raised to the integral of negative the additional force of mortality from 0 to 1. The additional force of mortality is the difference between the two lines in the figure to the right. Its integral is the area of the shaded trapezoid, which has legs of lengths 0.5 and 1 and height c. This area is 0.5 (0.5 + 1) c  0.75c. So p 0x  0.985e −0.75c . The answer is therefore (E). 3.27.

R

1 0 ( µ35+t −µ35+t ) dt 0

µ+c µ 35

By equation (3.7), 0.4

Z 0.50  exp −0.4F −

! e 2x dx

0

e 0.8 − 1 ln 0.50  −0.4F −  −0.4F − 0.61277 2 −0.61277 − ln 0.50 F  0.20094 (E) 0.4 3.28.

The survival probability to time x is x

Z 0 2

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0.1u du  exp −0.05x 2

exp − so 10p 0  e −0.05 (10 )  e −5  0.006738 .

!



µ035+t µ35+t 35.5

36

x

EXERCISE SOLUTIONS FOR LESSON 3

3.29.

The law of total probability says that P ( X )  2

Z 0

3.30.

57

R

P ( X | y ) f ( y ) dy, where

1 1 (1 − e −µ ) dµ  (2 + e −2 − 1)  0.5677 2 2

We’ll use monthly subscripts. Then we need 5 p x and 20 p x .

Z

5 p0

20 p 0

2 1 − 3 3 3.31.

(D)

5

!

1  exp − dx 0 10 + x 2  exp (ln 10 − ln 15)  3 ! Z 20 1  exp − dx 10 + x 0 1  exp (ln 10 − ln 30)  3 1 (E)   0.3333 3

We need 5 p30 and 6 p30 .

Z 5 p 30

35

 exp −

! 0.02x

0.5

dx

30

0.02  exp − (351.5 − 301.5 ) 1.5





!

 exp . −

*

0.02 (42.74603) +/  0.565555 1.5

,

!

6 p 30

 exp . −

*

0.02 (361.5 − 301.5 ) +/ 1.5

-

, !

 exp . −

*

0.02 (51.68323) +/  0.502023 1.5

,

-

0.565555 − 0.502023  0.063533

3.32.

(D)

Let 3 p be the probability of surviving 3 years; since µ is constant, 3 p does not vary with x. Then 3 p (1 3p

But µ  − ln p  − is 1.005 . (C)

ln 3 p 3

2

− 3 p )  0.0030

− 3 p + 0.0030  0

√ 1 ± 1 − 0.012  0.996991, 0.003009 3p  2

 0.001005, 1.935376. Since only the first solution is not greater than 1, the answer

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3. SURVIVAL DISTRIBUTIONS: FORCE OF MORTALITY

58

3.33. is

Let p 0x be p x for the other person. As discussed in the lesson, the adjustment to survival probabilities

3.34.

Doubling the force of mortality corresponds to squaring the survival rate. We need 5 p42 5 q 47 .

p 0x  e 0.05 p x

5 p 42


1.1

 e 0.05 (0.851.1 )  0.8792

 3 p42 2 p45 9,164,051  9,259,571

5 p 47

9,088,049 9,164,051

!2  0.973336

 3 p47 2 p50 

5|5 q 42

!

8,950,901 9,088,049

!2

!

8,840,770  0.958110 8,950,901

 0.973336 (1 − 0.958110)  0.040773

θ serves the same function as µ x . The probability of survival to time 0.5 for a given individual is  e −0.5θ . To obtain the probability of survival for a randomly chosen individual, by the Law of Total Probability, we must integrate this over the uniform distribution of the population, which has density 0.1 on the interval [1, 11]. 3.35.

0.5 p x

11

Z 0.1 1

11

e −0.5θ dθ  −0.2e −0.5θ

1





 0.2 e −0.5 − e −5.5  0.1205

3.36.

Use equation (3.3).

! ex + * . / ln S0 ( x )  ln 1 − 100 , ! −d ln S0 ( x ) ex 1 dx



100

1 − e x /100

e4  1.2026 100 − e 4 3.37.

! 

ex 100 − e x

(E)

R  1 − p x , and S  1 − p x e k , so 1 − px e k  ek 

2 (1 − p x ) 3 1 − 23 (1 − p x ) px

k  ln *

,

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1 − 32 (1 − p x ) 1 − qx

+ -

(E)

(E)

EXERCISE SOLUTIONS FOR LESSON 3

3.38.

59

By equation (3.7), R  1 − p x  q x . Also, 1 − S  e−

R

1 ( µ x+t +k ) dt 0

 e−

R

1 0

µ x+t dt −k

e

 (1 − R ) e −k

Substituting S  0.75R: 1 − 0.75R  (1 − R ) e −k 1 − 0.75R e −k  1−R k  ln

3.39.

1 − qx 1−R  ln 1 − 0.75R 1 − 0.75q x

(A)

Calculate the values of S0 ( x ) that we need: S0 (36)  0.641/2  0.8 S0 (51)  0.491/2  0.7 S0 (64)  0.361/2  0.6

So 15|13 q 36



0.7 − 0.6  0.125 0.8

(A)

3.40. The population contains equal numbers at birth, but the numbers are not equal at age 60 due to f m and q60 . different survivorship, so you can’t simply average q 60 Survivorship from birth to age 60 is e −6  0.002479 for males, e −4.8  0.008230 for females, so these are the relative proportions in the population. f m  1 − p60  1 − e −0.1  0.095163 and for females q 60  1 − e −0.08  0.076884. Weighting For males, q 60 these rates with the proportions in the population, q 60 

0.002479 (0.095163) + 0.008230 (0.076884)  0.081115 0.002479 + 0.008230

As an alternative, you can calculate weighted survivorship at 60 and 61: S0 (60)  0.5 ( e −0.10 (60) + e −0.08 (60) )  0.5 (0.010708) S0 (61)  0.5 ( e −0.10 (61) + e −0.08 (61) )  0.5 (0.009840) and then compute q 60  1 − 0.009840/0.010708  0.081115 . 3.41. 4|14q 50

 4p 50 14q 54



 e −0.05 (4) 1 − e −0.05 (6) −0.04 (8)  e −0.2 − e −0.82  0.378299

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 (A)

(B)

3. SURVIVAL DISTRIBUTIONS: FORCE OF MORTALITY

60

3.42. Calculate the probability of survival to age 80 and the probability of survival to age 81. The difference, divided by the probability of survival to age 80, is the probability that someone surviving to age 80 does not survive to age 81, which is the definition of q 80 . 51p 30

 0.5 ( e −51 (0.08) + e −51 (0.16) )  0.017193

50p 30

 0.5 ( e −50 (0.08) + e −50 (0.16) )  0.018651 0.018651 − 0.017193 50p 30 − 51p 30    0.078160 0.018651 50p 30

q 80 3.43.

(A)

By equation (3.7), 4p 71

 e−

Then 5p 70  3p 70 2p 73  0.95

R

75 71

µ x dx

4p 71

 e −0.107  0.898526

! 

2p 71

0.95 (0.898526)  0.889166 0.96

Quiz Solutions 3-1. We need 30p 20 − 40p 20 . We must integrate µ x from 20 to 50 and from 20 to 60.

Z t p 20

20+t

 exp −

! µ s ds

20 Z 20+t

 exp −

√

!

0.001 s ds 20

 0.001  (20 + t ) 1.5 − 201.5 1.5   0.001  1.5  exp − 50 − 201.5 1.5 −0.176074 e  0.838556   0.001  1.5  exp − 60 − 201.5 1.5  e −0.250210  0.778637 

 exp − 30p 20

40p 20

30|10q 20

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 0.838556 − 0.778637  0.05992



(E)

Lesson 4

Survival Distributions: Mortality Laws Reading: Models for Quantifying Risk (4th or 5th edition) 5.2 There are two approaches to defining S0 ( x ) , the survival function for the age at death. One is to build a life table defining the probability of survival to each integral age, and then to interpolate between integral ages. We shall learn how to interpolate in Lesson 7. The other approach is to define S0 ( x ) as a continuous function with parameters. Using a parametric function has the advantage of capturing the distribution with only a small number of parameters, which makes the table more portable. The function may have nice properties which simplify computations. Such parametric functions are called mortality laws. This lesson will discuss two types of mortality laws. The first type of mortality law is a parametric distribution that reasonably fits human mortality or some other type of failure over a wide range of ages. The second type of mortality law is a simple parametric distribution that is quite unrealistic for human mortality, but is easy to work with. They are useful for exam questions. Two of the mortality laws we discuss in this section are virtually the only ones which allow easy computation of insurances and annuities in closed form without numerical methods, and thus they appear again and again on exams. Most other parametric distributions do not allow closed-form solutions to the integrals needed to compute insurances and annuities.

4.1

Mortality laws that may be used for human mortality

To get some idea of what the survival curve looks like, we will look at graphs. Figure 4.1 shows the survival function for three starting ages: 0, 40, and 80. These functions are related; for example S40 ( t )  S0 (40 + t ) /S0 (40) . So each curve is a truncated, shifted, and scaled version of the previous one. S x ( t ) is the same as t p x , and if x is fixed, then l x+t  l x t p x is a constant multiple of t p x , so the curves for t p x and l x+t look the same as the curves for S x ( t ) . t px

1

3 8

0 50 Figure 4.1: Survival function for three ages LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

61

80

t

4. SURVIVAL DISTRIBUTIONS: MORTALITY LAWS

62

t 3 2 1

0.1

0.3

0.6

1.0

t qx

Figure 4.2: Probability density function for three ages

t 3 2 1

0.1

0.3

0.6

1.0

t qx

Figure 4.3: Force of mortality for males and females

Figure 4.2 shows the probability density function for three starting ages: 0, 40, 80. In this graph and the following ones, I used simple approximations for µ x . The three curves are related in the same way as the three S x ( t ) curves are related: truncate and scale to go from one curve to another. We see that the most likely age at death is approximately 85. If we had drawn a curve for f x ( t ) , x > 85, it would be monotonically decreasing. Since f x ( t )  t p x µ x+t , curves for t p x µ x+t and l x µ x would look the same. The most interesting graph is the one for force of mortality. Figure 4.3 graphs male and female forces of mortality. But a clearer way to see the pattern is to use a logarithmic scale, as in Figure 4.4. Characteristics of the ln µ x curve, based on this graph, are 1. Mortality is higher for males than for females at all ages. The lines appear to merge as age increases, t 3 2 1

0.1

0.3

0.6

1.0

t qx

Figure 4.4: Force of mortality for males and females with logarithmic scale LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

4.1. MORTALITY LAWS THAT MAY BE USED FOR HUMAN MORTALITY

63

and some believe male mortality may be lower than female at very high ages. 2. Mortality is very high at birth but quickly drops until about age 10. 3. For the male table only, there is a hump in the 20’s. For both sexes, there is a rapid increase in mortality in the late teens. Both of these are caused by high accident rates, for example from youth driving. 4. Most importantly, the graphs are virtually linear starting at about age 40. The last characteristic leads us to our first mortality law.

4.1.1

Gompertz’s law

If we assume ln µ x is a straight line, we can solve for the parameters (by linear regression or some other means): ln µ x  α + βx Exponentiating, we get Gompertz’s law:

µ x  Bc x

(4.1)

with appropriate parameters B and c > 1.1 As we saw from Figure 4.4, this law fits ages 40 and above fairly well. The survival function under this law is t

Z tpx

 exp −

! Bc

x+u

du

0

 exp −

Bc x ( c t − 1) ln c

! (4.2)

Since the law has two parameters, you can solve for all functions if you are given two values of mortality. Example 4A Mortality for a life age 20 follows Gompertz’s law. You are given µ30  0.001 and µ80  0.15. Determine 50p 20 . Answer: Set up two equations for ages 30 and 80. ln B + 30 ln c  ln 0.001 ln B + 80 ln c  ln 0.15 50 ln c  ln 150 c  e (ln 150)/50  1.105406

!

ln B  ln 0.15 − 80

ln 150  −9.914136 50

B  0.0000494704 Using formula (4.2), 0.0000494704 (1.10540620 )(1.10540650 − 1)  0.57937  exp − ln 1.105406

!

50p 20

1If c  1, this is exponential mortality. If c < 1, then the integral LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

R

x 0

µ s ds does not approach infinity as x → ∞.



4. SURVIVAL DISTRIBUTIONS: MORTALITY LAWS

64

4.1.2

Makeham’s law

Makeham improved Gompertz’s law by adding a third parameter A: µ x  A + Bc x

(4.3)

A represents the constant part of the force of mortality, mortality that is independent of age and is due to accidental causes. Bc x , with c > 1, represents mortality resulting from deterioration and degeneration, which increases exponentially by age. Makeham’s law provides a good fit for ages above 20. As we know, adding A to µ multiplies the survival function by e −At , so building on equation (4.2), Bc x ( c t − 1) t p x  exp −At − ln c

! (4.4)

Since the law has three parameters, you can solve for all functions if you are given three values of mortality. However, numerical methods may be needed to solve for the parameters. Example 4B Mortality follows Makeham’s law. You are given µ10  0.0014, µ20  0.0024, and µ30  0.0042. Determine 50p 20 . Answer: Write out the three equations for the three µ’s A + Bc 10  0.0014 A + Bc 20  0.0024 A + Bc 30  0.0042 Bc 10 ( c 10 − 1)  0.0010 Bc 20 ( c 10 − 1)  0.0018 c 10  1.8 c  e (ln 1.8)/10  1.060540 B (1.8)(0.8)  0.0010 B  0.000694444 A + 0.000694444 (1.8)  0.0014 A  0.00015 Therefore, the probability of surviving 50 years is (In the following expression, we use c 10  1.8, so c 20  ( c 2 ) 10  1.82 and c 50  ( c 5 ) 10  1.85 .) 0.000694444 (1.82 )(1.85 − 1)  0.50031 ln 1.060540

!

50p 20  exp −0.00015 (50) −

4.1.3

Weibull Distribution

The Weibull distribution is unlikely to appear on Exam LC, since it is not mentioned in the textbook. For the Weibull distribution, the cumulative distribution function is F ( x )  1 − e − ( x/θ )

τ

If you log, negate, and differentiate the survival function, you will obtain µ x  LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

τ x
τ−1 . θ θ



4.2. MORTALITY LAWS FOR EXAM QUESTIONS

65

The above parametrization is the one used in Exams ST and C. The more traditional parametrization of the Weibull is µ x  kx n , where n > −1 and need not be an integer. You can then derive probability functions in the usual manner. The Weibull distribution is flexible in that n (or τ) can be set to allow mortality or failure rate to have a reasonable pattern. If n  0 (τ  1), the Weibull reduces to an exponential distribution, which is not a reasonable mortality assumption. But if n > 0 (or τ > 1), the force of mortality increases with age, so it becomes more reasonable. Rather than using it for human mortality, it is more commonly used for other types of failure, like machine breakdown. After studying the constant force of mortality law in the next section, a confusion to avoid is that µ x  kx is the force of mortality for a Weibull, not an exponential, distribution. The reason this may be confusing is that when going from force of mortality to survival rate, you integrate the force of mortality, and for an exponential, the integral of the force of mortality has the form kx. Example 4C You are given the force of failure for a battery is 0.1x, with x measured in hours of use. Calculate the probability that the battery will last 10 hours. Answer: The survival probability to time x is x

Z

!



0.1u du  exp −0.05x 2

exp −



0 2

so 10p 0  e −0.05 (10 )  e −5  0.006738 .

4.2 4.2.1



Mortality laws for exam questions Exponential distribution, or constant force of mortality

If the force of mortality is the constant µ, the distribution of survival time is exponential. Survival probabilities are then independent of age; t p x does not depend on x. The constant force of mortality µ is the reciprocal of mean survival time; in other words, a life with constant force of mortality µ has expected future lifetime µ1 , regardless of the life’s current age. Here are the probability functions for Tx if Tx has constant force of mortality µ. Notice that none of these functions vary with x. Fx ( t )  1 − e −µt S x ( t )  e −µt f x ( t )  µe −µt µx  µ

(4.5)

e

(4.6)

tpx

−µt

(4.7)

4.2.2

Uniform distribution

The uniform distribution on [0, θ] has a mean of θ/2, which is its median and midrange. Its variance is θ 2 /12. It is traditional to use the letter ω to indicate the upper limit of a mortality table. For a uniform model, age at death T0 is uniformly distributed on (0, ω]. While the uniform distribution has memory—after all, you can’t live beyond ω, so the older one is, the less time until certain death—it has the following property that simplifies calculation: if age at death is uniform on (0, ω], then survival time for ( x ) is uniform on (0, ω − x]. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

4. SURVIVAL DISTRIBUTIONS: MORTALITY LAWS

66

Here are the probability functions for Tx if T0 is uniform on (0, ω]: After working out several problems, you should recognize the uniform distribution on sight. Fx ( t )  Sx ( t )  fx (t )  µx  tpx



tqx



t|u q x



E[Tx ]  Var (Tx ) 

t ω−x ω−x−t ω−x 1 ω−x 1 ω−x ω−x−t ω−x t ω−x u ω−x ω−x 2 (ω − x )2 12

(4.8) (4.9) (4.10) (4.11)

Example 4D The force of mortality for (30) is µx 

1 100 − x

x > 30

Calculate the probability of (30) dying in his 70’s. Answer: We recognize this mortality law as uniform on (0, 100]. For (30), remaining lifetime is uniform on (0, 70]. We need 40p 30 − 50p 30 . 40p 30 50p 30

So the answer is

3 7

−

2 7



1 7

70 − 40 3  70 7 70 − 50 2   70 7 

.

Alternatively, reason it out: mortality is uniform, and the interval [70, 80] is one-seventh the size of the interval of all possible ages at death [30, 100], so the probability of X being in that interval is 1/7.  When you work with a force of mortality like µx 

1 1 + 100 − x 50 − x

the principle we learned in the last lesson, that the survival probability is the product of the survival probabilities corresponding to each summand in the force, is useful. Mortality following a uniform distribution is sometimes called “de Moivre’s law”. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

4.2. MORTALITY LAWS FOR EXAM QUESTIONS

4.2.3

67

Beta distribution

A generalized version of the uniform distribution has two parameters α > 0 and ω and force of mortality: α 0≤x1

 exp −

Bc x ( c t

− 1) ln c

(4.1)

! (4.2)

Makeham’s law µ x  A + Bc x , tpx

c>1 Bc x ( c t

− 1) ln c

 exp −At −

(4.3)

! (4.4)

Weibull Distribution µ x  kx n S0 ( x )  e −kx

n+1 / ( n+1)

Constant force of mortality µx  µ

(4.5)

e

(4.6)

tpx

−µt

Uniform distribution 1 ω−x ω−x−t tpx  ω−x t tqx  ω−x u t|u q x  ω−x µx 

0≤x 10: t

Z t p 40

 exp −

! 0.02 du  e −0.02t

0 t p 30

 (10p 30 )( t−10p 40 )  e −0.1−0.02t+0.2  e 0.1−0.02t

Using formula (5.2), ∞

Z e˚30 

t p 30 dt 0 10

Z 

Z

∞

e 0.1−0.02t dt

e −0.01t dt + 0

10

 100 (1 − e −0.1 ) + 50e 0.1 ( e −0.2 )  54.7581

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5.1. COMPLETE

?

79

Quiz 5-1 You are given 1 x −1/2 µx  √ 2 10 − x

! x < 100

Calculate e˚50 .

5.1.2

Special mortality laws

Constant force of mortality If µ is constant, then future lifetime is exponential, so 1 µ 1 Var (Tx )  2 µ e˚x 

(5.8) (5.9)

The n-year temporary complete life expectancy is: n

Z e˚x:n 

e −µt dt  0

1 − e −µn µ

and the variance of temporary future lifetime can be calculated by calculating the second moment:

f

E min (Tx , n )

2g

n

Z 2

te −µt dt 0 n

n

Z

n te −µt 1 + e −µt dt µ 0 µ 0 ne −µn 1 − e −µn + − µ µ2 1 − (1 + µn ) e −µn  µ2

Z

te −µt dt  − 0





2g

f

2 and then calculating Var min (Tx , n )  E min (Tx , n ) − e˚x:n . We’ll discuss this integral again in Subsection 10.2.1. Temporary moments for exponentials do not appear often on exams.

Uniform and beta If age at death is uniformly distributed on (0, ω] then remaining lifetime for ( x ) follows a uniform distribution. The mean is ( ω − x ) /2, and is the same as the midrange and the median, and the variance is ( ω − x ) 2 /12. If survival follows a beta distribution with parameters α and ω, or µ x  α/ ( ω − x ) , then E[Tx ]  and Var (Tx ) 

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ω−x α+1

α (ω − x )2 ( α + 1) 2 ( α + 2)

(5.10)

(5.11)

5. SURVIVAL DISTRIBUTIONS: MOMENTS

80

Example 5B Future lifetime of (20) is subject to force of mortality µ x  0.5/ (100 − x ) , x < 100. Calculate e˚20 . Answer: We recognize the mortality distribution as beta with α  0.5, ω  100. As discussed above, ω−x 80   53 13 α+1 3/2

E[Tx ] 



Example 5C Kevin and Kira are age 30. • Kevin’s future lifetime is uniformly distributed with ω  100. • The force of mortality for Kira is α µ30+t  70 − t • Kira’s probability of survival to age 80 is twice as high as Kevin’s. Determine Kira’s expected future lifetime. Answer: Probability Kevin survives to 80 is 50p 30

Kira’s survival is beta, so t p y for her is



ω − 80 100 − 80 2   ω−x 100 − 30 7

ω−x−t
α , ω−x

70 − 50 70

!α

and here ω  100, x  30, and t  50, so from (iii), 2 2 7

!

α ln (2/7)  ln (4/7) α (−1.25276)  −0.55962 −0.55962 α  0.4467 −1.25276 Then since Kira’s survival is beta, the expected value is e˚30 

ω−y 70   48.39 α+1 1.4467



To calculate temporary life expectancy under a uniform distribution, it is not necessary to integrate. Instead, use the double expectation formula (equation (1.11) on page 7). The condition is whether Tx > n or not. For those lives surviving n years, min (Tx , n )  n. For those lives not surviving n years, average future lifetime is n/2, since future lifetime is uniformly distributed. Therefore, e˚x:n  n p x ( n ) + n q x ( n/2) ω−x−n n n  (n ) + ω−x ω−x 2

! (5.12)

Just remember the logic behind this equation, not the equation itself. Example 5D Future lifetime of (20) is subject to force of mortality µ x  Calculate e˚20:50 .

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1 100−x ,

x < 100.

5.1. COMPLETE

81

Answer: The survival function is uniform with ω  100. For those who survive 50 years, min (T20 , 50)  50. For those who don’t, the average future lifetime is 25. Therefore e˚20:50  50p 20 (50) + 50q 20 (25) 3 5 275  (50) + (25)   34.375 8 8 8



Another method for computing temporary life expectancies under a uniform distribution is the trapeRn zoidal rule. The idea of this rule is that to compute 0 t p x dt, since the function is linear, we evaluate the area of the trapezoid bounded horizontally by the t-axis and the slanted line t p x and vertically by the two lines t  0 and t  n. The area of this trapezoid is e˚x:n  0.5 (1 + n p x )( n ) . You can easily show that this formula is equivalent to equation (5.12). Let’s apply the trapezoidal rule to Example 5D. We need to integrate t p 20 from t  0 to t  50. The integral is the area of this trapezoid: t px

1 0.875 0.75 0.625 0.5 0.375 0.25 0.125 10

20

30

40

50

t

At t  0, 0p 20  1. At t  50, 50p 20  3/8. We multiply the average of these two heights, 0.5 (1 + 3/8)  11/16, by the width of the trapezoid, 50, to obtain 50 (11/16)  275/8  34.375. We will discuss the trapezoidal rule in more detail on page 123.

?

Quiz 5-2 Future lifetime for (20) follows a uniform distribution with ω  120. The n-year temporary complete life expectancy for (20) is 48. Determine n. We can calculate the variance of temporary future lifetime under uniform survival time using the conditional variance formula (equation (1.13) on page 9). Example 5E (Continuation of Example 5D) Future lifetime of (20) is subject to force of mortality µ x  1 100−x , x < 100.





Calculate Var min (T20 , 50) .

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5. SURVIVAL DISTRIBUTIONS: MOMENTS

82

Answer: The condition I is T20 > 50. Using this condition, E[min (Tx , 50) | T20 > 50]  50





Var min (T20 , 50) | T20 > 50  0 E[min (Tx , 50) | T20 ≤ 50]  25





Var min (T20 , 50) | T20 ≤ 50 

502 12

For the previous line, we used the fact that min (T20 , 50) | T20 ≤ 50 is uniformly distributed on [0, 50].





Var min (T20 , 50)  E[0, 502 /12] + Var (50, 25) 502 + 50p 20 50q 20 (50 − 25) 2 12

!

 50q 20

For the previous line, we used the Bernoulli shortcut (Section 1.2.1) to evaluate the variance. 5 Var min (T20 , 50)  8





!

!

2500 3 + 12 8

!

!

5 (625)  276.6927 8



These methods for calculating mean and variance of future lifetime work not only for uniform mortality, but for any case in which mortality is uniformly distributed throughout the temporary period. An important case, which will be discussed in Lesson 7, is when mortality is uniform for one year. Suppose mortality for ( x ) is uniformly distributed over the next year. Then formula (5.12) with n  1 becomes e˚x:1  p x + 0.5q x

(5.13)

Using conditional variance, you can obtain a formula for the variance of min (Tx , 1) in this case. Let I be the indicator variable for whether Tx ≥ 1. Then, since the mean of a uniform random variable on (0, 1] is 1/2 and the variance is 1/12,









f



Var min (Tx , 1)  Var E[min (Tx , 1) | I] + E Var min (Tx , 1) | I

g

 Var (1/2, 1) + E[1/12, 0] In each summand, the probability of the first case is q x and the probability of the second case is p x . The variance of 1/2 and 1, by the Bernoulli shortcut (page 3) is p x q x (1 − 1/2) 2 , and the expected value of 1/12 and 0 is (1/12) q x + (0) p x , so we have





Var min (Tx , 1) 

5.2

px qx qx + 4 12

Curtate

So far we have only dealt with the random variable Tx , the complete survival time. We now introduce K x , the random variable measuring future lifetime not counting the last fraction of a year. In other words, if time until death is 39.4, K x would be 39. With symbols: K x  bTx c LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

5.2. CURTATE

83

where bxc is the greatest integer less than or equal to x. K x is called the curtate lifetime (K stands for kurtate.) K x is a discrete random variable; in fact K x assumes integer values only. The expected value of K x is denoted by e x (without the circle on the e) and is called the curtate life expectation. From the definition of moments, formulas for the first and second moments of K x and min ( K x , n ) are:

ex 

∞ X

k k| q x

(5.14)

k k| q x + n n p x

(5.15)

k 2 k| q x

(5.16)

k 2 k| q x + n 2 n p x

(5.17)

k0

e x:n 

f

g

E K 2x 

n−1 X k0 ∞ X k0

 E

min ( K x , n )

2



n−1 X k0

In all of these formulas, the sum can be started at k  1 instead of k  0. The variance of K x can be f g calculated as E K 2x − e x2 . As with complete expectation, better formulas can be obtained through summation by parts: ex  e x:n 

g

2

k1 n X

f



min ( K x , n )

k1 n X k1 ∞ X

E K 2x 

E

∞ X



k px

(5.18)

k px

(5.19)

(2k − 1) k p x

(5.20)

(2k − 1) k p x

(5.21)

k1

To help you better understand the two alternative formulas for life expectancy, we shall demonstrate the formula graphically. Let’s use a simple example. Suppose mortality follows the following distribution: t

t−1| q x

t qx

t px

1 2 3 4

0.1 0.2 0.3 0.4

0.1 0.3 0.6 1.0

0.9 0.7 0.4 0.0

To calculate e x using formula (5.14), we add up k times the probability of living exactly (truncated to the next lowest integer) k years. This means e x  1 (0.2) + 2 (0.3) + 3 (0.4)  2.0

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5. SURVIVAL DISTRIBUTIONS: MOMENTS

84

t

t

3

3

2

2

1

1

0.1

0.3

0.6

1.0

t qx

0.1

0.3

(a) Using formula (5.14)

0.6

1.0

t qx

(b) Using formula (5.18)

Figure 5.2: Graphic explanation of equivalence of the two formulas for e x

This is illustrated in Figure 5.2a. The first rectangle from the left is 0.2 wide (the difference between q 2 x and 1 q x ) and 1 high, representing the first term in the sum, and so on. To calculate e x using formula (5.18), we add up the probabilities of living at least 1 year, 2 years, 3 years, and 4 years, or e x  0.9 + 0.7 + 0.4 + 0  2.0 This is illustrated in Figure 5.2b. Instead of splitting the area into vertical rectangles, we now split the area into horizontal rectangles. The first rectangle from the bottom represents the probability of living at least 1 year, and so on. The same graphical representation can be used to demonstrate the equivalence of the two formulas for complete expectation of life, equations (5.1) and (5.2). Instead of using intervals of 1 for t, let the size of the intervals go to zero and then the graph, instead of being a set of steps, turns into a continuous curve. Example 5F You are given the following mortality table: x

qx

90 91 92 93 94

0.10 0.12 0.15 0.20 1.00

Calculate the curtate life expectancy of (90). Answer: Calculate t p90 as cumulative products of 1 − q x : p90  0.90 2p 90

 (0.90)(0.88)  0.792

3p 90

 (0.792)(0.85)  0.6732

4p 90  (0.6732)(0.80)  0.53856 5p 90

0

Then e90  0.90 + 0.792 + 0.6732 + 0.53856  2.90376 .

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5.2. CURTATE

?

85

Quiz 5-3 You are given the following life table: x

lx

90 91 92 93

100 98 95 0

Calculate the variance of curtate future lifetime of (90). If future lifetime is exponential, the sum t p x is a geometric series. Example 5G The force of mortality for ( x ) is the constant µ  0.01. Calculate the curtate life expectancy of ( x ) . Answer: The probability of survival is k p x  e −0.01k . Then ex 

∞ X

e −0.01k 

k1

e −0.01  99.5008 1 − e −0.01



More generally, if future lifetime is exponential with constant force µ, then ex 

∞ X k1

e −kµ 

e −µ 1 − e −µ

(5.22)

In general, curtate future lifetime is about 1/2 less than complete future lifetime, since fraction of a year is 1/2. If future lifetime is uniform, this relationship is exact. In symbols:

More generally:

e˚x  e x + 0.5

(5.23)

e˚x:n  e x:n + 0.5n q x

(5.24)

The proof of this is by double expectation. Expand E[min (Tx , n ) ] into1

f

g

E E[min (Tx , n ) | Tx ]  n q x E[min (Tx , n ) | Tx < n] + n p x E[min (Tx , n ) | Tx ≥ n]  n q x E[Tx | Tx < n] + n p x E[n | Tx ≥ n] Now E[n | Tx ≥ n]  n, since the expectation of a constant is the constant. And E[Tx | Tx ≥ n]  E[K x | Tx < n] + 0.5, since with uniform mortality, Tx is on the average 0.5 higher than K x . So e˚x:n  n q x (E[K x | K x < n] + 0.5) + n p x ( n )  n q x (E[min ( K x , n ) | K x < n] + 0.5) + n p x E[min ( K x , n ) | K x ≥ n]  e x:n + 0.5n q x Example 5H (Same data as Example 5D) Future lifetime of (20) is subject to force of mortality µ x  x < 100.

1 100−x ,

1Strictly speaking, n q x  Pr (Tx ≤ n ) rather than Pr (Tx < n ) , but we’re assuming that Tx is a continuous random variable, so Pr (Tx  n )  0. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

5. SURVIVAL DISTRIBUTIONS: MOMENTS

86

1. Calculate e20 . 2. Calculate e20:50 .





3. Calculate Var K 20 . Answer: 1. For uniform mortality, e˚20  ( ω − 20) /2  (100 − 20) /2  40. Since e20  e˚20 − 0.5, the answer is e20  39.5 . 2. In Example 5D we computed e˚20:50  34.375. Then e20:50  e˚20:50 − 0.550 q20  34.375 − 0.5 (5/8)  34.0625 3. We will use (5.16) to calculate the second moment.

f

g

2  E K 20

79 X

k 2 k| q 20

k0 79



1 X 2 k 80 k0

!



Pn

because formula.

k1

1 (79)(80)(159)  2093.5 80 6

k 2  n ( n + 1)(2n + 1) /6. It is doubtful that an exam would expect you to know this

The variance is then 2093.5 − 39.52  533.25 .

?



Quiz 5-4 For a life age 20, µ20+t  1/ ( ω − 20 − t ) . The 24-year temporary curtate life expectancy is 20.25. Determine ω.

Exercises 5.1. A person age 70 is subject to the following force of mortality:

  0.01 µ70+t    0.02

t≤5 t>5

 Calculate e˚70 for this person. 5.2. [3-F01:1] You are given:

  0.04, µx    0.05,

0 < x < 40 x > 40

 Calculate e˚25:25 . A. 14.0

B. 14.4

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C. 14.8

D. 15.2

E. 15.6 Exercises continue on the next page . . .

EXERCISES FOR LESSON 5

87

Table 5.1: Formula summary for Survival Moments

Complete Future Life time

Curtate Future Lifetime

∞

Z e˚x 

t t p x µ x+t dt

(5.1)

ex 

Z0 ∞ e˚x 

tpx

dt

(5.2)

ex 

0

1 e˚x  for constant force of mortality µ ω−x e˚x  for uniform 2 ω−x e˚x  for beta α+1

 E

Tx

2

2

(5.8)

k k| q x

(5.14)

k px

(5.18)

e x  e˚x − 0.5

 E (5.10)

Kx

2



∞ X

for uniform

(2k − 1) k p x







Var K x  Var Tx −

(5.3)

0





Var Tx 

1 µ2

for constant force of mortality



Var Tx 

(ω −

e x:n 

x )2

for uniform 12   α (ω − x )2 Var Tx  for beta ( α + 1) 2 ( α + 2)

e x:n  (5.11)

n-year Temporary Complete Future Lifetime e˚x:n 

t t p x µ x+t dt + n n p x

(5.5)

Z0 n e˚x:n 

tpx

dt

1 12

for uniform

n−1 X k1 n X

k k| q x + n n p x

(5.15)

kpx

(5.19)

k1

e x:n  e˚x:n − 0.5 n q x



n

Z

(5.20)

n-year Temporary Curtate Future Lifetime

(5.9)



(5.23)

k1



t t p x dt

k1 ∞ X k1

∞

Z

∞ X

E

min ( K x , n )

2



for uniform n X

(2k − 1) k p x

(5.24) (5.21)

k1

(5.6)

0

e˚x:n  n p x ( n ) + n q x ( n/2) for uniform e˚x:1  p x + 0.5q x for uniform

 E

min (Tx , n )

2

(5.12) (5.13)

n

Z 2

t t p x dt

(5.7)

0

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5. SURVIVAL DISTRIBUTIONS: MOMENTS

88

5.3. [CAS4A-F97:13] (2 points) You are given the following information about two lives: Life

Future Lifetime Random Variable

x y

Constant force of mortality µ x  0.10 Constant force of mortality µ y  0.20

Determine the ratio of ( x ) ’s expected future lifetime between ages x and x + 10 to ( y ) ’s expected future lifetime between ages y and y + 10. A. B. C. D. E.

Less than 1.00 At least 1.00, but less than 1.25 At least 1.25, but less than 1.50 At least 1.50, but less than 1.75 At least 1.75

5.4. A life is subject to a constant force of mortality µ. You are given that e50  24. Determine µ. 5.5. [4-S86:22] You are given: • Age at death is uniformly distributed. • e˚30  30. Calculate q 30 . A. 1/30

B. 1/60

C. 1/61

D. 1/62

E. 1/70

D. 1/90

E. 1/100

5.6. [CAS3-F03:5] Given: • Age at death is uniformly distributed. • e˚20  30 Calculate q 20 . A. 1/60

B. 1/70

C. 1/80

5.7. [150-F97:1] For the current type of refrigerator, you are given: • •

S0 ( x )  1 − x/ω, 0 ≤ x ≤ ω e˚0  20

For a proposed new type, with the same ω, the new survival function is:

  1, S0∗ ( x )    ( ω − x ) / ( ω − 5) ,

0≤x≤5 5 0

For age y, 0 ≤ y < ω, you are given: • •

µ y  0.1 e˚ y  8.75

Calculate r. A. 1

B. 3

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C. 5

D. 7

E. 9

Exercises continue on the next page . . .

5. SURVIVAL DISTRIBUTIONS: MOMENTS

90

[150-81-94:8] You are given:

5.13.



S0 ( x )  1 −

x ω

α

,

0 ≤ x < ω, where α is a positive constant

Calculate µ x · e˚x . A.

α α+1

5.14.

B.

αω α+1

C.

α2 α+1

D.

α2 ω−x

E.

α (ω − x ) ( α + 1) ω

[150-83-96:25] You are given: √ k2 − x • S0 ( x )  , 0 ≤ x ≤ k2, k > 0 k • e˚40  2e˚80 .

Calculate e˚60 . A. 10

B. 20

C. 30

D. 40

E. 50

D. 55

E. 60

[150-S98:25] You are given:

5.15.

( k 3 − x ) 1/3

•

S0 ( x ) 

•

e˚40  2e˚80

k

, 0 ≤ x ≤ k3, k > 0

Calculate e˚60 . A. 40

B. 45

C. 50

5.16. [3-F00:31] For an industry-wide study of patients admitted to hospitals for treatment of cardiovascular illness in 1998, you are given: • Duration In Days

Number of Patients Remaining Hospitalized

0 5 10 15 20 25 30 35 40

4,386,000 1,461,554 486,739 161,805 53,488 17,384 5,349 1,337 0

• Discharges from the hospital are uniformly distributed between durations shown in the table. Calculate the mean residual time remaining hospitalized, in days, for a patient who has been hospitalized for 21 days. A. 4.4

B. 4.9

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C. 5.3

D. 5.8

E. 6.3

Exercises continue on the next page . . .

EXERCISES FOR LESSON 5

5.17.

91

[SOA3-F04:24] The future lifetime of (0) follows S0 ( x ) 

50 50 + x

!3 x>0

Calculate e˚20 . A. 5

B. 15

C. 25

5.18.

A life is subject to force of mortality µx 

D. 35

1 1 + 200 − x 100 − x

E. 45

x < 100

Calculate the complete life expectancy at age 0 for this life. 5.19.

For a life whose survival function is S0 ( x ) 

ω−x ω

you are given that e10:20  18. Determine ω. 5.20.

[CAS4-S90:2] (1 point) Mortality follows l x  1000 (1 −

x 100 )

for 0 ≤ x ≤ 100.

Calculate e90 A. B. C. D. E. 5.21.

Less than 4.2 At least 4.2, but less than 4.4 At least 4.4, but less than 4.6 At least 4.6, but less than 4.8 At least 4.8 The force of mortality for a life is

  0.01 µx    0.01 + 

1 100−x

0 ≤ x < 50 50 ≤ x < 100

Calculate e˚40 . 5.22.

Light bulbs have the following distribution for the amount of time until burning out: Time t in hours

F (t )

0–4800 4800–6000 6000

( t − 4800) /1200

0 1

Each bulb uses 0.015 kilowatt-hours of electricity per hour. Calculate the expected number of kilowatt-hours used by 50 bulbs in their first 5000 hours.

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5. SURVIVAL DISTRIBUTIONS: MOMENTS

92

[CAS4-S90:11] (2 points) You are given:

5.23.

• Future survival time is uniformly distributed. • e˚20  45 Calculate the variance of the future lifetime of a person age 20, Var (T20 ) , to the nearest integer. A. 108

B. 275

C. 350

D. 675

E. 700

D. 110

E. 114

[150-F87:11] You are given:

5.24.

• Age at death is uniformly distributed. • Var (T50 )  192. Calculate ω. A. 98

B. 100

C. 107

[CAS4A-S94:3] (2 points) You are given:

5.25.

S0 ( x )  1 −

x for 0 ≤ x ≤ 100 100

Determine Var (T20 ) . A. B. C. D. E. 5.26.

Less than 600 At least 600, but less than 800 At least 800, but less than 1,000 At least 1,000, but less than 1,200 At least 1,200 For a life with survival function S0 ( x ) 

ω−x ω

x≤ω

you are given • ω > 40 • For a life age 30, the variance of the number of years lived between 30 and 40 is 3.5755. Determine ω. 5.27. [150-S89:A1] You are given the survival function S0 ( x )  e −0.05x for x ≥ 0. Calculate each of the following. • 5|10q 30 • F (30) • e˚30 • Var (T30 )

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EXERCISES FOR LESSON 5

93

[150-F89:2] You are given:

5.28.

• Age at death is uniformly distributed. • Var (T15 )  675 Calculate e˚25 . A. 37.5

B. 40.0

C. 42.5

D. 45.0

E. 47.5

D. 178

E. 333

[3-S00:1] Given:

5.29.

• e˚0  25 • l x  ω − x, 0 ≤ x ≤ ω • Tx is the future lifetime random variable. Calculate Var (T10 ) . A. 65

B. 93

C. 133

Mortality follows the Illustrative Life Table.

5.30.

Calculate the variance of the number of complete years lived by (67) before reaching age 70. You are given that q x  0.3, q x+1  0.5, q x+2  0.7, and q x+3  1.

5.31.

Calculate the variance of curtate future lifetime, Var ( K x ) . You are given:

5.32.

• The force of mortality is the constant µ. • e35  49. Calculate the revised value of e35 if the force of mortality is changed to the constant µ + 0.01. [SOA3-F03:28] For ( x ) :

5.33.

K is the curtate future lifetime random variable. q x+k  0.1 ( k + 1) , k  0, 1, 2, . . . , 9 X  min ( K, 3)

• • •

Calculate Var ( X ) . A. 1.1

B. 1.2

C. 1.3

D. 1.4

E. 1.5

D. 392

E. 412

[M-S05:21] You are given:

5.34.

e˚30:40  27.692 t • S0 ( x )  1 − , 0≤t≤ω ω • Tx is the future lifetime variable for ( x ) . •

Calculate Var (T30 ) . A. 332

B. 352

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C. 372

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5. SURVIVAL DISTRIBUTIONS: MOMENTS

94

5.35. [M-F05:13] The actuarial department for the SharpPoint Corporation models the lifetime pencil sharpeners from purchase using S0 ( t )  (1 − t/ω ) α , for α > 0 and 0 ≤ x ≤ ω. A senior actuary examining mortality tables for pencil sharpeners has determined that the original value of α must change. You are given: The new complete expectation of life at purchase is half what it was originally. The new force of mortality for pencil sharpeners is 2.25 times the previous force of mortality for all durations. ω remains the same.

• • •

Calculate the original value of α. A. 1

B. 2

C. 3

D. 4

E. 5

D. 9.5

E. 10.0

[M-F05:14] You are given:

5.36.

• T is the future lifetime random variable. • µ x  µ, x ≥ 0 • Var (T )  100 • X  min (T, 10) Calculate E[X]. A. 2.6 5.37.

B. 5.4

C. 6.3

[M-F06:2] You are given the survival function S0 ( t )  1 − (0.01t ) 2

0 ≤ t < 100

Calculate e˚30:50 , the 50-year temporary complete expectation of life for (30). A. 27 5.38.

B. 30

C. 34

D. 37

E. 41

[M-F06:23] You are given 3 mortality assumptions:

• Illustrative Life Table (ILT) • Constant force model (CF), where S0 ( t )  e −µt , x ≥ 0 t • DeMoivre model (DM), where S0 ( t )  1 − , 0 ≤ t ≤ ω, ω ≥ 72. ω For the constant force and DeMoivre models, 2p 70 is the same as for the Illustrative Life Table. Rank e70:2 for these 3 models. A. B. C. D. E.

ILT < CF < DM ILT < DM < CF CF < DM < ILT DM < CF < ILT DM < ILT < CF

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EXERCISES FOR LESSON 5

95

5.39. [MLC-S07:21] You are given the following information about a new model for buildings with limiting age ω. • The expected number of buildings surviving at age x will be l x  ( ω − x ) α , x < ω. • The new model predicts a 33 31 % higher complete life expectancy (over the previous uniform model with the same ω) for buildings aged 30. • The complete life expectancy for buildings aged 60 under the new model is 20 years. Calculate the complete life expectancy under the previous uniform model for buildings aged 70. A. 8 B. 10 C. 12 D. 14 E. 16 Additional old CAS Exam 3/3L questions: F05:10, S07:6, S08:13, S09:1, F09:2, S10:3, S11:1, F11:2, F12:1, F13:2 Additional old CAS Exam LC questions: F14:2,3

Solutions 5.1. Use equation (5.2).

∞

Z e˚70 

t p 70 dt 0

To calculate t p 70 , we consider two cases: t ≤ 5 and t > 5. For t ≤ 5, t p 70  e −0.01t . For t > 5, t p 70 is the exponential of the negative integral of µ70+t . That integral is e −0.01t for t ≤ 5 and e −0.05−0.02 ( t−5) for t > 5. (See exercise 3.1 for an example of how to calculate 20p 70 .) So 5

Z e˚70 

∞

Z

e −0.05−0.02 ( t−5) dt

e −0.01t dt + 0

5









∞

Z

 100 1 − e −0.05 + e −0.05

e −0.02 ( t−5) dt 5

 100 1 − e −0.05 + e −0.05 (50)  4.8771 + 47.5615  52.4386 5.2. We want to use equation (5.6), so we have to compute t p 25 . t p 25

 e−

R

t 0

µ u du

 e −0.04t e

t ≤ 15

−0.04 (15) −

e

R

t 15

 e −0.6−0.05 ( t−15)

µ u du

t > 15

Now we calculate e˚25:25 . 15

Z e˚25:25 

Z

25

t p 25 dt + 0 15

Z 

25

Z e

0

t p 25 dt 15

−0.04t

e −0.6−0.05 ( t−15) dt

dt +

15 −0.6

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(1 − e −0.5 )

5. SURVIVAL DISTRIBUTIONS: MOMENTS

96

 25 − 5e −0.6 − 20e −1.1  25 − 5 (0.548812) − 20 (0.332871)  15.598520

(E)

5.3. We will use equation (5.6). For ( x ) , 10

Z e˚x:10 

e −0.1t dt 0

 10 (1 − e −1 )  6.3212 For ( y ) , 10

Z e˚ y:10 

e −0.2t dt 0

 5 (1 − e −2 )  4.3233 The ratio is 6.3212/4.3233  1.4621 . (C) 5.4. Use formula (5.18). e50 

∞ X

e −tµ  24

t1

Summing up the geometric series with ratio e −µ : e −µ  24 1 − e −µ 24 e −µ  25 µ  − ln 0.96  0.040822 5.5. Under uniform distribution, life expectancy is the midrange. If life expectancy at 30 is 30, then ω must be 30 + 2 (30)  90. Then q 30 

1 ω−x



1 60

. (B)

5.6. A repeat of the previous question with the age changed. Under uniform distribution, life expectation is half of remaining life, so ω  20 + 2 (30)  80. Then q 20 

1 80−20



1 60

. (A)

5.7. The current type is uniform, so ω  2e˚0  40. The new type is sure to survive 5 years, then survival is uniform over 35 years thereafter, so the conditional expectation after surviving 5 years is 12 (35)  17.5, for total expected survival time of 5 + 17.5  22.5, which is 2.5 years longer. (B) 5.8. This is a uniform distribution with ω  72 in months. Remaining time to ω (72) is 60, so life expectancy is 30. The number of eggs is then the number of hens times eggs per month times life expectancy, or (100)(30)(30)  90,000 . (E) 5.9. Under uniform distribution, life expectancy is half the limiting age, so if expectancy increases 4, the limiting age increases by 8. 108 (E)

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EXERCISE SOLUTIONS FOR LESSON 5

5.10.

97

Use equation (5.12). From the given information, we have two equations for ω − 20 and n:

( ω − 20 − 2n )(2n )

4n 2  25 ω − 20 2 ( ω − 20) 16n 2 ( ω − 20 − 4n )(4n ) +  40 ω − 20 2 ( ω − 20) +

or 2n ( ω − 20) − 2n 2  25 ( ω − 20) 4n ( ω − 20) − 8n 2  40 ( ω − 20) Solve for ω in both equations. ω − 20 

8n 2 2n 2  4n − 40 2n − 25

Solve this for n. 2 (4n − 40)  8 (2n − 25) −80 + 8n  −200 + 16n 8n  120 n  15 ω − 20  Then e˚20:45  5.11.

2 (152 ) 450   90 2 (15) − 25 5

1 1 (45) + (22.5)  33.75 2 2

By equation (5.5), the difference between e˚36:28 and 2828 p36

l 64  28 l 36

r  28 Hence

R

28 0

R

28 0

t t p 36 µ36+t dt is 2828p 36 . We calculate:

!

36  21 64

t t p36 µ36+t dt  24.67 − 21  3.67 . (A).

5.12. For the beta distribution, we know that µ y  we are given,

r ω−y

and e˚ y  ( ω − y ) / ( r + 1) . Therefore, using what

r  0.1 ω−y ω−y  8.75 r+1 Multiplying these two equations together: r  0.875 r+1 r 7 (D)

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5. SURVIVAL DISTRIBUTIONS: MOMENTS

98

5.13.

For the beta distribution, µ x  α/ ( ω − x ) and e˚x  ( ω − x ) / ( α + 1) . Multiplying together gets (A).

5.14.

k2 − x We are given S0 ( x )  k2

! 1/2 . This is a beta with ω  k 2 and α  1/2. Therefore, e˚x 

ω − x k2 − x  α+1 3/2

We are given that e˚40  2e˚80 , so k 2 − 40 2 ( k 2 − 80)  3/2 3/2 2 k  120 120 − 60  40 . (D) 3/2 We can write the denominator k as ( k 3 ) 1/3 , so

Then e˚60  5.15.

k3 − x ( k 3 − x ) 1/3  S0 ( x )  3 1/3 k3 (k )

! 1/3

This is a beta with ω  k 3 and α  1/3. Therefore, ω−x 3 3  (k − x ) α+1 4

!

e˚x  We are given that e˚40  2e˚80 , so

k 3 − 40  2 ( k 3 − 80) k 3  120 Then e˚60 

3
3 4 (k

− 60) 

3
4 (60)

 45 . (B)

5.16. The total number of patients hospitalized 21 days or longer is obtained by linear interpolation between 21 and 25: l21  0.8 (53,488) + 0.2 (17,384)  46,267.2 That will be the denominator. The numerator is the number of days past day 21 hospitalized times the number of patients hospitalized for that period. Within each interval of durations, the average patient released during that interval is hospitalized for half the period. So 46,267.2 − 17,384  28,883.2 patients are hospitalized for 2 days after day 21, 17,384 − 5,349  12,035 for 4 + 2.5  6.5 days, 5,349 − 1,337  4,012 for 11.5 days, and 1,337 for 16.5 days. Add it up: 28,883.2 (2) + 12,035 (6.5) + 4,012 (11.5) + 1,337 (16.5)  204,192.4 The mean residual time is 204,192.4/46,267.2  4.41333 . (A) 5.17. The question was asked when the Exam C distribution tables were part of this exam. The survival distribution is a two-parameter Pareto, and you could look up the mean and you also learned other properties of the Pareto. Then the question is easy: Complete life expectancy e˚x is ( θ + x ) / ( α − 1) for a two-parameter Pareto, as you learn in Exam C/4. From this formula, it immediately follows that e˚20  (50 + 20) / (3 − 1)  35 . LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

EXERCISE SOLUTIONS FOR LESSON 5

99

To solve it from basic principles: 

t p 20

S0 (20 + t ) S0 (20)

.



50 50 + (20 + t )

3

50/ (50 + 20) 3

70  70 + t

!3

From equation (5.2), ∞

Z e˚20 

t p 20 dt 0 ∞

Z  0

 5.18.

70 70 + t

!3 dt

703  35 (2)(702 )

(D)

The survival probability is tpx



(200 − t )(100 − t ) 20,000



(100 − t ) 2 + 100 (100 − t ) 20,000

where the second equality, based on 200 − t  100 − t + 100, is used to make the coming integration easier. Then 1 e˚0  20,000

100 

Z

2



(100 − t ) + 100 (100 − t ) dt

0

833,333 13 1003 1003 1 +   41 23  20,000 3 2 20,000

!

5.19.

Mortality is uniformly distributed. By equation (5.24), e10:20  e˚10:20 − 0.5 20q 10

However, by equation (5.12),

e˚10:20  2020p 1 0 + 1020q 10

Putting these two equations together, e10:20  2020p 10 + 9.520q 10 We are given that e10:20  18, so 18  20 20p 10 + 9.520q 10  20 − 10.5 20q 10 2 4  20q 10  10.5 21 For uniform mortality, t q x  t/ ( ω − x ) , so 20q 10  20/ ( ω − 10) . 20 4  ω − 10 21 ω  115

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!

5. SURVIVAL DISTRIBUTIONS: MOMENTS

100

5.20.

Survival time is uniformly distributed with ω  100. Therefore 10 5 2  e˚90 − 0.5  4.5

e˚90  e90 5.21.

(C)

Calculating t p 40  10p 40 t−10p 50 for t > 10, we get −0.01t    t p 40  e  t p 40  e −0.01(10) e −0.01( t−10) 

50− ( t−10)
50




e −0.01t 60−t 50

t ≤ 10 t ≥ 10

We integrate t p 40 to calculate e˚40 . 10

Z e˚40 

Z e

−0.01t

60

dt +

0

1 − e −0.1  + 1.2 0.01

!

e

60 − t dt 50

−0.01t

10 e −0.1

1 − e −0.6 − 0.01 50

!

60

Z

te −0.01t dt 10

The final integral can be integrated by parts.

Z

60 10

60 + 100

te −0.01t dt  −100te −0.01t

10

Z

60

e −0.01t dt 10

 −6000e −0.6 + 1000e −0.1 − 10000e −0.6 + 10000e −0.1  11000e −0.1 − 16000e −0.6 Putting everything together, e˚40  100 − 100e −0.1 + 120e −0.1 − 120e −0.6 − 220e −0.1 + 320e −0.6  100 − 200e −0.1 + 200e −0.6  28.7948

5.22.

We use equation (5.6): ∞

Z E[min ( X, 5000) ] 

4800

Z S0 ( x ) dx 

0

Z

5000

dx + 0

1− 4800

t − 4800 dx 1200





The first integral is 4800. The second integral is the integral
of a
trapezoid with heights 1 at 4800 and 1 5/6 at 5000, and width 200. The area of the trapezoid is 200 21 11 6  183 3 . The complete temporary life expectancy to 5000 is E[min ( X, 5000) ]  4800 + 183 13  4983 13 . The number of kilowatt-hours used by 50 bulbs is 50 (0.015)(4983 13 )  3737.5 . Here is a more intuitive way to solve the question. Survival for 4800 hours is definite. After 4800 hours, survival is uniform for 1200 hours. Therefore, 5/6 of the bulbs survive to time 5000 (1/6 of the interval [4800,6000]), and the remaining 1/6 of the bulbs survive for half of the interval [4800,5000], or to time 4900 on the average. Therefore, average survival for all bulbs is 5/6 (5000) + 1/6 (4900)  4983 13 , and the number of kilowatt-hours used by 50 bulbs is once again 50 (0.015)(4983 13 )  3737.5 . 5.23. Since e˚20  45, ω  20 + 2 (45)  110, and future lifetime is uniform over [0, 90]. Then the variance is the length of the interval squared over 12, or 902 /12  675 . (D) 5.24.

Variance is ( ω − 50) 2 /12  192, so ω − 50  48, ω  98 . (A)

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EXERCISE SOLUTIONS FOR LESSON 5

5.25.

101

Future lifetime is uniformly distributed on [0, 80], so the variance is 802 /12  533 13 . (A)

5.26. Let θ  ω − 30. The straightforward approach is to calculate the first and second limited moments of the minimum of 10 and future lifetime for (30). Since mortality is uniform, this may be calculated by conditioning on survival to time 40. The first moment is e˚30  10p 30 (10) + 10q 30 E[T30 | T30 ≤ 10] 10 10θ − 50 θ − 10 (10) + (5)  θ θ θ

!



!

To calculate the second moment, for a uniform distribution on [0, n], the second moment is n 2 /3, so given that death occurs within 10 years, the second moment of survival time is 100/3. θ − 10 10 E[min (T30 , 10) ]  (102 ) + θ θ

!

!

2

!

100 300θ − 2000  3 3θ

Alternatively, the first and second moments can be calculated using integration. The first moment is 10

Z E[min (T30 , 10) ] 

t p 30 dt 0 10

Z 

( θ − t ) dt θ

0

 10 −

50 θ

and the second moment, using formula (5.7) on page 78 is

 E

min (T30 , 10)

2

10

Z 2

t t p30 dt 0 10

t ( θ − t ) dt θ 0   2 1000  50θ − θ 3 2000  100 − 3θ

Z

2

Either way, the variance is





2000 1000 2500 − 100 + − 2 3θ θ θ 2500 1000 − 2 + 3θ θ

Var min (T30 , 10)  100 −

This is set equal to 3.5755 and the quadratic is solved. An alternative way to get the same quadratic which may be easier is to use the conditional variance formula (1.13) on page 9. We condition on surviving 10 years. Given that (30) survives 10 years, the amount of time (30) survives in the next 10 years is exactly 10 years, so the expected value of the amount of time survived in the next 10 years is 10 and the variance is 0. If (30) does not survive 10 years, death time is uniformly distributed on [0, 10]. The expected value of a uniform distribution is the midrange or 2 25 5, and the variance is the range squared divided by 12, or 10 12  3 . The probability of dying within 10 10 years is θ . Therefore: •

The expected value of the two variances is

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10
25
θ 3 .

5. SURVIVAL DISTRIBUTIONS: MOMENTS

102

•

The
variance of the expected values (by the Bernoulli shortcut. See Section 1.2.1 on page 3) is 10 θ−10
(25) . θ θ

So the variance of 10-year future lifetime for (30) is 10 θ

!

θ − 10 10 25 (25) +  3.5755 θ θ 3 250 2500 250 − 2 +  3.5755 θ 3θ θ 2500 1000 − + 3.5755  0 3θ θ2

!

!

!

Either way yields this quadratic. Using the quadratic formula for θ1 , we get the solutions 1  θ

− 1000 3 ±

q

1000
2 3

− 35,755

2 (2500) 1  0.011765, 0.121569 θ θ  85.00, 8.22579

Since we are given that ω > 40, the answer is 30 + 85.00  115.00 . Note that if ω were less than 40, then the variance in number of years lived between 30 and 40 is 2 variance of complete future lifetime. The variance of a uniform distribution on [0, θ] is θ12 . Since future lifetime is uniformly distributed, the variance is the maximum number of future years squared over 12. So we’d have θ2  3.5755 12 θ  6.5503 and ω would then be 36.5503. 5.27. This came from a written answer question. For (i), because of the lack of memory of an exponential distribution, we can evaluate it as 5|10q 0 : S0 (5) − S0 (15)  e −0.25 − e −0.75  0.306434 S0 ( 0 ) (ii) is 1 − S0 (30)  1 − e −1.5  0.776870 . (iii) is the reciprocal of the force of mortality, or 20 . For (iv), variance is the square of the mean for an exponential, or 400 . 5.28.

The variance of a uniform distribution is the square of time to ω divided by 12.

( ω − 15) 2

 675 12 ( ω − 15) 2  8100 ω  105

The complete expectation of life is half the maximum remaining life (ω − 25  80), or 40 . (B) LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

EXERCISE SOLUTIONS FOR LESSON 5

5.29.

103

Age at death is uniformly distributed with ω  2 e˚0  50. Then Var (T10 ) 

5.30.

(50 − 10) 2 12

 133 13

(C)

This is variance of curtate lifetime. We’ll use formula (5.19). Expected value is e67:3  p67 + 2p 67 + 3p 67 l 68 + l69 + l70  l67 7,018,432 + 6,823,367 + 6,616,155   2.840737 7,201,635

The second moment is

f

E min ( K 67 , 3)

2g

 p 67 + 3 2 p67 + 5 3 p67 l68 + 3l69 + 5l 70 l 67 7,018,432 + 3 (6,823,367) + 5 (6,616,155)   8.410493 7,201,635



The variance is 8.410493 − 2.8407372  0.340706 5.31.

The survival probabilities are p x  0.7 2p x

 (0.7)(0.5)  0.35

3p x  (0.35)(0.3)  0.105 4p x

0

The moments are E[K x ]  0.7 + 0.35 + 0.105  1.155





E[K 2x ]  2 0.7 + 0.35 (2) + 0.105 (3) − 1.155  2.275 Var ( K x )  2.275 − 1.1552  0.940975 5.32. For constant force of mortality, the curtate life expectancy is e −µ / (1−e −µ ) by equation (5.22). Setting e35  49 and solving for µ, we get e −µ 1 − e −µ −µ 49 − 49e  e −µ 49 e −µ  50 µ  0.020203 49 

Using primes for revised functions, adding 0.01, the revised µ0 is 0.030203. Then 0 e35 

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e −0.030203  32.61213 1 − e −0.030203

5. SURVIVAL DISTRIBUTIONS: MOMENTS

104

5.33. This could be done from the definition, and such a solution is provided in the official solutions. But we will use formulas (5.19) and (5.21). The survival probabilities are p x  0.9 2p x

 (0.9)(0.8)  0.72

3p x

 (0.9)(0.8)(0.7)  0.504

The moments are E[X]  0.9 + 0.72 + 0.504  2.124

f

g

E X 2  0.9 (1) + 0.72 (3) + 0.504 (5)  5.58 Var ( X )  5.58 − 2.1242  1.068624

(A)

5.34. Survival is uniform up to age 70. Complete temporary life expectancy to age 70 is therefore equal to the probability of death before age 70 times the median time to death (20), plus 40 times the probability of death at age 70. Setting this equal to 27.692, 20 40q 30 + 40 (1 − 40q 30 )  27.692 40 − 20 40q 30  27.692 40 − 27.692  0.6154 40q 30  20 But 40q 30  40/ ( ω − 30) , so ω − 30  40/0.6154  65. The variance of a uniform distribution is the interval 1 squared over 12. Here the interval for (30) is ω − 30  65, so the answer is 652 /12  352 12 . (B)

5.35. Mortality follows a beta distribution. In the paragraph before example 5B on page 80 we mention that if X follows a beta distribution, then E[X]  e˚0  µx 

ω α+1

α ω−x

Let α0 be the revised α. We are given ω ω  α0 + 1 2 ( α + 1 ) α0 2.25α  ω−x ω−x

from (i) from (ii)

From the second equation, α0  2.25α. From the first equation, 2.25α + 1  2α + 2 0.25α  1 α 4

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(D)

EXERCISE SOLUTIONS FOR LESSON 5

105

√ 5.36. Mortality is exponential with variance 100 and therefore mean 100  10, so we want the limited expected value of an exponential with mean 10. Integrating the survival function e −x/10 : 10

Z





e −x/10 dx  10 1 − e −1  6.3212

(C)

0

or if you prefer a non-integral solution, note that E[X] is the complete expectation now minus the probability of survival for 10 years times the complete expectation 10 years from now, or E[X]  e˚x:10  e˚x − 10p x e˚x+10  10 (1 − e −10 (0.1) )  6.3212 5.37.

We use equation (5.2) in conjunction with t p 30  S0 (30 + t ) /S0 (30) .

Z e˚30:50  

50

t p 30 dt 0  R 80  1 − (0.01t ) 2 dt 30

S0 (30)



S0 (30)  1 − 0.01 (30)

Z

80 



1 − (0.01t ) 2 dt  50 −

0.012 (803

33 56 0.91

 0.91

− 303 )

3

30

e˚30:50 

2

 37.179

 33 56

(D)

5.38. Since 2p 70 is equal for all three models and e70:2  p70 + 2p 70 , the answer depends purely on p70 . The curve t p 70 for t ∈ [0, 2] is a line for DM and a convex curve lying below the line for CF (since they intersect at t  0 and t  2), so CF < DM. For ILT, we find that d70  6,615,155 − 6,396,609  218,546 while DM ILT d71  6,396,609 − 6,164,663  231,946, so d x is increasing (whereas it is level for DM), so p70 > p 70 and DM > CF. Thus (C) is the correct choice. 5.39. The new model is beta, so the expected lifetime is e˚x  ( ω − x ) / ( α + 1) and can set up simultaneous equations to solve for ω: ω − 30 4 ω − 30  α+1 3 2 ω − 60  20 α+1

! from (ii) from (iii)

From the second equation, α + 1  ( ω − 60) /20 which we plug into the first equation. 20 ( ω − 30) 4 ω − 30  ω − 60 3 2 20 4 2   ω − 60 6 3 20 ω − 60   30 2/3 ω  90

!

Then in the previous uniform model, e˚70  (90 − 70) /2  10 . (B) It is not necessary to solve for α, but α  0.5. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

5. SURVIVAL DISTRIBUTIONS: MOMENTS

106

Quiz Solutions 5-1. Calculate t p 50 .

Z

50+t 50

√ 50+t µ x dx  − ln (10 − x ) 50    √  √  ln 10 − 50 − ln 10 − 50 + t √ ! Z 50+t 10 − 50 + t µ x dx  √ t p 50  exp − 10 − 50 50

Calculate e˚50 using formula (5.2). √ 10 − 50 + t dt √ 10 − 50 0 ! 50 2 1 3/2 500 − (50 + t ) √ 3 0 10 − 50   2 1 500 − (1003/2 − 503/2 ) √ 3 10 − 50 69.03559  23.57 2.92893

Z e˚50    

50

5-2. Set up the equation for e˚20:n . e˚20:n

100 − n n n  (n ) + 100 100 2

!

4800  100n − n 2 + 0.5n 2  −0.5n 2 + 100n We therefore have the quadratic n 2 − 200n + 9600  0. Solutions are n  80 and n  120. However, 120 is spurious since 20 + n must be less than ω. 5-3. First calculate the mean: Then the second moment:

e90  0.98 + .95  1.93

f

g





2 E K 90  2 0.98 + 0.95 (2) − 1.93  3.83

The variance is Var ( K 20 )  3.83 − 1.932  0.1051 . 5-4. Using e˚20:24  2424p 20 + 1224q 20 and e20:24  e˚20:24 − 0.524q 20 , we have 20.25  2424 p20 + 11.524q 20

24q 20

 24 − 2424q 20 + 11.524q 20 3.75   0.3 12.5

24  0.3 ω − 20 ω − 20  80 ω  100

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Lesson 6

Survival Distributions: Percentiles and Recursions Reading: SModels for Quantifying Risk (4th or 5th edition) 5.1, 5.3

6.1

Percentiles

A 100π percentile of survival time is the time t such that there is a 100π% probability that survival time is less than t. In other words, it is t such that t q x  π, or t p x  1 − π. Notice that high percentiles correspond to low probabilities of survival. A special case is the median remaining lifetime at age x, which is t such that t p x  t q x  0.50. If future survival time is uniformly distributed with limiting age ω, median remaining lifetime at age x is the midrange, half way from x to ω, or ( ω − x ) /2. If survival has a constant force of mortality µ, meaning that it follows an exponential distribution, we must find t such that e −µt  0.5. Solving this, we have that t  (ln 2) /µ. It doesn’t matter what x is in this case since the exponential distribution is memoryless. Here’s an example that’s a little harder: Example 6A A person age 70 is subject to the following force of mortality:

  0.1 µ70+t    0.2

t≤5 t>5

 Calculate median future lifetime for this person. Answer: We want t p x  0.5. For t ≤ 5, t p x  e −0.1t . Plugging in t  5, we get e −0.5  0.606531, which is greater than 0.5, so the median is greater than 5. For t > 5, t p x  e −0.5 e −0.2 ( t−5) . We solve: e −0.5 e −0.2 ( t−5)  0.5 −0.5 − 0.2t + 1  ln 0.5 − ln 0.5 + 0.5  5.9657 t 0.2



If you are using a life table, then the 100π percentile of future lifetime at age x is the age x + t at which l x+t is equal to (1− π ) l x . For example, if l x  1,000,000, then the 20th percentile is the t such that l x+t  800,000. Usually there will be no t satisfying this exactly, so you will only know the median is between two integral ages. To get an exact answer, it will be necessary to interpolate between the ages. Interpolation is discussed in the next lesson. Example 6B Mortality follows the Illustrative Life Table. Determine the age containing the 90th percentile of survival time for a person age (30). Answer: l30  9,501,381, so we need the age x such that l x  0.1l30  950,138.1. We see that l 90  1,058,491 is higher and l 91  858,676 is lower, so age 90 is the age containing the 90th percentile. 

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107

6. SURVIVAL DISTRIBUTIONS: PERCENTILES AND RECURSIONS

108

?

Quiz 6-1 Future lifetime is subject to force of mortality µ x+t 

1 t + 50

Determine the third quartile of future lifetime.

6.2

Recursive formulas for life expectancy

For both complete and curtate future lifetime, we have formulas expressing them as the sums or integrals of probabilities of survival. We can break the formula for the life expectancy of ( x ) up into a component for temporary life expectancy for a period of n years, plus a component involving the life expectancy of ( x + n ) times the probability of surviving n years. A special case is when n  1. For complete life expectancy, the decomposition looks like this: e˚x  e˚x:n + n p x e˚x+n

(6.1)

whereas for curtate life expectancy, the decomposition looks like this: e x  e x:n + n p x e x+n  e x:n−1 + n p x (1 + e x+n )

(6.2) (6.3)

and when n  1 e x  p x + p x e x+1  p x (1 + e x+1 )

(6.4)

These formulas allow recursive construction of a table of life expectancies at all ages. To construct such a table, start with the end of the table ω, at which e ω  0, and then calculate e x−1 from e x repeatedly starting with x  ω and ending with x  1. These formulas also allow fast computation of life expectancy when you are given a piecewise constant force of mortality, as the next example shows. Example 6C A person age 70 is subject to the following force of mortality:

  0.01 µ70+t    0.02

t≤5 t>5

 Calculate e˚70 for this person. Answer: This is the same as exercise 5.1, but we will now do it without integrals. By the recursive formula, e˚70  e˚70:5 + 5p 70 e˚75 Now, e˚75 is the life expectancy of someone with constant force of mortality 0.02, or exponential mortality, and we know that the life expectancy for exponential mortality is the reciprocal of the force, or e˚75  1/0.02  50. Also, 5 p70  e −0.01 (5)  e −0.05 . Now, consider a person age 70 subject to the constant force of mortality 0.01. We will use primes for this person’s mortality functions. By the recursive formula, 0 0 0 0 e˚70  e˚70:5 + 5 p70 e˚75 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

EXERCISES FOR LESSON 6

109





0 0 0 0 However, e˚70  100 1 − e −0.05 . And now the punch line:  e˚75  1/0.01  100 and 5 p70  e −0.05 , so e˚70:5 0 e˚70:5  e˚70:5 . Why? Because for the first five years, the forces of mortality for the person in our example and the person with constant force 0.01 are the same, and e˚70:5 is a function only of the force of mortality in the first five years. So we have





e˚70  100 1 − e −0.05 + 50e −0.05  4.8771 + 47.5615  52.4386



Another use of recursive computation for exam questions is calculating the effect on life expectancy of changing the mortality assumption for a year or for a period. You would start with the original e x , then advance to the age x + n beyond the change in mortality, then work back to e x using the revised mortality assumptions. Example 6D For ( x ) , standard curtate life expectancy is 72 years and standard q x  0.01. Because ( x ) has better underwriting characteristics, q x  0.005 for ( x ) , but mortality for ages x + 1 and higher is standard. Calculate curtate life expectancy for ( x ) . Answer: We calculate e x+1 by rearranging equation (6.4): ex − px px 72 − 0.99   71.72727 0.99

e x+1 

Then for ( x ) , we have e x  0.995 + 0.995 (71.72727)  72.36364



The recursive formulas also work for temporary life expectancies on the left side of the equation. For example, equations (6.2) and (6.3) become e x:n  e x:m + m p x e x+m:n−m  e x:m−1 + m p x (1 + e x+m:n−m )

m C.

The inequality is true if and only if e x+1 >

D.

px p x+1 q x+1

The inequality is true if and only if e x+1 >

E.

px + 1 qx

The inequality is true if and only if e x+1 >

px qx

6.14. [CAS4A-F97:20] (1 point) For a life age 50, the curtate expectation of life e50  20. For that same life, you are also given that p50  0.97. Determine e51 . A. B. C. D. E.

Less than 18.75 At least 18.75, but less than 19.00 At least 19.00, but less than 19.25 At least 19.25, but less than 19.50 At least 19.50

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Exercises continue on the next page . . .

EXERCISES FOR LESSON 6

113

6.15. [CAS4-S86:26] (2 points) Consider a subgroup of lives that have been exposed to a certain disease. It is estimated that this subgroup will have a higher than normal rate of mortality for two years following exposure to this disease. The mortality rate is 10% higher than normal during the first year and 5% higher during the second year. After that the mortality rate returns to normal. You are given: • • • •

q x  0.07 q x+1  0.10 q x+2  0.11 e x+3  5

Calculate the reduction in curtate life expectancy, in years, for a person age ( x ) who has just been exposed to this disease. A. B. C. D. E.

Less than 0.050 At least 0.050, but less than 0.075 At least 0.075, but less than 0.100 At least 0.100, but less than 0.125 At least 0.125 You are given

6.16. • • • •

e˚40  35 e˚40:10  9 10 p 40  0.85 t p 50  1 − 0.01t for 0 ≤ t ≤ 1.

Improvements in mortality at age 50 cause t p50 to change to 1 − 0.009t for 0 ≤ t ≤ 1. Calculate the revised value of e˚40 . 6.17.

You are given: • S0 (20)  0.9 • S0 (60)  y • The survival distribution function is linear between ages 20 and 60. • e˚20  60 • e˚60  25

Determine y. 6.18.

You are given: • e˚40  75 • e˚60  70 • The force of mortality µ x for x ∈ [40, 60] is 1/ ( k − x ) for some k.

Determine k. 6.19.

Mortality follows Gompertz’s law with B  0.001 and c  1.05.

You are given that e40  34.97. Determine e41 .

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Exercises continue on the next page . . .

6. SURVIVAL DISTRIBUTIONS: PERCENTILES AND RECURSIONS

114

6.20.

Mortality follows Gompertz’s law. You are given e80  5.665, e81  5.362 and e82  5.071.

Determine e83 . Additional old CAS Exam 3/3L questions: F08:13, S10:1, S12:2 Additional old CAS Exam LC questions: S14:2

Solutions 6.1.

Under Gompertz’s law, x

Z S0 ( x )  exp −

! t

Bc dt 0

−B ( c x − 1)  exp ln c

!

We set S0 (40)  0.9 (the 90th percentile of the survival function is the 10th percentile of future lifetime at birth) and S0 (80)  0.3. −B ( c 40 − 1)  0.9 ln c

!

exp

B ( c 40 − 1)  − ln 0.9 ln c ! −B ( c 80 − 1)  0.3 exp ln c B ( c 80 − 1)  − ln 0.3 ln c

(*)

(**)

Dividing (*) into (**) and using c 80 − 1  ( c 40 + 1)( c 40 − 1) , c 80 − 1 − ln 0.3  c 40 + 1   11.427173 40 − ln 0.9 c −1 √ 40 c  11.427173 − 1  1.060362 6.2. For this beta, t p 30 

√

(90 − t ) /90. Setting this equal to 0.5, ! 0.5

90 − t  0.5 90 90 − t  0.52  0.25 90 t  90 − 22.5  67.5

6.3. We want x such that S0 ( x ) 

1
1
2 1+y .

1 1  1 + x 2 (1 + y ) x  1 + 2y x is the median age at death for ( y ) . Median future lifetime for ( y ) is x − y  1 + y . (A) This survival function is a Pareto. You can see why a Pareto distribution is not a plausible distribution for human life: the older ( y ) is, the longer ( y ) ’s median future lifetime! LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

EXERCISE SOLUTIONS FOR LESSON 6

115

6.4. I.

II. III.

We can compare the complements, t+u p x and u p x+t . The former is the latter times t p x , and is therefore less than or equal to the latter. It follows the complements have the reverse relationship and I is true. ! t|u q x

 t p x u q x+t , and t p x ≤ 1, so II is true. !

Both mean and median are equal to the midrange, so III is true. !

(D) 6.5.

Survival is uniform with ω  100, so the median is the midrange,

100−10 2

 45 (C)

6.6. The conditional survival function is t

Z t p 20

 exp *−

, 0  √

r

1 du + 80 − 20 − u

 t  exp 2 60 − u 0  √ √  exp 2

60 − t − 60



We set this equal to 0.5 to get the median. 2

√

√  60 − t − 60  ln 0.5 √ √ ln 0.5 0.6931 60 − t  60 +  7.7460 −  7.3994 2 2 60 − t  7.39942  54.751 t  60 − 54.751  5.249

(A)

6.7. The population has a mixture distribution; the probability of lifetime greater than x is 0.3e −0.2x + 0.7e −0.1x . We want to set this equal to 0.25, so that there will be a 25% chance of living longer than x and therefore a 75% chance of living less. Let y  e −0.1x . Then we have 0.3y 2 + 0.7y − 0.25  0 √ −0.7 + 0.49 + 0.3 y  0.31470 0.6 e −0.1x  0.31470 x  −10 ln 0.31470  11.5614

(D)

6.8. We will do three recursions, one age at a time. Since e x  p x (1 + e x+1 ) , it follows that e x+1 

( e x /p x ) − 1.

e66  24/0.99 − 1  23.24242 e67  23.24242/0.985 − 1  22.59637 e68  22.59637/0.98 − 1  22.0575

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116

6.9. The complete expectation of life is the complete expectation bounded by t plus the probability of survival to t times the complete expectation of life after t years (formula (6.1)).

f

g

E[T]  E min (T, 40) + e˚40 (40 p0 )





62  40 − 0.005 (402 ) + e˚40 (0.6)

e˚40 6.10.

 32 + e˚40 (0.6) 62 − 32   50 0.6

(E)

The original 11-year temporary complete life expectancy is computed using equation (5.12): e˚25:11  11 p 25 (11) + 11 q25 (5.5) 64 (11) 11 (5.5)  +  10.193333 75 75

The 10-year temporary complete life expectancy for (26) is e˚26:10  10 p 26 (10) + 10 q26 (5) 64 (10) 10 (5) +  9.324324  74 74 We recursively develop the modified e˚25:11 using equation (6.1). We’ll use a prime to denote the modified functions. 0 0 0 e˚26:10 e˚25:11  e˚25:1 + p25 0 p25  e −0.1 1

Z 

0 e˚25:1



e −0.1t dt  10 1 − e −0.1



0





 10 1 − e −0.1 + e −0.1 (9.324324)

0 e˚25:11

 0.951626 + 8.436997  9.388623 The difference in mortality is 10.193333 − 9.388623  0.8047 (D) 6.11. As discussed in Table 3.1 at the bottom, adding an expression to µ x multiplies survival probabilities by the exponential of negative its integral, so 1

Z N p25



M p25

exp −

! 0.1 (1 − t ) dt

0



1

M  p25 exp −0.05 (1 − t ) 2



0

M  e −0.05 p25 M M N N Since t p 26  t p26 for all t, it follows that e26  e26 . We use recursive formula (6.4). N N N ) e25  p 25 (1 + e26 M M  e −0.05 p25 (1 + e26 ) M  e −0.05 e25

 10e −0.05  9.5123

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(D)

EXERCISE SOLUTIONS FOR LESSON 6

117

6.12. As discussed at the bottom of Table 3.1, doubling µ squares p35 . Using primes for revised values, 0 p35  0.9952 . Then using the recursion of formula (6.4) 1 + e36  e35 /p35  e35 /0.995 0 and since e36  e36 , 0 0 e35  p35 (1 + e36 )  0.9952 (1 + e36 ) 

so we conclude

6.13.

0.9952 e35 0.995

0 e35  49 (0.995)  48.755

By equation (6.4)

e x  p x + p x e x+1

So we want e x+1 > p x + p x e x+1  p x + e x+1 − q x e x+1 0 > p x − q x e x+1 px e x+1 > (E) qx 6.14.

Using formula (6.4), e50  p50 + p50 e51 20  0.97 + 0.97e51 20 e51  − 1  19.6186 0.97

(E)

6.15. Note that mortality rate refers to q x , not µ x . The latter, µ x , would be called force of mortality. Using primes to indicate modified functions for the subgroup of lives exposed to the disease, ex 

∞ X

kpx

 p x + 2p x + 3p x (1 + e x+3 )  p x + 2p x + 63p x

k1

so the difference between the two life expectancies is p x − p 0x + 2p x − 2p 0x + 6 (3p x − 3p 0x ) Let’s calculate the needed survival probabilities. q 0x  0.07 (1.1)  0.077 q 0x+1  0.10 (1.05)  0.105 p x  0.93 2p x

 (0.93)(0.9)  0.837

3p x

 (0.837)(0.89)  0.74493

p 0x+1  0.895

p 0x  0.923 0 2p x 0 3p x

p x − p 0x  0.007

 (0.923)(0.895)  0.826085

2p x

− 2p 0x  0.010915

 (0.826085)(0.89)  0.735216

3p x

− 3p 0x  0.009714

The answer is 0.007 + 0.010915 + 6 (0.009714)  0.07620 . (C)

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p 0x  0.923

6. SURVIVAL DISTRIBUTIONS: PERCENTILES AND RECURSIONS

118

6.16.

First we use the recursive formula to calculate e˚50 . e˚40  e˚40:10 + 10 p 40 e˚50 35  9 + 0.85 e˚50 26 e˚50   30.588235 0.85

Then we use the recursive formula to calculate e˚51 . By substituting t  1 into condition (iv), w see that p50  0.99. Since deaths are uniformly distributed over age 50, by either using the trapezoidal rule or the formula e˚50:1  0.5q50 + p50 (equation (5.12)), we get e˚50:1  0.5 (1 + 0.99)  0.995. Then e˚50  e˚50:1 + p50 e˚51 30.588235  0.995 + 0.99e˚51 e˚51  29.892157 0 0  0.991. Using  1 − 0.009t, we get p50 We use primes for the revised functions. Substituting t  1 into t p50 0 the trapezoidal rule or equation (5.12), e˚  0.5 (1 + 0.991)  0.9955. 50:1

0 e˚50  0.9955 + 0.991 (29.892157)  30.618627 0 e˚40  9 + 0.85 (30.618627)  35.025833

6.17. Split the universe into two groups, the ones that survive to age 60 and the ones who don’t. The ones who survive to age 60 have an expected lifetime of 40 plus e˚60 , or 65. The ones who don’t survive to age 60 have an expected lifetime of 20, since survival is uniform between ages 20 and 60. Expected lifetime at 20 is the weighted average of the expected lifetime of these two groups: e˚20  40p 20 (65) + (1 − 40p 20 )(20) 60  20 + 45 40p 20 40 8  40p 20  45 9 But 40p 20  S0 (60) /S0 (20) , so it follows that S0 (60)  (8/9) S0 (20)  0.8 . 6.18. Note that mortality has a uniform distribution between 40 and 60, since µ x has the form 1/ ( ω − x ) in that interval. (This is not saying that mortality is uniform globally; it may be non-uniform above 60. All we’re saying is that t q 40  tq 40 for t ≤ 20.) The life expectancies are related by ∞

Z e˚40 

20

Z t p 40 dt 

∞

Z t p 40 dt + 20p 40

0

0

t−20p 60 dt 20

20

Z 

t p 40 dt

+ 20p 40 e˚60

t p 40 dt

+ 70 20p 40

0 20

Z 75 

(*)

0

However, since mortality is uniform, t p 40 is linear, making the integral equal to the midpoint times the length of the range (60 − 40  20). The midpoint of the interval [40, 60] is 50, so the average of t p 40 on the interval t ∈ [0, 20] is 10p 40 , and the integral is 20 10p 40 . If we express 10p 40 and 20p 40 in terms of k, we’ll

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EXERCISE SOLUTIONS FOR LESSON 6

119

almost be done. t

Z t p 40  exp − 0

du k − (40 + u )

!



 exp ln ( k − 40 − t ) − ln ( k − 40)



k − 40 − t k − 40 k − 50  k − 40 k − 60  k − 40

 10p 40 20p 40

Plugging back into (*)

!

k − 50 k − 60 + 70 75  20 k − 40 k − 40

!

75k − 3000  20k − 1000 + 70k − 4200 15k  2200 2200 k  146 32 15 6.19.

Let’s calculate p40 .

0.001 (1.0540 )(0.05)  0.992811  exp − ln 1.05

!

p40 Then from the recursive formula,

e40  p40 (1 + e41 ) 34.97  0.992811 (1 + e41 ) 34.97 e41  − 1  34.22 0.992811 6.20.

Back out p 80 and p81 from the recursive formula. e80  p80 (1 + e81 ) 5.665 p80   0.8904 6.362 5.362 p81   0.8832 6.071

Now back out the B and c parameters of Gompertz’s law. Bc 80 ( c − 1)  − ln 0.8904 ln c Bc 81 ( c − 1)  − ln 0.8832 ln c ln 0.8832 c  1.07024 ln 0.8904 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

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120

0.11604 (0.06788)   0.11214 c−1 0.07024 ! 0.11214 (1.070242 )(0.07024)  exp −  0.87554 ln 1.07024 e82 5.071  −1 − 1  4.7918 p82 0.87554

Bc 80  p82 e83

(− ln 0.8904)(ln c )

Quiz Solutions 6-1. The third quartile is the 75th percentile. The survival probability is t

Z tpx

 exp − 0

du u + 50

!





 exp − ln ( t + 50) + ln (50)  This equals 0.25 at: 50  0.25 50 + t t  150 6-2. Since e0  p0 (1 + e1 ) , we have e1  70/p 0 − 1 > 70, so 70 > 71 p0 p0 <

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70 71

50 50 + t

Lesson 7

Survival Distributions: Fractional Ages Reading: Models for Quantifying Risk (4th or 5th edition) 6.6.1, 6.6.2 Life tables list mortality rates (q x ) or lives (l x ) for integral ages only. Often, it is necessary to determine lives at fractional ages (like l x+0.5 for x an integer) or mortality rates for fractions of a year. We need some way to interpolate between ages. The easiest interpolation method is linear interpolation, or uniform distribution of deaths between integral ages (UDD). This means that the number of lives at age x + s, 0 ≤ s ≤ 1, is a weighted average of the number of lives at age x and the number of lives at age x + 1: l x+s  (1 − s ) l x + sl x+1  l x − sd x The graph of l x+s is a straight line between s  0 and s  1 with slope −d x . The graph at the right portrays this for a mortality rate q 100  0.45 and l 100  1000. Contrast UDD with an assumption of a uniform survival function. If age at death is uniformly distributed, then l x as a function of x is a straight line. If UDD is assumed, l x is a straight line between integral ages, but the slope may vary for different ages. Thus if age at death is uniformly distributed, UDD holds at all ages, but not conversely. Using l x+s , we can compute s q x : s qx

 1 − s px l x+s 1−  1 − (1 − sq x )  sq x lx

(7.1) l 100+s 1000

550

0

0

1

s

(7.2)

That is one of the most important formulas, so let’s state it again: s qx

 sq x

(7.2)

More generally, for 0 ≤ s + t ≤ 1, s q x+t

l x+s+t l x+t sq x sd x l x − ( s + t ) dx 1−   l x − td x l x − td x 1 − tq x  1 − s p x+t  1 −

(7.3)

where the last equation was obtained by dividing numerator and denominator by l x . The important point to pick up is that while s q x is the proportion of the year s times q x , the corresponding concept at age x + t, s q x+t , is not sq x , but is in fact higher than sq x . The number of lives dying in any amount of time is constant, and since there are fewer and fewer lives as the year progresses, the rate of death is in fact increasing over the year. The numerator of s q x+t is the proportion of the year being measured s times the death rate, but then this must be divided by 1 minus the proportion of the year that elapsed before the start of measurement. For most problems involving death probabilities, it will suffice if you remember that l x+s is linearly interpolated. It often helps to create a life table with an arbitrary radix. Try working out the following example before looking at the answer. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

121

7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES

122

Example 7A You are given: • q x  0.1 • Uniform distribution of deaths between integral ages is assumed. Calculate 1/2 q x+1/4 . Answer: Let l x  1. Then l x+1  l x (1 − q x )  0.9 and d x  0.1. Linearly interpolating, l x+1/4  l x − 14 d x  1 − 41 (0.1)  0.975 l x+3/4  l x − 43 d x  1 − 43 (0.1)  0.925 l x+1/4 − l x+3/4 0.975 − 0.925   0.051282 1/2 q x+1/4  l x+1/4 0.975 You could also use equation (7.3) to work this example.



Example 7B For two lives age ( x ) with independent future lifetimes, Deaths are uniformly distributed between integral ages. Calculate the probability that both lives will survive 2.25 years.

k| q x

 0.1 ( k + 1) for k  0, 1, 2.

Answer: Since the two lives are independent, the probability of both surviving 2.25 years is the square of 2.25 p x , the probability of one surviving 2.25 years. If we let l x  1 and use d x+k  l x k| q x , we get q x  0.1 (1)  0.1

l x+1  1 − d x  1 − 0.1  0.9

1| q x  0.1 (2)  0.2

l x+2  0.9 − d x+1  0.9 − 0.2  0.7

 0.1 (3)  0.3

l x+3  0.7 − d x+2  0.7 − 0.3  0.4

2| q x

Then linearly interpolating between l x+2 and l x+3 , we get l x+2.25  0.7 − 0.25 (0.3)  0.625 l x+2.25  0.625 2.25p x  lx Squaring, the answer is 0.6252  0.390625 .



The probability density function of Tx , s p x µ x+s , is the constant q x , the derivative of the conditional cumulative distribution function s q x  sq x with respect to s. That is another important formula, since the density is needed to compute expected values, so let’s repeat it: s px

µ x+s  q x

(7.4)

µ100+s 1 0.45 0.55

0.45

It follows that the force of mortality is q x divided by 1 − sq x : µ x+s 

qx qx  p 1 − sq x s x

(7.5)

0

0 1 The force of mortality increases over the year, as illustrated in the graph for q 100  0.45 to the right.

?

Quiz 7-1 You are given: • µ50.4  0.01 • Deaths are uniformly distributed between integral ages. Calculate 0.6q 50.4 .

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s

7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES

123

Complete Expectation of Life Under UDD If the complete future lifetime random variable T is written as T  K + S, where K is the curtate future lifetime and S is the fraction of the last year lived, then K and S are independent. This is usually not true 1 . It if uniform distribution of deaths is not assumed. Since S is uniform on [0, 1) , E[S]  21 and Var ( S )  12 1 follows from E[S]  2 that e˚x  e x + 12 (7.6) More common on exams are questions asking you to evaluate the temporary complete expectancy of life under UDD. You can always evaluate the temporary complete expectancy, whether or not UDD is assumed, by integrating t p x , as indicated by formula (5.6) on page 78. For UDD, t p x is linear between integral ages. Therefore, a rule we learned in Lesson 5 applies for all integral x: e˚x:1  p x + 0.5q x

(5.13)

This equation will be useful. In addition, the method for generating this equation can be used to work out questions involving temporary complete life expectancies for short periods. The following example illustrates this. This example will be reminiscent of calculating temporary complete life expectancy for uniform mortality. Example 7C You are given • q x  0.1. • Deaths are uniformly distributed between integral ages. Calculate e˚x:0.4 . Answer: We will discuss two ways to solve this: an algebraic method and a geometric method. The algebraic method is based on the double expectation theorem, equation (1.11). It uses the fact that for a uniform distribution, the mean is the midpoint. If deaths occur uniformly between integral ages, then those who die within a period contained within a year survive half the period on the average. In this example, those who die within 0.4 survive an average of 0.2. Those who survive 0.4 survive an average of 0.4 of course. The temporary life expectancy is the weighted average of these two groups, or 0.4 q x (0.2) + 0.4 p x (0.4) . This is: 0.4 q x

 (0.4)(0.1)  0.04

0.4 p x

 1 − 0.04  0.96

e˚x:0.4  0.04 (0.2) + 0.96 (0.4)  0.392 An equivalent geometric method, the trapezoidal rule, is to draw the t p x function from 0 to 0.4. The integral of t p x is the area under the line, which is the area of a trapezoid: the average of the heights times the width. The following is the graph (not drawn to scale): t px

(0.4, 0.96)

1

(1.0, 0.9) A

0

B

0.4

Trapezoid A is the area we are interested in. Its area is 12 (1 + 0.96)(0.4)  0.392 .

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1.0

t



7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES

124

?

Quiz 7-2 As in Example 7C, you are given • q x  0.1. • Deaths are uniformly distributed between integral ages. Calculate e˚x+0.4:0.6 . Let’s now work out an example in which the duration crosses an integral boundary. Example 7D You are given: • q x  0.1 • q x+1  0.2 • Deaths are uniformly distributed between integral ages. Calculate e˚x+0.5:1 . Answer: Let’s start with the algebraic method. Since the mortality rate changes at x + 1, we must split the group into those who die before x + 1, those who die afterwards, and those who survive. Those who die before x + 1 live 0.25 on the average since the period to x + 1 is length 0.5. Those who die after x + 1 live between 0.5 and 1 years; the midpoint of 0.5 and 1 is 0.75, so they live 0.75 years on the average. Those who survive live 1 year. Now let’s calculate the probabilities. 0.5 q x+0.5 0.5 p x+0.5 0.5|0.5 q x+0.5 1 p x+0.5

0.5 (0.1) 5  1 − 0.5 (0.1) 95 5 90 1−  95 95 !  9 90 0.5 (0.2)   95 95 5 9 81 1− −  95 95 95 

These probabilities could also be calculated by setting up an l x table with radix 100 at age x and interpolating within it to get l x+0.5 and l x+1.5 . Then l x+1  0.9l x  90 l x+2  0.8l x+1  72 l x+0.5  0.5 (90 + 100)  95 l x+1.5  0.5 (72 + 90)  81 90 5  0.5 q x+0.5  1 − 95 95 90 − 81 9  0.5|0.5 q x+0.5  95 95 l x+1.5 81  1 p x+0.5  l x+0.5 95 Either way, we’re now ready to calculate e˚x+0.5:1 . e˚x+0.5:1 

5 (0.25) + 9 (0.75) + 81 (1) 89  95 95

For the geometric method we draw the following graph: LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES

125

t px +0.5

0.5, 90 95

1


1.0, 81 95

A




B

0 x + 0.5

0.5 x +1

t 1.0 x + 1.5

The heights at x + 1 and Then  x + 1.5 are as we computed above.   we compute each area separately. The 185 1 90 81 171 ( 0.5 )  . The area of B is + area of A is 21 1 + 90 95 95 (4) 2 95 95 (0.5)  95 (4) . Adding them up, we get 185+171 95 (4)

?



89 95

.



Quiz 7-3 The probability that a battery fails by the end of the k th month is given in the following table:

k

Probability of battery failure by the end of month k

1 2 3

0.05 0.20 0.60

Between integral months, time of failure for the battery is uniformly distributed. Calculate the expected amount of time the battery survives within 2.25 months. To calculate e˚x:n in terms of e x:n when x and n are both integers, note that those who survive n years contribute the same to both. Those who die contribute an average of 12 more to e˚x:n since they die on the average in the middle of the year. Thus the difference is 21 n q x : e˚x:n  e x:n + 0.5n q x

(7.7)

Example 7E You are given: • q x  0.01 for x  50, 51, . . . , 59. • Deaths are uniformly distributed between integral ages. Calculate e˚50:10 . Answer: As we just said, e˚50:10  e50:10 + 0.510 q50 . The first summand, e50:10 , is the sum of k p50  0.99k for k  1, . . . , 10. This sum is a geometric series: e50:10 

10 X

0.99k 

k1

0.99 − 0.9911  9.46617 1 − 0.99

The second summand, the probability of dying within 10 years is 10 q 50  1 − 0.9910  0.095618. Therefore e˚50:10  9.46617 + 0.5 (0.095618)  9.51398 The formulas are summarized in Table 7.1. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM



7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES

126

Table 7.1: Summary of formulas for fractional ages

l x+s  l x − sd x sqx

 sq x

s p x  1 − sq x sq x s q x+t  1 − tq x qx µ x+s  1 − sq x spx

µ x+s  q x e˚x  e x + 0.5 e˚x:n  e x:n + 0.5 n q x e˚x:1  p x + 0.5q x

Exercises 7.1. [CAS4-S85:16] (1 point) Deaths are uniformly distributed between integral ages. Which of the following represents 3/4p x + A.

3/4p x

7.2.

B.

3/4q x

C.

1 2 1/2p x

µ x+1/2 ?

1/2p x

D.

1/2q x

E.

1/4p x

[Based on 150-S88:25] You are given: • 0.25 q x+0.75  3/31. • Mortality is uniformly distributed within age x.

Calculate q x . Use the following information for questions 7.3 and 7.4: You are given: • Deaths are uniformly distributed between integral ages. • q x  0.10. • q x+1  0.15. 7.3. Calculate 1/2q x+3/4 . 7.4. Calculate 0.3|0.5 q x+0.4 . 7.5. You are given: • Deaths are uniformly distributed between integral ages. • Mortality follows the Illustrative Life Table. Calculate the median future lifetime for (45.5).

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Exercises continue on the next page . . .

EXERCISES FOR LESSON 7

127

7.6. [160-F90:5] You are given: • A survival distribution is defined by l x  1000 1 − • •



x 100

2!

, 0 ≤ x ≤ 100.

µ x denotes the actual force of mortality for the survival distribution. µ Lx denotes the approximation of the force of mortality based on the uniform distribution of deaths assumption for l x , 50 ≤ x < 51.

L Calculate µ50.25 − µ50.25 .

A. −0.00016

B. −0.00007

C. 0

D. 0.00007

E. 0.00016

7.7. A survival distribution is defined by • S0 ( k )  1/ (1 + 0.01k ) 4 for k a non-negative integer. • Deaths are uniformly distributed between integral ages. Calculate 0.4 q20.2 . 7.8. [Based on150-S89:15] You are given: • Deaths are uniformly distributed over each year of age. • x lx 35 36 37 38 39

100 99 96 92 87

Which of the following are true? I. II. III.

1|2q 36

 0.091

µ37.5  0.043 0.33q 38.5

 0.021

A. I and II only B. I and III only C. II and III only E. The correct answer is not given by A. , B. , C. , or D.

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D. I, II and III

Exercises continue on the next page . . .

7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES

128

7.9. [150-82-94:5] You are given: • Deaths are uniformly distributed over each year of age. • 0.75 p x  0.25. Which of the following are true? I.

0.25 q x+0.5

II.

0.5 q x

 0.5

 0.5

µ x+0.5  0.5

III.

A. I and II only B. I and III only C. II and III only E. The correct answer is not given by A. , B. , C. , or D. 7.10.

D. I, II and III

[3-S00:12] For a certain mortality table, you are given: • µ80.5  0.0202 • µ81.5  0.0408 • µ82.5  0.0619 • Deaths are uniformly distributed between integral ages.

Calculate the probability that a person age 80.5 will die within two years. A. 0.0782 7.11.

B. 0.0785

C. 0.0790

D. 0.0796

E. 0.0800

You are given: • Deaths are uniformly distributed between integral ages. • q x  0.1. • q x+1  0.3.

Calculate e˚x+0.7:1 . 7.12.

You are given: • Deaths are uniformly distributed between integral ages. • q45  0.01. • q46  0.011.







Calculate Var min T45 , 2 . 7.13.

You are given: • Deaths are uniformly distributed between integral ages. • 10 p x  0.2.

Calculate e˚x:10 − e x:10 .

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Exercises continue on the next page . . .

EXERCISES FOR LESSON 7

7.14.

129

[4-F86:21] You are given: • q60  0.020 • q61  0.022 • Deaths are uniformly distributed over each year of age.

Calculate e˚60:1.5 . A. 1.447 7.15.

B. 1.457

C. 1.467

D. 1.477

E. 1.487

[150-F89:21] You are given: • q70  0.040 • q71  0.044 • Deaths are uniformly distributed over each year of age.

Calculate e˚70:1.5 . A. 1.435 7.16.

B. 1.445

C. 1.455

D. 1.465

E. 1.475

[3-S01:33] For a 4-year college, you are given the following probabilities for dropout from all causes: q0  0.15 q1  0.10 q2  0.05 q3  0.01

Dropouts are uniformly distributed over each year. Compute the temporary 1.5-year complete expected college lifetime of a student entering the second year, e˚1:1.5 . A. 1.25 7.17.

B. 1.30

C. 1.35

D. 1.40

E. 1.45

You are given: • Deaths are uniformly distributed between integral ages. • e˚x+0.5:0.5  5/12.

Calculate q x . 7.18.

You are given: • Deaths are uniformly distributed over each year of age. • e˚55.2:0.4  0.396.

Calculate µ55.2 . 7.19.

[150-S87:21] You are given: • d x  k for x  0, 1, 2, . . . , ω − 1 • e˚20:20  18 • Deaths are uniformly distributed over each year of age.

Calculate 30|10 q30 . A. 0.111

B. 0.125

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C. 0.143

D. 0.167

E. 0.200 Exercises continue on the next page . . .

7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES

130

[150-S89:24] You are given:

7.20.

• Deaths are uniformly distributed over each year of age. • µ45.5  0.5 Calculate e˚45:1 . A. 0.4

B. 0.5

C. 0.6

D. 0.7

E. 0.8

7.21. [CAS3-S04:10] 4,000 people age (30) each pay an amount, P, into a fund. Immediately after the 1,000th death, the fund will be dissolved and each of the survivors will be paid $50,000. •

Mortality follows the Illustrative Life Table, using linear interpolation at fractional ages.

•

i  12% Calculate P.

A. B. C. D. E.

Less than 515 At least 515, but less than 525 At least 525, but less than 535 At least 535, but less than 545 At least 545

Additional old CAS Exam 3/3L questions: S05:31, F05:13, S06:13, F06:13, S07:24, S08:16, S09:3, F09:3, S10:4, F10:3, S11:3, S12:3, F12:3, S13:3, F13:3 Additional old CAS Exam LC questions: S14:4, F14:4

Solutions 7.1. In the second summand, 1/2 p x µ x+1/2 is the density function, which is the constant q x under UDD. The first summand 3/4 p x  1 − 34 q x . So the sum is 1 − 14 q x , or 1/4 p x . (E) 7.2. Using equation (7.3), 3  0.25 q x+0.75 31 3 2.25 − qx 31 31 3 31 qx



0.25q x 1 − 0.75q x

 0.25q x 10 qx 31  0.3 

7.3. We calculate the probability that ( x+ 34 ) survives for half a year. Since the duration crosses an integer boundary, we break the period up into two quarters of a year. The probability of ( x + 3/4) surviving for 0.25 years is, by equation (7.3), 1 − 0.10 0.9  1/4p x+3/4  1 − 0.75 (0.10) 0.925 The probability of ( x + 1) surviving to x + 1.25 is 1/4p x+1

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 1 − 0.25 (0.15)  0.9625

EXERCISE SOLUTIONS FOR LESSON 7

131

The answer to the question is then the complement of the product of these two numbers:

!

1/2q x+3/4  1 − 1/2p x+3/4  1 − 1/4p x+3/4 1/4p x+1  1 −

0.9 (0.9625)  0.06351 0.925

Alternatively, you could build a life table starting at age x, with l x  1. Then l x+1  (1 − 0.1)  0.9 and l x+2  0.9 (1 − 0.15)  0.765. Under UDD, l x at fractional ages is obtained by linear interpolation, so l x+0.75  0.75 (0.9) + 0.25 (1)  0.925 l x+1.25  0.25 (0.765) + 0.75 (0.9)  0.86625 l x+1.25 0.86625   0.93649 1/2 p 3/4  l x+0.75 0.925 1/2 q 3/4

7.4.

0.3|0.5q x+0.4

 1 − 1/2 p3/4  1 − 0.93649  0.06351

is 0.3p x+0.4 − 0.8p x+0.4 . The first summand is 0.3 p x+0.4



1 − 0.7q x 1 − 0.07 93   1 − 0.4q x 1 − 0.04 96

The probability that ( x + 0.4) survives to x + 1 is, by equation (7.3), 0.6p x+0.4



1 − 0.10 90  1 − 0.04 96

and the probability ( x + 1) survives to x + 1.2 is 0.2p x+1

 1 − 0.2q x+1  1 − 0.2 (0.15)  0.97

So

!

0.3|0.5q x+0.4 

93 90 − (0.97)  0.059375 96 96

Alternatively, you could use the life table from the solution to the last question, and linearly interpolate: l x+0.4  0.4 (0.9) + 0.6 (1)  0.96 l x+0.7  0.7 (0.9) + 0.3 (1)  0.93 l x+1.2  0.2 (0.765) + 0.8 (0.9)  0.873 0.93 − 0.873  0.059375 0.3|0.5 q x+0.4  0.96 7.5. Under uniform distribution of deaths between integral ages, l x+0.5  12 ( l x + l x+1 ) , since the survival function is a straight line between two integral ages. Therefore, l45.5  12 (9,164,051 + 9,127,426)  9,145,738.5. Median future lifetime occurs when l x  12 (9,145,738.5)  4,572,869. This happens between ages 77 and 78. We interpolate between the ages to get the exact median: l77 − s ( l 77 − l 78 )  4,572,869 4,828,182 − s (4,828,182 − 4,530,360)  4,572,869 4,828,182 − 297,822s  4,572,869 4,828,182 − 4,572,869 255,313 s   0.8573 297,822 297,822 So the median age at death is 77.8573, and median future lifetime is 77.8573 − 45.5  32.3573 . LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES

132

7.6.

x p0



lx l0

1−

x
2 100 .

The force of mortality is calculated as the negative derivative of ln x p0 : x 1 d ln x p0 2 100 2x 100  µx  − 
2 x dx 1002 − x 2 1 − 100 100.5 µ50.25   0.0134449 1002 − 50.252







For UDD, we need to calculate q 50 . l51 1 − 0.512   0.986533 l50 1 − 0.502  1 − 0.986533  0.013467

p 50  q 50 so under UDD, L µ50.25 

q50 0.013467   0.013512. 1 − 0.25q 50 1 − 0.25 (0.013467)

L is 0.013445 − 0.013512  −0.000067 . (B) The difference between µ50.25 and µ50.25

7.7.

S0 (20)  1/1.24 and S0 (21)  1/1.214 , so q 20  1 − (1.2/1.21) 4  0.03265. Then 0.4 q 20.2

0.4q20 0.4 (0.03265)   0.01315 1 − 0.2q 20 1 − 0.2 (0.03265)



7.8. I.

Calculate 1|2 q36 . 1|2q 36



2 d37

l 36



96 − 87  0.09091 99

!

This statement does not require uniform distribution of deaths. II.

By equation (7.5), µ37.5 

III.

q37 4/96 4    0.042553 1 − 0.5q37 1 − 2/96 94

!

Calculate 0.33 q38.5 . 0.33q 38.5



0.33 d38.5

l 38.5



(0.33)(5) 89.5

 0.018436

I can’t figure out what mistake you’d have to make to get 0.021. (A) 7.9. First calculate q x . 1 − 0.75q x  0.25 qx  1 Then by equation (7.3), 0.25 q x+0.5  0.25/ (1 − 0.5)  0.5, making I true. By equation (7.2), 0.5 q x  0.5q x  0.5, making II true. By equation (7.5), µ x+0.5  1/ (1 − 0.5)  2, making III false. (A)

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#

EXERCISE SOLUTIONS FOR LESSON 7

7.10.

133

We use equation (7.5) to back out q x for each age. µ x+0.5 

µ x+0.5 qx ⇒ qx  1 − 0.5q x 1 + 0.5µ x+0.5 0.0202 q 80   0.02 1.0101 0.0408  0.04 q 81  1.0204 0.0619 q 82   0.06 1.03095

Then by equation (7.3), 0.5 p80.5  0.98/0.99. p81  0.96, and 0.5 p82  1 − 0.5 (0.06)  0.97. Therefore

!

2 q 80.5  1 −

0.98 (0.96)(0.97)  0.0782 0.99

(A)

7.11. To do this algebraically, we split the group into those who die within 0.3 years, those who die between 0.3 and 1 years, and those who survive one year. Under UDD, those who die will die at the midpoint of the interval (assuming the interval doesn’t cross an integral age), so we have Group

Survival time

Probability of group

Average survival time

I II III

(0, 0.3] (0.3, 1] (1, ∞)

1 − 0.3 p x+0.7 p 0.3 x+0.7 − 1 p x+0.7 1 p x+0.7

0.15 0.65 1

We calculate the required probabilities. 0.3 p x+0.7 1 p x+0.7

1 − 0.3 p x+0.7

0.9  0.967742 0.93   0.9  1 − 0.7 (0.3)  0.764516 0.93  1 − 0.967742  0.032258 

0.3 p x+0.7 − 1 p x+0.7  0.967742 − 0.764516  0.203226

e˚x+0.7:1  0.032258 (0.15) + 0.203226 (0.65) + 0.764516 (1)  0.901452 Alternatively, we can use trapezoids. We already know from the above solution that the heights of the first trapezoid are 1 and 0.967742, and the heights of the second trapezoid are 0.967742 and 0.764516. So the sum of the area of the two trapezoids is e˚x+0.7:1  (0.3)(0.5)(1 + 0.967742) + (0.7)(0.5)(0.967742 + 0.764516)  0.295161 + 0.606290  0.901451 7.12.

For the expected value, we’ll use the recursive formula. (The trapezoidal rule could also be used.) e˚45:2  e˚45:1 + p45 e˚46:1  (1 − 0.005) + 0.99 (1 − 0.0055)  1.979555

We’ll use equation (5.7)to calculate the second moment. 2

Z E[min (T45 , 2) 2 ]  2

t t p x dt 0

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7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES

134

1

Z 2

2

Z t (1 − 0.01t ) dt +

0





!

t (0.99) 1 − 0.011 ( t − 1) dt 1

! 1 (1.011)(22 − 12 ) 23 − 13 ++ 1 * * // + 0.99 . − 0.011  2 . − 0.01 2 3 2 3 , -,  2 (0.496667 + 1.475925)  3.94518 !

So the variance is 3.94518 − 1.9795552  0.02654 . 7.13.

As discussed on page 125, by equation (7.7), the difference is 1 1 10 q x  (1 − 0.2)  0.4 2 2

7.14. Those who die in the first year survive 21 year on the average and those who die in the first half of the second year survive 1.25 years on the average, so we have p60  0.98 1.5 p 60





 0.98 1 − 0.5 (0.022)  0.96922

e˚60:1.5  0.5 (0.02) + 1.25 (0.98 − 0.96922) + 1.5 (0.96922)  1.477305

(D)

Alternatively, we use the trapezoidal method. The first trapezoid has heights 1 and p60  0.98 and width 1. The second trapezoid has heights p60  0.98 and 1.5 p 60  0.96922 and width 1/2. e˚60:1.5

1 1  (1 + 0.98) + 2 2  1.477305

!

!

1 (0.98 + 0.96922) 2

(D)

7.15. p70  1 − 0.040  0.96, 2p 70  (0.96)(0.956)  0.91776, and by linear interpolation, 1.5 p70  0.5 (0.96 + 0.91776)  0.93888. Those who die in the first year survive 0.5 years on the average and those who die in the first half of the second year survive 1.25 years on the average. So e˚70:1.5  0.5 (0.04) + 1.25 (0.96 − 0.93888) + 1.5 (0.93888)  1.45472

(C)

Alternatively, we can use the trapezoidal method. The first year’s trapezoid has heights 1 and 0.96 and width 1 and the second year’s trapezoid has heights 0.96 and 0.93888 and width 1/2, so e˚70:1.5  0.5 (1 + 0.96) + 0.5 (0.5)(0.96 + 0.93888)  1.45472 7.16.

(C)

First we calculate t p1 for t  1, 2. p 1  1 − q 1  0.90 2 p1

 (1 − q1 )(1 − q2 )  (0.90)(0.95)  0.855

By linear interpolation, 1.5 p1  (0.5)(0.9 + 0.855)  0.8775. The algebraic method splits the students into three groups: first year dropouts, second year (up to time 1.5) dropouts, and survivors. In each dropout group survival on the average is to the midpoint (0.5 years for the first group, 1.25 years for the second group) and survivors survive 1.5 years. Therefore e˚1:1.5  0.10 (0.5) + (0.90 − 0.8775)(1.25) + 0.8775 (1.5)  1.394375

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(D)

EXERCISE SOLUTIONS FOR LESSON 7

135

t p1

Alternatively, we could sum the two trapezoids making up the shaded area at the right.

1

(1, 0.9)

(1.5,0.8775)

e˚1:1.5  (1)(0.5)(1 + 0.9) + (0.5)(0.5)(0.90 + 0.8775)  0.95 + 0.444375  1.394375 7.17.

(D)

0 1 1.5 2 Those who die survive 0.25 years on the average and survivors survive 0.5 years, so we have 0.25 0.5 q x+0.5 + 0.5 0.5 p x+0.5 

!

5 12

!

1 − qx 0.5q x + 0.5  1 − 0.5q x 1 − 0.5q x

0.25

t

5 12

5 5 − 24 qx 0.125q x + 0.5 − 0.5q x  12   1 5 5 1 1 qx −  − + − 2 12 24 2 8 qx 1  12 6

qx 

1 2

Alternatively, complete life expectancy is the area of the trapezoid shown on the right, so

t px +0.5

1 5 12

5  0.5 (0.5)(1 + 0.5 p x+0.5 ) 12 Then 0.5 p x+0.5  32 , from which it follows

0 1 − qx 2  3 1 − 12 q x qx 

7.18.

1 2

Survivors live 0.4 years and those who die live 0.2 years on the average, so 0.396  0.40.4 p 55.2 + 0.20.4 q 55.2

Using the formula 0.4 q55.2  0.4q55 / (1 − 0.2q55 ) (equation (7.3)), we have

!

0.4

!

0.4q 55 1 − 0.6q55 + 0.2  0.396 1 − 0.2q55 1 − 0.2q 55 0.4 − 0.24q 55 + 0.08q 55  0.396 − 0.0792q 55 0.0808q 55  0.004 0.004 q 55   0.0495 0.0808 q 55 0.0495 µ55.2    0.05 1 − 0.2q 55 1 − 0.2 (0.0495)

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0.5 px +0.5

0.5

t

7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES

136

7.19. Since d x is constant for all x and deaths are uniformly distributed within each year of age, mortality is uniform globally. We back out ω using equation (5.12), e˚x:n  n p x ( n ) + n q x ( n/2) : 10 20 q 20 + 20 20 p 20  18 20 ω − 40 + 20  18 ω − 20 ω − 20

!

10

!

200 + 20ω − 800  18ω − 360 2ω  240 ω  120 x −20 p20

Alternatively, we can back out ω using the trapezoidal rule. Complete life expectancy is the area of the trapezoid shown to the right.



e˚20:20  18  (20)(0.5) 1 + ω − 40 ω − 20 0.8ω − 16  ω − 40

ω − 40 ω − 20

1



ω − 40 ω − 20

18

0.8 

20

40

x

0.2ω  24 ω  120

Once we have ω, we compute 30|10 q 30

7.20.



10 10   0.1111 ω − 30 90

(A)

We use equation (7.5) to obtain 0.5 

qx 1 − 0.5q x

q x  0.4





Then e˚45:1  0.5 1 + (1 − 0.4)  0.8 . (E) 7.21. According to the Illustrative Life Table, l30  9,501,381, so we are looking for the age x such that l x  0.75 (9,501,381)  7,126,036. This is between 67 and 68. Using linear interpolation, since l 67  7,201,635 and l 68  7,018,432, we have x  67 + This is 37.4127 years into the future.

3 4

7,201,635 − 7,126,036  67.4127 7,201,635 − 7,018,432 of the people collect 50,000. We need 50,000

540.32 per person. (D)

Quiz Solutions 7-1. Notice that µ50.4 

q 50 1−0.4q50

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while 0.6q 50.4 

0.6q50 1−0.4q50 ,

so 0.6q 50.4  0.6 (0.01)  0.006

3 4

!

1 1.1237.4127

! 

QUIZ SOLUTIONS FOR LESSON 7

137

7-2. The algebraic method goes: those who die will survive 0.3 on the average, and those who survive will survive 0.6. 0.6 q x+0.4 0.6 p x+0.4

e˚x+0.4:0.6

0.6 (0.1) 6  1 − 0.4 (0.1) 96 90 6  1− 96 96 6 90 55.8  (0.3) + (0.6)   0.58125 96 96 96 

The geometric method goes: we need the area of a trapezoid having height 1 at x + 0.4 and height 90/96 at x + 1, where 90/96 is 0.6 p x+0.4 , as calculated above. The width of the trapezoid is 0.6. The answer is therefore 0.5 (1 + 90/96) (0.6)  0.58125 . 7-3. Batteries failing in month 1 survive an average of 0.5 month, those failing in month 2 survive an average of 1.5 months, and those failing in month 3 survive an average of 2.125 months (the average of 2 and 2.25). By linear interpolation, 2.25 q 0  0.25 (0.6) + 0.75 (0.2)  0.3. So we have e˚0:2.25  q 0 (0.5) + 1| q 0 (1.5) + 2|0.25 q 0 (2.125) + 2.25 p0 (2.25)  (0.05)(0.5) + (0.20 − 0.05)(1.5) + (0.3 − 0.2)(2.125) + 0.70 (2.25)  2.0375

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138

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7. SURVIVAL DISTRIBUTIONS: FRACTIONAL AGES

Lesson 8

Insurance: Annual—Moments Reading: Models for Quantifying Risk (4th or 5th edition) 7.1, 7.2

8.1

Review of Financial Mathematics

Since we are going to deal with present values, let’s review the notation and formulas from Financial Mathematics that we will need. Let i be the interest rate. Then • d  i/ (1 + i ) is the rate of discount. • v  1/ (1 + i )  1 − d is the discounting rate. A formula relating i, d, and v is d  iv. • δ  ln (1 + i ) is the continuously compounded interest rate corresponding to i. We have e δ  1 + i, e −δ  v, and v t  e −δt . • a n is the present value of an n-year immediate certain annuity,1 one that pays 1 per year at the end of each year for n years. The formula for it is an 

1 − vn i

• a¨ n is the present value of an n-year certain annuity-due, one that pays 1 per year at the beginning of each year for n years. The formula for it is a¨ n 

1 − vn d

• a¯ n is the present value of an n-year continuous certain annuity, one that pays at a rate of 1 per year continuously. The formula for it is 1 − vn a¯ n  δ • s n is the cumulative value of an annuity, the value at the end of n years. The symbol may be decorated with double-dot or bar to represent “due” or “continuous”, but the value is taken at the end of n years regardless. Thus it is (1 + i ) n times the corresponding “a” value; s n  (1 + i ) n a n , for example. • The superscript ( m ) indicates payments on an mthly basis. For example, a¨ n(4) is an annuity-due paying once every three months at a rate of 1 per year, so it pays 0.25 every three months, starting immediately.

1In this course, “annuity” means life annuity when not specified otherwise. Life annuities will be discussed later in the course. The annuity you learned about in your Financial Mathematics course will be called a certain annuity or an annuity-certain to distinguish it from the annuities of this course. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

139

8. INSURANCE: ANNUAL—MOMENTS

140

8.2

Moments of annual insurances

A life insurance pays a benefit to the insured policyholder upon death. The benefit may be payable whenever the insured dies, or may be payable only if the insured dies within a fixed number of years, or may be payable only if the insured dies after a fixed number of years. The benefit may vary depending on when death occurs. We are interested in calculating the expected value of the present value of the benefit. This is the amount we would put aside today to fund the benefit. For a large group of independent policyholders, if we set aside the expected present value of the benefit for each policyholder, the law of large numbers tells us that we will have approximately enough money to pay all the benefits when they become due. We would also like to calculate the variance of the present value of the benefit. Then we can use the normal approximation to determine the size of the fund that has a high probability of being adequate to pay all the benefits. The expected present value is such an important concept, there are a lot of names for it. The expected present value of the benefit is sometimes called the actuarial present value of the insurance. Another name for the expected present value is the net single premium, since this is the lump sum the company would charge for such a contract if no provision for expenses and profit was made. Real life insurance pays the benefit soon after the death of the policyholder. However, in this lesson, we will discuss insurances that pay at the end of the year of death, rather than at the moment of death. This makes the random variable for the present value of the benefit discrete, and allows us to use life tables to calculate the expected value. We will sometimes call this insurance “discrete”. In the exercises, some questions refer to “fully discrete” insurances. For the meantime, ignore the word “fully”; we will define it when we discuss premiums. Later on, we will discuss how to adjust the expected value of a discrete insurance to take into account payment at the moment of death. The expected value of a discrete random variable is the sum of the probabilities of each possible outcome times the value of the random variable in each outcome. Here’s a simple example of the calculation of the actuarial present value of an insurance: An insurance contract on a person age 40 will pay $1000 at the end of a year to that person’s estate if that person dies during the year. Let the random variable Z be the present value of the payment made to the estate. It is equal to 1000v if death occurs within a year, 0 otherwise. The actuarial present value of the insurance is 1000v times the probability of death within a year; in other words, E[Z]  1000vq40 . This logic can be generalized. For an insurance on ( x ) , let b be the benefit paid at the end of year of death if death occurs at least m year from now but not later than n years from now. Let Z be the present value random variable for the insurance. Then E[Z]  b

n−1 X

v k+1 k| q x  b

km

E[Z 2 ]  b 2

n−1 X

n−1 X

v k+1 k p x q x+k

(8.1)

km

v 2 ( k+1) k| q x  b 2

km

n−1 X

v 2 ( k+1) k p x q x+k

(8.2)

km

Example 8A An insurance on (70) pays 1000 at the end of the year of death if (70) dies after 1 year but not after 3 years. You are given: • Mortality is based on the following table: LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

8.3. STANDARD INSURANCES AND NOTATION

141

x

qx

70 71 72 73

0.05 0.07 0.10 0.12

• i  0.04 • Z is the present value random variable for the insurance. Calculate E[Z]. Answer: The answer, by formula (8.1), is E[Z]  1000

 q 1| 70 1.042

+

2| q 70 1.043



We have 1| q 70

 p70 q 71  (0.95)(0.07)  0.0665

2| q 70

 p70 p71 q 72  (0.95)(0.93)(0.10)  0.08835

So the answer is E[Z]  1000



0.0665 0.08835 +  1000 (0.061483 + 0.078543)  140.03 . 1.042 1.043





To calculate the variance of the present value of an insurance, we use the formula Var ( Z )  E[Z 2 ] − E[Z]2 Example 8B Assume the same mortality and interest as the previous example. An insurance on [70] pays a benefit of 1000 at the end of the year if death occurs in the first year and 2000 at the end of the year if death occurs in the second year. Let Z be the payment random variable for this insurance. Calculate Var ( Z ) . Answer: The first moment is

!

!

0.05 0.0665 E[Z]  1000 + 2000  48.07692 + 122.96598  171.04290 1.04 1.042 When calculating the second moment, the benefit as well as the interest factor must be squared.

!

!

0.05 0.0665 E[Z ]  1000 + 20002  46,227.81 + 227,377.91  273,605.72 2 1.04 1.044 2

2

Var ( Z )  273,605.72 − 171.042902  244,350.05

8.3



Standard insurances and notation

We will discuss several standard types of insurance. For each one, if the benefit is 1, there is a standard symbol in International Actuarial Notation (IAN) for the actuarial present value. Here is a list of standard types of insurance coverage having standard notation to describe their actuarial present value: LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

8. INSURANCE: ANNUAL—MOMENTS

142

Whole life insurance A whole life insurance of 1 on ( x ) pays 1 whenever death occurs. The standard symbol for the actuarial present value of whole life insurance payable at the end of the year of death is A x . The random variable Z for a whole life insurance is v K x +1 . Term life insurance A term life insurance of 1 on ( x ) for n years pays 1 if death occurs within n years, 0 otherwise. The standard symbol for the actuarial present value of term life insurance payable at the end of the year of death is A x1:n . The 1 over the x indicates that the payment is made only if ( x ) expires before the n-year certain period. The random variable Z for a term life insurance is v K x +1 if K x < n, 0 otherwise. Deferred whole life insurance An n-year deferred life insurance on ( x ) pays 0 if death occurs within n years, 1 otherwise. The standard symbol for the actuarial present value of deferred insurance payable at the end of the year of death is n| A x . The random variable Z for a deferred life insurance is v K x +1 if K x ≥ n, 0 otherwise. The sum of a term life insurance for n years and an n-year deferred life insurance is a whole life insurance. Deferred term insurance It is possible to combine deferral and term. The actuarial present value of an insurance payable at the end of the year of death only if death occurs between times n and n + m would be denoted by n| A x1:m . The random variable Z is v K x +1 if n ≤ K x < n + m, 0 otherwise. Pure endowment An n-year pure endowment of 1 on ( x ) pays 1 at the end of n years if ( x ) survives to that point, 0 otherwise. There are two standard symbols for its actuarial present value. The insurance style symbol is A x:n1 . The 1 over the n indicates that the payment is made only if the n-year certain period expires before ( x ) . The other, annuity style, symbol is n E x . Get used to both symbols! The random variable Z for the actuarial present value of a pure endowment is v n if Tx ≥ n, 0 otherwise. Endowment insurance An n-year endowment insurance of 1 on ( x ) pays 1 at the earlier of the death of ( x ) or n years. It is the sum of a term insurance for n years and an n-year pure endowment. The standard symbol for the actuarial present value of it is A x:n . The random variable Z for endowment insurance is v K x +1 if K x < n, v n otherwise. The above descriptions are summarized in Table 8.1.

?

Quiz 8-1 Which symbol would be used for the item we calculated in Example 8A? Professor Geoffrey Crofts introduced the following notation: the present value random variable for an insurance by Z has the same decorations as the corresponding expected value. For example, Z30:10 is the present value random variable for a 10-year endowment insurance on (30). A useful method for calculating higher moments of standard insurances is that the k th moment of a standard insurance equals the first moment calculated at k times the force of interest. This works whenever the death benefit in all years is either 0 or 1. This method is called the rule of moments. The concept of the first moment calculated at k times the force of interest is so useful, we will define a symbol for it. Whenever the presubscript of an actuarial symbol is k, that means that the item is calculated at k times the force of interest. For example, 2A x is the actuarial present value of a whole life insurance of 1 on ( x ) calculated at twice the force of interest. Then if Z is the present value random variable for an n-year endowment insurance of 1, (8.3) Var ( Z )  2A x:n − ( A x:n ) 2 A presuperscript of 2 does not represent the second moment; it merely represents doubling δ. 2A happens to be the second moment of the random variable for which A is the first moment, x x but we will see later that a corresponding statement is not true for annuity symbols. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

8.4. ILLUSTRATIVE LIFE TABLE

143

Table 8.1: Actuarial notation for standard types of insurance

Name

a Another

Present value random variable

Symbol for actuarial present value

Whole life insurance

v K x +1

Term life insurance

v K x +1 0

Kx < n Kx ≥ n

A x1:n

Deferred life insurance

0 v K x +1

Kx < n Kx ≥ n

n| A x

Deferred term insurancea

0 v K x +1 0

Kx ≤ n n ≤ Kx < n + m Kx ≥ n + m

Ax

1 n| A x :m

Pure endowment

0 vn

Kx < n Kx ≥ n

A x:n1 or n E x

Endowment insurance

v K x +1 vn

Kx < n Kx ≥ n

A x:n

symbol that can be used is n|m A x .

Doubling the force of interest is not the same as doubling the rate of interest. It actually “squares” the rate of interest. For example, if δ is ln 1.06 (corresponding to i  0.06), then 2δ is 2 ln 1.06  ln 1.062  ln 1.1236, which corresponds to i  0.1236. In general, if δ, i, d, and v correspond to each other, and we use a presuperscript of 2 on i, d, and v to indicate doubling the force of interest, then 2

8.4

i  2i + i 2

2

d  2d − d 2

2

vv

(8.4)

2

Illustrative Life Table

On the exam, you may be asked to calculate insurance moments using the Illustrative Life Table and i  0.06. The Illustrative Life Table has the basic functions l x and 1000q x , which can be used for a short term example like the previous one regardless of the interest rate assumption. In addition, it has single life actuarial functions 1000A x , 1000 (2A x ) , and 1000 k E x for k  5, 10, 20, all at 6% interest, which lets you calculate long-term insurances without having to add up a lot of numbers. The A x column is used for first moments, and the 2A x column is used for second moments. The 2A x column can also be used for first moments at 1.062 − 1  0.1236 interest, but it’s rare you’d use it for that purpose. To calculate actuarial present values for term life insurances, deferred life insurances, and endowment insurances, use the pure endowment columns in conjunction with the insurance columns. You must relate these types of insurance to whole life insurance. An n-year deferred life insurance on ( x ) is an n-year pure endowment providing a whole life insurance on ( x + n ) . An n-year term life insurance on ( x ) is a whole life insurance on ( x ) minus an n-year deferred life insurance on ( x ) . An n-year endowment insurance on LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

8. INSURANCE: ANNUAL—MOMENTS

144

( x ) is an n-year term insurance on ( x ) plus an n-year pure endowment on ( x ) . In symbols: A x1:n  A x − n E x A x+n n| A x

 n E x A x+n

A x:n  A x − n E x A x+n + n E x

(8.5) (8.6) (8.7)

Exam questions will almost always set n equal to a multiple of 5. For n  5, 10, 20, you are given n E x . For n  15, use 5 E x 10 E x+5 or 10 E x 5 E x+10 . You can similarly obtain 25 E x , 30 E x , and 40 E x by multiplying two of the pure endowments in the table together. In the rare case of an exam question requiring an n not a multiple of 5, you will have to calculate n p x as a quotient l x+n /l x and then multiply by v n . These three formulas apply equally well to second moments as long as you square the benefits and double the force of interest. The variance can then be calculated from the first and second moments. To calculate the actuarial present value of a pure endowment at twice the force of interest, multiply the first moment by v n , where n is the period of the pure endowment. For example, if we are using the Illustrative Life Table with i  0.06, a 10-year pure endowment on (40) at twice the force of interest at i  0.06 is 10 10 E 40 /1.06 , where 10 E 40 from the Illustrative Life Table is 0.53667. The result is 0.29967. Example 8C For a 20-year endowment insurance on a life currently age 60 with face amount of 1000 and benefit payable at the end of the year of death: • Mortality follows the Illustrative Life Table. • i  0.06 Calculate the actuarial present value of the insurance. Answer: 1000A60:20  1000A60 − 100020 E60 A80 + 100020 E60  369.13 − 0.14906 (665.75) + 149.06  418.95



Example 8D A special insurance on (65) pays 1000 at the end of the year of death if death occurs between ages 70 and 80, and 600 at the end of the year of death if death occurs after age 80. Mortality follows the Illustrative Life Table and i  0.06. Calculate the net single premium for this insurance. Answer: There are several ways to view this insurance. One way is to view it as a 5-year deferred whole life insurance for 1000 minus a 15-year deferred whole life insurance for 400, and we will do it this way. Alternatively, it is a sum of a 5-year deferred 10-year term insurance of 1000 plus a 15-year deferred whole life insurance of 600. That alternative expresses it as a sum of two mutually exclusive insurances and is better for calculating variance (which is not required for this example). The actuarial present value of a 5-year deferred whole life insurance of 1000 is a whole life insurance of 1000 on 70 discounted 5 years with mortality and interest, or 1000 5 E65 A70 . The Illustrative Life Table provides k E x for k  5, 10, and 20. The 5-year deferred insurance’s actuarial present value is 1000 5| A65  1000 5 E65 A70  (656.23)(0.51495)  337.93 The actuarial present value of a 15-year deferred whole life insurance of 400 is 400 15 E65 A80 . To use the Illustrative Life Table’s values of k E x , we express 15 E65  5 E65 10 E70 . (You can also express it as 10 E65 5 E75 .) Whenever k is a multiple of 5 (as it will probably be on the exam), decompose k E x into components found in the Illustrative Life Table! It is faster to use the values for k E x in the ILT than to compute k E x using l x ’s and discounting at 6%. The 15-year deferred insurance’s actuarial present value is 400 15| A65  4005 E65 10 E70 A80  0.4 (656.23)(0.33037)(0.66575)  57.73 The net single premium for the insurance is 337.93 − 57.73  280.20 .

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8.5. CONSTANT FORCE AND UNIFORM MORTALITY

?

145

Quiz 8-2 For a 5-year term life insurance on (45), you are given: • A benefit of 1000 is paid at the end of the year of death. • Mortality follows the Illustrative Life Table. • i  0.06. Calculate the second moment of the present value of the benefit payment.

8.5

Moments for insurance under constant force of mortality and uniform mortality

We will derive formulas for moments of insurance under constant force of mortality and uniform mortality. However, it is rare that these situations will come up on exams, so you may skip this section if you wish. If mortality has constant force, or more generally if q x+k is constant for all integral k ≥ 0, then the sum to evaluate the moment is a geometric series. Example 8E For a whole life insurance on ( x ) , you are given • The valuation interest rate is i. • q x+k  q is constant for integral k ≥ 0. Calculate the expected value and variance of the present value of the insurance. Answer: The survival probabilities are k p x  (1 − q ) k . Let Z be the present value random variable. Using equation (8.1), Ax 

∞ X

(1 − q ) k qv k+1

k0

qv 1 − (1 − q ) v qv  d + qv q  q+i 

(8.8)

For the second moment, double the force of interest. E[Z 2 ] 

q q + 2i + i 2

Then the variance is 2A x − A2x . Var ( Z ) 

q q + 2i + i 2

−

q q+i

!2 

Since in this case A x  A x+n for any n (no memory), it follows that n| A x  n E x A x+n  n E x A x , so we can develop formulas for term and deferred insurances similar to the ones we developed for insurances payable at the moment of death. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

8. INSURANCE: ANNUAL—MOMENTS

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Suppose mortality is uniformly distributed, or more generally if k| q x  k p x q x+k is a constant c for all integral k ≥ 0. For an insurance with benefit 1 for death between times m and n, equation (8.1) becomes E[Z] 

n−1 X

v k+1 c  cv m a n−m

km

Note that c  1/ ( ω − x ) . For a whole life insurance, n  ω − x and m  0. Example 8F Age at death for (35) is uniformly distributed, with ω  100. You are given: • i  0.05 • Z is the present value random variable for a 20-year endowment insurance on this life. Calculate Var ( Z ) . Answer: ω − x  65. Z is the sum of a 20-year term insurance and a 20-year pure endowment. Let Z1 be the present value random variable for the term insurance and Z2 the present value random variable for the pure endowment. The expected value of Z1 is E[Z1 ] 

a 20 1 1 − (1/1.05) 20   0.19173 65 65 0.05

and the second moment is calculated using twice the force of interest, or squaring v. E[Z12 ] 

1 1 − (1/1.05) 40  0.12877 65 1.052 − 1

The expected value of Z2 is 20 p35 v 20 , and 20 p35  (100 − 55) / (100 − 35)  45/65. E[Z2 ] 

45 1  0.26092 65 1.0520

The second moment is calculated by squaring v. E[Z22 ] 

45 1  0.09834 65 1.0540

Therefore, the variance of Z is E[Z]  0.19173 + 0.26092  0.45265 E[Z 2 ]  0.12877 + 0.09834  0.22711 Var ( Z )  0.22711 − 0.452652  0.02222



Table 8.2 summarizes the results of this section.

?

Quiz 8-3 A whole life insurance on (60) pays a benefit of 1000 at the end of the year of death. You are given:

  0.02 • µ60+t    1/ (60 − t )  • i  0.06

t ≤ 10 10 < t < 60

Calculate the net single premium for this insurance.

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8.6. NORMAL APPROXIMATION

147

Table 8.2: Actuarial present value under constant force and uniform mortality for insurances payable at the end of the year of death

Type of insurance

Actuarial present value under constant force

Actuarial present value under uniform distribution

q q+i

a ω−x ω−x

 q  1 − ( vp ) n q+i

an ω−x

q ( vp ) n q+i

v n a ω− ( x+n )

Whole life n-year term n-year deferred life

( vp ) n

n-year pure endowment

8.6

ω−x vn



ω − (x + n )



ω−x

Normal approximation

For a large group of insureds, the distribution of the present value of benefit payments can be approximated as a normal distribution with mean and variance as calculated by the methods of this lesson. When working out such problems, remember that the variance of a coverage of x on one insured is x 2 times the variance of a coverage of 1, but the variance of a coverage of 1 on n insureds is only n times the variance of a coverage of 1 on one insured. Example 8G A group of 400 independent individuals age (40) each purchases a whole life insurance of 1000. You are given: • Mortality follows the Illustrative Life Table. • i  0.06. Using the normal approximation, calculate the size of the fund needed so that there is a 95% probability that the fund will be adequate to pay all benefits. Answer: Let Z be the present value random variable for one insurance of 1. E[Z]  A40  0.16132 E[Z 2 ]  2A40  0.04863 Var ( Z )  0.04863 − 0.161322  0.02261 Let Z p be the present value random variable for the entire portfolio of 400 insurances of 1000. For 400 individuals and 1000 of insurance, the expected value is multiplied by (400)(1000) and the variance is multiplied by (400)(106 ) . Then E[Z p ]  0.16132 (400,000)  64,528

p

Var ( Z p ) 

p

0.02261 (400)(106 )  3007

The 95th percentile of the standard normal distribution is 1.645. By the normal approximation, the fund needed is 64,528 + 1.645 (3007)  69,475 . 

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148

Note The concepts discussed in these lessons can be applied to non-life insurance situations. They can be applied to any situation in which a payment is made contingent on some event, with the present value of the payment depending on when the event occurs. A warranty that replaces a product if it fails within 5 years would be an example.

Exercises Expected value 8.1. You are given that i  0. Which of the following expressions equals A x:30 ? A.

30 p x

B.

30 q x

C.

30| q x

D. p x+30

E. 1

D. p x+30

E. 1

D.

E. 1

8.2. You are given that i  0. Which of the following expressions equals A x1:30 ? A.

30 p x

B.

30 q x

C.

30| q x

8.3. You are given that i  0. Which of the following expressions equals 10| A x ? A. 0

B.

10 p x

C.

10 q x

10| q x

8.4. [C3 Sample:8] You are given: •

Jim buys a new car for 20,000. He intends to keep the car for 3 years.

•

A device called Car-Tracker can be attached to Jim’s car. Car-Tracker will help police locate the car if stolen.

•

The probability that the car will be stolen without Car-Tracker is 0.2 in each year. (q k  0.2, k  0, 1, 2)

•

The probability that the car will be stolen with Car-Tracker is 0.1 in each year. (q CT  0.1, k  0, 1, 2) k

•

Theft is the only decrement.

•

Insurance against theft is purchased with a net single premium.

•

If the car is stolen in year j, the insurance company will pay 25,000 − 5000j at the end of the year, for j  1, 2, 3. After the third year, no benefit is paid.

•

i  0.06

Calculate the actuarial present value of the insurance benefit against theft assuming Jim does not purchase Car-Tracker. A. 3800

B. 4800

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C. 7000

D. 8000

E. 9000

Exercises continue on the next page . . .

EXERCISES FOR LESSON 8

149

8.5. A special term insurance policy pays 1000 at the end of the year of death for the first 10 years and 500 at the end of the year of death for the next 10 years. Mortality follows the Illustrative Life Table. i  0.06. Calculate the actuarial present value of a policy on (30). 8.6. [3-S00:8] (This exercise refers to the Weibull distribution. However, I have edited the question to provide all the information you need to solve it.) For a two-year term insurance on a randomly chosen member of a population: • 1/3 of the population are smokers and 2/3 are nonsmokers. • The future lifetimes follow a Weibull distribution with: 2 • Fx ( t )  1 − e − ( t/1.5) for smokers 2 • Fx ( t )  1 − e − ( t/2) for nonsmokers • The death benefit is 100,000 payable at the end of the year of death. • i  0.05 Calculate the actuarial present value of this insurance. A. 64,100

B. 64,300

C. 64,600

D. 64,900

E. 65,100

8.7. [CAS4A-S96:9] (2 points) Henry, age 30, is subject to a constant force of mortality, µ x  0.12. Henry wants to buy a 3-year endowment insurance, with a $1,000 benefit payable at the end of the year of death. δ  0.09. Determine the net single premium for this insurance. A. B. C. D. E.

Less than $788 At least $788, but less than $790 At least $790, but less than $792 At least $792, but less than $794 At least $794

8.8. [CAS3-S04:1] A car leasing company leases cars to customers for a three-year period. •

Each year 15% of the vehicles get into accidents.

•

Different accident years are independent.

•

At the end of the lease, 25% of the customers decide to keep their cars. This decision is made independently of their accident history.

•

The company pays a $1,000 bonus for each car returned that has not been in an accident.

•

i  10% What is the actuarial present value of the bonus payment at the time that the car is leased?

A. B. C. D. E.

Less than $250 At least $250, but less than $275 At least $275, but less than $300 At least $300, but less than $325 At least $325

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8.9. [CAS4-F92:18] (3 points) A group of 100 men age 40 set up a fund to pay 1000 at the end of the year of death of each member to his designated survivor. Their mutual agreement is to pay into the fund an amount equal to the whole life insurance net single premium calculated on the basis of a life table at 6% interest. From the life table that they used you are given the following: 1000A40  161.3242

1000q40  2.7812

1000A42  176.3572

1000q41  2.9818

l 40  93,131.64

1000q42  3.2017

l 42  92,595.69 The actual experience of the fund during the first two years is as follows: Year

Number of Deaths

Interest Rate

1 2

1 0

6% 7%

Determine the difference, at the end of the first two years, between the expected size of the fund as determined at the inception of the plan and the actual fund. A. B. C. D. E.

Less than 175 At least 175, but less than 350 At least 350, but less than 525 At least 525, but less than 700 At least 700

8.10. [3-F01:12] A fund is established by collecting an amount P from each of 100 independent lives age 70. The fund will pay the following benefits: •

10, payable at the end of the year of death, for those who die before age 72, or

•

P, payable at age 72, to those who survive. You are given: • Mortality follows the Illustrative Life Table. • i  0.08

Calculate P, using the equivalence principle. (Note: we will learn about the equivalence principle in Lesson 18. For now, all you need to know is that P is the number such that the fund equals the expected present value of the benefits.) A. 2.33

B. 2.38

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C. 3.02

D. 3.07

E. 3.55

Exercises continue on the next page . . .

EXERCISES FOR LESSON 8

8.11.

151

[Based on CAS4A-S98:15] (2 points) You are given the following excerpt from a mortality table: x

lx

30 31 32 33 34 35 36

10,000 9,000 8,000 7,500 7,000 6,300 5,200

You are also given that a 4-year pure endowment for $10,000 issued at age 30 has a net single premium of $5,051. Assume a constant rate of interest. Determine the net single premium for a 3-year pure endowment for $10,000 issued at age 31 to a life age 31. A. B. C. D. E. 8.12.

Less than $5,500 At least $5,500, but less than $5,700 At least $5,700, but less than $5,900 At least $5,900, but less than $6,100 At least $6,100 [CAS4A-S98:16] (2 points) You are given the following mortality information: x

lx

40 41 42 43 44

11,000 10,000 8,500 6,500 4,000

Assume the effective annual rate of interest is set as follows:

Year 1 2 3 4+

Interest Rate 6% 7% 8% 9%

Calculate the net single premium for a 3-year term life insurance with a benefit of 1 issued to a person age 41, with benefits payable at the end of the year of death. A. B. C. D. E.

Less than 0.515 At least 0.515, but less than 0.520 At least 0.520, but less than 0.525 At least 0.525, but less than 0.530 At least 0.530

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152

8.13. A service contract repairs machines that break down during the next 3 months, regardless of how many times they break down, at the end of the month in which they break down. A company has 10 newly installed machines. The probability that a machine installed or fixed at the beginning of month m breaks down during month m + t − 1 is Months (t)

Probability of breakdown in month t

1 2 3 Greater than 3

0.1 0.2 0.3 0.4

Each repair is worth 100. Calculate the actuarial present value of the contract at i  6%. 8.14. [3-F02:40] A maintenance contract on a hotel promises to replace burned out light bulbs at the end of each year for three years. The hotel has 10,000 light bulbs. The light bulbs are all new. If a replacement bulb burns out, it too will be replaced with a new bulb. You are given: • For new light bulbs, q 0  0.10, q 1  0.30, q2  0.50. • Each light bulb costs 1. • i  0.05 Calculate the actuarial present value of this contract. A. 6700

B. 7000

C. 7300

D. 7600

E. 8000

8.15. A 10-year term life insurance contract on (20) pays 1000 at the end of the year of death. You are given: • Mortality follows the Illustrative Life Table. •

  0.1236 i  0.06

05

 Calculate the actuarial present value of this insurance contract.

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EXERCISES FOR LESSON 8

153

[Based on 3-F02:3] You are given:

8.16.

• The following mortality table:

•

x

qx

60 61 62 63 64 65

0.09 0.11 0.13 0.15 0.16 0.17

i  0.03

Calculate 2|2 A60 , the actuarial present value of a 2-year deferred 2-year term insurance on (60) . A. 0.156

B. 0.160

C. 0.186

D. 0.190

E. 0.195

[CAS3-F03:2] For a special fully discrete life insurance on (45), you are given:

8.17.

• i  6% • Mortality follows the Illustrative Life Table. • The death benefit is 1,000 until age 65, and 500 thereafter. Calculate the actuarial present value of the benefit payment. A. B. C. D. E.

Less than 100 At least 100, but less than 150 At least 150, but less than 200 At least 200, but less than 250 At least 250

8.18. [CAS3-S04:4] For a single premium discrete 15-year term insurance of $100,000 on a person age (45), you are given: •

i  5.0%

•

Mortality is uniformly distributed with ω  100.

After 5 years, it is discovered that the insurance should have been calculated using ω  120. At this time, a one-time premium adjustment is made only for the remaining 10 years of this insurance. Calculate the one-time adjustment. A. B. C. D. E.

Refund of $7,091 Refund of $4,412 Refund of $2,679 Additional premium of $4,412 Additional premium of $7,091

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154

8.19. [150-S91:24] A special n-year endowment insurance on ( x ) pays a pure endowment of 1000 at the end of n years and pays only the net single premium at the end of the year of death if death occurs during the n-year period. The net single premium for this insurance is 600. The net single premium for an n-year endowment insurance of 1000 with death benefits payable at the end of the year of death of ( x ) is 800. Calculate the net single premium for an n-year pure endowment of 1000 on ( x ) . A. 100

B. 200

C. 300

D. 400

E. 500

[150-83-96:6] For a modified ten-year pure endowment of 1000 on (35), you are given:

8.20.

• A35:10  0.57 • This modified pure endowment includes a death benefit payable at the end of the year of death. • Z1 is the present value random variable if the death benefit is 1000 10 E35 . • Z2 is the present value random variable if the death benefit is 750 10 E35 . • Z3 is the present value random variable if the death benefit is 500 10 E35 . • E[Z1 ]/ E[Z2 ]  1.005 Calculate E[Z3 ]. A. 550

B. 555

C. 560

D. 565

E. 570

[SOA3-F04:2] For a group of individuals all age x, you are given:

8.21.

• 25% are smokers (s); 75% are non-smokers (ns). s • k q x+k 0 1 2 •

0.10 0.20 0.30

ns q x+k

0.05 0.10 0.15

i  0.02

Calculate 10,000A x1:2 for an individual chosen at random from this group. A. 1690 8.22.

B. 1710

C. 1730

D. 1750

E. 1770

[CAS3-S04:9] A company offers two special discrete life policies to a life age (30):

The first policy has a death benefit that starts at 5 and increases 5 per year until age (50). It then decreases 5 per year until age (70). From age (70) onward it is level at 10. The second policy has a death benefit that starts at 10 and increases 5 per year until age (50). It then decreases 5 per year until age (70). From age (70) onward it is level at 20. •

Mortality follows the Illustrative Life Table.

•

i  6% Compute the difference in the actuarial present value of these two death benefits at issue.

A. B. C. D. E.

Less than 0.500 At least 0.500, but less than 0.625 At least 0.625, but less than 0.750 At least 0.750, but less than 0.825 At least 0.825

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EXERCISES FOR LESSON 8

155

[150-F97:3] For a 20-year pure endowment of 1 on ( x ) , you are given:

8.23.

• Z is the present value random variable at issue of the benefit payment. • Var ( Z )  0.05 E[Z] • 20 p x  0.65 Calculate i. A. 0.102

B. 0.106

C. 0.110

D. 0.114

E. 0.118

[SOA3-F04:37] Z is the present value random variable for a 15-year pure endowment of 1 on ( x ) .

8.24.

• The force of mortality is constant over the 15-year period. • v  0.9 • Var ( Z )  0.065 E[Z] Calculate q x . A. 0.020

B. 0.025

C. 0.030

D. 0.035

E. 0.040

8.25. For a 20-year term insurance of 1000 on a life currently age 40 with benefit payable at the end of the year of death, you are given: • µ x  0.001 (1.05x ) • i  0.05 • The actuarial present value of the insurance is 124.60. Calculate the actuarial present value of a 21-year term insurance on the same life with benefit payable at the end of the year of death. You are given:

8.26.

• q65  0.02 • Deaths are uniformly distributed between integral ages. • v  0.95 (4)

Calculate A 1

65:1

.

You are given that

8.27. • •

i  0.06. µ x+t  ln 1.02 for t > 0

Calculate A x(4) . Variance Given:

8.28.

Z is the present value random variable for a whole life insurance of 1 payable at the end of the year of death. • t p x  0.94t for all t • i  0.07 •

Calculate Var ( Z ) . LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

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156

8.29. Z is the present value random variable for a 2-year endowment insurance of 1000 on (50) payable at the end of the year of death. You are given: • • •

p50  0.98 p51  0.96 i  0.06

Calculate Var ( Z ) . 8.30. Z is the present value random variable for a 2-year term insurance of 1000 on (50) payable at the end of the year of death. You are given: • • •

p50  0.98 p51  0.96 i  0.06

Calculate Var ( Z ) . 8.31. [CAS4A-S95:18] (2 points) You are given that the probability density function of future lifetime T for ( x ) is:   1/75 for 0 < t < 75 g (t )   0 otherwise

 The force of interest δ  0.05. Z is the present-value random variable for a whole life insurance of unit amount issued to ( x ) payable at the end of the year of death. Determine Var ( Z ) . A. B. C. D. E.

Less than 0.0640 At least 0.0640, but less than 0.0650 At least 0.0650, but less than 0.0660 At least 0.0660, but less than 0.0670 At least 0.0670 [150-S90:26] You are given:

8.32. • •

Mortality follows the Illustrative Life Table. i  0.06.

Calculate 2A55:10 . 8.33. [150-S88:6] Z is the present value random variable for a discrete one-year term insurance of 1000 issued to ( x ) . You are given: • E[Z]  19 • Var ( Z )  17,689 Calculate i. A. 0.0500

B. 0.0526

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C. 0.0550

D. 0.0582

E. 0.0600

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EXERCISES FOR LESSON 8

157

8.34. [CAS4-S90:17] (2 points) Z is the present-value random variable for a 2 year term insurance on ( x ) which pays a benefit of 1 at the end of the year of death. You are given: • q x  0.50 • i0 • Var ( Z )  0.16 Calculate q x+1 . A. 0.55

B. 0.60

C. 0.65

D. 0.70

E. 0.75

8.35. [CAS4A-F93:10] (2 points) You are given the following information about a 2-year term insurance policy issued to ( x ) which pays a benefit of 1 at the end of the year of death: q x  0.50 i  0% Var ( Z )  0.09

where Z is the random variable representing the present value of future benefits

Determine q x+1 A. B. C. D. E. 8.36.

Less than 0.55 At least 0.55, but less than 0.65 At least 0.65, but less than 0.75 At least 0.75, but less than 0.85 At least 0.85 [Based on 150-S98:30] For a 2-year term insurance of 10 on (60), you are given:

• Benefits are payable at the end of the year of death. • Z is the present-value random variable for this insurance. • i  0.05 • The following extract from a select-and-ultimate mortality table: x

lx

60 61 62 63 64

78,900 77,200 75,100 72,500 69.800

Calculate Var ( Z ) . A. 0.39

B. 0.52

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C. 3.58

D. 3.94

E. 5.22

Exercises continue on the next page . . .

8. INSURANCE: ANNUAL—MOMENTS

158

[3-S00:36] A new insurance salesperson has 10 friends, each of whom is considering buying a pol-

8.37. icy.

Each policy is a whole life insurance of 1000, payable at the end of the year of death. The friends are all age 22 and make their purchase decisions independently. Each friend has a probability of 0.10 of buying a policy. The 10 future lifetimes are independent. S is the random variable for the present value at issue of the total payments to those who purchase the insurance. • Mortality follows the Illustrative Life Table. • i  0.06 • • • • •

Calculate the variance of S. A. 9,200 8.38.

B. 10,800

C. 12,300

D. 13,800

E. 15,400

[3-F01:5] For a fully discrete 2-year term insurance of 1 on ( x ) : • 0.95 is the lowest premium such that there is a 0% chance of loss in year 1. • p x  0.75 • p x+1  0.80 • Z is the random variable for the present value at issue of future benefits.

Calculate Var ( Z ) . A. 0.15

B. 0.17

C. 0.19

D. 0.21

E. 0.23

Normal approximation 8.39.

You are given the following mortality rates:

q 49 q 50

Non-smoker

Smoker

0.01 0.02

0.02 0.03

i  0.06 Two-year term insurance paying 1000 at the end of the year of death is sold to a group of 100 lives age 49. The group is drawn from a population in which 80% are non-smokers and 20% are smokers. Using the normal approximation, calculate the size of the initial fund needed to have enough money to pay out all claims with 95% probability.

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Exercises continue on the next page . . .

EXERCISES FOR LESSON 8

159

8.40. [CAS3-S04:7] A fund is set up to pay a death benefit of 10 to each of 1,000 lives age (65). The fund is required to have an 80% chance that it will be able to pay all of its claims. •

Mortality is from the Illustrative Life Table

•

i  6%

•

Death benefits are payable at the end of the year of death. Using the normal approximation, calculate the initial amount of the fund.

A. B. C. D. E.

Less than 4,420 At least 4,420, but less than 4,430 At least 4,430, but less than 4,440 At least 4,440, but less than 4,450 At least 4,450

Miscellaneous 8.41. [C3 Sample:43] A worker is injured on 1/1/99, resulting in a workers compensation claim. The insurer establishes a reserve, on that date, of 274,000. You are given: •

Medical and indemnity payments will be made for 15 years.

•

The medical payments will be 10,000 in 1999, increasing by 10% per year thereafter.

•

The annual indemnity payment is P.

•

The medical and indemnity payments are discounted at an interest rate of 5% per year to determine the reserve.

•

Payments will be made at the end of each year. Calculate P.

A. 5300

B. 5800

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C. 6300

D. 6900

E. 7200

Exercises continue on the next page . . .

8. INSURANCE: ANNUAL—MOMENTS

160

8.42. [M-S05:38] A group of 1000 lives each age 30 sets up a fund to pay 1000 at the end of the first year for each member who dies in the first year, and 500 at the end of the second year for each member who dies in the second year. Each member pays into the fund an amount equal to the net single premium for a special 2-year term insurance, with: • Benefits: k

b k+1

0 1

1000 500

• Mortality follows the Illustrative Life Table. • i  0.06 The actual experience of the fund is as follows: k

Interest Rate Earned

Number of Deaths

0 1

0.070 0.069

1 1

Calculate the difference, at the end of the second year, between the expected size of the fund as projected at time 0 and the actual fund. A. 840 8.43.

B. 870

C. 900

D. 930

E. 960

[M-F06:26] Oil wells produce until they run dry. The survival function for a well is given by: t (years)

S0 ( t )

0 1 2 3 4 5 6 7

1.00 0.90 0.80 0.60 0.30 0.10 0.05 0.00

An oil company owns 10 wells age 3. It insures them for 1 million each against failure for two years where the loss is payable at the end of the year of failure. You are given: • R is the present-value random variable for the insurer’s aggregate losses on the 10 wells. • The insurer actually experiences 3 failures in the first year and 5 in the second year. • i  0.10 Calculate the ratio of the actual value of R to the expected value of R. A. 0.94

B. 0.96

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C. 0.98

D. 1.00

E. 1.02

Exercises continue on the next page . . .

EXERCISES FOR LESSON 8

8.44.

161

[Sample Question 286] You are given: • The force of mortality follows Gompertz’s law with B  0.000005 and c  1.2. • The annual effective rate of interest is 3%.

1 . Calculate A50 :2

A. 0.1024

B. 0.1018

C. 0.1009

D. 0.1000

E. 0.0994

Additional old CAS Exam 3/3L questions: S05:1,26, S06:20, S07:9, F09:15, S10:18, F10:13, S12:13 Additional old CAS Exam LC questions: S14:12 Additional old SOA Exam MLC questions: F12:14, S13:7, F13:13

Solutions 8.1. Payment of 1 is definite. (E) 8.2. Payment of 1 occurs if and only if ( x ) dies within 30 years. (B) 8.3. Payment of 1 occurs if and only if ( x ) lives 10 years. (B) 8.4. This was the first of a pair of questions, hence the irrelevant information about Car-Tracker. The actuarial present value is, by formula (8.1),

(0.2)

15,000 10,000 20,000 + (0.8)(0.2) + (0.82 )(0.2)  6984.29 1.06 1.062 1.063

(C)

8.5. The actuarial present value of the insurance is 1 1 500A30 :20 + 500A30:10

 500 ( A30 − 20 E30 A50 ) + 500 ( A30 − 10 E30 A40 )  1000A30 − 50020 E30 A50 − 50010 E30 A40  102.48 − 0.5 (0.29374)(249.05) − 0.5 (0.54733)(161.32)  102.48 − 36.58 − 44.15  21.75 2

8.6. For smokers, the probability of death in the first year is F (1)  1 − e − (1/1.5)  0.358820 and the 2 2 probability of death in the second year is Fx (2) − Fx (1)  e − (1/1.5) − e − (2/1.5)  0.472167. For non-smokers, 2 the probability of death in the first year is F (1)  1 − e − (1/2)  0.221199 and the probability of death in 2 2 the second year is F (2) − F (1)  e − (1/2) − e − (2/2)  0.410921. The actuarial present value is the weighted average of the value for smokers and nonsmokers, or

    * 1 0.358820 + 0.472167 + 2 0.221199 + 0.410921 +/ 3 1.05 3 1.05 1.052 1.052 , 

100,000 .

 100,000

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0.77000 2 (0.58338) +  64,559 3 3



(C)

8. INSURANCE: ANNUAL—MOMENTS

162

8.7. The net single premium is A30:3  vq 30 + v 2 p30 q 31 + v 3 2 p30  e −0.09 (1 − e −0.12 ) + e −0.18 ( e −0.12 )(1 − e −0.12 ) + e −0.27 e −0.24  0.103347 + 0.083771 + 0.600496  0.787614 The answer is therefore 1000 (0.787614)  787.61 . (A) 8.8. The bonus is a pure endowment. The probability of no accident each year is 0.85, or 0.853 for 3 years. The probability of returning the car is 0.75. The present value of the bonus is therefore 1000 (0.853 )(0.75) /1.13  346.05 . (E) 8.9. The fund starts out at 100,000A40  16,132. If all goes as expected for two years, the fund will have 1000A42 for each survivor at the end of two years. That is ! 92,595.69 l 42  17,534  17,635.72 1000A42 (100) l40 93,131.64 Actual experience is: after one year, the fund accumulates at 6% interest and pays 1000: 16,132 (1.06) − 1000  16,100. After the second year, the fund accumulates at 7% interest: 16,100 (1.07)  17,227. The difference is 17,534 − 17,227  307 . (B) 8.10.

We will need from the Illustrative Life Table q 70  0.03318 and q71  0.03626. Then 1| q 70

 p 70 q71  (1 − 0.03318)(0.03626)  0.03506

2 p 70  1 − q 70 − 1| q 70  1 − 0.03318 − 0.03506  0.93176

The present value of the payments per life is 10

1| q 70 1.082

P 2 p 70 1.082   P (0.93176) 0.03318 0.03506 +  10 + 2 1.08 1.08 1.082  0.30722 + 0.30058 + 0.79883P

q 70

1.08

+



+

 0.60780 + 0.79883P This is set equal to P, resulting in P  0.60780/ (1 − 0.79883)  3.0213 . (C) 8.11.

First back out the discount factor. 0.5051 

7000 4 v 10,000

0.5051 v 0.7

! 0.25  0.921658

Now compute the net single premium for the 3-year pure endowment.

!

10,000A31:31

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7000  10,000 (0.9216593 )  6089.27 9000

(D)

EXERCISE SOLUTIONS FOR LESSON 8

8.12.

163

Let A be the net single premium.

!

10,000 − 8,500 8,500 − 6,500 6,500 − 4,000 1 + + A 10,000 1.06 (1.06)(1.07) (1.06)(1.07)(1.08) 1  (1415.09 + 1763.36 + 2040.92) 10,000  0.52194



(C)

8.13.

In this question, the table gives the probability of machines breaking down during month m + t − 1 given that they were installed or fixed in month m. This is t−1| q 0 , not q t−1 . The probability that a machine installed at time 0 breaks down in the second month is 0.2, not (0.2)(0.9) . For 10 machines installed, 2 machines, not (0.2)(9) machines, break down. Notice how the four probabilities add up to 1. The expected numbers of machines initially installed breaking down in months 1, 2, and 3 are 1, 2, and 3 respectively. The expected numbers of machines breaking down in month 1 that break down in months 2 and 3 are 0.1 and 0.2. The expected number of machines breaking down in month 2 that break down in month 3 is 0.21. The following table summarizes this: Month of breakdown Month of Number installation/ installed/ 1 2 3 repair repaired 0 1 2 Total

10.0 1.0 2.1

1 — — 1

2 0.1 — 2.1

3 0.2 0.21 3.41

The actuarial present value is 100 .

*

1 1.06

! 1/12 (1) +

1 1.06

! 2/12 (2.1) +

,

1 1.06

! 3/12 (3.41) +/  643.5546 -

8.14. This exercise is complicated by the replacement bulbs, which burn out based on time from installation rather than time from beginning. Let’s make a table, by year of light installation, of the burn outs, using the given q’s. Year of installation

Number installed

0 1 2

10,000 1,000 2,800

Number of bulbs burned out Year 1 Year 2 Burned out Remaining Burned out Remaining 1000

9000 1000

2700 100

6300 900 2800

Year 3 Burned out 3150 270 280

Thus we see that 1000 bulbs burn out in year 1, 2800 in year 2, and 3700 in year 3. The actuarial present value is 1000 2800 3700 + +  6688.26 (A) 1.05 1.052 1.053 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

8. INSURANCE: ANNUAL—MOMENTS

164

8.15. Split the 10-year term into a 5-year term plus a deferred 5-year term. For i  0.1236, the force of interest is doubled, so we use the 1000 (2A x ) column of the Illustrative Life Table for a 5-year term. The 5-year pure endowment factor can be derived from the ILT’s factor by multiplying it by v 5  1/1.065 . 2 2 2 1 A20 :5  A20 − 5 E 20 A25

2

!

0.74316 (0.01875)  0.00389  0.0143 − 1.065 1 The actuarial present value of the 5-year deferred term insurance is A25 :5 at 6% discounted 5 years at 12.36%. 1 A25 :5  0.08165 − (0.74229)(0.10248)  0.00558

!

5|5 A20 

0.74316 (0.00558)  0.00310 1.065

The 10-year term insurance has actuarial present value 1000 (0.00389 + 0.00310)  6.99 . 8.16.

We need 2| q60 and 3| q 60 . 2| q 60

 2 p60 q62  (1 − 0.09)(1 − 0.11)(0.13)  0.105287

3| q 60

 3 p60 q63  (1 − 0.09)(1 − 0.11)(1 − 0.13)(0.15)  0.105692 0.105287 0.105692  0.190258 + (D)  1.033 1.034

2|2 A60

8.17. We decompose this into a whole life insurance of 1000 on (45) minus a 20-year deferred whole life insurance of 500 on (65). Then 1000A45  201.20 500 20| A45  500 20 E45 A65  (0.5)(256.34)(0.43980)  56.37 Answer  201.20 − 56.37  144.83 8.18.

(B)

At age x, the density of Tx is 1/ ( ω − x ) . Thus the single premium for term to 60 is 60−x X t1

1 1 1  a t ω − x 1.05 ω − x 60−x

Plugging in x  50, the point at which the error is found, this becomes 1 1 − (1/1.0510 ) 1  (7.72173) ω − 50 0.05 ω − 50

!

Originally ω is 100 so that the actuarial present value of the 10 year insurance of 100,000 is 1 (7.72173)(100,000) 100 − 50 When ω is changed to 120, the actuarial present value is 1 (7.72173)(100,000) 120 − 50 The difference in premiums is:



1 1 − (7.72173)(100,000)  −4412.42 120 − 50 100 − 50

It is negative, so a refund of 4,412 is due. (B) LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM



EXERCISE SOLUTIONS FOR LESSON 8

8.19.

165

We are given two equations in two unknowns: 600A x1:n + 1000A x:n1  600 1000A x1:n + 1000A x:n1  800

Subtracting one from the other yields 400A x1:n  200 A x1:n  0.5 1000A x:n1  800 − 1000A x1:n  300 8.20.

From (vi) we have

1 1000 10 E35 A35 :10 + 1000 10 E35 1 750 10 E35 A35 :10 + 1000 10 E35

(C)

 1.005

so 1 1 (1.005)(750) A35 :10 10 E35 + 1000 (1.005) 10 E35  1000A35:10 10 E35 + 100010 E35   1 1000 − (1.005)(750) A35 :10  5 1 A35 :10 

5  0.020305 1000 − 1.005 (750)

and from (i), 10 E35

 0.57 − 0.020305  0.549695

So the answer is 500 (0.020305)(0.549695) + 1000 (0.549695)  555.28 . (B) 8.21. We have 1| q xs  (0.90)(0.20)  0.18 and 1| q xns  (0.95)(0.10)  0.095. Thus the probability of death in the first year from a randomly chosen individual is (0.25)(0.10) + (0.75)(0.05)  0.0625 and the probability of death in the second year from a randomly chosen individual is (0.25)(0.18) + (0.75)(0.095)  0.11625. Discounting at 2%: 0.0625 0.11625 +  0.173010 1.02 1.022 Multiplying by the 10,000 death benefit, the answer is 1730.10 . (C) 8.22. It’s unusual that the answer ranges don’t all have the same size. The second policy pays 5 more than the first policy starting at age 30 and ending at age 70, and 10 more starting at age 70. The question is ambiguous about whether the increase in benefit is continuous or not, but it doesn’t matter for the difference. The difference is a whole life policy on age 30 plus a 40-year deferred whole life policy. So we need to calculate 5A30 + 5 40| A30 . 5A30  5 (0.10248)  0.5124 5 40| A30  5 40 E30 A70 To calculate 40 E30 , we break it down into 20 E30 20 E50 so that we can look it up in the Illustrative Life Table. 40 E30

 20 E30 20 E50  (0.29374)(0.23047)  0.06770

5 40| A30  5 (0.06770)(0.51495)  0.17431 Answer  0.5124 + 0.17431  0.68671 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

(C)

8. INSURANCE: ANNUAL—MOMENTS

166

The following table shows the amounts each policy pays for deaths at each age, assuming that the increases and decreases are annual rather than continuous, in case you are interested: Age

First policy

Second policy

Age

First policy

Second policy

30 31 32 33 34 35 36 37 38 39 40 41 42 43

5 10 15 20 25 30 35 40 45 50 55 60 65 70

10 15 20 25 30 35 40 45 50 55 60 65 70 75

44 45 46 47 48 49 50 51 52 53 54 55 56 57

75 80 85 90 95 100 105 100 95 90 85 80 75 70

80 85 90 95 100 105 110 105 100 95 90 85 80 75

Age

First policy

Second policy

65 60 55 50 45 40 35 30 25 20 15 10 10

70 65 60 55 50 45 40 35 30 25 20 15 20

58 59 60 61 62 63 64 65 66 67 68 69 70+

8.23. A pure endowment either pays or doesn’t, so Z is a Bernoulli-type random variable, with payments of v 20 (probability 20 p x ) and 0 (probability 1 − 20p x ). We use the Bernoulli shortcut (Subsection 1.2.1) for variance. We have E[Z]  20 p x v 20 Var ( Z )  20 p x (1 − 20 p x ) v 40 Var ( Z ) 0.05   v 20 (1 − 20 p x )  v 20 (0.35) E[Z] 1 v 20  7

r v i 8.24.

20

1  0.907288 7

1 − 1  0.102186 v

(A)

From (i), 15 p x  p 15 x . Using (iii): Var ( Z )  15 p x (1 − 15 p x ) v 30  0.065 E[Z]  0.06515 p x v 15 0.065 1 − p 15  0.315701 x  0.915 p 15 x  0.684299 p x  0.975027 q x  1 − p x  0.024973

8.25. 21 1 1 A40 :21  A40:20 + v 20p 40 q 60

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(B)

EXERCISE SOLUTIONS FOR LESSON 8

167

We need to calculate the mortality probabilities. 0.001 (1.0560 − 1.0540 )  e −0.238556  0.787764  exp − ln 1.05

!

20p 40

0.001 (1.0561 − 1.0560 )  1 − e −0.019142  0.018960 ln 1.05 1000  124.60 + (0.787764)(0.018960)  129.96 1.0521

!

q 60  1 − exp − 1 1000A40 :21

8.26.

Since the probability of death in each quarter year is 0.005, we can factor out 0.005: (4)

1  0.005 (0.950.25 + 0.950.5 + 0.950.75 + 0.95)  0.01937 A65 :1

8.27.

The unconditional mortality probabilities are: t|0.25q x

 t p x 0.25 q x+t  1.02−t (1 − 1.02−0.25 )  (1.02−t ) u

where u  0.004938. From basic principles, A x(4) 

∞ X

0.25k|0.25 q x

v 0.25 ( k+1)

k0

 1.06−0.25 u

∞ X

(1.02−0.25k )(1.06−0.25k )

k0



which is a geometric series. Let r  (1.02)(1.06) A (4) 

 −0.25

 0.980671.

u (1.06−0.25 )  0.2518 1−r

We could have also calculated this by adapting the q/ ( q + i ) formula to use quarterly interest i (4) /4  1.060.25 − 1  0.014674 instead of i and quarterly mortality 1 − 1.02−0.25  0.004938 instead of q. 8.28.

Use the formulas in Table 8.2. q 0.06   0.461538 q + i 0.06 + 0.07 q 0.06 E[Z 2 ]    0.292826 q + 2 i 0.06 + 0.14 + 0.072 E[Z] 

Var ( Z )  0.292826 − 0.4615382  0.079809 8.29. If death occurs at the end of year 1, the present value of a benefit of 1 is 1/1.06  0.943396, otherwise it is 1/1.062  0.889996. The variance is calculated using the Bernoulli shortcut: Var ( Z )  10002 (0.98)(0.02)(0.943396 − 0.889996) 2  55.89

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8. INSURANCE: ANNUAL—MOMENTS

168

8.30. If death occurs at the end of year 1, the present value of a benefit of 1 is 1/1.06  0.943396. If death occurs in year 2, it is 1/1.062  0.889996. Otherwise it is 0. E[Z]  0.02 (0.943396) + (0.98)(0.04)(0.889996)  0.053756 E[Z 2 ]  0.02 (0.9433962 ) + (0.98)(0.04)(0.8899962 )  0.048850 Var ( Z )  10002 (0.048850 − 0.0537562 )  45,960 8.31. Mortality is uniformly distributed with ω − x  75. We calculate the first two moments using the formulas in Table 8.2. a E[Z]  75 75 1 − e −0.05 (75)   0.253940 75 ( e 0.05 − 1) 2a E[Z 2 ]  75 75 1 − e −0.1 (75)  0.126708  75 ( e 0.1 − 1) Var ( Z )  0.126708 − 0.2539402  0.06222 8.32.

(A)

The prescript of 2 means everything is done at twice the force of interest: A55:10  2A55 − 102E55 2A65 + 102E55

2

We need 102E55 , which we can derive from the value of 10 E55 given in the ILT. 2 10 E 55

 v 10 10 E55 0.48686  0.27186  1.0610

Now, A55:10  2A55 − 102E55 2A65 + 102E55

2

 0.13067 − (0.27186)(0.23603) + 0.27186  0.33836 8.33. From (i), vq x  0.019. From (ii) using the Bernoulli shortcut (there are only two possible values for the present value of benefits, 0 and v) v 2 q x (1 − q x )  0.017689. Dividing the second expression by the first, v (1 − q x ) 

0.017689  0.931 0.019

Adding this to the expression from (i), v  0.019 + 0.931  0.950 So i  1/v − 1  1/0.950 − 1  0.052632 . (B) 8.34. Due to 0 interest, Z is a Bernoulli random variable whose probability is 2 q x . Since 2 q x (1−2 q x )  0.16,  0.8. We reject 2 q x  0.2 since q x  0.5. Then

2 qx

q x + p x q x+1  0.8 0.8 − 0.5 q x+1   0.6 (B) 0.5 An unusual CAS question in that they provided answer choices rather than ranges.

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EXERCISE SOLUTIONS FOR LESSON 8

8.35.

169

A repeat of the last exercise, but now 2 q x (1 − 2 q x )  0.09 so 2 q x  0.9. Since 2 q x  0.9 and q x  0.5, p x q x+1  0.9 − 0.5  0.4 0.4  0.8 q x+1  (D) 0.5

8.36. There are three possibilities: death in the first year, death in the second year, and death thereafter. The payment for death thereafter is 0, so we can ignore it when calculating first and second moments. The probability of death in the first year is q 60  1 −

77,200  0.021546, 78,900

and the probability of death in the second year is 1| q 60



77,200 − 75,100  0.026616. 78,900

The present value of the payment for death in the first year is 10/1.05  9.52381 and the present value of the payment for death in the second year is 10/1.052  9.07030. Therefore the moments are: E[Z]  0.021546 (9.52381) + 0.026616 (9.07030)  0.44662 E[Z 2 ]  0.021546 (9.523812 ) + 0.026616 (9.070302 )  4.1440 Var ( Z )  4.1440 − 0.446622  3.9445

(D)

8.37. Use the conditional variance formula, equation (1.13) on page 9. Let N be the random variable for the number of friends buying a policy. N is binomial with parameters 10 and 0.1, so E[N]  1 and Var ( N )  10 (0.1)(0.9)  0.9. Then Var ( S )  E[Var ( S | N ) ] + Var (E[S | N]) E[S | N]  1000NA22  71.35N Var (E[S | N])  71.352 (0.9)  4581.74 The variance of S given N is N times the variance of one policy, since the future lifetimes are independent: the variance of the sum of N independent identical random variables is N times the variance of one of them.





Var ( S | N )  N 10002 (2A22 ) − (1000A22 ) 2  (15,870 − 71.352 ) N  10,779.18N E[Var ( S | N ) ]  10,779.18 E[N]  10,779.18 (1)  10,779.18 Therefore, the variance of S is Var ( S )  4581.74 + 10,779.18  15,360.92

(E)

8.38. They give you the interest rate in a weird manner, by telling you the lowest premium such that there is no chance of a loss. Since the present value of the death benefit for death in the first year is v, it follows that v  0.95. The moments of Z are: E[Z]  q x (0.95) + p x q x+1 (0.952 ) + 2 p x (0)  0.25 (0.95) + 0.15 (0.9025)  0.372875 E[Z ]  0.25 (0.952 ) + 0.15 (0.90252 )  0.347801 2

Var ( Z )  0.347801 − 0.3728752  0.2088

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(D)

8. INSURANCE: ANNUAL—MOMENTS

170

8.39. Let Z be the present value random variable for insurance of 1 for the combined group, Z SM the present value random variable for insurance of 1 for smokers, Z NS the present value random variable for insurance of 1 for non-smokers. Since we don’t know exactly how many smokers and non-smokers there are in this group, mortality has a mixture distribution. Moments for a mixture (each individual has a 20% probability of being a smoker and 80% of being a non-smoker) are calculated using the proportions of the mixture, even though the variance of a mixture is not the weighted sum of the individual variances. 0.01 (0.99)(0.02)  0.0270559 + 1.06 1.062 0.01 (0.99)(0.02) E[ ( Z NS ) 2 ]  +  0.0245834 2 1.06 1.064 0.02 (0.98)(0.03) E[Z SM ]  +  0.0450338 1.06 1.062 (0.98)(0.03) 0.02 +  0.0410875 E[ ( Z SM ) 2 ]  1.062 1.064 E[Z]  0.8 (0.0270559) + 0.2 (0.0450338)  0.0306515 E[Z NS ] 

E[Z 2 ]  0.8 (0.0245834) + 0.2 (0.0410875)  0.0278842 Var ( Z )  0.0278842 − 0.03065152  0.02694472 For 1000 of insurance on 100 lives, the mean is (100)(1000)(0.0306515)  3065.15 and the variance is

(100)(10002 )(0.02694472)  2,694,472. The fund needed is

p

3065.148 + 1.645 2,694,472  5765.39 .

8.40. According to the Illustrative Life Table, 1000A65  439.80 and 10002A65  236.03. Therefore, the variance of an insurance of 1 is 0.23603 − 0.439802  0.042606 For 1000 lives and insurance of 10, the mean is (1000)(10)(0.43980)  4398 and the variance is

(1000)(102 )(0.042606)  4260.6. The standard normal distribution has its 80th percentile at 0.842. Therefore, the required fund is √ 4398 + 0.842 4260  4453 (E) 8.41. This question seems to belong on what was then Course 2 (now Course CAS 2/SOA FM) rather than Course 3, but is included here for completeness. The present value of the medical payments, since they’re made at the end of the year, is 10,000

14

15 X 1.1t−1 t1

10,000 X 1.1  1.05 1.05 1.05t

!t

t0

10,000  1.05

!

1− 1

1.1
15 ! 1.05 1.1 − 1.05

!



10,000 (21.19588)  201,865.56 1.05

The annuity certain for the indemnity payments is a15  (1−v 15 ) /0.05  10.379658 where v  1/1.05. Their present value is 274,000−201,865.56  72,134.44. Therefore the payments P must be 72,134.44/10.379658  6949.60 . (D) LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

QUIZ SOLUTIONS FOR LESSON 8

8.42.

171

The net single premium for one life is 1000vq 30 + 500v 2 p30 q 31 

1.53 (1 − 0.00153)(0.5)(1.61) +  2.158747 1.06 1.062

The expected size of the fund at the end of two years with this premium is zero, by the definition of the net single premium. The actual size of the fund is obtained by recursion. Let Ft be the fund at time t. F1  2158.747 (1.07) − 1000  1309.860 F2  1309.860 (1.069) − 500  900.240

(C)

There is an error in the official solution, where they use 1.065 instead of 1.069 in the last line. 3 8.43. Actual losses in millions are 1.1 + 1.15 2  6.8595. The conditional-on-survival-to-3 probability of failure in one year is q3  0.30/0.60  0.5 and the conditional-on-survival-to-3 of failure in two   probability 1/3 0.5 years is 1| q 3  (0.30 − 0.10) /0.60  1/3, so expected losses in millions are 10 1.1 + 1.12  7.3003. The ratio

is 6.8595/7.3003  0.9396 . (A) 8.44.

We’ll need q 50 and 1| q 50 .



p50  exp Bc

0.2 −1  exp −0.000005 (1.250 ) ln c ln 1.2

50 c



50 2p 50  exp −0.000005 (1.2 )

1.22 − 1 ln 1.2

!!  0.9513110

!!  0.8960032

q 50  1 − 0.9513110  0.048689 1| q 50 1 A50 :2

 0.9513110 − 0.8960032  0.055308 0.048689 0.055308  vq 50 + v 2 1| q 50  +  0.099404 1.03 1.032

(E)

Quiz Solutions 8-1.

1 1| A[70]:2

.

8-2. The formula is

2 2 2 1 A45 :5  A45 − 5 E 45 A50

2

While 52E45 is not provided in the ILT, it can be calculated as v 5 5 E45 . 2 5 E 45 2

1 A45 :5

0.72988  0.54541 1.065  0.06802 − (0.54541)(0.09476)  0.01634



For a benefit amount of 1000, we multiply by 10002 , so the answer is 16,340 . 8-3. Split the insurance into two components: a 10-year term insurance and a 10-year deferred insurance. In the initial 10-year period, p x  p  e −0.02 . 1 A60 :10 

 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

 q  1 − ( vp ) 10 q+i 1 − e −0.02 e −0.2 1− −0.02 1−e + 0.06 1.0610

!

8. INSURANCE: ANNUAL—MOMENTS

172

 (0.248133)(0.542825)  0.134693 In the ultimate period (that is, after 10 years), A70 is calculated using uniform survival with ω  120. However, A70 must be discounted to age 60 using the exponential mortality of the first 10 years. 10 10| A60  ( vp )

a50 50

!

e −0.2 1 − (1/1.06) 50  50 (0.06) 1.0610

!

 (0.457175)(0.315237)  0.144119 The net single premium for the insurance is 1000A60  1000 (0.134693 + 0.144119)  278.81 .

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Lesson 9

Insurance: Continuous—Moments—Part 1 Reading: Models for Quantifying Risk (4th or 5th edition) 7.3

9.1

Definitions and general formulas

In this and the next lesson, we will discuss the evaluation of moments of insurances payable at the moment of death. The present value random variable for such an insurance is continuous. We sometimes say that the insurance is continuous. In the exercises, some questions use the term “fully continuous”. At this point, ignore the word “fully”. We will explain that word when we discuss premiums. The IAN symbols for insurances payable at the moment of death are the same as the symbols for insurance payable at the end of the year of death, except that bars are placed on the A’s. Thus, the actuarial present value of: Whole life insurance is denoted by A¯ x , which represents v Tx . n-year term insurance is denoted by A¯ x1:n , which represents v Tx if Tx ≤ n, 0 otherwise. n-year deferred insurance is denoted by n| A¯ x , which represents 0 if Tx ≤ n, v Tx otherwise. n-year deferred m-year term insurance is denoted by n| A¯ x1:m , which represents v Tx if n < Tx ≤ n + m, 0 otherwise. n-year endowment insurance is denoted by A¯ x:n , which represents v Tx if Tx ≤ n, v n otherwise. Notice that pure endowments are never payable at death, so no bar would ever be put on A x:n1 . The first contract we’ll discuss is whole life insurance. Whole life insurance on ( x ) pays a benefit at the time of ( x ) ’s death, whenever ( x ) dies. Let’s consider a whole life insurance of 1 on ( x ) that pays 1 at the moment of death of ( x ) . When an insurance is payable at the moment of death, it is called continuous, since the random variables associated with its present value are continuous random variables. Let Z denote the present value of this whole life insurance. We have defined Tx as the time until death. Then Z  v Tx . The expected value of Z is obtained by integrating v Tx over the density function of Tx . So E[Z]  A¯ x 

∞

Z

v t f x ( t ) dt

(9.1)

0

By equation (3.8) on page 40, the density function of Tx is , f x ( t )  t p x µ x+t . Also, v t  e −δt . So the equation becomes Z ∞

A¯ x 

e −δt t p x µ x+t dt

(9.2)

0

We may also be interested in the variance, third moment, etc. of Z. To calculate these, we need to calculate higher moments of Z. Higher moments of Z can be calculated by raising v Tx to powers while LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

173

9. INSURANCE: CONTINUOUS—MOMENTS—PART 1

174

leaving the density function intact: ∞

Z E[Z ] 

e −nδt t p x µ x+t dt

n

0

Notice that E[Z n ] at force of interest δ equals E[Z] at force of interest nδ. This is the “rule of moments”, which we already discussed on page 142. The variance of Z can be calculated as Var ( Z )  E[Z 2 ] − E[Z]2  2A¯ x − A¯ 2x

(9.3)

To calculate the actuarial present value of a continuous insurance, one must carry out the integral of formula (9.2). This means a probability density function must be provided. In most cases, including plausible mortality functions such as Makeham, there is no closed form for the integral and it must be numerically approximated. On exams, you can only be expected to evaluate insurances for mortality functions that allow evaluation of the integrals in closed form. We will discuss two such mortality functions in detail: constant force of mortality and uniformly distributed age at death.

9.2

Constant force of mortality

Constant force of mortality leads to simple integrals for evaluating actuarial present values. By understanding concepts or memorizing formulas, you will never have to integrate in order to evaluate these expected values. Let’s first work out the simplest case: constant face amount of 1, constant force of interest, constant force of mortality. The exponential distribution has no memory, so the results will not depend on the age of the insured. Example 9A The force of mortality for ( x ) is µ. The force of interest is δ. 1. Let Z1 be the random variable for 1 unit of whole life insurance. Calculate E[Z1 ] and Var ( Z1 ) . 2. Let Z2 be the random variable for 1 unit of term insurance for 5 years. Calculate E[Z2 ] and Var ( Z2 ) . 3. Let Z3 be the random variable for 1 unit of 10-year deferred whole life insurance. Calculate E[Z3 ] and Var ( Z3 ) . 4. Let Z4 be the random variable for the present value of 20-year endowment insurance of 1. Suppose µ  0.01, δ  0.04. Calculate E[Z4 ] and Var ( Z4 ) . Answer:

1. Under constant force of mortality, t p x  e −µt . We have ∞

Z E[Z1 ] 

e −δt e −µt µ dt 0 ∞

Z

e − ( δ+µ ) t µ dt

 0

∞

µ − ( δ+µ) t e µ+δ 0 µ  µ+δ −

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9.2. CONSTANT FORCE OF MORTALITY

175

By the rule of moments, we calculate the second moment by doubling δ. E[Z12 ]  Var ( Z1 ) 

µ µ + 2δ µ µ − µ + 2δ µ+δ

!2

2. The term insurance is the same as the whole life insurance, except that we integrate from 0 to 5 instead of from 0 to ∞. 5

Z E[Z2 ] 

e −δt e −µt µdt 0

 µ  1 − e −5 ( µ+δ ) µ+δ  µ  2 1 − e −5 ( µ+2δ ) E[Z2 ]  µ + 2δ 

 * µ  + µ  Var ( Z2 )  1 − e −5 ( µ+2δ ) − . 1 − e −5 ( µ+δ ) / µ + 2δ µ+δ !

,

2

-

3. For the deferred insurance, we integrate from 10 to ∞. ∞

Z E[Z3 ] 

e −δt e −µt µdt 10

µ  −10 ( µ+δ )  e µ+δ µ  −10 ( µ+2δ )  E[Z32 ]  e µ + 2δ 

µ  −10 ( µ+2δ )  µ  −10 ( µ+δ )  Var ( Z3 )  e − e µ + 2δ µ+δ

!2

4. We are using numerical values for this part of the example to make it more concrete. For the endowment insurance the expected value is the sum of the expected values of term insurance and a pure endowment. Let the term insurance be Z5 and the pure endowment Z6 . For term insurance, by substituting 20 for 5, we can use what we calculated above: E[Z5 ] 

 1  µ  1 − e −20 ( δ+µ )  1 − e −1  0.2 (1 − 0.367879)  0.126424 µ+δ 5

For the pure endowment,

E[Z6 ]  20 p x v 20  e −20 ( µ+δ )  0.367879

Therefore E[Z4 ]  0.126424 + 0.367879  0.494304 . We now calculate the second moment. 20

Z E[Z42 ]



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e 0



Z −2δt −µt

e

∞

µ dt +

e −40δ e −µt µ dt 20

 µ  1 − e −20 ( µ+2δ ) + e −40δ e −20µ µ + 2δ

9. INSURANCE: CONTINUOUS—MOMENTS—PART 1

176

In the second integral, we pulled out the constant e −40δ and were left with the integral of the density from 20 to ∞, which is the survival function. For an exponential, the survival function at t is e −µt . Continuing,

 1 1 − e −1.8 + e −1.8 9 1  (1 − 0.165299) + 0.165299  0.258043 9 Var ( Z4 )  0.258043 − 0.4943042  0.013707 E[Z42 ] 



The above example was meant to show you how you can calculate insurances from first principles, but you would never do so much work. For routine questions with constant force of mortality, you will speed up your work with shortcuts. Let’s go through shortcuts for constant force of mortality only. 1. The actuarial present value of a whole life insurance of 1 with benefit paid at moment of death is A¯ x  µ/ ( µ + δ ) . Engrave that in your head. 2. An n-year deferred whole life insurance is a whole life insurance paid n years later if the insured survives. In other words, it’s an n-year pure endowment with a benefit equal to a whole life insurance at age x + n. So n| A¯ x  n E x A¯ x+n for any mortality assumption. For constant force, n E x  v n n p x  e − ( µ+δ ) n , and A¯ x+n  A¯ x since the actuarial present value is independent of age. It follows that ¯  e −n ( µ+δ )

n| A x

µ µ+δ

(9.4)

3. A term life insurance is an immediate whole life insurance minus a deferred whole life insurance. A¯ x1:n  A¯ x − n E x A¯ x+n for any mortality assumption. For constant force, since A¯ x+n  A¯ x , it follows that A¯ x1:n  A¯ x (1 − n E x ) 

 µ  1 − e −n ( µ+δ ) µ+δ

(9.5)

This equation works even if the force of mortality is not constant after the term period. In such a case, A¯ x+n , A¯ x , but you can still prove this formula by integrating over the n-year term period. Intuitively, the force of mortality after the term period cannot have any effect on the EPV of the term insurance, so you might as well assume that it is constant after the term period even if it isn’t. 4. A deferred term insurance is the difference of two deferred whole life insurances: ¯1 n| A x :m

 A¯ x ( n E x − m+n E x ) 

 µe −n ( µ+δ )  1 − e −m ( µ+δ ) µ+δ

(9.6)

5. An endowment insurance is a term life insurance plus a pure endowment. For constant force, it follows that  µ  A¯ x:n  1 − e −n ( µ+δ ) + e −n ( µ+δ ) (9.7) µ+δ 6. To calculate the j th moment, multiply δ by j in each of these formulas.

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9.2. CONSTANT FORCE OF MORTALITY

?

177

Quiz 9-1 An insurance contract on (30) will pay $1000 at the moment of death if (30) dies before reaching age 40. It will pay $2000 if the insured survives to age 40. You are given: • The force of mortality is constant and equal to 0.01 at all durations. • δ  0.05. Calculate the actuarial present value of this contract. You may split a whole life insurance up into a term insurance and a deferred insurance. This allows calculating the expected value, since E[Z1 + Z2 ]  E[Z1 ] + E[Z2 ]. This also allows calculating higher moments, because Z1 and Z2 are mutually exclusive. “Mutually exclusive” means that regardless of the time of death, only one of the two variables is nonzero. If death occurs during the the deferred f term period, g n insurance is 0; if death occurs afterwards, the term insurance is 0. Therefore E ( Z1 + Z2 )  E[Z1n ]+E[Z2n ] (all the cross terms are zero). However, the variance of Z1 + Z2 is not the sum of the individual variances of Z1 and Z2 . The variance should be calculated as second moment minus first moment squared. Example 9B An insurance contract on (70) will pay $5 if death occurs before age 90 and $1 if death occurs after age 90. The benefit is paid at the moment of death. You are given: • Z is the present value random variable for this contract. • µ70+t  0.02, t > 0. • δ  0.04 Calculate Var ( Z ) . Answer: Let Z1 and Z2 be the present value random variables for a 20-year term insurance and a 20-year deferred insurance respectively. Then Z  5Z1 + Z2 , and Z1 Z2  0. Using the shortcuts above,

 0.02   µ  1 − e −n ( µ+δ )  1 − e −1.2  0.232935 µ+δ 0.06   0.02   µ E[Z12 ]  1 − e −n ( µ+2δ )  1 − e −2  0.172933 µ + 2δ 0.10 µ 0.02 E[Z2 ]  e −n ( µ+δ )  e −1.2  0.100398 µ+δ 0.06 µ 0.02 E[Z22 ]  e −n ( µ+2δ )  e −2  0.027067 µ + 2δ 0.10 E[Z1 ] 

Therefore, the first and second moments and the variance of Z are E[Z]  5 E[Z1 ] + E[Z2 ]  5 (0.232935) + 0.100398  1.26507 E[Z 2 ]  E[ (5Z1 + Z2 ) 2 ]  25 E[Z12 ] + E[Z22 ]  25 (0.172933) + 0.027067  4.35039 Var ( Z )  4.35039 − 1.265072  2.7500



Similarly, an endowment insurance can be split into a term insurance and a pure endowment. Once again the two benefits are mutually exclusive, so that the k th moment is the sum of the individual k th moments. We could’ve saved work calculating the second moment in Example 9A part 4 using this method, and calculating E[Z62 ]  20 p x v 40  e −0.01 (20) −0.04 (40)  e −1.8 where the discount factor R ∞ is squared to double the force of interest. This would avoid the need to calculate the second integral 20 e −40δ t p x µ x+t dt. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

9. INSURANCE: CONTINUOUS—MOMENTS—PART 1

178

Splitting an insurance into a term and a deferred insurance will be especially helpful if a parameter changes at some duration. For example, if the force of mortality or the force of interest changes at time n, you would evaluate a whole life insurance as an n-year term life insurance plus an n-year deferred whole life insurance. To evaluate the moments of the n-year deferred insurance, we use: ¯  n E x A¯ x+n

(9.8)

n| A x

which is true for any mortality assumption (not just constant force). The next few examples illustrate the idea of splitting an insurance into term and deferred in order to handle varying assumptions. Example 9C An insurance company offers a retirement plan which pays a $1,000,000 lump sum to a person who survives to age 65. For a specific purchaser, you are given: • He is age 50. • His force of mortality is 0.005 at all durations. • δ  0.05. Calculate the actuarial present value of the retirement plan for this purchaser. Answer: The retirement plan is a pure endowment. The actuarial present value of a pure endowment under constant force of mortality is n E x  e − ( µ+δ ) n . Therefore, the actuarial present value of the pure endowment in this example is 1,000,000 15 E50  1,000,000e − (0.005+0.05) 15  438,235



Example 9D An insurance company offers a retirement insurance plan. This plan pays $1,000,000 at the time of death, if death occurs after age 65. For a specific purchaser, you are given: • He is age 50. • His force of mortality is

  0.005 µ50+t    0.02

t ≤ 15 t > 15

 • δ  0.05. Calculate the actuarial present value of the retirement insurance plan for this purchaser. Answer: In the previous example, we calculated the value of a 15-year pure endowment for this purchaser as 0.438235. This factor discounts to age 50 the value of any benefit the person receives at age 65. Once the purchaser reaches age 65, the value of the insurance is µ 0.02 2,000,000  1,000,000  µ+δ 0.02 + 0.05 7

!

1,000,000A¯ 65  1,000,000

Notice that at age 65, the lower force of mortality before age 65 is irrelevant to the calculation. The actuarial present value of the retirement insurance plan at age 50 is

!

2,000,000 1,000,00015| A¯ 50  15 E 50 (1,000,000A¯ 65 )  0.438235  125,210 7



Students sometimes mistakenly think that the µ of the second period (0.02) in the above example should be used for computing the discount factor 15 p50 to apply to the deferred insurance in the above LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

9.2. CONSTANT FORCE OF MORTALITY

179

example. If you understand the above two examples, you will avoid this mistake. The pure endowment discount factor uses the µ and δ of the discount period, regardless of what you are discounting. Now we’re ready for an example of a whole life insurance with varying force of mortality, in which you would split the whole life into a term and deferred insurance. Example 9E A whole life insurance on ( x ) provides a benefit of 1000 at the moment of death. You are given t 0 • δ  0.03. Calculate the net single premium. Answer: Let A¯ (with no subscript) be the net single premium. A¯  A¯ x (1 + A¯ ) µ (1 + A¯ )  µ+δ 1 + A¯  4 1 A¯  3



Table 10.1 on page 198 summarizes results for standard insurances (of 1) payable at the moment of death for constant force of mortality.

Exercises 9.1. You are given: • The benefit of 1 on a ten-year endowment insurance is payable at the moment of death, or at the end of 10 years if ( x ) survives 10 years. • µ x+t  0.01 for t > 0 • δ  0.06. Calculate the actuarial present value. 9.2. The actuarial present value of an n-year term insurance paying 1 at the moment of death to ( x ) is 0.05720. You are given: • •

µ x+t  0.007, t > 0. δ  0.05.

Determine n. 9.3. [150-F87:16] You are given: • • •

A¯ x1:n  0.4275 δ  0.055 µ x+t  0.045 for all t

Calculate A¯ x:n . A. 0.4600

B. 0.4775

C. 0.4950

D. 0.5245

E. 0.5725

1Forgot what net single premium means? See page 140. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

Exercises continue on the next page . . .

EXERCISES FOR LESSON 9

181

9.4. [SOA3-F03:2] For a whole life insurance of 1000 on ( x ) with benefits payable at the moment of death:

  0.04, • The force of interest at time t, δ t    0.05,

0 < t ≤ 10 10 < t

 •

µ x+t

  0.06, 0 < t ≤ 10   0.07, 10 < t 

Calculate the net single premium for this insurance. A. 379

B. 411

C. 444

D. 519

E. 594

9.5. A 3-year deferred insurance of 1 is payable at the moment of death. You are given that

  0.01 µ x+t    0.02

0≤t≤2 t>2

   0.05 0 ≤ t ≤ 2 δt    0.06 t > 2  Calculate the actuarial present value of this insurance. 9.6. [CAS4A-S93:16] (3 points) Dave wants to purchase a 5 year pure endowment with a net single premium. The amount of the endowment is $1,000. His insurance agent convinces him instead to use the same money to purchase a 5 year endowment insurance policy which pays at the moment of death or at the end of five years, whichever comes first. You are given that µ  0.04 and δ  0.06. Calculate the benefit amount for this 5 year endowment. A. B. C. D. E.

Less than $800 At least $800, but less than $850 At least $850, but less than $900 At least $900, but less than $950 At least $950

9.7. [CAS4A-F96:10] (2 points) Bryon, a nonactuary, estimates the net single premium for a continuous whole life policy with a benefit of $100,000 on (30) by calculating the present value of $100,000 paid at the expected time of death. (30) is subject to a constant force of mortality µ x  0.05, and the force of interest is δ  0.08. Determine the absolute value of the error of Bryon’s estimate. A. B. C. D. E.

Less than $15,000 At least $15,000, but less than $20,000 At least $20,000, but less than $25,000 At least $25,000, but less than $30,000 At least $30,000

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Exercises continue on the next page . . .

9. INSURANCE: CONTINUOUS—MOMENTS—PART 1

182

9.8. [CAS4A-S97:21] (2 points) For some length of term, n, the expected value of an n-year term insurance with benefit payable at the moment of death equals the expected value of an n-year pure endowment. The force of interest and the force of mortality are constant. Determine n. ln A. B. C.

µ
2µ+δ

µ+δ ln (2µ + δ ) − ln ( µ ) µ+δ ln ( δ ) − ln ( µ ) µ+δ

D.

1 µ+2δ

E.

None of A, B, C, or D is correct

9.9. [CAS4A-F97:10] (2 points) The benefit under an m-year deferred whole life policy, with benefit payable at the moment of death, is twice that of a similar nondeferred whole life insurance. The actuarial present value for these insurances are equal. You are given that µ  0.08 and δ  0.06. Determine m. A. B. C. D. E.

Less than 2 At least 2, but less than 4 At least 4, but less than 6 At least 6, but less than 8 At least 8

9.10. [CAS4A-F98:17] (2 points) Which of the following is an expression for the actuarial present value of a continuous whole life insurance, assuming a constant force of mortality µ, and that the effective rate of interest is unknown, but is known to be constant and uniformly distributed over the interval [0, 0.10]? ∞

Z

µe −µt e −t ln (1.05) dt

A. 0

B.

µ µ + ln (1.05) ∞ Z 0.10

Z C.

0

(1 + i ) −t µe −µt dt di

D. 0

0 0.10

Z E.

(1 + i ) −t µe −µt di dt

(0.10) 0 Z 0.10 Z ∞

(10µ ) 0

di µ + ln (1 + i )

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Exercises continue on the next page . . .

EXERCISES FOR LESSON 9

183

[CAS3-F03:7] Given:

9.11.

• i  5% • The force of mortality is constant. • e˚x  16.0 Calculate 20| A¯ x . A. B. C. D. E.

Less than 0.050 At least 0.050, but less than 0.075 At least 0.075, but less than 0.100 At least 0.100, but less than 0.125 At least 0.125 You are given:

9.12.

• Tx is the random variable for future lifetime for ( x ) . • µ x+t  0.02, t > 0. • Z is the present value random variable for an insurance on ( x ) . •

 

1

1.04 Z  0.5 


Tx

Tx < 30 Tx ≥ 30

Calculate the expected value of Z. Tx is the future lifetime random variable for ( x ) .

9.13.

An insurance pays 1000 at the moment of death if the integral part of Tx is odd. Otherwise no benefit is paid. You are given: • •

µ x  0.04 for all x. δ  0.02

Calculate the actuarial present value of the insurance. A deferred endowment insurance provides the following benefits:

9.14.

• A payment of 1 at the moment of death if death occurs between time t  5 and time t  12. • A payment of 2 if the insured survives 12 years. No benefit is paid if death occurs before time t  5. You are given: • •

  0.05 µ x+t    0.10  δ  0.10

t≤8 t>8

Calculate the actuarial present value of this endowment.

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Exercises continue on the next page . . .

9. INSURANCE: CONTINUOUS—MOMENTS—PART 1

184

You are given:

9.15.

1 • A¯ 40 :35  0.368 • 35p 40  0.85 • q75  0.01 • δ  0.04 • The force of mortality between integral ages is constant. 1 Calculate A¯ 40 :36 .

9.16. [CAS4A-S97:18] (2 points) A continuous whole life insurance is issued to a life age 30. This insurance pays an increasing benefit, starting at 1 at time of issue, increasing continuously at a rate of 10% per annum for the first 10 years, and increasing 5% per annum thereafter. You are given: µ  0.05 δ  0.06

• •

Determine the net single premium for this insurance. A. B. C. D. E.

Less than 1.25 At least 1.25, but less than 1.75 At least 1.75, but less than 2.25 At least 2.25, but less than 2.75 At least 2.75 [150-S87:15] For a continuous whole life insurance (Z  v T , T ≥ 0), E[Z]  0.25.

9.17.

Assume the forces of mortality and interest are each constant. Calculate Var ( Z ) . A. 0.04

B. 0.08

C. 0.11

D. 0.12

E. 0.19

9.18. [150-S90:5] Z is the present-value random variable for a whole life insurance of 1 payable at the moment of death of ( x ) . You are given: • •

µ x+t  0.05 for t ≥ 0 δ  0.10

Which of the following are true? I.

d ¯ dx A x

0

II. E[Z]  1/3 III.

Var ( Z )  1/5

A. I and II only B. I and III only C. II and III only E. The correct answer is not given by A. , B. , C. , or D.

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D. I, II and III

Exercises continue on the next page . . .

EXERCISES FOR LESSON 9

185

9.19. [150-S91:3] Z is the present-value random variable for a whole life insurance of b payable at the moment of death of ( x ) . You are given: • µ x+t  0.01 t ≥ 0 • δ  0.05 • The net single premium for this insurance is equal to Var ( Z ) . Calculate b. A. 1.36

B. 1.68

C. 2.00

D. 2.32

E. 2.64

9.20. [CAS4-S88:18] (2 points) A five-year deferred whole life insurance policy pays 10,000 at the moment of death for a person age 30. You are given: • The force of mortality is constant, µ  0.05. • δ  0.10 Calculate the standard deviation of the present value of the benefit payment. A. B. C. D. E.

Less than 1820 At least 1820, but less than 1840 At least 1840, but less than 1860 At least 1860, but less than 1880 At least 1880

9.21. [CAS4A-S97:11] (1 point) A whole life insurance pays a benefit of 10 at the moment of death. δ  0.05 and µ x  0.04 for all x. Determine the variance of the present value for this insurance. A. B. C. D. E.

Less than 2.0 At least 2.0, but less than 4.0 At least 4.0, but less than 6.0 At least 6.0, but less than 8.0 At least 8.0

9.22. [CAS3-F04:2] For a 5-year deferred whole life insurance of 1, payable at the moment of death of ( x ) , you are given: •

Z is the present value random variable of this insurance.

•

δ  0.10

•

µ  0.04 Calculate Var ( Z ) .

A. B. C. D. E.

Less than 0.035 At least 0.035, but less than 0.045 At least 0.045, but less than 0.055 At least 0.055, but less than 0.065 At least 0.065

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9. INSURANCE: CONTINUOUS—MOMENTS—PART 1

186

9.23. [150-F87:26] Z is the present value random variable for a special continuous whole life insurance issued to ( x ) . You are given for all t: • • •

µ x+t  0.01 δ t  0.06 b t  e 0.05t

Calculate Var ( Z ) . A. 0.033

B. 0.037

C. 0.057

D. 0.065

E. 0.083

[150-F96:11] For a special whole life insurance on ( x ) , you are given:

9.24.

• µ x+t  µ, t ≥ 0 • δ t  µ, t ≥ 0 • The death benefit, payable at the moment of death, is 1 for the first 10 years and 0.5 thereafter. • The net single premium is 0.3324. • Z is the present-value random variable at issue of the death benefit. Calculate Var ( Z ) . A. B. C. D. E.

Less than 0.07 At least 0.07, but less than 0.08 At least 0.08, but less than 0.09 At least 0.09, but less than 0.10 At least 0.10

9.25. [CAS4A-S94:17] (2 points) A life insurance policy has a death benefit of 20 payable at the moment of death. Z is the present value at policy issue of the benefit payments. You are given: µ x  0.06 for x ≥ 0 δ  0.12 Determine Var ( Z ) for this whole life insurance policy. A. B. C. D. E. 9.26.

Less than 22.5 At least 22.5, but less than 27.5 At least 27.5, but less than 32.5 At least 32.5, but less than 37.5 At least 37.5 [SOA3-F04:1] For a special whole life insurance on ( x ) , payable at the moment of death:

• µ x+t  0.05, t > 0 • δ  0.08 • The death benefit at time t is b t  e 0.06t , t > 0. • Z is the present value random variable for this insurance at issue. Calculate Var ( Z ) . A. 0.038

B. 0.041

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C. 0.043

D. 0.045

E. 0.048 Exercises continue on the next page . . .

EXERCISES FOR LESSON 9

187

9.27. [CAS4A-S92:12] (3 points) Z is the present value random variable for a whole life insurance with variable death benefit b t payable at the moment of death. For t > 0, you are given: • • •

µ x+t  0.03 δ t  0.05 b t  e 0.04t

Determine Var ( Z ) . A. B. C. D. E.

Less than 0.03 At least 0.03, but less than 0.04 At least 0.04, but less than 0.05 At least 0.05, but less than 0.06 At least 0.06

9.28. A special insurance pays a death benefit of 1 plus the net single premium with no interest at the moment of death. You are given that µ x+t  0.04 for all t and δ  0.03. Calculate the variance of the present value of this insurance. 9.29. [CAS4A-S99:16] (2 points) Your chief actuary is concerned about the skewness of your book of business. You are given: • The forces of mortality and interest are constant. • A¯ 20  0.25 Compute the third moment around the origin of the present value random variable for a whole life policy issued to (50). A. B. C. D. E.

Less than 0.04 At least 0.04, but less than 0.08 At least 0.08, but less than 0.12 At least 0.12, but less than 0.16 At least 0.16

9.30. [M-S05:7] Z is the present-value random variable for a whole-life insurance of b payable at the moment of death of ( x ) . You are given: • δ  0.04 • µ x+t  0.02, t ≥ 0 • The net single premium for this insurance is equal to Var ( Z ) . Calculate b. A. 2.75

B. 3.00

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C. 3.25

D. 3.50

E. 3.75

Exercises continue on the next page . . .

9. INSURANCE: CONTINUOUS—MOMENTS—PART 1

188

[M-F05:1] For a special whole life insurance on ( x ) , you are given:

9.31.

• Z is the present value random variable for this insurance. • Death benefits are paid at the moment of death. • µ x+t  0.02, t ≥ 0 • δ  0.08 • The death benefit at time t is b t  e 0.03t , t ≥ 0 Calculate Var ( Z ) . A. 0.075

B. 0.080

C. 0.085

D. 0.090

E. 0.095

9.32. [M-F06:12] For a whole life insurance of 1 on ( x ) with benefits payable at the moment of death, you are given: •

t < 12 t ≥ 12

  0.02, δ t , the force of interest at time t is δ t    0.03, 

•

µ x+t

  0.04, t < 5   0.05, t ≥ 5 

Calculate the actuarial present value of this insurance. A. 0.59

B. 0.61

C. 0.64

D. 0.66

Additional old CAS Exam 3/3L questions: S05:37,38, F05:35, S06:19, F06:33 Additional old CAS Exam LC questions: F14:13

Solutions 9.1. Using formula (9.7), A¯ x:10 

 µ  1 − e −n ( µ+δ ) + e −n ( µ+δ ) µ+δ !



 1  1 − e −0.7 + e −0.7 7

 0.071916 + 0.496585  0.5685 9.2. Using formula (9.5), µ µ+δ 0.007  0.057

0.05720 



1 − e −n ( µ+δ )



1 − e −0.057n





Solving for n, 0.465771  1 − e −0.057n 0.534229  e −0.057n ln 0.534229 n−  11 0.057 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

E. 0.68

EXERCISE SOLUTIONS FOR LESSON 9

189

9.3. The difference between A¯ x:n and A¯ x1:n is an n-year pure endowment, or e −0.1n . We have to back out e −0.1n . For constant force of mortality, by equation (9.5), A¯ x1:n  A¯ x (1 − n E x ) Therefore, 0.4275 

  0.045 1 − e − (0.045+0.055) n 0.045 + 0.055

0.4275  1 − e −0.1n 0.45 0.95  1 − e −0.1n 0.05  e −0.1n  A x:n1 A¯ x:n  A¯ x1:n + A x:n1  0.4275 + 0.05  0.4775

(B)

9.4. The net single premium for an insurance with a benefit of 1 is, A¯ x  A¯ x1:10 + 10 E x A¯ x+10  A¯ x (1 − 10 E x ) + 10 E x A¯ x+10 The pure endowment’s actuarial present value is 10 E x  e −10 (0.04+0.06)  e −1 .

  0.07 0.06 1 − e −1 + e −1 0.06 + 0.04 0.07 + 0.05 ! 7  0.6 (1 − 0.367879) + (0.367879) 12

A¯ x 

 0.379272 + 0.214596  0.593868 Therefore, the net single premium for a benefit of 1000 is 1000 (0.593868)  593.868 . (E) 9.5. Forces of interest and mortality are constant for A¯ x+3 , so A¯ x+3  0.02/ (0.02 + 0.06)  0.25. However, the 3-year pure endowment uses mixed rates of mortality and interest. ¯  3 E x A¯ x+3  0.25 3 E x

3| A x

3 Ex

 e −2 (0.01+0.05) − (0.02+0.06)  e −0.2

¯  0.25e −0.2  0.204683

3| A x

−0.5 ) + e −0.5  0.763918, so the benefit 9.6. A x:51  e −5 (0.04+0.06)  0.606531, whereas A x:5  0.04 0.1 (1 − e amount for endowment insurance is 1000 (0.606531/0.763918)  793.97 (A)

9.7. The expected value of an exponential distribution is the reciprocal of the force, or 1/0.05  20. The value of $100,000 paid in 20 years is 100,000e −20 (0.08)  100,000e −1.6  20,190. The true value of 100,000A¯ 30 0.05
is 100,000 0.05+0.08  38,462. The error is $18, 272 . (B)

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9. INSURANCE: CONTINUOUS—MOMENTS—PART 1

190

9.8. The n-year term insurance has expected value A¯ x1:n  A x 1 − e −n ( µ+δ ) 





 µ  1 − e −n ( µ+δ ) µ+δ

whereas the pure endowment has expected value e − ( µ+δ ) n . Setting them equal

 µ  1 − e − ( µ+δ ) n  e − ( µ+δ ) n µ+δ ! µ µ e − ( µ+δ ) n 1 +  µ+δ µ+δ µ e − ( µ+δ ) n  2µ + δ − ( µ + δ ) n  ln µ − ln (2µ + δ ) ln (2µ + δ ) − ln µ n µ+δ

(B)

9.9. Let the benefit under the deferred whole life policy be b d and the benefit under the nondeferred policy b n . Then b d  2b n and b d m| A¯ x  b n A¯ x . It follows that m| A¯ x  0.5A¯ x . ¯  0.5A¯ x e − ( µ+δ ) m A¯ x  0.5A¯ x m| A x

e − ( µ+δ ) m  0.5 1 e −0.14m  2 ln 2 m  4.9511 0.14 9.10.

(C)

For a fixed i, the actuarial present value of a whole life insurance is µ µ  µ + δ µ + ln (1 + i )

The uniform distribution on [0, 0.10] has density 10. Integrating the actuarial present value over this uniform distribution, Z 0.10 Z 0.10 µ di di 10  10µ (E) µ + ln ( 1 + i ) µ + ln (1 + i ) 0 0 9.11. The mean of an exponential distribution is the reciprocal of the force, as explained on page 79. Here mean survival time is 16, so the force of mortality is 1/16. The force of interest δ  ln (1 + i )  ln 1.05. Under constant force of mortality, A¯ x+20  A¯ x . Thus we have: ¯  20 E x A¯ x

20| A x

e

− (ln 1.05+1/16)(20)

 e −20 (0.0625+0.04879)

(1/16) (1/16) + ln 1.05

0.0625 0.0625 + 0.04879

 (0.1080)(0.5616)  0.06065

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! !

(B)

EXERCISE SOLUTIONS FOR LESSON 9

9.12.

191

The 30-year term insurance has δ  ln (1 + i )  ln 1.04, so its actuarial present value is 0.02 (1 − e −30(0.02+ln 1.04) )  0.280575 0.02 + ln 1.04

The pure endowment paid when T ≥ 30 has present value 0.5 if paid and actuarial present value 0.5e −30 (0.02)  0.274406. The total expected value of Z is E[Z]  0.280575 + 0.274406  0.55498 . 9.13. The actuarial present value of an insurance of 1 paid for death in ( k, k + 1] is the difference of a k-year and a k + 1-year deferred insurance, or



( k E x − k+1 E x ) A¯ x  e −(0.04+0.02) k − e −(0.04+0.02)( k+1)



 0.04 2e −0.06k   1 − e −0.06 0.04 + 0.02 3 !

Let A¯ be the actuarial present value of the insurance. Sum up the above expression over odd values of k. It is a geometric series.

! ∞ 2 (1 − e −0.06 ) X −0.06 (2k+1) ¯ A  1000 e 3 k0

e −0.06  38.8236 1 − e −0.12

!

 38.8236 (8.328335)  323.34

9.14.

We will calculate the actuarial present value of the endowment in three pieces:

(1)

the benefit for times 5–8,

(2)

the benefit for times 8–12, and

(3)

the pure endowment benefit. The actuarial present value of the first benefit is: 8

Z

e (−0.05−0.10) t 0.05 dt  5

 0.05  −0.75 e − e −1.2  0.05706 0.15

Because of the changing mortality, it is awkward to use shortcuts for the first benefit. The second benefit is a deferred term policy, or the difference between an 8-year deferred whole life insurance and a 12-year deferred whole life insurance. Keep in mind that 8 E x must be computed using µ x+t  0.05, while 12 E x  8 E x 4 E x+8 , with each pure endowment using a different µ. Since the force of mortality is constant for t ≥ 8, A¯ x+t is constant for t ≥ 8. 8Ex

A¯ x+8 − 12 E x A¯ x+12  8 E x (1 − 4 E x+8 ) A¯ x+8 e

−8 (0.05+0.10)





1−e

−4 (0.10+0.10)



0.10 0.10 + 0.10



 e −1.2 1 − e −0.8 (0.5)  0.08293 The actuarial present value of the third benefit is 2 8 E x 4 E x+8  2e −8 (0.15) −4 (0.20)  2e −2  0.27067 The net single premium is 0.05706 + 0.08293 + 0.27067  0.41066 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

!

9. INSURANCE: CONTINUOUS—MOMENTS—PART 1

192

9.15.

Decompose the 36-year insurance into 35 years and the 36th year. 1 ¯1 ¯1 A¯ 40 :36  A40:35 + 35 E 40 A75:1

Due to constant force of mortality between integral ages, µ75+s  − ln p75  − ln 0.99 for 0 < s < 1. Therefore 35 E 40

 0.85e −0.04 (35)  0.209607

(− ln 0.99)(1 − e −0.04+ln 0.99 ) 1  0.00980297 A¯ 75 :1  0.04 − ln 0.99 So the actuarial present value of the 36-year insurance is 0.368 + 0.209607 (0.00980297)  0.370055 . 9.16. Let A be the net single premium. The continuous rate of increase offsets the interest, so in effect we have δ  −0.04 in the first 10 years and δ  0.01 thereafter. Then A¯  9.17.

0.05 0.05 (1 − e −0.01(10) ) + e −0.01(10)  1.2298 0.05 − 0.04 0.05 + 0.01

Since E[Z]  µ/ ( µ + δ )  0.25, it follows that δ  3µ. Then E[Z 2 ] 

µ 1  µ + 2δ 7

1 1 Var ( Z )  − 7 4

!2  0.0804

(B)

9.18. I is true because µ and δ do not vary with x, so A¯ x does not vary with x. For constant force of mortality we know that E[Z]  and

µ 1  µ+δ 3

µ 1 1 − E[Z]2  − Var ( Z )  µ + 2δ 5 3

!2

so II is true and III is false. (A) 9.19.

For one unit, E[Z] 

µ µ+δ



1 6

and

µ 1 1 Var ( Z )  − E[Z]2  − µ + 2δ 11 6

!2

so we need b such that b 25b 2  6 396 396 b  2.64 (6)(25)

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(E)



25 396

(A)

EXERCISE SOLUTIONS FOR LESSON 9

9.20.

193

The mean for one unit is e −0.75 −0.15 (5) 0.05 ¯   0.157456 5 E 30 A35  e 0.15 3

!

and the second moment for one unit is (As usual, pre-superscript 2 both on E and on A¯ means double the force of interest.) ! e −1.25 2¯ −0.25 (5) 0.05 2   0.057301 5 E 30 A35  e 0.25 5 so the variance for 10,000 is 10,0002 (0.057301 − 0.1574562 )  3,250,900 and the standard deviation is the square root of the variance, or 1803 . (A) 9.21.

The variance for one unit is µ µ − µ + 2δ µ+δ

!2

0.04 0.04 − 0.04 + 2 (0.05) 0.04 + 0.05



!2  0.08818

So the variance of 10 units is 100 (0.08818)  8.818 . (E) 9.22.

We have 0.04 2 e −5 (0.04+0.10)  (0.496585)  0.141882 0.04 + 0.10 7 0.04 1 0.301194 2 2 2¯ −5 (0.04+0.20) E[Z ]  5 E x A x  e  e −1.2   0.050199 0.04 + 0.20 6 6 (A) Var ( Z )  0.050199 − 0.1418822  0.030068 E[Z]  5 E x A¯ x 

9.23. The increase in benefit rate can be treated as negative interest, so effectively the net force of interest is 0.06 − 0.05  0.01. We can use the usual formulas with δ  0.01. Var ( Z ) 

0.01 0.01 − 0.01 + 2 (0.01) 0.01 + 0.01

!2 

1 1 − 3 2

!2 

1  0.08333 12

(E)

9.24. An unusual SOA question in that it uses ranges for answer choices, something more typical for the CAS. The second moment is E[Z 2 ]  2A¯ x1:10 + 0.52 102E x 2A¯ x Note that the 0.5 benefit is squared when calculating the second moment.

 µ µ   1 − e −10 ( µ+2µ) + (0.52 ) e −10 ( µ+2µ) µ + 2µ µ + 2µ  1 1 1 1  1 − e −30µ + e −30µ  − e −30µ 3 12 3 4

!

Thus we must back out e −30µ from the given information. The expected value is 10

Z 0.3324  E[Z]  0 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

Z

∞

e (−µ−µ) t µ dt +

(0.5) e (−µ−µ) t µ dt 10

9. INSURANCE: CONTINUOUS—MOMENTS—PART 1

194

1 1 1 1 (1 − e −20µ ) + e −20µ  − e −20µ 2 4 2 4  4 (0.5 − 0.3324)  0.6704 

e −20µ

It is unnecessary to back out µ itself (although it was arranged that it come out even), since we can calculate directly e −30µ  ( e −20µ ) 1.5  0.67041.5  0.548910. Therefore 1 1 − (0.548910)  0.196106 3 4 Var ( Z )  0.196106 − 0.33242  0.0856161 E[Z 2 ] 

9.25.

For one unit, the variance is 0.06 0.06 − 0.06 + 2 (0.12) 0.06 + 0.12

so for 20 units it is 202 9.26.

(C)

4
45

!2 

4 45

 35 59 . (D)

Taking the death benefit rate of increase into account, the net force of interest is 0.08 − 0.06  0.02. 0.05 5  0.05 + 0.02 7 0.05 5 E[Z 2 ]   0.05 + 2 (0.02) 9 E[Z] 

Var ( Z ) 

5 5 − 9 7

!2  0.04535

(D)

9.27. The net force of interest after netting the rate of benefit increase is 0.05 − 0.04  0.01. The variance is then !2 0.03 0.03 Var ( Z )  −  0.0375 (B) 0.03 + 2 (0.01) 0.03 + 0.01 9.28.

Let A be the net single premium. Then µ A  (1 + A ) µ+δ 4  (1 + A ) 7 

4 7

A

4 3

3 7A

!

!

This means that the amount insured is 1 + A  73 . The variance is then of insurance, or 7 3

!2

7
2 3

times the variance for one unit

! !2 !2 !2 0.04 0.04 7 *4 4 + *. + / . − /  0.4 − 0.04 + 2 (0.03) 0.04 + 0.03 3 10 7 , , -

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EXERCISE SOLUTIONS FOR LESSON 9

195

µ

9.29. Since µ+δ  0.25, it follows that δ  3µ. The third moment around the origin (or the third raw moment) is obtained by tripling the force of interest (in order to cube the discount factor), so the third moment is µ µ   0.1 (C) µ + 3δ 10µ 9.30. If X is the random variable for an exponential insurance of 1 payable at the moment of death, the formulas are (see example 9A on page 174) µ µ+δ µ E[X 2 ]  µ + 2δ E[X] 

Here, Z  bX, and by (iii),

!2

µ µ µ +  b2 * − µ+δ µ + 2δ µ+δ

!

b

!

b

, 

2 1 8 1  b2 −  b2 3 10 9 90 15 1/3   3.75 b 8/90 4



!-

(E)

9.31. The increase rate in benefits of 0.03 offsets δ, since the integrand becomes e 0.03t e (−0.08−0.02) t (0.02) , and the increase rate gets doubled when calculating the second moment just like δ. So E[Z]  and E[Z 2 ] 

µ 0.02 2   µ + δ − 0.03 0.02 + 0.08 − 0.03 7

µ 0.02 1   µ + 2δ − 2 (0.03) 0.02 + 0.16 − 0.06 6


2

making the variance Var ( Z )  61 − 27  0.085034 . (C) 9.32. A¯ x is the actuarial present value of the insurance. Split it up into three time intervals: (0, 5) , (5, 12) , (12, ∞) . A¯ x  A¯ x1:5 + 5|7 A¯ x + 12| A¯ x

!

0.04 0.05 0.05  (1 − e −0.06(5) ) + e −0.06(5) (1 − e −0.07(7) ) + e −0.3−0.49 0.04 + 0.02 0.05 + 0.02 0.05 + 0.03  5  5  4  1 − e −0.3 + e −0.3 − e −0.79 + e −0.79 6 7 8  0.17279 + 0.20498 + 0.28365  0.66142 (D)

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9. INSURANCE: CONTINUOUS—MOMENTS—PART 1

196

Quiz Solutions 9-1. The actuarial present value of one unit of the 10-year term benefit is 1 A¯ 30 :10 

 0.01   µ  1 − e −10 ( µ+δ )  1 − e −0.6  0.075198 µ+δ 0.06

The actuarial present value of one unit of the pure endowment benefit is 1 A30:10  e −10 ( µ+δ )  e −0.6  0.548812

The actuarial present value for the contract is 1000 (0.075198) + 2000 (0.548812)  1172.82 .

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Lesson 10

Insurance: Continuous—Moments—Part 2 Reading: Models for Quantifying Risk (4th or 5th edition) 7.3

10.1

Uniform survival function

You can evaluate the present value integral for insurance in closed form when survival time has a uniform distribution with limiting age ω. Rather than using the expression t p x µ x+t in the integral, it is easier to use the density function f x ( t ) , which is constant 1/ ( ω−x ) , and then use equation (9.1) instead of equation (9.2). Example 10A A 20-year deferred insurance on (35) pays a benefit of 10 at the moment of death if death occurs no earlier than 20 years from now. You are given: • Mortality for (35) is uniformly distributed with ω  100. • δ  0.06 • Z is the random variable for the present value of the insurance. Calculate E ( Z ) and Var ( Z ) . Answer: The expected value is 65

Z E[Z] 

10e −0.06t 20

1 dt 65

Note that the density is 1/65 for t ≤ 65 and 0 for t > 65, so we set the upper bound of the integral equal to 65.

!

65

1 1 E[Z]  10 − e −0.06t 65 0.06 20  10  −20 (0.06) −65 (0.06)  e −e 3.9 10  (0.301194 − 0.020242)  0.72039 3.9 Z 65 1 E[Z 2 ]  102 e −0.12t dt 65 20 



!

  100 e −20 (0.12) − e −65 (0.12) 0.12 (65) 100  (0.090718 − 0.000410)  1.15780 7.8 Var ( Z )  1.15780 − 0.720392  0.63884 

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197



10. INSURANCE: CONTINUOUS—MOMENTS—PART 2

198

Table 10.1: Expected present value under constant force and uniform mortality for insurances payable at the moment of death

Type of insurance

Actuarial present value under constant force

Actuarial present value under uniform distribution

µ µ+δ

a¯ ω−x ω−x

 µ  1 − e −n ( µ+δ ) µ+δ

a¯ n ω−x

µ − ( µ+δ ) n e µ+δ

e −δn a¯ ω− ( x+n )

Whole life n-year term n-year deferred life

ω−x

 n-year pure endowment

e − ( µ+δ ) n

e −δn



ω − (x + n )



ω−x

If mortality is uniformly distributed, then expected insurance payments are level, at the rate of 1/ ( ω−x ) per year. It is therefore unnecessary to carry out integration. Under these assumptions, an insurance is equivalent to an annuity-certain of 1/ ( ω − x ) for the period of the insurance. To calculate the second moment of an insurance, double the force of interest in the annuity-certain and anywhere else it appears in the formula. The annuity-certain factor (1 − v n ) /δ becomes (1 − v 2n ) / (2δ ) . In the formulas for n-year deferred life and n-year pure endowment (see Table 10.1), e −δn becomes e −2δn . Table 10.1 summarizes results for standard insurances of 1 payable at the moment of death for constant force of mortality and uniform mortality assumptions. Example 10B A special whole life insurance on (30) pays 1000 at the moment of death if it occurs before age 60 and 500 at the moment of death if death occurs after age 60. You are given: • µ30+t  1/ (70 − t ) • δ  0.05 Let Z be the present value random variable for this insurance. Calculate Var ( Z ) . Answer: As in the previous lesson, we’ll split Z into 1000 units of a 30-year term insurance (Z1 ) and 500 units of a 30-year deferred whole life insurance (Z2 ). Since the two are mutually exclusive, E[Z 2 ]  E[Z12 ] + E[Z22 ]. Note that ω − x  70. For the 30-year term insurance: a¯30 1 − e −30 (0.05)   0.221963 70 70 (0.05) 2a¯ 1 − e −3 E[Z12 ]  30   0.135745 70 70 (0.1) E[Z1 ] 

For the 30-year deferred whole life insurance: e −30 (0.05) a¯40 e −1.5 − e −3.5   0.0551237 70 70 (0.05) e −30 (0.1) 2a¯ 40 e −3 − e −7 E[Z22 ]    0.0069822 70 70 (0.1) E[Z2 ] 

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10.2. OTHER MORTALITY FUNCTIONS

199

Combining 1000 units of term and 500 units of deferred: E[Z]  E[1000Z1 + 500Z2 ]  1000 (0.221963) + 500 (0.0551237)  249.525 E[Z 2 ]  10002 (0.135745) + 5002 (0.0069822)  137,490 Var ( Z )  137,490 − 249.5252  75,228

?



Quiz 10-1 A 20-year endowment insurance on (30) pays 1 at the moment of death or at age 50, whichever comes first. You are given • Mortality is uniformly distributed with ω  120. • δ  0.04 Calculate the expected present value of the endowment insurance.

10.2

Other mortality functions

Another mortality function that can be integrated, with technique, is a gamma distribution. Most other mortality functions can be integrated in closed form only if the benefit pattern is arranged to cancel out the objectionable parts of the mortality function, or if zero interest is assumed, or if the interest function is altered; for example, if simple interest is used. Using zero interest isn’t interesting (no pun intended), since then insurances reduce to probabilities; the only complication may be in varying benefits, as in the following example. Example 10C A special 10-year term life insurance on (55) pays a benefit of 1000 if death occurs in the first 5 years and 500 if death occurs in the second 5 years. The benefit is paid at the moment of death. You are given • µ55+t  2/ (50 − t ) , 0 < t < 50 • δ0 Calculate the expected present value of the insurance. Answer: With zero interest, the value of an insurance of 1 is the probability of death during the insured  period. The special 10-year insurance is 10005 q 55 + 5005|5 q 55 . For the given beta distribution, t p 55  (50 −

2

t ) /50 . So the expected present value of the insurance is

!2 !2 !2 45 + 45 * 40 + * / + 500 .1 − /  1000 (0.19) + 500 (0.17)  275 1000 .1 − 50 50 45 , , -

10.2.1



Integrating at n e −ct (Gamma Integrands)

Integrands of the form at n e −ct , a a constant, n usually an integer, and c a positive real number are unlikely to occur on the exam, so you may skip this material. You can integrate them by parts if n is an integer. If n  1, one integration by parts will be necessary, but if n  2, two integrations by parts will be necessary, and so on. We can factor out a, so let’s ignore it. Let’s first consider the easier case of integrating from 0 to ∞. An exponential distribution with mean θ has the following density function and moments: LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

10. INSURANCE: CONTINUOUS—MOMENTS—PART 2

200

• f (x ) 

1 −x/θ e . θ ∞

Z • E[X ]  k

xk 0

e −x/θ dx  k!θ k θ

Let c  1/θ. Then the integral for E[X n ] is c times the integral we’re interested in. So the integral we’re interested in is evaluated as Z ∞ n! t n e −ct dt  n+1 (10.1) c 0 Example 10D You are given: • Future lifetime for (40) has the following probability density function: f (t ) 

1 t 5 e −0.1t 120 × 106

• δ  0.06 Calculate A¯ 40 . Answer: The expected present value of the insurance is, by the general formula (9.1), 1 120 × 106

A¯ 40 

∞

Z

t 5 e −0.16t dt 0

where we’ve multiplied the density function by v t  e −0.06t . By formula (10.1), ∞

Z

t 5 e −0.16t dt  0

A¯ 40 

5! 0.166 1

(106 )(0.166 )

 0.059605



For the special cases of n  1 and n  2,

Z

∞

1 2 c Z 0∞ 2 t 2 e −ct dt  3 c 0 te −ct dt 

Now let’s consider formula for it is

R

u n −ct t e dt. 0

The most common case you will encounter is n  1, or u

Z

te −ct dt  0

 1  1 − (1 + cu ) e −cu 2 c

(10.2) (10.3)

R

u 0

te −ct dt. The (10.4)

As an alternative to memorizing it or rederiving it with integration by parts, you may have already memorized this formula as part of Financial Mathematics. The integral represents the present value of a continuously increasing annuity-certain for u years at force of interest δ, whose formula is

( I¯a¯ )u  LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

a¯ u − uv u δ

(10.5)

10.3. VARIANCE OF ENDOWMENT INSURANCE

10.3

201

Variance of endowment insurance

Any question using the techniques we’re about to discuss can also be done by calculating first and second moments. So you may skip this section if you wish. An endowment insurance is a sum of a term insurance and a pure endowment. If we let Z1 be the random variable for the term insurance, Z2 for the pure endowment, and Z3 for the endowment insurance, then Z3  Z1 + Z2 By the formula for the variance of a sum, Var ( Z3 )  Var ( Z1 ) + Var ( Z2 ) + 2 Cov ( Z1 , Z2 ) Since the term insurance and the pure endowment are mutually exclusive, E[Z1 Z2 ]  0, and therefore Cov ( Z1 , Z2 )  E[Z1 Z2 ] − E[Z1 ] E[Z2 ]  − E[Z1 ] E[Z2 ] Thus if we know the means and variances for the term insurance and the pure endowment, we can calculate the variance of the endowment insurance. A pure endowment’s present value only has two values—0 or the discounted value of the maturity benefit. To calculate the moments of a pure endowment, the Bernoulli shortcut is helpful. See Subsection 1.2.1. An n-year pure endowment Z2 has present value 0 or v n , with probabilities n q x and n p x respectively. Therefore, Var ( Z2 )  v 2n n q x n p x . Example 10E You are given: • A¯ x1:15  0.1 • 2A¯ x1:15  0.09 • 15 p x  0.9 • i  0.04 • Z is the present value random variable for the payment under a 15-year endowment insurance. Calculate Var ( Z ) . Answer: The mean of the term insurance is given as 0.1, and the variance of the term insurance is 0.09 − 0.12  0.08. Let Z2 be the present value random variable for the pure endowment. The mean of the pure endowment is 0.9 15 p x 1 E[Z2 ]  A x:15    0.499738 (1 + i ) 15 1.0415 As discussed above, the variance of the pure endowment is Var ( Z2 )  v 2 (15) 15p x 15q x 

(0.9)(0.1) 1.0430

 0.027749

So the variance of the endowment insurance is Var ( Z )  0.08 + 0.027749 − 2 (0.1)(0.499738)  0.007801



This method can be used to calculate the variance of a whole life insurance from the means and variances of an n-year term insurance and an n-year deferred insurance. We learned a different method for calculating the variance of an insurance broken down into two components, by using first and second moments, on page 177 (see Example 9B). Either method may be used. Example 10F Redo Example 10E using the method of Example 9B. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

10. INSURANCE: CONTINUOUS—MOMENTS—PART 2

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Answer: The second moment of the pure endowment can be computed directly as 15 p x

v 2 (15) 

0.9  0.277487 1.0430

Then E[Z]  E[Z1 ] + E[Z2 ]  0.1 + 0.499738  0.599738 E[Z 2 ]  E[Z12 ] + E[Z22 ]  0.09 + 0.277487  0.367487 Var ( Z )  0.367487 − 0.5997382  0.007801

?



Quiz 10-2 Z1 is the present value random variable for a 10-year term life insurance ( x ) , and Z2 is the present value random variable for a 10-year endowment insurance on ( x ) . Each insurance pays at the moment of death. You are given: • E[Z1 ]  0.0820 • E[Z2 ]  0.6518 • Var ( Z1 )  0.0625 • Var ( Z2 )  0.0158 Determine the continuously compounded interest rate δ used to compute these values.

10.4

Normal approximation

For a large group of insureds, the distribution of the present value of benefit payments can be approximated as a normal distribution with mean and variance as calculated by the methods of this lesson. When working out such problems, remember that the variance of the present value of benefits for a coverage of x on one insured is x 2 times the variance of the present value of benefits for a coverage of 1, but the variance of the present value of benefits for a coverage of 1 on n insureds is only n times the variance of the present value of benefits for a coverage of 1 on one insured. Example 10G A company sells an appliance with a 5-year warranty. The warranty pays 100 at the time of failure if the appliance fails within 5 years. You are given: • The force of failure is 0.1. • δ  0.05. • 250 appliances are sold at one time. Using the normal approximation, calculate the size of the fund needed at the time of sale in order to have a 95% probability of having sufficient funds to pay for failures under the warranty. Answer: Let the present value of payment random variable be Z. Then

!

  0.1 E[Z]  (100) 1 − e −5 (0.1+0.05) 0.1 + 0.05   200 200  1 − e −0.75  (1 − 0.472367)  35.1756 3 3 !     0.1 1 − e −5 (0.1+0.1) E[Z 2 ]  1002 0.1 + 0.1 



 5000 1 − e −1  5000 (1 − 0.367879)  3160.6028 Var ( Z )  3160.6028 − 35.17562  1923.2826 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

EXERCISES FOR LESSON 10

203

Table 10.2: Formula summary for this lesson

Moments for Insurances under Uniform Survival See Table 10.1 for a summary of expected present value formulas for standard insurances under uniform survival. To calculate the second moment, double the force of interest δ, which also squares v in the annuity-certain included in the formula. Gamma Integrands

Z

∞

n! n+1 δ 0 Z u  1  te −δt dt  2 1 − (1 + δu ) e −δu δ 0 a¯ u δ − uv u  δ t n e −δt dt 

(10.1) (10.4) (10.5)

Variance If Z3  Z1 + Z2 and Z1 , Z2 are mutually exclusive, then Var ( Z3 )  Var ( Z1 ) + Var ( Z2 ) − 2 E[Z1 ] E[Z2 ]. The 95th percentile of the standard normal distribution is 1.645. Therefore, the fund needed is

p

250 (35.1756) + 1.645 (250)(1923.2826)  9934.55 .



Exercises 10.1. [CAS4A-F99:15] (2 points) If l x  103 − x for 0 ≤ x ≤ 103, and the force of interest is δ  0.06, calculate A¯ 45:20 . A. B. C. D. E.

Less than 0.24 At least 0.24, but less than 0.30 At least 0.30, but less than 0.36 At least 0.36, but less than 0.42 At least 0.42

10.2. [CAS4A-S94:19] (2 points) You are given that a life is subject to a uniform survival function with terminal age 100 and force of interest δ  0.06. Calculate A¯ 1 . 40:10

A. B. C. D. E.

Less than 0.125 At least 0.125, but less than 0.126 At least 0.126, but less than 0.127 At least 0.127, but less than 0.128 At least 0.128

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10. INSURANCE: CONTINUOUS—MOMENTS—PART 2

204

10.3. [Based on CAS4A-S95:3] (2 points) The survival distribution is uniform with ω  80. The force of interest is δ  0.05. Determine the net single premium for a 5-year term insurance with benefit $1000 for a life age 40. A. B. C. D. E.

Less than $75 At least $75, but less than $85 At least $85, but less than $95 At least $95, but less than $105 At least $105 [150-S87:5] A continuous whole life insurance is issued to (50).

10.4.

Z is the present value random variable for this insurance. You are given: • Mortality follows a uniform distribution with ω  100. • Simple interest with i  0.01. • b t  1000 − 0.1t 2 Calculate E[Z]. A. 250

B. 375

C. 500

D. 625

E. 750

10.5. Mortality for (40) follows a uniform distribution with ω  100. The random variable Z represents the present value of a whole life insurance payable at the moment of death. δ  0.06. Calculate Var ( Z ) . 10.6. [CAS4A-S95:18] (2 points) You are given that the probability density function of future lifetime T for ( x ) is:   1/75 for 0 < t < 75 g (t )   0 otherwise

 The force of interest δ  0.05. Z is the present-value random variable for a whole life insurance of unit amount issued to ( x ) payable at the moment of death. Determine Var ( Z ) . A. B. C. D. E.

Less than 0.0640 At least 0.0640, but less than 0.0650 At least 0.0650, but less than 0.0660 At least 0.0660, but less than 0.0670 At least 0.0670

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EXERCISES FOR LESSON 10

205

10.7. [150-S89:11] Z is the present-value random variable for a special increasing whole life insurance with benefits payable at the moment of death of (50). You are given: b t  1 + 0.1t v t  (1 + 0.1t ) −2 t p 50 µ50+t  0.02

• • •

0 ≤ t < 50

Calculate Var ( Z ) . A. 0.01

B. 0.02

C. 0.03

D. 0.04

E. 0.05

[150-82-94:20] A special 3-year endowment insurance is issued on (60).

10.8.

The death benefit is 100 during policy year 1 and 200 thereafter. Death benefits are payable at the moment of death. The pure endowment benefit is 200. You are given: • Mortality follows a uniform distribution with ω  70. • i0 Calculate the variance of this insurance. A. 0

B. 900

C. 3,600

D. 6,500

E. 15,700

A special 3-year term insurance is issued on (60).

10.9.

The death benefit is 100 during policy year 1 and 200 thereafter. Death benefits are payable at the moment of death. You are given: • Mortality follows a uniform distribution with ω  70. • i0 Calculate the variance of this insurance. 10.10. [150-F97:20] For a 25-year term insurance of 1 on (25), you are given: • Benefits are payable at the moment of death. • Z is the present value random variable at issue of the benefit payment. • s ( x )  (100 − x ) /100, 0 ≤ x ≤ 100 • i  0.10 Calculate 1000 Var ( Z ) . A. 30

B. 47

C. 53

D. 87

E. 122

10.11. For a special whole life insurance policy on (55) with benefits payable at the moment of death, you are given • • •

b t  50/ (50 − t ) , t < 50. µ x  2/ (105 − x ) , x < 105. δ  0.04.

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10. INSURANCE: CONTINUOUS—MOMENTS—PART 2

206

10.12. You are given that the net single premium for a 20-year term insurance on a person currently age 40 with benefit of 1 payable at the moment of death is 0.065. You are also given • δ  0.04 • 20p 40  0.88 • p60  0.99 • Deaths occurring between ages 60 and 61 are uniformly distributed. Calculate the net single premium for a 21-year endowment insurance on a person currently age 40 with benefit of 1 payable at the moment of death. 10.13. A 5-year warranty on refrigerators pays 500 upon the breakdown of a refrigerator. There are 1000 refrigerators having this warranty. The probability that a refrigerator will not break down before time x is: e −x/25

0≤x≤3

!

20 − x −3/25 e 17

x>3

δ  0.05. Calculate the size of the initial fund needed to satisfy claims 95% of the time, using the normal approximation. 10.14. [CAS4A-S94:10] (2 points) 100 independent lives, each age 35, wish to establish a fund that will pay a benefit of 10 at the moment of death of each life. They will each contribute an equal amount at the inception of the fund and will then make no further payments to the fund. You are given: µ x  0.05 for x ≥ 0 δ  0.03 Using the normal approximation, determine the minimum amount that must be contributed to the fund by the group to have 95% confidence that all claims will be paid. A. B. C. D. E.

Less than 600 At least 600, but less than 625 At least 625, but less than 650 At least 650, but less than 675 At least 675

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EXERCISES FOR LESSON 10

207

10.15. [C3 Sample:1, CAS4A-F97:9] A manufacturer offers a warranty paying 1000 at the time of failure for each machine that fails within 5 years of purchase. One customer purchases 500 machines. The manufacturer establishes a fund for warranty claims from this customer. The manufacturer wants to have at least a 95% probability that the fund is sufficient to pay the warranty claims. You are given: •

The constant force of failure for each machine is µ  0.02.

•

The force of interest is δ  0.02.

•

The times until failure of the machines are independent. Using the normal approximation, determine the minimum size of the fund.

A. 27,900

B. 55,600

C. 78,200

D. 86,400

E. 90,600

10.16. A 5-year endowment insurance of 1000 is payable at the moment of death. You are given: • •

δ  0.05 µ x+t  0.03 for all t

An initial fund of 800 per insured is set up. Using the normal approximation, calculate the number of insureds needed so that there is a 95% chance that the fund is sufficient. 10.17. [3-S00:13] An investment fund is established to provide benefits on 400 independent lives age x. • On January 1, 2001, each life is issued a 10-year deferred whole life insurance of 1000, payable at the moment of death. • Each life is subject to a constant force of mortality of 0.05. • The force of interest is 0.07. Calculate the amount needed in the investment fund on January 1, 2001, so that the probability, as determined by the normal approximation, is 0.95 that the fund will be sufficient to provide these benefits. A. 55,300

B. 56,400

C. 58,500

D. 59,300

E. 60,100

10.18. [CAS3-F04:4] A fund will pay death benefits of $10,000 on each of 900 independent lives age 30. You are given: •

δ  0.04

•

µ  0.01

•

The death benefits are payable at the moment of death.

The initial amount of the fund is established so that the probability is 0.95 that sufficient funds will be on hand to withdraw the benefit payment at the death of each individual. Calculate the initial fund amount. A. B. C. D. E.

Less than $1.90 million At least $1.90 million, but less than $1.95 million At least $1.95 million, but less than $2.00 million At least $2.00 million, but less than $2.05 million At least $2.05 million

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10. INSURANCE: CONTINUOUS—MOMENTS—PART 2

208

Use the following information for questions 10.19 and 10.20: A builder offers lifetime-of-the-house fire protection on houses that he builds. The force of failure due to fire is a constant equal to 0.0005. • δ  0.10. • Any fire loss will be a total loss and coverage remains in effect regardless of a change in ownership. • There are no other causes of loss. • Payments are made at the time of the fire. 10.19. [CAS4-S88:19] (2 points) Calculate the expected present value of this benefit for a $100,000 house. A. B. C. D. E.

Less than $350 At least $350, but less than $400 At least $400, but less than $450 At least $450, but less than $500 At least $500

10.20. [CAS4-S88:20] (2 points) The builder has 150 $100,000 houses on which he plans to offer this protection. In addition to putting the expected present value of the benefit in a reserve, the builder will set aside additional funds so that the probability is 0.95 that the funds will cover all losses. Calculate the amount of additional funds needed for each house. A. B. C. D. E.

Less than $600 At least $600, but less than $800 At least $800, but less than $1000 At least $1000, but less than $1200 At least $1200

10.21. [3-F01:8] Each of 100 independent lives purchase a single premium 5-year deferred whole life insurance of 10 payable at the moment of death. You are given: • • •

µ  0.04 δ  0.06 F is the aggregate amount the insurer receives from the 100 lives.

Using the normal approximation, calculate F such that the probability the insurer has sufficient funds to pay all claims is 0.95. A. 280

B. 390

C. 500

D. 610

E. 720

10.22. [CAS4A-F92:4] (1 point) Let Z be the present value random variable for a 10-year term insurance with a benefit of 10 payable at the moment of death of (30). You are given that E[Z]  5 and Var ( Z )  375. Determine 2A¯ 1 . 30:10

A. B. C. D. E.

Less than 3.6 At least 3.6, but less than 3.7 At least 3.7, but less than 3.8 At least 3.8, but less than 3.9 At least 3.9

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EXERCISES FOR LESSON 10

10.23.

209

Z1 is the present value of a 10-year term insurance payable at the moment of death.

Z2 is the present value of a 10-year deferred insurance payable at the moment of death. Z3 is the present value of a whole life insurance payable at the moment of death. You are given: • µ x+t  0.01 for 0 < t ≤ 10. • δ  0.04. • E[Z2 ]  0.35. • Var ( Z2 )  0.23. Calculate Var ( Z3 ) . 10.24. [CAS3-F03:6] Let Z1 be the present value random variable for an n-year term insurance of 1 on ( x ) , and let Z2 be the present value random variable for an n-year endowment insurance of 1 on ( x ) . Claims are payable at the moment of death. Given: • v n  0.250 • n p x  0.400 • E[Z2 ]  0.400 • Var ( Z2 )  0.055 Calculate Var ( Z1 ) . A. 0.025

B. 0.100

C. 0.115

D. 0.190

E. 0.215

10.25. Let X be the present value random variable for an n-year term insurance with unit face amount. Let Y be the present value random variable for an n-year endowment insurance with unit face amount. Let Z be the present value random variable for an n-year pure endowment with unit face amount. You are given: • • • •

E[X]  0.04 Var ( X )  0.12 E[Y]  0.65 Var ( Y )  0.08

Determine Var ( Z ) . Additional old CAS Exam 3/3L questions: S08:21, S09:11 Additional old SOA Exam MLC questions: S13:8

Solutions 10.1. Recognize this as uniform survival with ω  103. As we mentioned, a (term or whole life) insurance then equals a continuous annuity-certain divided by length of time to ω, here 103 − 45  58. For an endowment insurance, we must add a pure endowment. a¯ A¯ 45:20  20 + 20 E45 58 1 − e −1.2 38e −1.2  + (58)(0.06) 58 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

10. INSURANCE: CONTINUOUS—MOMENTS—PART 2

210

 0.200806 + 0.197334  0.398140 10.2.

(D)

Time to ω is 100 − 40  60. a¯10 60 1 − e −0.6  (60)(0.06)

1 A¯ 40 :10 

 0.125330 10.3.

(B)

Time to ω is 80 − 40  40. a¯5 1 1000A¯ 40 :5  1000 40 ! 1 − e −0.25  1000 (40)(0.05)  110.60

(E)

R

10.4. We use the general formula for an insurance, b t v t fT ( t ) dt, where v t is the appropriate discounting factor for time t. Simple interest means money accumulates linearly so that 1 becomes 1 + it at time t. Therefore with 1% simple interest, v t  1/ (1 + 0.01t ) . The density function is a constant fT ( t )  1/50. 50

Z E[Z]  0



1000 − 0.1t 2 1 dt 1 + 0.01t 50

1000 50

Z  20

!

50

Z 0 50

1 − 0.0001t 2 dt 1 + 0.01t

(1 − 0.01t ) dt 0

502 + /  750 2

!

 20 .50 − 0.01

* ,

10.5.

(E)

-

Time to ω is 100 − 40  60. The mean of Z is a¯60 0.06 60 1 − e −3.6   0.270188 (60)(0.06)

E[Z] 

and the second moment is computed the same way but doubling the force of interest: a¯60 0.12 60 1 − e −7.2   0.138785 (60)(0.12)

E[Z 2 ] 

So Var ( Z )  0.138785 − 0.2701882  0.06578 .

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EXERCISE SOLUTIONS FOR LESSON 10

10.6.

211

Time to ω is 75. We calculate the first 2 moments. The second moment is calculated using 2δ. a¯75 0.05 75 1 − e −0.05 (75)  0.260395  75 (0.05) a¯ E[Z 2 ]  75 0.10 75 1 − e −0.10 (75)   0.133260 75 (0.10) E[Z] 

Var ( Z )  0.133260 − 0.2603952  0.06545

(C)

Does the exercise look familiar? It is the continuous version of exercise 8.31, page 156. The original exam question did not specify whether the insurance paid at the moment of death or at the end of the year of death, so the CAS had to give credit for both A and C. 10.7.

The density function is constant, 0.02. The mean is 50

Z E[Z]  0.02 0

1 dt 1 + 0.1t

50 0.02  ln (1 + 0.1t ) 0.1 0  0.2 ln 6 The second moment is

f

Z

g

50

E Z 2  0.02

1

dt

(1 + 0.1t ) 2 0 ! ! 50 0.02 1 − 0.1 1 + 0.1t 0    0.2 1 −

Var ( Z ) 

1 1  6 6

1 − (0.2 ln 6) 2  0.03825 6

(D)

10.8. Because interest is zero, there are only two possible present values: 100 if death occurs in the first year and 200 otherwise. So the present value random variable is a Bernoulli variable—a variable which can only assume one of two values. By the Bernoulli shortcut (Subsection 1.2.1), the variance of such a variable is the product of the probabilities of each of the two values, times the square of the difference 1 , and the probability of not dying in the between the values. The probability of dying in the first year is 10 9 first year is 10 . So the variance is 1 9 (1002 ) 10 (B) 10  900 10.9. Unlike the previous question, the Bernoulli shortcut cannot be used, since there are three possible values for the present value: 0 (after year 3), 100 (year 1), and 200 (years 2 and 3). So we calculate the first and second moments. We have (calling the term insurance Z) Pr ( Z  0)  3 p60  0.7 Pr ( Z  100)  q60  0.1 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

10. INSURANCE: CONTINUOUS—MOMENTS—PART 2

212

Pr ( Z  200)  1|2 q 60  0.2 Therefore, E[Z]  0.1 (100) + 0.2 (200)  50 E[Z 2 ]  0.1 (1002 ) + 0.2 (2002 )  9000 Var ( Z )  9000 − 502  6500 10.10. Age at death is uniformly distributed with ω  100, and time to ω at age 25 is 75. δ  ln (1 + i )  ln 1.1. Therefore a¯ E[Z]  25 ln 1.1 75   1 1  1− 75 (0.095310) 1.125  (0.139894)(0.907704)  0.126982 a¯ E[Z 2 ]  25 2 ln 1.1 75   1 1 1−  150 ln 1.1 1.150  (0.069947)(0.991481)  0.069351 1000 Var ( Z )  1000 (0.069351 − 0.1269822 )  53.2267

(C)

10.11. This is a beta with ω  105 and α  2, so 50 − t 50

t p 55  50

Z

Z

50

50 50 − t e −0.04t 50 − t 50

Z

50

!

b t v t t p x µ x+t dt  0

0

1  25 

!2 !2

!

2 dt 50 − t

e −0.04t dt 0

1

(25)(0.04)

(1 − e −2 )  0.864665

10.12. Decompose the 21-year endowment insurance: 1 ¯1 A¯ 40:21  A¯ 40 :20 + 20 E 40 A60:1 + 21 E 40

For the second component, note that the probability density function over age 60 is q60  0.01, so 1 A¯ 60 :1 

1

Z

0.01e −0.04t dt 0

1 − e −0.04  0.009803 0.04

!

 0.01

Also, the pure endowments are 20 E 40  e −0.04 (20) (0.88)  0.395409 and 0.376106. The net single premium for a 21-endowment insurance is

21 E 40

 e −0.04 (21) (0.88)(0.99) 

A¯ 40:21  0.065 + (0.395409)(0.009803) + 0.376106  0.444982

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EXERCISE SOLUTIONS FOR LESSON 10

213

10.13. Let Z be the value of 1 for one refrigerator that breaks down. Note that the density is constant, e −0.12 17 , between 3 and 5. Although it’s worked out from first principles below, you could use exponential shortcuts for [0, 3] and uniform shortcuts for [3, 5]. 3

Z E[Z] 

5

Z e − (0.04+0.05) t 0.04 dt +

e −0.12 dt 17

e −0.05t

0

3

4 20  (1 − e −0.27 ) + e −0.12 ( e −0.15 − e −0.25 ) 9 17  0.105165 + 0.085465  0.19063 3

Z E[Z 2 ] 

5

Z e − (0.04+0.10) t 0.04 dt +

e −0.12 dt 17

e −0.10t

0

3

4 10  (1 − e −0.42 ) + e −0.12 ( e −0.30 − e −0.50 ) 14 17  0.097987 + 0.070060  0.168047 Var ( Z )  0.168047 − 0.190632  0.131707 For 1000 refrigerators and 500 per refrigerator, the mean is 1000 (500)(0.19063)  95,315 and the variance is

1000 (5002 )(0.131707)  32,926,804

so the fund needed for 95% confidence is

p

95,315 + 1.645 32,926,804  104,754 10.14. The expected value of the payments is 100 (10)

µ 5000   625 µ+δ 8

and the variance is

!2

100 (102 ) .

*

!2

µ µ +/  10,000 * 5 − 5 +  639.2045 − µ + 2δ µ+δ 11 8

,

-

,

-

So the fund needed for 95% confidence is √ 625 + 1.645 639.2045  666.59

(D)

10.15. Let Z be the expected value of 1 at the time of failure of one machine. 0.02 (1 − e −5(0.04) )  0.090635 0.02 + 0.02 0.02 E[Z 2 ]  (1 − e −5(0.06) )  0.086394 0.02 + 2 (0.02) Var ( Z )  0.086394 − 0.0906352  0.078179 E[Z] 

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214

The expected value of 1000 for 500 machines is 1000 (500)(0.090635)  45,317.5 and the variance is

10002 (500)(0.078179)  39,089,500

so the fund needed for 95% confidence is

p

45,317.5 + 1.645 39,089,500  55,602

(B)

10.16. Let Z be the present value of a 5-year endowment insurance of 1. We will calculate the first moment, the second moment, and then the variance.

  0.03 1 − e −5 (0.03+0.05) + e −5 (0.03+0.05) 0.03 + 0.05  3  1 − e −0.4 + e −0.4  0.793950 8   0.03 1 − e −5[0.03+2 (0.05) ] + e −5[0.03+2 (0.05) ] E[Z 2 ]  0.03 + 2 (0.05)  3   1 − e −0.65 + e −0.65  0.632343 13 Var ( Z )  0.632343 − 0.7939502  0.001986 E[Z] 

The fund needed for 95% confidence for n insureds is √

(1000n )(0.793950) + 1.645 (1000) 0.001986n Set this equal to 800 per insured. For one insured, it is 793.950 + 73.3087 n 800 − 793.950

73.3087  800 √ n

!2  12.11722  146.825

147 insureds are needed for 95% confidence. 10.17. Let Z be the present value of a 10-year deferred insurance of 1. 0.05 e −10 (0.12)  0.125498 0.05 + 0.07 0.05 e −10 (0.19)  0.039360 E[Z 2 ]  0.05 + 2 (0.07) Var ( Z )  0.039360 − 0.1254982  0.023611 E[Z] 

The fund for 95% confidence is

p (400)(1000)(0.125498) + 1.645 (1000) (400)(0.023611)  50,199 + 5,055  55,254

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(A)

EXERCISE SOLUTIONS FOR LESSON 10

215

10.18. Let Z be the present value of insurance of 1 for 1 life. 0.01  0.2 0.01 + 0.04 1 0.01 2¯  A30  0.01 + 2 (0.04) 9 1 Var ( Z )  − 0.22  0.071111 9 A¯ 30 

If S is the present value of aggregate death benefits, E[S]  900 (10,000)(0.2)  1,800,000 Var ( S )  900 (108 )(0.071111)  64 × 108

p

Var ( S )  80,000

So the answer is 1,800,000 + 1.645 (80,000)  1,931,600 . (B) 10.19. The expected present value is µ 0.0005 $100,000  $100,000  $497.51 µ+δ 0.1005

!

!

(D)

10.20. The variance for a single house is

!2

µ µ +/  24,690,137 − 10 . µ + 2δ µ+δ 10 *

-

,

For 150 houses, the variance is 150 times this. The additional funds needed for all 150 houses equal the 1.645 times the square root of the variance for 150 houses, or

p

1.645 150 × 24,690,137  100,109 Per house, this comes out to 100,109/150  667.39 . (B) 10.21. Let Z be the present value of the insurance for each life. The moments for each life is

!

0.04 E[Z]  10 e −0.1 (5) 0.04 + 0.06  4e −0.5  2.426123

!

E[Z 2 ]  102

0.04 e −0.16 (5) 0.04 + 2 (0.06)

 25e −0.8  11.233224 Var ( Z )  11.233224 − 2.4261232  5.347153 Using the normal approximation, the insurer would need

p

100 (2.426123) + 1.645 100 (5.347153)  280.6511

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(A)

10. INSURANCE: CONTINUOUS—MOMENTS—PART 2

216

10.22. 375 5 1 A¯ 30 + :10  100 10

!2

2

 4.0

(E)

10.23. 0.01 (1 − e −10(0.05) )  0.078694 0.01 + 0.04 0.01 E[Z12 ]  (1 − e −10(0.09) )  0.065937 0.01 + 0.08 Var ( Z1 )  0.065937 − 0.0786942  0.059744 E[Z1 ] 

Cov ( Z1 , Z2 )  E[Z1 Z2 ] − E[Z1 ] E[Z2 ]  0 − (0.078694)(0.35)  −0.027540 Var ( Z3 )  Var ( Z1 ) + Var ( Z2 ) + 2 Cov ( Z1 , Z2 )  0.059744 + 0.23 + 2 (−0.027540)  0.2347 10.24. Let Z3 be an n-year pure endowment. Then we have, as discussed in Section 10.3, Var ( Z2 )  Var ( Z1 ) + Var ( Z3 ) − 2 E[Z1 ] E[Z3 ] Z3 is a Bernoulli-type variable, with mean n p x v n  (0.400)(0.250)  0.1 and variance n p x (1

− n p x ) v 2n  (0.400)(0.600)(0.2502 )  0.015.

Since Z2  Z1 + Z3 , E[Z1 ]  E[Z2 ] − E[Z3 ]  0.4 − 0.1  0.3. So we have 0.055  Var ( Z1 ) + 0.015 − 2 (0.3)(0.1) Var ( Z1 )  0.055 − 0.015 + 2 (0.3)(0.1)  0.1

(B)

10.25. Since X and Z are mutually exclusive and Y  X + Z, E[Z]  E[Y] − E[X]  0.65 − 0.04  0.61 Var ( Y )  Var ( X ) + Var ( Z ) + 2 Cov ( X, Z )  Var ( X ) + Var ( Z ) − 2 E[X] E[Z] 0.08  0.12 + Var ( Z ) − 2 (0.04)(0.61) Var ( Z )  0.08 + 2 (0.04)(0.61) − 0.12  0.0088

Quiz Solutions 10-1.

Add up a term insurance and a pure endowment. Note that ω − x  120 − 30  90. a¯20 1 − e −0.04 (20) 1 A¯ 30   0.152964 :20  90 90 (0.04) ! 70 1 A¯ 30:20  e −0.04 (20)  0.349478 90 A¯  0.152964 + 0.349478  0.5024 30:20

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QUIZ SOLUTIONS FOR LESSON 10

10-2.

217

Let Z3 be a 10-year pure endowment. Then E[Z3 ]  E[Z2 ] − E[Z1 ]  0.6518 − 0.0820  0.5698  v 10 10 p x

(*)

and Var ( Z2 )  Var ( Z1 ) + Var ( Z3 ) − 2 E[Z1 ] E[Z3 ] 0.0158  0.0625 + Var ( Z3 ) − 2 (0.0820)(0.5698) Var ( Z3 )  0.0158 − 0.0625 + 2 (0.0820)(0.5698)  0.046747  v 20 10 p x (1 − 10 p x ) We divide (**) by (*) to get v 10 (1 − 10 p x ) 

0.046747  0.082041 0.5698

Adding this result to (*), we get v 10  0.082041 + 0.5698  0.651841 ln 0.651841  0.04280 δ  − ln v  − 10

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(**)

218

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10. INSURANCE: CONTINUOUS—MOMENTS—PART 2

Lesson 11

Insurance: Probabilities and Percentiles Reading: Models for Quantifying Risk (4th or 5th edition) 7.3 Whenever you deal with a random variable, you may study the probability that the random variable is greater than or less than something, and you may study percentiles of the random variable. In this course, three types of random variables that we discuss are 1. Present value of insurance benefits 2. Present value of annuity payments 3. Present value of future loss We will discuss present values of insurances here.

11.1

Introduction

Since students sometimes have difficulty with probabilities, let’s discuss simple examples. Whenever you have a random variable X and a function g ( x ) , you can create a new random variable Y  g ( X ) . You can then study the probability of Y being in a range. Usually this will involve applying the inverse function g −1 ( y ) to the range and studying the probability of that inverse range under the original distribution. (Let’s not immediately be concerned about whether a well-defined inverse exists.) Let’s do a few examples to illustrate the point. Example 11A You invest 10,000. The compound annual effective rate of return over a five year period is uniformly distributed on [0.01, 0.09]. Calculate the probability that at the end of 5 years you will have at least 13,000. Answer: The higher the rate of return, the more you will have. Therefore, we need to identify the minimum rate of return resulting in 13,000. If I is the interest rate, then the amount of money you have at the end of five years is 10,000 (1 + I ) 5 . Set this equal to 13,000 and solve for I. 10,000 (1 + I ) 5  13,000

(1 + I ) 5  1.3 I

√5

1.3 − 1  0.053874

So you have at least 13,000 if and only if I ≥ 0.053874. What is the probability of that? I is uniform on [0.01, 0.09], so the probability of that is Pr ( I ≥ 0.053874)  LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

0.09 − 0.053874  0.4516 0.09 − 0.01 219



11. INSURANCE: PROBABILITIES AND PERCENTILES

220

In this example, the transforming function was y  g ( i )  (1 + i ) 5 . That is a one-to-one monotonically increasing function. Higher values of y correspond to higher values of i. To determine the probability that y is higher than something, you determine the probability that i is greater than f -inverse of the something. Now consider the case of a monotonically decreasing function. Example 11B The price per ounce of gold at the end of one year is uniformly distributed on [1000, 3000]. You will have 5000 at the end of one year. Calculate the probability that you will have enough money to buy at least 2 ounces of gold. Answer: While the answer is fairly obvious, let’s go through the steps anyway. The higher the price, the fewer ounces of gold that you will be able to buy. Let’s identify the price that will allow you to buy 2 ounces; then we calculate the probability that the actual price is lower than this price. Since you will have 5000, the price of gold allowing you to buy 2 ounces is 2500. 2500−1000 The probability that the price of gold is 2500 or less, under the uniform distribution, is 3000−1000  0.75 .  In this example, the transforming function was y  5000/x. When x ≤ 2500, then y ≥ 2. When the transforming function is one-to-one monotonically decreasing, then higher values of y correspond to lower values of x. Most of the functions we’ll deal with are monotonic, which makes them simple. A somewhat more complicated case that we’ll deal with is a function like the following: Example 11C You play a gamble. A random number uniformly distributed between 0 and 30 inclusive is picked. If the number is 10 or lower, you pay 10. If the number is greater than 10, you receive 25 minus the number; if the result is negative, it represents an amount that you pay. For example, if the number is 15, you receive 10, while if the number is 27, you pay 2. Calculate the probability that you will receive at least 5. Answer: If the number is in the interval [0, 10], you receive nothing. Otherwise we want to know when X − 25 (where X is the uniform random number) is less than or equal to −5. This happens when X ≤ 20. So to receive at least 5, the random number must be in the interval (10, 20]. The probability of X being in that interval of length 10 is 10/30  1/3 .  The complication in this example is that the function is monotonic on [0, 10]—in fact, it’s constant—and then jumps and is monotonically decreasing afterwards. We had to take care of the [0, 10] range separately, and then find the number between 10 and 30 for which the gain is greater than 5. That number, and anything lower, but not lower than 10, is the range we need the probability of. If you understand these three examples, you’re ready for almost any probability question on this exam.

11.2

Probabilities for Continuous Insurance Variables

If Z is the present value random variable for an insurance payable at the moment of death, it is continuous. It is a function of Tx . Moreover, for many insurances, it is monotonically decreasing: the later one dies, the lower the present value of the insurance benefit. Thus a lower Z will correspond to a higher Tx . Let’s look at graphs of common insurances. See Figure 11.1. Whole life insurance is monotonically decreasing. The transforming function is FZ ( z )  v z . Higher Z’s correspond to lower Tx ’s. Suppose you wanted to know the probability that the present value of a whole life insurance Z is greater than 0.6. You would draw Figure 11.1a, start at Z  0.6, and move to the right until you hit the graph, which would happen at the point ( t, v t ) for some t. But since v t  0.6, it follows that t  (ln 0.6) / ln v. Since Z is above 0.6 only to the left of this point, the probability of Z > 0.6 is the probability that Tx < (ln 0.6) / ln v, or (ln 0.6)/ ln v q x . LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

11.2. PROBABILITIES FOR CONTINUOUS INSURANCE VARIABLES

Z

221

Z

1

1

Z = vT

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2 10

20

30

40

50

Z=

T

10

(a) Whole life



20

vT 0

30

T ≤ 20 T > 20

40

50

T

(b) 20-year term

Z

Z

1

Z=

0.8



vT v 20

1

T ≤ 20 T > 20

0.6

0.6

0.4

0.4

0.2

0.2 10

20

30

40

(c) 20-year endowment insurance

Z=

0.8

50

T

10

20



0 vT

30

T ≤ 20 T > 20

40

50

T

(d) 20-year deferred whole life

Figure 11.1: Relation of present value random variable to survival time random variable. The graphs were drawn with δ  0.04.

Term and endowment insurances are also monotonically decreasing, so the same graphic technique would work for term and endowment insurance, and in fact would give the same answer. Notice that if you wanted to know the probability that a term insurance is greater than 0.4, and you move right along the line z  0.4, since you hit the graph at t  20, the answer is 20q x . If you wanted to know the probability that an endowment insurance is greater than 0.4, since Z is above 0.4 at all points, the probability is 1. A deferred insurance is like Example 11C. We have to handle the deferral period separately. Suppose you wanted to know the probability that a deferred whole life insurance is greater than 0.4. As you can see from Figure 11.1d, Z is above 0.4 starting at Tx  20 up to a point slightly higher than 20. It follows that the required probability is the probability of Tx being greater than 20 but less than t  (ln 0.4) / ln v, or 20p x − t p x . Conversely, if you want to calculate the probability that Z < 0.4, you automatically get the deferral period, and you also get the late period beyond the point at which v t  0.4, so Pr ( Z < 0.4)  20q x + t p x . Example 11D For a life age 30: • Z is the present value random variable for a 10-year deferred whole life insurance of 1 payable at the moment of death. • µ30+t  1/ (70 − t ) , 0 < t < 70 • δ  0.06 Calculate Pr ( Z < 0.25) .

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11. INSURANCE: PROBABILITIES AND PERCENTILES

222

Answer: Use a graph like Figure 11.1d, which may not be scaled correctly for this v and deferral period but gives us the general idea. We need to include the first 10-year period plus an ultimate period in which Z < 0.25, or v T30 < 0.25, or T30 > − (ln 0.25) /δ  − (ln 0.25) /0.06  23.105. So Pr ( Z < 0.25)  10q 30 + 23.105p 30 . Mortality is uniformly distributed with ω − 30  70, so Pr ( Z < 0.25)  10q 30 + 23.105p 30 

10 46.895 +  0.8128 70 70



If the force of mortality is constant, you can skip the logging of the bound, since you end up exponentiating it when calculating the mortality probabilities, as the next example demonstrates. Example 11E For a life age 30: • Z is the present value random variable for a 10-year deferred whole life insurance of 1 payable at the moment of death. • µ30+t  0.01, t > 0 • δ  0.06 Calculate Pr ( Z < 0.25) . Answer: Once again we want the sum of 10q x and Pr ( v T30 < 0.25) . The latter is



Pr ( e −0.06T30 < 0.25)  Pr T30 > − (ln 0.25) /0.06



 e (0.01 ln 0.25)/0.06  0.251/6  0.793701 Adding 10q x  1 − e −0.1 , the final answer is Pr ( Z < 0.25)  1 − e −0.1 + 0.793701  0.8889 .



By similar reasoning, if Z is the random variable for whole life and the force of mortality is the constant µ, then Pr ( Z < z )  z µ/δ . In particular, the probability that Z is less than its mean is µ/ ( µ + δ )



?

 µ/δ

.

Quiz 11-1 For a life age 60: • Z is the present value random variable for a 10-year deferred 10-year term life insurance of 1 payable at the moment of death.

.



• µ60+t  1 2 (50 − t ) , 0 < t < 50 • δ  0.04 Calculate Pr ( Z > 0.4) .

11.3

Probabilities for Discrete Variables

Present value random variables for insurances payable at the end of the year of death are discrete random variables. They jump in value at every integer. Calculating probabilities for them involves logic similar to continuous calculations, but you need to determine the integer corresponding to the real number you’re working with. Example 11F For a 30-year term life insurance on (60) paying a benefit at the end of the year of death: • Z is the present value random variable for the insurance. • Mortality is uniformly distributed with ω  100. • i  0.05 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

11.4. PERCENTILES

223

Calculate the probability that Z is greater than its mean. Answer: First we have to calculate E[Z]. Without loss of generality, we can set the benefit equal to 1. Note that ω − x  100 − 60  40. a 1 − (1/1.05) 30 E[Z]  30   0.384311 40 40 (0.05) Since the function is monotonically decreasing, Z less than its mean means T60 is greater than something. Let’s first solve the question for a continuous insurance, one that pays at the moment of death. Then we want 1 1.05

! T60 > 0.384311 T60 < −

ln 0.384311  19.60 ln 1.05

For our discrete insurance, if death occurs after the 19th anniversary, the present value of the payment will be v 20 , which is less than v 19.60 . So Z is greater than its mean only if death occurs within the first 19 years. The probability of that is 19q 60  19/40  0.475 .  Although I doubt the exam will involve itself in the subtlety of strict versus non-strict inequalities, here’s an example where the strictness of the inequality matters. Example 11G For a whole life insurance on (40) paying 1 at the end of the year of death: • Z is the present value random variable for the insurance. • µ40+t  0.05 • v  0.9 Calculate Pr ( Z > 0.6561) . Answer: Notice that 0.6561  v 4 , so Z > 0.6561 if and only if it is discounted for less than 4 years. Thus death must occur within 3 years, and Pr ( Z > 0.6561)  3 q40  1 − e −0.05 (3)  0.1393 . If you had been asked for Pr ( Z ≥ 0.6561) , then a death in the 4th year would also count since Z  0.6561 in that year, and the answer would be 4 q 40  1 − e −0.05 (4)  0.1813. In fact, Pr ( Z  0.6561)  0.1813 − 0.1393  0.0420. 

?

Quiz 11-2 For a 10-year deferred whole life insurance on a life currently age x payable at the end of the year of death, you are given • Z is the present value random variable for the insurance. √ • t p x  (60 − t ) /60, t < 60 • v  0.9 Calculate Pr ( Z < 0.1) .

11.4

Percentiles

Percentiles are the inverses of probabilities. For a continuous random variable, the 100p th percentile is the number π p such that the probability that the random variable is less than or equal to π p is p. We won’t deal with percentiles of discrete variables, which aren’t always well-defined. We will only discuss insurances payable at the moment of death. For an insurance random variable that is monotonically decreasing, higher percentiles correspond to lower survival times. To find the 100p th percentile, find the 100 (1−p ) th percentile of Tx . In other words, set LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

11. INSURANCE: PROBABILITIES AND PERCENTILES

224

 1− p and solve for t. Then, the percentile of the insurance random variable will be v t . However, if the resulting t is greater than the period of the insurance—for example, past the maturity of an endowment insurance or the expiry of a term insurance—then the percentile is 0 for term and v n for endowment insurance. To repeat, for a level benefit whole life insurance, the 100p th percentile of the present value is based on the 100 (1 − p ) th percentile of the time to death random variable. So if Z is the present value random variable for a whole life insurance on ( x ) and you wanted to calculate the 30th percentile of Z, you would determine A¯ x+t for a certain t. What percentile of the survival distribution would t be? 1 A geometric approach is to use graphs like Figure 11.1. In each graph, observe where Z attains the bottom p of its values, and define the boundary. The height of that boundary point ( t, v t ) is the 100p th percentile of Z. tqx

Example 11H For a life age 30: • µ30+t  0.0001t 2 • δ  0.05 • Z is the present value random variable for a 35-year endowment insurance of 1 on (30) . Calculate the 40th percentile of Z. Answer: The 40th percentile of Z corresponds to the 60th percentile of T30 . We calculate t p 30 : t

Z t p 30  exp −

! 0.0001u 2 du  exp −

0

0.0001t 3 3

!

The t such that t p 30  0.40 is the 60th percentile of T30 , since there is a 40% chance of surviving t years therefore a 60% chance of not surviving that long. So 0.0001t 3 exp −  0.4 3

!

0.0001t 3  − ln 0.4  0.916291 3 t 3  27488.72 t  30.1799 The corresponding value of Z is e −0.05 (30.1799)  0.2211 . If t had been 35 or greater, the answer would have been e −0.05 (35) .



As we mentioned for probabilities, if the force of mortality is constant, you do not have to log v, since you’ll end up exponentiating it when calculating v T . We also found that for whole life, Pr ( Z < z )  z µ/δ . By setting this equal to p, we can quickly calculate the percentile. This also works for n-year term and endowment insurance if the resulting time is in the n-year period, but it may be necessary to log to verify this. In the next example, we illustrate this. Example 11I For a life age 30, • µ30+t  0.02 • δ  0.06 • Z is the present value random variable for a 20-year term insurance of 1 on (30) payable at the moment of death. 1. Determine the median of Z. The 70th percentile

1

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225

Z 0.5 0.4 (t , v t )

0.3 0.2 0.1 10

t

20

30

40

50

T

Figure 11.2: Deferred insurance example, example 11J

2. Determine the 80th percentile of Z. Answer: 1. Using the shortcut, z 0.02/0.06  0.5, so z  0.53  0.125. The corresponding time is e −0.06t  0.125, or t  − (ln 0.125) /0.06  34.66 > 20, so the median is 0 . 2. This time, Z  0.83  0.512 , and − (ln 0.512) /0.06  11.16 < 20, so 0.512 is really the 80th percentile.

 A trickier case is when Z is not a monotonic function of Tx . For example, for a deferred insurance, the benefit is worth 0 if Tx is less than the deferral period, but is positive if Tx is greater. If the deferral period is n, to calculate the 100p th percentile, you must consider the point x > n such that the probability that Tx is between x and n is 1 − p. This is illustrated in Figure 11.1d. Example 11J For a life currently age 30, • µ30+t  0.0001t 2 • δ  0.05 • Z is the present value random variable for a 20-year deferred insurance of 1 on (30) . (µ30+t and δ are the same as in Example 11H.) Calculate the 80th percentile of Z Answer: We have to calculate t > 20 such that Pr (20 ≤ T30 ≤ t )  0.20. This is illustrated in Figure 11.2, where t is chosen so that the interval between 20 and t is the interval on which Z realizes the highest 20% of its possible values. The 80th percentile of Z is then v t . To compute t, we need to make Pr (20 ≤ T30 ≤ t )  0.2, or 20p 30 − t p 30  0.2. 0.0001 (203 )  exp −  e −0.266667  0.765928 3

!

20p 30

So we want t such that t p 30  0.565928. 0.0001t 3  0.565928 3

!

exp −

0.0001t 3  − ln 0.565928  0.569290 3 √3 t  17078.63  25.752400 Then Z  e −0.05 (25.752400)  0.275927 .

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Table 11.1: Summary of probability and percentile concepts

• To calculate Pr ( Z ≤ z ) for continuous Z, draw a graph of Z as a function of Tx . Identify the parts of the graph that are below the horizontal line Z  z, and the corresponding t’s. Then calculate the probability of Tx being in the range of those t’s. • In the case of constant force of mortality µ and interest δ, Pr ( Z ≤ z )  z µ/δ . • For discrete Z, identify Tx and then identify K x + 1 corresponding to that Tx . • To calculate percentiles of continuous Z, draw a graph of Z as a function of Tx . Identify where the lower parts of the graph are, and how they vary as a function of T. For example, for whole life, higher T lead to lower Z. For n-year deferred whole life, both Tx < n and higher Tx lead to lower Z. Write an equation for the probability Z is less than z in terms of mortality probabilities expressed in terms of t. Set it equal to the desired percentile, and solve for t or for e kt for any k. Then solve for z (which often is v t ).

?

Quiz 11-3 A 10-year deferred 20-year term insurance on (60) pays a benefit of 1 at the moment of death. You are given: • µ x  1/ (110 − x ) • δ  0.02. • Z is the present value random variable for the insurance. Calculate the 90th percentile of Z.

Exercises For a whole life insurance of 1000 on ( x ) payable at the moment of death, you are given

11.1. •

µ x+t  0.04

•

  0.06 δt    0.03

•

Z is the present value random variable for the insurance.

t < 10 t ≥ 10



Calculate Pr ( Z ≤ 400) . 11.2. Z is the present value random variable for a 30-year term life insurance on ( x ) payable at the moment of death. You are given •

  0.01 µ x+t    0.02

•

δ  0.04

t < 10 t ≥ 10



Calculate Pr ( Z > 0.25) .

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EXERCISES FOR LESSON 11

227

For a 10-year deferred whole life insurance on ( x ) payable at the moment of death:

11.3. • • •

Z is the present value random variable.

  0.01 µ x+t    0.02  δ  0.05.

t ≤ 10 t > 10

Calculate Pr ( Z > 0.5) . [4-S86:21] You are given the following for (40):

11.4.

• Mortality is uniformly distributed with ω  100. • Z  the present value random variable for a 5-year deferred life insurance of 1. • M  Mode of Z. Which of the following are true? FZ ( 0 ) > 0

I.

M  v5

II.

Pr ( Z < v 30 )  0.5

III.

A. I and II only B. I and III only C. II and III only E. The correct answer is not given by A. , B. , C. , or D.

D. I, II and III

11.5. Z is the present value random variable for a whole life insurance on (20) paying a benefit of 1000 at the end of the year of death. You are given: • •

µ x  1/ (100 − x ) , x < 100 i  0.06

Calculate Pr ( Z > 500) . 11.6. For a 20-year endowment insurance on (45) paying a benefit of 1000 at the end of the year of death or at the end of the 20th year if earlier, you are given: • Mortality follows the Illustrative Life Table. • i  0.06. • Z is the present value random variable for the endowment insurance. Calculate Pr ( Z ≤ 600) . For a unit face amount 10-year deferred whole life on a life age x payable at the moment of death:

11.7. • •

µ x+t  0.01 for t > 0 δ  0.06

Calculate the 90th percentile of the present value random variable for the insurance.

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You are given:

11.8.

( x ) is subject to a constant force of mortality, µ x+t  0.05. Z is the present value random variable of a 20-year deferred whole life insurance of 1000 on ( x ) with benefit payable at the moment of death. • δ  0.06.

• •

Calculate the 60th percentile of Z. 11.9. [C3 Sample:44] For a 10-year deferred whole life insurance of 1 payable at the moment of death on a life age 35, you are given: •

The force of interest is δ  0.10.

•

The force of mortality is µ  0.06.

•

Z is the present value random variable for this insurance. Determine the 90th percentile of Z.

A. 0.1335

B. 0.1847

C. 0.2631

D. 0.4512

E. 0.4488

11.10. [3-S00:21] A risky investment with a constant rate of default will pay: • principal and accumulated interest at 16% compounded annually at the end of 20 years if it does not default; and • zero if it defaults. A risk-free investment will pay principal and accumulated interest at 10% compounded annually at the end of 20 years. The principal amounts of the two investments are equal. The expected present values of the two investments are equal at time zero. Calculate the median time until default or maturity of the risky investment. A. 9

B. 10

C. 11

D. 12

E. 13

11.11. [CAS4A-F96:6] (2 points) A whole life insurance has a present value benefit function Z  v T . The force of interest is twice the force of mortality, and both are constant. Determine the ratio of the mean of Z to the median of Z. A. B. C. D. E.

Less than 0.80 At least 0.80, but less than 0.95 At least 0.95, but less than 1.10 At least 1.10, but less than 1.25 At least 1.25

11.12. A 20-year endowment insurance on (40) is payable at the moment of death. You are given: • Mortality is uniformly distributed with ω  100. • δ  0.04. Calculate the median of the present value random variable for the insurance.

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EXERCISES FOR LESSON 11

229

11.13. Z is the present value random variable for a 40-year continuously decreasing insurance on (40) payable at the moment of death. You are given: • •

 1/60 for 0 ≤ k ≤ 59 δ  0.05

k| q 40

Calculate the 60th percentile of Z. 11.14. Z is the present value random variable for a whole life insurance on ( x ) payable at the moment of death. You are given: • •

µ x+t  0.01t δ  0.04

Calculate the 80th percentile of Z. 11.15. Z is the present value random variable for a whole life insurance on ( x ) payable at the moment of death. You are given: • •

µ x+t  0.3/ (40 − t ) , t < 40 δ  0.04

Calculate the 30th percentile of Z. 11.16. Z is the present value random variable of a 10-year deferred whole life insurance on ( x ) payable at the moment of death. You are given: • •

µ x+t  0.01t δ  0.04

Calculate the 70th percentile of Z. 11.17. You are given: • •

µ x  0.001 (1.05x ) . δ  0.04.

Calculate the third quartile of the present value of a whole life insurance of 1 on (25) payable at the moment of death. 11.18. Z is the present value random variable for a whole life insurance on ( x ) payable at the moment of death. You are given • •

µ x+t  0.01 (1.07t ) i  0.05

Calculate the 40th percentile of Z. Additional old CAS Exam 3/3L questions: S08:22 Additional old SOA Exam MLC questions: F12:15

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Solutions 11.1. Lower Z corresponds to higher Tx , since Z  1000v t . Let’s determine the t for which v t  400/1000  0.4. For t  10, v 10  e −0.06 (10)  0.548812. For t > 10, v t  e −0.6−[0.03 ( t−10) ]  0.548812e −0.03 ( t−10) , so 0.548812e −0.03 ( t−10)  0.4 0.548812e 0.3 e −0.03t  0.4 e −0.03t  0.539943 Now, Pr ( Z ≤ 400)  e −0.04t , so the answer is Pr ( Z ≤ 400)  0.5399434/3  0.4397 . 11.2.

Z  e −0.04t , so Z > 0.25 ⇒ e −0.04Tx > 0.25 − ln 0.25  34.66 ⇒ Tx < 0.04

However, since it is a 30-year term insurance, Z  0 unless Tx ≤ 30, so the probability of Z > 0.25 is 30 q x . We calculate it: −0.01 (10) −0.02 (20)  1 − e −0.5  0.3935 30q x  1 − e 11.3. Refer to Figure 11.1d. For 10-year deferred insurance, Z > 0.5 only if T > 10 and T ≤ t where v t  0.5, so t  − (ln 0.5) /0.05  13.8629. The survival probabilities are 10 p x 13.8629 p x

 e −0.01 (10)  0.904837  e −0.01 (10) −0.02 (3.8629)  0.837564

Pr ( Z > 0.5)  0.904837 − 0.837564  0.0673 11.4. I

Let T40 be time to death for (40). Since Z can never be less than zero, FZ (0)  Pr ( Z  0) . A 5-year deferred life insurance pays nothing if death occurs within 5 years, so Pr ( Z  0)  Pr (T40 ≤ 5) > 0. !

II

Tricky. v 5 is the maximum of Z, but the mode is the point of maximum probability or maximum density of Z. For a mixed distribution, it is not clear how the mode should be defined. The most likely value of Z, is 0, since Z  0 whenever T40 < 5, and that has probability of 5/60  1/12. Some would argue that this is the mode, since the value of 0 is more likely than any value from the continuous portion of the distribution. Others would argue that if the density of the continuous part of the distribution is higher than the probability of all point masses, then that point is the mode. To calculate the density function of the continuous portion of the distribution, we need to transform T40 to Z. The density function of T40 is 1/60, and Z  v T40 for 5 < T40 ≤ 60, so the transforming function is z  g ( x )  v x . Then ln z  x ln v ln z x ln v

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EXERCISE SOLUTIONS FOR LESSON 11

231

Z 0.6 0.5 0.4 (t , v t )

0.3 0.2 0.1 20 t

10

30

40

50

T

Figure 11.3: Graph of Z on Tx : solution to exercise 11.7

Therefore, the density function of Z is



 dg −1 dz

f Z ( z )  f40 g −1 ( z ) 

1 1 60 zδ

for g (60) ≤ z < g (5) , or v 60 ≤ z < v 5 . The density is a decreasing function of z. It is maximized at the smallest value of z, or v 60 . For δ sufficiently low, f Z ( v 60 ) may be higher than 1/12. If it is, some would argue that v 60 , rather than 0, is the mode. III

In any case, it is clear that v 5 is not the mode. # Z < v 30 if and only if death occurs within 5 years (so that Z  0) or T40 > 30 (so that Z < v 30 due to discounting 30 years). The probability of death within 5 years is 1/12 and the probability of death 30 after 30 years is 60  0.5. Therefore, Pr ( Z < v 30 )  0.5 + 1/12 , 0.5. #. (E)

11.5. If v T20  0.5, then T20  (ln 0.5) / (ln v )  − (ln 0.5) / (ln 1.06)  11.8957. Thus Z > 500 if and only if K 20 +1 ≤ 11, which means death occurs within the first 11 years. Since mortality is uniform, the probability of this is 11q 20  11/80  0.1375 . 11.6. If v T45  0.6, then T45  (ln 0.6) / (ln v )  − (ln 0.6) / (ln 1.06)  8.7667. Thus Z ≤ 600 if and only if K 45 + 1 ≥ 9, which means survival for 8 years. Using the ILT, 8p 45  l 53 /l 45  8,779,128/9,164,051  0.95800 . Notice that the answer would’ve been the same for a whole life policy or a 20-year term policy. 11.7. See Figure 11.3. Z is highest for Tx > 10. To get the highest 10% of values for Z, we need t for which 10p x − t p x  0.1. 10p x tpx

e

−0.01t

 e −0.1  0.904837  0.904837 − 0.1  0.804837  0.804837

Then z  v t  e −0.06t  0.8048376  0.2718 . 11.8.

This is similar to the previous exercise. Z is highest for Tx > 20. We need t for which 20p x − t p x  0.4. 20p x

 e −0.05 (20)  0.367879

Thus the probability of Z  0 is 1 − 20p x  0.632121 > 0.6. The 60th percentile of Z is therefore 0 . LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

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11.9.

This is similar to the previous exercise. Once again, we want t for which 10 p x − t p x  0.1.  e −0.6  0.548812

10 p 35

 0.548812 − 0.1  0.448812

t p 35

e

 0.448812

−0.06t

Then z  e −0.1t  0.4488120.1/0.06  0.2631 . (C) This question is unusual in that the solutions are not in increasing order, but it was not a real exam question. Answer (E) would result if by mistake you discounted at δ  0.06 instead of at δ  0.1. 11.10. First we calculate the default rate. Notice that the question does not indicate the valuation interest rate, i.e., the interest rate used to compute present values. Just because the bonds pay 10% or 16% does not imply we should use one of those as a valuation rate. The valuation rate doesn’t matter! Since the two investments have equal present value, they also have equal accumulated values (multiply both sides of the equality by (1 + i ) 20 , regardless of what i is), and the accumulated values at the end of 20 years are independent of the valuation rate. So we equate the accumulated values. 1.120  e −20µ 1.1620 1.1 e −µ  1.16 We want t such that e −µt  21 , so 1.1 1.16

!t  0.5 t

ln 0.5  13.0512 ln (1.1/1.16)

11.11. E[Z] 

(E)

µ 1  µ+δ 3

The median occurs at the point m such that e −mµ  0.5, so m  (ln 2) /µ and the median of Z is e

−mδ

e

−δ ln 2/µ

e

−2 ln 2

1  2

!2  0.25

Then the ratio of mean to median is (1/3) /0.25  4/3 (E) 11.12. The present value is a monotonically decreasing function of time, so the median is assumed at the median time of death t  30, in which case the endowment is paid after 20 years. The answer is therefore e −20 (0.04)  0.449329 . 11.13. is 0.6.

Z decreases in value as time increases, so we need t for which the probability of surviving past t t p 40

1−

60 − t  0.6 60

t  0.6 60 t  60 (0.4)  24

The value of Z is (40 − 24) e −24 (0.05)  4.8191 . LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM



EXERCISE SOLUTIONS FOR LESSON 11

233





11.14. Z  v Tx is highest when Tx , time to death, is lowest. We want t such Pr e −0.04T < e −0.04t  0.8, or Pr (T > t )  0.8, so we want to solve t p x  0.8 for t. Survival follows a Weibull with t px

 e−

R

t 0

0.01t dt

 e −0.005t

2

so ln t p x  −0.005t 2  ln 0.8 ln 0.8  44.6287 t2  − 0.005 t  6.68047 Then Z  e −0.04 (6.68047)  0.7655 . 11.15. Z is highest when Tx is lowest. We want t such that the probability of living beyond t is 30%, or 40−t
0.3 . t p x  0.3.. For this beta distribution of mortality, t p x  40 40 − t 40

! 0.3

1−

 0.3 t  0.31/0.3  0.018075 40 t  40 (1 − 0.018075)  39.2770

Then Z  e −0.04 (39.2770)  0.2078 . 11.16. Z  0 for Tx ≤ 10, and is then a decreasing positive function. We  want t such that the probability of Z being greater than v t is 0.3, so we need t such that Pr 10 < Tx ≤ t  0.3. 10 p x

− t p x  0.3 t

Z t px

 exp −

! 0.01tdt  e −0.005t

0 10 p x t px

e

−0.005t 2

e

−0.005 (102 )

 0.606531

 0.606531 − 0.3  0.306531  0.306531

0.005t  − ln 0.306531  1.18244 2

r t

1.18244  15.3782 0.005

So Z  e −0.04 (15.3782)  0.54057 . 11.17. The third quartile is the 75th percentile. The survival function is calculated as follows: t

Z 0

1.0525+u µ25+u du  0.001 ln 1.05

! t 0

0.001 (1.0525 ) (1.05t − 1) ln 1.05  0.069407 (1.05t − 1)



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234

Now,

t

Z

!

t p x  exp −



µ25+u du  exp −0.069407 (1.05t − 1)



0

We want to solve for t. Note that the present value is a monotonically decreasing function of survival time, so the 25th percentile of survival time (the t for which the probability of living less than that amount of time is 25%, or for which t p 25  0.75) corresponds to the 75th percentile of the present value of the insurance. t

e −0.069407 (1.05 −1)  0.75 −0.069407 (1.05t − 1)  ln 0.75  −0.28768 0.287682  4.144886 1.05t − 1  0.069407 ln 5.144886 1.638003   33.57241 t ln 1.05 0.048791 Then e −0.04 (33.57241)  0.261088 .



11.18.



Z decreases in value as Tx increases, so we want t such that t p x  Pr Tx > t  0.4.

Z

! t !  0.01 1.07t − 1 ln 1.07 0 0 ! 1.07t − 1 + * /  0.4 t p x  exp . − 0.01 ln 1.07 ,   0.01 1.07t − 1  (− ln 0.4)(ln 1.07)  0.06200 t

1.07u 0.01 (1.07 ) du  0.01 ln 1.07 u

1.07t  1 + 100 (0.06200)  7.2 ln 7.2 t  29.176 ln 1.07 Then Z  1.05−29.176  0.24087 .

Quiz Solutions 11-1.

The graph of Z on T60 looks like this:

Z 1 0.8 0.6 0.4 0.2 10

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20

30

40

50

T

QUIZ SOLUTIONS FOR LESSON 11

235

Let’s check that v 20 > 0.4 by evaluating it: v 20  e −0.04 (20)  0.4493. So Z > 0.4 whenever 10 < T60 ≤ 20 and at no other time. Therefore Pr ( Z > 0.4)  10 p 60 − 20 p60

40  50

! 0.5

30 − 50

! 0.5  0.1198

11-2. We know that Z and Tx are related as in Figure 11.1d. First let’s check whether Z is ever higher than 0.1. Z is at its maximum when K x  10, and is then equal to v 11 , or 0.911  0.313811 > 0.1 So Pr ( Z < 0.1) is 10q x plus the probability that K x is high enough to make v K x +1 < 0.1. Solving that for Kx , 0.9K x +1 < 0.1 K x + 1 > (ln 0.1) / (ln 0.9)  21.85 So K x > 20, which means at least 21 years of survival. Z is less than 0.1 if survival is less than 10 years or at least 21 years. So the answer is



p



p

Pr ( Z < 0.1)  10q x + 21p x  1 − 5/6 + 39/60  0.8934

11-3.

The graph of Z on T60 looks like this:

Z 1 (t , v t )

0.8 0.6 0.4 0.2 10

t

20

30

40

50

T

We need to find t such that 10p 60 − t p 60  0.1. Mortality is uniform with ω  110, so 10p 60  0.8, so we need 15  e −0.02 (15)  0.7408 . t p 60  0.7, or 1 − t/50  0.7, or t  15. Then v

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11. INSURANCE: PROBABILITIES AND PERCENTILES

Lesson 12

Insurance: Recursive Formulas Reading: Models for Quantifying Risk (4th or 5th edition) 7.4, 7.7, exercise 7-6 Actuarial present values of insurance on ( x ) can be related to Actuarial present values of insurance on ( x +1) in a straightforward manner. This enables building insurance function tables recursively by starting at the end of the table and working backwards. You should not memorize these formulas by rote. You should understand them and be able to reconstruct them. We will write formulas only for insurance payable at the end of the year of death. The reason we use such insurances is they can be calculated for integral x using life tables, and recursive formulas allow you to build insurance tables from life tables by starting with the end of the table, and working backwards. Whole life insurance Whole life on ( x ) can be split into 1-year term insurance on ( x ) plus 1-year deferred whole life on ( x ) . In other words A x  vq x + vp x A x+1 (12.1) The recursion at the end of the table for x + 1  ω will have q ω−1  1, p ω−1  0, so formula (12.1) becomes A ω−1  ( v )(1) + ( v )(0) A ω  v Endowment insurance An n-year endowment insurance on ( x ) can be split into 1-year term insurance on ( x ) plus 1-year deferred n − 1-year endowment insurance on ( x ) . In other words A x:n  vq x + vp x A x+1:n−1

(12.2)

At age x + n − 1, the recursion will be A x+n−1:1  vq x+n−1 + vp x+n−1  v since a payment will definitely be made at the end of the year. Term life insurance An n-year term insurance on ( x ) can be split into 1-year term insurance on ( x ) plus 1-year deferred n − 1-year term insurance on ( x ) . In other words 1 A x1:n  vq x + vp x A x+1 :n−1

(12.3)

1 1 The recursion at age x + n − 1 will have A x+n−1 :1  vq x+n−1 , because A x+n :0  0.

Deferred whole life insurance An n-year deferred whole life insurance on ( x ) can be expressed as a 1-year deferred n − 1-year deferred whole life insurance, or n| A x

 vp x

n−1| A x+1

At the end of the deferral period, 0| A x+n  A x+n . LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

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(12.4)

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238

Although the recursive formulas were expressed in terms of expected value symbols, they apply to the present value random variables themselves and therefore to all moments of them. Example 12A You are given: • Z1 is the present value random variable for a 10-year term insurance of 1 on (80) payable at the end of the year of death. • Z2 is the present value random variable for a 9-year term insurance of 1 on (81) payable at the end of the year of death. • q80  0.1 • i  0.04 • E[Z1 ]  0.5 • Var ( Z1 )  0.2 Calculate Var ( Z2 ) . Answer: All the recursion formulas apply equally well when the force of interest is doubled, so we can use them to get second moments as well as first moments, and then to get variances. So let’s first get 2A 1 80:10 . 2 1 A80:10  Var ( Z1 ) + E[Z1 ]2  0.2 + 0.52  0.45 Now let’s use recursion to get first and second moments of Z2 . 1 1 A80 :10  vq 80 + vp 80 A81:9 1 0.9A81 0.1 :9 0.5  + 1.04 1.04 0.42 1 A81  0.46667 :9  0.9 2 1 1 A80:10  v 2 q 80 + v 2 p 80 2A81 :9 1 0.9 (2A81 0.1 :9 ) + 1.042 1.042 (0.45)(1.042 ) − 0.1 0.38672 2 1 A81:9    0.42969 0.9 0.9 Var ( Z2 )  0.42969 − 0.466672  0.21191

0.45 



The recursive equations can be applied multiple times. Applying the whole life recursive equation twice yields A x  vq x + v 2 p x q x+1 + v 2 2 p x A x+2 As with recursive formulas for life expectancy, the recursive formulas for insurance allow you to modify an insurance for a short-term change in interest rate or mortality assumption. Example 12B For a whole life insurance on Jane (age 35), you are given: • A benefit of 10,000 is payable at the end of the year of death. • Jane’s mortality follows the Illustrative Life Table, except that her mortality rate is one-half of the Table’s rate at ages 35 and 36. • An interest rate of 6% is used to value the policy. Calculate the net single premium for the insurance. Answer: The recursive formula, applied twice, is A35  vq 35 + v 2 p35 q 36 + v 2 2 p35 A37 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

EXERCISES FOR LESSON 12

239

Substitute the values from the Illustrative Life Table, with halved mortality at 35 and 36. q 35  0.5 (0.00201)  0.001005 q 36  0.5 (0.00214)  0.00107 2 p 35

 (1 − 0.001005)(1 − 0.00107)  0.997926

!

A35 

0.001005 (0.997926)(0.14094) (0.998995)(0.00107) + + 1.06 1.062 1.062

 0.127075 The net single premium is 1270.75 .

?



Quiz 12-1 A life table has the following values: A50  0.12970

A51  0.13032

q50  0.006

Determine the effective annual interest rate used to build this table. Recursion can be used for lengths of time greater than 1. A whole life insurance is equal to an n-year term insurance plus an n-year deferred insurance, no matter what n is. Example 12C You are given • A85 − A65  0.15 • A65:20  0.7 1 • A65:20  0.5 Determine A65 . Answer: Split A65 into 20-year term and deferred insurances. 1 A65  A65 :20 + 20 E65 A85

 A65:20 − 20 E65 + 20 E65 ( A65 + 0.15)  0.7 + 20 E65 (−0.85 + A65 )  0.7 + 0.5 (−0.85 + A65 ) 0.5A65  0.275 A65  0.55



Exercises You are given:

12.1. • • • •

A40  0.3 A40:20  0.45 20 p 40  0.9 i  0.04

Calculate A60 .

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12. INSURANCE: RECURSIVE FORMULAS

240

Table 12.1: Summary of Formulas in this Lesson

Recursive Formulas A x  vq x + vp x A x+1 A x  vq x + v p x q x+1 + v 2

(12.1) 2

2 px

A x+2

A x:n  vq x + vp x A x+1:n−1

(12.2)

1 A x1:n  vq x + vp x A x+1 :n−1

(12.3)

 vp x n−1| A x+1

(12.4)

n| A x A x:n1



1 vp x A x+1:n−1

[4-S86:15] You are given:

12.2. • • • •

i  0.02 p50  0.98 A51 − A50  0.004 2A − 2A  0.005 51 50

Let Z be the random variable representing the present value of a whole life insurance of 1 with death benefit payable at the end of the year of death. Calculate Var ( Z ) for x  51. A. 0.055

B. 0.060

C. 0.065

D. 0.260

E. 0.265

12.3. [3-S01:34] Lee, age 63, considers the purchase of a single premium whole life insurance of 10,000 with death benefit payable at the end of the year of death. The company calculates net single premium using: • mortality based on the Illustrative Life Table, • i  0.05 The company calculates gross premiums at 112% of net single premiums. The single gross premium at age 63 is 5233. Lee decides to delay the purchase for two years and invests the 5233. Calculate the minimum annual rate of return that the investment must earn to accumulate to an amount equal to the single gross premium at age 65. A. 0.030 12.4.

B. 0.035

C. 0.040

D. 0.045

E. 0.050

You are given: 1 • For a standard life, A45 :10  0.15. • A standard life has mortality rate q45  0.01. • v  0.95.

Due to an extra hazard at age 45, a certain life has q 45  0.02, but has standard mortality at all other ages. 1 Calculate A45 :10 for this life.

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EXERCISES FOR LESSON 12

241

12.5. You are developing a table of net single premiums for whole life insurance with unit face amount and benefits payable at the end of the year of death. The assumed interest rate is i  0.04. You have calculated that A68  0.3285, and are given the following life table: l65  9852

l 66  9742

l67  9625

l68  9500

Calculate A65 . [4-F86:32] You are given:

12.6.

A x:n  u A x1:n  y A x+n  z

• • •

Determine the value of A x . A. B. C. D. E.

(1 − z ) y + uz (1 − z ) u + yz (1 + z ) y − uz (1 + z ) u − yz (1 + z ) u − y [150-S88:16] You are given:

12.7.

A60  0.58896 A60:11  0.9506 A61  0.60122

• • •

Calculate q60 . A. 0.017

B. 0.018

C. 0.021

D. 0.032

E. 0.033

C. 1.18

D. 1.30

E. 1.56

[150-S89:27] You are given:

12.8.

A x+1 − A x  0.015 i  0.06 q x  0.05

• • •

Calculate A x + A x+1 . A. 0.60

B. 0.86

[CAS4-F92:14] (2 points) You are given the following:

12.9. • • •

A x  0.25 A x+15  0.40 A x:15  0.50

Determine A x1:15 . A. B. C. D. E.

Less than 0.050 At least 0.050, but less than 0.060 At least 0.060, but less than 0.070 At least 0.070, but less than 0.080 At least 0.080

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12.10. [CAS4-S86:21] (2 points) You are given: • • •

A x  0.632 A x+1  0.644 i  3%

Calculate q x . A. B. C. D. E. 12.11. • • •

Less than 0.013 At least 0.013, but less than 0.015 At least 0.015, but less than 0.017 At least 0.017, but less than 0.019 At least 0.019 [CAS4A-F96:12] (2 points) You are given that: A20  0.30 q20  0.005 i  0.07

Determine A21 . A. B. C. D. E.

Less than 0.30 At least 0.30, but less than 0.31 At least 0.31, but less than 0.32 At least 0.32, but less than 0.33 At least 0.33

12.12. [SOA3-F03:22] For a whole life insurance of 1 on (41) with death benefit payable at the end of the year of death, you are given: • • • •

i  0.05 p40  0.9972 A41 − A40  0.00822 2A − 2A  0.00433 41 40

Calculate Var ( Z ) . A. 0.023

B. 0.024

C. 0.025

D. 0.026

E. 0.027

Additional old CAS Exam 3/3L questions: F06:34, F09:13 Additional old SOA Exam MLC questions: F14:4

Solutions 12.1.

We split the whole life into a 20-year term insurance and a 20-year deferred insurance, 1 A40  A40 :20 + 20 E40 A60

 A40:20 − 20 E40 + 20 E40 A60

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EXERCISE SOLUTIONS FOR LESSON 12

243

The 20-year pure endowment’s value is computed with the given information (iii) and (iv). 0.9  0.410748 1.0420 0.3  0.45 − 0.410748 + 0.410748A60 0.3 − 0.45 + 0.410748 0.260748 A60    0.634813 0.410748 0.410748

20 E40

12.2.



By recursive equation (12.1), A50  vq 50 + vp50 A51

so 0.02 0.98 + A51 1.02 1.02   0.98 0.02 1−  + 0.004 1.02 1.02 A51  0.602

A51 − 0.004  A51

We repeat this with twice the force of interest to derive 2A51 . When the force of interest is doubled, v is squared. 0.02 0.98 2 + A51 2 1.02 1.022   0.02 0.98  + 0.005 1− 1.022 1.022 2 A51  0.417252 A51 − 0.005 

2 2

A51

Var ( Z )  0.417252 − 0.6022  0.054848 12.3.

The net single premium is

so A63

5233  4672.32, 1.12  0.467232. We use two iterations of the recursive equation

(A)

10,000A63 

A63  vq 63 + v 2 p63 q 64 + v 2 2 p63 A65 and the values of q 63  0.01788, q 64  0.01952, l 63  7,823,879, l65  7,533,964 to obtain: l 65 7,533,964   0.962945 l 63 7,823,879 0.01788 (1 − 0.01788)(0.01952) 0.962945 0.467232  + + A65 1.05 1.052 1.052 0.467232  0.017029 + 0.017389 + 0.873420A65 0.467232 − 0.017029 − 0.017389  0.49554 A65  0.873420 2 p 63



The gross premium at 65 is 1.12 (10,000)(0.49554)  5550. The earnings rate needed is (5550/5233) 1/2− 1  0.030 . (A)

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12. INSURANCE: RECURSIVE FORMULAS

244

12.4.

We’ll use primes for the life with extra hazard. 1 1 A45 :10  vq 45 + vp 45 A46:9 1 0.15  (0.01)(0.95) + (0.99)(0.95) A46 :9 1 A46:9  0.149389 0 1 (A45 :10 )  (0.02)(0.95) + (0.98)(0.95)(0.149389)

 0.158081 12.5. 1 A65 :3 

9852−9742 1.04

+

9742−9625 1.042

+

9625−9500 1.043

9852

 0.032995 A65  0.032995 +

12.6.

9500

(1.043 )(9852)

(0.3285)  0.3146

As usual, split the whole life into a term insurance and a deferred insurance. A x  A x1:n + n E x A x+n n Ex

 A x:n − A x1:n  u − y

Ax  y + (u − y ) z  y (1 − z ) + uz 12.7.

(A)

We use the recursive formula and the fact that A60:11  vp 60 . A60  vq 60 + A60:11 A61 0.58896  vq 60 + (0.9506)(0.60122) vq 60  0.58896 − (0.9506)(0.60122)  0.01744 vp60  A60:11  0.9506 v  vp 60 + vq 60  0.9506 + 0.01744  0.96804 vq 60 0.01744 q60    0.018016 v 0.96804

(B)

12.8. We set up a relationship between A x and A x+1 using the recursive equation (12.1), and then we can solve for A x+1 . 0.05 0.95 + A x+1 1.06 1.06 0.05 0.95 A x+1 − 0.015  + A x+1 1.06 1.06 (1.06 − 0.95) A x+1  1.06 (0.015) + 0.05 0.0659 A x+1   0.59909 0.11 A x + A x+1  2A x+1 − 0.015  1.18318 Ax 

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(C)

QUIZ SOLUTIONS FOR LESSON 12

12.9.

245

Split the whole life into 15-year term and deferred insurances. 1 (0.40) 0.25  A x1:15 + A x:15

1 A x:15

A x1:15

1 1  0.5 − A x:15 + A x:15 (0.40) 0.25  0.6 0.25 1  0.50 −   0.08333 0.6 12

(E)

12.10. Direct application of the recursive equation. A x  vq x + vp x A x+1 qx 1 − qx 0.632  + (0.644) 1.03 1.03 (1.03)(0.632)  0.644 + q x (1 − 0.644) 0.00696  0.01955 qx  1 − 0.644

(E)

12.11. Use the recursive equation (12.1). A20  vq 20 + vp20 A21 0.005 0.995 0.30  + A21 1.07 1.07 0.30 (1.07)  0.005 + 0.995A21 0.30 (1.07) − 0.005  0.31759 A21  0.995

(C)

12.12. We will show how to calculate A41 , and the exact same formulas can be used with double the force of interest for 2A41 . We know from recursion that A40  vq 40 + vp40 A41 , so A40  A41 − 0.00822  vq 40 + vp 40 A41 A41 (1 − vp 40 )  vq 40 + 0.00822 vq 40 + 0.00822 0.010887 A41    0.216496 1 − vp 40 0.050286 A41 

2

v 2 q 40 + 0.00433 0.006870   0.071926 0.095510 1 − v 2 p40

Var ( Z )  0.071926 − 0.2164962  0.025056

Quiz Solutions 12-1.

By the recursive formula, 0.12970  0.006v + (0.994v )(0.13032)  0.13554v 0.13554 1+i   1.045 0.12970 i  0.045

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(C)

246

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12. INSURANCE: RECURSIVE FORMULAS

Lesson 13

Annuities: Continuous, Expectation Reading: Models for Quantifying Risk (4th or 5th edition) 8.1.3, 8.2.3, 8.3.3 You are familiar with annuities-certain from the Financial Mathematics course. The formula for an n-year certain continuous annuity is: 1 − vn a¯ n  δ Notice that we said certain annuity. In this course, we discuss life annuities. These annuities pay only while the owner is alive. In fact, we may occasionally omit the word “life”; when not otherwise stated, an annuity is a life annuity, not a certain one. In this lesson, we discuss continuous life annuities. A standard continuous annuity pays a level amount continuously at a rate of 1 per year. This is an abstraction which is useful as a model for annuities with frequent payments, such as once a day. Standard types of annuities, and their International Actuarial Notation symbols, are Whole life annuity A continuous whole life annuity pays at a rate of 1 per annum until the death of the annuitant. Thus if the annuitant lives T years, the total payments are T, where T does not have to be an integer. The symbol for the actuarial present value of a continuous life annuity on ( x ) is a¯ x . Temporary life annuity A continuous n-year temporary life annuity pays at a rate of 1 per annum until the earliest of death and time n, where n is almost always an integer, although it doesn’t have to be. The symbol for its actuarial present value is a¯ x:n . Deferred whole life annuity A continuous n-year deferred life annuity pays at a rate of 1 per annum starting at time n and continuing until the death of ( x ) . If death occurs before time n, there are no payments. The symbol for its actuarial present value is n| a¯ x . Deferred temporary life annuity An n-year deferred m-year temporary life annuity pays at a rate of 1 per annum starting at time n and continuing to time n + m, assuming the annuitant is alive. The symbol for its actuarial present value1 is n| a¯ x:m . n-year certain-and-life annuity A continuous n-year certain-and-life annuity pays at a rate of 1 per annum until the latest of death and time n. Thus the payments for the first n years are certain. This annuity is a sum of an n-year certain annuity and an n-year deferred life annuity. The symbol for its actuarial present value is a¯ x:n . The above descriptions, along with the present values of each annuity, are summarized in Table 13.1. As for insurance, traditional actuarial notation is available only for the actuarial present value, not for the random variable itself. The random variable for the present value is usually denoted by Y, just as it was denoted by Z for insurance. Professor Crofts introduced the following notation: the present value random variable for an annuity by Y has the same decorations as the corresponding expected value. For example, Y40:20 is the present value random variable for a 20-year temporary life annuity on (40).

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247

13. ANNUITIES: CONTINUOUS, EXPECTATION

248

Table 13.1: Actuarial notation for standard types of annuity

Payment per annum at time t

Name

Present value

Symbol for expected present value

Whole life annuity

1

t≤T

a¯T

Temporary life annuity

1 0

t ≤ min (T, n ) t > min (T, n )

a¯T a¯ n

T≤n T>n

a¯ x:n

Deferred life annuity

0 1

t ≤ n or t > T nn

n| a¯ x

Deferred temporary life annuity

0 1 0

t ≤ n or t > T n < t ≤ n + m and t ≤ T T > n+m

Certain-and-life

1 0

t ≤ max (T, n ) t > max (T, n )

13.1

0 a¯T − a¯ n a¯ n+m − a¯ n a¯ n a¯T

a¯ x

T≤n n n+m T≤n T>n

n| a¯ x:m

a¯ x:n

Whole life annuity

A whole life annuity is an annuity for Tx years, so using the certain annuity formula, it equals a¯Tx 

1 − v Tx δ

(13.1)

Notice that although the certain annuity formula was used, a¯Tx is still a random variable since Tx is random. Equation (13.1) expresses a whole life annuity in terms of v Tx , a whole life insurance. Taking expected values of both sides, 1 − A¯ x a¯ x  (13.2) δ A¯ x  1 − δ a¯ x (13.3) These formulas for going between whole life annuities and whole life insurances are very important. For an intuitive interpretation, rearrange (13.3) as A¯ x + δ a¯ x  1 This statement says that the value of 1 paid immediately is equal to the present value of continuous interest paid on 1 while ( x ) is alive plus the present value of a payment of 1 at the death of ( x ) . The direct way of calculating the actuarial present value of a whole life annuity is to integrate the value of the payments over the density function t p x µ x+t : ∞

Z a¯ x  0

a¯ t t p x µ x+t dt

(13.4)

This is not the normal way you would evaluate the expected value. Here’s an artificial example of its use: Example 13A A life age x is subject to a constant force of mortality of 0.01. For an annuity on this life, you have determined that the present value of the annuity if the life lives exactly T years is 10 (1 − e −0.02T ) . Calculate the actuarial present value of this annuity. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

13.1. WHOLE LIFE ANNUITY

249

FT (t )

FT (t )

1

1

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2

t

0

t

0

(a) a¯ x calculated from definition

(b) a¯ x calculated from alternative formula

Figure 13.1: Comparison of 2 methods of calculating a¯ x

Answer: Let Y be the present value random variable. The density function t p x µ x+t  e −0.01t (0.01) . From equation (13.4), ∞

Z



E[Y] 



10 1 − e −0.02t e −0.01t 0.01 dt 0 ∞

Z  0.1

e

dt −

0



∞

Z −0.01t

! e

−0.03t

dt

0

0.1 0.1 −  6 32 0.01 0.03



By plugging in a¯T  (1 − v T ) /δ and integrating by parts, equation (13.4) is transformed into ∞

Z a¯ x 

v t t p x dt

(13.5)

0

The actuarial present value of the annuity is the integral of the present value of each payment (v t ) times the probability that the payment is made (t p x ). This is the formula you’ll usually use for calculating a¯ x , assuming you can’t use (13.2). It is a very useful formula, but can only be used for the expected value, not for higher moments. For higher moments, you must use a version of (13.4) with powers of the certain annuities in the integral. Formula (13.5) is analogous to a similar formula for calculating life expectancy discussed on page 83 and illustrated in Figure 5.2. In Figure 13.1, the shaded areas are the value of the payments on the annuity. Compare the subfigures: • In Subfigure 13.1a, each rectangle is the value of the payments until time t (the horizontal axis) times the probability of living exactly t (the vertical axis; the difference between Fx ( t ) at the top of the rectangle and at the bottom of the rectangle). This is the calculation based on the definition of expected value. • In Subfigure 13.1b, each rectangle is the payment in the interval of time (the horizontal axis) times the probability of receiving the payment, the probability of survival to time t (the vertical axis). This is the calculation using the alternative formula. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

13. ANNUITIES: CONTINUOUS, EXPECTATION

250

The alternative formula will be called the “current payment technique”, but that term is not in the textbooks and you’re not responsible for it. Even the alternative formula’s integral has an exponential variable in it, so it is difficult to calculate in closed form unless the survival function is exponential or uniform. So exam questions on continuous annuities will mostly use just those two survival functions. Example 13B (Exponential mortality) A life is subject to a constant force of mortality µ. The force of interest is δ. Evaluate a¯ x . Answer: By now you’ve memorized that A¯ x  µ/ ( µ + δ ) , so you can apply formula (13.2) to immediately get µ δ 1 − µ+δ µ+δ 1 ¯a x    (13.6) δ δ µ+δ Alternatively, you can evaluate the integral in (13.5) ∞

Z a¯ x 

e −δt e −µt dt  0

1 µ+δ



Example 13C (Uniform mortality) Mortality is uniformly distributed with limiting age ω. The force of interest is δ. Evaluate a¯ x . Answer: It is not that easy to directly calculate annuities under uniform mortality. It is easier to calculate an insurance and use formula (13.2) to convert it into an annuity. 1 − A¯ x δ ¯ a ω−x A¯ x  ω−x 1 − a¯ ω−x / ( ω − x ) a¯ x  δ .  1 − (1 − v ω−x ) δ ( ω − x )  δ 1 1 − v ω−x  − 2 δ δ (ω − x ) a¯ x 

13.2



Temporary and deferred life annuities

The random variable for an n-year temporary life annuity is a¯min (Tx ,n ) 

1 − v min (Tx ,n ) δ

But v min (Tx ,n ) is an n-year endowment insurance, so taking expected values on both sides, we get a nice relationship between the annuity and an endowment insurance:

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1 − A¯ x:n δ  1 − δ a¯ x:n

a¯ x:n 

(13.7)

A¯ x:n

(13.8)

13.2. TEMPORARY AND DEFERRED LIFE ANNUITIES

?

251

Quiz 13-1 You are given the following values: A¯ 50  0.28

1 A¯ 50 :10  0.05

A¯ 60  0.46

a¯50  12

Calculate a¯50:10 . The random variable for an n-year deferred whole life annuity is v n a¯max (0,Tx −n ) 

v n − v max ( n,Tx ) δ

There is no similar concept for insurance where the maximum of two times is used, so we don’t have a formula of the form (13.8) relating a deferred annuity to an insurance. However, a deferred annuity is the difference between a whole life annuity and a temporary life annuity, so that the expected values obey: n| a¯ x

 a¯ x − a¯ x:n

and we can calculate the right hand side from insurances: 

n| a¯ x

1 − A¯ x 1 − A¯ x:n A¯ x:n − A¯ x −  δ δ δ

(13.9)

A similar formula can be developed for a deferred temporary life annuity using n| a¯ x:m

 a¯ x:n+m − a¯ x:n

The integrals for evaluating the actuarial present value of a temporary life annuity are: n

Z a¯ x:n  0

n

Z a¯ t t p x µ x+t dt + n p x a¯ n 

v t t p x dt 0

If you need to integrate, you’ll almost always use the latter expression. The integrals for evaluating the actuarial present value of a deferred whole life annuity are ∞

Z n| a¯ x  n

∞

Z ( a¯ t − a¯ n ) t p x µ x+t dt 

v t t p x dt n

Once again, you’ll almost always use the latter expression. Whole life, temporary, and deferred annuities have formulas that express their actuarial present value   − δt+

R

t

µ x+u du

0 as integrals of pure endowments. The actuarial present value of a pure endowment is e . In this expression, adding a constant to µ x+t has the same effect as adding a constant to δ. Thus, for these types of annuities (whole life, temporary, deferred), adding a constant to µ x+s has the same effect as adding a constant to δ. Adding a constant to µ x+s and subtracting the same constant from δ will result in no change in the actuarial present value. This property is true even if benefits vary, but it is not true for certain-and-life annuities. It is only true for expected values, not for higher moments. A whole life annuity can be decomposed into n-year temporary and deferred life annuities:

a¯ x  a¯ x:n + n| a¯ x and the deferred whole life annuity can be expressed as n| a¯ x  n E x a¯ x+n , so a¯ x  a¯ x:n + n E x a¯ x+n LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

13. ANNUITIES: CONTINUOUS, EXPECTATION

252

For constant-force mortality, a¯ x  a¯ x+n for any n. We’ve already computed a¯ x  1/ ( µ + δ ) in Example 13B. As a result, we have formulas for deferred and temporary annuities with constant force of mortality: n| a¯ x

 n E x a¯ x 

e − ( µ+δ ) n µ+δ

a¯ x:n  a¯ x (1 − n E x ) 

(13.10)

1 − e − ( µ+δ ) n µ+δ

(13.11)

For uniform mortality, use formula (13.7) to express the annuity in terms of endowment insurances. Example 13D For (35) , you are given: • Mortality is uniformly distributed with ω  100. • δ  0.05 Evaluate a¯35:20 . Answer: The easier way is to use equation (13.7). The actuarial present value of 20-year temporary life insurance is a¯20 1 A¯ 35 :20  65  20  1 − e −1  65 and the actuarial present value of the pure endowment is 20 p35 v 20 

so

45 1 − e −1 + e −1 65 65 20 25 −1  + e . 65 65

A¯ 35:20 

20 

45 −1 65 e ,



Then using equation (13.7), a¯35:20  

1 − A¯ 35:20 δ 45 25 −1 − 65 65 e

0.05 900 500 −1  − e  11.0163 65 65

?



Quiz 13-2 A continuous whole life annuity on ( x ) makes payments at a rate of 100 per year. You are given:

  0.03 • µ x+t    0.06

t < 10 t ≥ 10



• δ  0.04 Calculate the actuarial present value of this annuity. Example 13E A continuous deferred life annuity pays at the rate of 1.04t at time t starting 5 years from now. You are given:

  0.01 • µ x+t    0.02  • δ  0.07

t 10

Calculate a¯40 . 13.4.

[3-F02:39] For a whole life insurance of 1 on ( x ) , you are given: • The force of mortality is µ x+t . • The benefits are payable at the moment of death. • δ  0.06 • A¯ x  0.60

Calculate the revised actuarial present value of this insurance assuming µ x+t is increased by 0.03 for all t and δ is decreased by 0.03. A. 0.5

B. 0.6

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C. 0.7

D. 0.8

E. 0.9

Exercises continue on the next page . . .

13. ANNUITIES: CONTINUOUS, EXPECTATION

256

[150-F88:17] You are given:

13.5. • • •

A¯ x and a¯ x are based on force of interest δ and force of mortality µ x+t . A¯ 0x and a¯0x are based on force of interest k + δ and force of mortality µ x+t . A¯ 00x is based on force of interest δ and force of mortality k + µ x+t .

Determine A¯ 00x − A¯ x . A. B. C. D. E.

k a¯ x A¯ 0x − A¯ x A¯ 0x + k a¯0x ( k − δ ) a¯0x + δ a¯ x δ ( a¯ x − a¯0x )

13.6. [CAS4A-F93:11] (2 points) You are given the following values for the net single premium of a whole life insurance policy: 1. 2. 3.

A¯ x with force of mortality µ x and constant force of interest δ. A¯ 0x with µ0x  µ x + c, δ0  δ, c is a constant and c > 0. A¯ 00x where µ00x  µ x , δ00  δ + c, c is the same constant as in item 2 above.

Determine which statement is true. A. 1. > 2. > 3.

B. 2. > 1. > 3.

C. 3. > 1. > 2.

D. 2.  3. > 1.

E. 1.  2.  3.

[150-S90:1, CAS4A-S92:9] Y is the present-value random variable for a benefit based on ( x ) such

13.7. that:

  a¯ n , Y  a¯ , T

0 ≤ Tx ≤ n Tx > n

 Determine E[Y]. A. B. C. D. E.

a¯ x:n a¯ x:n + n| a¯ x a¯ n + n| a¯ x a¯ n + v n · a¯ x+n n q x · a¯ n + n| a¯ x [CAS4A-S90:9] (2 points) You are given:

13.8. • • •

µ x  0.03 for 2A ¯ x  0.25

all x > 0.

δ is constant

Calculate a¯ x . A. B. C. D. E.

Less than 10.0 At least 10.0, but less than 12.0 At least 12.0, but less than 14.0 At least 14.0, but less than 16.0 At least 16.0

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EXERCISES FOR LESSON 13

257

For a continuous 30-year temporary life annuity of 1 per year on (40), you are given

13.9.

• Mortality is uniformly distributed with ω  120. • δ  0. Calculate the actuarial present value of this annuity. 13.10. For a continuous 10-year certain-and-life annuity of 1 per year on (50), you are given: • Mortality is uniformly distributed with ω  120. • δ0 Calculate the actuarial present value of this annuity. 13.11. [CAS4A-S97:5] (2 points) A retiree is offered the choice of two continuous annuities: 1. 2.

a life annuity, paying $30,000 per year a life annuity with a 5-year certain period, i.e., an annuity paying benefits for life, but with a minimum payout period of 5 years.

The benefit amount for the second annuity is chosen so that the actuarial present value of the two annuities is the same. δ  0.02 and µ  0.07 Determine the annual benefit amount on the second annuity. A. B. C. D. E.

Less than $26,000 At least $26,000, but less than $27,000 At least $27,000, but less than $28,000 At least $28,000, but less than $29,000 At least $29,000

13.12. [CAS4A-S97:19] (2 points) George, age 50, is considering purchasing two annuities: •

a 20-year annuity certain with an annual benefit of $10,000 payable continuously, or

•

a 20-year temporary life annuity with an annual benefit of $10,000 payable continuously. The cost for either annuity is its net single premium.

George opts for the temporary life annuity, and he invests the savings (the difference in net single premiums) immediately. George survives 20 years. You are given that µ  0.04 and δ  0.06. Determine the value of the invested savings after 20 years. A. B. C. D. E.

Less than $80,000 At least $80,000, but less than $105,000 At least $105,000, but less than $130,000 At least $130,000, but less than $155,000 At least $155,000

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13. ANNUITIES: CONTINUOUS, EXPECTATION

258

13.13. [CAS4A-S98:17] (2 points) A man age 35 has a choice between two continuous life annuities: 1.

A whole life annuity priced assuming a constant force of mortality µ1  0.06 and a constant force of interest δ 1  0.07.

2.

A two-year deferred whole life annuity priced assuming a constant force of mortality µ2  0.06 and a constant force of interest δ2 . Both annuities pay benefits at the same annual rate. Determine δ2 such that the actuarial present values of the two annuities are equal.

A. B. C. D. E.

Less than 0.04 At least 0.04, but less than 0.05 At least 0.05, but less than 0.06 At least 0.06, but less than 0.07 At least 0.07

13.14. [3-F02:4] You are given: • • •

µ x+t  0.01, 0 ≤ t < 5 µ x+t  0.02, 5 ≤ t δ  0.06

Calculate a¯ x . A. 12.5

B. 13.0

C. 13.4

D. 13.9

E. 14.3

13.15. [CAS4A-S93:5] (3 points) You are given the following: δ  0.05

  0.05 µx    0.10

for 0 ≤ x ≤ 50 for 50 ≤ x ≤ 110

 Determine the actuarial present value of a 50 year continuous term life annuity of 1 for an insured age 30. A. B. C. D. E.

Less than 8.25 At least 8.25, but less than 8.75 At least 8.75, but less than 9.25 At least 9.25, but less than 9.75 At least 9.75

13.16. You are given: •

  0.04 δ  0.03

•

µ  0.01

0≤t≤5 t>5



Calculate a¯ x:10 .

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EXERCISES FOR LESSON 13

259

13.17. For a 10-year certain-and-life annuity on ( x ) paying at a continuous rate of 1 per year, you are given: •

  0.01t µ x+t    0.05

•

δ  0.02

t≤5 t>5



Calculate the actuarial present value for this annuity.



13.18. [CAS4-F82:18] Mortality is uniformly distributed with ω  100; that is, l x  l0 1 −

x 100



.

1 Which of the following expressions represents A¯ 40 :10 ?

A.

a¯40 − v 10 a¯50 60

B.

a¯40:10 40

C.

a¯40:10 60

D.

a¯10 60

E. The correct answer is not given by A. , B. , C. , or D. 13.19. Mortality is uniformly distributed with ω  100. The force of interest, δ, is 0.05. Calculate a¯35 . 13.20. A continuous temporary life annuity on (30) pays at a rate of 10 per year for 20 years. You are given • •

µ30+t  1/ (80 − t ) , t < 80 δ  0.08

Calculate the actuarial present value of this annuity. 13.21. [M-F05:20] For a group of lives age x, you are given: • Each member of the group has a constant force of mortality that is drawn from the uniform distribution on [0.01, 0.02] . • δ  0.01 For a member selected at random from this group, calculate the actuarial present value of a continuous lifetime annuity of 1 per year. A. 40.0

B. 40.5

C. 41.1

D. 41.7

E. 42.3

13.22. [M-F06:5] Your company sells a product that pays the cost of nursing home care for the remaining lifetime of the insured. • Insureds who enter a nursing home remain there until death. • The force of mortality, µ , for each insured who enters a nursing home is constant. • µ is uniformly distributed on the interval [0.5, 1]. • The cost of nursing home care is 50,000 per year payable continuously. • δ  0.045. Calculate the actuarial present value of this benefit for a randomly selected insured who has just entered a nursing home. A. 60,800

B. 62,900

C. 65,100

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D. 67,400

E. 69,800

13. ANNUITIES: CONTINUOUS, EXPECTATION

260

Additional old SOA Exam MLC questions: F13:5

Solutions 13.1. We express the certain-and-life annuity as a certain annuity plus a life annuity minus a temporary annuity for the certain period. a¯ x:20  a¯ x + a¯20 − a¯ x:20 1 − A¯ x 1 − 0.3   14.347154 δ ln 1.05 1 − 1/1.0520   12.771232 ln 1.05 1 − A¯ x:20 1 − 0.4    12.297561 δ ln 1.05  14.347154 + 12.771232 − 12.297561  14.820826

a¯ x  a¯20 a¯ x:20 a¯ x:20 13.2.

Back out δ from A¯ 30 . 0.03 0.03 + δ δ  0.05

0.375  A¯ 30 

Calculate a¯30 using equation (13.2). a¯30 

1 − A¯ 30 1 − 0.375   12.50 δ 0.05

(D)

13.3. For the first 10 years interest is constant so a¯40:10  (1 − A¯ 40:10 ) /δ  0.3/0.05  6 We can value a¯50 using A¯ 50 , since the interest rate is constant after the tenth year. 1 ¯ A¯ 40  A¯ 40 :10 + 10 E 40 A50  A¯ 40:10 − 10 E40 + 10 E40 A¯ 50 0.4  0.7 + 10 E40 ( A¯ 50 − 1)

0.3 A¯ 50  1 − 10 E 40 1 − A¯ 50 7.5 a¯50   0.04 10 E 40 The original annuity a¯40 is then a¯40  a¯40:10 + 10 E40 a¯50  6 + 7.5  13.5

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EXERCISE SOLUTIONS FOR LESSON 13

261

13.4. Although this is an insurance question, annuity properties play a role, so this question belongs here. One could cheat and assume µ x+t has a constant value of µ, since whatever the answer to the question is, it has to work for any µ x+t , including the constant one. If µ x+t  µ, then µ  0.6 µ+δ µ  0.6 µ + 0.06

A¯ x 

µ  0.09 The revised values are µ0  µ + 0.03  0.09 + 0.03  0.12 and δ0  δ − 0.03  0.06 − 0.03  0.03, resulting in A¯ 0x  0.12/ (0.12 + 0.03)  0.8 . (D) Without cheating, we should not assume µ x+t is constant. As we mentioned on page 251, whole life annuities are unchanged by increasing µ and decreasing δ by the same amount. So it is logical to use the formula A¯ x  1 − δ a¯ x and compare annuities instead of comparing insurances. The original insurance is 1 − δ a¯ x . The revised insurance is 1 − ( δ − 0.03) a¯ x . In both cases, a¯ x  (1 − A¯ x ) /δ  (1 − 0.60) /0.06  20/3. The revision to the insurance is +0.03 a¯ x  0.03 (20/3)  0.2. The revised value of the insurance is 0.60 + 0.2  0.8 . (D) 13.5. You could cheat by assuming constant force of mortality and see which of the five choices works. But we will do the exercise in a general fashion. Define a¯00x to be based on force of interest δ and force of mortality k + µ x+t . Then A¯ x  1 − δ a¯ x A¯ 00x  1 − δ a¯00x A¯ 00 − A¯ x  δ ( a¯ x − a¯00 ) x

x

However, by the property discussed on page 251, adding a constant to the force of mortality has the same effect on a whole life annuity as adding a constant to the force of interest. So a¯00x  a¯0x , even though A¯ 00x , A¯ 0x . We conclude that A¯ 00 − A¯ x  δ ( a¯ x − a¯ 0 ) . (E) x

x

13.6. Intuitively, increasing interest rates lowers the value of insurance and increasing mortality increases the value of insurance, so 2. > 1. > 3. . (B) But let’s prove it. First consider adding a constant to δ. Since A¯ x  E[v T ], A¯ 00x  E[v T e −cT ]. For any two functions g1 ( t ) and g2 ( t ) of a random variable T, if g1 ( t ) < g 2 ( t ) always, then E[g1 ( t ) ] < E[g2 ( t ) ]. Here, g1 ( t )  e −ct v t and g2 ( t )  v t , and g1 ( t ) < g2 ( t ) since e −ct < 1. So E[v T e −cT ] < E[v T ] and we have proved that A¯ 00x < A¯ x . Now R ∞ consider adding a constant to µ. For a¯ x , adding a constant to µ results in a lower value, since a¯ x  0 v t t p x dt, and adding a constant to µ lowers t p x . However, A¯ x  1 − δ a¯ x , so making a¯ x lower results in making A¯ x higher. 13.7. This is a¯ x:n , and thus (C) is the correct answer. 13.8. Back out δ from 2A¯ x . 0.25  2A¯ x 

µ 0.03  µ + 2δ 0.03 + 2δ

δ  0.045 1 1 a¯ x    13 13 µ + δ 0.075

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(C)

13. ANNUITIES: CONTINUOUS, EXPECTATION

262

13.9. The question is asking for e˚40:30 . We use the methods we learned in Lesson 5. Condition on surviving 30 years. Let Y be the present value random variable. Then E[Y]  E[Y | T40 ≤ 30] Pr ( Y ≤ 30) + E[Y | T40 > 30] Pr ( Y > 30) Now, the expected value of Y given that T40 > 30 is 30, since 30 years of payments with no interest are made. And the expected value of Y given T40 ≤ 30 is 15, since the average of a uniform distribution on [0, x] is x/2. The probability of survival 30 years is (80 − 30) /80  5/8. So E[Y] 

5 3 (15) + (30)  24 38 8 8

Alternatively, you can use formula (13.5) with v t  1: 30

Z a¯40:30 

30

Z t p 40 dt



0

0

!

80 − t dt 80

Since (80 − t ) /80 is a straight line, the integral is the average value, assumed at t  15, times the length of the integration range, 30, or (65/80)(30)  24 83 . 13.10. As in the previous exercise, we condition on surviving 10 years, which has probability (70 − 10) /70  6/7. Let Y be the present value random variable. If T50 > 10, then it is between 10 and 70, and the midpoint is 40. E[Y]  E[Y | T50 ≤ 10] Pr (T50 ≤ 10) + E[Y | T50 > 10] Pr (T50 > 10)

!

!

1 6  10 + 40  35.7143 7 7 13.11. The actuarial present value of the life annuity is 1/ ( µ + δ )  1/ (0.02 + 0.07)  11.111111. The actuarial present value of the 5-year certain-and-life annuity is a¯ x:5  a¯5 + 5| a¯ x and 1 − e −5 (0.02)  4.758129 0.02 e −5 (0.02+0.07)  7.084757 5| a¯ x  0.02 + 0.07 a¯ x:5  4.758129 + 7.084757  11.842886 a¯5 

So the annual benefit amount of the second annuity is

!

11.111111  $28,146.29 $30,000 11.842886

(D)

13.12. The two annuities are worth: 1 − e −20 (0.06)  10,000  116,467.63 0.06

!

10,000 a¯20

1 − e − (0.06+0.04)(20)  86,466.47 0.06 + 0.04

!

10,000 a¯50:20  10,000 The accumulated value of the difference is then

(116,467.63 − 86,466.47) e 20(0.06)  99,607.36

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(B)

EXERCISE SOLUTIONS FOR LESSON 13

263

13.13. The first annuity has present value 1/0.13, while the second annuity has present value 

2| a¯ 35

e −2 (0.06+δ2 ) e −0.12−2δ2  0.06 + δ2 0.06 + δ2

One can equate the two, to get

6 δ2 + 13 0.13 Apparently the intended method of solution was trial and error, since the calculators allowed on the exam in 1998 did not have equation solvers. Trying δ2  0.04 (right side higher than left side) and δ2  0.05 (right side lower than left side) shows that equality occurs between these values. However, if you want to be fancy, you can estimate the solution by estimating from the first terms of a Taylor series e −0.12−2δ2 

e −0.12−2δ2 ≈ 1 − 0.12 − 2δ 2  0.88 − 2δ 2 and solve 100δ2 6 + 13 13 11.44 − 26δ2 ≈ 6 + 100δ2 0.88 − 2δ2 ≈

126δ2 ≈ 5.44 δ2 ≈ 0.043175

(B)

13.14. Split the annuity into 5-year temporary life and 5-year deferred and use the formulas for constant force for each. The temporary life annuity’s actuarial present value is a¯ x:5 

1 − e −5 (0.01+0.06)  4.21874 0.01 + 0.06

The deferred annuity must be discounted from year 5 to year 0 using µ  0.01. 5| a¯ x



e −5 (0.01+0.06)  8.80860 0.02 + 0.06

The answer is 4.21874 + 8.80860  13.0273 . (B) 13.15. Since we’re starting at age 30, the switch in mortality at age 50 is at duration 20. We’ll evaluate 20-year term and deferred term annuities. 1 − e − (0.05+0.05)(20)  8.64665 0.05 + 0.05 e −2 − e −2− (0.10+0.05)(30)   0.89221 0.10 + 0.05  8.64665 + 0.89221  9.53886 (D)

a¯30:20  20|30 a¯ 30

a¯30:50

13.16. Express the annuity as a 5-year temporary life annuity plus a 5-year deferred temporary life annuity. a¯ x:10  a¯ x:5 + 5 E x a¯ x+5:5 1 − e − (0.01+0.04) 5 e − (0.01+0.04) 5 (1 − e − (0.01+0.03) 5 ) + 0.01 + 0.04 0.01 + 0.03 −0.25 −0.25  20 (1 − e ) + 25 ( e − e −0.45 )



 7.953300

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13. ANNUITIES: CONTINUOUS, EXPECTATION

264

13.17. We evaluate the 10-year certain-and-life annuity as a 10-year annuity certain plus a 10-year deferred life annuity. 1 − e −0.02 (10)  9.06346 0.02 1   14.28571 0.05 + 0.02

a¯10  a¯ x+10

Z 10 p x

 exp − e

10 E x

5

! 0.01t dt − 5 (0.05)

0 −0.125−0.25

 e −0.375

 v 10 10 p x  e −0.2 e −0.375  e −0.575

a¯ x:10  9.06346 + e −0.575 (14.28571)  17.1021 13.18. The mortality density function is 1/60, so 1 A¯ 40 :10 

10

Z 0

a¯ vt dt  10 60 60

(D)

13.19. Calculate an insurance and use formula (13.2). a¯ A¯ 35  65 65 1 − e −3.25   0.295762 65 (0.05) 1 − A¯ 35 a¯35  δ 1 − 0.295762   14.084764 0.05 13.20. a¯ A¯ 30:20  20 + 20 E30 80 1 − e −1.6  + 0.75e −1.6  0.124704 + 0.151422  0.276126 80 (0.08) ! 1 − A30:20 10 (1 − 0.276126) 10a¯30:20  10   90.484 δ 0.08 1 1 13.21. Given µ, an annuity’s conditional actuarial present value is µ+δ  µ+0.01 . To calculate the overall expected value, we use the double expectation theorem, formula (1.11). We calculate the expected value of the conditional expected value by integrating it over the uniform distribution on [0.01, 0.02], which has density 100.

Z

0.02

!

100 0.01

dµ 0.02  100 ln 0.03 − ln 0.02  40.5465  100 ln ( µ + 0.01) ( ) 0.01 µ + 0.01

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(B)

QUIZ SOLUTIONS FOR LESSON 13

265

13.22. A continuous annuity with constant force of mortality has actuarial present value 1/ ( µ + δ ) . Here, we must integrate this expression, with δ  0.045, over a uniform distribution on [0.5, 1]. This uniform distribution has density 2, so the overall expected value is

Z

1 0.5

2 dµ  2 (ln 1.045 − ln 0.545) µ + 0.045

Multiplying by 50,000, we get 100,000 (ln 1.045 − ln 0.545)  65,099 . (C)

Quiz Solutions 1 ¯ ¯ 13-1. Since A¯ 50  A¯ 50 :10 + 10 E 50 A60 , we have 10 E 50  (0.28 − 0.05) /0.46  0.5. Then A50:10  0.05 + 0.5  ¯ 0.55. Also, since A x  1 − δ a¯ x , we have δ  (1 − 0.28) /12  0.06. Therefore

a¯50:10 

1 − A¯ 50:10 1 − 0.55   7 12 δ 0.06

13-2. Split the annuity into 10-year temporary and deferred annuities. The actuarial present value of the temporary life annuity is 1 − e − (0.03+0.04)(10) a¯ x:10   7.19164 0.03 + 0.04 The actuarial present value of the whole life annuity at time 10 is 1/ ( µ + δ )  1/ (0.06 + 0.04) . However, this value is discounted from time 10 to time 0 using µ  0.03, the force of mortality that is applicable during the first 10 years. Therefore, the actuarial present value of the 10-year deferred annuity is 10| a¯ x



e − (0.03+0.04)(10)  4.96585 0.06 + 0.04

The whole life annuity’s actuarial present value is 100 (7.19164 + 4.96585)  1215.75 .

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13. ANNUITIES: CONTINUOUS, EXPECTATION

Lesson 14

Annuities: Discrete, Expectation Reading: Models for Quantifying Risk (4th or 5th edition) 8.1.1, 8.1.2, 8.2.1, 8.2.2, 8.3.1, 8.3.2 A discrete annual life annuity1 pays some amount at the beginning or at the end of each year. An annuitydue pays at the beginning of the year; an annuity-immediate at the end of the year. The same set of standard products exist as for continuous annuities. The formulas you learned in Financial Mathematics for annuity-due and annuity-immediate: 1 − vn d 1 − vn  i

a¨ n  an will be useful here.

14.1

Annuities-due

There are whole life, temporary life, deferred whole life, deferred term, and certain-and-life annuitiesdue, just as there are continuous annuities of these forms. In each case, the IAN symbol for the actuarial present value of the annuity is the same as the corresponding symbol for a continuous annuity except that a double-dot is put on top of the a instead of a bar. Some students like drawing timelines. The conventions used in timelines in this lesson are given in Table 14.1. A standard whole life annuity-due pays 1 per year at the beginning of every year the annuitant is alive: $1

$1

$1

$1

$1

0

1

2

3

4

a¨x

Table 14.1: Conventions used in timelines in this lesson

• Payments are preceded by a dollar sign. • Payments are only made if the annuitant is alive, unless the payment is in boldface in which case it is certain. • An arrow at the right indicates payments continue forever, as long as the annuitant is alive. • A circle instead of a tick mark indicates the valuation date. • The numbers on the bottom are either durations or ages. They are ages if they have x in them.

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267

14. ANNUITIES: DISCRETE, EXPECTATION

268

The present value of a whole life annuity-due is a¨ K x +1 , where K x is curtate lifetime. The actuarial present value of a whole life annuity is denoted by a¨ x . Since a¨ K x +1 

1 − v K x +1 , d

and v K x +1 is a whole life insurance payable at the end of the year of death, taking expected values yields 1 − Ax d A x  1 − d a¨ x a¨ x 

(14.1) (14.2)

which correspond to formulas (13.2) and (13.3) for continuous life annuities. An n-year temporary life annuity-due pays 1 per year at the beginning of the first n years, with the last payment at n − 1, if the annuitant is alive: $1

$1

$1

$1

$1

$1

0

1

2

3

n −2

n −1

a¨x :n n

n +1

It has present value

 1 − v K x +1   ¨  a  K +1  d  x  1 − vn    a¨ n   d

Kx ≤ n − 1 Kx ≥ n

and the actuarial present value is denoted by a¨ x:n . Since v min ( K x +1,n ) is the present value of an n-year endowment insurance, we have the following formulas after taking present values: 1 − A x:n d  1 − d a¨ x:n

a¨ x:n 

(14.3)

A x:n

(14.4)

which correspond to formulas (13.7) and (13.8) for continuous life annuities. An n-year deferred whole life annuity-due pays 1 per year starting at n if the annuitant is alive: $1

$1

n

n +1

n | a¨ x

0

1

2

3

n −2

n −1

It has present value 0     n K x +1   a¨ K +1 − a¨ n  v − v  x d

Kx ≤ n − 1 Kx ≥ n

and the actuarial present value is denoted by n| a¨ x . An n-year certain-and-life annuity-due pays 1 per year at the beginning of the first n years regardless of whether the annuitant is alive, and after that (starting at time n) 1 per year if the annuitant is alive: $1

$1

$1

$1

$1

$1

$1

$1

0

1

2

3

n −2

n −1

n

n +1

a¨x :n

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14.1. ANNUITIES-DUE

269

It has present value

1 − vn   ¨   a n   d  K x +1     a¨ K +1  1 − v  x d

Kx ≤ n − 1 Kx ≥ n

and the actuarial present value is denoted by a¨ x:n . Relationships between these annuities are similar to the ones for continuous annuities: a¨ x:n  a¨ n + n| a¨ x a¨ x:n  a¨ x − n E x a¨ x+n n| a¨ x  n E x a¨ x+n

(14.6)

a¨ x  a¨ x:n + n| a¨ x

(14.8)

(14.5) (14.7)

The definition of expected value says that the expected value of the annuity is the sum of the products of the probability of living a certain number of years times the present value of the amount paid if the annuitant lives exactly that long. In the case of an n-year temporary life annuity-due, for k < n, the probability of living k − 1 full years but not k full years is k−1 p x q x+k−1 , and if the annuitant lives that long, we pay a¨ k . The remaining possibility is that the annuitant lives n years or longer, and then we pay a¨ n . So the actuarial present value is a¨ x:n 

n X

a¨ k

k−1 p x

q x+k−1 + a¨ n

n px

k1

For an annuity-due, the annuitant gets a¨ n after surviving n −1 years, so the formula’s sum can be reduced by one term: a¨ x:n 

n−1 X

a¨ k

k−1 p x

q x+k−1 + a¨ n

n−1 p x

(14.9)

k1

Example 14A You are given the following information regarding mortality rates and present values of annuities-due certain: k

k| q x

a¨ k+1

0 1 2 3

0.10 0.15 0.20 0.25

1 1.92 2.75 3.45

Calculate a¨ x:3 . Answer: The probabilities of death in each of the first two years is given in the second column of the table, since k| q x  k p x q x+k . The probability of survival for 2 years is 2 p x  1 − 0.10 − 0.15  0.75. Using formula (14.9), the actuarial present value of the 3-year temporary life annuity is a¨ x:3  0.10 (1) + 0.15 (1.92) + 0.75 (2.75)  2.4505



You will more commonly use the alternative technique for calculating the actuarial present value of an annuity. The alternative formula for an n-year temporary life annuity-due is a¨ x:n 

n−1 X k0

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v k k px

(14.10)

14. ANNUITIES: DISCRETE, EXPECTATION

270

Similarly, for an n-year deferred whole-life annuity-due, n| a¨ x



∞ X

v k k px

kn

Example 14B You are given the following mortality rates: x

qx

75 76 77

0.01 0.02 0.04

i  0.05 Calculate a¨75:3 . Answer: Using equation (14.10) a¨75:3  1 +

0.99 (0.99)(0.98) +  2.8229 1.05 1.052



Example 14C Mortality follows the Illustrative Life Table. The interest rate is i  0.06. Calculate a¨35:30 . Answer: We express the temporary life annuity as a whole life annuity minus a deferred life annuity. We look up in the Illustrative Life Table the two values a¨35  15.3926 and a¨65  9.8969. a¨35:30  a¨35 − 30| a¨ 35  15.3926 − 30 E35 a¨65 As usual, we express 30 E35  10 E35 20 E45 . 30 E35

 (0.54318)(0.25634)  0.13924

a¨35:30  15.3926 − (0.13924)(9.8969)  14.0146



If the mortality rate q x is constant for all integers x, we can derive the following simple formula for the actuarial present value of a whole life annuity: a¨ x     

1 − Ax d 1 − q/ ( q + i ) d (q + i ) − q d (q + i ) i/d q+i 1+i q+i

from formula (8.8)

(14.11)

The actuarial present value of a temporary life annuity, a¨ x:n , and other annuities can be calculated using the fact that a¨ x is constant over x for a geometric distribution, so that n| a¨ x  n E x a¨ x . LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

14.2. ANNUITIES-IMMEDIATE

?

271

Quiz 14-1 You are given

  0.1k • q x+k−1    0.3  • d  0.05 Calculate a¨ x .

14.2

k≤3 k>3

Annuities-immediate

An annual whole life annuity-immediate pays 1 per year at the end of each year as long as the annuitant is alive at the end of the year: ax 0

$1

$1

$1

$1

1

2

3

4

Letting K x be curtate lifetime, the present value is aKx 

1 − v Kx i

The actuarial present value is denoted by a x . This annuity is not as convenient to express in terms of insurance; we’d have to multiply by v to get v K x +1 , a whole life insurance. Taking expected values: v − Ax i 1 − (1 + i ) A x ax  i ia x  1 − (1 + i ) A x

va x 

Rather than using this formula, use a x  a¨ x − 1 to convert it into an annuity-due, then use the easier annuity-due formula. An n-year temporary life annuity-immediate pays 1 per year from time 1 to time n if the annuitant is alive: $1 $1 $1 $1 $1 $1 a x :n n +1 0 3 n −2 n −1 n 1 2 The actuarial present value is denoted by a x:n . To relate it to insurance, start with the insurance relationship for temporary annuities-due, equation (14.3): 1 − A x:n a¨ x:n  d and then note that a¨ x:n  a x:n + 1 − n E x since there is an extra payment at 0 and a missing payment at n. a x:n + 1 − n E x  a x:n

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(1 + i ) − (1 + i ) A x:n

i 1 − (1 + i ) A x:n + i n E x  i 1 − A x:n − iA x1:n  i

14. ANNUITIES: DISCRETE, EXPECTATION

272

or

1  ia x:n + A x:n + iA x1:n

(*)

The interpretation of this is: the value of 1 right now is equal to interest on 1 at the end of the year if alive at that point for the next n years, plus the return of 1 with interest for that year if death occurs before the end of year n, plus the return of 1 only if ( x ) lives n years. There is no need to add interest if ( x ) lives for n years, since the annuity already pays interest for year n in that case. Another way of putting it is that the right side pays i at the end of each year if alive and up to year n − 1, plus 1 + i at the end of the earliest of the year of death and n years. Rather than memorizing this formula, you’re probably better off translating the annuity-due formula as needed. A useful formula to relate an n-year term insurance to annuities-due and immediate is A x1:n  v a¨ x:n − a x:n

(14.12)

An n-year deferred whole life annuity-immediate makes payments starting at time n+1 if the annuitant is alive: $1 n |a x n +1 0 3 n −2 n −1 n 1 2 The actuarial present value is denoted by n| a x . An n-year certain-and-life annuity-immediate makes certain payments from time 1 to time n and then starting at time n + 1 payments contingent on the annuitant being alive. a x :n 0

$1

$1

$1

$1

$1

$1

$1

1

2

3

n −2

n −1

n

n +1

The actuarial present value is denoted by a x:n . Some relationships between annuities-due and annuities-immediate were already mentioned. Here are a couple: a¨ x  a x + 1 a¨ x:n  a x:n−1 + 1 n| a¨ x

(14.13) (14.14)

 a x:n + 1 − n E x

(14.15)

 n| a x + n E x

(14.16)

You should be able to reason these relationships out. Example 14D You are given: • Mortality follows the Illustrative Life Table. • i  0.06. Calculate a65:20 . Answer: The ILT provides annuities due. To get this annuity immediate, use a65:20  a¨65:21 − 1 We’ll calculate 21 E65  20 E65 (1 − q85 ) /1.06. 21 E 65  (0.09760)(1 − 0.12389) /1.06  0.080668 a65:20  a¨65 − 21 E 65 a¨86 − 1

 9.8969 − 0.080668 (4.4742) − 1  8.5360 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM



14.3. ACTUARIAL ACCUMULATED VALUE

273

Example 14E You are given: • a40:10  6.7 1 • A40 :10  0.1 • d  0.06 Calculate 10 E40 . Answer: We have a¨40:10  6.7 + 1 − 10 E40 , so A40:10  1 − 0.06 (7.7 − 10 E40 )  0.538 + 0.06 10 E40 1 The other equation we have is A40:10  A40 :10 + 10 E 40 :

A40:10  0.1 + 10 E40 Substituting that into the previous equation, 0.1 + 10 E40  0.538 + 0.06 10 E40 0.438  0.46596 10 E 40  0.94



The direct way to calculate the expected value of an immediate annuity is to add up the payments times their probabilities: a x:n 

n X

v t t px

t1

This formula can be generalized to any series of payments b t ; the sum then becomes n X

bt v t t px

t1

?

Quiz 14-2 You are given: • µ x+t  1/ (50 − t ) , t < 50 • i  0.05. Calculate a x .

14.3

Actuarial Accumulated Value

The actuarial accumulated value s¨ x:n is the expected value of the accumulated value of an annuity with payments of 1 at the beginning of each year as of the end of the last year, or one year after the last payment, per survivor. Therefore a¨ x:n (14.17) s¨ x:n  n Ex Similar definitions apply for continuous annuities and immediate annuities. For an immediate annuity, the valuation date is the date of the last payment. Example 14F You are given: • A40  0.3 • A50  0.4 • 10 p 40  0.94074 • i  0.03 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

14. ANNUITIES: DISCRETE, EXPECTATION

274

Calculate s¨40:10 Answer: We calculate A40:10 and then use equation (14.3) 1 A40  A40 :10 + 10 E40 A50 0.94074 1 (0.4) 0.3  A40 :10 + 1.0310 1  A40 :10 + (0.7)(0.4) 1 A40 :10  0.3 − 0.28  0.02 1 A40:10  A40 :10 + 10 E40

a¨40:10 s¨40:10

 0.02 + 0.7  0.72 1 − 0.72   9.61333 0.03/1.03 9.61333  13.73333  0.7

An alternative is to first compute a¨40 and a¨50 using equation (14.1):

a¨40:10

1 − A40 (1 − 0.3)(1.03)   24.03333 d 0.03 1 − A50 (1 − 0.4)(1.03)    20.6 d 0.03  a¨40 − 10 E40 a¨50

s¨40:10

 24.03333 − 0.7 (20.6)  9.61333 9.61333  13.73333  0.7

a¨40  a¨50



Exercises Expected value of annual annuities 14.1.

[150-F89:1] You are given: k

a¨ k

k−1| q x

1 2 3 4

1.00 1.93 2.80 3.62

0.33 0.24 0.16 0.11

Calculate a¨ x:4 . A. 1.6

B. 1.8

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C. 2.0

D. 2.2

E. 2.4

Exercises continue on the next page . . .

EXERCISES FOR LESSON 14

275

Table 14.2: Summary of annuity expected value formulas

Relationships between annuities and insurances 1 − Ax d A x  1 − d a¨ x 1 − A x:n a¨ x:n  d A x:n  1 − d a¨ x:n a¨ x 

(14.1) (14.2) (14.3) (14.4)

 v a¨ x:n − a x:n

(14.12)

a¨ x:n  a¨ n + n| a¨ x a¨ x:n  a¨ x − n E x a¨ x+n n| a¨ x  n E x a¨ x+n

(14.6)

a¨ x  a¨ x:n + n| a¨ x

(14.8)

A x1:n Relationships between annuities-due

(14.5) (14.7)

Relationships between annuities-due and annuities-immediate a¨ x  a x + 1 a¨ x:n  a x:n−1 + 1 n| a¨ x

(14.13) (14.14)

 a x:n + 1 − n E x

(14.15)

 n| a x + n E x

(14.16)

Other annuity equations a¨ x:n 

n−1 X

a¨ k

k−1 p x

q x+k−1 + a¨ n

n−1 p x

(14.9)

k1

a¨ x:n 

n−1 X

v k k px

(14.10)

k0

a¨ x 

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1+i q+i

if q x is constant

(14.11)

Exercises continue on the next page . . .

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276

14.2. [3-S00:5] An insurance company has agreed to make payments to a worker age x who was injured at work. • The payments are 150,000 per year, paid annually, starting immediately and continuing for the remainder of the worker’s life. • After the first 500,000 is paid by the insurance company, the remainder will be paid by a reinsurance company. • •

  0.7t ,   0,  i  0.05

tpx

0 ≤ t ≤ 5.5 5.5 < t

Calculate the actuarial present value of the payments to be made by the reinsurer. A. B. C. D. E. 14.3.

Less than 50,000 At least 50,000, but less than 100,000 At least 100,000, but less than 150,000 At least 150,000, but less than 200,000 At least 200,000 [CAS3-S04:12] You are given the following:

•

The probability that a newborn lives to be 25 is 70%

•

The probability that a newborn lives to be 35 is 50%

•

The following annuities-due each have an actuarial present value equal to 60,000: a life annuity-due of 7,500 on (25) a life annuity-due of 12,300 on (35) a life annuity-due of 9,400 on (25) that makes at most 10 payments What is the interest rate?

A. 8.0%

B. 8.1%

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C. 8.2%

D. 8.3%

E. 8.4%

Exercises continue on the next page . . .

EXERCISES FOR LESSON 14

277

14.4. [Based on CAS4A-F96:7] (2 points) A life, age 65, is subject to mortality as described in the following excerpt from a mortality table: x

lx

65 66 67 68 69 70 71

5,000 4,750 4,500 4,200 3,800 3,300 2,800

For some interest rate i, the actuarial present value of a 2-year temporary life annuity-immediate for this life equals the actuarial present value of a 2-year deferred, 3-year temporary life annuity-immediate for the same life. Determine the range in which the interest rate, i, lies. A. B. C. D. E. 14.5.

Less than 7% At least 7%, but less than 8% At least 8%, but less than 9% At least 9%, but less than 10% At least 10% The number of full years of future lifetime of ( x ) has the following distribution: Full years of future lifetime

Probability

0 1 2 3 4

0.1 0.1 0.2 0.2 0.4

i  0.03 Calculate a x:3 .

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Exercises continue on the next page . . .

14. ANNUITIES: DISCRETE, EXPECTATION

278

14.6. [Based on CAS4A-F97:21] (2 points) Two lives, ages 50 and 51, were selected at their current ages and are subject to mortality as described in the following excerpt from a mortality table: x

lx

50 51 52 53 54 55 56

5,000 4,700 4,500 4,200 3,775 3,350 2,625

For some annual interest rate i, the actuarial present value of a 1-year deferred, 2-year temporary life annuity-immediate for the life age 50 equals the actuarial present value of a 2-year deferred, 3-year temporary annuity-immediate for the life age 51. Determine the annual rate of interest, i. A. B. C. D. E.

Less than 12.0% At least 12.0%, but less than 12.5% At least 12.5%, but less than 13.0% At least 13.0%, but less than 13.5% At least 13.5%

14.7. [CAS4-F82:25] An annuity of 1, issued at age 35, is payable at the beginning of each year until age 65. The annuity payments are certain for the first 15 years. You are given: a¨15  11.94 a¨30  19.60

a¨35:15  11.62 a¨35:30  18.13

a¨35  21.02 a¨50  15.66 a¨65  9.65

Calculate the present value of the annuity. A. B. C. D. E.

Less than 18.00 At least 18.00, but less than 18.50 At least 18.50, but less than 19.00 At least 19.00, but less than 19.50 At least 19.50

[CAS4A-F98:18] (2 points) Mortality follows the Illustrative Life Table. i  0.06. Determine a¨ .

14.8.

65:10

A. B. C. D. E.

Less than 10.0 At least 10.0, but less than 10.5 At least 10.5, but less than 11.0 At least 11.0, but less than 11.5 At least 11.5

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Exercises continue on the next page . . .

EXERCISES FOR LESSON 14

279

14.9. [CAS4A-S97:2] (2 points) John (age ( x ) ) and Paul (age ( y ) ) are coworkers who are injured in a common occurrence. John’s first-year medical expenses are $200,000, with subsequent years’ expenses increased by 12% annual inflation. Paul’s first-year indemnity payment is $50,000, with subsequent years’ payments increased by 7% annual inflation. Their expected mortality is described in the following table: Expected Mortality for: John Paul t l x+t l y+t 0 1 2 3 4

1,000 800 700 600 450

1,000 900 800 650 500

John receives no indemnity payments and Paul receives no medical benefits. Payments are made at the end of each year the claimant survives. Determine the insurer’s expected total payment at the end of the third year after the accident (t  3). A. B. C. D. E.

Less than $175,000 At least $175,000, but less than $190,000 At least $190,000, but less than $205,000 At least $205,000, but less than $220,000 At least $220,000

14.10. For a 10-year deferred annuity-due on (55), you are given • Survival time is uniformly distributed with limiting age 110. • i  0. Calculate the actuarial present value of the annuity. 14.11. [150-S91:13] You are given: • •

µ x+t  µ t≥0 δt  δ t≥0

Which of the following is a correct expression for a¯ x:1 ? I. II. III. IV.

1 − e − ( µ+δ ) µ + δ2 a¨ x:1 + a x:1 2 ¨a x:1 − a x:1 − ln a x:1 1 − e − ( µ+δ ) ln a x:1

A. None

B. I only

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C. II only

D. III only

E. IV only

Exercises continue on the next page . . .

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280

14.12. [CAS4A-S99:18] (2 points) Mike is 30 years old. Beginning one year from today, he will receive $5,000 annually for as long as he lives. p x  k for all x. The present value of the annuity is $22,500. Using i  0.10, calculate k A. B. C. D. E.

Less than 0.800 At least 0.800, but less than 0.825 At least 0.825, but less than 0.850 At least 0.850, but less than 0.875 At least 0.875

14.13. [CAS4A-F93:3] (2 points) You are given the following insurance and annuity options for a single insured: 1. 2. 3. 4.

10-year pure endowment insurance of 1. Whole life annuity immediate with annual payments of 1. Whole life annuity due with annual payments of 1. 10-year temporary life annuity immediate with annual payments of 1.

Rank the actuarial present value of each option from the largest to the smallest. A. B. C. D. E.

4. 3. 2. 2. 3.

> 3. > 2. > 4. > 3. > 2.

> 2. > 1. > 3. > 4. > 4.

> 1. > 4. > 1. > 1. > 1.

14.14. [CAS3-F04:1] A special 30-year annuity-due on a person age 30 pays 10 for the first 10 years, 20 for the next 10 years and 30 for the last 10 years. Given: m

•

20 E30

•

a¨30:10  u

•

a¨30:30  v

•

a¨50:10  w Which of the following represents the actuarial present value of this annuity?

A. B. C. D. E.

20v + 10w (1 − m ) 10u + 20v + 10w 10u + 20v + 10mw −10u + 20v + 10mw −20u + 20v + 10w

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Exercises continue on the next page . . .

EXERCISES FOR LESSON 14

281

14.15. [CAS4-S87:18] (2 points) Which of the following is equal to n| a¨ x−n n E x−n

− a¨ x:n − a x+n n E x ?

A. −n E x B. n E x C. n+1 E x E. The correct answer is not given by A. , B. , C. , or D.

D. −n+1 E x

14.16. [150-F96:4] Which of the following is equivalent to: ia x:n + (1 + i ) A x:n − 1? A. 0

B. i n E x

C. i

D.

E. 1

n Ex

14.17. [4-F86:27] Simplify n−1 X * (1 − v k+1 ) k p x q x+k + + d n p x a¨ n . , k1

d a¨ x:n d a¨ x:n − q x d a¨ x:n + vq x d ( a¨ x:n − q x ) d a¨ x:n+1 − v n n p x

A. B. C. D. E.

14.18. [CAS4A-S98:18] (2 points) Which of the following is NOT an expression for a¨ x ?

P∞

v k k px 1 − (1 + i ) A x −1 P∞ i ¨ k0 ( k p x )( q x+k ) a k+1 1 − Ax d All of A, B, C, and D are correct

A.

k0

B. C. D. E.

14.19. [CAS4A-F99:16] (2 points) Which of the following identities are correct?  a x − a x:n−1

1.

n| a¨ x

2.

1 + a x:n−1  (1 + i ) a x:n

3.

A x:n  v a¨ x:n − a x:n−1

A. 1

B. 1, 2

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C. 1, 3

D. 2, 3

E. 1, 2, 3

Exercises continue on the next page . . .

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282

14.20. [SOA3-F04:11] Your company is competing to sell a life annuity-due with an actuarial present value of 500,000 to a 50-year old individual. Based on your company’s experience, typical 50-year old annuitants have a complete life expectancy of 25 years. However, this individual is not as healthy as your company’s typical annuitant, and your medical experts estimate that his complete life expectancy is only 15 years. You decide to price the benefit using the issue age that produces a complete life expectancy of 15 years. You also assume: • For typical annuitants of all ages, l x  100 ( ω − x ) , 0 ≤ x ≤ ω. • i  0.06 Calculate the annual benefit that your company can offer to this individual. A. 38,000

B. 41,000

C. 46,000

D. 49,000

E. 52,000

C. 10.4

D. 10.9

E. 11.1

C. 11.7

D. 12.0

E. 12.3

14.21. [150-S98:9] You are given: • • • •

A x  0.22 A x+25  0.46 1  0.20 A x:25 i  0.06

Calculate a x:25 . A. 9.8

B. 10.1

14.22. [3-F01:26] You are given: • • • •

A x  0.28 A x+20  0.40 1 A x:20  0.25 i  0.05

Calculate a x:20 . A. 11.0

B. 11.2

14.23. You are given: • • • •

a¨ x:29  17.1 a¨ x:30  17.4 a¨ x:31  17.6 d  0.03

Calculate A x1:30 .

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Exercises continue on the next page . . .

EXERCISES FOR LESSON 14

283

14.24. [150-S90:6] For a 30-year deferred, annual life annuity-due of 1 on (35), you are given: R is the net single premium for this annuity if the net single premium is refunded at the end of the year of death for death during the deferral period. • N is the net single premium for this annuity if the net single premium is not refunded. •

Which of the following correctly expresses R − N? A35:30 · 30| a¨35

I.

1 1 − A35 :30 1 A35 :30 ( A35:30 − A35 )

II.

1 d (1 − A35 :30 )

(1 − d a¨35:30 ) 30| a¨35

III.

d a¨35:30

A. None B. I only C. II only E. The correct answer is not given by A. , B. , C. , or D.

D. III only

14.25. You are given • • •

A25:10  0.84 A25:20  0.69 i  0.02

Calculate 10| a¨ 25:10 . 14.26. [150-F87:1] You are given: • • •

 0.35 a30:9  5.6 i  0.10

10 E30

1 Calculate A30 :10 .

A. 0.05

B. 0.10

C. 0.15

D. 0.20

E. 0.25

14.27. [150-81-94:19] Consider the following present-value random variables, where K is the curtate future lifetime of ( x ) : Y  a¨ K+1

  a¨ K+1 Z  a¨ n

K≥0 0≤K 0 i  0.11

• •

Calculate the actuarial present value of this annuity. A. 264

B. 266

C. 268

D. 270

E. 272

14.38. [MLC-S07:29] For a special fully discrete, 30-year deferred, annual life annuity-due of 200 on (30), you are given: • The net single premium is refunded without interest at the end of the year of death if death occurs during the deferral period. • Mortality follows the Illustrative Life Table. • i  0.06 Calculate the net single premium for this annuity. A. 350

B. 360

C. 370

D. 380

E. 390

14.39. [Sample Question 285] You are given: • The force of mortality follows Makeham’s law where A  0.00020, B  0.000003, and c  1.10000. • The annual effective rate of interest is 5%. Calculate 1| a 70:2 . A. 1.73 B. 1.76 C. 1.79 D. 1.82 E. 1.85 Additional old CAS Exam 3/3L questions: S05:5, F08:20, S09:13, F09:12, S10:15, S11:12, F12:12, S13:12,13, F13:1 Additional old SOA Exam MLC questions: F13:14

Solutions 14.1.

We add up the probability of living exactly k years times a¨ k , as indicated by equation (14.9). a¨ x:4  0.33 (1.00) + 0.24 (1.93) + 0.16 (2.80) + (1 − 0.33 − 0.24 − 0.16)(3.62)  2.2186

(D)

The last probability is not 0.11, since if the insured survives three years he gets four payments regardless of whether or not he dies in the fourth year. 14.2. The insurer pays 150,000 for 3 years followed by 50,000 in the fourth year. The reinsurer pays 100,000 in the fourth year followed by 150,000 afterwards, but the worker cannot survive past the fifth year (the sixth payment). The actuarial present value of the payments made by the reinsurer is 100,000 3 E x + 150,000 4 E x + 150,000 5 E x 0.73  0.296296 1.053 0.74  0.197531 4 Ex  1.054 0.75  0.131687 5 Ex  1.055

3 Ex

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EXERCISE SOLUTIONS FOR LESSON 14

287

APV of payments  100,000 (0.296296) + 150,000 (0.197531) + 150,000 (0.131687)  79,012 14.3. Translating the first two bullets into actuarial symbols, s (25)  0.7 and S0 (35)  0.5, so 10p 25  5 7 . Then we have 3 annuities-due:

(B) 0.5 0.7



60,000 8 7,500 60,000   4.878 12,300 60,000   6.383 9,400

a¨25  a¨35 a¨25:10

Now we use the fact that the sum of a temporary and deferred annuity is a whole life annuity. a¨25  a¨25:10 + v 10 10 p25 a¨35

!

8  6.383 + v v 10 

10

8 − 6.383 4.878

!

5 (4.878) 7 7 5

!

v  0.9261 1 i  − 1  0.0798 v 14.4.

(A)

The equation can be set up as 4750v + 4500v 2  4200v 3 + 3800v 4 + 3300v 5

where we’ve multiplied t p x through by l65  5000. I don’t see how to solve this quartic equation (after dividing out v) with the calculators allowed at the time of this exam other than by trial and error; you’re not expected to know the quartic equation, and it’s pretty messy to use anyway! Plugging in 7%, 8%, and 9% shows that the solution is between 8% and 9%. (C) 14.5.

There is a 0.6 probability of three payments and a 0.4 probability of four payments. a x:3  0.6a3 + 0.4a4

 







0.6 1 − (1/1.03) 3 + 0.4 1 − (1/1.03) 4



0.03 0.6 (0.084858) + 0.4 (0.111513)   3.184 0.03

14.6.

This time, the equation is 4500v 2 + 4200v 3 3775v 3 + 3350v 4 + 2625v 5  5000 4700

which is “only” a cubic. Trial and error demonstrates that the interest rate that solves this is higher than 13.5%. (E) I wonder what the CAS had in mind in these two questions (14.4 and this one), whether they really expected students to plug in four possibilities or whether there is some shortcut that I am not aware of. With modern calculators like the BA-II, however, you can solve for a rate of return. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

14. ANNUITIES: DISCRETE, EXPECTATION

288

14.7. This annuity is the sum of a 15-year annuity-certain and a 15-year deferred 15-year temporary life annuity on (35). The former is 11.94, and the latter is equal to a 30-year annuity minus a 15-year annuity on (35), or 18.13 − 11.62. The sum of the two annuities is 11.94 + 18.13 − 11.62  18.45 . (B) 14.8. We use a¨65:10  10 E65 a¨75 + a¨10 . The annuity factor a¨75 and the pure endowment factor 10 E65 at 6% can be read right off the Illustrative Life Table. 1 − (1/1.0610 )  7.801692 0.06/1.06  10 E65 a¨75 + a¨10

a¨10  a¨65:10

 (0.39994)(7.2170) + 7.801692  10.68806 14.9.

(C)

There are two years of inflation by the third year. The answer is 200,000 (1.122 )(0.6) + 50,000 (1.072 )(0.65)  187,737.25

(B)

14.10. Conditioning on 10-year survival, the annuity will make a minimum of 1 and a maximum of 45 payments if the annuitant survives to age 65. Due to uniform distribution, the average number of payments is 23. Letting Y be the actuarial present value of the annuity, 45 (23)  18.8182 55

E[Y]  10p 55 (23) + 10q 55 (0) 

14.11. For expressions III and IV: ln a x:1  ln e − ( µ+δ )  − ( µ + δ ) The correct expression for a¯ x:1 is

1 − e − ( µ+δ ) µ+δ

I should not have a square on δ. II is only an approximation. IV is no good because of the missing negative sign in the denominator. Since a¨ x:1  1 and a x:1  vp x  e − ( µ+δ ) , III is correct. (D) 14.12. a x  22,500/5000  4.5. You may use equation (14.11), by adding 1 to a x to get a¨ x . 1+i  5.5 q+i 1.1  5.5 1.1 − k 1.1  6.05 − 5.5k 4.95 k  0.9 5.5

(E)

If you prefer to work it out from basic principles: 4.5  a x 

∞ X n1

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k 1.1

!n

k/1.1 1 − k/1.1

EXERCISE SOLUTIONS FOR LESSON 14

289

k 1.1 − k 5.5k  4.95 

k  0.9

(E)

14.13. The annuity-due provides one more payment than the whole life annuity immediate, which provides more payments than a 10-year annuity. The pure endowment provides the least payments, only 1 at duration 10. (E) 14.14. We have 10u for the first 10 payments and 30mw for the last 10. The middle 10 are harder. We decompose v: a¨30:30  a¨30:20 + 20 E30 a¨50:10  a¨30:10 + 10| a¨ 30:10 + 20 E30 a¨50:10 v  u + 10| a¨ 30:10 + mw So we need 20 ( v − u − mw ) for the present value of the middle 10 payments. The total actuarial present value is then 10u + 30mw + 20 ( v − u − mw )  −10u + 20v + 10mw . (D) 14.15. n| a¨ x−n  n E x−n a¨ x , so the first fraction is a¨ x . a¨ x:n + n E x a¨ x+n  a¨ x , so the last two summands are lower than a¨ x by the present value of the n th payment, n E x . So the expression reduces to a¨ x − ( a¨ x − n E x )  n E x

(B)

14.16. We worked this out on page 272; see the equation with the star on that page. However, here’s how you could reason if you didn’t remember that formula. Use the formula a¨ x:n  (1 − A x:n ) /d. a x:n  a¨ x:n − 1 + n E x   1 − A x:n ia x:n  i − 1 + n Ex d  (1 + i )(1 − A x:n ) − i + i n E x ia x:n + (1 + i ) A x:n − 1  1 + i − i + i n E x − 1  i n E x

(B)

14.17. Each term of the first summand pays a certain annuity-due of d with k +1 payments if one survives exactly k years, 0 < k < n. The second summand pays a certain annuity-due of d with n payments if one survives at least n years. So the sum of the two summands pays d a¨ x:n , except if one survives zero years. The probability of surviving zero years is q x , and in that case payments included in d a¨ x:n are d. So the answer is (D). 14.18.

(B) would be correct if 1 were added rather than subtracted; then it would be a x + 1.

14.19. 1 provides payments of 1 starting at duration n on both sides of the equation, and so is correct. 2 is not correct. Even though both sides provide payments of 1 from duration 0 to duration n − 1, the left side conditions the payments on being alive at the beginning of the year whereas the right side conditions the payment at the beginning of each year on being alive at the end of the year. 3 is correct, and can be derived from A x:n  1 − d a¨ x:n  1 − (1 − v ) a¨ x:n  v a¨ x:n − ( a¨ x:n − 1)  v a¨ x:n − a x:n−1 (C) LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

14. ANNUITIES: DISCRETE, EXPECTATION

290

14.20. Since the complete life expectancy for typical annuitants is 25, under uniform mortality this must be half of the maximum lifetime, which is therefore 50, and ω  100. For a life expectancy of 15, you’d make the issue age 100 − 2 (15)  70 so that maximum future lifetime would be twice expectancy. We’ve mentioned that for uniform mortality, it is easier to calculate insurance and then to use formula (14.1) to convert it to an annuity-due. The density is 1/30, so 30

A70

1 1 X 1  a  30 1.06t 30 30 t1

1 1 − (1/1.0630 ) 30 0.06 0.8259  0.458828 (30)(0.06) 1 − A70 d 1 − 0.458828  9.5607 0.06/1.06

!

  a¨70  

!

The benefit is then 500,000/9.5607  52,297 . (E) 14.21. First let’s calculate A x:25 . To calculate this endowment insurance, we take whole life, remove the discounted value of insurance after 25 years, and add in a pure endowment. A x:25  0.22 − 0.20 (0.46) + 0.20  0.328 Then we convert it into an annuity-due. a¨ x:25 

1 − 0.328  11.872 0.06/1.06

Finally we convert that into an annuity-immediate by removing the initial payment of 1 and adding a payment at the end of the 25th year. a x:25  11.872 − 1 + 0.20  11.072

(E)

14.22. Use equation (14.3). A x:20  0.28 − 0.25 (0.40) + 0.25  0.43 1 − 0.43 a¨ x:20   11.97 0.05/1.05 a x:20  11.97 − 1 + 0.25  11.22

(B)

14.23. Use formula (14.12). a x:30  a¨ x:31 − 1  16.6 A x1:30  v a¨ x:30 − a x:30  (1 − 0.03)(17.4) − 16.6  0.278

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EXERCISE SOLUTIONS FOR LESSON 14

291

14.24. The annuity with a refund is 1 R  A35 :30 R + 30| a¨35 30| a¨35 R 1 1 − A35 :30

The difference is then

1 A35 :30 1 1 − A35:30

30| a¨35

and 30| a¨35 can be expressed as 30| a¨35

 a¨35 − a¨35:30 A − A35  35:30 d

Putting the last two expressions together gets II. The other expressions don’t work. I is obviously mistaken. III is wrong because there is no general way to convert a term insurance into d times an annuity. (C) 14.25. A 10=year deferred 10-year temporary life annuity-due makes payments at the beginning of years 10, 11, . . . , 19, so it can be expressed as a 20-year temporary life annuity-due minus a 10-year temporary life annuity-due. 10| a¨ 25:10

 a¨25:20 − a¨25:10 1 − A25:10 1 − A25:20  − d! d
1.02  A25:10 − A25:20 0.02 (1.02)(0.15)   7.65 0.02

14.26. a¨30:10  5.6 + 1  6.6

!

1 A30 :10

14.27.

0.1 1− 6.6 − 0.35  0.05 1.1

(A)

E[Y] is a¨ x , whereas E[Z] is a¨ x:n . 1 − Ax (1 − 0.20755)(1.06)   14 d 0.06  a x:n−1 + 1  7

a¨ x  a¨ x:n So the difference is 14 − 7  7 . (D)

14.28. The annuity is like a 10-year deferred annuity due except that it also pays a¨10 if (45) survives 10 years. Therefore E[Y]  10 E45 a¨55 + 10 p45 a¨10 8,640,861  (0.52652)(12.2758) + 9,164,051

!

1 − (1/1.0610 ) 0.06/1.06

 6.46345 + (0.942908)(7.801692)  13.82

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!

14. ANNUITIES: DISCRETE, EXPECTATION

292

14.29. To convert n| a¨ x to n| a x , use n| a x  n| a¨ x − n E x . 1 − 0.129  15.280702 0.057 1 − 0.563 a¨ x:n   7.666667 0.057 n| a¨ x  15.280702 − 7.666667  7.614035 a¨ x 

n| a x

 7.614035 − 0.543  7.071035

14.30. We need a¨40:10 . a¨40:10  10| a¨ 40 + a¨10 1 1 A40  A40 :10 + A40:10 A50 1 (0.35) 0.30  0.09 + A40:10 1 A40:10  0.6

A40:10  0.09 + 0.6  0.69 A40:10 − A40 10| a¨ 40  d (0.69 − 0.3)(1.04)  10.14  0.04 (1 − 1/1.0410 )(1.04)  8.435332 a¨10  0.04 a¨40:10  10.14 + 8.435332  18.575332 Thus K  10,000/18.575332  538.35 . (A) 14.31. Let Y be the present value random variable for the annuity. 1 E[Y]  30| a¨35 + E[Y]A35 :30
1  A35:30 a¨65 + 0.07 E[Y]

 (0.21 − 0.07)(9.90) + 0.07 E[Y] (0.14)(9.90) E[Y]  1 − 0.07  1.490323 (C) 14.32. n| a x

 n E x a x+n

!

1 − (1 + i ) A x+n i E − ( 1 + i ) n x n E x A x+n  i i n| a x  n E x − (1 + i ) n| A x (E)  n Ex

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(A)

EXERCISE SOLUTIONS FOR LESSON 14

293

14.33. 1 is a well-known formula. ! 2 is backwards. n| a¯ x

 a¯ x − a¯ x:n 

(1 − A¯ x ) − (1 − A¯ x:n ) δ

so the order of the numerator should be reversed. # 3 is true, since A x  1 − d a¨ x  1 − (1 − v ) a¨ x  v a¨ x + 1 − a¨ x !

 v a¨ x − a x (D) 14.34. a¨50:10 

1 − A50:10 (1 − 0.5713)(1.06)   7.5737 d 0.06

(D)

14.35. The 15-year annuity on (25) can be broken down into the first 10 years and the last 5 years: a¨25:15  a¨25:10 + v 10 10 p 25 a¨35:5 9.868  a¨25:10 + 0.9510 (0.87)(4.392) a¨25:10  9.868 − 2.288  7.580 Then we convert this into a 10-year endowment on (25): A25:10  1 − (1 − v )(7.580)  1 − 0.05 (7.580)  0.621 Finally we remove the pure endowment at (35). Apparently the ranges in the solution were spread out a lot to allow someone forgetting to do this to answer (D). 10 1 A25 :10  0.621 − 0.95 (0.87)  0.100

(A)

14.36. We’ll first calculate a monthly term insurance. The time to the limiting age is 110 − 65  45. (12)

1 − v 20 45i (12) 1 − 1/1.0520   0.283229 45 (12)(1.051/12 − 1)

1 A65  :20

We add a pure endowment to obtain an endowment insurance. (12) A65:20  0.283229 + v 20 20p 65  0.283229 +

25/45  0.283229 + 0.209383  0.492612 1.0520

Then we translate this into an annuity-due (12) a¨65:20 

(12) 1 − A65:20

d (12)







d (12)  12 1 − (1 − d ) 1/12  12 1 − LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

1  0.048691 1.051/12



14. ANNUITIES: DISCRETE, EXPECTATION

294

(12) a¨65:20 

1 − 0.492612  10.4206 0.048691

To turn it into an annuity-immediate paying 12 per year, we multiply by 12, subtract 1, and add a 20-year pure endowment. (12)  12 (10.4206) − 1 + 0.209383  124.26 a65:20 14.37. The actuarial present value of the annuity is p, we have from equation (3.7)

P2

k0 ( k p 75 ) /1.11

(equation (14.10)). To go from µ to

x+k

Z * k p x  exp − ,

k

µ t dt + x

-

Therefore 76

Z p75  exp −

! µ t dt

75





 exp −0.01 761.2 − 751.2



 e −0.028495  0.97191 2p 75

100 a¨75:3





 exp −0.01 771.2 − 751.2



 e −0.057065  0.94453   0.97191 0.94453  100 1 + +  264.22 1.11 1.112

(A)

14.38. Let P be the premium. Then we need 1 P  PA30 :30 + 20030| a¨ 30 30 E30 1 A30 :30

 10 E30 20 E40  (0.54733)(0.27414)  0.15005

 A30 − 30 E30 A60  0.10248 − (0.15005)(0.36913)  0.04709 30| a¨ 30  30 E30 a¨60  (0.15005)(11.1454)  1.6723 20030| a¨ 30 (200)(1.6723) P  (A)  350.99 1 1 − 0.04709 1 − A30 :30

14.39. The annuity we need the expected value for pays 1 at times 2 and 3, so we need v 2 2p 70 + v 3 3p 70 . 2p 70

 exp −2A − Bc

70 c

−1 1.12 − 1  exp −0.00040 − 0.000003 (1.170 ) ln c ln 1.1

2

!

70 3p 70  exp −0.00060 − 0.000003 (1.1 ) 1| a 70:2



1.13 − 1 ln 1.1

0.9943955 0.9912108 +  1.758191 1.052 1.053

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!!

(B)

 0.9912108

!!  0.9943955

QUIZ SOLUTIONS FOR LESSON 14

295

Quiz Solutions 14-1. By the formula for constant mortality rates, a¨ x+2  (1 + 0.05/0.95) / (0.3 + 0.05/0.95)  2.985075. The 2-year temporary life annuity’s APV is calculated using the formula (14.10): a¨ x:2  1 + 0.95 (0.9)  1.855 The pure endowment factor for 2 years is 2Ex

 0.952 (0.9)(0.8)  0.6498

Therefore, the APV of the whole life annuity is 1.855 + 0.6498 (2.985075)  3.7947 . 14-2. We’ll calculate A x , then a¨ x , and then subtract 1. a50 1 − (1/1.0550 )   0.365119 50 50 (0.05) 1 − Ax (1 − 0.365119)(1.05) a¨ x    13.3325 d 0.05 a x  13.3325 − 1  12.3325

Ax 

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296

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14. ANNUITIES: DISCRETE, EXPECTATION

Lesson 15

Annuities: Variance Reading: Models for Quantifying Risk (4th or 5th edition) 8.1–8.3

15.1

Whole Life and Temporary Life Annuities

The second moment of a life annuity is not so easy to calculate directly. For a continuous whole life annuity, the definition of second moment is ∞

Z E[ a¯T2 ]  x

a¯ 2t t p x µ x+t dt

0 ∞

Z  0

1 − vt δ

(15.1)

!2 t px

µ x+t dt

(15.2)

so you’d have to expand the square and then you’d get an expression involving v 2t and v t , which you could express in terms of first and second moments of insurance. It’s doable, but takes work. Integrating by parts would reduce the square to a first power but remove µ x+t and would get us something which is neither an insurance nor easy to integrate, so it isn’t too useful. However, for whole life annuities we can calculate the variance directly without going through the second moment by using the relationship to insurance (equation (13.2)): 1 − v Tx δ Var ( v Tx ) ) δ2 2A ¯ x − ( A¯ x ) 2  δ2

a¯Tx  Var ( a¯Tx

(15.3)

A similar formula applies for temporary annuities. Let Y be the random variable for a continuous n-year temporary life annuity on ( x ) . Then Var ( Y ) 

2A ¯

x:n

− ( A¯ x:n ) 2 δ2

(15.4)

The above formulas express the variance of an annuity in terms of moments of an insurance. But using the formula relating annuities to insurance, equation (13.3), but at twice the force of interest (2δ): A¯ x  1 − (2δ ) 2a¯ x

2

(15.5)

This allows expressing the variance of whole life annuities in terms of their first moments at two interest LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

297

15. ANNUITIES: VARIANCE

298

rates: 1 − (2δ ) 2a¯ x − (1 − δ a¯ x ) 2 δ2 2 1 − (2δ ) a¯ x − 1 + 2δ a¯ x − δ2 ( a¯ x ) 2  δ2 2 2 ( a¯ x − a¯ x )  − ( a¯ x ) 2 δ

Var ( Y ) 

By starting with

A¯ x:n  1 − (2δ ) 2a¯ x:n

2

(15.6)

(15.7)

and applying it to equation (15.4), we derive a similar formula relating temporary life annuities to their first moments at two interest rates: 1 − (2δ ) 2a¯ x:n − (1 − δ a¯ x:n ) 2 δ2 2 1 − (2δ ) a¯ x:n − 1 + 2δ a¯ x:n − δ 2 ( a¯ x:n ) 2  δ2 2 2 ( a¯ x:n − a¯ x:n )  − ( a¯ x:n ) 2 δ

Var ( Y ) 

(15.8)

Note that 2a¯ x is simply the first moment of an annuity at twice the force of interest. It is not the second moment of an annuity.

The above formulas can be adapted to annuities-due by replacing δ with d. However, d at twice the force of interest is 2d − d 2 . Formula (15.3) becomes Var ( a¨ K x +1 ) 

2A

x

− (A x ) 2 d2

(15.9)

If Y is the random variable for an n-year temporary life annuity-due, then Var ( Y )  Equation (15.5) becomes

2A

x:n

− ( A x:n ) 2 d2

A x  1 − 2 d 2a¨ x  1 − (2d − d 2 ) 2a¨ x

2

(15.10)

(15.11)

This results in the following version of equation (15.6): Var ( Y ) 

2 ( a¨ x − 2a¨ x ) 2 + a¨ x − ( a¨ x ) 2 d

(15.12)

Equations (15.11) and (15.12) are also true for temporary life annuities-due if the subscripts x are changed to x : n . Example 15A For ( x ) , the force of mortality is µ. The force of interest is δ. Calculate the variance of a continuous 30-year temporary life annuity on ( x ) .

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15.2. OTHER ANNUITIES

299

Answer: As we’ve calculated many times: 1 A¯ x:30  A¯ x1:30 + A¯ x:30 µ (1 − e −30( δ+µ) ) + e −30( δ+µ)  µ+δ

At double the force of interest: A¯ x:30 

2

 µ  1 − e −30 (2δ+µ) + e −30 (2δ+µ) µ + 2δ

So the variance of a 30-year temporary life annuity is µ µ+2δ





1 − e −30 (2δ+µ) + e −30 (2δ+µ) −



µ µ+δ



1 − e −30 ( δ+µ) + e −30 ( δ+µ )



δ2

?

2 

Quiz 15-1 A continuous 20-year temporary life annuity on (40) pays at a rate of 100 per year. You are given: • • • • •

Y is the present-value random variable for the annuity. 1 A¯ 40 :20  0.04 2A ¯ 40:20  0.4 2a¯ 40:20  12 p 20 40  0.95

Calculate Var ( Y ) .

15.2

Other Annuities

A life annuity-immediate is a life annuity-due minus a certain payment of 1 at time 0, so it has the same variance. The variance of an n-year temporary annuity-immediate is the same as the variance of an n + 1 year temporary life annuity-due, since they differ only by a time-0 certain payment of 1. Example 15B A 5-year temporary life annuity-immediate on ( x ) pays 10 per year. You are given: • • • •

 0.04  0.03 5p x  0.94 i  0.05

A x1:5 2A 1 x :5

Calculate the variance of the present value of the payments under this annuity. Answer: We’ll calculate the variance of a 6-year temporary life annuity-due, whose present value we’ll call Y. Notice that a 6-year endowment insurance pays 1 at the end of the sixth year if the insured lives LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

15. ANNUITIES: VARIANCE

300

more than 5 years, so it is unnecessary to know q x+5 to compute it. A x:6  A x1:5 + v 6 5 p x 0.94  0.04 +  0.741442 1.056 0.94 2 A x:6  0.03 +  0.553427 1.0512 0.553427 − 0.7414422 Var ( Y )   1.627625 (0.05/1.05) 2 The variance of the 5-year annuity immediate is 102 (1.627625)  162.7625 .



The variance of a deferred annuity is complicated, so I doubt you would be expected to know it. If you need the formula, you can derive it using conditional variance, conditioning on death before or after time n. Here’s the formula for the variance of a continuous n-year deferred whole life annuity, if you are curious: 2 Var ( Y )  v 2n n p x ( a¯ x+n − 2a¯ x+n ) − ( n| a¯ x ) 2 δ

15.3

Typical Exam Questions

Exam questions may ask you to calculate variances of whole life and temporary annuities by relating them to insurances. Let’s discuss other types of questions you may encounter.

Discrete annuities from first principles This type of question asks you to calculate the variance of a discrete annuity from first principles. That means calculating the first moment, then calculating the second moment using the discrete version of formula (15.1), and then subtracting the square of the first moment from the second moment. For annuitiesdue, these would be the versions of equation (15.1):1 E[Y¨ x2 ] 

∞ X

a¨ 2k

k−1| q x

a¨ 2k

k−1| q x

(15.13)

k1 2 ] E[Y¨ x:n

n X

+ n p x a¨ 2n 

k1

n−1 X

a¨ 2k

k−1| q x

+ n−1p x a¨ 2n

(15.14)

k1

The critical mistake to avoid is do not apply an imitation of equation (13.5) to calculate the second moment. The following equation is incorrect: 2 E[Y¨ x:n ]

n X

v 2k k p x

WRONG!

k0 k p x is not the probability function for the random variable, so squaring the other factor in the sum is not going to get you the second moment.

1These equations use Crofts notation, decorating Y the same way a is decorated to denote the present value random variable for which the corresponding a is the expected value. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

15.3. TYPICAL EXAM QUESTIONS

301

Example 15C For a 3-year temporary life annuity-due on (60) , you are given: • The annuity pays 1 in the first year, 2 in the second year, and 3 in the third year. • x qx 60 61 62 63

0.011 0.014 0.018 0.025

• i  0.05 • Y is the present-value random variable for the annuity. Calculate Var ( Y ) . Answer: We calculate the first and second moments. The first moment can be calculated by the indicator variable technique. However, it will be necessary to calculate a k-year annuity-certain for each k in order to calculate the second moment. Therefore, let’s calculate both moments directly from the definition of moments. The present value of the annuity is 1 if death occurs in the first year. It is 1 + 2/1.05  2.90476 if death occurs in the second year. It is 2.90476 + 3/1.052  5.62585 if death occurs after the second year. The probability of death in the first year is 0.011. The probability of death in the second year is (1 − 0.011)(0.014)  0.013846. The probability of death after the second year is 1 − 0.011 − 0.013846  0.975154. The moments are E[Y]  0.011 (1) + 0.013846 (2.90476) + 0.975154 (5.62585)  5.53729 E[Y 2 ]  0.011 (12 ) + 0.013846 (2.904762 ) + 0.975154 (5.625852 )  0.011 (1) + 0.013846 (8.43764) + 0.975154 (31.65019)  30.99164 Var ( Y )  30.99164 − 5.537292  0.33006 You can verify that 1 + (0.989)(22 ) /1.052 + (32 )(0.989)(0.986) /1.054 is not the second moment.

?



Quiz 15-2 A special annuity on (70) will pay 10 at the end of every tenth year, starting from today, if (70) is alive at that point. You are given: • 10 p 70  0.8 • 20 p 70  0.5 • 30 p 70  0 • i  0.04 Calculate the variance of the present value of the payments on this annuity.

Normal approximation Using the normal approximation, you can calculate the amount of a fund needed to have a certain probability that the funds will suffice to pay all annuity benefits for a large population, as we did in Section 8.6. Example 15D Each of a group of 500 people age (65) receives an annuity-due of 100 per year. Mortality is uniformly distributed with limiting age 100. i  0.03. Using the normal approximation, calculate the size of the fund needed so that there is a 95% probability that the funds will be sufficient to pay all benefits. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

15. ANNUITIES: VARIANCE

302

Answer: We’ll calculate the first and second moments of insurances, since that’s easier than calculating annuity moments. 35

A65 

a 1 X 1  35 t 35 35 1.03 t1

1 1 − (1/1.03) 35   0.613921 35 0.03

!

35

A65 

2

1 X 1 35 1.032t t1



1 1 − (1/1.03) 70  0.409900 35 1.032 − 1

Let Y be the present value random variable for the annuity. 1 − A65 1 − 0.613921   13.255394 d 0.03/1.03 2A − A2 65 0.409900 − 0.6139212 65 Var ( Y )    38.901907 d2 (0.03/1.03) 2 E[Y] 

The fund needed is √ √ 100 (500)(13.255394) + 1.645 (100) 500 38.901901  662,769.68 + 1.645 (13,946.67)  685,712

?



Quiz 15-3 A company has 100 retirees each receiving a continuous pension of 5000 per year for life, starting immediately. The company creates a fund to pay all future pensions for these retirees. You are given: • For each retiree, µ  0.02 • δ  0.08 • The amount of the fund is determined, using the normal approximation, so that the probability that the fund is sufficient to make all payments is 95%. Calculate the initial amount of the fund.

15.4

Combinations of Annuities and Insurances with No Variance

If Y is a whole life continuous annuity and Z is a whole life insurance payable at the moment of death, then Z  v T and Y  (1 − v T ) /δ, so δY + Z  1. This implies that Var ( δY + Z )  0. The same conclusion applies if Y is an n-year temporary continuous annuity and Z an n-year endowment payable at the moment of death. If Y is a whole life annuity-due and Z is a whole life insurance payable at the end of the year of death, then Z  v K+1 and Y  (1 − v K+1 ) /d, so dY + Z  1. This implies that Var ( dY + Z )  0. The same conclusion applies if Y is an n-year temporary life annuity-due and Z an n-year endowment payable at the end of the year of death.   For annuities-immediate, the corresponding statement is that Var iY + (1 + i ) Z  0. Example 15E A life annuity pays 100 per year at the end of each year. The interest rate i  0.06. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

EXERCISES FOR LESSON 15

303

1. You would like to add a death benefit payable at the end of the year of death to the annuity. The amount of the death benefit should be chosen to minimize the variance of the present value of the combination of annuity and insurance. Calculate the amount of the death benefit which would achieve this. 2. You are given that the variance of a whole life insurance of 1 on ( x ) payable at the end of the year of death is 0.04. If a death benefit of 500 is combined with the annuity, calculate the variance of the combination. Answer: 1. The minimum possible variance is 0. We achieve that by making the combination a multiple of ia x + (1 + i ) A x . We’re given an annuity of 100a x , and i  0.06, so we have (100/0.06) ia x . Therefore, the coefficient of A x must be (100/0.06)(1 + i )  106/0.06  1766.67 . 2. The random variable for the package is 1 − vK 100 v − v K+1 + 500v K+1  + 500v K+1 i v i   100 100 (1.06)  + v K+1 − + 500 0.06 0.06

!

100

!

The variance is

 −

100 (1.06) + 500 0.06

2





Var v K+1  (−1266.6667) 2 (0.04)  64,177.78



Exercises Whole life and temporary annuities [4-F86:23] You are given:

15.1.

• Rδ  0 ∞ • 0 t t p x dt  g • Var ( a¯Tx )  h, where Tx is the future lifetime random variable for ( x ) . Express e˚x in terms of g and h. A. h − g

B.

p

h−g

C.

p

g−h

D.

p

2g − h

E.

p

2h − g

15.2. [150-F87:12] Y is the present value random variable for a 30-year temporary life annuity of 1 payable at the beginning of each year while ( x ) survives. You are given: • • • •

i  0.05 30 p x  0.7 2A 1 x :30  0.0694 1 A x :30  0.1443

Calculate E[Y 2 ]. A. 35.6

B. 47.1

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C. 206.4

D. 218.0

E. 233.6 Exercises continue on the next page . . .

15. ANNUITIES: VARIANCE

304

Table 15.1: Formula summary for annuity variance

In the following formulas, each decorated Y represents the present value random variable whose expected value is a with the same decorations. General formulas for second moments E[Y¯ x2 ]  E[Y¨ x2 ] 

Z

∞

0 ∞ X

a¯ 2t t p x µ x+t dt a¨ 2k

k−1| q x

a¨ 2k

2 k−1| q x + n p x a¨ n 

(15.1) (15.13)

k1 2 E[Y¨ x:n ]

n X k1

n−1 X

a¨ 2k

k−1| q x

+ n−1p x a¨ 2n

(15.14)

k1

Special formulas for variance of whole life annuities and temporary annuities 2A ¯ x − ( A¯ x ) 2 Var ( Y¯ x )  δ2 2 ( a¯ x − 2a¯ x )  − ( a¯ x ) 2 δ 2A ¯ x:n − ( A¯ x:n ) 2 Var ( Y¯ x:n )  δ2 2 ( a¯ x:n − 2a¯ x:n ) − ( a¯ x:n ) 2  δ 2A − ( A ) 2 x x Var ( Y¨ x )  Var ( Yx )  d2 2 ( a¨ x − 2a¨ x ) 2  + a¨ x − ( a¨ x ) 2 d 2A 2 x:n − ( A x:n ) Var ( Y¨ x:n )  Var ( Yx:n−1 )  2 d

(15.3) (15.6) (15.4)

(15.10) (15.12) (15.10)

15.3. [150-F88:7] Z is the present-value random variable for an n-year endowment insurance of 1 issued to ( x ) . The death benefit is payable at the end of the year of death. Y is the present-value random variable for a special n-year temporary life annuity issued to ( x ) . A payment of 1 is made at the end of each year for n years if ( x ) is alive at the beginning of that year. You are given: • Var ( Z )  0.02 • i  0.05 Calculate Var ( Y ) . A. 8.0

B. 8.4

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C. 8.8

D. 9.2

E. 9.6

Exercises continue on the next page . . .

EXERCISES FOR LESSON 15

305

[150-S89:18] You are given:

15.4.

• Var ( a¯Tx )  100/9 • µ x+t  k for all t • δ  4k Calculate k. A. 0.005

B. 0.010

C. 0.015

D. 0.020

E. 0.025

[150-82-94:3] Y is the present value random variable for a continuous whole life annuity of 1 on

15.5. (x ).

You are given: • • •

µ x+t  µ a¯ x  5 2a¯  4 x

t≥0

Calculate Var ( Y ) . A. 15 15.6.

B. 25

C. 35

D. 45

E. 55

You are given: • The second moment of a whole life continuous annuity of 1 on ( x ) is 400. • The second moment of a whole life insurance of 1 on ( x ) with payment at the moment of death is 0.4. • 2a¯ x  15.

Determine a¯ x . 15.7.

You are given: • Y is the present value random variable for a whole life continuous annuity on ( x ) . • E[Y]  15. • E[Y 2 ]  240. • δ  0.05.

Calculate 2A¯ x . 15.8.

You are given: • Y is the present value random variable for a whole life annuity-due on ( x ) . • E[Y]  15. • E[Y 2 ]  288. • d  0.05.

Calculate 2A x .

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Exercises continue on the next page . . .

15. ANNUITIES: VARIANCE

306

Y is the present value random variable for a whole life continuous annuity of 1 on ( x ) .

15.9.

Z is the present value random variable for a whole life insurance of 1 on ( x ) with payment at the moment of death. You are given: • µ x+t  µ t>0 • δ  0.06 • E[Y 2 ]  125 Calculate E[Z 2 ]. 15.10. You are given: • • • •

d  0.05 a¨ x  12. 2a¨  8. x Y is the present value random variable for a whole life annuity-due of 1 on ( x ) .

Calculate Var ( Y ) . 15.11. [CAS4A-S92:11] (3 points) You are given: • a¯ x  15 • 2a¯ x  9 • δ  0.05 • Tx is the random variable denoting the future lifetime of ( x ) . Determine E[ a¯T2 ]. x

A. B. C. D. E.

Less than 100 At least 100, but less than 150 At least 150, but less than 200 At least 200, but less than 250 At least 250

15.12. [150-83-96:1] You are given: • Tx is the random variable for the future lifetime of ( x ) . • µ x+t  µ, t ≥ 0 is the constant force of mortality. • δµ Determine Var ( a¯Tx ) . 1 A. 12µδ

B. µ + δ

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1 C. 6µδ

µ 1 D. 2 δ 4δ + 1

! E.

1 6µδ

!2

Exercises continue on the next page . . .

EXERCISES FOR LESSON 15

307

15.13. [150-F96:10] For a 3-year temporary life annuity-due on (30), you are given: S0 ( x )  1 − x/80, 0 ≤ x ≤ 80 i  0.05   a¨ K30 +1 , K30  0, 1, 2 Y  a¨3 , K 30  3, 4, 5, . . .

• • •

 Calculate Var ( Y ) . A. 0.08

B. 0.29

C. 0.36

D. 0.60

E. 0.93

15.14. [CAS4A-S93:14] (1 point) You are given the following: • 2A¯ x  0.08 • δ  0.10 • Var ( a¯T )  4 Determine the expected present value of a continuous whole life annuity of 1 payable while ( x ) survives. A. B. C. D. E.

Less than 1 At least 1, but less than 4 At least 4, but less than 7 At least 7, but less than 10 At least 10

Use the following information for questions 15.15 through 15.18: For ( x ) , you are given: • Tx is the random variable for the future lifetime of ( x ) . • µ x+t  0.04, t ≥ 0 • δ  0.06 15.15. [150-83-96:43] Determine A. −0.1ne −0.1n

∂ ( E ). ∂n n x

B. −0.1e −0.1n

C. −0.1e −0.1

D. 0.1e −0.1n

E. 0.1ne −0.1n

D. 25.0

E. 42.3

D. 0.33e −0.1n

E. 2e −0.1n

15.16. [150-83-96:44] Calculate the standard deviation of a¯T . A. 5.0

B. 6.0

C. 6.5

15.17. [150-83-96:45] Determine a¯ x:n . A. B. C. D. E. 15.18.

10 (1 − e −0.1 ) 10 (1 − e −0.1n ) 10 (1 + e −0.1n ) 10 (1 − 0.67e −0.1n ) 10 (1 + 0.67e −0.1n ) [150-83-96:46] Determine

A. −2e −0.1n

∂ ¯ ( a x:n ∂x

B. −0.33e −0.1n

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). C. 0

Exercises continue on the next page . . .

15. ANNUITIES: VARIANCE

308

15.19. [CAS4A-F97:12] (2 points) Y is the expected present value of an n-year temporary life annuity of 1 per annum, payable continuously while ( x ) survives during the next n years. That is to say E[Y]  a¯ x:n . You are given that a¯ x:n  4.9 a¯ x:n  3.6

2

δ  0.095 Determine Var ( Y ) . A. B. C. D. E.

Less than 2.40 At least 2.40, but less than 3.00 At least 3.00, but less than 3.60 At least 3.60, but less than 4.20 At least 4.20

15.20. [3-F01:3] For a continuous whole life annuity of 1 on ( x ) : • Tx is the future lifetime random variable for ( x ) . • The force of interest and force of mortality are constant and equal. • a¯ x  12.50 Calculate the standard deviation of a¯Tx . A. 1.67

B. 2.50

C. 2.89

D. 6.25

E. 7.22

D. 18

E. 20

15.21. [MLC-S07:2] You are given: • µ x+t  c, t ≥ 0 • δ  0.08 • A¯ x  0.3443 • Tx is the future lifetime random variable for ( x ) . Calculate Var ( a¯Tx ) . A. 12

B. 14

C. 16

15.22. [3-F01:17] For a group of individuals all age x, you are given: • 30% are smokers and 70% are non-smokers. • The constant force of mortality for smokers is 0.06. • The constant force of mortality for non-smokers is 0.03. • δ  0.08



Calculate Var a¯Tx A. 13.0



for an individual chosen at random from this group.

B. 13.3

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C. 13.8

D. 14.1

E. 14.6

Exercises continue on the next page . . .

EXERCISES FOR LESSON 15

309

15.23. [SOA3-F03:15] For a group of individuals all age x, of which 30% are smokers and 70% are nonsmokers, you are given: • δ  0.10 • A¯ smoker  0.444 x non-smoker ¯ • Ax  0.286 • Tx is the future  lifetime of ( x ) • Var a¯Tsmoker  8.818 x

 • Var

a¯Tnon-smoker x



Calculate Var a¯Tx A. 8.5





 8.503

for an individual chosen at random from this group.

B. 8.6

C. 8.8

D. 9.0

E. 9.1

15.24. [CAS4A-F99:17] (2 points) A company has just assumed responsibility for paying pensions to a group of 100 people who have just turned 65. The pension will pay each of them $25,000 annually, payable continuously while they are alive. Assume that all the lives are independent, and that the same mortality table applies to all of them. In addition, you are given: •

δ  0.06

•

The expected present value of the portfolio of 100 pensions is $30,000,000.

•

2A ¯

65

 0.2

Calculate the standard deviation of the total present value of the payments for these 100 pensions. A. B. C. D. E.

Less than $1,000,000 At least $1,000,000, but less than $1,250,000 At least $1,250,000, but less than $1,500,000 At least $1,500,000, but less than $1,750,000 At least $1,750,000

15.25. A special life annuity on (35) pays 1 per year at the end of each year until five years after (35)’s death. In other words, the annuity makes five annual payments after (35)’s death in addition to any payments made during (35)’s life. You are given: • Y is the present value random variable for this annuity. • Mortality follows the Illustrative Life Table. • i  0.06 Calculate Var ( Y ) .

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Exercises continue on the next page . . .

15. ANNUITIES: VARIANCE

310

15.26. You are given: • • • •

A x  0.3 2A  0.16 x

d  0.04 Z is the present value random variable for a unit whole life insurance on ( x ) with benefits paid at the end of the year of death. • Y is the present value random variable for a unit whole life annuity-due on ( x ) .

Calculate Cov ( Y, Z ) . 15.27. [150-F89:13] Tx is the random variable for the future lifetime of ( x ) . Determine Cov ( a¯Tx , v Tx ) . A. B. C. D. E.

(A¯ 2x − 2A¯ x ) /δ (A¯ 2x − 2A¯ x ) 0

(2A¯ x − A¯ 2x ) (2A¯ x − A¯ 2x ) /δ

Variance from first principles 15.28. [3-F00:20] Y is the present-value random variable for a special 3-year temporary life annuity-due on ( x ) . You are given: • • •

 0.9t , t ≥ 0 K x is the curtate-future-lifetime random variable for ( x ) .  1.00, K x  0     Y   1.87, K x  1   2.72, K  2, 3, . . . x  tpx

Calculate Var ( Y ) . A. 0.19

B. 0.30

C. 0.37

D. 0.46

E. 0.55

15.29. [SOA3-F04:40] For a special 3-year temporary life annuity-due on ( x ) , you are given: •

•

t

Annuity Payment

p x+t

0 1 2

15 20 25

0.95 0.90 0.85

i  0.06

Calculate the variance of the present value random variable for this annuity. A. 91

B. 102

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C. 114

D. 127

E. 139

Exercises continue on the next page . . .

EXERCISES FOR LESSON 15

311

15.30. A deferred 3-year temporary life annuity on (55) will pay 1 per year at the beginning of each year with the first payment made when the annuitant reaches age 65. You are given • • • •

 0.92 p 11 55  0.90 12 p 55  0.88 v  0.95 10 p 55

Calculate the variance of the present value random variable for this annuity. Normal approximation 15.31. Fifty individuals purchase whole life annuities-due paying 100 per year. The force of mortality is a constant 0.01 and the force of interest is 0.05. Using the normal approximation, calculate the size of the initial fund needed so that there is a 90% probability that the fund will be sufficient to make all annuity payments. 15.32. [3-S01:39] A government creates a fund to pay this year’s lottery winners. You are given: • • • • • •

There are 100 winners each age 40. Each winner receives payments of 10 per year for life, payable annually, beginning immediately. Mortality follows the Illustrative Life Table. The lifetimes are independent. i  0.06 The amount of the fund is determined, using the normal approximation, such that the probability that the fund is sufficient to make all payments is 95%.

Calculate the initial amount of the fund. A. 14,800

B. 14,900

C. 15,050

D. 15,150

E. 15,250

15.33. [3-F00:26] A fund is established to pay annuities to 100 independent lives age x. Each annuitant will receive 10,000 per year continuously until death. You are given: • • •

δ  0.06 A¯ x  0.40 2A ¯ x  0.25

Calculate the amount (in millions) needed in the fund so that the probability, using the normal approximation, is 0.90 that the fund will be sufficient to provide the payments. A. 9.74

B. 9.96

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C. 10.30

D. 10.64

E. 11.10

Exercises continue on the next page . . .

15. ANNUITIES: VARIANCE

312

15.34. [M-F05:11] For a group of 250 individuals age x, you are given: • The future lifetimes are independent. • Each individual is paid 500 at the beginning of each year, if living. • A x  0.369131 • 2A x  0.1774113 • i  0.06 Using the normal approximation, calculate the size of the fund needed at inception in order to be 90% certain of having enough money to pay the life annuities. A. 1.43 million

B. 1.53 million

C. 1.63 million

D. 1.73 million

E. 1.83 million

15.35. [M-F06:4] For a pension plan portfolio, you are given: • 80 individuals with mutually independent future lifetimes are each to receive a whole life annuitydue. • i  0.06 • Number of Annual annuity 2A Age annuitants payment a¨ x Ax x 65 75 •

50 30

2 1

9.8969 7.2170

0.43980 0.59149

0.23603 0.38681

X is the random variable for the present value of total payments to the 80 annuitants.

Using the normal approximation, calculate the 95th percentile of the distribution of X. A. 1220

B. 1239

C. 1258

D. 1277

E. 1296

Combinations of insurance and annuity 15.36. A life annuity-due pays 1 per year. i  0.06. You are considering adding a death benefit, payable at the end of the year of death, to this annuity. The amount of the death benefit is chosen so as to minimize the variance of the combined annuity and insurance payments. Calculate the amount of the death benefit. 15.37. [SOA3-F03:32] Your company currently offers a whole life annuity product that pays the annuitant 12,000 at the beginning of each year. A member of your product development team suggests enhancing the product by adding a death benefit that will be paid at the end of the year of death. Using a discount rate, d, of 8%, calculate the death benefit that minimizes the variance of the present value random variable of the new product. A. 0

B. 50,000

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C. 100,000

D. 150,000

E. 200,000

Exercises continue on the next page . . .

EXERCISES FOR LESSON 15

313

15.38. [150-S98:27] You are given: •

Z is the present-value random variable for a special life annuity on (25) which provides the following benefits: (a) 2 per year payable continuously while (25) survives; and (b) 10 payable at the moment of death of (25).

• •

µ25+t  0.02 δ  0.06

Calculate Var ( Z ) . A. 22.46

B. 43.75

C. 66.67

D. 84.37

E. 102.08

15.39. A life annuity-immediate on ( x ) pays 1 per year. The annuity has a death benefit, payable at the end of the year of death, in the amount b. Let Z be the present-value random variable for this annuity. You are given: • t p x  0.99t for all t. • i  0.05. • Var ( Z )  1. Determine the two possible values of b. Additional old CAS Exam 3/3L questions: F05:36, F08:21, S09:12 Additional old SOA Exam MLC questions: F12:19

Solutions 15.1.

Since δ  0:

•

The first integral g is the formula for half the second moment of future lifetime; see equation (5.3).

•

The payout on a continuous annuity paying 1 per year for life is Tx . Since the interest rate is 0, the present value of the payout is also Tx . The variance of a¯Tx is the variance of future lifetime Tx , which f g is E Tx2 − e˚x2 .

Therefore, h  2g − e˚x2 e˚x  15.2.

q 2g − h

(D)

We need A x:30 and 2A x:30 . 0.7  0.306264 1.0530 (1 − 0.3063)(1.05) E[Y]  a¨ x:30   14.5685 0.05 0.7 2 A x:30  0.0694 +  0.106875 1.0560 2 2A −A Var ( Y )  x:30 2 x:30 d A x:30  0.1443 +

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15. ANNUITIES: VARIANCE

314

0.106875 − 0.3062642  5.7671 (0.05/1.05) 2 E[Y 2 ]  Var ( Y ) + E[Y]2 

 5.7671 + 14.56852  218.01 15.3. then

(D)

If the temporary life annuity Y 0 were not special but paid benefits at the beginning of each year, Y0 

1−Z d

Since Y  vY 0 v (1 − Z ) 1 − Z  d i 0.02 Var ( Z )   8.0 Var ( Y )  i2 0.052 Y

15.4.

(A)

If Z is the random variable for an insurance, Var ( Z ) δ2 2A ¯ x − A2 x  δ2

Var ( a¯T ) 



 

µ/ ( µ + 2δ ) − µ/ ( µ + δ ) δ2 1 9

−

1
2 5

16k 2 100 1  9 225k 2 1  0.02 k 50 15.5.

2

(D)

We’ll express Y in terms of insurance, then use A¯ x  1 − δ a¯ x . First we need δ. 1 µ+δ 1 4  2a¯ x  µ + 2δ 5  a¯ x 

0.2  µ + δ 0.25  µ + 2δ δ  0.05 µ  0.15 Now we can calculate Var ( Y ) . − A¯ 2x δ2 (1 − 2δ 2a¯ x ) − (1 − δ a¯ x ) 2  δ2

Var ( Y ) 

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2A ¯

x

EXERCISE SOLUTIONS FOR LESSON 15

315





1 − 2 (0.05)(4) − 1 − 0.05 (5)

2

0.052

 15

(A)

If you memorized equation (15.6) you could calculate Var ( Y ) directly, but I don’t think it’s worth memorizing it. 15.6. Since 2a¯ x  (1 − 2A¯ x ) /2δ (formula (15.5)), we can back out δ: 1 − 0.4  15 2δ 0.6  30δ δ  0.02 By a version of formula (15.6) without the n ’s, with Y being the random variable for our annuity, Var ( Y ) 

2 ( a¯ x − 2a¯ x ) − ( a¯ x ) 2 δ

so the second moment of Y, the variance plus the mean squared, is the first summand of this expression, and we have 2 ( a¯ x − 15)  400 0.02 2 ( a¯ x − 15)  8 a¯ x  19

15.7.

The variance of the annuity is Var ( Y )  E[Y 2 ] − E[Y]2  240 − 152  15. Using equation (15.3), − A¯ 2x 0.052 2¯ 0.0375  A x − A¯ 2 15 

2A ¯

x

x

However, A¯ x  1 − δ a¯ x  1 − 0.05 (15)  0.25. Therefore, 2A¯ x  0.0375 + 0.252  0.1 . Alternatively, using formula (15.6): 2 ( a¯ x − 2a¯ x ) − a¯ 2x δ 2 ( a¯ x − 2a¯ x ) E[Y 2 ]  δ 2 (15 − 2a¯ x ) 240  0.05 12  2 (15 − 2a¯ x )

Var ( Y ) 

a¯ x  9

2

Then using the relationship 2A¯ x  1 − 2δ 2a¯ x , we get A¯ x  1 − 0.1 (9)  0.1

2

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15. ANNUITIES: VARIANCE

316

15.8.

The variance of the annuity is Var ( Y )  E[Y 2 ] − E[Y]2  288 − 152  63. Using equation (15.10), − A2x 0.052 2 0.1575  A x − A2x 63 

2A

x

However, A x  1 − d a¨ x  1 − 0.05 (15)  0.25. Therefore, 2A x  0.1575 + 0.252  0.22 . Alternatively, using formula (15.12): 2 ( a¨ x − 2a¨ x ) 2 + a¨ x − a¨ 2x d 2 ( a¨ x − 2a¨ x ) 2 E[Y 2 ]  + a¨ x d 2 (15 − 2a¨ x ) 2 288  + a¨ x  600 − 39 (2a¨ x ) 0.05 312 2 a¨ x  8 39

Var ( Y ) 

Then using the relationship 2A x  1 − 2 d 2a¨ x , we get





A x  1 − 2 (0.05) − 0.052 (10)  0.22

2

15.9.

By formula (15.3), the variance of Y is Var ( Y ) 

2A ¯

x

− A¯ 2x δ2

and the numerator of that is µ µ − µ + 2δ µ+δ

!2 

µ ( µ + δ ) 2 − µ2 ( µ + 2δ ) ( µ + 2δ )( µ + δ ) 2



µ3 + 2µ2 δ + µδ 2 − µ3 − 2µ2 δ ( µ + 2δ )( µ + δ ) 2



µδ2 ( µ + 2δ )( µ + δ ) 2

Divide this by δ2 , and add the square of the mean of Y to obtain the second moment. µ

1 E[Y ]  + 2 µ+δ ( µ + 2δ )( µ + δ ) µ + µ + 2δ  ( µ + 2δ )( µ + δ ) 2 2  ( µ + 2δ )( µ + δ ) 2

Substitute δ  0.06 and set the second moment equal to 125. 2

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 125

!2

EXERCISE SOLUTIONS FOR LESSON 15

317

1  0.016 62.5 µ2 + 0.18µ − 0.0088  0

µ2 + 0.18µ + 0.0072 

p

−0.18 + 0.182 + 4 (0.0088) µ  0.04 2 Alternatively, use formula (15.6): 2 ( a¯ x − 2a¯ x ) δ 1/ (0.06 + µ ) − 1/ (0.12 + µ ) 62.5  0.06

E[Y 2 ] 

where in the numerator we used that for constant force of mortality, a¯ x  1/ ( µ + δ ) . Solving for µ, 3.75 

1 1 0.06 −  0.06 + µ 0.12 + µ (0.06 + µ )(0.12 + µ ) 0.06 µ2 + 0.18µ + (0.06)(0.12) − 0 3.75 µ2 + 0.18µ − 0.0088  0 −0.18 + 0.26 µ  0.04 2

The second moment of a whole life insurance is µ/ ( µ + 2δ )  0.04/0.16  0.25 . 15.10. This exercise is most easily done with formula (15.12). Var ( Y ) 

2 (12 − 8) + 8 − 122  24 0.05

If you didn’t memorize that formula, you could calculate insurance moments: A x  1 − 0.05 (12)  0.4 2

d  2d − d 2  0.1 − 0.052  0.0975

A x  1 − 0.0975 (8)  0.22

2

Var ( Y ) 

0.22 − 0.42  24 0.052

15.11. We will calculate the variance of a¯Tx . The second moment is the sum of the variance and the mean (15) squared. We will express the annuity in terms of insurance. Var ( a¯Tx ) 

2A ¯

− A¯ 2x 0.052 x





1 − 0.1 (9) − 1 − 0.05 (15) 0.052

0.0375  15 0.052 ]  15 + 152  240 

E[ a¯T2

x

Once again, it’s slightly faster with equation (15.6). LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

(D)

2

15. ANNUITIES: VARIANCE

318

15.12. Var ( a¯Tx )  

 2 ( µ/ ( µ + 2δ ) − µ/ ( µ + δ ) ) δ2 1 3

−

1
2 2

µ2 1  12µ2 which is the same as (A). 15.13. Since this annuity is for only 3 years, there are only 3 alternatives for the value of the payment, so this could be worked out directly, and the official solution did so. However, it looks a little easier to use the formula relating temporary annuities to endowment insurances. We calculate the moments of a 3-year endowment insurance when t| q 30  1/50. 3

A30:3 

1 X 1 47 1 + 50 1.05t 50 1.053

!

t1

1 1 − (1/1.05) 3 0.94  + 50 0.05 1.053

!

 0.02 (2.723248) + 0.94 (0.863838)  0.866472 3

2

A30:3

1 X 1 47 1  + 2t 50 50 1.05 1.056

!

t1

1 1 − (1/1.05) 6 0.94  + 2 50 1.05 − 1 1.056

!

 0.02 (2.475947) + 0.94 (0.746215)  0.750961 Now we can use the formula for variance of annuities in terms of endowment insurances. − ( A30:3 ) 2 d2 0.750961 − 0.8664722  (0.05/1.05) 2 1.052 (0.00018717)   0.082541 0.052

Var ( Y ) 

2A

30:3

15.14. − A¯ 2x 0.102 0.04  0.08 − A¯ 2

4  Var ( a¯T ) 

2A ¯

x

x

A¯ x  0.2 1 − 0.2 a¯ x   8 0.1

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(D)

(A)

EXERCISE SOLUTIONS FOR LESSON 15

319

15.15. n Ex

 e − (0.04+0.06) n

∂ ( n E x )  −0.1e −0.1n ∂n

(B)

15.16. 0.04
2 0.04+0.06 0.062

0.04 0.04+2 (0.06)

Var ( a¯T ) 

−

 25

√ The standard deviation is 25  5 . (A) 15.17. a¯ x:n  

1 − A¯ x:n δ 1−



0.04
0.04+0.06 (1

− e −0.1n ) + e −0.1n



0.06 (B)

 10 − 10e −0.1n

15.18. Based on the previous answer, 0 . (C) But even if you didn’t get the last answer, since µ and δ are independent of x, a¯ x:n must also be. 15.19. The exercise would have the same solution if all the n ’s were removed. We use A¯ x:n  1 − δ a¯ x:n . Doubling the force of interest, 2A¯ x:n  1 − 2δ (2a¯ x:n ) . By formula (15.4), Var ( Y ) 

2A ¯

x:n

− A¯ 2x:n

δ2 1 − 2δ (2a¯ x:n ) − (1 − δ a¯ x:n )  δ2



 

1 − 0.19 (3.6) − 1 − 0.095 (4.9) 0.0952 0.030310  3.3584 0.0952

(C)

15.20. The mean of an exponential annuity is 1/ ( µ + δ ) and δ  µ, so 1  12.5 2µ µ  δ  0.04 µ 1 A¯ x   2µ 2 µ 1 2¯ Ax   3µ 3 Var ( a¯Tx )  √ The standard deviation is 25/ 12  7.2169 . (E) LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

1 3

1 4 0.042

−



252 12

2

15. ANNUITIES: VARIANCE

320

15.21. We will use formula (15.3) on page 297. First we back out c. For an exponential insurance, A¯ x  µ/ ( µ + δ ) . c c + 0.08 0.08 1− c + 0.08

0.3443 

0.08  1 − 0.3443  0.6557 c + 0.08 0.08 c + 0.08  0.6557 0.08 c − 0.08  0.0420 0.6557 Now we can calculate the variance. µ 0.0420   0.2079 µ + 2δ 0.0420 + 0.16  2A¯ x − A¯ 2x 0.2079 − 0.34432    13.962 δ2 0.082

A¯ x 

2



Var a¯Tx

(B)

15.22. We’ll use the formula relating annuities to insurances. The first and second moments are weighted averages of the moments for smokers and the moments for non-smokers. For non-smokers, A¯ x  3/11 and 2A ¯ x  3/19, while for smokers A¯ x  6/14 and 2A¯ x  6/22. So overall

!

!

!

!

6 3 A¯ x  0.3 + 0.7  0.319481 14 11 6 3 A¯ x  0.3 + 0.7  0.192344 22 19

2



Var a¯Tx





0.192344 − 0.3194812  14.1056 0.082

(D)

Note that the variance of the annuity is not the weighted average of the variances of the components. See Section 1.6 for a discussion of the conditional variance formula, the correct method of calculating the variance using the moments of the components. 15.23. The official solution has many methods for doing this. I prefer the method they call the shortest method, the one using the conditional variance formula. Unlike them, I would also use the Bernoulli shortcut and shorten the solution even more! The conditional variance formula says that



Var a¯Tx



f



 E Var a¯Tx | S

g

 f

+ Var E a¯Tx | S

g

where S is smoking habit. We calculate the expected values of the annuities using formula (13.2). 1 − 0.444  5.56 0.1 1 − 0.286   7.14 0.1

a¯ smoker  x a¯ non-smoker x

By the Bernoulli shortcut, the variance of these expected values over the two smoking habits is

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 (7.14 − 5.56) 2 (0.3)(0.7)  0.5242.

EXERCISE SOLUTIONS FOR LESSON 15

321

The expected value of the variances of the annuities over the two smoking habits is 0.3 (8.818) +0.7 (8.503)  8.5975. Therefore, the answer is 8.5975 + 0.5242  9.1217 . (E) The second method they present uses the fact that any moment of a mixture2 is the weighted average of the individual moments. We can obtain the second moments of each component by adding the variance and mean squared, and at the end we can get the variance by subtracting the mean squared from the second moment. In the following, a’s and A’s without superscripts are for a randomly drawn individual. 1 − A¯ smoker 1 − 0.444 x   5.56 x δ 0.10   1 − A¯ non-smoker 1 − 0.286 x E a¯Tnon-smoker  a¯ non-smoker    7.14 x x δ 0.10 g f E a¯Tx  0.3 (5.56) + 0.7 (7.14)  6.666





 E a¯Tsmoker  a¯ smoker x

" E

a¯Tsmoker

2#

x

" E

a¯Tnon-smoker x

 E

a¯Tx



2# 2

Var a¯Tx



 8.818 + 5.562  39.7316  8.503 + 7.142  59.4826  0.3 (39.7316) + 0.7 (59.4826)  53.5573  53.5573 − 6.6662  9.1217

Check the official solution if you are interested in a third method. 15.24. Let Y be the random variable for a pension of 1. Note that by the second bullet, a¯65  12. 2A ¯ 65 − A¯ 2 65 Var ( Y )  0.062



 

0.2 − 1 − 0.06 (12)

30,000,000

(25,000)(100)



2

0.062 33 79

The standard deviation of total payments is

q

25,000 (100)(33 79 )  1,452,966

(C)

15.25. This annuity can be viewed as a 4-year certain annuity-immediate followed by a whole life annuity-due on which each payment is made five years later than usual, minus the payment at time 0. For example, if the annuitant dies in the first year, payments are made at times 1, 2, 3, 4, 5. The first 4 of these payments are a 5-year certain annuity-immediate, and the fifth payment is a 5-year postponed life annuity-due. If the annuitant dies in the second year, the same five payments are made, and in addition a payment is made at time 6, so the 5-year postponed life annuity-due pays at times 5 and 6. The annuity-immediate certain has no variance. The postponed whole life annuity-due is v 5 times a whole life annuity-due starting at time 0, so its variance is v 10 times the variance of a whole life annuitydue on (35), which is 2A − A2 35 0.03488 − 0.128722 35   5.71512 2 d (0.06/1.06) 2 2A mixture distribution is a weighted average of two or more distributions. Here, the distribution is 30% of the smoker distribution plus 70% of the non-smoker distribution. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

15. ANNUITIES: VARIANCE

322

The answer is Var ( Y )  5.71512/1.0610  3.19129 . Another method for solving this question is to express this annuity as a whole life annuity-immediate plus a whole life insurance. The whole life insurance pays a 5-year certain annuity-due at the end of the year of death. Let Y be a whole life annuity-immediate and Z a whole life insurance paying at the end of the year of death. Then we want Var ( Y + Z a¨5 )  Var ( Y ) + ( a¨5 ) 2 Var ( Z ) + 2 a¨5 Cov ( Y, Z ) First, a¨5  (1 − 1/1.065 ) / (0.06/1.06)  4.465106. Then using the Illustrative Life Table, Var ( Z )  2A35 − A235  0.03488 − 0.128722  0.018311 Var ( Y ) 

Var ( Z ) 0.018311 (1.062 )   5.715117 d2 0.062

Now, Y  (1 − Z ) /d, so E[YZ]  E[Z − Z 2 ]/d. So E[Z]  0.12872 E[Y]  15.3926 (0.12872 − 0.03488)(1.06) E[YZ]   1.657840 0.06 Cov ( Y, Z )  1.657840 − (15.3926)(0.12872)  −0.323495 So the variance we want to calculate is Var ( Y + Z a¨5 )  5.715117 + (4.4651062 )(0.018311) + 2 (−0.323495)(4.465106)  3.19131 The difference between this and the previous answer is due to rounding in the Illustrative Life Table. 15.26. 1 − v Tx Cov ( Y, Z )  Cov v , d   1  − Cov v Tx , v Tx d 2A − A2 x x − d 0.16 − 0.32 −  −1.75 0.04

!

Tx

15.27. 1 − vT T Cov ( a¯T , v )  Cov ,v δ   1  − Cov v T , v T δ Var ( v T ) − δ 2 ¯ A − 2A¯ x  x (A) δ

!

T

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EXERCISE SOLUTIONS FOR LESSON 15

323

15.28. We calculate the first and second moments. Note that q x  0.1, 1| q x  0.09, and 2 p x  0.81. E[Y]  0.1 (1) + 0.09 (1.87) + 0.81 (2.72)  2.4715 E[Y 2 ]  0.1 (12 ) + 0.09 (1.872 ) + 0.81 (2.722 )  6.407425 Var ( Y )  6.407425 − 2.47152  0.299113

(B)

15.29. Let Y be the present value random variable. It’s easy to calculate the mean with this data:

!

!

(0.95)(0.90) 0.95 + 25  51.9482 E[Y]  15 + 20 1.06 1.062 but as mentioned in example 15C, calculating the second moment requires using the definition. The annuity is worth 15 if death occurs in the first year, 15 + 20/1.06  33.8679 if death occurs in the second year, and 33.8679 + 25/1.062  56.1178 if death occurs later. The probabilities are q x  0.05 for death in the first year, 1| q x  (0.95)(0.1)  0.095 for death in the second year, and 1 − 0.05 − 0.095  0.855 for death after the second year. So E[Y 2 ]  0.05 (152 ) + 0.095 (33.86792 ) + 0.855 (56.11782 )  2812.794 Var ( Y )  E[Y 2 ] − E[Y]2  2812.794 − 51.94822  114.18

(C)

15.30. The possible present values of the annuity are 0 (probability 0.08), v 10 (probability 0.02), v 10 + v 11 (probability 0.02), and v 10 + v 11 + v 12 (probability 0.88). We calculate the moments. Let Y be the random variable. v 10  0.598737 v 10 + v 11  1.167537 v 10 + v 11 + v 12  1.707897 E[Y]  0.02 (0.598737 + 1.167537) + 0.88 (1.707897)  1.538275 E[Y 2 ]  0.02 (0.5987372 + 1.1675372 ) + 0.88 (1.7078972 )  2.601316 Var ( Y )  2.601316 − 1.5382752  0.23503 15.31. The mean of an annual insurance of 1 is Ax 

∞ X

v k+1 k p x q x+k

k0



∞ X



e −0.05 ( k+1) e −0.01k 1 − e −0.01



k0



 e −0.05 1 − e −0.01

∞ X

e −0.06k

k0



e −0.05



1−

e −0.01



1 − e −0.06

 0.162528

The second moment of an annual insurance of 1 is the same formula at twice the interest rate, or



Ax 

2

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e −0.10 1 − e −0.01 1 − e −0.11

  0.086432

15. ANNUITIES: VARIANCE

324

We need d, and δ  ln (1 + i ) , so 1 + i  e 0.05  1.051271 1 d 1−  0.048771 1+i So the mean and variance of an annuity-due of 1, random variable Y, are 1 − A x 1 − 0.162528   17.172 d 0.048771 2A − A2 x x Var ( Y )  d2 0.086432 − 0.1625282  25.232  0.0487712 E[Y] 

Therefore, the fund needed is for 50 annuities of 100 per year is

p (50)(100)(17.172) + (100)(1.282) (50)(25.232)  90,412 . 15.32. Apparently an annuity-due was intended. We calculate the mean and variance of an annuity of 1. The mean is a¨40  14.8166. By formula (15.10), the variance is 2A

The fund is

40

− A40 2 0.04863 − 0.161322   7.0565 d2 0.056602

p (100)(10)(14.8166) + (10)(1.645) (100)(7.0565)  15,254

(E)

15.33. For an annuity of 1, the mean is (1 − 0.40) /0.06  10 and the variance is (0.25 − 0.402 ) /0.062  25. The fund is p (100)(10,000)(10) + 10,000 (1.282) (100)(25)  10,641,000 (D) 15.34. Let Y be an annuity of 1 on one individual. By formula (14.1), E[Y] 

1 − Ax (1 − 0.369131)(1.06)   11.14535 d 0.06

By formula (15.3), Var ( Y ) 

2A

x

− A2x (0.1774113 − 0.3691312 )(1.062 )   12.84450 d2 0.062

The mean of 250 annuities of 500 is 250 (500)(11.14535)  1,393,169. The standard deviation of 250 √ √ annuities of 500 is 500 250 12.84450  28,333. The normal approximation of the 90th percentile is 1,393,169 + 1.282 (28,333)  1,429,492 . (A) 15.35. The annuity and insurance values are taken from the Illustrative Life Table. Let Yx be the present value random variable for an annuity to ( x ) . For the variance, use formula (15.3), Var ( Yx )  (2A x − A2x ) /d 2 . E[Y65 ]  a¨65  9.8969 Var ( Y65 )  LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

0.23603 − 0.439802  13.29779 (0.06/1.06) 2

EXERCISE SOLUTIONS FOR LESSON 15

325

E[Y75 ]  a¨75  7.2170 Var ( Y75 ) 

0.38681 − 0.591492  11.53237 (0.06/1.06) 2

For the entire portfolio:

X  50 30 X 2Y65 + Y75   100 (9.8969) + 30 (7.2170)  1206.20  i1  i1

E 

Var *

50 X

2Y65 +

, i1

The

95th

30 X

Y75 +  50 (4)(13.29779) + 30 (11.53237)  3005.53

i1

-

√ percentile is approximated as 1206.2 + 1.645 3005.53  1296.38 . (E)

15.36. Since 1  d a¨ x + A x , variance is 0 if the death benefit were 1/d  1.06/0.06  17 32 . 15.37. We want y ( d a¨ x + A x )  12,000 a¨ x + yA x . So 12,000  0.08y, y  150,000 . (D) 15.38. Let T  T30 be the future lifetime random variable. 1 − vT Z2 + 10v T δ

!



Var ( Z )  −

2 + 10 δ

2



Var v T



v T is a whole life insurance. Under exponential mortality, the first moment is µ/ ( µ + δ ) and the second moment is µ/ ( µ + 2δ ) . 0.02 1  0.02 + 0.06 4 0.02 1 E[v 2T ]   0.02 + 2 (0.06) 7     2 2 1 1 Var ( Z )  10 − − 0.06 7 16  43.75 (B) E[v T ] 

15.39. 1 − v Kx + bv K x +1 i v − v K x +1 + bv K x +1  iv   v 1  + v K x +1 b − d d   1 2 Var ( Z )  b − Var ( v K x +1 ) d Z



 ( b − 21) 2 2A x − A2x ∞ X

0.99k Ax  (0.01) 1.05k+1 k0

!

!





1 1 1  105 1 − 0.99/1.05 6

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15. ANNUITIES: VARIANCE

326

Ax 

2

∞ X

(0.01)

k0

0.99k 1.052 ( k+1)

!

!

1 0.01 0.01   0.0888889  1.052 1 − 0.99/1.052 1.052 − 0.99 Var ( Z )  ( b − 21) 2 (0.0888889 − 0.16666672 )  ( b − 21) 2 (0.0611111) 1  ( b − 21) 2 (0.0611111) b  21 ± 4.0452  16.9548, 25.0452

Quiz Solutions 15-1.

Let’s back out the force of interest. A¯ 40:20  1 − (2δ ) 2a¯ 40:20

2

0.4  1 − 2 (12) δ δ  0.025 Then a 20-year endowment insurance has present value 20 1 A¯ 40:20  A¯ 40 :20 + v 20 p 40

 0.04 + e −20 (0.025) (0.95)  0.616204 The variance of a continuous 20-year temporary life annuity that pays 1 per year (which is 1/100 of Y) is Var ( Y/100) 

2A ¯

40:20

− A¯ 240:20

δ2 0.4 − 0.6162042   32.47 0.0252

so the variance of Y is 324,700 . 15-2. Since 30 p70  0, there are three possibilities: no payments (probability 10 q70  0.2), one payment (probability 10 p 70 − 20 p70  0.3), or two payments (probability 20 p70 − 30 p70  0.5). The values of the annuities in these three cases are 0, 10v 10 , and 10 ( v 10 + v 20 ) . Let Y be the present value. Then 3 1 1 +5 +  7.68645 10 10 1.0420 1.04 1.04  2  2 30 1 1 + 50  77.75727 E[Y 2 ]  0.3 (10v 10 ) 2 + 0.5 10 ( v 10 + v 20 )  + 1.0420 1.0410 1.0420 Var ( Y )  77.75727 − 7.686452  18.67579





E[Y]  0.3 (10v 10 ) + 0.5 10 ( v 10 + v 20 ) 





15-3. For each retiree, the mean present value of 1 per year is 1/ (0.02 + 0.08)  10. Let Y be the variance of 1 per year for one retiree. The variance is Var ( Y )  LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

2A ¯

x

− A¯ 2x δ2

QUIZ SOLUTIONS FOR LESSON 15

327

.







0.02 0.02 + 2 (0.08) − 0.02/ (0.02 + 0.08)

2

0.082

1/9 − 1/25 100   9 (2/25) 2 For 100 retirees receiving 5000 apiece, the mean is (100)(5000)(10)  5,000,000 and the standard deviation √ is (10)(5000) 100/9  166,667. The fund needed is 5,000,000 + 1.645 (166,667)  5,274,167 .

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328

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15. ANNUITIES: VARIANCE

Lesson 16

Annuities: Probabilities and Percentiles Reading: Models for Quantifying Risk (4th or 5th edition) 8.1–8.3

16.1

Probabilities for continuous annuities

Unlike insurances, annuity random variables always increase as a function of survival time. They are always monotonically increasing. The longer one lives, the more annuity payments one receives. Therefore, to calculate the probability that the present value of an annuity Y is less than something, you calculate the probability that Tx is less than something. The probability that the present value of an annuity is less than a value corresponds to the probability that Tx is less than some value. An algebraic method for relating the distribution of Y to the distribution of Tx is shown in Table 16.1. But it is easier to reason out the relationship for each problem. To reason out the relationship, it may help to graph Y on Tx to get a better feel for what is going on. The graphs of Figure 16.1 graph Y on Tx for the four standard annuities. Most exam questions will be on whole life annuities or deferred life annuities. Example 16A For a continuous whole life annuity on (40) : • Y is the present value random variable of the annuity. • The probability of survival for (40) is t p 40

60 − t  60

! 0.5

• δ  0.05 Calculate Pr ( Y ≤ 10) . Answer: Based on Figure 16.1a, Y ≤ 10 means Tx ≤ t in that figure, which means we want t q 40 for the t satisfying a¯ t  10. Solve for t: a¯ t  10 1 − vt  10 0.05 t 1 − v  0.5 ln 0.5 ln 2 t−   13.8629 δ 0.05 The desired probability is t q 40

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60 − 13.8629 1− 60 329

! 0.5  0.1231



16. ANNUITIES: PROBABILITIES AND PERCENTILES

330

Y

Y

15

15

(t , a¯t )

10

10

5

0

5

0

5

10

15

20

T

0

0

(a) Whole life annuity

15

20

T

Y

15

15

10

10

5

5

0

10

(b) 10-year temporary life annuity

Y

0

5

5

10

15

20

T

0

0

(c) 10-year deferred life annuity

5

10

15

20

T

(d) 10-year certain-and-life annuity

Figure 16.1: Graphs of Y on Tx , assuming δ  0.05

As with probabilities and percentiles for insurances, if mortality is exponential, you can skip logging, because you’ll end up exponentiating back. If Pr (Tx > y )  e −µ y , then Pr ( v Tx < v y )  e −µ y , and v y  e −δ y , so Pr ( v Tx < u )  u µ/δ . Example 16B For a continuous whole life annuity on (40) : • Y is the present value random variable of the annuity. • µ40+t  0.02 for all t > 0. • δ  0.04 Calculate Pr ( Y > 16) .

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16.1. PROBABILITIES FOR CONTINUOUS ANNUITIES

331

Table 16.1: Expressing FY ( y ) in terms of Fx ( t )

For the continuous whole life annuity present-value random variable Y, you can relate FY ( y ) to Fx ( t ) as follows: FY ( y )  Pr ( Y ≤ y ) 1 − v Tx ≤y δ

 Pr

!

 Pr ( v Tx ≥ 1 − δ y )



 Pr Tx ln v ≥ ln (1 − δ y )



 Pr −Tx δ ≥ ln (1 − δ y ) ln (1 − δ y )  Pr Tx ≤ − δ  Fx

ln (1 − δ y ) − δ





!

!

Answer: We need t p 40 where t satisfies a¯ t  16. 1 − vt  16 0.04 1 − v t  0.64 v t  0.36 e −0.04t  0.36 Then Pr ( Y > 16)  e −µt  e −0.02t , the square root of e −0.04t , or 0.6 .



Example 16C For a continuous whole life annuity on ( x ) , µ x  µ is constant. Derive a formula for the probability that the present value of the payments will be greater than the actuarial present value. Answer: We want the probability of the event a¯Tx > a¯ x  1/ ( µ + δ ) , equivalent to: 1 − e −δt 1 > δ µ+δ e −δt < 1 −

The probability of this event is

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e −µt ,

or

µ µ+δ

µ δ  µ+δ µ+δ

! µ/δ .



16. ANNUITIES: PROBABILITIES AND PERCENTILES

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?

Quiz 16-1 For a continuous 10-year deferred life annuity on (70) : • Y is the present value random variable of the annuity. • µ70+t  0.02 for all t > 0. • δ  0.03 Calculate Pr ( Y > 5) .

16.2

Distribution functions of annuity present values

The distribution function for a continuous annuity present-value random variable Y, in other words the function FY ( y )  Pr ( Y ≤ y ) , will depend on mortality and interest. Typical patterns are shown in Figures 16.2 and 16.3. Let’s discuss these patterns so that you’ll understand them. For a continuous whole life annuity, the present value of the annuity keeps increasing as the annuitant lives longer. However, the present value of later payments is less than the present value of earlier payments. For example, it may take six years to accumulate a present value of 5, but fourteen years to accumulate a present value of 10. If Tx were uniformly distributed, then in this case FY (10)  (14/5) FY (5) , so the distribution function grows more rapidly for higher values. However, Tx is not uniform for realistic mortality. As a result, FY ( y ) decelerates for high y. Figure 16.2a has a typical pattern. Now let’s discuss the distribution function of the present-value random variable Yx:n of a continuous n-year temporary life annuity. The probability of Y¯ x:n ≤ y is the same as the corresponding probability of a whole-life annuity random variable if y < a¯ n . However, Y¯ x:n can never be worth more than a¯ n no matter how long ( x ) lives. So the distribution function rises to n q x at y  a¯ n and then jumps to 1. Figure 16.2b has a typical pattern for a 15-year temporary life annuity. Now let’s discuss the distribution function of the present-value random variable n| Y¯ x of a continuous n-year deferred life annuity. This annuity pays nothing when Tx ≤ n, and its present value is the present value of a whole-life annuity minus the value of an n-year temporary annuity if Tx > n. Accordingly, the probability of 0 is n q x and F (0)  n q x . The graph jumps to n q x at 0 and afterwards is parallel to the graph

FY ( y ) 1

FY ( y ) 1

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2

0

0

2

4

6

8

10 12 14 16 18 20

(a) Whole life annuity. FY ( y ) hits 1 at 1/δ  20.

y

0

0

2

4

6

8

10 12 14 16 18 20

(b) 15-year temporary life annuity. Vertical part of graph occurs at a¯15  10.55267. At that point, FY (10.55267− )  15 q x  0.11750.

Figure 16.2: Life annuity distribution functions—Part I. Assumes δ  0.05 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

y

16.3. PROBABILITIES FOR DISCRETE ANNUITIES

333

FY ( y ) 1

FY ( y ) 1

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2

0

0

2

4

6

8

10 12 14 16 18 20

(a) 15-year deferred life annuity. FY (0)  the graph hits 1 at 1δ − a¯15  9.44733.

15 q x

y

 0.11750, and

0

0

2

4

6

8

10 12 14 16 18 20

y

(b) 15-year certain & life annuity. The vertical part of the graph is at y  a¯15  10.55267, and FY (10.55267)  15 q x  0.11750.

Figure 16.3: Life annuity distribution functions—Part II. Assumes δ  0.05

of a whole life annuity shifted left by a¯ n . See Figure 16.3a for a typical pattern for a 15-year deferred life annuity. Finally let’s discuss the distribution function of the present value random variable Y¯ x:n of a continuous n-year certain and life annuity. This random variable is never less than a¯ n , because an n-year annuitycertain is guaranteed. If Tx > n, then Yx:n is the same as Y, the random variable for a whole-life annuity. So the graph of the distribution function is 0 up to a¯ n , then jumps to n q x , then is the same as the graph of Y to the right of that point. Figure 16.3b has a typical pattern of a 15-year certain and life annuity.

16.3

Probabilities for discrete annuities

I am using the term “discrete” annuity to mean an annuity that is not continuous. So we’re discussing annuities-due and annuities-immediate. For discrete annuities, the solution may begin with the methods discussed for continuous annuities, but then you must round the answer to an integer in a way consistent with what the question is asking for. Example 16D For a whole life annuity-due on (50) : • Y is the present value random variable. • Mortality follows the Illustrative Life Table. • d  0.04. Calculate Pr ( Y > 10) . Answer: Y is a¨ K50 +1 , and we must compute the probability that this annuity has present value greater than 10. 1 − v K50 +1 > 10 0.04 v K50 +1 < 0.6 ln 0.6 K 50 + 1 > − δ LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

16. ANNUITIES: PROBABILITIES AND PERCENTILES

334





and δ  ln (1 + i )  ln 1 + d/ (1 − d )  ln (1 + 1/24)  0.040822. So K 50 + 1 > − (ln 0.6) /0.040822  12.51. So when K 50 ≥ 12, the annuity’s present value is greater than 10. From the Illustrative Life Table, 12 p 50



l 62 7,954,179   0.8886 l 50 8,950,901



Many old exam questions ask for the probability that the sum of the annuity payments not discounted at interest is greater than the actuarial present value. Example 16E For a whole life annuity-immediate on (55) : • µ55+t  1/ (50 − t ) for 0 < t < 50. • i  0.06 Calculate the probability that the sum of the payments on the annuity is greater than the actuarial present value of the annuity. Answer: Let’s calculate the actuarial present value of the annuity. Since mortality is uniform, it is easier to work with insurances. a 50 50 1 − v 50 1 − 1/1.0650    0.315237 50 (0.06) 3 1 − 0.315237  12.0975  0.06/1.06  12.0975 − 1  11.0975

A55 

a¨55 a55

The payments are greater if there are at least 12 payments, or if (55) survives 12 years. But (50 − 12) /50  0.76 .

?

12 p 55





Quiz 16-2 For a 10-year certain-and-life annuity-due on (60) : • µ60+t  1/ (40 − t ) for 0 < t < 40. • i  0.05 Calculate the probability that the sum of the payments on the annuity is greater than the actuarial present value of the annuity.

16.4

Percentiles

Since annuities monotonically increase in value as survival time increases, calculating an annuity percentile reduces to calculating the percentile of Tx and then calculating a¯Tx for that value. Example 16F A continuous whole life annuity on (40) pays at a rate of 2 per year until age 65 and 1 per year thereafter. You are given: • Y is the present value random variable for the annuity. √ • µ40+t  0.01 t. • δ  0.05 Determine the third quartile of Y.

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16.4. PERCENTILES

335

Answer: We need the third quartile for T40 , the t for which t q x  0.75, or t p x  0.25. t

Z

√ 0.01 u du  (0.02/3) t 3/2

0 t p 40





 exp −0.02t 3/2 /3  0.25

0.02t 3/2  − ln 0.25 3 t

−3 ln 0.25 0.02

! 2/3  35.09925

The value of the annuity-certain paying 2 per year for 25 years and 1 per year from t  25 to t  35.09925 is 1 − e −0.05 (25) 1 − e −0.05 (35.09925) +  14.2699 + 16.5417  30.8116  0.05 0.05 p

For annuities with annual (instead of continuous) payments, you can determine πTx , the 100p th perp

centile of Tx , and then calculate the present value of an annuity-certain for someone who dies at time πTx . Example 16G An annuity-due on (35) pays 1 per annum until age 65, and then 2 per annum starting on the 65th birthday. You are given: • Mortality follows the Illustrative Life Table. • i  0.06 Calculate the 60tth percentile of the present value of this annuity. Answer: Let’s determine the 60th percentile of survival. We need t such that t q 35

 0.6

or 1−

l35+t  0.6 l35 l35+t  0.4 l35 l 35+t  0.4l 35

l 35  9,420,657 and 0.4 (9,420,657)  3,768,263. Looking up the Illustrative Life Table, we see that l 80  3,914,365 and l 81  3,600,038, so the 60th percentile is somewhere between ages 80 and 81. By the above discussion, this means that the 60th percentile of the annuity-due will be the annuity-certain that makes all payments up to and including the one at age 80. The present value of that is a¨30 + 230| a¨16 

1 − (1/1.06) 30 2 + 0.06/1.06 1.0630

!

1 − (1/1.06) 16 0.06/1.06

 14.5907 + 0.3482 (10.7122)  18.321

!



Example 16H For a whole life immediate annuity, you are given: • S0 ( x )  1 − x/105, x ≤ 105 • i  0.04. Calculate the amount of time (45) must live so that the present value of annuity benefits exceeds the actuarial present value of the annuity. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

16. ANNUITIES: PROBABILITIES AND PERCENTILES

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Answer: The actuarial present value of a whole life insurance with benefits paid at the end of the year of death is 60

Ax 

1 X 1 60 1.04t t1



1 60

!

1 − 1/1.0460  0.377058 0.04

!

Therefore

1 − (1 + i ) A x 1 − 1.04 (0.377058)   15.19649 i 0.04 We want the present value of a t-year annuity immediate certain to exceed 15.19649. ax 

a t > 15.19649 1 − (1/1.04) t > 15.19649 0.04 !t 1 1− > 0.60789 1.04 1.04t > 2.55011 t ln 1.04 > ln 2.55011 t > 23.8684 Therefore, (45) must live at least 24 years, to the 69th birthday.

?



Quiz 16-3 For a 10-year temporary life annuity-due on (40), • Y is the present value random variable. • Mortality is uniformly distributed with ω  120. • i  0.06 Calculate the 25th percentile of this annuity.

Exercises Probabilities [150-S98:15] For a 5-year certain-and-life annuity of 1 on ( x ) , payable continuously, you are given:

16.1. • • • •

Y is the present-value random variable. FY is the distribution function for Y. µ x+t  0.01, t≥0 δ  0.04

Calculate FY (30) − FY (5) . A. 0.71

B. 0.88

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C. 0.91

D. 0.95

E. 1.00

Exercises continue on the next page . . .

EXERCISES FOR LESSON 16

337

Table 16.2: Summary of formulas and concepts for this lesson

Probabilities and Percentiles • To calculate a probability for an annuity, calculate the t for which a¯ t has the desired property. Then calculate the probability t is in that range. • To calculate a percentile of an annuity, calculate the percentile of Tx , then calculate a¯Tx . • Some adjustments may be needed for discrete annuities or non-whole-life annuities, as discussed in the lesson. • If forces of mortality and interest are constant, then the probability that the present value of payments on a continuous whole life annuity will be greater than its actuarial present value is Pr ( a¯Tx > a¯ x ) 

µ µ+δ

! µ/δ

[150-S88:12] A 10-year deferred life annuity-due of 1 per annum is issued to (50).

16.2.

You are given: • Mortality is uniformly distributed with ω  100. • i0 Calculate the probability that the sum of the payments made under the annuity will exceed the actuarial present value, at issue, of the annuity. A. 0.46

B. 0.48

C. 0.50

D. 0.52

E. 0.54

[150-S91:12] For a 10-year deferred life annuity-due of 1 per year on (60), you are given:

16.3.

• Mortality is uniformly distributed with ω  100. • i0 Calculate the probability that the sum of the payments made under the annuity will exceed the actuarial present value, at issue, of the annuity. A. 0.475

B. 0.500

C. 0.525

D. 0.550

E. 0.575

[3-F02:31] For a 20-year deferred whole life annuity-due of 1 per year on (45) , you are given:

16.4. • •

l x  10 (105 − x ) , 0 ≤ 0 ≤ 105 i0

Calculate the probability that the sum of the annuity payments actually made will exceed the actuarial present value at issue of the annuity. A. 0.425

B. 0.450

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C. 0.475

D. 0.500

E. 0.525

Exercises continue on the next page . . .

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16.5. Y is the present value of a 10-year deferred 30-year temporary life annuity-immediate on (30). You are given: • Mortality is uniformly distributed with ω  100. • i  0. Calculate the probability that the sum of the payments will exceed E[Y]. 16.6. For an increasing whole life annuity-due on (40) paying k at the beginning of every year k for k ≥ 1, you are given: • Mortality is uniformly distributed with ω  120. • i0 Pn 2 • j1 j  n ( n + 1)(2n + 1) /6. Calculate the probability that the sum of the annuity payments actually made will exceed the actuarial present value at issue of the annuity. For a 10-year deferred whole life annuity-due paying 1 per year on (40) , you are given:

16.7.

µ40+t  0.01 i0

• •

Calculate the probability that the sum of the annuity payments actually made will exceed the actuarial present value at issue of the annuity. [150-83-96:12] For a 5-year deferred whole life annuity-due of 1 on ( x ) , you are given:

16.8.

• µ x+t  0.01, t ≥ 0 • i  0.04 • a¨ x:5  4.542 • The random variable S denotes the sum of the annuity payments. Calculate Pr ( S > 5| a¨ x ) . A. 0.81

B. 0.82

C. 0.83

D. 0.84

E. 0.85

16.9. Y is the present value random variable for a continuous 10-year temporary life annuity on ( x ) . You are given: • •

µ x+t  0.02 for all t. δ  0.04.

Calculate the probability that total payments on this annuity will exceed E[Y]. 16.10. • •

Y is the present value of a whole life annuity-due on (20). You are given: µ20+t  1/ (80 − t ) , t < 80 i  0.06.

Calculate the probability that the sum of the payments will exceed E[Y].

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Exercises continue on the next page . . .

EXERCISES FOR LESSON 16

339

16.11. Kira purchases a 3-year deferred whole life annuity with continuous payments. You are given: • •

µ x  0.02 for all x. δ  0.05

Calculate the probability that the present value of the payments on this annuity will exceed the actuarial present value of the annuity. Use the following information for questions 16.12 through 16.15: You are given: • δ  0.05 • Mortality is uniformly distributed with ω  100. 16.12. [150-S98:31] Calculate 2A¯ 35 . A. 0.1536

B. 0.1597

C. 0.1648

D. 0.1701

E. 0.1755

C. 14.96

D. 15.35

E. 15.74

C. 0.63

D. 0.65

E. 0.69

C. 28

D. 30

E. 32

16.13. [150-S98:32] Calculate a¯35 . A. 14.08

B. 14.57

16.14. [150-S98:33] Calculate Pr ( a¯T35 > a¯35 ) . A. 0.54

B. 0.59

16.15. [150-S98:34] Calculate Var ( a¯T35 ) . A. 24

B. 26

16.16. [3-S00:39] For a continuous whole life annuity of 1 on ( x ) : • Tx , the future lifetime of ( x ) , follows a constant force of mortality 0.06. • The force of interest is 0.04. Calculate Pr ( a¯Tx > a¯ x ) . A. 0.40

B. 0.44

C. 0.46

D. 0.48

E. 0.50

16.17. [CAS4A-F93:8] (3 points) You are given that ( x ) is subject to a constant force of mortality, µ x  0.06, and that the force of interest is also a constant, δ  0.04. Tx is the time until death random variable. Calculate the probability that a¯Tx > 1.50 a¯ x . A. B. C. D. E.

Less than 0.20 At least 0.20, but less than 0.40 At least 0.40, but less than 0.60 At least 0.60, but less than 0.80 At least 0.80

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Exercises continue on the next page . . .

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340

16.18. [150-F97:26] You are given: • µ x+t  0.01, t ≥ 0 • δ  0.05 • Tx is the future lifetime of ( x ) . Calculate the probability that a¯Tx will exceed a¯ x:20 . A. 0.70

B. 0.77

C. 0.84

D. 0.91

E. 0.98

D. 0.68

E. 0.79

16.19. [M-S05:26] You are given: • µ x+t  0.03, t ≥ 0 • δ  0.05 • Tx is the future lifetime random variable. • g is the standard deviation of a¯Tx . Calculate Pr ( a¯Tx > a¯ x − g ) . A. 0.53

B. 0.56

C. 0.63

Percentiles 16.20. • •

Y is the present value random variable for a continuous whole life annuity on (60). You are given: S0 ( x )  1 − x/110, x ≤ 110. δ  0.06.

Calculate the 90th percentile of Y. 16.21. You are given: • • •

Y is the present value random variable for a whole life annuity-due. q x  0.1 for all x. i  0.05

Calculate the 60th percentile of Y. 16.22. [CAS4A-S96:12] (2 points) A life, age x, wants to fund a life annuity with a deposit of amount P. This life is subject to a constant force of mortality, µ  0.06. δ  0.08. Determine P such that the probability is 95% that P will be sufficient to fund a life annuity of $100 per year payable continuously. A. B. C. D. E.

Less than $1,000 At least $1,000, but less than $1,100 At least $1,100, but less than $1,200 At least $1,200, but less than $1,300 At least $1,300

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Exercises continue on the next page . . .

EXERCISES FOR LESSON 16

341

16.23. You are given: • •

µ x  0.01 for all x. δ  0.045.

Let Y be the present value random variable for a 5-year deferred continuous life annuity. Calculate the median of Y. 16.24. A 20-year deferred whole life annuity on ( x ) pays 1 per year continuously. If ( x ) dies within 20 years, the net single premium for the annuity with interest accumulated at δ  0.06 is refunded. You are given: • • •

µ x+t  0.02 for all t > 0. δ  0.06 Y is the present value random variable for the annuity.

Determine the median of Y. 16.25. [M-F06:33] You are given: Y is the present value random variable for a continuous whole life annuity of 1 per year on (40). t • Mortality follows S0 ( t )  1 − , 0 ≤ t ≤ 120. 120 • δ  0.05 •

Calculate the 75th percentile of the distribution of Y. A. 12.6

B. 14.0

C. 15.3

D. 17.7

E. 19.0

Additional old CAS Exam 3/3L questions: S06:21, F11:12 Additional old CAS Exam LC questions: F14:11 Additional old SOA Exam MLC questions: S12:15, F12:20

Solutions 16.1. We must calculate the probability that the present value of the annuity is between 5 and 30, or Pr (5 < Y ≤ 30)  Pr ( Y ≤ 30) − Pr ( Y ≤ 5) . Note that a perpetuity would have present value 1δ  25, so the present value is always less than 30, and Pr ( Y ≤ 30)  1. Hence we need 1 − Pr ( Y ≤ 5)  Pr ( Y > 5) The annuity would have present value less than 5 if death occurs during the certain period, so the probability of the present value being greater than 5 is less than 1. We need the time t such that the present value of the annuity is exactly 5, or t such that a¯ t  5, and then the answer will be t p x  e −0.01t . 1 − vt 5 δ 1 − e −0.04t  5δ  0.2 e −0.04t  0.8 t px

 e −0.01t  0.80.25  0.9457

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(D)

16. ANNUITIES: PROBABILITIES AND PERCENTILES

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16.2. With an interest rate of zero, the annuity-due’s actuarial present value is the expected sum of its payments. If (50) lives less than 10 years, that is 0, and if (50) lives more than 10 years, the average number of payments is 20.5, since it ranges uniformly from 1 to 40, so the average is the midpoint or 20.5. Therefore, the expected sum of the payments is 10 q 50 (0)

+ 10 p50 (20.5)  0.8 (20.5)  16.4

To exceed this, 17 payments are needed. To get 17 payments, the annuitant must live beyond age 76. The probability of that, 26 p50 , is 0.48 . (B) 16.3. Virtually the same as the previous question. With an interest rate of zero, the annuity-due’s actuarial present value is the expected sum of its payments. If (60) lives less than 10 years, that is 0, and if (60) lives more than 10 years, the average number of payments is 15.5, since it ranges uniformly from 1 to 30, so the average is the midpoint or 15.5. Therefore, the expected sum of the payments is 10 q 60 (0)

+ 10 p60 (15.5)  0.75 (15.5)  11.625

To exceed this, 12 payments are needed. To get 12 payments, the annuitant must live beyond age 81. The probability of that, 21 p60 , is 19/40  0.475 . (A) 16.4. With an interest rate of zero, the annuity-due’s actuarial present value is the expected sum of its payments. If (45) lives less than 20 years, that is 0, and if (45) lives more than 20 years, the average number of payments is 20.5, since it ranges uniformly from 1 to 40, so the average is the midpoint or 20.5. Therefore, the expected sum of the payments is 41 3 To exceed this, 14 payments are needed. To get 14 or more payments requires living at least to age 45 + 20 + 14 − 1  78, or 33 years, and the probability of that is (60 − 33) /60  0.45 . (B) 20 q 45 (0)

+ 20 p 45 (20.5)  (2/3)(20.5) 

16.5. First calculate the expected sum of payments. The probability of 0 (survival less than 11 years) is 11/70. The probability of 30 (survival at least 40 years) is 3/7. Otherwise, the payment is between 1 and 29, average 15, with probability 29/70. So E[Y]  (29/70)(15) + (3/7)(30)  19.07143 To achieve 20 payments, the annuitant must survive 30 years. The probability of that is 4/7 . 16.6.

The expected sum of the payments on the annuity is 80 X k (81 − k ) k1

80

 

81

P80 k1

k−

P80 k1

k2

80 81 (80)(81) /2 − (80)(81)(161) /6  1107 80

We want k ( k +1) /2 > 1107. Rather than solving a quadratic, it’s easier to do this by estimating k 2 > 2 (1107) so k > 47. Trying k  47 gets (47)(48) /2  1128, so 47 payments exceeds 1107 and 46 payments doesn’t. So the desired probability is 46 p40  (80 − 46) /80  0.425 . 16.7.

The actuarial present value of the deferred annuity is, ∞ X k10

e −0.01k 

e −0.1  90.937 1 − e −0.01

The probability of more than 90 payments is the probability of surviving 100 years, or e −0.01 (100)  0.3678 .

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EXERCISE SOLUTIONS FOR LESSON 16

16.8.

343

By formula (14.11),

1+i 1.04   20.821 −0.01 q+i 1−e + 0.04 If you didn’t remember the formula, you could calculate a¨ x from first principles: a¨ x 

a¨ x 

∞ X e −0.01k k0

1.04k



1 1−

e −0.01 /1.04

 20.821

Either way, 5| a¨ x  20.821 − 4.542  16.279. To get 17 annuity payments, the annuitant must survive 21 years. The probability of that is 21 p x  e −0.21  0.8105 (A) 16.9. E[Y] 

  1 1 − e −10 (0.02+0.04)  7.519806 0.02 + 0.04

The probability of payments over a period of at least 7.519806 years is e −0.02 (7.519806)  0.860367 . 16.10. It is easier to calculate the expected value of A20 and then to calculate a¨20 from it. A20 

79 X 1 1 80 (1.06) 1.06t t0

1  80 (1.06)

a¨20

!

1 − (1/1.0680 ) 1 − (1/1.06)

!

 0.011792 (17.49968)  0.206364 1 − A20 1 − 0.206364   14.0209  d 0.06/1.06

The probability of 15 payments is the probability of surviving 14 years, or 1 − 14/80  0.825 . 16.11. The actuarial present value of the annuity is 3| a¯ x



e −3 ( µ+δ )  e −0.21 /0.07  11.5798 µ+δ

We want the t for which a 3-year deferred continuous annuity with payments until time t has 11.5798 as its present value, and then the probability that the present value of the payments will exceed 11.5798 will be t p x  e −0.02t . 3| a¯ t−3

 11.5798

v3 − v t  11.5798 δ e −0.15 − e −0.05t  0.05 (11.5798)  0.578989 e −0.05t  e −0.15 − 0.578989  0.28172 e −0.02t  0.281722/5  0.6025 16.12. As usual with uniform survival, this is an annuity-certain for period ω − x divided by ω − x. a¯ A¯ 35  65 0.10 65  10   1 − e −0.10 (65)  0.1536 65

2

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(A)

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344

16.13. Calculate A¯ 35 and then transform it with equation (13.2). a¯ A¯ 35  65 0.05 65  20  1 − e −0.05 (65)  0.295762  65 1 − 0.295762 a¯35   14.0848 (A) 0.05 16.14. We need t such that a¯ t  a¯35  14.0848 (from the previous exercise), and then the required probability will be t p35 . 1 − vt  14.0848 0.05 v t  1 − 0.05 (14.0848)  0.29576 t ln v  ln 0.29576  −1.218207 −1.218207 1.218207 1.218207    24.3642 t ln v δ 0.05 24.3642 (C)  0.6252 24.3642p 35  1 − 65 16.15. From the previous exercises we already have the first and second moments for a whole life insurance, so we’ll use Var ( a¯T35 ) 

2A ¯

35

− A¯ 235

δ2 0.1536 − 0.2957622   26.4559 0.052

(B)

16.16. For a continuous whole life annuity, we derived in Example 16C that the probability is µ µ+δ 16.17.

! µ/δ 

0.06 0.06 + 0.04

! 1.5  0.4648

(C)

a¯ x  1/ ( µ + δ )  10. We want the probability t p x , where t satisfies: 1 − e −δt  15 δ 1 − e −0.04t  15 0.04 e −0.04t  0.4

The probability is t px

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 e −0.06t  0.43/2  0.2530

(B)

EXERCISE SOLUTIONS FOR LESSON 16

345

16.18. First we calculate a¯ x:20 . 20

Z a¯ x:20  

e − (0.05+0.01) t dt 0

1 − e −0.06 (20)  11.646763 0.06

Now let’s find the t such that a¯ t  11.646763. 1 − v Tx  11.646763 0.05  1 − 0.05 (11.646763)  0.417662

a¯ t  e −0.05t We need Pr (Tx > t )  e −0.01t .

e −0.01t  0.4176620.2  0.839779

(C)

16.19. Let’s calculate a¯ x and g. a¯ x  Var ( a¯Tx )  

1 1   12.5 µ + δ 0.03 + 0.05 2A ¯ x − A¯ 2 x

δ2

µ µ+2δ

−

µ
2 µ+δ

0.052 (3/13) − (3/8) 2  36.0577  0.052 q √ g  Var ( a¯Tx )  36.0577  6.0048 a¯ x − g  12.5 − 6.0048  6.4952 a¯Tx  (1 − v Tx ) /δ. Let’s see which Tx makes the annuity greater than a¯ x − g. 1 − v Tx > 6.4952 0.05 v Tx < 1 − 6.4952 (0.05)  0.675240 e −0.05Tx < 0.675240









Since Pr Tx > y  e −0.03y , we have Pr e −0.05Tx < a  a 0.03/0.05 , so the probability we want is





Pr a¯Tx > a¯ x − g  0.6752403/5  0.7901

(E)

16.20. The 90th percentile of T60 for this uniform distribution on [0, 50] is 45. The present value of a 45-year continuous annuity is 1 − e −0.06 (45) a¯45   15.546575 0.06

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16. ANNUITIES: PROBABILITIES AND PERCENTILES

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16.21. The survival probability k p x  0.9k . We want k q x  0.6, so set k p x equal to 1 − 0.6  0.4 0.9k  0.4 ln 0.4 k  8.69672 ln 0.9 The 60th percentile occurs  when time to  . death is 8.69672 years, in which case there are 9 payments (times  0 through 8). Then a¨9  1 − (1/1.059 ) (0.05/1.05)  7.4632 . This is the 60th percentile because Pr K x <







8 < 0.6 and Pr K x ≤ 8 > 0.6. 16.22. We want the 95th percentile of Tx . We want t such that t q x  1 − e −0.06t  0.95, or e −0.06t  0.05. Then P will be 100 (1 − e −0.08t ) P  100 a¯ t   1250 (1 − e −0.08t ) 0.08



However, since e −0.06t  0.05, it follows that e −0.08t  e −0.06t

 4/3

 0.054/3 , so

P  1250 (1 − 0.054/3 )  1226.975

(D)

16.23. From the fact that µ x  0.01 for all x, Pr (Tx > t )  t p x  e −0.01t . We want to determine t such that Pr (Tx > t )  0.5, or e −0.01t  0.5. Then we’ll calculate the median of Y as a¯ t − a¯5 



However, e −0.045t  e −0.01t

 4.5

e −5 (0.045) − e −0.045t 0.045

 0.50.045/0.01  0.54.5  0.044194, so the median is

e −0.225 − 0.044194 0.798516 − 0.044194   16.7627 0.045 0.045 16.24. Let Tx be survival time. The present value of this annuity is equal to the net single premium if Tx ≤ 20. It drops to 0 at Tx  20 and gradually increases afterwards. The pattern is shown in Figure 16.4. Let the net single premium be P. The median of Y occurs when Tx is in one of three ranges: 1.

Tx ∈ (20, y ) , where y is the time for which Y  P. The median will be in this range if Pr (20 < T < y ) > 0.5.

2.

Tx ∈ (0, 20) . The median will be in this range if Pr (20 < Tx < y ) < 0.5 but Pr (0 < Tx < y ) > 0.5.

3.

Tx ∈ ( y, ∞) . The median will be in this range if Pr (0 < Tx < y ) < 0.5.

Let’s calculate the net single premium P. We equate P to the probability of dying in the first 20 years times P, plus the probability of surviving 20 years and receiving payments. P  20q x P + v 20 20p x a¯ x+20 20q x

 1 − e −0.02 (20)

v 20 20p x  e − (0.06+0.02)(20) 1 1 a¯ x+20   µ + δ 0.06 + 0.02





P  1 − e −0.02 (20) P +

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e − (0.06+0.02)(20) 0.08

EXERCISE SOLUTIONS FOR LESSON 16

347

Y 5 4 3 2 1 0

0

10

20

30

40

x

50

60

T

Figure 16.4: Present value of deferred annuity Y of exercise 16.24 as function of survival time Tx .

e −1.6 0.08 e −1.2  3.76493 P 0.08

Pe −0.4 

Let’s find the y corresponding to Y  P. a¯ y − a¯20 

e −1.2 − e −0.06y 0.06

Setting this equal to 3.76493, e −1.2 − e −0.06y  0.06 (3.76493)  0.225896 e −0.06y  e −1.2 − 0.225896  0.075299 ln 0.075299 y−  43.1049 0.06 The probability of survival to 43.1049 is e −0.02 (43.1049)  0.422275. The probability of survival to 20 is e −0.02 (20)  0.670320. Looking at the three ranges discussed above, Pr (20 < Tx < y )  0.670320−0.422275 < 0.5, but Pr (0 < Tx < y )  0.670320 > 0.5. We conclude that the median of Y is assumed when Tx < 20 and is equal to the present value of the refund of net single premium with interest, or the net single premium. The median of Y is P  3.76493 . 16.25. The longer survival (T40 ), the higher the annuity value, so we need an annuity-certain for the 75th percentile of T40 . With uniform survival time, that is 75% of the distance to ω, or 0.75 (120 − 40)  60. a¯60 

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1 − e −60 (0.05)  19.0043 0.05

(E)

16. ANNUITIES: PROBABILITIES AND PERCENTILES

348

Quiz Solutions 16-1.

The actuarial present value of the payments if death occurs at t is a¯ t − a¯10 

v 10 − v t δ

and we want the t making this equal to 5. The answer will be t p 70  e −0.02t . v 10 − v t  0.03 (5)  0.15 v t  v 10 − 0.15  e −0.3 − 0.15  0.590818 e −0.03t  0.590818 Then e −0.02t  0.5908182/3  0.70410 16-2. We have a¨60:10  a¨10 + 10| a¨ 60 . Since mortality is uniform, we’ll calculate the actuarial present value of the deferred annuity using insurances. A60  A60:10   10| a¨ 60

 

a40 1 − (1/1.0540 )   0.428977 40 40 (0.05) a10 + 10 E60 40 1 − (1/1.0510 ) 0.75  0.653478 + 40 (0.05) 1.0510 A60:10 − A60 d 0.653478 − 0.428977  4.71452 0.05/1.05

The 10-year certain annuity has actuarial present value a¨10 

1 − (1/1.0510 )  8.10782 0.05/1.05

The actuarial present value of the 10-year certain-and-life annuity-due is 4.71452 + 8.10782  12.82. Therefore, 13 payments are needed to top this value, which happens when the annuitant survives 12 years. In other words, the answer is 12 p60  (40 − 12) /40  0.7 . 16-3. The 25th percentile of T40 is 20. The annuity assumes its maximum value at 20 (or at any value of T40 ≥ 10), but that doesn’t matter; a¨10 is any percentile from 12.5th to 100th . a¨10 

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1 − (1/1.0610 )  7.80169 0.06/1.06

Lesson 17

Annuities: Recursive Formulas Reading: Models for Quantifying Risk (4th or 5th edition) exercise 8-6

We can relate annuities at age x to those at age x +1. The following states the rules for the four standard annuities with three versions apiece: annuities-due, annuities-immediate, and continuous annuities. Whole life annuities For an annuity-due a¨ x is a¨ x+1 discounted one year, with an immediate payment of 1 added, so a¨ x  vp x a¨ x+1 + 1 (17.1) (m ) For an mthly annuity due, a¨ x( m ) is a¨ x+1/m discounted for 1/m years, plus an immediate payment of 1/m: (m ) (17.2) a¨ x( m )  v 1/m 1/m p x a¨ x+1/m + m1

For an annuity-immediate a x is a x+1 discounted one year with a payment of 1 at the end of the year added: a x  vp x a x+1 + vp x For a continuous annuity, a¯ x is a¯ x+1 discounted one year with a¯ x:1 added: a¯ x  vp x a¯ x+1 + a¯ x:1 Temporary annuities For an annuity-due, a¨ x:n is a¨ x+1:n−1 discounted one year with an immediate payment of 1 added: a¨ x:n  vp x a¨ x+1:n−1 + 1 For an immediate annuity, a x:n is a x+1:n−1 discounted one year with a payment of 1 at the end of the year added: a x:n  vp x a x+1:n−1 + vp x For a continuous annuity, a¯ x:n is a¯ x+1:n−1 discounted one year plus a 1-year continuous life annuity: a¯ x:n  vp x a¯ x+1:n−1 + a¯ x:1 Deferred life annuities When bringing these back through the deferral period, the next year’s amount is discounted, but nothing is added. n| a¨ x

 vp x n−1| a¨ x+1

 vp x n−1| a x+1 ¯ n| a x  vp x n−1| a¯ x+1 n| a x

n-year certain-and-life annuities During the certain period, there is an immediate payment of 1. If death occurs, there is still an annuity-certain benefit. The formula for an annuity-due is a¨ x:n  1 + vq x a¨ n−1 + vp x a¨ x+1:n−1 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

349

17. ANNUITIES: RECURSIVE FORMULAS

350

Table 17.1: Summary of formulas for this lesson

Recursive Formulas a¨ x  vp x a¨ x+1 + 1

n| a¨ x

 vp x n−1| a¨ x+1

a x  vp x a x+1 + vp x a¯ x  vp x a¯ x+1 + a¯ x:1 a¨ x:n  vp x a¨ x+1:n−1 + 1

n| a x  vp x n−1| a x+1 n| a¯ x  vp x n−1| a¯ x+1 a¨ x:n  1 + vq x a¨ n−1 + vp x a¨ x+1:n−1

a x:n  vp x a x+1:n−1 + vp x

a x:n  v + vq x a n−1 + vp x a x+1:n−1

a¯ x:n  vp x a¯ x+1:n−1 + a¯ x:1

a¯ x:n  a¯1 + vq x a¯ n−1 + vp x a¯ x+1:n−1

For an annuity-immediate, the formula is a x:n  v + vq x a n−1 + vp x a x+1:n−1 For a continuous annuity, the formula is a¯ x:n  a¯1 + vq x a¯ n−1 + vp x a¯ x+1:n−1 Recursion formulas are useful if a change is made to an assumption for a short period. Example 17A For a 10-year temporary life annuity-due on ( x ) , you are given • q x  0.01 • i  0.07 • a¨ x:10  5 Calculate the change in the actuarial present value of this annuity due if q x is increased by 0.03. Answer: a¨ x:10  1 + vp x a¨ x+1:9 , so 0.99 a¨ 1.07 x+1:9  4.32323

51+ a¨ x+1:9 If we change q x , the revised present value is 1+

0.96 (4.32323)  4.87879 1.07

The change is therefore 4.87879 − 5  −0.12121 .



A whole life annuity on ( x ) may be split into an n-year temporary life annuity plus an n-year pure endowment factor times an annuity on x + n. a¨ x  a¨ x:n + n E x a¨ x+n and we used this formula in previous lessons.

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EXERCISES FOR LESSON 17

351

Exercises [3-S00:29] For a whole life annuity-due of 1 on ( x ) , payable annually:

17.1.

q x  0.01 q x+1  0.05 i  0.05 a¨ x+1  6.951

• • • •

Calculate the change in the actuarial present value of this annuity-due if p x+1 is increased by 0.03. A. 0.16

B. 0.17

C. 0.18

D. 0.19

E. 0.20

C. 0.537

D. 0.636

E. 0.737

[150-S88:5] You are given:

17.2. • •

a¨ x  8 for all integral x. i  0.08

Calculate 8 q 30 . A. 0.263

B. 0.364

[CAS4A-S99:17] (1 point) You are given:

17.3. • • •

a¨30  15.85 a30:4  3.45 i  0.06

Calculate the actuarial present value of a 5-year deferred whole life annuity-due issued to (30). A. B. C. D. E.

Less than 10.0 At least 10.0, but less than 10.5 At least 10.5, but less than 11.0 At least 11.0, but less than 11.5 At least 11.5

17.4. [CAS4-S84:21] An annuity is issued to Sally, who is age x. a x is the actuarial present value of an annuity assuming standard mortality, but Sally’s mortality in the n + 1st year, q 0x+n , is equal to q x+n + c, where q x+n is standard mortality. Which of the following expressions must be added to a x to obtain the actuarial present value of Sally’s annuity? A. B. C. D. E.

−cv n · n p x · a¨ x+n+1 −cv n+1 · n p x · a¨ x+n+1 −cv n · n p x · a x+n+1 −cv n+1 · n p x · a x+n+1 The correct answer is not given by (A), (B), (C), or (D).

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Exercises continue on the next page . . .

17. ANNUITIES: RECURSIVE FORMULAS

352

[CAS4A-F92:21] (2 points) You are given the following life table data:

17.5.

x 50 51 52

lx

qx

dx 508

0.00600 91,365

You are given that i  0.06 and a51  11.888. Determine a50 . A. B. C. D. E.

Less than 12.083 At least 12.083, but less than 12.089 At least 12.089, but less than 12.095 At least 12.095, but less than 12.101 At least 12.101 [CAS4-S85:18] (2 points) Given:

17.6. • • •

a¨19  25.04 a¨20  24.85 i  0.03

Calculate p19 . A. B. C. D. E.

Less than 0.9900 At least 0.9900, but less than 0.9925 At least 0.9925, but less than 0.9950 At least 0.9950, but less than 0.9975 At least 0.9975 For an annuity-immediate on ( x ) , you are given:

17.7. • • •

a x  18.4. E x  0.96. q x+1  0.012.

The mortality rate assumption q x+1 is then changed to 0.009. Calculate the revised value of a x . 17.8. The actuarial present value of a 5 year deferred continuous whole life annuity evaluated at δ  0.06 is 10. The interest assumption for the first 5 years only is then changed to δ0. Determine the value of δ0 for which the actuarial present value of this annuity is 10.2. For ( x ) , you are given

17.9. • • •

a¨ x:10  15 q x  0.01 v  0.95

The mortality assumption for age x only is then changed to q x  0.03. Calculate the revised value of a¨ . x:10

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Exercises continue on the next page . . .

EXERCISES FOR LESSON 17

353

17.10. [SOA3-F04:29] At interest rate i: • a¨ x  5.6 • The actuarial present value of a 2-year certain-and-life annuity-due of 1 on ( x ) is a¨ x:2  5.6459. • e x  8.83 • e x+1  8.29 Calculate i A. 0.077

B. 0.079

C. 0.081

D. 0.083

E. 0.084

Solutions 17.1.

We use primes for the revised values. 1 − 0.05 a¨ x+2 1.05 6.57742  a¨ x+2  a¨0x+2 0.98 a¨0x+1  1 + (6.57742)  7.13893 1.05 6.951  1 +

The original a¨ x is 1 + (0.99/1.05)(6.951)  7.5538. The revised a¨0x is 1 + (0.99/1.05)(7.13893)  7.730988. The difference is 7.730988 − 7.5538  0.1772 (C) 17.2. We can relate a¨ x to a¨ x+1 with recursion, then use that they’re both 8. a¨ x  1 + vp x a¨ x+1

!

px (8) 1.08 (7)(1.08) px   0.945 8 8 8 q 30  1 − 0.945  0.36400 81+

(B)

17.3. a¨30  a¨30:5 + 5| a¨30 15.85  1 + 3.45 + 5| a¨30 5| a¨30

 11.4

(D)

The interest rate is extraneous. 17.4. a x  a x:n + v n+1 n p x (1 − q x+n )(1 + a x+n+1 ) a 0x  a x:n + v n+1 n p x (1 − q 0x+n )(1 + a x+n+1 ) The difference is −cv n+1 n p x (1 + a x+n+1 )  −cv n+1 n p x a¨ x+n+1 , which is (B).

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17. ANNUITIES: RECURSIVE FORMULAS

354

17.5. a50  vp 50 + vp 50 a51 91,365 l51   91,916.5 0.994 l50  91,916.5 + 508  92,424.5 91,916.5 p50   0.9945 92,424.5 0.9945 a50  (1 + 11.888) 1.06 (C)  12.092 17.6. a¨19  1 + vp19 a¨20 p19 (24.85) 25.04  1 + 1.03 (24.04)(1.03) p 19   0.996427 24.85 17.7.

(D)

We use primes for revised values. 18.4  0.96 + 0.96v (0.988)(1 + a x+2 ) 0.96v (0.988)(1 + a x+2 )  17.44 17.44 (0.991) 0.96v (0.991)(1 + a x+2 )   17.4930 0.988 a 0x  0.96 + 17.4930  18.4530

17.8. 5 px

e −5 (0.06) a¯ x+5  10 0

e −5δ a¯ x+5  10.2 0 10.2 e −5 ( δ −0.06)   1.02 10 −5δ0 + 0.3  ln 1.02  0.019803 0.3 − 0.019803 δ0   0.056039 5

5 px

17.9. We use primes for revised values. We can express the annuity as the value of the current payment of 1 plus the values of the certain annuity if death occurs or the 9-year-certain and life annuity if survival occurs. 15  a¨ x:10  1 + vq x a¨9 + vp x a¨ x+1:9 1 − 0.959  7.395012 1 − 0.95 15  1 + (0.95)(0.01)(7.395012) + (0.95)(0.99) a¨ x+1:9

a¨9 

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EXERCISE SOLUTIONS FOR LESSON 17

355

14.81100  a¨ x+1:9 a¨0

x:10

 1 + (0.95)(0.03)(7.395012) + (0.95)(0.97)(14.81100)  1 + 0.21076 + 13.64834  14.8591

Alternatively, we can express the annuity as a 10-year annuity certain plus a deferred life annuity, and use recursion on the deferred annuity: 15  a¨ x:10  a¨10 + 10| a¨ x 1 − 0.9510  8.025261 1 − 0.95 10| a¨ x  15 − 8.025261  6.974739 10| a¨ x  vp x 9| a¨ x+1 a¨10 

 (0.95)(0.99) 9| a¨ x+1 6.974739  7.41599 9| a¨ x+1  (0.95)(0.99) 0 10| a¨ x  (0.95)(0.97)(7.41599)  6.833835 a¨0

x:10

 8.025261 + 6.833835  14.8591

17.10. First we note that by the recursive equation for e x , equation (6.4) on page 108: 8.83  p x (1 + 8.29) 883 px  929 Comparing a life annuity to a 2-year certain-and-life annuity, we find that a¨ x  1 + vp x +

∞ X

v t t px

t2

a¨ x:2  1 + v +

∞ X

v t t px

t2

so the difference is v (1 − p x ) . Then we have



5.6459 − 5.6  v 1 −

883 929



46v 929 v  0.92698 1 i  − 1  0.0788 v

0.0459 

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(B)

356

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17. ANNUITIES: RECURSIVE FORMULAS

Lesson 18

Premiums: Net Premiums for Fully Continuous Insurances Reading: 9.2, 9.4.2 In return for potential future benefits from an insurance company, the purchaser of a policy pays the company premiums. In the following few lessons, we will discuss how premiums can be calculated. Premiums may be paid in a lump sum upon purchase of the policy. Such a premium is called a single premium. We’ve already used the term “net single premium” in the sense of the actuarial present value of a policy. But usually premiums are paid on a regular basis—annually, semi-annually, quarterly, monthly. In the case of an annuity, it wouldn’t make much sense for money to be going in both directions at the same time, so an immediate annuity would usually be purchased with a single premium. A deferred annuity is often paid for with regular premiums over the deferral period. How can the premium be determined? We will discuss one principle in the next few lessons. But before we define a principle, let’s discuss future loss.

18.1

Future loss

The premium principles we will discuss in this lesson will be based on future loss. Future loss is the present value of the benefits, and possibly expenses, paid under the contract minus the present value of the premiums collected. It is a random variable that depends on the time of death,1 or Tx . For a whole life insurance contract with premiums payable regularly, the higher Tx is, the lower the present value of the death benefit and the higher the present value of the premiums collected, so higher Tx leads to lower future loss. The present value of future benefits minus the present value of future premiums, ignoring expenses, is called the net future loss. Often we will leave out the word “net”. We will use the letter “L” for future loss. This letter is not part of International Actuarial Notation, so there is no standard subscripting convention. Typically there is a 0 subscript to indicate that this is the future loss at time 0, the time of issue of the policy, but the 0 may either precede or succeed the symbol: L 0 or 0 L. The exam may use either symbol, or not use a subscript at all, but will define the symbol when it uses it. Example 18A A whole life contract provides for a benefit of 1000 at the moment of death. The policyholder pays a premium of 20 at the beginning of each year for the contract. The assumed interest rate is 0.05. The policyholder dies at the end of 25.8 years. Calculate the future loss at issue Answer: The present value of benefits is 1000v 25.8  1000/1.0525.8  284.00. 26 premiums are paid, so the present value of the premiums is 20 times a 26-year certain annuity-due, or 20a¨26 

20 (1 − (1/1.05) 26 )  20 (15.09394)  301.88 0.05/1.05

1If expenses are included, we assume that they are fixed and their annual amount is known in advance. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

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18. PREMIUMS: NET PREMIUMS FOR FULLY CONTINUOUS INSURANCES

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Then the future loss at issue is 0 L  284.00 − 301.88  −17.88 . Note that we did not need a mortality assumption to calculate the future loss. 

?

Quiz 18-1 For a whole life insurance on (40) paying 1000 at the moment of death, • • • •

Mortality is uniformly distributed with ω  100. δ  0.04. The annual premium, payable continuously, is 30. The insured person dies at exact age 60.

Calculate the net future loss.

18.2

Net premium

The method we will use most of the time for calculating premium is the equivalence principle. Under the equivalence principle, the actuarial present value of the premiums is set equal to the actuarial present value of the insurance company’s payments. In other words, the future loss at issue is set equal to 0. Let’s ignore expenses and only calculate the amount of premium needed to fund the benefit—the death benefit or the annuity payments. When we set the expected value of future premiums equal to the expected value of future benefits, the premium that results is called the net premium. ’‘Net single premium” is an example of this terminology. They can also be called “benefit premiums”. The rest of this lesson discusses continuous premiums for continuous insurances, or fully continuous insurances. A fully continuous insurance is defined as an insurance funded with continuous premiums and paying a benefit at the moment of death. A fully continuous annuity is a continuous annuity that is funded with continuous premiums. Let’s begin with the simple case of level premiums and a face amount of 1. The letter P will denote the annual net premium. If A¯ is the APV of the insurance and a¯ is the APV of a unit continuous life annuity paid for the period of the insurance coverage, then by the equivalence principle, the premium P is P a¯  A¯ which implies P

A¯ a¯

For whole life insurance on ( x ) , P  A¯ x /a¯ x . Formulas (13.2) and (13.3) allow expressing the premium in terms of a¯ x or A¯ x alone: 1 1 − δ a¯ x  −δ (18.1) P a¯ x a¯ x A¯ x δ A¯ x P  (18.2) (1 − A¯ x ) /δ 1 − A¯ x For an n-year endowment insurance on ( x ) , P  A¯ x:n /a¯ x:n . Formulas (13.7) and (13.8) allow expressing the premium in terms of a¯ x:n or A¯ x:n alone: 1 −δ a¯ x:n δ A¯ x:n P 1 − A¯ x:n P

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(18.3) (18.4)

18.2. NET PREMIUM

359

Formulas for net premiums for other forms of insurance require both an insurance and an annuity function; they cannot be simplified into a form depending only on insurances or only on annuities. For n-year term insurance, the formula is A¯ x1:n P a¯ x:n and for n-year deferred insurance, if premiums are payable only during the deferral period, the net premium is ¯ n| A x P a¯ x:n while if premiums are payable for life, the net premium is P

¯

n| A x

a¯ x

For an n-year deferred annuity, the premium is the APV of the annuity benefit divided by the APV of an n-year temporary life annuity of 1 per year: P

n| a¯ x

a¯ x:n

Example 18B For fully continuous insurances of 1, you are given: • Premiums are payable for life. • The force of mortality, µ, is constant. • The force of interest is δ. Calculate the net premium for (a) whole life insurance and (b) n-year term insurance. Answer: For whole life insurance, A¯ x  µ/ ( µ + δ ) and a¯ x  1/ ( µ + δ ) , so P  µ . For term insurance, A¯ x1:n  µ 1 − e −n ( µ+δ ) ( µ + δ ) and a¯ x:n  1 − e −n ( µ+δ ) ( µ + δ ) , so once again the premium is µ . Both of these results are logical. With constant force of mortality, the system has no memory. Thus the net premium for an insurance with no endowment benefit is the current mortality cost. 



?

.



.

Quiz 18-2 You are given: • µ  0.02 • δ  0.04 Calculate the annual net premium for a fully continuous 20-year endowment insurance of 1. Example 18C For a fully continuous whole life insurance on (40) of 1, you are given: • Mortality is uniformly distributed with ω  120. • Premiums are payable for life. • δ  0.05 Calculate the annual net premium for the insurance. Answer: Calculating APV’s of insurances under uniform survival is easier than calculating APV’s of annuities, so we’ll use formula (18.2). a¯ 1 − e −0.05 (80) A¯ 40  80   0.245421 80 80 (0.05) δ A¯ 40 0.05 (0.245421) P   0.016262 ¯ 1 − 0.245421 1 − A40

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?

Quiz 18-3 For a fully continuous 20-year term insurance on (50) with unit face amount, you are given: • Premiums are payable for the entire duration of the policy. • µ50+t  1/ (50 − t ) . • δ  0.03 Calculate the annual net premium for the term insurance. Sometimes premiums are only payable for a limited period, as in the following example. Example 18D For a fully continuous whole life insurance of 1 on (50), premiums are payable for at most 20 years. You are given: • µ  0.01 • δ  0.04 Calculate the annual net premium. Answer: The equivalence principle gives A¯ 50  P a¯50:20 . Using the usual constant force formulas for A¯ 50 and a¯50:20 : A¯ 50 

µ  0.2 µ+δ

1 − e −20 ( µ+δ ) 0.632121   12.6424 µ+δ 0.05 0.2 P  0.01582 12.6424

a¯50:20 



Example 18E Kevin, age 55, is purchasing a fully continuous whole life annuity that will pay 1000 per year starting at age 65. He will pay continuous premiums starting immediately for 10 years. You are given: • µ  0.01 • δ  0.05 Calculate the premium determined by the equivalence principle. Answer: The actuarial present value of one unit of the deferred whole life annuity is e −10 (0.06) /0.06  9.14686. A 10-year temporary life annuity has APV (1 − e −10 (0.06) ) /0.06  7.51981. Therefore, the net  premium is 1000 (9.14686) /7.51981  1216.37 .

18.3

Expected value of future loss

The expected value of the future loss at issue is the expected value of benefits minus the expected value of premiums. For the standard insurances, it has the form A − Pa. For whole life, both the insurance and the annuity are functions of v Tx , simplifying matters. For a fully continuous whole life insurance on ( x ) , if b is the face amount and P is the premium, then 1 − v Tx P P  v Tx b + − δ δ δ

!

Tx − P a¯Tx  bv Tx − P 0 L  bv





(18.5)

A similar formula with Tx replaced with min (Tx , n ) can be written for an n-year endowment insurance. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

18.3. EXPECTED VALUE OF FUTURE LOSS

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Applying expected values to equation (18.5), we obtain the following formula for the expected loss at issue on a whole life insurance: P P − E[0 L]  b A¯ x − P a¯ x  A¯ x b + δ δ





(18.6)

A similar formula with A¯ x replaced with A¯ x:n can be written for an n-year endowment insurance. For other types of insurance or for limited-pay whole life, the future loss is the present value of the death benefit minus P times an appropriate annuity. For example, for n-year term insurance, the formula would be ! 1 − v min (Tx ,n ) L  Z − P 0 δ

  bv Tx Tx ≤ n where Z   . 0 Tx > n  The equivalence principle determines the premium such that E[0 L]  0. Example 18F For a fully continuous 10-year term insurance of 1000 on (60) , you are given: • Premiums are payable in all years. • δ  0.06 • µ60+t  1/ (40 − t ) , 0 < t < 40. 1. The gross premium is 25. Calculate the expected value of the net future loss. 2. Determine the annual continuous premium so that the expected value of the future loss will be −1. Answer:

1 1. The actuarial present value of the term insurance is 1000A¯ 60 :10 1 A¯ 60 :10 

a¯10 1 − e −0.6   0.187995 40 (40)(0.06)

For the actuarial present value of the premiums, we need a¯60:10 . For uniform mortality it is easier to first calculate a 10-year endowment to get a 10-year temporary life annuity than to do it directly. A¯ 60:10  0.187995 + 10 E60  0.187995 + 34 e −10 (0.06)  0.187995 + 0.411609  0.599604 1 − 0.599604 a¯60:10   6.673269 0.06 We’re now ready to calculate E[0 L]. 1 E[0 L]  1000A¯ 60 :10 − 25 a¯60:10

 187.995 − 25 (6.67327)  21.163 2. Notice that we are explicitly told not to use the equivalence principle, but rather to set the expected future loss to a nonzero number. We want the premium P such that 0 L  187.995 − P (6.67327)  −1. P

188.995  28.321 6.67327

Example 18G For a fully continuous whole life insurance, the force of mortality is constant. Determine the annual gross premium so that the expected value of the net future loss will be l. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM



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Answer: The expected future loss for an exponential fully continuous whole life insurance with premium P is µ P E[0 L]  − µ+δ µ+δ Setting this equal to l, we get µ − P  l (µ + δ) P  µ − l (µ + δ)

18.4



International Actuarial Premium Notation

International Actuarial Notation defines symbols for premiums on standard insurances of 1. The base symbol is P. For insurances payable at the end of the year of death, subscripts on P are the same as the corresponding subscripts on the A which relates to the insurance. If there are no other decorations, the premium is annual and payable throughout the duration of the policy. Thus • Px is the premium for a fully discrete whole life insurance, or A x /a¨ x . • Px1:n is the premium for a fully discrete n-year term insurance, or A x1:n /a¨ x:n . • Px:n1 is the annual premium for an n-year pure endowment, or A x:n1 /a¨ x:n . • Px:n is the premium for a fully discrete n-year endowment insurance, or A x:n /a¨ x:n . A presubscript indicates the premium payment period if it is not the same as the insurance period. Thus 20 P x indicates a 20-pay whole life. If the insurance is payable at the moment of death, the insurance must be shown separately, with a bar. For example, P ( A¯ x ) indicates a premium payable annually on a whole life insurance payable at the moment of death. A bar on the P indicates continuous premiums. A parenthesized A is also necessary for deferred insurances.

Exercises Future loss at issue 18.1.

For a 10-year deferred fully continuous whole life insurance of 1000: • µ  0.04 • δ  0.02 • Premiums are payable for life. • The premium is determined by the equivalence principle.

Calculate the net future loss at issue if the policyholder dies at the end of 20 years. 18.2.

For a whole life insurance policy on ( x ) : • A benefit of 100,000 is payable at the moment of death. • Premium of 3000 is payable annually at the beginning of each year for 20 years. • i  0.04

Calculate the net future loss if Tx  34. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

Exercises continue on the next page . . .

EXERCISES FOR LESSON 18

363

Table 18.1: Summary of formulas for this lesson

For whole life: 1 −δ P¯ ( A¯ x )  a¯ x δ A¯ x P¯ ( A¯ x )  1 − A¯ x

(18.1) (18.2)

For n-year endowment insurance: 1 −δ a¯ x:n δ A¯ x:n ) 1 − A¯ x:n

P¯ ( A¯ x:n ) 

(18.3)

P¯ ( A¯ x:n

(18.4)

For constant force of mortality: The net premiums for fully continuous whole life and fully continuous term are equal to µ. Future loss formulas for whole life with face amount b: P P − δ δ   P P ¯ E[0 L]  A x b + − δ δ



0L

 v Tx b +



(18.5) (18.6)

Similar formulas are available for endowment insurances. 18.3. A 20-year endowment insurance pays b t at the end of the year of death if death occurs at time t, where   2000 t ≤ 10 bt    1000 10 < t ≤ 20

 The benefit upon survival for 20 years is 1000. You are given: • A premium of 40 per year is payable quarterly. • i  0.06 Calculate the net future loss if the insured dies at time 14.5, right before the quarterly premium is paid. 18.4.

For a 10-year deferred annual annuity-due of 1 per year: • Premiums of 0.6 per year are payable at the beginning of each year for 10 years. • i  0.05.

Calculate the net future loss if the annuitant dies at time 25.1.

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[3-S01:2] On January 1, 2002, Pat, age 40, purchases a 5-payment, 10-year term insurance of 100,000:

18.5.

• Death benefits are payable at the moment of death. • Gross premiums of 4000 are payable annually at the beginning of each year for 5 years. • i  0.05 • L is the loss random variable at time of issue. Calculate the value of L if Pat dies on June 30, 2004. A. 77,100

B. 80,700

C. 82,700

D. 85,900

E. 88,000

Net premiums for fully continuous insurances [150-82-94:1] For a fully continuous 2-year deferred whole life insurance of 1 on ( x ) , you are given:

18.6. • •

δ  0.05 µ x+t  0.15, t ≥ 0

Calculate the fully continuous net premium, 2| A¯ x /a¯ x . A. 0.031

B. 0.041

C. 0.101

D. 0.123

E. 0.150

18.7. [CAS4A-F93:15] (2 points) An insured, age 25, purchases a 10-year continuous payment, continuous whole life insurance policy with a benefit of 1. You are given that the insured is subject to a constant force of mortality equal to 0.025 and a constant force of interest equal to 0.075. Determine the net annual premium for this policy. A. B. C. D. E.

Less than 0.050 At least 0.050, but less than 0.150 At least 0.150, but less than 0.250 At least 0.250, but less than 0.350 At least 0.350 [150-S89:29] You are given:

18.8.

• Fully continuous whole life insurances of 1 are issued to both Matthew and Zachary. • Fully continuous net level annual premiums for each insurance are determined in accordance with the equivalence principle. • Matthew is subject to a constant force of mortality µ1 . • Because of his hanggliding hobby, Zachary is subject to a constant force of mortality µ1 + µ2 . • The force of interest is δ for both insurances. Determine the excess of Zachary’s premium over Matthew’s premium. A. B. C. D. E.

µ1 µ2 µ1 − µ2

δµ2 ( µ1 + µ2 + δ )( µ1 + δ ) µ1 µ2 + µ22 − δµ1

( µ1 + µ2 )( µ1 + µ2 + δ )

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EXERCISES FOR LESSON 18

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[CAS4A-S92:21] (2 points) For a fully continuous 2 year term insurance on (80), you are given:

18.9.

• The benefit is 1000. • µ  0.4 • δ  0.1 • The premium is calculated using the equivalence principle. Determine the annual net premium. A. B. C. D. E.

Less than 275 At least 275, but less than 325 At least 325, but less than 375 At least 375, but less than 425 At least 425

18.10. [CAS4A-F97:14] (2 points) A person age 25 is subject to a constant force of mortality, µ  0.03. This person wishes to purchase a 10-year endowment insurance with a benefit of $10,000, payable at the moment of death. δ  0.07 Using the equivalence principle, determine the continuous net annual premium. A. B. C. D. E.

Less than $825 At least $825, but less than $850 At least $850, but less than $875 At least $875, but less than $900 At least $900

Use the following information for questions 18.11 and 18.12: You are given: • Premiums are calculated using the equivalence principle. • •

0.01 µ  0.02  δ  0.05.

 

0 ≤ t ≤ 10 t > 10

18.11. Calculate the annual net premium for a fully continuous whole life insurance. 18.12. Calculate the annual net premium for a fully continuous 15-year endowment insurance. 18.13. For a fully continuous whole life insurance on ( x ) : • The death benefit is 1000 (1.025K x ) . • δ  0.05 • µ  0.02 • Level premiums are determined using the equivalence principle. Determine the annual net premium.

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18.14. You are given: • Premiums are calculated using the equivalence principle and are continuous. • µ x  1/ (100 − x ) • δ  0.05. Calculate the 10-year annual net premium for a 10-year deferred fully continuous life annuity of 1 per year on (55). 18.15. [CAS4A-S95:10] (2 points) A life, aged 35, purchases a whole life insurance policy with a benefit of $10 payable immediately upon death. Premiums are paid continuously for life. The mortality for the life is uniform with l x  100 − x. The effective interest rate is 6%. Assume that there are no expenses. Determine the continuous annual premium for the insurer to have an expected loss of 0. A. B. C. D. E.

Less than 0.22 At least 0.22, but less than 0.28 At least 0.28, but less than 0.34 At least 0.34, but less than 0.40 At least 0.40

18.16. [CAS4A-S97:6] (2 points) For a life age 50, mortality is uniformly distributed with ω  100. This life purchases a fully continuous $100,000 10-year term life insurance policy with level annual premiums determined by the equivalence principle. δ  0.08. Calculate the net premium. A. B. C. D. E.

Less than $2,000 At least $2,000, but less than $2,500 At least $2,500, but less than $3,000 At least $3,000, but less than $3,500 At least $3,500

18.17. [CAS4A-S98:19] (2 points) A person age 30 wishes to purchase a 20-year endowment policy with a benefit of $100,000. The premiums are fully continuous and the benefit is payable at the moment of death. Assume that the force of interest is constant δ  0.06, and the net single premium for a 20-year endowment policy with a benefit of $1 payable at the moment of death for the same person is 0.125. Determine the continuous net annual premium for this endowment. A. B. C. D. E.

Less than $600 At least $600, but less than $675 At least $675, but less than $750 At least $750, but less than $825 At least $825

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18.18. [CAS4A-F92:9] (1 point) You are given the following: • •

A¯ x  0.40 δ  0.06

Determine the annual net premium for a fully continuous whole life insurance on ( x ) . A. B. C. D. E.

Less than 0.036 At least 0.036, but less than 0.037 At least 0.037, but less than 0.038 At least 0.038, but less than 0.039 At least 0.039

18.19. For a fully continuous 20-year term life insurance on ( x ) with face amount 1000, you are given: • The net single premium is 150. • The annual net premium is 15.15. • δ  0.06 Determine 20p x . 18.20. [3-S00:35] The distribution of Jack’s future lifetime is a two-point mixture: • With probability 0.60, Jack’s future lifetime follows the Illustrative Life Table, with deaths uniformly distributed over each year of age. • With probability 0.40, Jack’s future lifetime follows a constant force of mortality with µ  0.02. A fully continuous whole life insurance of 1000 is issued on Jack at age 62. Calculate the net premium for this insurance at i  0.06. A. 31

B. 32

C. 33

D. 34

E. 35

18.21. For a fully continuous whole life insurance of 1000 on (45), • The probability of survival is t p 45

•

 0.6e −0.01t + 0.4e −0.03t ,

t>0

δ  0.06

Calculate the net premium for this policy. A. B. C. D. E.

Less than 16.90 At least 16.90, but less than 17.00 At least 17.00, but less than 17.10 At least 17.10, but less than 17.20 At least 17.20

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18.22. [3-S01:32 and 150-S88:7] For a special fully continuous whole life insurance on ( x ) : • The level premium is determined using the equivalence principle. • Death benefits are given by b t  (1 + i ) t where i is the interest rate. • L is the loss random variable at t  0 for the insurance. • T is the future lifetime random variable of ( x ) . Which of the following expressions is equal to L? A. ( v T − A¯ x ) / (1 + A¯ x ) B. ( v T − A¯ x )(1 − A¯ x ) C. v T − A¯ x D. ( v T − A¯ x )(1 + A¯ x ) E. ( v T − A¯ x ) / (1 − A¯ x ) 18.23. For a 20-pay 30-year term life insurance of 1000 on (50): • Benefits are payable at the moment of death. • µ x  1/ (100 − x ) , x < 100. • δ  0.04. Determine the continuous premium per year resulting in an expected loss at issue of −10. 18.24. [M-S05:37] Company ABC sets the gross premium for a continuous life annuity of 1 per year on ( x ) equal to the net single premium calculated using: • •

δ  0.03 µ x+t  0.02, t ≥ 0

However, a revised mortality assumption reflects future mortality improvement and is given by

  0.02 µ x+t    0.01

for t ≤ 10 for t > 10

 Calculate the expected loss at issue for ABC (using the revised mortality assumption) as a percentage of the gross premium. A. 2%

B. 8%

C. 15%

D. 20%

E. 23%

18.25. [CAS4A-S99:20] (2 points) Your company has just sold a fully continuous whole life policy of $1 million to (40), but you have charged her the rate for (30) by accident. Assuming that the error cannot be corrected, how much money has this error cost you in present value? Assume A¯ t  t/ (60 + t ) , for t > 20. A. B. C. D. E.

Less than $80,000 At least $80,000, but less than $85,000 At least $85,000, but less than $90,000 At least $90,000, but less than $95,000 At least $95,000

Additional old CAS Exam 3/3L questions: F08:22

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EXERCISE SOLUTIONS FOR LESSON 18

369

Solutions 18.1.

The continuous net premium is 1000 (0.04e −0.06 (10) )  365.874 0.04 + 0.02 1 a¯ x  0.06 P  365.874 (0.06)  21.952

100010| A¯ x 

The net future loss at issue for death at 20 years is 1 − e −0.4  308.46 0.02

!

1000v 20 − P a¯20  1000e −0.4 − 21.952

18.2. The present value of the benefit is 100,000/1.0434  26,355.21. The present value of the 20-year annuity due of premiums is 1 − (1/1.0420 )  3000 (14.13394)  42,401.82 3000 0.04/1.04

!

The net future loss is 26,355.21 − 42,401.82  −16,046.61 . 18.3.

The present value of the benefit is 1000/1.0615  417.27. The present value of the premiums is d (4)  4 (1 − 1/1.060.25 )  0.057847 1 − (1/1.0614.5 )  9.860559 0.057847  394.42

a¨ (4)  14.5

40a¨ (4)

14.5

The net future loss is 417.27 − 394.42  22.85 . 18.4. The present value of the benefits, which are a 10-year deferred certain annuity-due for 16 years (16 payments are made) is 1/1.0510 − 1/1.0526  6.98612 0.05/1.05 The present value of the premiums, which are a 10-year temporary life annuity-due, is 1 − (1/1.0510 ) 0.6  0.6 (8.10782)  4.86469 0.05/1.05

!

The net future loss is 6.98612 − 4.86469  2.12143 . 18.5. 100,000 1 − (1/1.05) 3 L − 4000 0.05/1.05 1.052.5

!

 88,517.01 − 4000 (2.85941)  77,079.37

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(A)

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18.6.

As the notation indicates, the premiums are payable for life, not just for the deferral period. 0.15 ( e −2(0.20) )  0.502740 0.05 + 0.15 1 5 a¯ x  0.05 + 0.15 ¯ 0.502740 2| A x   0.100548 (C) a¯ x 5 ¯ 

2| A x

18.7. We calculate a 10-year temporary annuity for the denominator and a whole-life insurance for the numerator, using the usual constant force formulas.

  1 − e −10 ( µ+δ )  10 1 − e −1 µ+δ µ 0.025 1   A¯ 25  µ + δ 0.025 + 0.075 4 1/4 P (A)
 0.03955 10 1 − e −1

a¯25:10 

18.8. For fully continuous insurances with constant force of mortality, the net level annual premium is the force of mortality, so the premium for Matthew is µ1 , the premium for Zachary is µ1 + µ2 , and the excess of Zachary’s premium over Matthew’s is µ2 . (B) 18.9. (D)

For constant force term, the net premium for 1 is µ, so the premium for 1000 is 1000 (0.4)  400 .

18.10. For an endowment insurance, use formula (18.3): P

1 a¯ x:n

−δ

Therefore

  1 1 − e − (0.03+0.07) 10  6.321206 0.03 + 0.07 1 P − 0.07  0.088198 6.321206

a¯ x:10 

The answer is 10,000 (0.088198)  881.98 .

(D)

18.11. There are many formulas we could use. We’ll use 1 −δ a¯ x a¯ x  a¯ x:10 + 10 E x a¯ x P

1 − e −0.6 e −0.6 + 0.06 0.07  7.519806 + 7.840166  15.359972 1 P − 0.05  0.015104 15.359972 

A plausible answer, since we know that if µ were constant 0.01, the answer would be 0.01, and if µ were constant 0.02, the answer would be 0.02. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

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18.12. We’ll do this similarly to the previous exercise. 1 −δ a¯ x:15  a¯ x:10 + 10 E x a¯ x:5

P a¯ x:15

1 − e −0.6 1 − e −0.35  + e −0.6 0.06 0.07

!

 7.519806 + 2.315294  9.8351 1 P − 0.05  0.051677 9.8351 18.13. We will first calculate the value of a unit 1-year term insurance. From Table 10.1, 0.02 (1 − e −0.07 ) A¯ x1:1   0.019316 0.07 Our insurance provides 1-year term insurances geometrically increasing at the effective rate 0.025. Each year is discounted for interest and mortality at the continuous rate 0.02 + 0.05  0.07. Therefore, the APV of the insurance is ∞ X



1000 (0.019316) 1.025e −0.07

k0

k

!

 19.316

1  19.316 (22.5752)  436.06 1 − 1.025e −0.07

A unit continuous annuity has APV 1/ (0.02 + 0.05) , so the annual net premium is 436.06 (0.07)  30.52 . 18.14. Mortality is uniform with ω  100. The easiest way to do this exercise is to calculate A¯ 55 and A¯ 55:10 , and to use those to calculate the two annuities we need. a¯45 45 1 − e −45 (0.05)   0.397600 0.05 (45) 1 − 0.397600  12.048 a¯55  0.05 a¯10 1 A¯ 55 :10  45 4  (1 − e −0.5 )  0.174875 9 7 −10 (0.05)  0.471746 10 E55  e 9 A¯ 55:10  0.174875 + 10 E55  0.174875 + 0.471746  0.646621 1 − 0.646621 a¯55:10   7.06758 0.05 10| a¯ 55  12.048 − 7.06758  4.98042 4.98042 P  0.70469 7.06758 A¯ 55 

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18. PREMIUMS: NET PREMIUMS FOR FULLY CONTINUOUS INSURANCES

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18.15. We are being asked for the equivalence-principle premium, P  10A¯ 35 /a¯35 . For uniform survival, it is easier to compute insurances than annuities, so we use the formula δ A¯ 35 P  10 1 − A¯ 35 Since ω  100, the density function for survival is i is given, we convert it to δ using δ  ln (1 + i ) .

!

1 1 1   . Since an effective rate of interest ω − x 100 − 35 65

a¯ A¯ 35  65 65 1 − e −65δ  65δ 1 − (1/1.0665 )  65δ δ  ln 1.06  0.058269 0.977347  0.258046 A¯ 35  65 (0.058269) (2.58047)(0.058269) P 1 − 0.258046  0.2027 (A)

18.16. The present value of the insurance is

!

1 A¯ 50 :10 

  a¯10 1  1 − e −0.8  0.137668 50 (50)(0.08)

To calculate the value of the annuity, we use 1 − A¯ 50:10 δ ¯ A50:10  0.137668 + 10 E50 4  0.137668 + e −0.8  0.497131 5 1 − 0.497131 a¯50:10   6.285863 0.08 0.137668 P  0.0219012 6.285863 a¯50:10 

The answer is $100,000 (0.0219012)  $2190.12 . (B) 18.17. We’re given A¯  0.125. We use formula (18.2). x:20

P

δ A¯ x:20 1 − A¯



(0.125)(0.06)

x:20

The answer is 100,000 (0.00857143)  857.14 . (E) LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

1 − 0.125

 0.00857143

EXERCISE SOLUTIONS FOR LESSON 18

373

18.18. We use formula (18.2). P

(0.40)(0.06) δ A¯ x   0.040 ¯ 1 − 0.40 1 − Ax

(E)

18.19. The annual net premium is 1000A¯ x1:20 1000δ A¯ x1:20 150 (0.06)    15.15 ¯ a¯ x:20 1 − A x:20 1 − A¯ x:20 and therefore A¯ x:20  1 −

9  0.40594 15.15

1  0.40594 − 0.15  0.25594  20 p x e −20 (0.06) , so Then A x:20 20 p x

 0.25594e 1.2  0.8498

18.20. We have here a mixture of two distributions. The expected value of a mixture—or any of the moments of a mixture—can be calculated as the weighted average of the expected values of each of the components of the mixture. However, the premium of the mixture is not the weighted average of the premiums of each component!

!

!

i 0.02 A¯ 62  0.6 (0.3966965) + 0.4  0.3473008 δ 0.02 + δ δ A¯ 62 P¯ ( A¯ 62 )   0.0310048 1 − A¯ 62 The answer is 1000 (0.0310048)  31 . (A) 18.21. It is easier to calculate the EPV of an annuity than it is to calculate the EPV of an insurance for this survival function. ∞

Z a¯45 

v t t p 45 dt

Z0 ∞ 

(0.6e −0.07t + 0.4e −0.09t ) dt 0

0.6 0.4  +  13.0159 0.07 0.09   1 P  1000 −δ a¯45   1  1000 − 0.06 13.0159  16.83 (A)

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18.22. The present value of the death benefit is (1 + i ) T v T  1 regardless of T, so the actuarial present value of the insurance is 1. The annual net premium is 1/a¯ x . The loss random variable is 1 − vT 1 a¯T  1 − 1− a¯ x δ a¯ x

!

1 − vT 1 − A¯ x v T − A¯ x  1 − A¯ x

1−

(E)

18.23. The actuarial present values of a 30-year insurance and a 20-year annuity are a¯30 1 − e −0.04 (30) 1   0.349403 A¯ 50 :30  50 50 (0.04) a¯20 1 − e −0.04 (20) 1   0.275336 A¯ 50 :20  50 50 (0.04) A¯  0.275336 + 0.6e −0.04 (20)  0.544933 50:20

a¯50:20 

1 − 0.544933  11.3767 0.04

The expected loss at issue for premium π is 1 1000A¯ 50 :30 − π a¯50:20  349.403 − 11.3767π

Setting this equal to −10, π 18.24. The gross premium is

1 0.03+0.02

359.403  31.59 11.3767

 20. The revised net single premium is

a¯0x  a¯ x:10 + 10| a¯ x  20 (1 − e −10 (0.03+0.02) ) +

e −10 (0.05)  23.0327 0.03 + 0.01

The expected future loss at issue as a percentage of the gross premium is 23.0327/20 − 1  15.16% . (C) 18.25.

A¯ 30 

1 3

and A¯ 40  0.4. Thus the premiums charged per unit are δ A¯ 30 δ  ¯ 1 − A30 2

and the present value of these premiums is δ 2

!

1 − A¯ 40  0.3 δ

!

so the present value of the premiums your company receives is 1,000,000 (0.3)  300,000. However, the present value of the insurance is 1,000,000A¯ 40  400,000. Thus your company loses $100, 000 . (E)

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QUIZ SOLUTIONS FOR LESSON 18

375

Quiz Solutions 18-1. Given the time of death, the present value of the death benefit is 1000e −0.04 (20) . The present value of the premiums is a continuous 20-year annuity-certain of 30. The net future loss at issue is therefore 1 − e −0.8  36.3257 − 30 0.04

!

0L

 1000e

−0.8

Note that the uniform assumption played no role. Once the time of death T40 is known, the distribution of T40 no longer matters. 18-2. By using formula (18.3), we only need to calculate a¯ , not A¯ . x:20

x:20

1 − e − (0.02+0.04)(20)  11.64676 0.02 + 0.04 1 − 0.04  0.04586 P 11.64676

a¯ x:20 

18-3.

Let’s calculate the APV of the insurance. a¯20 1 − e −0.03 (20) 1 A¯ 50    0.300792 :20 50 50 (0.03)

Calculate the APV of an endowment insurance and then use A¯ 50:20  1 − δ a¯50:20 to derive the APV of a temporary life annuity. Note that 20 p50  0.6. A¯ 50:20  0.300792 + 0.6e −0.03 (20)  0.630079 1 − 0.630079 a¯50:20   12.3307 0.03 0.300792 P  0.02439 12.3307

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376

18. PREMIUMS: NET PREMIUMS FOR FULLY CONTINUOUS INSURANCES

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Lesson 19

Premiums: Net Premiums for Discrete Insurances Calculated from Life Tables Reading: Models for Quantifying Risk (4th or 5th edition) 9.1 except 9.1.4, 9.2, 9.4.1 This lesson will deal with discrete net premiums. Often the insurance will also be non-continuous. A fully discrete insurance is one for which premiums are payable at the beginning of each year and benefits are paid at the end of the year of death. A semicontinuous insurance is one for which premiums are payable at the beginning of each year and benefits are paid at the moment of death. A fully discrete annuity has premiums payable at the beginning of each year and the benefit is either an annuity-due or an annuityimmediate. For a fully discrete insurance, it is convenient to calculate premiums from a life table. A spreadsheet is helpful for this calculation. On an exam, they may give you a small mortality table and ask you to calculate the premium directly. Example 19A A special annual 3-year endowment insurance on ( x ) pays a death benefit of 1000t at the end of the year if death occurs in year t, t  1, 2, 3, and pays 1500 if the insured survives 3 years. You are given • Premiums are payable at the beginning of the first 2 years only. The premium in the second year is half the premium of the first year. • Age qx x x+1 x+2

0.10 0.15 0.25

• d  0.05 Calculate the premium payable in the first year using the equivalence principle. Answer: Let P be the first year premium. The actuarial present value of the premiums is





P (1 + 0.5vp x )  P 1 + (0.5)(0.95)(1 − 0.10)  1.4275P We use the fact that v  1 − d. The actuarial present value of the benefits is 1000vq x + 2000v 2 1| q x + 3000v 3 2| q x + 1500v 3 3p x We must calculate these mortality probabilities.

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1| q x

 p x q x+1  (0.9)(0.15)  0.135

2| q x

 2 p x q x+2  (0.9)(0.85)(0.25)  0.19125

3p x

 (0.9)(0.85)(0.75)  0.57375 377

19. PREMIUMS: NET PREMIUMS FROM LIFE TABLES

378

The actuarial present value of the death benefits is therefore 1000 (0.95)(0.10) + 2000 (0.952 )(0.135) + 3000 (0.953 )(0.19125) + 1500 (0.953 )(0.57375)  95.0000 + 243.6750 + 491.9189 + 737.8784  1568.4723 The premium is therefore 1568.4723/1.4275  1098.75 .

?



Quiz 19-1 A special 3-year term insurance on ( x ) pays a death benefit of 1000 if death occurs in the first two years and 5000 if death occurs in the third year. The benefit is paid at the end of the year of death. Level premiums are payable at the beginning of each year. The ultimate mortality table has q x  0.1, q x+1  0.2, and q x+2  0.3. The selection period is two years, with q [x]+k  0.92−k q x+k , k  0, 1. Calculate the net premium at 5% effective annual interest. An exam question for a long-duration contract can use the Illustrative Life Table. Example 19B A deferred life annuity on (45) pays benefits of 120 at the beginning of each year with the first payment at age 65. Premiums for this annuity are paid at the beginning of each year for 20 years. You are given: • Mortality follows the Illustrative Life Table. • i  0.06 Calculate the annual net premium. Answer: The actuarial present value of the deferred annuity is 120 20 E45 a¨65  (120)(0.25634)(9.8969)  304.44 Let P be the premium. The actuarial present value of the premiums is





P ( a¨45 − 20 E45 a¨65 )  P 14.1121 − (0.25634)(9.8969)  11.5751P The premium is therefore P  304.44/11.5751  26.301 .



Exercises 19.1.

For a 30-year fully discrete term life insurance with face amount 1000 on (35), you are given: • Premiums are calculated using the equivalence principle. • Premiums are payable for 15 years. • Mortality follows the Illustrative Life Table. • i  0.06.

Calculate the annual net premium.

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Exercises continue on the next page . . .

EXERCISES FOR LESSON 19

379

19.2. [CAS4-F82:19] Experience on certain $20,000 luxury sports cars shows probabilities, q x , of theft within a one year period to be as follows: Age x

qx

0 1 2 3

0.15 0.10 0.08 0.03

The car neither appreciates nor depreciates in value and, once stolen, is never recovered. A 3 year non-cancelable (except for theft of the vehicle) theft policy that provides for the payment of an annual net level premium P is issued on one of these cars when purchased new. If the claim is payable at the end of the year of theft and interest is earned at an effective annual rate of 10%, what is P? A. B. C. D. E. 19.3.

Less than $2,050 At least $2,050, but less than $2,150 At least $2,150, but less than $2,250 At least $2,250, but less than $2,350 At least $2,350 [CAS3-S04:13] An insurer offers a three-year warranty for a certain piece of equipment.

•

A premium payment for the warranty is made at the start of each year.

•

All premium payments are the same.

•

Once the machine breaks, a benefit payment is made at the end of that year and the policy is canceled (no further payments, neither premium nor benefit, are made).

•

Age(x)

Probability of breakdown (q x )

Year-end benefit payment

0 1 2

10% 20% 20%

100 100 50

i  12%

•

Calculate the annual net premium. A. B. C. D. E.

Less than 5.00 At least 5.00, but less than 7.50 At least 7.50, but less than 10.00 At least 10.00, but less than 12.50 At least 12.50

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[SOA3-F04:39] For a special fully discrete 3-year term insurance on ( x ) :

19.4.

• The death benefit payable at the end of year k + 1 is

 0 b k+1    1,000 (11 − k )

for k  0 for k  1, 2

 k

•

q x+k

0 0.200 1 0.100 2 0.097 i  0.06

•

Calculate the level annual net premium for this insurance. A. 518 19.5.

B. 549

C. 638

D. 732

E. 799

[SOA3-F03:8] For a special fully discrete whole life insurance of 1000 on (40): • The level net premium for each of the first 20 years is π. • The net premium payable thereafter at age x is 1000vq x , x  60, 61, 62, . . . • Mortality follows the Illustrative Life Table. • i  0.06

Calculate π. A. 4.79 19.6.

B. 5.11

C. 5.34

D. 5.75

E. 6.70

For a fully discrete 10-year deferred whole life insurance of 1 on (25), you are given: • Mortality follows the Illustrative Life Table. • i  0.06.

Premiums are payable for the entire lifetime of (25). Calculate the annual premium for which the expected value of the net future loss is −0.2.

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EXERCISES FOR LESSON 19

381

19.7. [CAS4A-S96:22] (2 points) Smith, age 40, purchases a continuous 5-year deferred whole life annuity that will pay Jones, currently age 30, $1,000 per year as long as Jones is alive. Jones’s mortality is given by µ  0.04 and Smith’s is given by the life table below: Age

lx

39 40 41 42 43 44 45

40,000 39,000 38,000 37,000 36,000 35,000 34,000

Smith will make 3 equal payments, each at the end of the year, if he is alive at that time, such that the actuarial present values of the benefits and premiums are equal. Assume i  0.06. Calculate the annual premium that Smith pays. A. B. C. D. E. 19.8.

Less than $2,400 At least $2,400, but less than $2,700 At least $2,700, but less than $3,000 At least $3,000, but less than $3,300 At least $3,300 [3-F01:24] For a special 2-payment whole life insurance on (80):

• Premiums of π are paid at the beginning of years 1 and 3. • The death benefit is paid at the end of the year of death. • There is a partial refund of premium feature: If (80) dies in either year 1 or year 3, the death benefit is 1000 + π/2. Otherwise, the death benefit is 1000. • Mortality follows the Illustrative Life Table. • i  0.06 Calculate π, using the equivalence principle. A. 369

B. 381

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C. 397

D. 409

E. 425

Exercises continue on the next page . . .

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382

19.9. [3-S00:31] Company ABC issued a fully discrete three-year term insurance of 1000 on Pat whose stated age at issue was 30. You are given: • x

qx

30 31 32 33

0.01 0.02 0.03 0.04

• i  0.04 • Premiums are determined using the equivalence principle and are payable at the beginning of each year. During year 3, Company ABC discovers that Pat was really 31 when the insurance was issued. Using the equivalence principle, Company ABC adjusts the death benefit to the level death benefit it should have been at issue, given the premium charged. Calculate the adjusted death benefit. A. 646

B. 664

C. 712

D. 750

E. 963

19.10. A special fully discrete 3-year term insurance on ( x ) has a face amount of 1000. If the insured survives 3 years, the net premiums are refunded without interest. You are given • •

q x  0.02, 1| q x  0.03, and 2| q x  0.05. i  0.08

Calculate the annual net premium. Additional old CAS Exam 3/3L questions: F06:36, S13:14 Additional old CAS Exam LC questions: S14:14 Additional old SOA Exam MLC questions: S12:3, S13:18, F14:9

Solutions 19.1. By the equivalence principle, the annual net premium is the quotient of the insurance over the premium annuity, or 1 A35 :30 1 15 P 35:30  a¨35:15 The actuarial present value of the insurance is 1 A35 :30  A35 − 30 E35 A65 30 E35

can be calculated using the Illustrative Life Table as 10 E35 20 E45 . 30 E35 1 A35 :30

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 (0.54318)(0.25634)  0.13924  0.12872 − (0.13924)(0.43980)  0.067482

EXERCISE SOLUTIONS FOR LESSON 19

383

The actuarial present value of the annuity is a¨35:15  a¨35 − 15 E35 a¨50 15 E35

can be calculated as 10 E35 5 E45 . 15 E35

 (0.54318)(0.72988)  0.39646

a¨35:15  15.3926 − (0.39646)(13.2668)  10.13284 The annual net premium is

!

0.067482  1000  6.6597 . 10.13284

1 1000 15 P 35 :30

19.2.

The actuarial present value of the insurance is 0.15 (0.85)(0.1) (0.85)(0.9)(0.08) + + 1.1 1.12 1.13  $20,000 (0.13636 + 0.07024 + 0.04598)  $5051.84





$20,000

The value of an annuity of 1 per year for 3 years is 1+

0.85 (0.85)(0.9) +  1 + 0.772727 + 0.632231  2.404959 1.1 1.12

P is $5051.84/2.404959  $2100.59 . (B) 19.3.

The actuarial present value of the premiums is the annual premium times 1 + vp 0 + v 2 2 p0  1 +

0.9 (0.9)(0.8) +  2.3776 1.12 1.122

The actuarial present value of the benefits is 100vq 0 + 100v 2 1| q 0 + 50v 3 2| q 0 

100 (0.1) 100 (0.9)(0.2) 50 (0.9)(0.8)(0.2) + +  28.4029 1.12 1.122 1.123

The premium is therefore 28.4029/2.3776  11.946 . (D) 19.4.

The actuarial present value of the death benefits is 10,000 (0.8)(0.1) 9,000 (0.8)(0.9)(0.097) +  1239.7482 1.062 1.063

The actuarial present value of the premiums is the level annual premium times 1+

0.8 0.72 +  2.3955 1.06 1.062

So the level annual premium is 1239.7482/2.3955  517.53 . (A).

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19. PREMIUMS: NET PREMIUMS FROM LIFE TABLES

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19.5. The equivalence principle says that the actuarial present value of premiums must equal the actuarial present value of benefits. Splitting up each side of the equality into “before age 60” and “after age 60”, we have APV (premiums before 60) + APV (premiums after 60)  APV (benefits before 60) + APV (benefits after 60) However, note that the APV of premiums after 60 equals the APV of benefits after 60, on a year-by-year basis no less! So we can cancel them from both sides, and get APV (premiums before 60)  APV (benefits before 60) 1 π a¨40:20  A40 :20 π

1 A40 :20 a¨40:20

We can calculate this in many ways. We’ll calculate a¨40:20 , derive A40:20 from A x:n  1 − d a¨ x:n , and then subtract a 20-year pure endowment. The values from the Illustrative Life Table we’ll need are a¨40  14.8166

a¨60  11.1454

20 E40

 0.27414

Then: a¨40:20  14.8166 − 0.27414 (11.1454)  11.7612 A40:20  1 − (0.06/1.06)(11.7612)  0.33427 1 A40 :20  0.33427 − 0.27414  0.06013

and finally we calculate π: π

1000 (0.06013)  5.113 11.7612

(B)

In this exercise, using the value 0.05660 for d given in the tables (rather than computing 0.06/1.06 more accurately, as was done above) results in a final answer that rounds to 5.12 instead of 5.11. 19.6.

We need P such that − P a¨25  −0.2 0.2 + 10| A25 P a¨25

10| A25

Let’s calculate 10| A25 and look up a¨25 . 10| A25

 10 E25 A35  (0.54997)(0.12872)  0.070792

a¨25  16.2242 0.2 + 0.070792 P  0.016691 16.2242 19.7. The value of the whole life annuity on Jones is ($1,000e −5 (0.04+ln 1.06) ) / (0.04 + ln 1.06)  $6,225.81. The value of the annuity Smith pays is 1 38,000 37,000 36,000  2.53860 + + 39,000 1.06 1.062 1.063





The annual premium Smith pays is $6,225.81/2.53860  $2,452.46 . (B) LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

EXERCISE SOLUTIONS FOR LESSON 19

385

19.8. 2p 80



l 82 3,284,542   0.83910 l80 3,914,365

The actuarial present value of the premiums is π times 1 + v 2 2 p80  1 +

1 1.062

!

!

!

l 82 0.83910 1+  1.746796 l 80 1.062

To calculate the actuarial present value of the insurance, we need 2| q80 . 2| q 80

 2p 80 q 82  (0.83910)(0.09561)  0.08023

The actuarial present value of the insurance is





1000A80 + 0.5π vq 80 + v 3 2| q 80  665.75 + 0.5π



0.08030 0.08023  665.75 + 0.071559π + 1.06 1.063



Equating the two, 1.746796π  665.75 + 0.071559π 665.75 π  397.41 1.675237

(C)

19.9. The fact that discovery occurs in year 3 is irrelevant, since the premium is adjusted based on what it should have been at issue. We calculate the value of the term insurance, and then calculate the value of the 3-year endowment insurance in order to calculate the value of the 3-year temporary life annuity. The value of the annuity could also be calculated directly. Let Px be the correct net premium at age x. 0.01 (0.99)(0.02) (0.99)(0.98)(0.03) +  0.053797 + 1.04 1.042 1.043 0.02 (0.98)(0.03) (0.98)(0.97)(0.04) 1 A31 + +  0.080216 :3  1.04 1.042 1.043 (0.99)(0.98)(0.97) A30:3  0.053797 +  0.053797 + 0.836629  0.890426 1.043 1 − A30:3 a¨30:3   2.848928 0.04/1.04 0.053797 P30   0.018883 2.848928 (0.98)(0.97)(0.96) A31:3  0.080216 +  0.080216 + 0.811277  0.891493 1.043 1 − A31:3 a¨31:3   2.821191 0.04/1.04 0.080216 P31   0.028433 ! 2.821191 0.018883 1000  664.12 (B) 0.028433 1 A30 :3 

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19.10. The actuarial present value of a unit of premium annuity is 1 + 0.98/1.08 + 0.95/1.082  2.721879. By the equivalence principle, we need

 1000

0.02 0.03 0.05 0.9 (3P ) + + + − 2.721879P  0 1.08 1.082 1.083 1.083



Let’s solve for P 83.9303 + 2.143347P − 2.721879P  0 83.9303  145.07 P 2.721879 − 2.143347

Quiz Solutions 19-1.

First we need to calculate the mortality rates. q [x]  0.92 (0.1)  0.081

q [x]+1  0.9 (0.2)  0.18

p[x]  0.919

2p [x]

q x+2  0.3

 0.919 (0.82)  0.75358

The APV of the insurance is 1000 (0.081) 1000 (0.919)(0.18) 5000 (0.75358)(0.3) + +  77.1429 + 150.0408 + 976.4561  1203.64 1.05 1.052 1.053 The APV of a unit annuity-due is 1+

0.919 0.75358 +  2.55876 1.05 1.052

The net premium is 1203.64/2.55876  470.40 .

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Lesson 20

Premiums: Net Premiums for Discrete Insurances Calculated from Formulas Reading: Models for Quantifying Risk (4th or 5th edition) 9.1 except 9.1.4, 9.2, 9.4.1 In this lesson, we’ll discuss questions using formulas relating net premiums to insurances or annuities. Under the equivalence principle, the net premium is calculated by equating actuarial present values of premiums and benefits. The relations between a¨ and A for whole life and endowment insurances correspond to the ones in the continuous case, namely for whole life 1 −d a¨ x dA x Px  1 − Ax Px 

(20.1) (20.2)

and for endowment insurances 1 −d a¨ x:n dA x:n  1 − A x:n

Px:n 

(20.3)

Px:n

(20.4)

Just as there was a simple formula for the fully continuous net premium for whole life and term insurances when the force of mortality µ is constant (see Example 18B on page 359), there is a simple formula for the fully discrete net premium for annual whole life and term insurances, as derived in the following example. Example 20A q x is constant for all integral x. Develop formulas for the net premiums for whole life and n-year term insurances on ( x ) . Answer: For whole life, by formula (8.8), A x  q/ ( q + i ) , and by formula (14.11), a¨ x  (1+ i ) / ( q + i ) . Taking the quotient, Px  vq x



For term insurance, A x1:n  q/ ( q + i ) i)







(20.5)



1 − ( vp ) n as indicated in Table 8.2, and a¨ x:n  (1 + i ) / ( q +



1 − ( vp ) n , so once again Px1:n  vq x .

These formulas are logical. Due to lack of memory, the net premium is the cost of one year’s term insurance.  These formulas are not helpful for the following quiz, which should be done from first principles, since the insurance is semicontinuous.

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20. PREMIUMS: NET PREMIUMS FROM FORMULAS

388

?

Quiz 20-1 For a whole life insurance of 1 on ( x ) : • • • •

Premiums are payable annually at the beginning of each year. Benefits are paid at the moment of death. µ  0.01 δ  0.04

Calculate the annual net premium. Example 20B A fully discrete whole life insurance for 1000 on (30) was priced using A30  0.2 and 5% interest, but it turns out that A30  0.3. Calculate the expected future loss if the incorrect premium is not changed. Answer: Of course the answer is not 1000 (0.3 − 0.2) . The problem is that in addition to a higher present value of death benefits, there’s a lower present value of premiums. The premium we’re charging, P, can be calculated from equation (20.2):

!

P  1000

The correct annuity factor for the premiums is a¨30  value of the loss is

!

dA x 0.2 (0.05)  1000  11.9048 1 − Ax (1 − 0.2)(1.05) 1 − A30 (1 − 0.3)(1.05)   14.70. The expected d 0.05

E[0 L]  1000 (0.3) − 11.9048 (14.70)  125



There’s no pleasant recursion relationship between premiums at different ages, but for fully discrete whole life and endowment insurances, premiums can be expressed in terms of insurances or annuities, and those have recursive formulas. Example 20C For a standard insured, the net single premium for an annual whole life insurance is 0.4 and q 50  0.01. For an applicant you are considering, there will be extra mortality risk in the first year so that q 50  0.05, but mortality will be standard at all other ages. You are given that i  0.05 Calculate the annual net premium for a fully discrete whole life insurance for this applicant. Answer: Using the recursive formula twice, with A50 denoting the net single premium for the applicant, 0.01 + 0.99A51 1.05 1.05 (0.4) − 0.01 0.41   0.99 0.99 95
0.05 + 0.95A51 0.05 + 99 (0.41)    0.422318 1.05 1.05

0.4  A51 A50

The annual net premium is then P

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dA x 0.422318 (0.05)   0.034812 1 − Ax (1 − 0.422318)(1.05)



EXERCISES FOR LESSON 20

?

389

Quiz 20-2 For a fully discrete whole life insurance policy of 1000 on (45) with annual premium 24, the expected future loss is −20. You are given: • q45  0.01 • d  0.04 Calculate the annual premium for a similar policy on (46) that would result in an expected future loss of −20. For fully discrete whole life and endowment insurances, formulas for net future loss at issue are similar to those for fully continuous insurances, with δ replaced by d. The counterparts of formulas (18.5) and (18.6): P P − d d   P P − E[0 L]  A x b + d d



0L

 v K x +1 b +



(20.6) (20.7)

Example 20D For a unit fully discrete whole life insurance policy on (45): • A45  0.250 • i  0.05 • The gross premium is 0.018. Calculate the expected net future loss at issue. Answer:



E[0 L]  0.250 1 +

0.018 0.018 −  −0.0335 0.05/1.05 0.05/1.05





Exercises 20.1. [CAS4A-S94:15] (2 points) For (0), you are given that k| q 0  1/3 for k  0, 1, 2. An insurance policy pays 1 at the end of the year of death. The insured will pay net annual premiums at the beginning of the year for each year that he is alive. You are also given that the annual effective interest rate is i  0.05. Determine P, the annual premium necessary for the insurer to have an expected loss of 0. A. B. C. D. E.

Less than 0.450 At least 0.450, but less than 0.465 At least 0.465, but less than 0.480 At least 0.480, but less than 0.495 At least 0.495

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Table 20.1: Formula and concept summary for discrete premiums

For fully discrete whole life: 1 −d a¨ x dA x Px  1 − Ax Px 

(20.1) (20.2)

For fully discrete endowment insurance: Px:n  Px:n

1

(20.3)

−d a¨ x:n dA x:n  1 − A x:n

(20.4)

For fully discrete whole life and term insurances: If q x is constant, then P  vq x . Future loss at issue formulas for fully discrete whole life with face amount b: P P − d d   P P E[0 L]  A x b + − d d



0L

 v K x +1 b +



(20.6) (20.7)

Similar formulas are available for endowment insurances.

20.2. [CAS4A-F99:19] (2 points) A fully discrete whole life policy with a face amount of $180,000 on (30) can be paid for in any of three ways: 1. 2. 3.

A net single premium of $45,000 paid when the policy is issued Equal annual net premiums of $3,000 paid at the beginning of the year Annual net premiums of $2,000 paid at the beginning of each year through age 39, and annual net premiums of $3,875 at the beginning of each year at age 40 and above.

Calculate the actuarial present value of a 10-year temporary life annuity-due of $5,000 issued to (30). A. B. C. D. E.

Less than $33,500 At least $33,500, but less than $34,500 At least $34,500, but less than $35,500 At least $35,500 Not enough information is given.

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EXERCISES FOR LESSON 20

391

[150-F97:32] For 140 special 3-payment 10-year term insurances of 1000, you are given:

20.3.

80 insurances are on lives age 40. 60 insurances are on lives age 50. All 140 lives are independent. µ40+t  0.04, t ≥ 0 µ50+t  0.06, t ≥ 0 δ  0.06 Premiums are payable at the beginning of the year, and are determined using the equivalence principle. • For each issue age, the premium for each of years 1 and 2 is twice the premium for year 3. • Death benefits are payable at the moment of death. • • • • • • •

Calculate the total first year premium for these insurances. A. 16,700

B. 17,900

C. 18,800

D. 21,900

E. 23,500

20.4. [CAS4A-S95:8] (2 points) A life, aged 60, purchases a 4-year-deferred continuous whole life insurance with a benefit of $10,000. The insured will pay 5 equal annual premiums, of amount P, beginning immediately. The final premium coincides with the end of the deferral period. You are given the survival function: x , 100 δ  0.05

S0 ( x )  1 −

for 0 ≤ x ≤ 100, and

Determine P. A. B. C. D. E. 20.5.

Less than 800 At least 800, but less than 840 At least 840, but less than 880 At least 880, but less than 920 At least 920 [CAS3-F04:5] For a special decreasing 15-year term life insurance on a person age 30, you are given:

•

µ30+t  1/ (70 − t ) , 0 ≤ t ≤ 70

•

The benefit payment is 2,000 for the first 10 years and 1,000 for the last 5 years.

•

The death benefit is payable at the end of the year of death.

•

v  0.95 Calculate the level annual net premium for this insurance.

A. 14.90

B. 18.60

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C. 19.58

D. 25.96

E. 27.33

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20.6. [150-F89:16] For a special fully discrete whole life insurance of 1000 issued on the life of (75), increasing premiums, π k , are payable at time k, for k  0, 1, 2, . . .. You are given: • π k  π0 ( 1 + i ) k • Mortality is uniformly distributed with ω  105 • i  0.05 • Premiums are calculated in accordance with the equivalence principle. Calculate π0 . A. 33.1

B. 39.7

C. 44.3

D. 51.2

E. 56.4

[150-82-94:17] You are given:

20.7.

• k| q x  (0.90) k+1 /9 k  0, 1, 2, . . . • i  0.08 • The force of mortality, µ, is constant. Let P¯ ( A¯ x ) be the annual net premium for a unit fully continuous whole life insurance on ( x ) .   Calculate 1000 P¯ ( A¯ x ) − Px . A. 11.34

B. 11.94

C. 12.77

D. 13.17

E. 13.76

[M-S05:14] For a special fully discrete 2-year endowment insurance of 1000 on ( x ) , you are given:

20.8.

• The first year net premium is 668. • The second year net premium is 258. • d  0.06 Calculate the level annual premium using the equivalence principle. A. 469

B. 479

C. 489

D. 499

E. 509

[150-83-96:24] For ( x ) , you are given:

20.9. • • • • •

The premium for a 20-year endowment insurance of 1 is 0.0349. The premium for a 20-year pure endowment of 1 is 0.0230. The premium for a 20-year deferred whole life annuity-due of 1 per year is 0.2087. All premiums are fully discrete net level premiums. i  0.05.

Calculate the premium for a 20-payment whole life insurance of 1. A. 0.0131

B. 0.0198

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C. 0.0245

D. 0.0250

E. 0.0317

Exercises continue on the next page . . .

EXERCISES FOR LESSON 20

393

20.10. [CAS4-S90:21] (2 points) An insured age 40 buys a whole life policy with the death benefit initially set at 1. The interest rate is 5%. The insured’s net premiums and death benefits increase each year at a compound rate of 5%. The death benefit is payable at the end of the year of death. e40  35. Calculate the net premium payable at the beginning of the first year. A. B. C. D. E.

Less than 0.025 At least 0.025, but less than 0.026 At least 0.026, but less than 0.027 At least 0.027, but less than 0.028 At least 0.028

20.11. [M-F05:9] For a special fully discrete 30-payment whole life insurance on (45), you are given: • The death benefit of 1000 is payable at the end of the year of death. • The net premium for this insurance is equal to 1000P45 for the first 15 years followed by an increased level annual premium of π for the remaining 15 years. • Mortality follows the Illustrative Life Table. • i  0.06 Calculate π. A. 16.8

B. 17.3

C. 17.8

D. 18.3

E. 18.8

20.12. [M-F06:18] For a special fully discrete 20-year term insurance on (30): • The death benefit is 1000 during the first ten years and 2000 during the next ten years. • The net premium is π for each of the first ten years and 2π for each of the next ten years. • a¨30:20  15.0364 • x 1000A x1:10 a¨ x:10 30 40

8.7201 8.6602

16.66 32.61

Calculate π. A. 2.9

B. 3.0

C. 3.1

D. 3.2

E. 3.3

20.13. [M-F05:37] For a 10-payment, 20-year term insurance of 100,000 on Pat: • Death benefits are payable at the moment of death. • Gross premiums of 1600 are payable annually at the beginning of each year for 10 years. • i  0.05 • L is the loss random variable at the time of issue. Calculate the minimum value of L as a function of the time of death of Pat. A. −21,000

B. −17,000

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C. −13,000

D. −12,400

E. −12,000

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20.14. [150-S90:28] A fully discrete whole life insurance with annual premiums payable for 10 years is issued to (30). You are given: • The death benefit is equal to 1000 plus the refund of the net level annual premiums paid without interest. • Premiums are calculated in accordance with the equivalence principle. Determine the net annual premium for this insurance. A. B. C. D. E.

1000A30 a¨30:10 + 10 · 10| A30 1000A30 a¨30:10 − 10 · 10| A30 1000A30 1 a¨30:10 − ( IA )30 :10 1000A30 1 a¨30:10 − ( IA )30 :10 + 10 · 10| A30 1000A30 1 a¨30:10 − ( IA )30 :10 − 10 · 10| A30

20.15. [150-S87:23] P A is the net annual premium for a fully discrete whole life insurance of 1 calculated using mortality table A and interest rate i. P B is the net annual premium for a fully discrete whole life insurance of 1 calculated using mortality table B and interest rate i. For all ages, the probability of survival from age x to age x + 1 has the following relationship: p xA  (1 + c ) p xB , where the superscript identifies the table. Determine an expression for P A − P B in terms of functions based on table B. A. B. C. D. E. 20.16.

A x (at interest i) − A x (at interest ( i − c ) / (1 + c ) ) A x (at interest ( i − c ) / (1 + c ) ) − A x (at interest i) 1/a¨ x (at interest i) − 1/a¨ x (at interest ( i − c ) / (1 + c ) ) 1/a¨ x (at interest ( i − c ) / (1 + c ) ) − 1/a¨ x (at interest i) Px (at interest i) − Px (at interest ( i − c ) / (1 + c ) ) [150-F87:3] You are given:

• The annual net premium for a 20-pay annual whole life on (25) is 0.046. • The annual net premium for a 20-year annual endowment insurance on (25) is 0.064. • A45  0.640 Calculate the annual net premium for a 20-year annual term insurance on (25). A. 0.008

B. 0.014

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C. 0.023

D. 0.033

E. 0.039

Exercises continue on the next page . . .

EXERCISES FOR LESSON 20

395

20.17. [150-S89:10] You are given: • The annual net premium for a 15-pay annual whole life on (30) is 0.030. • The annual net premium for a 15-year annual endowment insurance on (30) is 0.046. • The annual net premium for a 15-year annual term insurance on (30) is 0.006. Calculate A45 . A. 0.462

B. 0.600

C. 0.692

D. 0.785

E. 0.900

20.18. [CAS4-S85:22] (2 points) Which of the following represents the annual net premium for an n-pay annual whole life insurance on ( x ) ? Ax A. 1 − dA x:n dA x B. 1 − A x:n A x1:n C. 1 − dA x:n dA x1:n D. 1 − A x:n dA x E. 1 − A x1:n 20.19. [CAS4-S86:23] (2 points) Given: • • • •

A x  0.6 n| A x  0.4 Px  0.1 Px+n  0.2

Calculate Px1:n . A. B. C. D. E.

Less than 0.06 At least 0.06, but less than 0.07 At least 0.07, but less than 0.08 At least 0.08, but less than 0.09 At least 0.09

20.20. [CAS4-S87:22] (2 points) Which of the following statements are true? (1) (2) (3) A. 1

Px ( 1 − A x ) d Ax Px i A x − Px ( 1 − A x ) A x − Px ( 1 − A x ) v Ax B. 1,2

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C. 1,3

D. 2,3

E. 1,2,3

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20.21. [CAS3-F03:8] Given: • i  6% • 10 E40  0.540 • 1000A40  168 • 1000A50  264 Calculate the annual net premium for a 10-payment annual whole life insurance of 1000 on (40). A. 21.53

B. 21.88

C. 22.19

D. 22.51

E. 22.83

20.22. Premiums are calculated using the equivalence principle. The annual net premium for a fully discrete whole life insurance of 1 on (20) is 0.023, based on p20  0.99. d  0.05. Determine the revised annual net premium if p20 is decreased to 0.98. 20.23. [3-F02:11] For a fully discrete whole life insurance of 1000 on (60), the annual net premium was calculated using the following: • i  0.06 • q60  0.01376 • 1000A60  369.33 • 1000A61  383.00 A particular insured is expected to experience a first-year mortality rate ten times the rate used to calculate the annual net premium. The expected mortality rates for all other years are the ones originally used. Calculate the expected loss for this insured, based on the original net premium. A. 72

B. 86

C. 100

D. 114

E. 128

20.24. [3-F02:26] For a fully discrete whole life insurance of 1000 on (60), you are given • i  0.06 • Mortality follows the Illustrative Life Table, except that there are extra mortality risks at age 60 such that q60  0.015. Calculate the annual net premium for this insurance. A. 31.5

B. 32.0

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C. 32.1

D. 33.1

E. 33.2

Exercises continue on the next page . . .

EXERCISES FOR LESSON 20

397

20.25. [150-F96:14] You are an actuary reviewing a tentative pricing calculation for a fully discrete whole life insurance of 1000 on (60). The tentative net premium seems surprisingly high. You discover that an incorrect value for q 60 was used. You are given: • • • • • •

i  0.06 The incorrect value of q 60  0.10 was used. Based on that incorrect value, 1000A60  405.54. The correct value is q60  0.01. For all other ages, the correct mortality rates were used. L∗ denotes the insurer’s future loss random variable, assuming the tentative net premium is not corrected.

Calculate E[L∗ ]. A. −110

B. −105

C. −100

D. −95

E. −90

Additional old CAS Exam 3/3L questions: F05:37, S08:23, S10:16, S12:12 Additional old SOA Exam MLC questions: F12:22, S13:1, F13:15,16

Solutions 20.1. Note that this is a whole life insurance since there is no survival past 3 years. We will use formula (20.2), although (20.1) could be used as well. A0 

3 X

k−1| q 0

vk

k1 3



1X 1  0.907749 3 1.05k k1

dA0 (0.05/1.05)(0.907749) P0    0.468573 1 − A0 1 − 0.907749

(C)

20.2. From 45,000  3000a¨30 , we get a¨30  15. From 3000 a¨30  2000a¨30 + 187510| a¨30 and a¨30  15, we get 10| a¨30  15,000/1875  8. Then a¨30:10  a¨30 − 10| a¨30  15 − 8  7 and the answer is 5000 (7)  $35,000 . (C) 20.3. This insurance is semicontinuous: the death benefit is payable at the moment of death but the premiums are discrete. The actuarial present value of each insurance is

 µ  1 − e −10 ( µ+δ ) µ+δ !

0.04 So the insurances on age 40 have present value 1000 (1 − e −1 )  252.848, and the insurances 0.04 + 0.06 ! 0.06 on age 50 have present value 1000 (1 − e −1.2 )  349.403. An annuity-due of 1 for two years 0.06 + 0.06 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

20. PREMIUMS: NET PREMIUMS FROM FORMULAS

398

and 0.5 for the third year imitates the premium pattern; the premiums are the first year premium times this pattern. This annuity-due has present value 1 + vp x + 0.5v 2 2 p x which is

1 + e −0.06 e −0.04 + 0.5e −2 (0.06) e −2 (0.04)  2.314203

for lives age 40 and 1 + e −0.06 e −0.06 + 0.5e −2 (0.06) e −2 (0.06)  2.280234 for lives age 50. Therefore, the first year premium is 252.848/2.314203  109.259 for lives age 40 and 349.403/2.280234  153.231 for lives age 50. Adding up, 80 (109.259) + 60 (153.231)  17,935 20.4.

(B)

The continuous 4-year deferred insurance under uniform mortality is computed as ¯

4| A60

 e −0.05 (4)

a¯36 e −0.2 − e −2   0.341698 40 40 (0.05)

We need the 5-year annuity-due of premiums. For uniform mortality, it is easier to compute a 5-year endowment and then use the annuity-to-insurance formula. A 5-year discrete term insurance is, by the usual formula for the sum of a geometric progression, a5 1 − e −0.25  0.107858  40 40 ( e 0.05 − 1)

1 A60 :5 

A 5-year pure endowment is 7e −0.25  0.681451 8  0.107858 + 0.681451  0.78931. Using the annuity-to-insurance

5 p 60

So a discrete 5-year endowment is A60:5 formula,

e −5δ 

1 − A60:5 d d  1 − v  1 − e −0.05  1 − 0.951229  0.048771 1 − 0.78931 a¨60:5   4.3201 0.048771

a¨60:5 

It follows that P

10,0004| A¯ 60 3,416.98  791  a¨60:5 4.3201

(A)

20.5. This can be treated as a sum of a 15-year level term insurance for 1000 and a 10-year level term insurance for 1000. Under uniform mortality, the actuarial present values of these are 1 A30 :n 

n X t1

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1 1 − vn 1 vt  an  ω − 30 70 70i

1 − 1  0.052632 0.95

EXERCISE SOLUTIONS FOR LESSON 20

399

1 − 0.9510  0.108914 70 (0.052632) 1 − 0.9515   0.145678 70 (0.052632)

1 A30 :10  1 A30 :15

So the actuarial present value of the insurance is 1000 (0.108914 + 0.145678)  254.592. To calculate the actuarial present value of the premium annuity, we know that with uniform mortality it is easier to calculate an endowment insurance and then convert it, especially since we’ve already calculated the term insurance! So

!

A30:15  0.145678 + 15 E30 a¨30:15 

55  0.509693  0.145678 + (0.95 ) 70 15

1 − 0.509693  9.806147 0.05

The premium is 254.592/9.806147  25.96 . (D) 20.6. The actuarial present value of one unit of the insurance is 30

1 X 1 30 1.05 k1

!k 

1 − (1/1.05) 30  0.512415 30 (0.05)

Notice that the present value of each premium is π, since the (1 + i ) k factor cancels the v k discounting factor. The actuarial present value of the premiums is therefore (1+e75 ) π0 , which under uniform mortality is   (1 + e˚75 − 0.5) π0  0.5 + ( ω − 75) /2 π0  15.5π0 Setting these equal and multiplying by 1000 units, π0  1000 (0.512415) /15.5  33.05903 . (A) 20.7. For a fully continuous insurance with constant force, P¯ ( A¯ x )  µ. Since q x  0.9/9  0.1, p x  0.9, so e −µ  0.9 and µ  − ln 0.9  0.105361. Therefore, 1000P¯ ( A¯ x )  105.36. For the annual insurance, use formula (20.5). Px  vq x 

0.1 1.08

Therefore 1000Px  92.59. The difference is 105.36 − 92.59  12.77 . (C) 20.8. We need to back out vp x . Equating the APV of the net premiums to the APV of the benefits, and using that v  1 − d  0.94, 668 + 258vp x  v (1000q x + 1000vp x )  940 − 1000vp x + 1000v 2 p x  940 − 60vp x

(258 + 60) vp x  940 − 668 vp x  0.855346 The APV of the benefits, the right side of the third line above, is 940 − 60vp x  940 − 60 (0.855346)  888.68 The APV of the level premiums is 1 + vp x  1.855346. The answer is 888.68/1.855346  478.98 . (B) LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

20. PREMIUMS: NET PREMIUMS FROM FORMULAS

400

20.9. We calculate 1/a¨ x:20 from (i). Let P1 , P2 and P3 be the three premiums in (i), (ii),and (iii), and P the premium we’re calculating. 1 −d a¨ x:20 0.05  0.0349 +  0.0825190 1.05

0.0349  P1  1 a¨ x:20 Now we use (iii):

a¨ x  a¨ x:20 + 20| a¨ x a¨ x  1 + P3 a¨ x:20 1 − Ax  1.2087 d a¨ x:20 0.0825190 − P  1.2087d P  0.0825190 − 0.0575571  0.0249619

(D)

(ii) is not needed. Interestingly, the official solution provided by the SOA used (ii), so the ones who wrote this question thought you needed it! 20.10. The actuarial present value of the death benefit is 1/1.05 regardless of which year it is paid in, since it grows each year at 5% and is then discounted at the same 5%. The actuarial present value of the premiums, per 1 of initial premium, is 1 + 35  36, since the present value of each premium payment is 1, and a premium payment is received for each full year of future lifetime, plus a payment at policy issue. Algebraically, let a¨ be the actuarial present value of the premiums. a¨ 

∞ X

t p 40

v t (1.05) t  1 + e40  1 + 35  36

t0

.



The answer is therefore 1 1.05 (36)  0.026455 . (C) 20.11. Using the equivalence principle, 1000A45  1000P45 a¨45:15 + π 15 E45 a¨60:15 However, by the definition of P45 , it is also true that 1000A45  1000P45 a¨45  1000P45 a¨45:15 + 1000P4515 E45 a¨60 Subtracting one of these from the other, we get π a¨60:15  1000P45 a¨60 So let’s calculate 1000P45 and then π. 1000A45 201.20   14.2573 a¨45 14.1121  a¨60 − 15 E60 a¨75

1000P45  a¨60:15

 11.1454 − 5 E60 10 E65 (7.2170)  11.1454 − (0.68756)(0.39994)(7.2170)  9.16085 (14.2573)(11.1454)  17.346 (B) π 9.16085 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

EXERCISE SOLUTIONS FOR LESSON 20

401

20.12. We’re missing 10 E30 , but can back it out from the annuity values. a¨30:20  a¨30:10 + 10 E30 a¨40:10 15.0364  8.7201 + 10 E30 (8.6602) 15.0364 − 8.7201  0.72935 10 E 30  8.6602 We can now solve for π.





π 8.7201 + 2 (15.0364 − 8.7201)  16.66 + 2 (0.72935)(32.61) 21.3527π  64.2281 64.2281 π  3.008 21.3527

(B)

20.13. The minimum value of future loss is if death occurs after 20 years. Then the insurance company receives 10 payments of 1600 without paying anything, so the loss is 1 − (1/1.05) 10  −1600  −1600 (8.10782)  −12,973 0.05/1.05

!

−1600 a¨10

(C)

20.14. Let P be the annual net premium. The equation for P is



1 P a¨30:10  1000A30 + P ( IA )30 :10 + 10 · 10| A30



Solving for P immediately results in (E). 20.15. The most promising formula for premium is P  1/a¨ x − d, since the d’s will cancel out when we take the difference of the two premiums. Then P B  1/a¨ x − d, and a¨ xA 

∞ X

A k px

vk

B k px



k0



∞ X

(1 + c ) v

k

k0



(1 + c ) v

k

1  (1 + i ) / (1 + c )

!k

1  1 + ( i − c ) / (1 + c )

!k

So a¨ xA is an a¨ x on table B with interest rate ( i − c ) / (1 + c ) . Answer (D) follows. 20.16. We’ll use International Actuarial Notation for the premiums. Use the insurance recursion formula A25  A25:20 − 20 E25 + 20 E25 A45 Divide through by a¨25:20 to obtain premiums. 20 P 25

1  P25:20 − P25:20 (1 − A45 )

1 0.046  0.064 − P25:20 (1 − 0.640) 1 P25:20  0.05 1 P25 :20  0.064 − 0.05  0.014

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(B)

20. PREMIUMS: NET PREMIUMS FROM FORMULAS

402

20.17. Use International Actuarial Notation for the premiums. As in the previous question, use the insurance recursion formula. 1 A30  A30 :15 + 15 E 30 A45

Divide both sides by a¨30:15 . 15 P 30

1 1  P30 :15 + ( P30:15 − P30:15 ) A45

0.030  0.006 + (0.046 − 0.006) A45 A45  0.600

(B)

20.18. Let P be the premium of the question. P

dA x Ax  a¨ x:n 1 − A x:n

(B)

20.19. The premium we seek is Px1:n  A x1:n /a¨ x:n . The numerator is A x1:n  A x − n| A x  0.6 − 0.4  0.2. We need to calculate a¨ x:n , which is the denominator. A x+n  a¨ x+n 0.4 0.2  n| a¨ x

Px+n 

n| A x /n E x n| a¨ x /n E x



n| A x n| a¨ x

2 A x 0.6  6 a¨ x  Px 0.1 a¨ x  a¨ x:n + n| a¨ x 6  a¨ x:n + 2

n| a¨ x

a¨ x:n  4 A x1:n 0.2 Px1:n    0.05 a¨ x:n 4 20.20. Use Px  dA x / (1 − A x ) to expand P in all of these. (1) follows immediately. For (2): Px dA x  A x − Px (1 − A x ) (1 − A x )( A x − dA x ) d i   (1 − A x ) v 1 − A x So (2) is false. For (3): A x − Px (1 − A x ) A x − dA x  v Ax Ax so (3) is true. (C) 20.21. The premium P  1000A40 /a¨40:10 . We need a¨40:10 . We get this from A40:10 A40:10  A40 − 10 E40 A50 + 10 E 40  0.168 − 0.54 (0.264) + 0.54  0.56544 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

EXERCISE SOLUTIONS FOR LESSON 20

403

1 − A40:10 d (1 − 0.56544)(1.06)   7.67723 0.06 168 P  21.8829 (B) 7.67723

a¨40:10 

20.22. Let P be the original premium. Since P  1/a¨20 − d, 1 − 0.05 a¨20 1 0.073  a¨20 1 a¨20   13.69863 0.073 0.023 

Using primes for the revised mortality rate, a¨20  1 + vp20 a¨21

!

0 a¨20

13.69863 − 1  13.57036  1 + v (0.98) 0.99v

The answer is P0 

1 − 0.05  0.02369 13.57036

20.23. Let’s use primes for the special functions for this particular insured. We use the recursive formula, equation (12.1) on page 237, to get the revised A060 . 0 0 A61 + vp60 A060  vq 60

!

0.1376 0.8624  0.383  0.44141 + 1.06 1.06 The revised annuity (which we’ll need for the premiums) is 0 a¨60 

1 − A060 d



1.06 (1 − 0.44141)  9.86835 0.06

The premium we are charging is based on the regular (unprimed) functions: P

1000dA60 0.06 (369.33)   33.14804 1 − A60 1.06 (1 − 0.36933)

The expected value of the loss is therefore 0 E[L]  1000A060 − P a¨60  441.41 − 33.14804 (9.86835)  114.298

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(D)

20. PREMIUMS: NET PREMIUMS FROM FORMULAS

404

20.24. The fastest way to do this is to use annuities rather than insurances (although using insurances doesn’t take too long either). We’ll use primes for the special actuarial functions with this mortality. The annuity recursive formula is equation (17.1) on page 349:

!

0 a¨60

 1 + vp 60 a¨61

0.985 (10.9041)  11.1326 1+ 1.06

The premium is then

!

1 1 0.06 1000 0 − d  1000 −  33.22 a¨60 11.1326 1.06





(E)

If you prefer to use insurances, you would use the recursive formula for insurances, equation (12.1). Then 0.015 0.985 (0.38279) A060  vq 60 + vp60 A61  +  0.36986 1.06 1.06 The premium is then ! ! dA060 0.06 (0.36986)  1000  33.22 1000 1 − A060 1.06 (1 − 0.36986) 20.25. We use the recursive formula for A x , (12.1) on page 237, to get the corrected value which we’ll call c . A60 0.1 + 0.9A61 1.06 0.01 + 0.99A61  1.06
0.01 + 0.99 0.9 (1.06 (0.40554) − 0.1)  0.35175  1.06

A60  c A60

The premium we’re charging is P

1000dA60 (0.06)(405.54)   38.615. 1 − A60 1.06 (1 − 0.40554)

Therefore E[L ]  ∗

c 1000A60

−P

c 1 − A60

!

d

 1000 (0.35175) − 38.615

(1 − 0.35175)(1.06) 0.06

!  −90.478

Quiz Solutions 20-1.

Using the usual formulas for the net single premiums under constant force, A¯ x 

µ 0.01   0.2 µ + δ 0.05

1+i e 0.04   20.50417 −0.01 q+i 1−e + e 0.04 − 1 0.2 P  0.009754 20.50417

a¨ x 

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(E)

QUIZ SOLUTIONS FOR LESSON 20

405

20-2. Clearly the premium given in this question is not the net premium. The future loss is

!

1000A45 − 24 a¨45  1000A45 − 24

1 − A45 24 24  1000 + A45 −  1600A45 − 600 d 0.04 0.04





Setting this equal to −20, A45  580/1600  0.3625. Using insurance recursion, equation (12.1), A45  vq 45 + vp 45 A46 0.3625  0.96 (0.01 + 0.99A46 ) 0.3625/0.96 − 0.01  0.371317 A46  0.99 Setting the future loss equal to −20 for age 46 and solving for the premium P,

!

1 − 0.371317  −20 0.04 20 + 371.317 391.317 P   24.90 (1 − 0.371317) /0.04 15.71707 1000 (0.371317) − P

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406

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20. PREMIUMS: NET PREMIUMS FROM FORMULAS

Lesson 21

Premiums: Variance of Future Loss, Continuous Reading: Models for Quantifying Risk (4th or 5th edition) 9.4.2 Calculating the variance of 0 L, the future loss at issue random variable, is pretty much the same for continuous and discrete insurances, but I broke it up into two lessons because of the large number of old exam questions on the topic. If P is the continuous premium charged for continuous whole life insurance on ( x ) , not necessarily the premium determined from the equivalence principle, the 0 L random variable is P P 1 − v Tx  v Tx 1 + − −P δ δ δ

!

0L

v

Tx





(18.5)

as mentioned on page 360. Since P/δ is constant,



Var (0 L )  Var v Tx 





A¯ x − A¯ 2x

2

1+



P δ

1+

2 P δ

2 (21.1)

This is easier to calculate than the second moment of 0 L; if you were asked for the second moment, your best course of action would be to calculate the variance and add the square of the mean. The above formula can be generalized to a continuous whole life insurance of face amount b (instead of 1) on ( x ) with continuous premiums. If P is the premium for b units of insurance, then P P 1 − v Tx −P  v Tx b + − δ δ δ

!

0L

 bv

Tx





so Var (0 L ) 



A¯ x − A¯ 2x

2



b+

P δ

2 (21.2)

The reason we got such nice formulas for variance of future loss is because the function of the random variable in the insurance, v Tx , was the same as the function of the random variable in the annuity. It’s easy to find the variance of a linear function of a single random variable. There are also nice formulas for the variance of endowment insurances, where you’d have v min (Tx ,n ) in both spots instead of v Tx , and equations (21.1) and (21.2) become Var (0 L ) 



2

Var (0 L ) 



2

A¯ x:n − ( A¯ x:n

respectively. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

P 2 δ 2  P )2 b + δ

A¯ x:n − ( A¯ x:n ) 2

407





1+

(21.3) (21.4)

21. PREMIUMS: VARIANCE OF FUTURE LOSS, CONTINUOUS

408

On the other hand, calculating the variance of a term insurance or an n-pay life would involve two random variables, one for the insurance and one for the annuity, and would involve calculating covariance as well. As an alternative, you could calculate  the second moment, which would be messy (you’d have to  square the term corresponding to v Tx 1 + Pδ and work out the sum). Let’s work with equation (21.3) with the understanding that by dropping the n ’s the equation applies to whole life as well. If P is the net premium determined by the equivalence principle, then we can simplify equation (21.3) further, since P  δ A¯ x:n / (1 − A¯ x:n ) : Var (0 L )  



A¯ x:n − ( A¯ x:n ) 2

2

2A ¯



δ A¯ x:n 1+ δ (1 − A¯ x:n )

!2

− ( A¯ x:n ) 2 (1 − A¯ x:n ) 2 x:n

(21.5)

When the equivalence principle net premium is used, the variance of the future loss equals the second moment of the future loss, since E[0 L]  0. This allows computation of the second moment of the loss of issue from equation (21.5). Example 21A The force of mortality is µ and the force of interest is δ. They are both constant. For a fully continuous whole life insurance with premiums calculated using the equivalence principle, calculate Var (0 L ) . Answer: Using equation (21.5),

Var (0 L ) 



µ µ+2δ

−

µ
2 µ+δ

δ
2 µ+δ µ ( µ+δ ) 2 µ+2δ

− µ2

δ2 µ ( µ + δ ) 2 − µ2 ( µ + 2δ )  δ2 ( µ + 2δ ) µ  µ + 2δ

(21.6)

So for constant force of mortality only, the variance of the future loss on whole life equals the second moment of the insurance. This may be a useful shortcut for exam questions.  Example 21B For a fully continuous 20-year endowment • µ  0.02 • δ  0.05 • The annual premium is set to make the expected future loss equal to −0.1. Calculate the variance of the future loss. Answer: We have

 2 1 − e −1.4 + e −1.4  0.461855 A¯ x:20  7 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

21. PREMIUMS: VARIANCE OF FUTURE LOSS, CONTINUOUS

409

and we want a premium P such that A¯ x:20 − P a¯ x:20  −0.1 + 0.1 A¯ P  x:20 a¯  x:20  δ A¯ x:20 + 0.1  1 − A¯ x:20 0.05 (0.461855 + 0.1)  0.052203  1 − 0.461855 The variance of the loss is then Var (0 L ) 



A¯ x:20 − A¯ x:20

2

2



1+

P δ

2

 2  1 − e −2.4 + e −2.4  0.242265 A¯ x:20  12   0.052203 2 Var (0 L )  (0.242265 − 0.4618552 ) 1 + 0.05 2  (0.028955)(2.044059 )  0.120979 2

?



Quiz 21-1 For a fully continuous whole life insurance of 1 on (70), you are given: • Mortality is uniformly distributed with ω  110. • δ  0.05 • The premium is set so that the expected future loss at issue is 0. Calculate the variance of the future loss. An exam question can have you compare the variances for two different premiums. If we are dealing with the same insurance, then the first factor 2A¯ x − A¯ 2x is the same, so it’s a matter of comparing the second factor (1 + P/δ ) 2 . Example 21C For a fully continuous whole life insurance with face amount 1000, you are given: • If the annual premium is 60, the variance of the future loss is 375,000. • δ  0.04 Calculate the variance of the future loss if the annual premium is 80. Answer: With either premium, the variance of the future loss is 10002 (2A¯ x − A¯ 2x ) times (1 + P/δ ) 2 , where P is expressed per unit of insurance. So the relative variance between premiums of P 0 and P is 1 + P 0/δ 1 + P/δ

!2 

which in our case is 0.04 + 0.08 0.04 + 0.06 and 1.44 (375,000)  540,000 .

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δ + P0 δ+P

!2

!2  1.44



21. PREMIUMS: VARIANCE OF FUTURE LOSS, CONTINUOUS

410

Table 21.1: Formula summary for variance of future loss—continuous

The following equations are for fully continuous whole life and endowment insurances of 1. For whole life, drop n . Var (0 L ) 



A¯ x:n − ( A¯ x:n ) 2

2

P δ

1+

− ( A¯ x:n ) 2 (1 − A¯ x:n ) 2 µ Var (0 L )  µ + 2δ Var (0 L ) 

2A ¯



x:n

2 (21.3) if equivalence principle premium is used

(21.5)

for whole life with equivalence principle and constant force of mortality only

(21.6)

For whole life and endowment insurance with face amount b: Var (0 L ) 



2

Var (0 L ) 



2

A¯ x − A¯ 2x



b+

P δ

A¯ x:n − ( A¯ x:n ) 2

2



(21.2) b+

P δ

2 (21.4)

For two whole life or endowment insurances, one with b 0 units and total premium P 0 and the other with b units and total premium P, the relative variance of future loss of the first to the second is ( b 0 δ +

2

P 0 ) / ( bδ + P ) .

?

Quiz 21-2 For a fully continuous 20-year endowment insurance of 1, you are given: • If the annual premium is 0.06, then the variance of the future loss is 0.578. • If the annual premium is 0.07, then the variance of the future loss is 0.722. Calculate the variance of the future loss if the annual premium is 0.08.

Exercises 21.1.

You are given: • The force of mortality, µ, and the force of interest, δ, are constant. • A¯ x  0.2. • Premiums are determined according to the equivalence principle. • 0 L is the present value of the loss at issue on a fully continuous whole life insurance on ( x ) .

Calculate Var (0 L ) .

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Exercises continue on the next page . . .

EXERCISES FOR LESSON 21

411

[150-81-94:10] For a fully continuous whole life insurance of 1 on ( x ) , you are given:

21.2.

• µ x+t  0.04, t ≥ 0 • A¯ x  0.4 • Premiums are determined according to the equivalence principle. Calculate the variance of the loss L. A. 0.09

B. 0.16

C. 0.25

D. 0.30

E. 0.36

[3-F01:34] For a fully continuous whole life insurance of 1:

21.3.

µ  0.04, x > 0 δ  0.08 L is the loss-at-issue random variable based on the net premium.

• • •

Calculate Var ( L ) . A. 1/10 21.4.

B. 1/5

C. 1/4

D. 1/3

E. 1/2

[SOA3-F04:3] For a fully continuous life insurance of 1 on ( x ) , you are given: • The forces of mortality and interest are constant. • 2A¯ x  0.20 • The net premium is 0.03. • 0 L is the loss-at-issue random variable based on the net premium.

Calculate Var (0 L ) . A. 0.20 21.5.

B. 0.21

C. 0.22

D. 0.23

E. 0.24

[150-81-94:22] For a fully continuous whole life insurance of 1 on ( x ) , you are given: • µ x+t  0.06, t ≥ 0 • δ  0.09 • Premiums are determined by the equivalence principle.

Which of the following are true? I.

The fully continuous net annual premium for this insurance is equal to 0.06.

II.

The variance of the loss at issue, L, is equal to 0.25.

III.

The variance of the loss under this insurance is smaller than the variance of the loss under a single premium whole life insurance on ( x ) with benefit of 1 payable at the moment of death.

A. I and II only B. I and III only C. II and III only E. The correct answer is not given by A. , B. , C. , or D. 21.6.

D. I, II and III

For a fully continuous 20-year endowment insurance of 1 on ( x ) , you are given: • 2A¯ x:20  0.35 • The net premium is 0.055. • δ  0.045 • 0 L is the loss-at-issue random variable based on the net premium.

Calculate Var (0 L ) . LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

Exercises continue on the next page . . .

21. PREMIUMS: VARIANCE OF FUTURE LOSS, CONTINUOUS

412

21.7. [CAS4A-S96:3] (2 points) L is the net future loss random variable for a fully continuous whole life insurance of 2 on ( x ) . This insurance has a total annual premium rate of 0.14. You are given: •

µ x+t  0.03 for t ≥ 0

•

δ  0.05 Determine Var ( L ) .

A. B. C. D. E.

Less than 1.5 At least 1.5, but less than 2.5 At least 2.5, but less than 3.5 At least 3.5, but less than 4.5 At least 4.5

21.8. [CAS4A-F96:13] (2 points) L is the net future loss random variable for a fully continuous whole life insurance of 2 on ( x ) . This insurance has a total annual premium rate of 0.09. You are given: µ x+t  0.04 for t ≥ 0 δ  0.06

• •

Calculate Var ( L ) . A. B. C. D. E.

Less than 1.0 At least 1.0, but less than 1.2 At least 1.2, but less than 1.4 At least 1.4, but less than 1.6 At least 1.6 [150-F97:23] For a fully continuous whole life insurance of 1 on (50), you are given:

21.9. •

The net future loss random variable L is defined as follows: L  v Tx − ( P + c ) a¯Tx ,

Tx ≥ 0,

c is a constant

where P is the net continuous premium for the whole life insurance. • µ50+t  0.06, t ≥ 0 • δ  0.10 • E[L]  −0.1875 Calculate E[L2 ] A. 0.21

B. 0.26

C. 0.33

D. 0.36

E. 0.40

21.10. For a fully continuous whole life insurance on (45) of 1, you are given: • Premiums are calculated according to the equivalence principle. • Mortality is uniformly distributed with ω  110. • δ  0.04. Calculate Var ( L ) .

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Exercises continue on the next page . . .

EXERCISES FOR LESSON 21

413

21.11. [150-81-94:21] The random variable 0 L is the future loss at issue for a fully continuous whole life insurance of 1 on (40). You are given: • δ  0.02 • Mortality follows l x  100 − x, 0 ≤ x ≤ 100. • 2A¯ 40  0.379 • Premiums are determined by the equivalence principle. Calculate Var (0 L ) . A. 0.040

B. 0.095

C. 0.150

D. 0.184

E. 0.228

21.12. For a fully continuous whole life insurance of 1 on ( x ) , you are given • If premiums are determined according to the equivalence principle, Var (0 L )  0.1073. • 2A¯ x  0.1. • For the actual premium P, E[0 L]  −0.1 Calculate P/δ. 21.13. [4-F86:26] You are given the following values calculated at δ  0.08 for two fully continuous whole life policies issued to ( x ) :

Policy #1 Policy #2

Death Benefit

Premium

4 6

0.18 0.22

Variance of Loss 3.25

Calculate the variance of the net future loss for policy #2. A. 4.33

B. 5.62

C. 6.37

D. 6.83

E. 9.74

21.14. [150-S87:13] L is the loss-at-issue random variable for a fully continuous whole life insurance of 1 issued to ( x ) . You are given: • The premium has been determined by the equivalence principle. • Var ( v Tx )  0.0344 • E[v Tx ]  0.166 Calculate Var ( L ) . A. 0.0239

B. 0.0495

C. 0.4896

D. 0.8020

E. 1.2470

21.15. [150-S87:18] An insurance company decides to waive all future premiums on a fully continuous whole life insurance of 1000 issued to ( x ) . The variance of the loss random variable after the change is 81% of the variance of the net future loss random variable before the change. The force of interest is 0.03. Calculate the net premium for the insurance. A. 3.00

B. 3.33

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C. 3.67

D. 3.90

E. 4.20 Exercises continue on the next page . . .

21. PREMIUMS: VARIANCE OF FUTURE LOSS, CONTINUOUS

414

21.16. [150-F88:2] L is the loss-at-issue random variable for a fully continuous whole life insurance of 1 with premiums based on the equivalence principle. You are given: • E[v 2Tx ]  0.34 • E[v Tx ]  0.40 Calculate Var ( L ) . A. 0.080

B. 0.300

C. 0.475

D. 0.500

E. 1.125

21.17. [150-F96:8] For a fully continuous whole life insurance of 1 on ( x ) , you are given; • Z is the present-value random variable at issue of the death benefit. • 0 L is the net future loss-at-issue random variable. • Premiums are determined using the equivalence principle. • Var ( Z ) / Var (0 L )  0.36 • a¯ x  10 Calculate the net premium. A. 0.03

B. 0.04

C. 0.05

D. 0.06

E. 0.07

21.18. [3-F00:22 and 150-S89:3] For a continuous whole life insurance of 1 on ( x ) : • π is the net premium. • L is the loss-at-issue random variable with the premium equal to π. • L∗ is the loss-at-issue random variable with the premium equal to 1.25π. • a¯ x  5.0 • δ  0.08 • Var ( L )  0.5625 Calculate the sum of the expected value and the standard deviation of L∗ . A. 0.59 21.19.

B. 0.71

C. 0.86

D. 0.89

E. 1.01

[M-F06:28] For a fully continuous whole life insurance of 1 on ( x ) :

• A¯ x  1/3 • δ  0.10 • L is the loss at issue random variable using the premium based on the equivalence principle. • Var ( L )  1/5. • L0 is the loss at issue random variable using the premium π. • Var ( L0 )  16/45. Calculate π. A. 0.05

B. 0.08

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C. 0.10

D. 0.12

E. 0.15

Exercises continue on the next page . . .

EXERCISES FOR LESSON 21

415

21.20. For a fully continuous whole life insurance of 1 on ( x ) , you are given: • • • •

A¯ x  0.3 2A ¯ x  0.2 δ  0.06 0 L is the net future loss random variable.

Determine the annual premium that minimizes E[0 L2 ]. 21.21. For a fully continuous whole life insurance of 1000, you are given: • A¯ x  0.515 • 2A¯ x  0.305 • δ  0.05 • Premiums are payable continuously at a rate of 60 per year. Let α be the probability that the future loss on a portfolio of these insurances is greater than zero. Using the normal approximation, calculate the minimum number of policies that must be sold so that α ≤ 0.05. Additional old CAS Exam 3/3L questions: S05:3

Solutions 21.1. 0.2  A¯ x 

µ ⇒ δ  4µ µ+δ

Using the special shortcut for exponential whole life, Var (0 L )  21.2. 0.4  A¯ x 

µ µ 1   µ + 2δ 9µ 9 µ ⇒ δ  1.5µ µ+δ

Using the special shortcut for exponential whole life, Var (0 L )  21.3.

(C)

Using the special shortcut for exponential whole life, Var ( L ) 

21.4.

µ µ   0.25 µ + 2δ 4µ

µ 0.04 1   µ + 2δ 0.04 + 2 (0.08) 5

(B)

From the special shortcut for exponential whole life, we know that Var (0 L )  2A¯ x  0.2 . (A)

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21. PREMIUMS: VARIANCE OF FUTURE LOSS, CONTINUOUS

416

21.5. I. II. III.

21.6.

(A) True, since for constant force of mortality the premium is µ. ! True, since the variance of the future loss at issue is

µ µ+2δ



 0.25. !

0.06 0.06+2 (0.09)

False, since the variance of a single premium whole life insurance would be the first factor only of equation (21.1) The annual premium adds variance since the amount of premium collected is no longer a constant. # To use formula (21.5), we must back out the net single premium. P

δ A¯ x:20 1 − A¯

x:20

0.045A¯ x:20 0.055  1 − A¯ x:20

0.055 − 0.055A¯ x:20  0.045A¯ x:20 A¯ x:20  0.55 Now we can calculate the variance of the future loss. Var (0 L ) 

2A ¯

− ( A¯ x:20 ) 2 0.35 − 0.552   0.2346 (1 − 0.55) 2 (1 − A¯ x:20 ) 2

x:20

21.7. Since we are not working with the equivalence principle premium, we cannot use a shortcut. We use equation (21.2). Var ( L ) 



A¯ x − A¯ 2x

2



2+

0.14 0.05

2

!2 * 3 − 3 +/ (4.82 )  2.0769 13 8 , -

.

21.8.

(B)

A variation of the previous exercise, and again we use equation (21.2).



2

0.09 2  3.52 0.06   Var ( L )  3.52 2A¯ x − A¯ 2x

b+

P δ



 2+



!2 4 4 + * /  1.1025  12.25 . − 16 10 , -

(B)

21.9. Based on (i), the gross premium used in calculating the loss L is P + c. The second moment of L is the first moment squared plus the variance: E[L 2 ]  E[L]2 + Var ( L ) and we’re given E[L]  −0.1875, so we only need to calculate the variance: P+c Var ( L )  (2A¯ 50 − A¯ 250 ) 1 + δ



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2

EXERCISE SOLUTIONS FOR LESSON 21

417

Consider the expected value of L, which is E[L]  A¯ 50 − ( P + c ) a¯50

(*)

By the usual formulas for constant force of mortality, A¯ 50  and a¯50 

1 0.06+0.10



1 0.16 .

µ 0.06 3   µ + δ 0.06 + 0.10 8

Plugging into (*), 3 P+c − 8 0.16 P + c  0.09

−0.1875 

Then Var ( L )  *

!2 

3 + 0.09 0.06 − 1+ 0.06 + 2 (0.10) 8 0.10

2

 0.325421

-

,

E[L2 ]  0.325421 + 0.18752  0.3606

(D)

21.10. We’ll use the formula (21.5). a¯ 1 − e −2.6 A¯ 45  65   0.356049 65 2.6 a¯ 1 − e −5.2 2¯ A45  65 0.08   0.191247 65 5.2 0.191247 − 0.3560492  0.155487 Var ( L )  (1 − 0.356049) 2 21.11. We’ll use formula (21.5). a¯ 1 − e −1.2 A¯ 40  60   0.582338 60 1.2 0.379 − 0.5823382 Var (0 L )   0.228629 (1 − 0.582338) 2

(E)

Although the answer rounds to 0.229 rather than to 0.228, if a less rounded 2A¯ 40 were used (like 0.378868), the answer would round to 0.228. 21.12. Let y  A¯ x . Using (i) to solve for A¯ x , 0.1073  Var (0 L ) 

2A ¯

− y2 (1 − y ) 2 x

0.1 − y 2  0.1073 − 0.2146y + 0.1073y 2 1.1073y 2 − 0.2146y + 0.0073  0

p

0.2146 ± 0.21462 − 4 (1.1073)(0.0073)  0.150, 0.044 y 2 (1.1073)

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21. PREMIUMS: VARIANCE OF FUTURE LOSS, CONTINUOUS

418

But A¯ x can only be 0.150, since it must be greater than 2A¯ x  0.1.



−0.1  E[0 L]  (0.150) 1 + −0.1  0.150 − 0.850

P P − δ δ



P δ

5 P  δ 17 21.13. As we discussed, the relative variance is the ratio of ( P + δ ) 2 for each unit, where P is the premium per unit. The premium per unit for policy #1 is 0.18/4  0.045 and the premium per unit for policy #2 is 0.22/6. The variance for y units is y 2 times the variance for one unit, so we’ll also have to multiply by (6/4) 2 . Putting these factors together, the answer is 6 4

!2

0.22/6 + 0.08 0.045 + 0.08

!2  1.96 (3.25)  6.37

(C)

If you prefer, you could derive a relationship from equation (21.2). The ratio of variances is b 0 + P 0/δ S + P/δ

!2

b0 δ + P0  bδ + P

!2

where P is the total premium (not the premium per unit). So in this question, the appropriate ratio would be !2 !2 6 (0.08) + 0.22 0.7   1.96 4 (0.08) + 0.18 0.5 leading to the same answer. 21.14. We use the formula Var ( L ) 

0.0344 Var ( v T )   0.0494568 2 (1 − 0.166) 2 (1 − A¯ x )

(B)

21.15. For a whole life of 1000 without future premiums, Var (0 L )  10002 Var ( v T ) , while for a whole life of 1000 with future premiums, Var (0 L )  10002 Var ( v T )(1 + P/δ ) 2 . Dividing the first expression into the second, P 2 1  δ 0.81 P 1 1+  0.03 0.9 1 P¯ ( A¯ x )  300 10 1000P¯ ( A¯ x )   3.33 3





1+

(B)

21.16. Var ( L )  

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2A ¯

− A¯ 2x (1 − A¯ x ) 2 x

0.34 − 0.402  0.5 0.602

(D)

EXERCISE SOLUTIONS FOR LESSON 21

419

21.17. By equation (21.5), Var (0 L )  Var ( Z ) / (1 − A¯ x ) 2 . Then Var ( Z )  (1 − A¯ x ) 2  0.36 Var ( L ) 1 − A¯ x  0.6 A¯ x  0.4 A¯ x 0.4  0.04  P a¯ x 10

(B)

21.18. A¯ x  1 − 0.08 (5)  0.6 0.6 π  0.12 5 E[L∗ ]  0.6 − 1.25 (0.12)(5)  −0.15 0.12 2 0.08    0.15 2 Var ( L∗ )  Var v Tx 1 + 0.08 2.8752  (0.5625) 2.52 p 2.875 Var ( L∗ )  (0.75)  0.8625 2.5



0.5625  Var ( L )  Var v Tx





1+

The answer is −0.15 + 0.8625  0.7125 . (B) 21.19. The equivalence-principle premium P by formula (18.2) is P

0.1 (1/3) δ A¯ x   0.05 2/3 1 − A¯ x

By formula (21.1),

π 2 δ so the ratio of Var ( L0 ) to Var ( L ) , which is given as (16/45) / (1/5)  16/9, is



Var ( L )  (2A¯ x − A2x ) 1 +

1 + π/0.1 1 + 0.05/0.1

!2 



16 9

Solving for π, 0.1 + π 4  0.1 + 0.05 3 4 π  (0.15) − 0.1  0.1 3

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(C)

21. PREMIUMS: VARIANCE OF FUTURE LOSS, CONTINUOUS

420

21.20. Let P be the annual premium and y  P/δ. We want to minimize



Var (0 L ) + E[0 L]2  (2A¯ x − A¯ 2x )(1 + y ) 2 + A¯ x (1 + y ) − y

2

 (0.2 − 0.32 )(1 + 2y + y 2 ) + (0.3 − 0.7y ) 2  0.11 + 0.22y + 0.11y 2 + 0.09 − 0.42y + 0.49y 2  0.6y 2 − 0.2y + 0.2 The minimum of a quadratic ax 2 + bx + c is assumed at −b/2a, which here is 0.2/1.2  1/6. So P/δ  1/6 and P  0.01 . 21.21. Let P be the premium E[L]  1000A¯ x − 60

1 − A¯ x δ

!

!

0.485  515 − 60  −67 0.05 Var ( L ) 



A¯ x − A¯ 2x

2



1000 +



 0.305 − 0.5152



60 δ

2 

22002  192,511

We need n such that

r 1.645

192,511  67 n 192,511 67  n 1.645 n

!2

(192,511)(1.6452 ) 672

 116.05

Hence 117 policies are needed.

Quiz Solutions 21-1.

We’ll use formula (21.5). a¯40 1 − e −2   0.432332 40 2 a¯ 1 − e −4 2¯ A70  40 0.10   0.245421 40 4 2A ¯ 70 − A¯ 2 70 Var (0 L )  ¯ (1 − A70 ) 2 A¯ 70 



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0.245421 − 0.4323322  0.1816 (1 − 0.432332) 2

QUIZ SOLUTIONS FOR LESSON 21

21-2.

421

The relative variance of future loss for premiums of 0.07 to 0.06 is

!2

0.07 + δ 0.722  1.249135  0.06 + δ 0.578 0.07 + δ √  1.249135  1.117647 0.06 + δ 0.07 + δ  1.117647 (0.06) + 1.117647δ 0.07 − 1.117647 (0.06) δ  0.025 0.117647 The variance of the future loss if the continuous premium is 0.08 is 0.722 times the relative variance to 0.07: !2 0.08 + 0.025 0.722  0.882 0.07 + 0.025

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422

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21. PREMIUMS: VARIANCE OF FUTURE LOSS, CONTINUOUS

Lesson 22

Premiums: Variance of Future Loss, Discrete Reading: Models for Quantifying Risk (4th or 5th edition) 9.2 The formulas for the variance of loss at issue for fully discrete whole life insurance and endowment insurance are analogs of the formulas for fully continuous insurances, equations (21.1) and (21.5). For whole life insurance of 1: Var (0 L )  



2

A x − A2x

2A

P 2 d if the equivalence principle premium is used



− A2x (1 − A x ) 2 x



1+

(22.1) (22.2)

For endowment insurance of 1: Var (0 L )  



A x:n − ( A x:n ) 2

2

2A

P 2 d if the equivalence principle premium is used



− ( A x:n ) 2 (1 − A x:n ) 2 x:n



1+

(22.3) (22.4)

Parallel to Example 21A, let’s develop a formula for the variance of net future loss for whole life insurance of 1 if q x is constant. Refer to Table 8.2 for the formula for A x . Var (0 L ) 

2A

− A2x (1 − A x ) 2 x





q/ ( q + 2 i ) − q/ ( q + i )



i/ ( q + i )

2

2

Multiply numerator and denominator by ( q + i ) 2 , and use 2 i  2i + i 2 . Var (0 L ) 

q ( q + i ) 2 − q 2 ( q + 2i + i 2 ) i 2 ( q + 2i + i 2 )

q 3 + 2q 2 i + qi 2 − q 3 − 2iq 2 − q 2 i 2 i 2 ( q + 2i + i 2 ) q (1 − q )  q + 2i 

(22.5)

In addition to exam questions on variance of net future loss for whole life insurances and endowment insurances, there can be questions involving insurances with periods so short that you can calculate the variance directly. In addition, questions with normal approximations of percentiles are possible. (They’re also possible with continuous insurances, but for some reason are more common with discrete ones.) Here’s an example of a combination of these two ideas: LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

423

22. PREMIUMS: VARIANCE OF FUTURE LOSS, DISCRETE

424

Example 22A For a 3-year term insurance policy with face amount 10,000, you are given: • t q x+t−1 1 2 3

0.10 0.15 0.20

• i  0.05 • The premium is 1500 Using the normal approximation, calculate the minimum number of policies that must be sold so that the probability that the net future loss is positive is no more than 5%. Answer: The present value of the net future loss is Year of death 1 2 3 >3

Present value of loss 10,000 − 1500  8023.8095 1.05   1 10,000 − 1500 1 +  6141.7234 1.05 1.052   1 10,000 1 − 1500 1 +  4349.2603 + 1.05 1.052 1.053   1 1 −1500 1 +  −4289.1156 + 1.05 1.052

We calculate the first and second moments of 0 L. 2

Year of death

Probability

0L

1 2 3 >3 Probabilityweighted total

0.100 0.135 0.153 0.612

8023.8095 6141.7234 4349.2603 −4289.1156

64,381,519 37,720,766 18,916,066 18,396,513

−327.99

25,683,279

0L

In the above table, the totals are weighted by probability. So the variance of the net future loss is 25,683,279 − 327.992  25,575,703. To have a 95% chance of being non-positive, if n is the number of policies, we need p −327.99n + 1.645 25,575,703n  0 √ Dividing out n, this means √ √ 1.645 25,575,703 n  25.364 327.99 n  25.3642  643.336 So the number of policies that must be sold is 644 .

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EXERCISES FOR LESSON 22

425

Table 22.1: Formula summary for variance of future loss—discrete

The following equations are for fully discrete whole life and endowment insurances of 1. For whole life, drop n . Var (0 L ) 



2

A x:n − ( A x:n ) 2

1+

P d

2

− ( A x:n ) 2 (1 − A x:n ) 2 q (1 − q ) Var (0 L )  q + 2i Var (0 L ) 

2A



x:n

(22.3) if equivalence principle premium is used

(22.4)

For whole life with equivalence principle and constant rate of mortality only

(22.5)

In (22.3), if the face amount is b instead of 1, then Var (0 L ) 



2

A x:n − ( A x:n ) 2



b+

P d

2

For two whole life or endowment insurances, one with b 0 units and total premium P 0 and the other with b units and total premium P, the relative variance of net future loss of the first to the second is



?

2 ( b 0 d + P 0 ) / ( bd + P ) .

Quiz 22-1 A 2-year pure endowment on ( x ) has benefit 1000. You are given: • k

k−1| q x

1 2

0.1 0.2

• d  0.05 • annual net premiums determined by the equivalence principle are paid at the beginning of each year. Calculate the variance of the net future loss.

Exercises 22.1.

For a fully discrete whole life insurance on (50), you are given: • Mortality follows the Illustrative Life Table. • i  0.06.

Premiums are calculated with the equivalence principle. Calculate Var (0 L ) .

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Exercises continue on the next page . . .

22. PREMIUMS: VARIANCE OF FUTURE LOSS, DISCRETE

426

For a fully discrete 3-year term insurance with a benefit of 1, you are given:

22.2. •

t

q x+t−1

1 2 3

0.04 0.06 0.10

• The annual premium is 0.07. • i  0.06. Calculate the variance of the net future loss at issue. [MLC-S07:4] For a fully discrete whole life insurance of 150,000 on ( x ) , you are given:

22.3.

• 2A x  0.0143 • A x  0.0653 • The annual premium is determined using the equivalence principle. • L is the future loss random variable. Calculate the standard deviation of L. A. 14,000

B. 14,500

C. 15,100

D. 15,600

E. 16,100

[150-F87:17] L is the loss random variable for a fully discrete 2-year term insurance of 1 issued to

22.4. (x ).

The net level annual premium is calculated using the equivalence principle. You are given: • • •

q x  0.1 q x+1  0.2 v  0.9

Calculate Var ( L ) . A. 0.119 22.5.

B. 0.143

C. 0.160

D. 0.187

E. 0.202

[150-S88:17] For a fully discrete whole life insurance of 1000 issued to ( x ) , you are given: • 2A x  0.08 • A x  0.2 • The annual premium is determined using the equivalence principle.

S is the sum of the net future loss-at-issue random variables for 100 such independent policies. Calculate the standard deviation of S. A. 1,000

B. 2,500

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C. 7,000

D. 10,000

E. 25,000

Exercises continue on the next page . . .

EXERCISES FOR LESSON 22

427

[150-F88:12] A fully discrete whole life insurance of 1 with a level annual premium is issued to ( x ) .

22.6.

You are given: L is the loss-at-issue random variable if the premium is determined in accordance with the equivalence principle. • Var ( L )  0.75. • L∗ is the net future loss-at-issue random variable if the premium is determined such that E[L∗ ]  −0.5. •

Calculate Var ( L∗ ) . A. 0.3333

B. 0.5625

C. 0.7500

D. 1.1250

E. 1.6875

[150-S91:16] 0 L is the loss-at-issue random variable for a fully discrete whole life insurance of 1 on

22.7. (49).

You are given: • Mortality follows the Illustrative Life Table. • i  0.06 • Var (0 L )  0.10 Calculate E[0 L] [CAS4A-S92:20] (2 points) For a fully discrete whole life insurance on ( x ) , you are given:

22.8.

• d  0.06 • A x  0.4 • 2A x  0.2 • The annual premium is 5% of the amount of insurance. • L is the net future loss random variable at the time of issue for a policy of this type with amount of insurance 1. Determine E[L]. A. B. C. D. E.

Less than −1.0 At least −1.0, but less than −0.75 At least −0.75, but less than −0.50 At least −0.50, but less than −0.25 At least −0.25

22.9. [150-81-94:24] 0 L is the loss-at-issue random variable for a fully discrete n-year endowment insurance of 1 on ( x ) with premium Px:n . You are given: • •

x:n  0.1774 Px:n /d  0.5850

2A

Calculate Var (0 L ) . A. 0.065

B. 0.103

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C. 0.140

D. 0.174

E. 0.201

Exercises continue on the next page . . .

22. PREMIUMS: VARIANCE OF FUTURE LOSS, DISCRETE

428

22.10. [4-S86:27] You are given K| q x  0.5K+1 for K  0, 1, 2, . . .. P is the annual net premium for a fully discrete whole life insurance of 1. Determine the variance of the net future loss random variable v K+1 − P a¨ K+1 for i > 0. 1 2 B. · 2 2−v

v 1 A. · 2 2−v

!2

v 1 C. · 2 2−v

!2 D.

v2 1 · 2 2 − v2

E.

v2 2 − v2

22.11. L ( x ) is the net future loss random variable for a fully discrete whole life policy of 1 on ( x ) for which the annual premium is determined from the equivalence principle. You are given: • •

A60  0.5 q60  0.01





• Var L (60)  0.2 • i  0.05





Calculate Var L (61) . 22.12. You are given: • • •

A50  0.2 2A  0.06 50 d  0.05

The annual premium for a fully discrete whole life insurance on (50) with face amount 1000 is 13. The probability of a net future loss on a portfolio of n policies, using the normal approximation, is 5%. Determine n. 22.13. [150-F97:9] A fully discrete 40-year endowment insurance of 1 is issued to each of n people age x. Premiums are payable at the beginning of each year. You are given: • • • • • • •

The premium mistakenly charged is the net premium for a fully discrete 40-year term insurance of 1 on ( x ) . d  0.057 A x1:40  0.257 A x:40  0.268 2A x:40  0.0745 Losses are independent and the total portfolio of losses is assumed to follow the normal distribution. The probability of a positive loss on the total portfolio is 95%.

Calculate n A. 110

B. 120

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C. 130

D. 140

E. 150

Exercises continue on the next page . . .

EXERCISES FOR LESSON 22

429

Use the following information for questions 22.14 through 22.17: An insurance company issued the following fully discrete n-year endowment insurances, all at issue age x: Face amount

Annual Premium

Number of policies

2 4

0.16 0.32

100 100

You are given: • d  0.05 • A x:n  0.60 • 2A x:n  0.40 • Losses are independent. • There are no expenses. 22.14. [150-82-94:38] Calculate the expected value of the loss-at-issue random variable of one insurance with a face amount of 2. A. −0.08

B. −0.04

C. −0.02

D. −0.01

E. 0.00

22.15. [150-82-94:39] Calculate the variance of the loss-at-issue random variable of one insurance with a face amount of 2. A. 0.25

B. 0.27

C. 0.71

D. 1.00

E. 1.08

22.16. [150-82-94:40] Calculate the variance of the total expected loss-at-issue of the insurance company. A. 160

B. 430

C. 540

D. 650

E. 1420

22.17. [150-82-94:41] Using the normal approximation, calculate the probability that the present value at issue of total gains for the insurance company will exceed 0. A. 0.50

B. 0.69

C. 0.85

D. 0.90

E. 0.98

22.18. [150-S98:12, CAS4-S90:18] For a fully discrete whole life insurance of 1 on ( x ) , you are given: • A x  0.190 • 2A x  0.064. • d  0.057 • The level annual premium payable is π x  0.019. • The insurer issues N such insurances. • The losses are independent. • N is the smallest number of such insurances for which the probability, using the normal approximation, of a positive aggregate loss is less than 0.05. Calculate N. A. 15

B. 16

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C. 18

D. 20

E. 21

Exercises continue on the next page . . .

22. PREMIUMS: VARIANCE OF FUTURE LOSS, DISCRETE

430

22.19. [SOA3-F03:31] For a block of fully discrete whole life insurances of 1 on independent lives age x, you are given: • i  0.06 • A x  0.24905 • 2A x  0.09476 • π  0.025, where π is the gross premium for each policy. • Losses are based on the gross premium. Using the normal approximation, calculate the minimum number of policies the insurer must issue so that the probability of a positive total loss on the policies issued is less than or equal to 0.05. A. 25

B. 27

C. 29

D. 31

E. 33

Additional old SOA Exam MLC questions: S07:8, S13:15, F13:12

Solutions 22.1. This is an immediate application of formula (22.2). We look up A50  0.24905 and 2A50  0.09476 in the Illustrative Life Table. 2A

Var (0 L ) 

50

− A250

(1 − A50 ) 2 0.09476 − 0.249052  0.05805 (1 − 0.24905) 2



22.2. We will calculate the first moment and the second moment of the loss. The following table tabulates the loss for each possible value of K: the present value of the insurance (which is 0 if K ≥ 3) minus the present value of the annuity (0.07 each year for K + 1 years). Please note that p k in the following table merely means Pr ( K  k ) and is not standard actuarial notation. k

Pr ( K  k ) pk

Present value of insurance

Present value of annuity

Future loss lk

pk × lk

p k × l 2k

0 1 2 ≥3

0.04 0.0576 0.09024 0.81216

0.943396 0.889996 0.839619 0

0.07 0.136038 0.198337 0.198337

0.873396 0.753959 0.641282 −0.198337

0.034936 0.043428 0.057869 −0.161081

0.030513 0.032743 0.037111 0.031949

−0.024848

0.132315

Total The variance is therefore 0.132315 − (−0.024848) 2  0.131697 . 22.3.

Use formula (22.2). 2 − A2x 2 0.0143 − 0.0653 Var ( L )  150,000  150,000  150,0002 (0.011487) (1 − A x ) 2 (1 − 0.0653) 2 p √ Var (0 L )  150,000 0.011487  150,000 (0.107178)  16,077 2

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2A

x

!

!

EXERCISE SOLUTIONS FOR LESSON 22

22.4.

431

The present value of the insurance is





A x1:2  (0.1)(0.9) + (0.9)(0.2) 0.92  0.2358 The present value of the annuity is a¨ x:2  1 + (0.9)(0.9)  1.81 So the premium is 0.2358/1.81  0.130276. The expected loss is 0, so the variance is the second moment. The following table calculates the second moment:

k

Pr ( K  k ) pk

Present value of insurance

Present value of annuity

Future loss lk

p k × l 2k

0 1 ≥2

0.10 0.18 0.72

0.90 0.81 0

0.130276 0.247525 0.247525

0.769724 0.562475 −0.247525

0.059247 0.056948 0.044113

Total

0.160309

Var ( L )  0.160309 . (C) 22.5.

We can use the formula − A2x (1 − A x ) 2 0.08 − 0.22   0.0625 (1 − 0.2) 2

Var (0 L ) 

2A

x

The standard deviation for one policy for 1 is 0.25. This is multiplied by 1000 since the face amount of each policy is 1000, and by 10, the square root of 100, since the variance of 100 policies is 100 times the √ variance of one policy (so the standard deviation of 100 policies is 100 times the standard deviation of one policy). The answer is 10,000 (0.25)  2500 . (B) 22.6.

If we let P ∗ denote the revised premium determined such that E[L∗ ]  −0.5, then A x − P ∗ a¨ x  −0.5 A x  P ∗ a¨ x − 0.5 P∗ 

A x + 0.5 1.5  −d a¨ x a¨ x

Now we calculate the variance of L∗ . Var ( L∗ ) 



2

A x − A2x



 Ap2x −



A2x

1+

P∗ d

2

 1.5 ! 2 d a¨ x

 2.25 Var ( L )  1.6875 22.7.

We back out P, the gross premium.



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2

A49 − A249



1+

P d

2

 0.10

(E)

22. PREMIUMS: VARIANCE OF FUTURE LOSS, DISCRETE

432



0.08873 − 0.238822

 0.031695 1 +



1+

2

P d

P d

2

 0.10

 0.10

P p  0.10/0.031695  1.7763 d P  0.7763 d

1+

Now we calculate E[L]. E[L]  A49 − P a¨49

!

P  0.23882 − (1 − A49 ) d  0.23882 − 0.7763 (1 − 0.23882)  −0.3521 The answer choices for the original question are not provided, since the original question was based on the Illustrative Life Table at 5%. 22.8. E[L]  A x − P a¨ x 1 − A x 1 − 0.4 a¨ x    10 d 0.06  0.4 − 0.05 (10)  −0.1

(E)

The information on 2A x is extraneous. Perhaps the CAS question writers were copying the next exercise and forgot to omit this when they removed the part about variance. 22.9.

We back out A x:n from Px:n /d. Px:n A x:n A x:n   d a¨ x:n d 1 − A x:n  A x:n

0.5850  0.5850 − 0.5850A x:n

A x:n  0.369085 Px:n 2 d 2  (0.1774 − 0.369085 )(1.58502 )  0.103444



Var (0 L )  (2A x:n − A2x:n ) 1 +



22.10. Ax 

∞ X k0

(0.5v ) k+1 

0.5v v  1 − 0.5v 2 − v

0.5v 2 v2  1 − 0.5v 2 2 − v 2 2 (1 − v ) 1−v 1 − Ax   1 − 0.5v 2−v 2A − A2 x x Var ( L )  (1 − A x ) 2 Ax 

2

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(B)

EXERCISE SOLUTIONS FOR LESSON 22

433

v2 v2 −  2 − v 2 (2 − v ) 2

! (2 − v ) 2 4 (1 − v ) 2 ! !   v2 1  (2 − v ) 2 − (2 − v 2 ) 2 2 2−v 4 (1 − v ) ! ! 2 v 1  (4 − 4v + v 2 − 2 + v 2 ) 2 − v 2 4 (1 − v ) 2 ! ! 2 



1 2

v 2 − v2

!

(D)



22.11. Back out 2A60 from Var L (60) and A60 .





Var L (60) 

A60

60

− A260

(1 − A60 ) 2 2A

− 0.25 0.25  0.25 (0.2) + 0.25  0.3

0.2  2

2A

60

Use recursion to calculate A61 and 2A61 . A60  v ( q60 + p60 A61 ) 0.5 (1.05)  0.01 + 0.99A61 0.5 (1.05) − 0.01 A61   0.520202 0.99 2 A60  v 2 ( q 60 + p60 2A61 ) 0.3 (1.052 )  0.01 + 0.992A61 A61 

2



0.3 (1.052 ) − 0.01  0.323990 0.99



Calculate Var L (61) using equation (22.2).





Var L (61) 

2A

61

− A261

(1 − A61

)2



0.323990 − 0.5202022  0.2319 (1 − 0.520202) 2

22.12. The APV of the premium annuity is P (1 − 0.2) /0.05  16P. The expected future loss is 200 − 16 (13)  −8. The variance of future loss is (0.06 − 0.22 )(1000 + 20P ) 2  31,752. Therefore we want to solve the following for n: 1.645 p −8 + √ 31,752  0 n Let’s proceed.

p √ 1.645 31,752  8 n √ 293.124 n  36.64 8 n  1343

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22. PREMIUMS: VARIANCE OF FUTURE LOSS, DISCRETE

434

22.13. The expected value of the loss for each person is 0.268 − 0.257  0.011, since the probabilities of premium payments don’t change. This is in contrast to a situation in which a mistake is made in the survival probability and the expected value of the loss is therefore greater than the difference in single premiums since the probabilities of premium payments also change. The variance of the loss for each person is



Var (0 L )  1 +

P d

2 

2

A x:40 − A2x:40

0.057A x1:40  1+ (1 − A x:40 ) 0.057



!2 

0.0745 − 0.2682



0.257 2 (0.002676)  1+ 1 − 0.268  (1.3510922 )(0.002676)  0.004885





√ √ We want n such that 0.011n − 1.645 0.004885n  0. Divide through by n to obtain √ √ 0.011 n  1.645 0.004885 √ n  10.4521 n  109.25

(A)

22.14.

!

1 − 0.60 E[L]  2 (0.60) − 0.16  −0.08 0.05

(A)

22.15. As discussed in the previous lesson, you can use the premium per unit (0.08) as P and multiply the resulting variance by units squared, or 22 :



Var ( L )  4 0.40 − 0.602



1+

0.08 0.05

2

 1.0816

(E)

Alternatively, you can use the factor ( S+bP/d ) 2  (2+0.16/0.05) 2 instead of (1+0.08/0.05) 2 in the formula. 22.16. Let S be the total loss. Doubling the face amount (to 4) quadruples the variance. Var ( S )  (1.0816)(100) + 22 (1.0816)(100)  540.8

(C)

22.17. The expected value of the loss is −0.08 for a policy of 2 and −0.16 for a policy of 4. The total expected value is E[S]  (100)(−0.08) + 100 (−0.16)  −24 The probability that the present value at issue of total gains exceeds 0, or in other words that the present value of losses is less than 0, is Φ √

24 540.8

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!  Φ (1.03)  0.8485

(C)

EXERCISE SOLUTIONS FOR LESSON 22

435

22.18. The expected value of the premium payments is

!

1 − 0.190  0.27 π x a¨ x  0.019 0.057 and the expected value of the insurance is A x  0.19, so the expected net future loss is 0.19 − 0.27  −0.08. The variance of 0 L is: πx 2 2 ( A x − A2x ) d   0.19 2  1+ (0.064 − 0.1902 )  0.0496 0.57 √ We want N such that −0.08N + 1.645 0.0496N  0. Then √ √ 0.08 N  1.645 0.0496 √ 1.645 √ N 0.0496 0.08



Var (0 L )  1 +



!2

1.645 (0.0496)  20.9717 N 0.08

(E)

22.19. The mean loss per policy is E[0 L]  A x − π a¨ x  0.24905 − 0.025

(1 − 0.24905)(1.06) 0.06

 0.24905 − (0.025)(13.26678)  −0.08262

The variance of the loss is Var (0 L ) 



2

A x − A2x



1+

π d

2

0.025 (1.06) 0.06  (0.032734)(2.078403)  0.068035



 (0.09476 − 0.249052 ) 1 +

We need n policies such that

p

n E[0 L] + 1.645 n Var (0 L ) 1.645 (0.260835) −0.08262 + √ n 0.42907 −0.08262 + √ n √ n

0 0 0

0.42907 0.08262 n  26.9709

Thus at least 27 policies are needed. (B)

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2

22. PREMIUMS: VARIANCE OF FUTURE LOSS, DISCRETE

436

Quiz Solutions 22-1.

Note that 2p x  1 − 0.1 − 0.2  0.7. The net premium is a¨ x:2  1 + (1 − 0.1)(0.95)  1.855 1000Px:21 

1000 (0.952 )(0.7)  340.5660 1.855

The present value of the net future loss is −340.5660 if death occurs in the first year. −340.5660 (1 + 0.95)  −664.1038 if death occurs in the second year. 1000 (0.952 ) − 664.1038  238.3962 if the insured survives two years. Since the equivalence principle is used, the variance of the net future loss equals the second moment, which is Var (0 L )  0.1 (−340.5660) 2 + 0.2 (−664.1038) 2 + 0.7 (238.3962) 2  139,588

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Lesson 23

Premiums: Probabilities and Percentiles of Future Loss Reading: Models for Quantifying Risk (4th or 5th edition) 9.2 It is unclear whether this lesson is on the LC syllabus. The syllabus mentions that you should be able to apply a principle to associate a pattern of costs with a set of future contingent cash flows, and that principles include percentiles, so you should be able to discuss percentiles of future losses. On the other hand, the section of the textbook that discusses percentile premiums, Section 9.3, is not on ths syllabus. You may skip this lesson if you wish.

23.1

Probabilities

23.1.1

Fully continuous insurances

First we discuss calculating the probability that the future loss at issue is less than a given number for a fully continuous insurance. The present value of benefits for whole life insurance, term insurance, and endowment insurance decreases with time, whereas the present value of the premium annuity increases. Therefore, the probability that the future loss is less than c is the probability that the time of death is greater than g ( c ) for some function g. To help you picture typical patterns, see Figure 23.1 for graphs of future loss against time at death for whole life and term insurances. In these graphs, the premium is determined by the equivalence principle. For whole life insurance, future loss decreases monotonically. If the whole life insurance is limited

0L

0L

1

1

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2

0

0

−0.2

−0.2

−0.4

10

20

30

40

50

T

−0.4

(a) Whole life insurance

10

20

30

40

50

T

(b) 20-year term insurance

Figure 23.1: Graphs of future loss as a function of time at death. Assumes fully continuous insurance, δ  0.05 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

437

23. PREMIUMS: PROBABILITIES AND PERCENTILES OF FUTURE LOSS

438

0L

0L

1

1

0.8

0.8

0.6

0.6

0.4

0.4

0.2

0.2

0

0

−0.2

−0.2

−0.4

10

20

30

40

50

T

−0.4

(a) 20-year endowment insurance

10

20

30

40

50

T

(b) 20-year deferred whole life insurance

Figure 23.2: Graphs of future loss as a function of time at death. Assumes fully continuous insurance, δ  0.05

pay, the future loss still decreases monotonically, but less rapidly after the premium payment period. For 20-year term insurance, future loss decreases monotonically to year 20. At that point the premiums and benefits stop, and for all values of T higher than 20 the future loss is negative the discounted value of the premiums paid. Figure 23.2 has graphs of the future loss for endowment insurance and deferred whole life insurance. For 20-year endowment insurance, the future loss monotonically decreases to year 20. After year 20, the future loss is the discounted value of the endowment paid minus the premiums collected during the 20 year endowment period. Example 23A For a fully continuous whole life insurance of 1000 on ( x ) : • The annual premium is 15. • µ  0.01 • δ  0.05 Calculate the probability that the loss at issue is greater than −14. Answer: The loss at issue in terms of Tx is −0.05Tx − 15 0 L  1000e

1 − e −0.05Tx 0.05

!

 1300e −0.05Tx − 300 We want the probability that this is greater than −14, or Pr (1300e −0.05Tx − 300 > −14)  Pr ( e −0.05Tx > 286/1300) and 286/1300  0.22. Due to exponential mortality and the fact that we’re dealing with whole life (so premiums and death benefits are payable at all times), we don’t have to solve for Tx , since Pr (Tx < w )  1 − e −0.01w  1 − ( e −0.05w ) 0.2 , or more generally Pr ( e −δTx > u )  1 − u µ/δ . So Pr ( e −0.05Tx > 0.22)  1 − 0.221/5  0.2613 . 

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23.1. PROBABILITIES

439

Example 23B For a fully continuous 20-year endowment insurance of 1000 on (60): • µ60+t  1/ (40 − t ) , 0 ≤ t < 40 • δ  0.06 • The annual premium is 36. Calculate the probability that the future loss is less than 0. Answer: If death occurs before year 20, the future loss is 1 − e −0.06T60 − 36  1600e −0.06T60 − 600 0.06

!

0L

 1000e

−0.06T60

Then the probability that the future loss is less than 0 is Pr (1600e −0.06T60 − 600 < 0)  Pr ( e −0.06T60 < 0.375)  Pr (−0.06T60 < ln 0.375)

! ln 0.375 + * /  Pr .T60 > − 0.06 , and − (ln 0.375) /0.06  16.3472. Since 16.3472 ≤ 20, the probability that the future loss is less than 0 is  therefore 16.3472 p60  23.6528/40  0.59132 .

?

Quiz 23-1 For a fully continuous 15-year term life insurance of 1000 on (80): • µ  0.04 • δ  0.02 • The premium is determined so that the expected future loss is 0. Calculate the probability that the future loss is greater than 10. A special shortcut is available for level-premium level-benefit fully continuous whole life and endowment insurance when the premium is the net premium. In that case, the time of death t for which the loss at issue is 0 is the t such that v t equals the EPV of the insurance. Here is the proof for fully continuous whole life, but the same logic works for endowment insurance. The loss at issue for death at time t is 1 − vt v − P a¯ t  v − P δ t

!

t

If v t  A¯ x , then this expression is 1 − A¯ x A¯ x A¯ x − P  A¯ x − a¯ x  0 δ a¯ x

!

!

This shortcut also works for gross loss at issue (which will be discussed later in this lesson) if the equivalence principle is used to calculate the gross premium. For a deferred insurance calculating probabilities is trickier. As Figure 23.2b shows, the future loss is negative and declining throughout the deferral period, then jumps, then declines thereafter. Example 23C For a fully continuous 10-year deferred whole life insurance of 1000 on ( x ) : • µ x+t  0.001t • δ  0.04 • The annual premium is 35 and is payable for the first 10 years only. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

23. PREMIUMS: PROBABILITIES AND PERCENTILES OF FUTURE LOSS

440

Calculate the probability that the future loss is greater than 5. Answer: The future loss is negative if death occurs before time 10. So we want the probability of surviving at least 10 years, but not longer than a certain time for which the future loss is 5. That point is defined by 0L

 1000e −0.04Tx − 35 a¯10  5

and a¯10  (1 − e −0.4 ) /0.04  8.24200. So we want Tx  t for which 1000e −0.04t − 35 (8.24200)  5 5 + 35 (8.24200) e −0.04t   0.29347 1000 ln 0.29347 t−  30.6495 0.04 The probability of survival to 10 but not to 30.6495 for the Weibull mortality that we are given is t

Z t px

 exp −

!



0.001u du  exp −0.0005t 2



0 10 p x

− 30.6495 p x  e

−0.0005 (102 )

2

− e −0.0005 (30.6495 )  0.951229 − 0.625192  0.32604



A little more work would be required in the previous question to calculate the probability that the loss is greater than a negative number, since there are two ways that can happen: survival for a short time and survival more than 10 years but less than some larger number. Example 23D (Version of previous example) For a fully continuous 10-year deferred whole life insurance of 1000 on ( x ) : • µ x+t  0.001t • δ  0.04 • The annual premium is 35 and is payable for the first 10 years only. Calculate the probability that the future loss at issue is greater than −5. Answer: First let’s repeat the steps in the solution to the previous example to calculate the probability of death after 10 years but before the time when the future loss equals −5. 1000e −0.04t − 35 (8.24200)  −5 e −0.04t  0.28347 t  31.5162 10 p x

2

− 31.5162 p x  e −0.0005 (10 ) − e −0.0005 (31.5162

2)

 0.951229 − 0.608574  0.34266 In addition, we need to calculate the probability that the premium payments in the first few years have present value less than 5. This means 35 (1 − e −0.04T ) /0.04 < 5, or 0.04 7 0.04 e −0.04T > 1 −  0.994286 7 ln 0.994286 T 156)  Pr 1.04

− ( K20 +1)

!

156 ln (156/1156) >  Pr K 20 + 1 < −  Pr ( K 20 + 1 < 51.07) 1156 ln 1.04



K 20 < 50.07 if the insured does not survive 51 years. The probability of not surviving 51 years is 51q 20

?

1−

l 71 6,396,609 1−  0.33492 l 20 9,617,802



Quiz 23-2 For a fully discrete whole life insurance of 1 on (50): • l x  100 (100 − x ) . • i  0.1 • The gross premium is 0.03. Calculate the probability that the loss at issue is greater than −0.1.

23.1.3

Annuities

Now let’s discuss annuities. For annuities, the future loss usually grows as Tx grows. Subfigure 23.3a shows the pattern for a single-premium annuity with continuous payments beginning immediately. The loss is initially negative the premium, and gradually grows. Subfigure 23.3b shows the pattern for a single premium deferred annuity, and is similar to Figure 6.2 in Actuarial Mathematics for Life Contingent Risks. In the deferral period, the loss is the negative of the single premium; afterwards it grows. Figure 23.4 shows the pattern for a deferred annuity with continuous premiums payable during the deferral period. Future loss declines during the premium payment period and increases monotonically thereafter. Example 23F For a 20-year deferred whole life annuity-due on (45) paying 1000 per year: • Net premiums are payable at the beginning of each year for 20 years, and are determined using the equivalence principle. • Mortality is uniformly distributed with limiting age 120. • i  0.04 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

23. PREMIUMS: PROBABILITIES AND PERCENTILES OF FUTURE LOSS

442

L0

L0 5

4 2

0

0 −2

−5

−4 −6

−10

−8 −15

10

20

30

40

50

Tx

−10

(a) Single premium annuity starting immediately

10

20

30

40

(b) Single premium deferred annuity

Figure 23.3: Future loss for single premium annuities

L0 4 2 0 −2 −4 −6 −8 −10

10

20

30

40

50

Tx

Figure 23.4: Deferred annuity with continuous premiums during deferral period

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50

Tx

23.1. PROBABILITIES

443

L0 2000 1000 0 −500 −1000 −2000 −3000 −4000 −5000 −6000

0

5

10

15

20

25

30

35

40

45

50

T45

Figure 23.5: Pattern of future losses in Example 23F

Calculate the probability that the future net loss will be less than −500. Answer: First let’s calculate the net premium. A whole life annuity at age 65 has actuarial present value 1 − (1/1.0455 )  0.401975 55 (0.04) 1 − 0.401975  15.54866  0.04/1.04

A65  a¨65 Discounting for 20 years,

!

20 E 45  20| a¨ 45

55 1  0.334684 75 1.0420

 (0.334684)(15.54866)  5.203883

A premium annuity-due of 1 per year has actuarial present value 1 − (1/1.0420 ) + 0.334684  0.515888 75 (0.04) 1 − 0.515888   12.58691 0.04/1.04

A45:20  a¨45:20 The premium is therefore

1000 (5.203883)  413.44 12.58691 The future losses are graphed in Figure 23.5. After collecting two premiums, the loss will be less than −500. After that, the future loss won’t be higher than −500 until the present value of the annuity P

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23. PREMIUMS: PROBABILITIES AND PERCENTILES OF FUTURE LOSS

444

payments is equal to the present value of all 20 premiums minus 500. Let’s calculate the present value of 20 premiums, and then solve for K 45 such that the present value of annuity payments is high enough. 1 − 1/1.0420  413.44 (14.1339)  5843.48 0.04/1.04

!

PV of 20 premiums  413.44 a¨20  413.44 and for the loss to be less than −500, we need

100020| a¨ K45 −19 − 5843.48 < −500 1000

v 20 − v K45 +1 < 5343.48 0.04/1.04

Let’s solve this.

!

v 20 − v K45 +1 < (5.34348)

0.04  0.205518 1.04

1 − 0.205518  0.250868 1.0420 ln 0.250868 K 45 + 1 <  35.26 − ln 1.04 v K45 +1 >

Thus the loss will be greater than −500 if the annuitant survives 35 years. The probability of the annuitant surviving at least 1 year but not 35 years, under a uniform distribution, is (35 − 1) /75  0.45333 . 

23.2

Percentiles

Since future loss for whole life, term, and endowment insurances is a monotonically decreasing function, the 100p th percentile of the future loss corresponds to the 1 − 100p th percentile of the survival distribution. Example 23G For a fully continuous insurance of 1 on Kevin: • δ  0.06   0.01 t ≤ 20 • µ  0.02 t > 20



Calculate the annual premium so that the probability of a positive future loss is 25%. Answer: We want the future loss to be less than 0 for 75% of the time, which means that the 75th percentile of the future loss should be 0. The 75th percentile of the future loss is the loss that occurs if Kevin survives up to the 25th percentile of his future lifetime. Let’s calculate that. Using ( x ) for Kevin, the 25th percentile of future lifetime is the t for which t q x  0.25 or t p x  0.75. Survival probability for 20 years is 20 p x

 e −0.01 (20)  0.8187

which is higher than 0.75, so t must be greater than 20. For t > 20, t px

 20 p x t−20 p x+20  e −0.2 e −0.02 ( t−20)  0.75

e −0.02t  0.75e −0.2 0.02t  0.2 − ln 0.75 t  10 − 50 ln 0.75  24.3841 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

EXERCISES FOR LESSON 23

445

Table 23.1: Summary of concepts in this lesson

• For level benefit or decreasing benefit insurance, the loss at issue decreases with time for whole life, endowment, and term insurances. To calculate the probability that the loss at issue is less than something, calculate the probability that survival time is greater than something. • For level benefit or decreasing benefit deferred insurance, the loss at issue decreases during the deferral period, then jumps at the end of the deferral period and declines thereafter. – To calculate the probability that the loss at issue is greater than a positive number, calculate the probability that survival time is less than something minus the probability that survival time is less than the deferral period. – To calculate the probability that the loss at issue is greater than a negative number, calculate the probability that survival time is less than something that is less than the deferral period, and add that to the probability that survival time is less than something that is greater than the deferral period minus the probability that survival time is less than the deferral period. • For a deferred annuity with a single premium, the loss at issue is a negative constant during the deferral period, then increases. If regular premiums are payable during the deferral period, the loss at issue decreases until the end of the deferral period and increases thereafter. Let P be the annual premium we are trying to calculate. The loss for T  24.3841 is the present value of the benefit paid if death occurs at that time minus present value of the premiums collected. Note that µ is not used in this calculation. 0L

 v 24.3841 − P a¯24.3841  e −0.06 (24.3841) − P

1 − e −0.06 (24.3841) 0.06

!

 0.231530 − P (12.807835) Setting this expression equal to 0, we get P  0.231530/12.807835  0.018077 .



Exercises 23.1.

For a fully continuous whole life insurance of 1000 on (40), you are given • l x  1 − x/100, 0 ≤ x ≤ 100 • δ  0.03 • Premiums are payable for 20 years. • The annual premium is 25.

Calculate the probability that the future loss is greater than 0.

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Exercises continue on the next page . . .

23. PREMIUMS: PROBABILITIES AND PERCENTILES OF FUTURE LOSS

446

23.2. For a fully continuous 20-year deferred 40-year term life insurance policy of 1000 on (35), you are given • µ35+t  0.02 • δ  0.05 • The premium is payable for 60 years and is determined by the equivalence principle. Calculate the probability that the future loss is greater than −10. 23.3. The Economical Life Insurance Company has designed a whole life insurance product with the following features: • Death benefits for each unit purchased are 2000 if death occurs during the first 20 years, 1000 if death occurs thereafter. • Death benefits are payable at the moment of death. • Premiums are level and payable at the beginning of each year. • The premium has been set so that the expected value of the future loss is −10. • It is assumed that µ  0.01 in all years. • δ  0.04 Calculate the probability that the future loss will be no more than −10. For a fully discrete 20-year endowment insurance of 1 on (65), you are given:

23.4.

• Mortality follows the Illustrative Life Table. • Premiums are calculated using the equivalence principle. • i  0.06. Calculate the 75th percentile of 0 L. For a fully discrete 20-year term insurance of 1 on (65), you are given:

23.5.

• Mortality follows the Illustrative Life Table. • Premiums are calculated according to the equivalence principle. • i  0.06. • 0 L is the present value of the loss at issue. Calculate the 75th percentile of 0 L. For a fully continuous whole life insurance of 1, you are given:

23.6. •

µ x  0.01 for all x.

•

  0.05 δ  0.04

0 ≤ t ≤ 20 t > 20

 Calculate the lowest annual premium required so that the probability of the future loss being above zero is no more than 25%. 23.7.

For a 10-year deferred annuity-due of 1000 per year on (55): • l x  1000 (120 − x ) , x < 100. • Premiums of 1200 are payable at the beginning of each year for ten years. • d  0.04

Calculate the probability that the future loss on the annuity will be greater than 0. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

Exercises continue on the next page . . .

EXERCISES FOR LESSON 23

447

For a single premium 20-year deferred whole life annuity-due on (40):

23.8.

• Mortality is uniformly distributed with ω  120. • The annuity pays 1 per year starting at age 60. • The net single premium is determined using the equivalence principle. • i  0.06. Calculate the probability that the loss at issue is greater than 0. For a 10-year deferred continuous annuity of 100 per year on (55):

23.9.

• Continuous premiums of 200 per year are paid during the first 10 years. • µ x+t  0.01, t > 0 • δ  0.04 Calculate the probability that the net future loss is greater than 0. 23.10. [CAS4A-F98:20] (2 points) Consider a 100,000 whole life insurance policy issued to (30). The benefit is payable at the time of death. The premium is π b , payable continuously. The force of mortality is constant, µ  0.015, and the force of interest is constant, δ  0.05. For this individual policy, determine the premium such that the probability is 0.5 that the loss, L ( π b ) , is positive. A. B. C. D. E.

Less than 500 At least 500, but less than 833 At least 833, but less than 1,167 At least 1,167, but less than 1,500 At least 1,500

23.11. [3-F00:16, 150-83-96:4] For a fully discrete whole life insurance of 10,000 on (30): π denotes the annual premium and L ( π ) denotes the loss-at-issue random variable for this insurance. • Mortality follows the Illustrative Life Table. • i  0.06 •

Calculate the lowest premium, π0, such that the probability is less than 0.5 that the loss L ( π0 ) is positive. A. 34.6

B. 36.6

C. 36.8

D. 39.0

E. 39.1

23.12. [150-S98:2] For a fully continuous 30-year term insurance of 1 on (50), you are given: • • •

δ  0.05 µ50+t  1/ (70 − t ) , 0 ≤ t < 70. P is the least annual premium such that the insurer’s probability of a positive financial loss is at most 0.20.

Calculate P. A. 0.046

B. 0.047

C. 0.048

Additional old CAS Exam 3/3L questions: S06:22, F10:14

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D. 0.049

E. 0.050

23. PREMIUMS: PROBABILITIES AND PERCENTILES OF FUTURE LOSS

448

Solutions 23.1.

The future loss is −0.03T40 − 25 0 L  1000e

1 − e −0.03 min (T40 ,20) 0.03

!

Note that a¯20  (1−e −0.6 ) /0.03  15.0396. If T40  20, the future loss is 1000e −0.6 −25 (15.0396)  172.82 > 0, so the T40 that makes the future loss equal to 0 is greater than 20. Therefore, we want T40 to satisfy the following equation: 1000e −0.03T40 − 25 (15.0396)  0 1000e −0.03T40  375.99 ln 0.37599 T−  32.6064 0.03 The probability that T40 is less than 32.6064 is Pr (T40 < 32.6064)  32.6064q 40 

32.6064  0.5434 60

23.2. Based on the equivalence principle, the actuarial present value of a 20-year deferred term life insurance policy is   0.02 ¯ e −20 (0.07) − e −60 (0.07)  0.066172 20|40 A35  0.02 + 0.05 and the actuarial present value of a premium annuity of 1 per year is a¯35:60 

1 − e −60 (0.07)  14.07149 0.02 + 0.05

so the net premium is 1000 (0.066172) /14.07149  4.70255. The future loss starts off at 0 and decreases until T35  20, then jumps above 0 and decreases thereafter. We need to calculate the time t1 < 20 such that the premium annuity is less than 10. 4.70255a¯ t1  10 1 − e −0.05t1  10 0.05

!

4.70255

0.05 (10)  0.893675 4.70255 < t1 )  1 − e −0.02t1

e −0.05t1  1 − Pr (T35

 1 − 0.8936750.02/0.05  0.04397 We need to calculate the time t2 > 20 such that the future loss is equal to −10. 1 − e −0.05t2 − 4.70255  −10 0.05

!

1000e

Pr (20 < T35

−0.05t2

1094.05e −0.05t2 − 94.05  −10 84.05 e −0.05t2   0.076825 1094.05 < t2 )  e −20 (0.02) − 0.0768250.02/0.05  0.31208

The answer is 0.04397 + 0.31208  0.35603 . LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

EXERCISE SOLUTIONS FOR LESSON 23

449

23.3. The insurance can be treated as a sum of a whole life insurance of 1000 and a 20-year term insurance of 1000. The APV of one unit of insurance is A  1000



  0.01 0.01 + 1 − e −20 (0.01+0.04)  326.42 0.01 + 0.04 0.01 + 0.04

The actuarial present value of a life annuity-due under constant force of mortality is 1+i e 0.04  20.5042  −0.01 q+i 1−e + e 0.04 − 1 The premium P satisfies 326.42 − 20.5042P  −10 336.42 P  16.4074 20.5042 The future loss decreases monotonically, so the future loss will be no more than −10 if Tx is high enough. Let’s calculate the future loss if Tx  20 and the benefit is 2000 so that we’ll know which benefit to use. Note that d  ( e 0.04 − 1) /e 0.04  0.039211. 1 − v 20 16.4074 − 16.4074  2000e −0.8 − (1 − e −0.8 )  668.24 d 0.039211

!

2000v

20

Therefore to get a future loss of −10 we must have Tx > 20, and in fact, if Tx is an integer, it must satisfy 1000v Tx − P a¨Tx  −10 1 − e −0.04Tx − 16.4074  −10 0.04/1.04

!

1000e

−0.04Tx





1000e −0.04Tx − 418.44 1 − e −0.04Tx  −10 408.44  0.287950 1418.44 ln 0.287950  31.12 Tx  − 0.04

e −0.04Tx 

We assumed Tx was an integer, but we got a non-integral Tx . However, this calculation does show that the loss at issue is greater than −10 for Tx  30, since v 30 is greater than v 31.12 and a¨31 is less than (1 − 31.12 e −0.04 ) /d. Thus we only need to verify that for Tx  31, the loss at issue is less than −10. At Tx  31, 32 premiums are collected, so 1 − e −0.04 (32) − 16.4074  −18.5996 0.04/1.04

!

1000e

−0.04 (31)

Thus the loss at issue is less than −10 as soon as the 32nd premium is collected, or whenever Tx ≥ 31. The probability Pr (Tx ≥ 31)  e −0.01 (31)  0.7334 . 23.4. The later the endowment is paid, the lower the loss. So we must find T65 such that there is a 75% chance that survival will be past T. We must find x such that l x  0.75l65  0.75 (7,533,964)  5,650,473. Death occurs between age 74 and age 75. The loss is then the present value of 1 paid 10 years from now, minus a 10-year certain annuity-due of premiums. We calculate the premium and then the loss: a¨65:20  a¨65 − 20 E65 a¨85 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

23. PREMIUMS: PROBABILITIES AND PERCENTILES OF FUTURE LOSS

450

 9.8969 − 0.09760 (4.6980)  9.4384 1 1 P −d  − 0.05660  0.04935 a¨65:20 9.4384 1 − (1/1.0610 )  7.80169 0.06/1.06 1 − P a¨10 0L  1.0610  0.55839 − (0.04935)(7.80169)  0.1734

a¨10 

23.5. As in the previous exercise, the loss decreases as survival time increases, so the 75th percentile will occur for deaths at age 74 (i.e., between 74 and 75). However, the premium is different. Refer to the previous solution for the derivation of the 10-year annuity certain and the 20-year temporary life annuity. 1 A65 :20  A65 − 20 E65 A85

 0.43980 − (0.0976)(0.73407)  0.36815 A1 0.36815 P  65:20   0.03901 a¨65:20 9.4384 0L

 v 10 − P a¨10  0.55839 − 0.03901 (7.80169)  0.25405

23.6. The future loss monotonically decreases as a function of Tx , time at death. Thus if we set the annual premium such that at the 25th percentile of Tx we break even, then the probability of a future loss will be 25%. The future loss is greater than 0 only when survival time is less than the 25th percentile of Tx . The 25th percentile of Tx is solved for: e −0.01Tx  0.75 0.01Tx  − ln (0.75) Tx  28.7682 For this Tx , the present value of the benefit is e −20 (0.05) −8.7682 (0.04)  0.259052 The present value of the annuity can be computed using annuity-certain formulas or by integration. By integration: 20

Z

8.7682

Z e

−0.05t

dt + e

0





e −0.04t dt  20 (1 − e −1 ) + 25e −1 1 − e −0.04 (8.7682)  15.36311

−1 0

Alternatively, by annuity-certain formulas, we must split the annuity into a 20-year annuity and a 20-year deferred 8.7682-year annuity. The first annuity has interest rate 0.05: a¯20

0.05



1 − e −0.05 (20)  12.64241 0.05



1 − e −0.04 (8.7682)  7.39562 0.04

The second annuity has interest rate 0.04: a¯8.7682 LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

0.04

EXERCISE SOLUTIONS FOR LESSON 23

451

The second annuity is deferred for 20 years at a 5% interest rate, so the total value of the 28.7682-year annuity is 12.64241 + e −20 (0.05) (7.39562)  15.36311. If you mistakenly thought that a 4% interest rate should be used for discounting, review the material around Examples 9C through 9E. The premium to make the loss zero when Tx  28.7682 is 0.259052/15.36311  0.016862 . 23.7. The future loss is negative during the first 10 years and is positive only after enough payments are made. First let’s calculate how many payments are needed to make the future loss positive. 1000 ( a¨ t − a¨10 ) − 1200a¨10  0 2200

0.9610

1− 0.04

! − 1000

0.96t

1− 0.04

t > 10

! 0

Multiplying through by 0.04, 737.3682 − 1000 + 1000 (0.96t )  0 t  (ln 0.26263) / (ln 0.96)  32.75 So the future loss is 0 at time 32.75. This means that if the annuitant receives the payment associated with the 33rd year, the one paid at the beginning of the 33rd year, the future loss will be positive. The probability of that is the probability of surviving 32 years, or 1 − 32/ (120 − 55)  0.5077 . 23.8.

Calculate the net single premium. 1 − 1/1.0660  0.269357 60 (0.06) 1 − 0.269357   12.90802 0.06/1.06 0.75  20p 40 v 20   0.233854 1.0620  (0.233854)(12.90802)  3.018587

A60  a¨60 20 E 40 20| a¨ 40

We need survival time t such that the payments on the annuity have present value 3.018587. v 20 − v t  3.018587 0.06/1.06

!

0.06  0.170863 0.311805 − v  3.018587 1.06 t

v t  0.311805 − 0.170863  0.140942 ln 0.140942  33.63 t− ln 1.06 The loss at issue is greater than 0 if the annuitant survives 33 years. The probability of that is 1 − 33/80  0.5875 . 23.9. The present value of the premiums (not the actuarial present value; rather, the present value given that the annuitant survives 10 years) is 1 − e −0.4  1648.40 0.04

!

200 a¯10  60

We need to calculate survival time t for which the annuity payments have this present value. v 10 − v t  1648.40 0.04

!

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23. PREMIUMS: PROBABILITIES AND PERCENTILES OF FUTURE LOSS

452

e −0.4 − e −0.04t  0.659360 e −0.04t  0.010960 ln 0.010960 t−  112.83733 0.04 The probability of surviving 112.8373 years is e −0.01 (112.8373)  0.3236 . 23.10. The 50th percentile of survival time is the t for which e −0.015t  0.5 ln 0.5 t−  46.2098 0.015 The loss at this time t is 1 − e −0.05 (46.2098)  0.0992126 − 18.015749P 0.05

!

e −0.05 (46.2098) − P

So the answer is 100,000 (0.992126) /18.015749  550.70 . (B) 23.11. In the Illustrative Life Table, l30  9,501,381. The median age to which ( x ) survives is the age x at which l x  0.5 (9,501,381)  4,750,691. This happens at age 77 (between 77 and 78). The probability of a death before age 78 is greater than a half, but the probability of a death before age 77 is less than a half. To make the probability less than 0.5, we must set the premium so that the loss is zero if death occurs between ages 77 and 78. The present value of the death benefit is then 10,000/1.0648  609.98. The present value of 1 − (1/1.06) 48 the premium annuity-due is then  16.5890. The premium is 609.98/16.5890  36.77 . (C) 0.06/1.06 23.12. The earlier the death, the greater the loss. Therefore, we set the premium to break even at the 20th percentile of survival time T50 . Since mortality is uniformly distributed, this occurs 20% of the way from age 50 to 120, or after 14 years, at age 64. To break even if death occurs at age 64: v 14  P a¯14 e

−0.05 (14)

1 − e −0.05 (14) P 0.05

!

0.05e −0.7 1 − e −0.7 0.05 (0.496585)   0.04932 1 − 0.496585

P

(D)

Quiz Solutions 23-1. Since the equivalence principle is used and mortality is exponential, the premium is 1000µ  40. (See the fifth line of Table 18.1). If death occurs by year 15, the future loss is 1 − e −0.02T80 − 40  3000e −0.02T80 − 2000 0.02

!

0L

 1000e

−0.02T80

The probability that the future loss is greater than 10 is Pr (3000e

−0.02T80

> 2010)  Pr ( e

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−0.02T80

! ln 0.67 + * /  Pr (T80 < 20.02) > 0.67)  Pr .T80 < − 0.02 , -

QUIZ SOLUTIONS FOR LESSON 23

453

However, 20.02 > 15, is beyond the term period. So the future loss is greater than 10 if and only if death occurs at time 15 or earlier. The probability of that is 1 − e −0.04 (15)  0.4512 . 23-2.

The future loss is

1 − v K x +1 − 0.03  1.33v K x +1 − 0.33 0.1/1.1

!

0L

v

K x +1

The probability that this expression is greater than −0.1 is 0.23 1.33 ! ln (0.23/1.33)  Pr K x + 1 < − ln 1.1



Pr (1.33v K x +1 − 0.33 > −0.1)  Pr 1.1− ( K x +1) >



 Pr ( K x + 1 < 18.4120) For K x < 17.4120, death must occur before time 18. Under uniform mortality, 18q 50  18/50  0.36 .

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454

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23. PREMIUMS: PROBABILITIES AND PERCENTILES OF FUTURE LOSS

Lesson 24

Multiple Lives: Joint Life Probabilities Reading: Models for Quantifying Risk (4th or 5th edition) 12.1.1–12.1.6

24.1

Introduction

An actuary sometimes has to deal with collections of lives, multiple lives. Some examples are: • A whole life insurance policy may pay a benefit at the second death of a husband and a wife. • A whole life annuity immediate may pay a fixed amount, e.g. 1000, per month to a set of two people (such as husband and wife) until they are both dead. • Complicated estate provisions may involve several lives. For example, a will may call for the payment, after death of the parent, of 1000 per month to each of the children until they turn 18. If ( x ) and ( y ) are two lives age x and y respectively, then we define the joint life status ( x y ) , sometimes called the joint status for short, as the situation of both lives being alive. The joint life status fails as soon as one life dies. The joint life status can apply to any number of lives. For example, the status ( wx yz ) fails whenever any of the four lives indicated in the status dies. However, exam questions involving more than two lives are rare. When letters in the joint status symbol are replaced with numbers, a colon is placed between the numbers; for example, the joint status of (30) and (40) is (30:40) . We have used the concept of status previously in this course, although we didn’t call it a status. A single life, e.g. (65) , is a status, which fails when that life dies. A fixed period, e.g. 10 , is a status. which fails exactly at the end of ten years. The status (65:10 ) is a joint status which fails upon the failure of either status within it: either (65) dying or ten years passing. We define the complete future lifetime random variable Tx y as the time of failure of the joint status; in other words, the time at which the first death of the two lives occurs. We define the curtate future lifetime random variable K x y as the greatest integer less than or equal to Tx y . In other words, it is the number of full years until the first death of the two lives occurs. Clearly Tx y  min (Tx , T y ) ≤ Tx and K x y  min ( K x , K y ) ≤ K x . The joint status may be used as a subscript in the same way the single life status is. For example t p x y is the probability that the joint status survives at least t years, or that both lives survive at least t years. 3|5 q 40:50 is the probability that the first death of (40) and (50) occurs at least three years but no more than eight years from now. The joint status is a simple status in the sense that either it survives or it fails. There is no intermediate state. Therefore, every equation we discussed in Lesson 2 still applies. For example: t px y t|u q x y

 t p x y u q x+t:y+t  t p x y − t+u p x y t+u p x y

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+ t qx y  1

 t p x y u p x+t:y+t 455

(24.1)

24. MULTIPLE LIVES: JOINT LIFE PROBABILITIES

456

The random variable Tx y represents the time to failure of the joint status, and has a distribution and a survival function with the usual meanings: Fx y ( t )  Pr (Tx y ≤ t ) S x y ( t )  Pr (Tx y > t )

24.2

Independence

A complete survival model would require a joint distribution function for all lives involved. While such a model could be built, it would be hard to gather data to estimate such a model. And such models are complicated mathematically. Therefore, it has been traditional to assume all lives are independent. This means that the joint survival probability function is the product of the individual survival probability functions: t p x y  t p x t p y . The independence assumption is usually false in real life situations. When one is interested in the survival of multiple lives, these lives are never a random group; they are always related to each other, for example as a married couple. At the very least, such a couple would be more likely to be in the same auto accidents. Many would argue that a couple would have a certain lifestyle which would increase or decrease both of their life expectancies. Nevertheless, actuaries usually assume that the error implicit in the independence assumption is small and the simplification of the model is more important. Exams rarely ask questions on dependent lives. Unless otherwise stated, we will be assuming that all lives in our joint life models are independent. However, exam questions will state the independence assumption every time it is used, at least when human lives are involved. Although independence simplifies matters considerably, it still will take you a little work to get the hang of calculating joint-life probabilities. You should review single life probabilities, the material of lessons 2–5, if necessary. With the assumption of independence, t p x y  t p x t p y , and this generalizes to any number of lives. This means that 1 − t q x y  (1 − t q x )(1 − t q y ) t qx y

 1 − (1 − t q x )(1 − t q y )  t q x + t q y − t q x t q y

(24.2)

In particular, t q x y , t q x t q y . For joint life statuses, it is easier to deal with p’s than with q’s, so convert q’s to p’s whenever feasible. The single-life future lifetime probability functions for the first and second lives of a joint life status do not have to be the same. Example 24A You are given the following mortality table:

x

Male qx

Female qx

80 81 82 83 84

0.10 0.12 0.14 0.16 0.18

0.07 0.09 0.11 0.13 0.15

Kevin and Kira are a couple. Kevin, a male, is age 82, and Kira, a female is age 80. Kevin and Kira have independent future lifetimes. Calculate the probability that the first death for this couple will occur within the third year from the current date. LC Study Manual—1st edition 3rd printing Copyright ©2014 ASM

24.2. INDEPENDENCE

457

Answer: The symbol for what we’re trying to calculate is 2| q 82:80 . The way to calculate this is by using equation (24.1). Namely, calculate the probability that the joint status will survive two years, and then subtract the probability that joint status will survive three years. For Kevin 2 p 82

 (1 − 0.14)(1 − 0.16)  0.7224

3 p 82

 (0.7224)(1 − 0.18)  0.592368

2 p 80

 (1 − 0.07)(1 − 0.09)  0.8463

3 p 80

 (0.8463)(1 − 0.11)  0.753207

For Kira

For the joint status, we therefore have 2 p 82:80

 (0.7224)(0.8463)  0.611367

3 p 82:80

 (0.592368)(0.753207)  0.446176

So the probability we seek is 2 p 82:80

− 3 p 82:80  0.611367 − 0.446176  0.165191



Let’s discuss the force of mortality for the joint status. Recall that µ x+t  −d (ln t p x ) /dt, a version of equation (3.3). For two independent lives, this becomes µ x+t:y+t  −

d (ln t p x y ) dt

−

d (ln t p x + ln t p y ) dt

 µ x+t + µ y+t

(24.3)

If all lives have constant forces of mortality, then the joint status will have a constant force of mortality which is the sum. Example 24B Three lives have independent future lifetimes. The forces of mortality are µ x+t  0.02 for the first life, µ y+t  0.03 for the second life, and µ z+t  0.04 for the third life. Calculate the probability that the first death from the three lives will occur at least five years from the current date, but no later than ten years from the current date. Answer: The joint status µ x+t:y+t:z+t for the joint status of the three lives is 0.02 + 0.03 + 0.04  0.09. Therefore,

5 p x yz

5 p x yz

 e −5 (0.09)  0.637628

10 p x yz

 e −10 (0.09)  0.406570

− 10 p x yz  0.637628 − 0.406570  0.231058



Recall that for uniform mortality, µx  and for beta,

1 ω−x

α ω−x If the mortality for each life is uniform with the same ω − x, the joint life mortality will be beta with the same ω − x, but not uniform. More generally, if the mortality for each life follows beta, each with its own α but all with the same ω − x, then the joint life mortality will be beta with the same ω − x and with α the sum of the individual α’s. µx 

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24. MULTIPLE LIVES: JOINT LIFE PROBABILITIES

458

Example 24C A flashlight requires three batteries. The force of failure for a fresh battery is µt 

1 8 − 4t

t