ARCH MLC

ARCH-MLC

Fall 2009 SOA MLC by Yufeng Guo, ASA

This electronic book is intended for individual buyer use for the sole purpose of preparing for Exam MLC. This book can NOT be resold to others. No part of this publication may be reproduced for resale or multiple copy distribution without the express written permission of the author.

Copyright Notice and Disclaimers This manual is ©2009 by Yufeng Guo. All Rights Reserved. This manual (and any portion thereof) may not be reproduced in any form without express written permission from the author. The author of this manual makes no warrant that this manual is free from errors. I will post an errata on my website (www.archactuarial.com). Please let me know of any errors you may find at [email protected]. Under no theory of law will the author be liable for any amount greater than the purchase price of this manual.

Arch MLC: Spring 2009 Yufeng Guo December 17, 2008

Chapter 0 Feature of Arch MLC Manual • Cover everything in the SOA MLC syllabus. • Thorough explanation of the core concepts with worked out problems • Reference to Actuarial Mathematics and Models for Quantifying Risk. You can use either textbook as a companion to go with Arch MLC. • Low Cost: $99. For $99, you bring a quality study manual home. • No shipping charge. Good saving for international exam candidates. • Convenience. After you buy Arch MLC PDF, you can print a hard copy. You can also install the PDF in your computer. • PDF has detailed bookmarks for quick reference. • PDF has a clickable table of contents for quick reference.

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Contents 0 INTRODUCTION The origin of Arch Manual . . . . . . . . . . . . . . . . Praises of the Arch Manual originally written by Nathan About Yufeng Guo . . . . . . . . . . . . . . . . . . . . . How to use this manual . . . . . . . . . . . . . . . . . .

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1 ACTUARIAL MATHEMATICS: CHAPTER TIONS AND LIFE TABLES 3.2.1 The Survival Function . . . . . . . . . . . 3.2.2 Time-until-Death for a Person Age x . . . 3.2.3 Curtate-Future-Lifetime . . . . . . . . . . 3.2.4 Force of Mortality . . . . . . . . . . . . . 3.3-3.5 Life Tables . . . . . . . . . . . . . . . . 3.5.2 Recursion Formulas . . . . . . . . . . . . 3.6 Assumptions for Fractional Ages . . . . . . 3.7 Some Analytical Laws of Mortality . . . . . Modified DeMoivre’s Law . . . . . . . . . . . . 3.8 Select and Ultimate Tables . . . . . . . . . Conclusion . . . . . . . . . . . . . . . . . . . . CHAPTER 3 Formula Summary . . . . . . . . Past SOA/CAS Exam Questions: . . . . . . . . Problems from Pre-2000 SOA-CAS exams . . . Solutions to Chapter 3 . . . . . . . . . . . . . .

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3 SURVIVAL DISTRIBU. . . . . . . . . . . . . . .

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2 ACTUARIAL MATHEMATICS CHAPTER 4 – LIFE INSURANCE 4.2 Insurances Payable at the Moment of Death . . . . . . . . . . . . . . . . . TYPES OF INSURANCE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Level Benefit Insurance . . . . . . . . . . . . . . . . . . . . . . . 4.2.2 Endowment Insurance . . . . . . . . . . . . . . . . . . . . . . . . 4.2.3 Deferred Insurance . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.4 Varying Benefit Insurance . . . . . . . . . . . . . . . . . . . . . . 4.3 Insurances Payable at the End of the Year of Death . . . . . . . . . . . . 4.4 Relationships between Insurances Payable at the Moment of death and the End of the Year of Death . . . . . . . . . . . . . . . . . . . . . . . . . . . . . CHAPTER 4 Formula Summary . . . . . . . . . . . . . . . . . . . . . . . . . Past SOA/CAS Exam Questions: . . . . . . . . . . . . . . . . . . . . . . . . . 3

9 9 10 10 11 13 15 16 19 20 21 27 28 33 35 35 38 39 43 56 58 61 62 63 63 66 68 70 73 79 81 83

Chapter 0 Problems from Pre-2000 SOA-CAS exams . . . . . . . . . . . . . . . . . . . . Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 ACTUARIAL MATHEMATICS: CHAPTER 5 LIFE ANNUITIES 5.2 Continuous Life Annuities . . . . . . . . . . . . . . . . . . . . . . . . The most important equation so far(!) . . . . . . . . . . . . . . . . . . . 5.3 Discrete Life Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Life Annuities with m-thly Payments . . . . . . . . . . . . . . . . . . CHAPTER 5 Formula Summary . . . . . . . . . . . . . . . . . . . . . . Continuous Annuities: . . . . . . . . . . . . . . . . . . . . . . . . Discrete annuities: . . . . . . . . . . . . . . . . . . . . . . . . . . Past SOA/CAS Exam Questions: . . . . . . . . . . . . . . . . . . . . . . Problems from Pre-2000 SOA-CAS exams . . . . . . . . . . . . . . . . . Solutions to Pre-2000 Problems: Chapter 5 . . . . . . . . . . . . . . . .

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4 ACTUARIAL MATHEMATICS: CHAPTER 6 BENEFIT 6.2 Fully Continuous Premiums . . . . . . . . . . . . . . . . . 6.3 Fully Discrete Premiums . . . . . . . . . . . . . . . . . . . 6.4 True m-thly Payment Premiums . . . . . . . . . . . . . . CHAPTER 6 Formula Summary . . . . . . . . . . . . . . . . Past SOA/CAS Exam Questions: . . . . . . . . . . . . . . . . Problems from Pre-2000 SOA-CAS exams . . . . . . . . . . . Solutions to Pre-2000 Exam Questions: Chapter 6 . . . . . .

PREMIUMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

135 136 142 147 151 152 168 170

5 ACTUARIAL MATHEMATICS: CHAPTER 7 BENEFIT 7.2 Fully Continuous Benefit Reserves . . . . . . . . . . . . . 7.3 Other Methods for Calculating the Benefit Reserve . . . . 1) Prospective Formula . . . . . . . . . . . . . . . . . 2) Retrospective Formula . . . . . . . . . . . . . . . . 3) Premium Di↵erence Formula . . . . . . . . . . . . . 4) Paid-Up Insurance Formula . . . . . . . . . . . . . 5) Other Reserve Formulas . . . . . . . . . . . . . . . 7.4 Fully Discrete Reserves . . . . . . . . . . . . . . . . . . . 7.5 Benefit Reserves on a Semi-Continuous Basis . . . . . . . 7.6 Benefit Reserves Based on True m-thly Benefit Premiums CHAPTER 7 Formula Summary . . . . . . . . . . . . . . . . Continuous Reserve Formulas: . . . . . . . . . . . . . . . . . . Discrete Reserves: . . . . . . . . . . . . . . . . . . . . Past SOA/CAS Exam Questions: . . . . . . . . . . . . . . . . Problems from Pre-2000 SOA-CAS exams . . . . . . . . . . . Solutions to Pre-2000 Questions: Chapter 7 . . . . . . . . . .

RESERVES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

173 174 177 177 177 180 180 181 183 186 187 188 188 188 190 194 196

6 ACTUARIAL MATHEMATICS: CHAPTER 8 ANALYSIS OF RESERVES 8.2 Benefit Reserves for General Insurances . . . . . . . . . . . . . 8.3 Recursion Relations for Fully Discrete Benefit Reserves . . . . 8.4 Benefit Reserves at Fractional Durations . . . . . . . . . . . . . Arch MLC, Spring 2009

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BENEFIT 197 . . . . . . 197 . . . . . . 202 . . . . . . 205 4

Chapter 0 8.5 The Hattendorf Theorem . . . . . . . . CHAPTER 8 Formula Summary . . . . . . Chapter 8 More Formulas . . . . . . . . . . Past SOA/CAS Exam Questions: . . . . . . Problems from Pre-2000 SOA-CAS exams . Solutions to Pre-2000 Problems: Chapter 8

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7 ACTUARIAL MATHEMATICS: CHAPTER 9 MULTIPLE TIONS 9.2 Joint Distributions of Future Lifetimes . . . . . . . . . . . . 9.3 Joint Life Status . . . . . . . . . . . . . . . . . . . . . . . . The following is important! . . . . . . . . . . . . . . . . 9.4 Last Survivor Status . . . . . . . . . . . . . . . . . . . . . . 9.5 More Probabilities and Expectations . . . . . . . . . . . . . 9.6 Dependent Lifetime Models . . . . . . . . . . . . . . . . . . 9.6.1 Common Shock –(Non-Theoretical Version) . . . . 9.7 Insurance and Annuity Benefits . . . . . . . . . . . . . . . . 9.7.1 Survival Statuses . . . . . . . . . . . . . . . . . . . 9.7.2 Special Two-Life Annuities . . . . . . . . . . . . . 9.7.3 Reversionary Annuities . . . . . . . . . . . . . . . 9.9 Simple Contingent Functions . . . . . . . . . . . . . . . . . CHAPTER 9 Formula Summary . . . . . . . . . . . . . . . . . Past SOA/CAS Exam Questions: . . . . . . . . . . . . . . . . . Problems from Pre-2000 SOA-CAS exams . . . . . . . . . . . . Solutions to Pre-2000 Exam Questions: Chapter 9 . . . . . . .

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208 212 213 214 234 237

LIFE FUNC241 . . . . . . . . 241 . . . . . . . . 243 . . . . . . . . 243 . . . . . . . . 248 . . . . . . . . 250 . . . . . . . . 253 . . . . . . . . 253 . . . . . . . . 256 . . . . . . . . 256 . . . . . . . . 262 . . . . . . . . 263 . . . . . . . . 265 . . . . . . . . 268 . . . . . . . . 270 . . . . . . . . 283 . . . . . . . . 285

8 ACTUARIAL MATHEMATICS: CHAPTER 10 MULTIPLE DECREMENT MODELS 287 10.2 Two Random Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 Probability density functions: . . . . . . . . . . . . . . . . . . . . . . . 289 10.3 Random Survivorship Group . . . . . . . . . . . . . . . . . . . . . . . . . 292 10.4 Deterministic Survivorship Group . . . . . . . . . . . . . . . . . . . . . . 292 10.5 Associated Single Decrement Tables . . . . . . . . . . . . . . . . . . . . . 296 10.5.1 Basic Relationships . . . . . . . . . . . . . . . . . . . . . . . . . 297 10.5.4 Uniform Distribution Assumption for Multiple Decrements . . . 298 10.6 Construction of a Multiple Decrement Table . . . . . . . . . . . . . . . . 300 CASE I : Two decrements that are uniformly distributed in the associated single decrement table . . . . . . . . . . . . . . . . . . . . . . . 300 CASE II : Three decrements that are uniformly distributed in the associated single decrement table . . . . . . . . . . . . . . . . . . . . . . 300 CASE III : Multiple decrements – some are uniformly distributed in the associated single decrement table and some are not. . . . . . . . . 300 CHAPTER 10 Formula Summary . . . . . . . . . . . . . . . . . . . . . . . . . 305 Past SOA/CAS Exam Questions: . . . . . . . . . . . . . . . . . . . . . . . . . 307 Problems from Pre-2000 SOA-CAS Exams . . . . . . . . . . . . . . . . . . . . 317 Solutions to Pre-2000 Problems: Chapter 10 . . . . . . . . . . . . . . . . . . . 320 Arch MLC, Spring 2009

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Chapter 0 9 ACTUARIAL MATHEMATICS: CHAPTER 11 APPLICATIONS OF MULTIPLE DECREMENT THEORY 323 11.2 Actuarial Present Values and Their Numerical Estimation . . . . . . . . 323 11.3 Benefit Premiums and Reserves . . . . . . . . . . . . . . . . . . . . . . . 324 CHAPTER 11 Formula Summary . . . . . . . . . . . . . . . . . . . . . . . . . 327 ARCH Sample Exam Problem . . . . . . . . . . . . . . . . . . . . . . . . . . 327 Solution: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327 Past SOA/CAS Exam Questions: . . . . . . . . . . . . . . . . . . . . . . . . . 329 10 ACTUARIAL MATHEMATICS: CHAPTER 15 INSURANCE MODELS INCLUDING EXPENSES 335 15.2 Expense Augmented Models . . . . . . . . . . . . . . . . . . . . . . . . . 335 15.4 More Expenses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338 15.6.1 Asset Shares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341 CHAPTER 15 Formula Summary . . . . . . . . . . . . . . . . . . . . . . . . . 343 Asset Shares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343 Past SOA/CAS Exam Questions: . . . . . . . . . . . . . . . . . . . . . . . . . 343 SOLUTIONS to Past SOA-CAS Exam Problems: . . . . . . . . . . . . . . . . 347 11 DANIEL CHAPTER 1 - MULTI-STATE TRANSITION MODELS FOR ACTUARIAL APPLICATIONS 351 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351 1.2 Non-homogeneous Markov Chains . . . . . . . . . . . . . . . . . . . . . . 354 CHAPTER 2 – CASH FLOWS AND THEIR ACTUARIAL PRESENT VALUES359 Section 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . 359 Cash Flows while in states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359 Cash Flows upon transitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364 Actuarial Present Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364 ARCH Warm-up Questions: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372 Past SOA/CAS Exam Questions: . . . . . . . . . . . . . . . . . . . . . . . . . 375 12 DANIEL STUDY NOTE ON POISSON PROCESS 5.3 The Poisson Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.1 Counting Processes . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.2 Definition of the Poisson Process . . . . . . . . . . . . . . . . . . 5.3.3 Interarrival and Waiting Time Distributions . . . . . . . . . . . . 5.3.4 Further Properties of Poisson Processes . . . . . . . . . . . . . . 5.3.5 Conditional Distribution of the Arrival Times . . . . . . . . . . . 5.4 Generalizations of the Poisson Process . . . . . . . . . . . . . . . . . . . . 5.4.1 Nonhomogeneous Poisson Process . . . . . . . . . . . . . . . . . 5.4.2 Compound Poisson Process . . . . . . . . . . . . . . . . . . . . . 5.4.3 Conditional or Mixed Poisson Processes: Gamma-Poisson Model CHAPTER 5 Formula Summary . . . . . . . . . . . . . . . . . . . . . . . . . ARCH Warm-up Problems: . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Past SOA/CAS Exam Questions: . . . . . . . . . . . . . . . . . . . . . . . . . Arch MLC, Spring 2009

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Chapter -1 13 ARCH Practice Exam 417 Answer Key for Practice Exam . . . . . . . . . . . . . . . . . . . . . . . . . . 430 14 SOLUTION TO MAY 2007 MLC

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Chapter -1

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Chapter 0

INTRODUCTION The origin of Arch Manual The Arch Manual was originally written by two gifted actuaries, Nathan Hardiman and Robin Cunningham. In the late 90’s, Nathan Hardiman and Robin Cunningham worked full-time at the former Je↵erson Pilot Financial. Nathan and Robin, like other exam candidates, faced the daunting challenge of plowing through difficult textbooks and mastering the fundamentals of life contingency theories and probability models to pass the Course 3 exam, the most difficult of the four preliminary actuarial exams and the exam with the highest failure rate. The difficulty of Course 3 was mainly due to its enormous scope. Candidates were required to read encyclopedia-like textbooks such as Actuarial Mathematics and Probability Models, gain sophisticated understanding of complex concepts such as multiple decrements, Markov Chain, Brownian motion, and be ready to tackle tricky word problems on the exam. Since Nathan and Robin both already had families and full-time jobs when they began studying for exams, they created their own study framework and philosophy for quickly passing Course 3. After passing the exam in one sitting using their unique study methods, Nathan and Robin decided to jointly write a new study guide that would enable candidates to build a core body of knowledge for Course 3 quickly. They wanted their manual to use straight talk and down-to-earth examples to explain difficult fundamental concepts intuitively and simply. Nathan and Robin published the first edition of their study manual for Course 3 in 2001. They named their study manual the Arch-3. Since its publication, Arch has been a popular study manual for Course 3 and Exam M. Arch’s power lies in its simplicity. While textbooks talk fancy, Arch talks simple. While textbooks rigorously prove theorems, Arch explains the intuition. While textbooks demand attention to everything, Arch separates the critical from the trivial. While Arch sells well, Nathan and Robin climbed corporate ladder higher and higher. With 9

Chapter 0 each day passing, they have less and less time to keep up with the SOA’s syllabus changes. Finally, in fall, 2006, Nathan and Robin decided to withdraw from the Arch manual business and passed on the copyright and ownership of Arch M manual to Yufeng Guo, who is the author of Deeper Understanding manuals for Exam P, FM, and M. Nathan and Robin’s contribution to actuarial education was not just the Arch manual but more importantly the Arch’s e↵ective teaching style. Before Arch was published, many thought that learning difficult things such as Course 3 ought to be slow and painful. Arch’s straight talk and down-to-earth examples showed the actuarial community that learning difficult actuarial theories can indeed by fast-paced and enjoyable.

Praises of the Arch Manual originally written by Nathan and Robin -I am a huge fan of ARCH. It is by far the best study manual and I am recommending it to all of my friends. I actually bought [several other manuals]. Now I think I wasted a whole set of money on the others since they always end up confusing me and I always have to come back to ARCH for clarification. -I start a seminar on Friday, and I never would have been able to finish and understand the material without your study guide. -I want to personally thank both of you for the fantastic and brilliant work that you did on ARCH. Seeing as it’s not my first time tackling this exam, I’ve had the chance to use [several other manuals]; however this is by far superior to all of those products. I have and will continue to recommend it to others in my company. -I would first like to say that I am very happy with your manual so far. I feel that I am progressing through the syllabus much faster than I would have without it, and the depth of understanding that I am on my own giving up due to my not using the texts themselves is more than compensated for by the excellent coverage of the important topics in your manual.

About Yufeng Guo Yufeng Guo was born in central China. After receiving his Bachelor’s degree in physics at Zhengzhou University, he attended Beijing Law School and received his Masters of law. He was an attorney and law school lecturer in China before immigrating to the United States. He received his Masters of accounting at Indiana University. He has pursued a life actuarial career and passed exams 1, 2, 3, 4, 5, 6, and 7 in rapid succession after discovering a successful study strategy. Mr. Guo’s exam records are as follows: Arch MLC, Spring 2009

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Chapter 0 Fall 2002 Spring 2003 Fall 2003 Spring 2004 Fall 2004 Spring 2005

Passed Passed Passed Passed Passed Passed

Course Course Course Course Course Course

1 2,3 4 6 5 7

Study guides by Mr. Guo: • Deeper Understanding, Faster Calc: P • Deeper Understanding, Faster Calc: FM • Deeper Understanding, Faster Calc: MLC • Deeper Understanding, Faster Calc: MFE • Deeper Understanding, Faster Calc: C • Guo’s Solution to Derivatives Markets: Exam FM • Guo’s Solution to Derivatives Markets: Exam MFE

In addition, Mr. Guo teaches online classes for Exam P, FM, MFE, and MLC. For details see http://www.guo.coursehost.com and http://www.myactuaryexam.com. If you have questions, you can email Mr. Guo at yufeng [email protected]. FAQ I notice you have two study guides for MLC: Arch MLC and Deeper Understanding MLC. What’s the di↵erence? Which one should I buy? Di↵erence: Arch MLC focuses on thoroughly explaining the core concepts. Deeper Understanding MLC focuses on teaching conceptual insights and calculation shortcuts. Which one to buy: If money is not an issue, consider buying both. If you want to buy only one guide, choose the one that better fits your need. For example, if you already have a study guide and want to learn calculation shortcuts, buy Deeper Understanding: MLC. If your goal, on the other hand, is to master basic concepts, buy Arch MLC.

How to use this manual The ARCH manual is designed and written in such a way as to help you learn the material as efficiently as possible. The material for the course is broken down into di↵erent chapters from the textbooks. The chapters are presented using down-to-earth explanations. In addition, I point out the critical concepts and formulas most likely to be tested. Each chapter of this manual contains plenty of examples with solutions. You are likely to benefit a great deal if every time you get to an example, you cover up the solution and Arch MLC, Spring 2009

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Chapter 0 attempt to work it. You will get many of them wrong, especially the first time you see them. But the problem-solving experience will be extremely valuable! On the exam, you will not be asked to explain anything. You will be asked to calculate numerical answers. Therefore, much of our explanation of the material is done by way of numerical examples and practice questions. Examples range from very simple ones (to make sure you know the basic concepts), to thought provoking ones (to help you think about what you’ve learned and really understand it), to exam questions from prior exams (to get you ready for exam day). I also suggest problems from the texts for you to work. Many of the problems in the text are not transferable to the exam. Some, however, provide useful insight and practice. Solutions to these suggested problems are available on the Download Samples page at www.archactuarial.com. A formula summary for each chapter is included. These summaries are intended to serve as a reference as you familiarize yourself with the syllabus material. Finally, there’s a full length practice exam of new questions. This practice exam is designed to be used in conjunction with the prior Course 3 and Exam M problems in the SOA website at www.soa.org. Make sure you work all of these exams! All materials contained herein are copyrighted by Yufeng Guo. This PDF study manual is for individual use for the sole purpose of taking Exam MLC. Reselling this manual is prohibited. Redistribution of this manual in any form is prohibited. You can purchase this manual at http://www.actuarialbookstore.com or http://www. actexmadriver.com Please check www.archactuarial.com for errata and answers to suggested text exercises.

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Chapter 1

ACTUARIAL MATHEMATICS: CHAPTER 3 SURVIVAL DISTRIBUTIONS AND LIFE TABLES • Option A reference: Actuarial Mathematics Chapter 3 • Option B reference: Models for Quantifying Risk Chapter 5,6 This text forms the heart and soul of the exam syllabus. The basic principles of life insurance (and annuities) are explained throughout the book. You need to have a solid understanding of this material in order to pass the exam. However, you do not need to understand the majority of the underlying theory in this text. The key points that a student must learn from this text are: KEYPOINTS: 1. Notation – much of this notation is new. While it can be confusing at first, there is some logic to it. It will help you to remember and understand the many symbols if you regularly translate the notation into words as you read. 2. Basic ideas – for example, chapter four introduces a variety of types of insurance. You will want to make sure you have an understanding of these di↵erent products and their benefit designs. Another key point is that there are many parallels. Again in Chapter 4, the first part of the chapter considers products which pay a benefit immediately upon death. The second part of the chapter considers the same products except that the benefit is paid at the end of the year in which death occurs. It is helpful to realize that you are really learning only one set of products, with a couple of benefit options, rather than two sets of products. These parallels run throughout the text (e.g., continuous vs curtate functions). 13

Chapter 1 3. Learn key formulas – there is no substitute for being able to recall the formula for, say, a net level premium reserve for term insurance. If you can do this for most of the formulas, you will be ready to answer questions quickly. This manual has tools to help you learn these formulas, so don’t feel overwhelmed! To the text!!!

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Chapter 1 Chapter 3 is all about notation, definitions, and a few basic ideas that are essential to life contingencies. If you can make yourself comfortable with the symbols and methods of Chapter 3, the rest of Actuarial Mathematics will be easier to absorb.

3.2.1 The Survival Function • Option A reference: Actuarial Mathematics Chapter 3.2.1 • Option B reference: Models for Quantifying Risk Chapter 5.1 Consider a newborn (i.e. a person whose attained age = 0). Definitions X = newborn’s age at death You can also think of X as “the future lifetime of a newborn”. Define F (x) = Pr (X  x), where x 0. Read as “the probability that death will occur prior to (or at) age x”. In statistics, F (x) is the cumulative distribution function for the future lifetime of a newborn. If y > x, it is always true that F (y) > F (x). This makes sense. For a newborn, F (98), the probability of dying before age 98, is greater than F (94), the probability of dying before age 94. Define s(x) = 1 F (X) = 1 Pr(X  x). The function s(x) is a survival function. Read it as “the probability that death does not occur by age x” or “the probability of attaining (surviving to) age x”. Pr(x < X  z) = probability that a newborn dies between ages x and z = F (z) = [1 = s(x)

F (x)

[1

s(z)]

s(x)]

s(z) s(x)

F(x)

the pdf y=f(x)

O

x

z

The figure shows the probability distribution function f (x) for death at age x. For any value of x, F (x) is equal to the area under the curve y = f (x) and to the left of x. Similarly s(x) is equal to the area under the curve and to the right of x. By the way, you may have noticed that in our discussion, we dropped the subscript X in FX (x) ... you can ignore it. I don’t know if the authors realize it but they are being a little Arch MLC, Spring 2009

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15

Chapter 1 inconsistent in their treatment of F and s! If two di↵erent random variables, say X and Y , referred to the future lifetimes of two di↵erent newborns, then you would need to keep the F and s straight for each kid. That’s all the subscript is indicating.

3.2.2 Time-until-Death for a Person Age x • Option A reference: Actuarial Mathematics Chapter 3.2.2 • Option B reference: Models for Quantifying Risk Chapter 5.3 Newborns are great, but if our pension and insurance companies are going to make money we need to be able to deal with people who are older than 0. So ... Consider a person with attained age = x. The simple F (x) and s(x) functions no longer work, since we are now dealing with a person who has already survived to age x. We are facing a conditional probability situation. Pr(x < X  z|X > x) = probability that person living at age x will die between ages x and z = the probability that an x-year-old will die before turning z [F (z) F (x)] [1 F (x)]

= =

[s(x) s(z)] [s(x)]

Why is this a conditional probability? Because it is the probability that a newborn will die before age z given that the newborn survives to age x. EXAMPLE: 1. Write two expressions (one with F only and one with s only) for the probability that a newborn dies between 17 and 40, assuming the newborn dies between 10 and 40. 2. Interpret the following expression in English (or the language of your choice!). S(20) S(35) 1 S(80) SOLUTION: 1.

Arch MLC, Spring 2009

F (40) F (40)

F (17) F (10)

or

s(17) s(10)

c Yufeng Guo

s(40) s(40) 16

Chapter 1 2. “The probability of death between ages 20 and 35, given that the newborn will not attain age 80.” } Now, let the symbol “(x)” represent a person age x and let T (x) be the future lifetime of a person age x. (So T (25) is the future lifetime of (25), a twenty-five-year-old.) Two basic probability functions exist regarding T (x):

t qx

= probability that person age x will die within t years = Pr[T (x)  t] where t

t px

0

= probability that person age x will survive at least t years = Pr[T (x) > t] where t

0

q t x x

tp x

Age

x+t

In the figure, t qx is the probability that (x)’s death will occur in the age-interval (x, x+t), and t px is the probability that (x)’s death will occur in the age interval (x + t, !). (! represents the oldest possible age to which a person may survive.) Useful notes: • t p0 is just s(t). • If t = 1, the convention is to drop the symbol 1, leaving us with either px or qx . Remember, these are the two basic functions. The formulas that follow are simply take-o↵s on t px or t qx which you will learn with practice. The symbol t|u qx

represents the probability that (x) (that is, a person age x) survives at least t more years, but dies before reaching age x + t + u. This is equal to each of the following expressions, each of which you want to be able to put into words: Pr[t < T (x)  t + u] t+u qx t px

t qx t+u px

(As with qx and px , if u = 1, we drop it, leaving t| qx , the probability that (x) will survive t years but not t + 1 years.) Arch MLC, Spring 2009

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17

Chapter 1 Useful formulas: t px

t qx

t|u qx

x+t p0

=

x p0

=

s(x + t) s(x)

s(x + t) s(x)

=1

s(x + t + u) s(x) s(x + t) s(x + t) s(x + t + u) = ⇤ s(x) s(x + t) = t px ⇤ u qx+t =

s(x + t)

This last equation makes sense. It says “The probability of (x) dying between t and t + u years from now (t|u qx ) is equal to the probability that (x) will first survive t years (t px ) and then die within u years (u qx+t ).” If you don’t remember anything else from the above, remember the following! t px

=

s(x + t) s(x)

CONCEPT REVIEW: 1. Write the symbol for the probability that (52) lives to at least age 77. 2. Write the symbol for the probability that a person age 74 dies before age 91. 3. Write the symbol for probability that (33) dies before age 34. 4. Write the symbol for probability that a person age 43 lives to age 50, but doesn’t survive to age 67. 5. Write

5|6 qx

in terms of F and then in terms of p.

SOLUTIONS: 1.

25 p52

5.

5|6 qx

2.

=

17 q74

3. q33

4.

7|17 q43

s(x + 5) s(x + 11) F (x + 11) F (x + 5) = s(x) 1 F (x)

= 5 px (1

6 px+5 )

or = 5 px

11 px .

}

To help memorize symbols, practice translating symbols into words and express words in symbols. You can also make flash cards and quiz yourself. Arch MLC, Spring 2009

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18

Chapter 1

3.2.3 Curtate-Future-Lifetime • Option A reference: Actuarial Mathematics Chapter 3.2.3 • Option B reference: Models for Quantifying Risk Chapter 5.3.6 Suppose a person born on Jan 1, 1900 died on Sept 30, 1990. How old was he at death? The true age was about 90.75 years old. The curtate age was 90. To find the curtate age, first find the true age. Next, throw away all the decimals and keep the integer. If there’s no decimal, then the curtate age is equal to the continuous (true) age. For example, if T (x) = 90, then K(x) = 90. (This book and others use ‘Curtate’ and ‘Discrete’ interchangeably.) Previously, we defined T (x) to be the future lifetime of (x). This is a continuous function. Now we define K(x) = curtate future lifetime of (x) = greatest integer in T (x) = number of future years completed by (x) prior to death = number of future birthdays (x) will have the opportunity to celebrate

A couple of formulas apply: Pr(K(x) = k) = Pr(k  T (x) < k + 1) = Pr(k < T (x)  k + 1) =

k px

=

k px

=

k| qx

k+1 px

⇤ qx+k

(Remember, the 1 in front of q has been dropped.) EXAMPLE: If s(x) =

100 x 100

for every x, what is the probability that K = 19 for (18)?

SOLUTION: Pr(K(18) = 19) = 19| q18 = =

Arch MLC, Spring 2009

63

62 82

=

s(37) s(38) s(18)

1 . 82

c Yufeng Guo

}

19

Chapter 1

3.2.4 Force of Mortality • Option A reference: Actuarial Mathematics Chapter 3.2.4 • Option B reference: Models for Quantifying Risk Chapter 5.1.4 The force of mortality can be thought of as the probability of death at a particular instant given survival up to that time. This is an instantaneous measure, rather than an interval measure. There is good bit of theory in this section, but the most important items are the following formulas and the table of relationships. f (x) s0 (x) = 1 F (x) s(x)

µ(x) =

(3.2.13)

It is very important to know the relationships and requirements given in Table 3.2.1. These will probably be tested on the exam. Below is a summary of the useful information in this table. Each row shows 4 ways to express the function in the left column. F (x) F (x)

F (x) 1

s(x)

f (x)

s(x) 1

F (x)

Rx

s(x)

0

R1

s(x)

f (u) du

F 0 (x)

s0 (x)

µ(x)

F 0 (x) 1 F (x)

s0 (x) s(x)

1

f (u) du

x

f (x)

µ(x) e e

f (x)

Rx 0

Rx

f (x) s(x)

µ(t) dt

0

µ(x) e

µ(t) dt

Rx 0

µ(t) dt

µ(x)

EXAMPLE: Constant Force of Mortality If the force of mortality is a constant µ for every age x, show that 1. s(x) = e

µx

2. t px = e

µt

SOLUTION: 1. s(x) = e 2. t px

Arch MLC, Spring 2009

=

Rx 0

µ dt

=e

s(x + t) =e s(x)

c Yufeng Guo

µx

µt

.

.

}

20

Chapter 1

3.3-3.5 Life Tables • Option A reference: Actuarial Mathematics Chapter 3.3-3.5 • Option B reference: Models for Quantifying Risk Chapter 6 Life Table is widely used actuarial practice. Even today, Life Tables are often loaded into systems for calculating reserves, premium rates, and the surrender cash value of an insurance policy. Learning Life Tables will not only help you pass Exam MLC, it also helps you when you become an actuary. Definitions: l0 = number of people in cohort at age 0, also called the “radix” li = number of people in cohort at age i (those remaining from the original l0 ) ! = limiting age at which probability of survival = 0 (s(x) = 0 for all x n dx

!)

= number alive at age x who die by age x + n

Relationships: lx = l0 ⇤ s(x) lx lx+1 qx = lx lx lx+n n qx = lx lx+n n px = lx d = l lx+n n x x

Illustrative Life Table: Basic Functions Age lx dx 1,000 qx 0 100,000.0 2,042.2 20.4 1 97,957.8 131.6 1.4 2 97,826.3 119.7 1.2 3 97,706.6 109.8 1.1 .. .. .. .. . . . . 40 93,131.6 259.0 2.8 41 92,872.6 276.9 3.0 42 92,595.7 296.5 3.2 43 92,299.2 317.8 3.4 Arch MLC, Spring 2009

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21

Chapter 1 EXAMPLE: Life Table Mortality Above is an excerpt from the Illustrative Life Table in the book. The following questions are all based on this excerpt. 1. Find s(42). 2. Find

40 d2 .

3. Find

38 q3 .

4. Find 2| q40 . SOLUTION: 1. s(42) = 92,595.7 100,000 = 0.925957. 2. 40 d2 = l2 l42 = 5230.6 38 q3

=1

4.

2| q40

= 2 p40 · q42 = 92,595.7 93,131.6 (0.0032) = 0.003182.

38 p3

=1

l41 l3

=1

92,872.6 = 0.04947. 97,706.6

3.

}

Concepts which follow from the Life Table: Based on Equation (3.2.13) on an earlier page, we can determine that the probability density function f (t) for T (x) is given by f (t) = t px µ(x + t). This says that the probability that (x) will die at age x + t, symbolized by f (t), is equal to the probability that (x) will survive t years and then be hit at that instant by the force of mortality. Among other things, this tells us that Z 1 0

t px µ(x + t)dt =

Z 1

f (t)dt = 1.

0

The complete-expectation-of-life is the expected value of T (x) (or E[T (x)] for fans of Statistics) and is denoted ex . If you remember how to find the expected value of a continuous random variable, you can figure out that ex = E[T (x)] = Var[T (x)] = 2

Z 1 0

Z 1 0

t px dt

t · t px dt

2

ex

(3.5.4)

The book shows how to figure both of these formulas out with integration by parts in Section 3.5.1. I suggest that you memorize these two expressions. The median future lifetime of (x) is denoted m(x) and simply represents the number m such that m px = m qx . In other words, it is the number of years that (x) is equally likely to survive or not survive. It can be found by solving any of the following: Pr[T (x) > m(x)] = Arch MLC, Spring 2009

c Yufeng Guo

1 2 22

Chapter 1 or

s[x + m(x)] 1 = s(x) 2

or

1 = . 2

m px

The curtate-expectation-of-life is E[K(x)] and is denoted ex (no circle). To remember the di↵erence between ex and ex , remember “life is a continuous circle.” So a circle means continuous. ex = E[T (x)] and ex = E[K(x)]. Here are the formulas, note the Continuous/Curtate parallel: ex = E[K(x)] =

1 X

k px

1

1 X

Var[K(x)] =

(2k

1) · k px

1

e2x

EXAMPLE: Constant Force of Mortality Find e0 and e50 if the force of mortality is a constant µ. SOLUTION: e0 =

Z 1

e50 =

Z 1

0

0

t p0 dt =

t p50 dt

Z 1

e

µt

dt =

0

=

Z 1

µt

e



dt =

0

1 e µ



µt

1

=

0

1 e µ

µt

1 0

1 µ

=

1 µ

If the force of mortality is constant, your future expected lifetime is the same whether you are 0 (a newborn) or 50. } EXAMPLE: DeMoivre’s Law for Mortality (We’ll learn DeMoivre later in this chapter.) If s(x) =

(

0 < x < 50 Otherwise

50 x 50

0

for all x between 0 and 50, find e0 and e45 . SOLUTION: e0 =

50 X 1

Arch MLC, Spring 2009

t p0

=

50 X 50 1

50

t

= 50

c Yufeng Guo

50 1 X t 50 1

23

Chapter 1 1 (50)(51) = 24.5 50 2

= 50

e45 =

5 X

t p45 =

1

=

5 X s(45 + t)

s(45)

1

=

5 X 5 1

t 5

4+3+2+1+0 = 2. 5

}

More Life Functions: The expression Lx denotes the total expected number of years, full or fractional, lived between ages x and x + 1 by survivors of the initial group of l0 lives. Those who survive to x + 1 will live one year between x and x + 1, contributing one full year to Lx . Those who die during the year will contribute a fraction of a year to Lx . Lx =

Z 1 0

lx+t dt

The expression mx is the central death rate over the interval x to x + 1. Make sure not to confuse mx with m(x), the median future lifetime! mx =

(lx

lx+1 ) Lx

Lx and mx can be extended to time periods longer than a year:

n Lx

=

n mx

=

Z n 0

lx

lx+t dt

lx+n L n x

The remaining of Section 3.5.1 has obscure symbols Tx and ↵(x). They rarely show up in the exam. Don’t spend too much time on them. Let Tx be the total number of years lived beyond age x by the survivorship group with l0 initial members (i.e. the lx people still alive at age x). Be careful with notation. This is not T (x), the future lifetime of (x). Tx =

Z 1 0

(3.5.16)

lx+t dt

Note from the definitions that you can think of Tx as

1 Lx .

The final symbol is ↵(x). It’s the expected death time given x dies next year. Arch MLC, Spring 2009

c Yufeng Guo

24

Chapter 1 R 1 lx+t R1 R1 R1 t· l µ(x+t)dt tf (t)dt tt px µ(x+t)dt t·lx+t µ(x+t)dt 0 0 0 x ↵ (x) = E [T |T < 1] = R 1 = R1 = R 1 lx+t = R0 1 0

f (t)dt

0

t px µ(x+t)dt

0

lx

µ(x+t)dt

0

lx+t µ(x+t)dt

R1 R1 R t·cdt tdt 0 If UDD, then f (t) = qx = c is a constant. Then ↵(x) = R 1 = R01 = 01 tdt = 12 . cdt dt 0

0

EXAMPLE: Constant Force of Mortality If l0 = 1000 and the force of mortality is a constant µ = 0.1, find (A) L5 (B) m5 (C) T5

Arch MLC, Spring 2009

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25

Chapter 1 SOLUTION: (A) L5 =

Z 1 0

l5+t dt =

Z 1 0

t p5 l5 dt.

Since l5 = l0 e we have L5 = 606.5

Z 1

µ·5

= 1000e

e

0.1t

h

(B) l5

l6 L5

=

= 606.5,

dt = 606.5

0

= 606.5 10(1

m5 =

0.5

e

0.1

i

h

10e

0.1t

i1 0

) = 577.16.

606.5 548.8 = 0.10 577.16

This approximates the rate at which people were dying between the 5th and 6th years. (C) T5 =

Z 1

1000e

0.1(5+t)

dt = 606.5

0

Z 1

e

0.1t

dt = 6065

0

So if we add up all of the time lived by each of the people alive at t = 5, we expect to get a total of 6065 years, or 10 years per person. } Relationship: Tx = ex lx This relationship makes sense. It says that the average number of years lived, ex , by the members of lx is equal to the total number of years lived by this group divided by lx . We can determine the average number of years lived between x and x + n by the lx survivors at age x as: n Lx

lx n Lx

lx

=

Z n 0

t px dt

= n-year temporary complete life expectancy of (x) = ex:n (p.71)

Arch MLC, Spring 2009

c Yufeng Guo

26

Chapter 1

3.5.2 Recursion Formulas • Option A reference: Actuarial Mathematics Chapter 3.5.2 • Option B reference: Models for Quantifying Risk Chapter 6 These are basically ways to avoid working integrals. They are based on the Trapezoid Rule for integration – maybe you remember the trapezoid rule from calculus. Backward: u(x) = c(x) + d(x) ⇤ u(x + 1) Forward: u(x + 1) =

u(x) c(x) d(x)

Note that the Forward Method is simply an algebraic recombination of the Backward Method. Note also that this Forward formula is di↵erent from the book – work out the formulas yourself to convince yourself of their equivalence. Then, learn whichever form you find more straightforward. The text shows how to use these formulas to compute ex and ex starting with e! and e! and working backward. For ex , using the recursion once will produce e! 1 , the second iteration will produce e! 2 , etc. until you get all the way back to e0 , when you will have produced a list of ex for every x between 0 and !. The formulas are: for ex , u(x) = ex c(x) = px d(x) = px Starting Value = e! = u(!) = 0 So to start, set x + 1 = ! and the recursion will produce u(x) = u(!

1).

For ex , u(x) =ex c(x) =

Z 1 0

s px ds

d(x) = px Starting Value =e! = u(!) = 0

Arch MLC, Spring 2009

c Yufeng Guo

27

Chapter 1

3.6 Assumptions for Fractional Ages • Option A reference: Actuarial Mathematics Chapter 3.6 • Option B reference: Models for Quantifying Risk Chapter 6.5 (OK, you can start paying attention again ....) The random variable T is a continuous measure of remaining lifetime. The life table has been developed as an approximation of T , using a curtate variable K. As we’ve discussed, K is only defined at integers. So, we need some way to measure between two integer ages. Three popular methods were developed. For all of the methods that follow, let x be an integer and let 0  t  1. Suppose that we know the value of s(x) for the two integers x and x + 1 and we want to approximate s at values between x and x+1. In other words, we want to approximate s(x+t) where 0  t  1. Method 1: Linear Interpolation: s(x + t) = (1

t)s(x) + t · s(x + 1)

This method is also known as “Uniform Distribution of Deaths”, or UDD. Under UDD, s(x + t) and t px are both straight lines between t = 0 to t = 1. This method assumes that the deaths occurring between ages x and x + 1 are evenly spread out between the two ages. As you might imagine, this is usually not quite correct, but is a pretty good approximation. (Please note: the linearity of s(x + t) and t px is only assumed to hold up to t = 1!) One key formula for UDD you might want to memorize is: f (t) = qx To see why, please note that the number of deaths from time zero to time t is a fraction of the total deaths in a year s(x)

s(x + t) = t[s(x)

s(x + 1)]

Here for convenience we interpret s(x + t) as the number of people alive at age x + t. For example, if s(x + 0.5) = 0.9, we say that for each unit of people at age x, we have 0.9 unit of people at age x + 0.5, with one unit being one billion, one million, or any other positive constant. t qx

=

s(x)

s(x + t) s(x) s(x + 1) =t = tqx s(x) s(x) f (t) =

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d t qx = qx dt

c Yufeng Guo

28

Chapter 1 You can also come up with f (t) = qx by intuitive thinking. Under UDD, death occurs at a constant speed. If 12 people died in one year, then one person died each month. So f (t) must be a constant. Then: qx =

Z 1

f (t)dt = f (t)

0

Z 1

dt = f (t)

0

Method 2: Exponential Interpolation: Forget about the complex formula: log s(x + t) = (1

t) log s(x) + t · log s(x + 1)

All you need to know is that under the constant force of mortality, µ(x + t) = µ for 0  t  1. Method 3: Harmonic Interpolation: This method is more commonly called the “Balducci assumption” or the “Hyperbolic assumption.” Forget about the complex formula 1 1 t t = + s(x + t) s(x) s(x + 1) All you need to know about Balducci assumption is this: 1 t qx+t

= (1

t)qx

The above formula says that if you are x + t years old (where 0  t  1), then your chance of dying in the remainder of the year is a fraction of your chance of dying in the whole year. You can derive all the other formulas using 1 t qx+t = (1 derive other formulas in Balducci assumptions.

Function t qx

Uniform Distribution tqx

µ(x + t)

qx 1 tqx

1 t qx+t

(1 t)qx 1 tqx

y qx+t

yqx 1 tqx

1

t px t px µ(x

+ t)

Arch MLC, Spring 2009

tqx qx

Constant Force 1 ptx log px 1 1

p1x

t

pyx ptx

ptx log px c Yufeng Guo

t)qx . Later I’ll show you how to

Hyperbolic tqx 1 (1 t)qx qx 1 (1 t)qx

(1

t)qx

yqx 1 (1 y t)qx px 1 (1 t)qx qx px [1 (1 t)qx ]2

29

Chapter 1 Table 3.6.1 Table 3.6.1 summarizes UDD, constant force of mortality, and the Balducci assumption. Don’t try to memorize the whole table. Learn basic formulas and derive the rest on the spot. Under UDD, for 0  t  1: t qx

f (t) = qx = constant =

Z t

f (t)dt =

0

Z t 0

t px

=1

qx dt = qx

t qx

µ(x + t) = µx (t) =

=

=1

dt = tqx

tqx

f (t) qx = 1 tqx t px

s(x + t + y) s(x)t+y px = = s(x + t) s(x)t px y qx+t

0

+ t) = f (t) = qx

t px µ(x y px+t

Z t

=1

y px+t

1 t qx+t

=

t+y px

t px yqx = 1 tqx

(1 1

=

1

(t + y)qx 1 tqx

t)qx tqx

Under constant force of mortality, for 0  t  1: µx (t) = µ t px

=e

Rt 0

µdt

px = e µ= t px

=e

y px+t =

Finally, let’s derive log s(x + t) = (1

µt

t+y px t px

=e

µt

µ

ln px

= (e =

µ t

) = (px )t

(px )t (t+y)

(px )

= (px )y

t) log s(x) + t · log s(x + 1) s(x + t) = s(x)e

µt

) log s(x + t) = log s(x) s(x + 1) = s(x)e

) (1

µ

) log s(x + 1) = s(x)

t) log s(x) + t log s(x + 1) = (1 ) log s(x + t) = (1

Arch MLC, Spring 2009

µt

µ

t) log s(x) + t log s(x)

µt = log s(x)

µt

t) log s(x) + t log s(x + 1)

c Yufeng Guo

30

Chapter 1 Under Balducci assumption: The starting point of Balducci assumption is 1 t qx+t

= (1

t)qx

You can derive all the other formulas from this starting point. For example, let’s derive the formula for t px . T (x) Age Number of people alive This is how to derive s(x + t) =

0 x s(x) = 1

t x+t s(x + t) = 1

px (1 t)qx

1 x+1 s(x + 1) = px

px 1 (1 t)qx .

First, we set the starting population at s(x) = 1 for convenience. You can set it to any positive constant and get the same answer. After setting s(x) = 1, we’ll have s(x + 1) = px . This is because px = s(x+1) s(x) . Next, let’s find s(x + t) using the formula 1 t qx+t

=1

1 t qx+t

s(x + 1) =1 s(x + t) s(x + t) =

However, t px =

s(x+t) s(x)

1

= (1

t)qx .

px = (1 s(x + t) px (1 t)qx

= s(x + t). This gives us: t px =

t)qx

px 1 (1 t)qx .

Next, let’s derive the formula f (t) = t px µ(x + t) = t px µx (t) f (t) =

d t px = dt

d dt 1

Derive µ(x + t): µ(x + t) =

f (t) = 1 t px

Derive y qx+t : y qx+t

px = (1 t)qx [1

=1

q x px (1 t)qx ]2

qx (1 t)qx

s(x + t + y) s(x + t)

Use the formula: s(x + t) = 1 (1px t)qx Replace t with t + y, assuming 0  t + y  1: s(x + t + y) = Arch MLC, Spring 2009

1

(1

px t

c Yufeng Guo

y)qx 31

Chapter 1

y qx+t

=1

Finally, let’s derive

s(x + t + y) =1 s(x + t)

1 s(x+t)

=

1 t s(x)

+

1 1

px (1 t

y)qx

t s(x+1)

s(x + t) = 1 1 = s(x + t)

(1 t)qx = (1 t y)qx 1

(1 t)qx 1 = px

(1

1

px (1 t)qx

t)(1 px

px )

=

(1

t)px + t = (1 px

t) +

t px

Since s(x) = 1 and s(x + 1) = px , we have: 1 t t + =1 s(x) s(x + 1) )

t+

t px

1 1 t t = + s(x + t) s(x) s(x + 1)

Now you see that you really don’t need to memorize Table 3.6.1. Just memorize the following: • Under UDD, f (t) = qx is a constant. • Under constant force of mortality, µ(x + t) = µ. • Under Balducci,

1 t qx+t

= (1

t)qx .

A couple of time-saving formulas that are valid under UDD only!! ex = ex +

1 2

Var(T ) = Var(K) +

1 12

EXAMPLE: You are given that qx = 0.1. Find (A)

0.5 qx ,

(B)

0.5 qx+0.5

under each of

• Uniform Density of Deaths (UDD)

• Exponentially distributed deaths (constant force) • Harmonic Interpolation (Balducci, hyperbolic) SOLUTION: (A)

• UDD:

Arch MLC, Spring 2009

0.5 qx

= (0.5)(0.1) = 0.05

c Yufeng Guo

32

Chapter 1 • CF:

0.5 qx

• Balducci: (B)

0.5 qx

=

• UDD: 0.5 qx+0.5

• CF:

=1

(0.5)(0.1) = 0.0526. 1 (0.5)(0.1) =

0.5 qx+0.5

• Balducci:

(0.9)0.5 = 0.0513

(0.5)(0.1) = 0.0526 1 (0.5)(0.1)

=1

0.5 qx+0.5

(0.9)0.5 = 0.0513

= (0.5)(0.1) = 0.05.

}

3.7 Some Analytical Laws of Mortality • Option A reference: Actuarial Mathematics Chapter 3.7 • Option B reference: Models for Quantifying Risk Chapter 5.2 Although computers have rendered Analytical Laws of Mortality less imperative to our profession, they are still important for understanding mortality and particularly for passing the exam. The book describes four basic analytical laws/formulas. Of these 4 laws, De Moivre’s Law is frequently tested in the exam. The other 3 laws are rarely tested in the exam (I would skip these 3 laws). De Moivre’s Law: µ(x) = (!

x)

1

and s(x) = 1

x , !

where 0  x < !

Gompertz’ Law: µ(x) = Bcx and s(x) = exp[ m(cx where B > 0, c > 1, m =

1)]

B , x log c

0

Makeham’s Law: µ(x) = A + Bcx and s(x) = exp[ Ax where B > 0, A

B, c > 1, m =

m(cx B , x log c

1)] 0

Weibull’s Law: µ(x) = kxn and s(x) = exp( uxn+1 ) Arch MLC, Spring 2009

c Yufeng Guo

33

Chapter 1 k , x (n + 1)

where k > 0, n > 0, u =

0

Notes: • Gompertz is simply Makeham with A = 0. • If c = 1 in Gompertz or Makeham, the exponential/constant force of mortality results. • In Makeham’s law, A is the “accident hazard” while Bcx is the “hazard of aging”. We will be seeing more of these laws later in the book. To be ready for the exam, you should become intimately familiar with De Moivre’s Law. DeMoivre’s Law says that at age x, you are equally likely to die in any year between x and !. Here are some key life-functions for DeMoivre’s Law in the form of an example. If you don’t read the proofs, still make an e↵ort to understand the formulas and what they mean. EXAMPLE: Under DeMoivre’s law, show that 1. t px

=

!

x

2. =

t qx

t

e0 = 4. e0 =

!

ex =

!

5.

ex =

Arch MLC, Spring 2009

!

x

!

3.

6.

t

!

x

! 2 1 2 x 2 x 2

c Yufeng Guo

1

34

Chapter 1 SOLUTION: 1. t px

=

s(x + t) ! x t = s(x) ! x

2. t qx

3. e0 = 4. e0 =

! X

Z ! ! 0

k p0

=

1

5. ex =

=1

t !

dt =

! X ! 1

Z ! x ! 0

t px

k !

"

!

(!

x

t x

#

t)2 ! = 2! 2 ! 1X ! 1 k= ! 1 2

=!

x !

t

=

dt =

!

x 2

6. Since mortality is uniform over all years under DeMoivre’s law, it is uniform over each individual year, so UDD applies. Therefore ex =ex

1 ! = 2

x 2

1

We could have done (4) this way also.

}

Modified DeMoivre’s Law Often on the exam, a modified version of DeMoivre’s Law arises. This occurs, for example, when c µ(x) = ! x where c is a positive constant. This gives rise to a set of formulas for each of the quantities found for standard DeMoivre’s Law (c = 1) in the example above. All of the formulas for Modified DeMoivre’s Law are listed in the formula summary at the end of this chapter.

3.8 Select and Ultimate Tables • Option A reference: Actuarial Mathematics Chapter 3.8 • Option B reference: Models for Quantifying Risk Chapter 6.6 Arch MLC, Spring 2009

c Yufeng Guo

35

Chapter 1 Suppose you are trying to issue life insurance policies and two 45 year-old women apply for policies. You want to make sure you charge appropriate premiums for each one to cover the cost of insuring them over time. One of the women is simply picked from the population at large. The second women was picked from a group of women who recently passed a comprehensive physical exam with flying colors – significantly healthier than the general population. Is it equitable to charge the same premium to the two women? No – because you have additional information about the second woman that would cause you to expect her to have better “mortality experience” than the general population. Thus, to her premiums, you might apply a “select” mortality table that reflects better the mortality experience of very healthy 45 year olds. However, after 15 years, research might show that being very healthy at 45 does not indicate much of anything about health at age 60. So, you might want to go back to using standard mortality rates at age 60 regardless of status at age 45. After all, 15 years is plenty of time to take up smoking, eat lots of fried foods, etc. This simple scenario illustrates the idea behind select and ultimate tables. For some period of time, you expect mortality to be di↵erent than that for the general population – the “select period”. However, at some point, you’re just not sure of this special status anymore, so those folks fall back into the pack at some point – the “ultimate” table. Consider table 3.8.1. The symbol [x] signifies an x-yr old with “select” status. Note that for the first two years (columns 1,2), select mortality applies with q[x] and q[x]+1 . However, at duration 3 (column 3), it’s back to standard mortality, qx+2 . This table assumes the “selection e↵ect” wears o↵ in just 2 years. Excerpt from the AF80 Select-and-Ultimate Table in Bowers, et al. (1) (2) (3) (4) (5) (6) (7) [x] 1,000 q[x] 1,000 q[x]+1 1,000 qx+2 l[x] l[x]+1 lx+2 x+2 30 0.222 0.330 0.422 9,906.74 9,904.54 9,901.27 32 31 0.234 0.352 0.459 9,902.89 9,900.58 9,897.09 33 32 0.250 0.377 0.500 9,898.75 9,896.28 9,892.55 34 33 0.269 0.407 0.545 9,894.29 9,891.63 9,887.60 35 34 0.291 0.441 0.596 9,889.45 9,886.57 9,882.21 36 Table 3.8.1 Here are a few useful formulas and relationships. In general,

q[x] < q[x

1]+1

< qx

Why would this hold? The expression qx represents a pick from the general population. The expression q[x] indicates special knowledge about the situation – for example, recently passing a physical exam (This formula assumes we are trying to select out healthy people, of course). The expression q[x 1]+1 indicates special knowledge about the situation as before – for example, recently passing a physical exam, but this time the applicant has had a year Arch MLC, Spring 2009

c Yufeng Guo

36

Chapter 1 for health to deteriorate since she was examined at age x age x.

1 (one year ago) rather than at

EXAMPLE: Select and Ultimate Life Table Using the select and ultimate life table shown above find the value of 1000

Arch MLC, Spring 2009

⇣

3 q32

⌘

3 q[32] .

c Yufeng Guo

37

Chapter 1 SOLUTION: deals only with the ultimate table so I am only interested in the values of lx + 2 in Column (6). 3 q32

3 q32

=1

3 p32

=1

9,888 = 0.001313 9,901

l35 =1 l32

For 3 q[32] , we need l[32] and l[32]+3 which is just l35 since the select period is only 2 years. So l35 9,888 =1 = 0.001111 3 q[32] = 1 l[32] 9,899 So the answer is 0.202.

}

Two important points regarding this example: • The probability that [32] will die in the next 3 years is lower if [32] is taken from a select group. People you are sure are healthy should be less likely to die than someone drawn from the general population. • To follow the people alive from the 9898.75 ‘selected’ at age 32, you first follow the numbers to the right until you hit the ultimate column and then proceed down the ultimate column. This is useful! You can quickly evaluate that 5 p[31]

=

9882 9903

by counting o↵ 5 years – 2 to the right and then 3 down.

Conclusion Chap. 3 introduces a lot of new concepts and notation. Make sure you understand the notation in Table 3.9.1 – this is the foundation for the rest of the text. Chapter 3 Suggested Problems: 1(do first row last), 5, 6, 7, 9, 12, 18abc, 20, 28, 30, 36, 39 There are lots for this chapter, some chapters in this book will have very few. (Solutions are available at archactuarial.com on the Download Samples page.)

Arch MLC, Spring 2009

c Yufeng Guo

38

Chapter 1

CHAPTER 3 Formula Summary s(x) = 1

F (X) = 1

Pr(x < X  z) = F (z) Pr(x < X  z|X > x) =

t px

h R t

=e

t|u qx

0

µ(x+s)ds

i

t px

= t px ⇤ u qx+t

t|u qx

=

Pr(X  x)

F (x) = s(x)

[F (z) F (x)] [s(x) s(z)] = [1 F (x)] [s(x)]

s(x + t) s(x)

= t+u qx

s(z)

=1

t qx

t qx

t|u qx

Pr(K(x) = k) =

k px

=

k px

=

k| qx

t px

= t px

t+u px

k+1 px

⇤ qx+k

Life Tables: lx = l0 ⇤ s(x) qx =

n qx

lx

=

lx+1 lx

lx

lx+n lx

n dx

= lx

px =

n px

lx+n

lx+1 lx =

lx+n lx

Constant Force of Mortality If the force of mortality is a constant µ for every age x, s(x) = e ex =

Arch MLC, Spring 2009

1 µ

µx

=e

µt

Var[T ] =

1 µ2

t px

c Yufeng Guo

39

Chapter 1 Expected Future Lifetime ex = E[T (x)] =

Z 1

ex = E[K(x)] =

Var[T (x)] = 2

t px dt

0

1 X

Z 1 0

Var[K(x)] =

k px

1

1 X

(2k

1

t · t px dt

2

ex

1) · k px

e2x

Under UDD only, 1 2

ex = ex +

Var(T ) = Var(K) +

1 12

Median future lifetime Pr[T (x) > m(x)] =

1 2

s[x + m(x)] 1 = s(x) 2

m px

=

1 2

Make sure not to confuse mx with m(x), the median future lifetime! Z 1

Lx =

0

mx = Tx =

n Lx =

lx+t dt

(lx

lx+1 )

n mx

Lx

Z 1 0

Tx = ex lx

Z n 0

=

lx+t dt

lx

lx+n n Lx

lx+t dt n Lx

lx

=

Z n 0

t px dt

=ex:n

a(x) is the average number of years lived between age x and age x + 1 by those of the survivorship group who die between age x and age x + 1. With the assumption of uniform distribution of deaths over the interval (x, x + 1), a(x) = 1/2. Without this assumption, R1

R

1 t · lx+t µ(x + t)dt t · t px µ(x + t)dt a(x) = R 1 = 0R 1 0 lx+t µ(x + t)dt 0 t px µ(x + t)dt 0

By the way, a(x) is a minor concept in MLC. I’ll be surprised if SOA tests this obscure concept. I would skip it. Arch MLC, Spring 2009

c Yufeng Guo

40

Chapter 1 De Moivre’s Law: µ(x) = (!

x)

1

x , !

and s(x) = 1

Gompertz’ Law: µ(x) = Bcx and s(x) = exp[ m(cx where B > 0, c > 1, m =

where 0  x < !

1)]

B , x log c

Makeham’s Law: µ(x) = A + Bcx and s(x) = exp[ Ax where B > 0, A

B, c > 1, m =

0

m(cx

1)]

B , x log c

0

Weibull’s Law: µ(x) = kxn and s(x) = exp( uxn+1 ) where k > 0, n > 0, u =

k , x (n + 1)

0

DeMoivre’s Law and Modified DeMoivre’s Law: If x is subject to DeMoivre’s Law with maximum age !, then all of the relations on the left below are true. The relations on the right are for Modified DeMoivre’s Law with c > 0. DeMoivre µ(x) = ! 1 x

Modified DeMoivre µ(x) = ! c x ! x c ! ! x c !

s(x) =

! x !

s(x) =

lx = l0 ·

! x !

l0

ex = E[T ] = Var[T ] = t px

=

(! x)2 12

! x t ! x

µx (t) =

ex =

! x 2

1 ! x t

Var[T ] = t px

=

⇣

! x c+1 (! x)2 c (c+1)2 (c+2) ! x t ! x

µx (t) =

⌘c

c ! x t

Be careful on the exam - Modified DeMoivre problems are often disguised in a question that starts something like You are given • s(x) = 1 • ···

x 2 80

In cases like this, you have to recognize that the question is just a modified DeMoivre written in a di↵erent algebraic form. Arch MLC, Spring 2009

c Yufeng Guo

41

Chapter 1 Note that in this table, the function listed at left are given in terms of the functions across the top row! F (x) F (x) s(x)

F (x) 1

F (x)

1

µ(x)

F 0 (x) 1 F (x)

s0 (x) s(x)

1 t qx+t

(1 t)qx 1 tqx

y qx+t

yqx 1 tqx

1

t px

+ t)

Arch MLC, Spring 2009

tqx qx

R1

µ(x)

f (u) du

1

f (u) du

x

s0 (x)

qx 1 tqx

0

s(x)

F 0 (x)

µ(x + t)

Rx

s(x)

f (x)

Assumptions for fractional ages. Uniform Function Distribution tqx t qx

t px µ(x

f (x)

s(x)

e

f (x)

log px 1 1

p1x

t

pyx ptx

ptx log px

c Yufeng Guo

0

Rx 0

µ(x) e

f (x) s(x)

Constant Force 1 ptx

Rx

e

µ(t) dt

µ(t) dt

Rx 0

µ(t) dt

µ(x)

Hyperbolic tqx 1 (1 t)qx qx 1 (1 t)qx

(1

t)qx

yqx 1 (1 y t)qx px 1 (1 t)qx qx px [1 (1 t)qx ]2

42

Chapter 1

Past SOA/CAS Exam Questions: All of these questions have appeared on SOA/CAS exams between the years 2000 and 2005. You will find that they often involve some ‘clever’ thinking in addition to knowledge of actuarial math. These questions are used with permission. 1. Given: (i) e0 = 25 (ii) lx = !

x, 0  x  !

(iii) T (x) is the future lifetime random variable. Calculate Var[T (10)]. (A) 65

(B)93

(C) 133

(D) 178

Z !✓

t dt = ! !

Solution: e0 =

Var [T (x)] = 2

◆

1

0

e10 =

Z 40 ✓

Z 40 ✓

!2 ! = = 25 ) ! = 50 2! 2

◆

t dt = 40 40

1

0

t 1

0

(E) 333

◆

t dt 40

402 = 20 (2)(40) "

t2 (20) = 2 2 2

Key: C

t3 3 · 40

#40

(20)2 = 133

0

2. For a certain mortality table, you are given: (i) µ(80.5) = 0.0202 (ii) µ(81.5) = 0.0408 (iii) µ(82.5) = 0.0619 (iv) Deaths are uniformly distributed between integral ages. Calculate the probability that a person age 80.5 will die within two years. (A) 0.0782

(B) 0.0785

Arch MLC, Spring 2009

(C) 0.0790

(D) 0.0796

c Yufeng Guo

(E) 0.0800

43

Chapter 1 Solution: 0.0408 = µ(81.5) =

q81 ) q81 = 0.0400 (1/2)q81

1

Similarly, q80 = 0.0200 and q82 = 0.0600 2 q80.5

h

= 1/2 q80.5 + 1/2 p80.5 q81 + p81 · 1/2 q82

i

0.01 0.98 + [0.04 + 0.96(0.03)] = 0.0782 0.99 0.99

= Key: A

3. Mortality for Audra, age 25, follows De Moivre’s law with ! = 100. If she takes up hot air ballooning for the coming year, her assumed mortality will be adjusted so that for the coming year only, she will have a constant force of mortality of 0.1. Calculate the decrease in the 11-year temporary complete life expectancy for Audra if she takes up hot air ballooning. (A) 0.10

(B) 0.35

(C) 0.60

(D) 0.80

(E) 1.00

t )dt = t 75

t2 |11 = 10.1933 2 ⇥ 75 0

Solution: STANDARD: e25:11 =

Z 11

(1

0

MODIFIED: p25 = e e25:11

Z 1

R1 0

0.1ds

=e

0.1

Z 10

= 0.90484

t )dt 74 0 0 Z 1 Z 10 t = e 0.1t dt + e 0.1 (1 )dt 74 0 0 ! 1 e 0.1 t2 = + e 0.1 t |10 0.1 2 ⇥ 74 0 =

t p25 dt

+ p25

(1

= 0.95163 + 0.90484(9.32432) = 9.3886

Here we use a recursive formula: ex:n = ex:1 + 1 px ⇥ ex+1:n

1

The di↵erence is 0.8047. Key: D

Arch MLC, Spring 2009

c Yufeng Guo

44

Chapter 1 4. You are given the following extract from a select-and-ultimate mortality table with a 2-year select period:

x 60 61 62

l[x] 80,625 79,137 77,575

l[x]+1 79,954 78,402 76,770

lx+2 78,839 77,252 75,578

x+2 62 63 64

Assume that deaths are uniformly distributed between integral ages. Calculate

0.9 q[60]+0.6 .

(A) 0.0102

(B) 0.0103

(C) 0.0104

(D) 0.0105

(E) 0.0106

Solution: l[60]+0.6 = (0.6)(79,954) + (0.4)(80,625) = 80,222.4 l[60]+1.5 = (0.5)(79,954) + (0.5)(78,839) = 79,396.5

0.9 q[60]+0.6

=

80,222.4 79,396.5 = 0.0103 80,222.4

Key: B

5. Given: (i) µ(x) = F + e2x , x (ii)

0.4 p0

0

= 0.50

Calculate F . (A) -0.20

(B) -0.09

Arch MLC, Spring 2009

(C) 0.00

(D) 0.09

c Yufeng Guo

(E) 0.20

45

Chapter 1 Solution: = 0.5 = e

0.4 p0

=e ) 0.5 = e Key: E

)

0.4F

h

e2x 2

0.4F 0.6128

0.6931 =

i0.4 0

0.4F

R 0.4 0

(F +e2x )dx

0.4F

=e

⇣

) ln(0.5) =

0.6128

e0.8 2

1

⌘

0.4F

0.6128

) F = 0.20

6. An actuary is modeling the mortality of a group of 1000 people, each age 95, for the next three years. The actuary starts by calculating the expected number of survivors at each integral age by l95+k = 1000 k p95 , k = 1, 2, 3 The actuary subsequently calculates the expected number of survivors at the middle of each year using the assumption that deaths are uniformly distributed over each year of age. This is the result of the actuary’s model: Age 95 95.5 96 96.5 97 97.5 98

Survivors 1000 800 600 480 288

The actuary decides to change his assumption for mortality at fractional ages to the constant force assumption. He retains his original assumption for each k p95 . Calculate the revised expected number of survivors at age 97.5. (A) 270

(B) 273

Arch MLC, Spring 2009

(C) 276

(D) 279

(E) 282

c Yufeng Guo

46

Chapter 1 Solution: From UDD,

Likewise, from

l96.5 =

l96 + l97 . 2

480 =

600 + l97 2

! l97 = 360

l97 = 360 and l97.5 = 288, we get l98 = 216.

For constant force,

e 0.5 px

=e

0.5µ

µ

l98 216 = = 0.6 l97 360

=

1

= (0.6) 2 = 0.7746

l97.5 = (0.7746)l97 = (0.7746)(360) = 278.86 Key: D

7. 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, e1:1.5 . (A) 1.25

(B) 1.3

(C) 1.35

(D) 1.4

(E) 1.45

Solution: e1:1.5 = = =

Z 1 0

Z 1 0

(1

"

= t

Z 1.5 0

t p1 dt + 1 p1

0.1t)dt + 0.9 0.1t2 2

#1 0

t p1 dt

Z 0.5

Z

0 0.5

x p2 dx

(1

"

+ 0.9 x

0.05x2 2

= 0.95 + 0.444 = 1.394

Arch MLC, Spring 2009

0.05x)dx

0

c Yufeng Guo

#0.5 0

Key: D

47

Chapter 1 8. For a given life age 30, it is estimated that an impact of a medical breakthrough will be an increase of 4 years in e30 , the complete expectation of life. Prior to the medical breakthrough, s(x) followed de Moivres law with ! = 100 as the limiting age. Assuming de Moivres law still applies after the medical breakthrough, calculate the new limiting age. (A) 104

(B) 105

(C) 106

(D) 107

Solution: For deMoivre’s law, e30 = Prior to medical breakthrough

30

! 2

(E)108

.

! = 100 )e30 =

100 30 2

= 35.

0

After medical breakthrough e30 =e30 +4 = 39. 0

)e30 = 39 =

30

!0 2

) ! 0 = 108.

Key E

9. For a select-and-ultimate mortality table with a 3-year select period: (i) x 60 61 62 63 64

q[x] q[x]+1 q[x]+2 qx+3 x + 3 0.09 0.11 0.13 0.15 63 0.10 0.12 0.14 0.16 64 0.11 0.13 0.15 0.17 65 0.12 0.14 0.16 0.18 66 0.13 0.15 0.17 0.19 67

(ii) White was a newly selected life on 01/01/2000. (iii) White’s age on 01/01/2001 is 61. (iv) P is the probability on 01/01/2001 that White will be alive on 01/01/2006. Calculate P . (A) 0  P < 0.43

(D) 0.47  P < 0.49

Arch MLC, Spring 2009

(B) 0.43  P < 0.45

(E) 0.49  P  1.00

(C) 0.45  P < 0.47

c Yufeng Guo

48

Chapter 1 Solution: 5 p[60]+1

⇣

= 1

q[60]+1

⌘⇣

⌘

1

q[60]+2 (1

q63 ) (1

q64 ) (1

q65 )

= (0.89)(0.87)(0.85)(0.84)(0.83) = 0.4589

Key: C

10. You are given: µ(x) =

(

0.04, 0.05,

0 < x < 40 x > 40

Calculate e25:25 . (A) 14.0

(B) 14.4

(C) 14.8

Solution: e25:25 = =

Z 15

e

(D) 15.2

Z 15

0.04t

0

t p25 dt

dt + e

0

1 ⇣ = 1 0.04

✓

0.60

e

⌘

+ 15 p25

R 15

+e

(D) 15.6

0

0.04 ds

0.60

Z 10 0

t p40 dt

◆ Z 10

e

0.05t

dt

0



1 ⇣ 1 0.05

e

= 11.2797 + 4.3187 = 15.60

0.50

⌘

Key: E

11. T , the future lifetime of (0), has a spliced distribution. (i) f1 (t) follows the Illustrative Life Table. (ii) f2 (t) follows DeMoivre’s Law with ! = 100. (iii) fT (t) = Calculate (A) 0.81

(

k f1 (t), 0  t  50 1.2 f2 (t), 50 < t

10 p40 .

(B) 0.85

Arch MLC, Spring 2009

(C) 0.88

(D) 0.92

c Yufeng Guo

(E) 0.96 49

Chapter 1 SOLUTION: From the Illustrative Life Table:

1=

Z 1 0

l50 = 0.8951 l0

l40 = 0.9313 l0

Z 50

Z 1

fT (t) dt =

0

kf1 (t) dt +

= kF1 (50) + 1.2 (F2 (1) = k (1

50 p0 )

= k(1 )k= For x  50,

FT (x) =

Z x 0

50

1.2 f2 (t) dt

F2 (50))

+ 1.2(1

0.5)

0.8951) + 0.6 1

1

0.6 = 3.813 08951

3.813f1 (t) dt = 3.813F1 (x)

This gives us the following two results: ✓

FT (40) = 3.813 1 ✓

FT (50) = 3.813 1 10 p40

=

1 1

FT (50) 1 = FT (40) 1

l40 l0 l50 l0

◆ ◆

= 0.262 = 0.400

0.400 = 0.813 0.262

12. For a double decrement table, you are given: (1)

(⌧ )

(i) µx (t) = 0.2µx (t), t > 0 (⌧ )

(ii) µx (t) = kt2 , t > 0 0 (1)

(iii) qx

= 0.04 (2)

Calculate 2 qx . (A) 0.45

(B) 0.53

Arch MLC, Spring 2009

(C) 0.58

(D) 0.64

c Yufeng Guo

(E) 0.73

50

Chapter 1 SOLUTION: We can use the exponential formulation for p0 (x) 0

0

0.04 = qx(1) = 1 px(1) = 1 e

R1 0

(1)

µx (t) dt

)e

=1 e

0.2k/3

(⌧ ) µ(1) x (t) = 0.2µx (t)

(2) 2 qx

=

Z 2 0

(⌧ ) t px

·

µ(2) x (t)dt

=

R1 0

(⌧ )

0.2µx (t) dt

=1 e

= 0.96

R1 0

0.2kt2 dt

=1 e

(⌧ ) ) µ(2) x (t) = 0.8µx (t)

Z 2 0

(⌧ ) t px

) (⌧ ) · (0.8)µ(⌧ x (t)dt = 0.82 qx

(⌧ )

To get 2 qx , we use (⌧ ) 2 qx

=1

(⌧ ) 2 px

=1

e

=1

⇣

R2 0

e

(⌧ )

µx (t) dt

0.2k/3

⌘40

=1

=1

e

R2 0

kt2 dt

=1

e

8k/3

(0.96)40

) 2 qx(⌧ ) = 0.8046

) 2 qx(2) = (0.8)(0.8046) = 0.644

13. You are given: (i) e30:40 = 27.692 (ii) s(x) = 1

x !,

x!

(iii) T (x) is the future lifetime random variable for (x). Calculate Var(T (30)). (A) 332 (B) 352 (C) 372 (D) 392 (E) 412

SOLUTION: R

e30:40 = 040 t p30 dt R = 040 !! 3030 t dt

Arch MLC, Spring 2009

c Yufeng Guo

51

0.2k/3

Chapter 1 h

t2 2(! 30) 800 ! 30

= t

= 40 = 27.692 ) ! = 95 t p30

=

i

|40 0

65 t 65

Now, realize (after getting ! = 95) that T (30) is uniformly on (0, 65), its variance is just the variance of a continuous uniform random variable: 2 V ar = (65120) = 352.08 Key: B

14. For a life table with a one-year select period, you are given: x l[x] d[x] lx+1 e[x] (i) 80 1000 90 8.5 81 920 90 (ii) Deaths are uniformly distributed over each year of age. Calculate e[81] . (A) 8.0 (B) 8.1 (C) 8.2 (D) 8.3 (E) 8.4

SOLUTION: Complete the table: l81 = l[80] d[80] = 910 l82 = l[81] d[81] = 830 910 p[80] = 1000 = 0.91 830 p[81] = 920 = 0.902 p81 = 830 910 = 0.912

Use the recursive formula: ex:n = ex:1 + 1 px ⇥ ex+1:n

1

We have: e[80] = e80:1 + 1 p[80] ⇥ e81 Arch MLC, Spring 2009

c Yufeng Guo

52

Chapter 1 Notice under UDD, t px is a straight line. Since ex:1 is the area of the function t px bounded by t = 0 and t = 1, under UDD we have ex:1 = 0.5(0 px + 1 px ) = 0.5(1 + 1 px ) You can also derive the above equation using the standard formula: ex:1 = R1 R1 tqx )dt t qx )dt = 0 (1 0 (1

R1

0 t px dt

=

Since qx is a constant, you can verify that: R1 0

(1

tqx )dt = 1

0.5qx = 0.5(1 + 1 px )

Hence: ex:n = 0.5(1 + 1 px ) + 1 px ⇥ ex+1:n 0.51 qx + 1 px (1 + ex+1:n 1 ).

1

= 0.5(1 qx + 1 px + 1 px ) + 1 px ⇥ ex+1:n

1

=

Set n = 1, we have: e˙ x = 12 qx + px (1 + e˙ x ) e˙ [80] = 12 q[80] + p[80] (1 + e˙ 81 ) 8.5 = 12 (1 0.91) + (0.91) (1 + e˙ 81 ) e˙ 81 = 8.291 e˙ 81 = 21 q81 + p81 (1 + e˙ 82 ) e˙ 82 = 8.043 e˙ [81] = 12 q[81] + p[81] (1 + e˙ 82 ) = 12 (1 0.912) + (0.912)(1 + 8.043) = 8.206 Key: C

15. You are given: µ(x) =

(

Calculate

0.05 50  x < 60 0.04 60  x < 70

4|14 q50 .

(A) 0.38 (B) 0.39 (C) 0.41 (D) 0.43 (E) 0.44

SOLUTION: = e (0.05)(4) = 0.8187 (0.05)(10) = 0.6065 10 p50 = e (0.04)(8) = 0.7261 8 p60 = e 18 p50 = (10 p50 )(8 p60 ) = 0.6065 ⇥ 0.7261 = 0.4404 4 p50

Arch MLC, Spring 2009

c Yufeng Guo

53

Chapter 1 = 4 p50 Key: A 4|14 q50

18 p50

Arch MLC, Spring 2009

= 0.8187

0.4404 = 0.3783

c Yufeng Guo

54

Chapter 1 16. For a population which contains equal numbers of males and females at birth: (i) For males, µm (x) = 0.10, x (ii) For females,

µf (x)

0

= 0.08, x

0

Calculate q60 for this population. (A) 0.076 (B) 0.081 (C) 0.086 (D) 0.091 (E) 0.096

SOLUTION: s(60) = P [T (0) > 60] (i.e. the probability that a newborn survives to age 60) Using the double expectation: s(60) = P [T (0) > 60] = P (M aleN ewborn)⇤P [T (M aleN ewborn) > 60]+P (F emalenewborn)⇤ P [T (F emaleN ewborn) > 60] = 0.5e (0.1)(60) + 0.5e (0.08)(60) = 0.005354 Similarly, s(61) = 0.5e q60 = 1 Key: B

(0.1)(61)

0.00492 0.005354

+ 0.5e

(0.08)(61)

= 0.00492

= 0.081

Arch MLC, Spring 2009

c Yufeng Guo

55

Chapter 1

Problems from Pre-2000 SOA-CAS exams 1. You are given: µ(x) =

s

1 80

x

, 0  x < 80.

Calculate the median future lifetime of (20). (A) 5.25

(B) 6.08

(C) 8.52

(D) 26.08

(E) 30.00

2. You are given: • t px = (0.8)t ,

t

0

• lx+2 = 6.4

Calculate Tx+1 . (A) 4.5

(B) 7.2

(C) 28.7

(D) 35.9

(E) 44.8

Use the following information for the next 4 questions: You are given: • T (x) is the random variable for the future lifetime of (x). • The p.d.f. of T is fT (t) = 2e

2t ,

t

0.

3. Calculate ex (A) 0.5

(B) 2.0

(C) 10.0

(D) 20.0

(E) 40.0

4. Calculate Var[T ]. (A) 0.25

(B) 0.50

(C) 1.00

(D) 2.00

(E) 4.00

5. Calculate m(x), the median future lifetime of (x). (A)

4

e 2

(B)

2

e 2

Arch MLC, Spring 2009

(C)

ln 2 2

(D)

ln 4 2

c Yufeng Guo

(E) 1

56

Chapter 1 6. Calculate mx , the central-death-rate at age x. (A)

2

e 2

(B) e

(C) 2e

2

2

7. You are given: ✓ ◆ x ↵ s(x) = 1 , 0  x < !, ! Calculate µx · ex . (A)

↵ ↵+1

(B)

↵! ↵+1

(D) 1

(E) 2

where ↵ > 0 is a constant.

(C)

↵2 ↵+1

(D)

↵2 !

x

(E)

↵(! x) (↵ + 1)!

8. You are given: • q60 = 0.3 • q61 = 0.4

• f is the probability that (60) will die between ages 60.5 and 61.5 under the uniform distribution of deaths assumption. • g is the probability that (60) will die between ages 60.5 and 61.5 under the Balducci assumption. Calculate 10,000(g

f ).

(A) 0

(C) 94

(B) 85

Arch MLC, Spring 2009

(D) 178

(E) 213

c Yufeng Guo

57

Chapter 1

Solutions to Chapter 3 1. Key: A

We need to find t such that t p20 = 0.5 Rt

) 0.5 = e

0

= e2[(60

µ(20+s) ds

=e p

s)0.5 ]0 t

= e2[

hp

Rt 0

(60 s)

60 t

p

0.5

ds

60]

i

p ln 0.5 = 2 60 t 60 p ) 60 t = 7.40 ) t = 5.25

2. Key: D Tx+1 = lx+1 =

6.4 =8 0.8

Tx+1 = 8

Z 1

Z 1 0

lx+1+t dt

) lx+1+t = 8(0.8)t

(0.8)t dt =

0

=

8 (0.8)t |1 0 ln 0.8

8 = 35.9 ln 0.8

3. Key: A This is constant force of mortality (CFM) with µ = 2. E[T ] =

4. Key: A

(CFM) Var[T ] =

1 µ2

1 = 0.5 =ex µ

= 0.25

5. Key: C

Arch MLC, Spring 2009

0.5 = t px = e

µ·t

)t=

ln 2 2

c Yufeng Guo

=e

2t

58

Chapter 1 6. Key: E mx = =

lx

lx+1 Lx

lx (1 e µ ) = R1 0 lx · t px dt

lx (1 e 2 ) 1 e 2 == 1 =2 R1 e 2) lx 0 e µt dt 2 (1

7. Key: A This is a Modified DeMoivre problem with constant equal to ↵. So µ(x) =

↵ !

x

ex =

,

! x ↵+1

Multiplying these two quantities gives ↵ ↵+1

8. Key: B

In both cases the probability is given by 0.5 p60

UDD: [1

Balducci: 1

· 0.5 q60.5 + 1 p60 · 0.5 q61

(0.5)(0.3)] 1 (0.5)(0.3) (0.5)(0.3) + (0.7) · (0.5)(0.4) = 0.2900 = f

(0.5)(0.3) 0.7 1 (0.5)(0.3)

+ (0.7) ·

(0.5)(0.4) 1 (0.5)(0.4)

10,000(g

Arch MLC, Spring 2009

= 0.2985 = g

f ) = 85

c Yufeng Guo

59

Chapter 1

Arch MLC, Spring 2009

c Yufeng Guo

60

Chapter 2

ACTUARIAL MATHEMATICS CHAPTER 4 – LIFE INSURANCE • Option A reference: Actuarial Mathematics Chapter 4 • Option B reference: Models for Quantifying Risk Chapter 9 This chapter introduces a variety of types of life insurance. Keep in mind that all of these “products” have a basic structure of some benefit paid at some point in the future if certain conditions are met. Don’t lose sight of the fact that all of the products mentioned have this same structure – they vary only by the conditions to be met. The text considers products that pay a “unit benefit” of 1. Sometimes this is easy to forget because the 1 does not show up in multiplicative formulas. The text also considers products that pay the benefit “at the moment of death”. Clearly, barring a very annoying life insurance agent, benefits are not paid precisely at this point. The text then considers products paying a benefit at the end of the year of death. This too is not realistic, but products that pay o↵ at the moment of death or at the end of the year of death have the most convenient formulas to work with. Also, they are what you need to know to pass the exam. To the text!

61

Chapter 2

4.2 Insurances Payable at the Moment of Death • Option A reference: Actuarial Mathematics Chapter 4.2 • Option B reference: Models for Quantifying Risk Chapter 9.3 Let bt be a “benefit function” – a benefit payable at time t. Let v t be a discount function that discounts the benefit back to the time of policy issue. Unless stated otherwise, we will assume the policy is issued at time t = 0. The “present value function” for the benefit is zt = bt v t . So, zt is the present value of the benefit payment at the time of policy issue.

zt is the value of the benefit at time 0.

issue t=0

bt paid here

t - death

time

Recall from Chapter 3 that for (x) (an individual age x), the future lifetime is a random variable T = T (x), so associated with zt is a random variable ZT . Generally the text uses the notation Z = bT v T . The expected value of Z is called the Actuarial Present Value of the benefit – this is a critical concept. The usual present value that you are familiar with from Interest Theory discounts for interest only. Actuarial Present Value (APV) discounts for interest and mortality. Mortality a↵ects when and if the benefit is received. Consider the following two scenarios: 1. A payment of 10,000 is payable to you (or your heirs if you are dead) at the end of t years from now. 2. A payment of 10,000 is payable to you at the end of t years from now only if you are alive. Actuarially, we would say the value of the payment in Scenario (1) is 10, 000v t , while the value of the payment in Scenario (2) is 10, 000v t t px . That, in a nutshell, is the di↵erence between interest theory from earlier actuarial courses and life contingencies in this course. Arch MLC, Spring 2009

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62

Chapter 2

TYPES OF INSURANCE 4.2.1 Level Benefit Insurance • Option A reference: Actuarial Mathematics Chapter 4.2.1 • Option B reference: Models for Quantifying Risk none Name: n-year Term Life Insurance Description: A flat benefit is paid at death if the insured dies within an n-year “term” starting at issue. (

bt =

1 t  n 1 if death within n years 0 t > n 0 otherwise

vt = v t t Z=

(

vT 0

0 T  n Note this is bt v t in each case of T. T >n

Since the random variable Z represents the present value of future benefits, E(Z) represents the expected present value of the benefits. This expected value takes into account the probability of actually receiving benefits. The expected value E(Z) is called the “actuarial present value” since it contains a mortality factor. The following is important notation: E(Z) = A 1

x:n

=

Z n 0

v t t px µx (t)dt

OK, there is a lot contained in that line that needs explaining so don’t go anywhere for a couple of minutes. First, let’s look at the components of the integral more carefully, then we will discuss the A 1 notation. x:n

A1

x:n

=

discount Z n interest z}|{ v

t

0

prob of death between times t and t + dt ·

t px

|{z}

survival from 0 to t

·

z }| {

µx (t)dt

The expression inside the integral can be thought of as the (very small) value of a death benefit of 1 payable only if you die exactly at time t. What should such a benefit be worth? It should equal (1 discounted to time t)⇥(the probability that you survive to time t) ⇥ (the probability that you then die at time t). That is ! v t t px µx (t). Now to get the value of A 1 x:n

we just have to integrate this benefit over all the time that the insurance is in force; that is, from t = 0 to t = n. I realize this may be a lot to take in right now, but it will become more intuitive as we go through the di↵erent types of insurance. Arch MLC, Spring 2009

c Yufeng Guo

63

Chapter 2 Now, what about the symbol A 1 ? Obviously this symbol was designed by a committee (no x:n

kidding!). This odd-looking notation will become second nature well before test day. 1. The bar over the A tells us that the death benefit is payable at the moment of death. This is continuous insurance. 2. The x tells the age at which the expected value of the benefit is calculated. Except in the case of deferred insurance (which we’ll see later), this is the age of the person when coverage begins. 3. The n is the term of the policy. When the policy is whole life (i.e. when n = 1), the “: n ” is omitted. 4. The 1 over the x tells us that death must occur for benefits to be paid. So much for E(Z), now for Var [Z]. Recall from Statistics that Var [Z] = E(Z 2 )

[E(Z)]2 .

Throughout the book, E(Z 2 ) will be calculated using the rule of moments. This rule says that as long as the force of interest is deterministic (i.e. not a random variable or something weird like that), E(Z j ) at the force of interest is equal to E(Z) at the force of interest j · . In other words, to find E(Z j ), just multiply the interest at each point in time by j and calculate the expected value. The symbol for “E(Z) at 2 times the force of interest” is 2A 1 x:n

(as if the notation weren’t bad enough already). This leads to Var(Z) = 2A 1

x:n

✓

A1

x:n

◆2

.

Note: The rule of moments holds as long as the benefit is always 0 or 1. EXAMPLE: (try this by covering up the solution first) Let the force of interest be a constant (x) be a constant µ = 0.01 for all t,

= 0.03 and let the force of mortality for

1. Find the actuarial present value (APV) of a 5-year term insurance with a death benefit of 1. 2. Find Var(Z) SOLUTION: 1. The APV equals A1

x:n

=

Z n 0

v t t px µx (t)dt =

Z 5

e

t

e

µt

µdt .

0

Remember, we saw in Chapter 3 that in this situation t px = e equals (0.01)

Z 5 0

Arch MLC, Spring 2009

e

0.04t

dt = 0.25

⇣

e

0.04t

⌘5 0

c Yufeng Guo

⇣

= 0.25 1

e

0.2

⌘

µt .

This

= 0.0453. 64

Chapter 2 2. The variance equals 2A 1

x:n

✓

A1

x:n

◆2

. To find 2A 1 , we take the expression x:n

for E(z) and double the force of interest every time it appears. This leads to Z 5 2 A1 = e 2 t e µt µdt . x:n

(0.01)

Z 5

e

0.07t

0

dt =

0

1⇣ 1 7

) Var(Z) = 0.0422

e

0.35

⌘

= 0.0422.

(0.0453)2 = 0.0401

}

Name: Whole Life Insurance Description: A flat benefit is paid at the moment of death at any time in the future (over the whole life of the insured) bt = 1 vt = v t Z = vT

0 1 will definitely be paid at some point. 0 0 Note this equals bt v t .

t t T

The symbol A x is used to represent E(Z) and A x = E(Z) =

Z 1 0

v t t px µx (t)dt

Notes: 1. This is term insurance with n = 1. 2. The bar over the A indicates the benefit is payable at the moment of death (continuous) 3. The x alone indicates whole life – no limit to the term of coverage. Finally, Var[Z] = 2A x

⇣

Ax

⌘2

Before some examples, note that Example 4.2.1 in the text is a good example to study and understand. Note: For certain mortality and interest assumptions, A x takes a particularly simple form that absolutely should be memorized. For example you can show for yourself (do a little integration or read through Example 4.2.3 in the text.) that if the force of mortality is a constant µ and the force of interest is a constant , then Ax =

µ µ+

You must know this formula and be comfortable using it on exam day. There will likely be several problems that will require it. Arch MLC, Spring 2009

c Yufeng Guo

65

Chapter 2 EXAMPLE: Constant Force of Mortality Assuming a constant force of mortality µ and constant force of interest , find Var[Z] in terms of µ and . SOLUTION:

⇣

⌘2

We know that Var[Z] = 2A x A x . To find 2A x , we remember that it is just the expression for E[Z] with the force of interest doubled every time it appears. So µ 2 Ax = µ + (2 ) and

✓

µ Var[Z] = µ + (2 )

◆2

µ µ+

.

Don’t bother memorizing this complicated formula. Instead, memorize the rule of moments and the formula for E[Z]. } EXAMPLE: DeMoivre’s Law (x) is subject to DeMoivre’s law for mortality with ! = 100. If are 1. A40 , and 2. A 1 ?

= 0.05, what

40:25

(You should definitely try this one without looking at the solution!) SOLUTION: 1. Ax = For DeMoivre, t px

=

!

x !

t x

Z 60 0

v t t px µx (t)dt.

and

µx (t) =

) t px µx (t) = Ax = 2. For A 1

40:25

Z 60 0

e

0.05t

1 dt = 60

20 h e 60

0.05t

!

1 x

t

,

1 . 60

i60 0

=

1h 1 3

e

3

i

= 0.3167.

, we do exactly the same thing except the limits of integration are

0 and 25. The resulting answer is A 1

40:25

= 0.2378.

}

4.2.2 Endowment Insurance • Option A reference: Actuarial Mathematics Chapter 4.2 • Option B reference: Models for Quantifying Risk Chapter 9.1.4 Arch MLC, Spring 2009

c Yufeng Guo

66

Chapter 2 Unlike term and whole life insurance, endowment insurances pay you something if you live. Name: n-year Pure Endowment Description: Provides a benefit at the end of n years if and only if the insured is still alive at the end of the n-year period. (You must survive to get the benefit.) bt =

(

0 t  n 0 if death within n years 1 t > n 1 if survive n years from issue

vt = v n t Z=

(

0 If payment occurs, we know it will be at t = n.

0 T  n Note this is bt v t in each case of T. vn T > n

The symbol for E(Z) is A

1. x: n

Note where the 1 is. If 1 is over the n it indicates you get

paid only if you survive n years. If it is over the x, you only get paid if you die within n years. Anyway, A 1 = E(Z) = v n n px x: n

and Var[Z] = A 2

1 x: n

✓

A

1 x: n

◆2

.

Applying the rule of moments (and remembering that doubling the force of interest turns v n into v (2n) ), we can see that Var[Z] = v (2n) n px

(v n n px )2 = v (2n) n px n qx .

Again, I wouldn’t memorize anything but the expression for A

1 x: n

and the rule of moments.

One important note before we move on: When we get to annuities - there will be a new symbol for pure endowments, n Ex . For now just remember that n Ex = A 1 and they mean x: n

the same thing: the present value of 1 paid to (x) in exactly n years if (x) survives n years. Name: n-year Endowment Insurance Description: The benefit is paid at the earlier of death or n years. You receive the benefit either way, only the timing is a↵ected by death or survival. bt = 1 t

0 A benefit of 1 is certain to be paid.

vt =

(

vt t  n vn t > n

Z=

(

vT vn

T n t>n

Arch MLC, Spring 2009

c Yufeng Guo

67

Chapter 2

E(Z) = A x:n (no 1 over either the x or the n ), ⇣

Var[Z] = 2A x:n

A x:n

⌘2

.

Note: This insurance is exactly the same as an n-yr term policy plus an n-yr pure endowment. Therefore, A x:n = A 1 + A 1 . x:n

x: n

This is a good relationship for the SOA-CAS to use in a question. It comes up somewhere on every SOA or CAS exam I have seen for this course. It also tells us that A x:n =

Z n 0

v t t px µx (t)dt + v n n px .

EXAMPLE: DeMoivre’s Law (x) is subject to DeMoivre’s law for mortality with ! = 100. If A 40:25 ?

= 0.05, what is

SOLUTION: We saw in an earlier example that A 1

40:25

A 40:25 = A 1

= 0.2378 for this scenario, and since

40:25

+A

1 , 40: 25

we only need to find the value of A

1 40: 25

= v 25

25 p40 = e

(0.05)(25)

✓

35 60

◆

= 0.1671.

) A 40:25 = 0.2378 + 0.1671 = 0.4049.

}

4.2.3 Deferred Insurance • Option A reference: Actuarial Mathematics Chapter 4.2.3 • Option B reference: Models for Quantifying Risk none Name: m-year Deferred Insurance Description: Provides for a benefit following death of the insured only if the insured dies more than m years after issue. Basically, this is a whole life policy in which “coverage” doesn’t begin until m years have elapsed. Benefit paid if death occurs in this time period.

issue t=0

Arch MLC, Spring 2009

t=m

c Yufeng Guo

68

Chapter 2 bt =

(

1 t>m 0 t  m no benefit unless insured lives at least m years.

vt = v t t Z=

(

vT 0

0 T > m Note this is bt v t in each case of T . T m

The symbol for the APV of this insurance is m|A x (although this is lousy notation since the omitted symbol to the right of the | is 1 instead of 1). Also m|A x

= E(Z) =

Z 1 m

v t t px µx (t)dt .

EXAMPLE: Constant Force of Mortality 1. If µ and 2. Write

are constant, find

m|A x

m|A x

in terms of µ, , and m.

in terms of two types of insurance that we have already seen.

SOLUTION: 1. m|A x =

Z 1 m

v t t px µx (t)dt =

Z 1

e

m

t

e

µt

µdt =

µ e µ+

m(µ+ )

It’s up to you to decide whether to derive the formula on the spot or memorize the formula. Robin thinks that he can derive the formula from scratch quickly in the exam. Nathan wants to memorize it to avoid deriving the formula in the exam. Yufeng will do both. 2.

m|A x

= Ax

A1

x:m

This makes sense if you say it in English – “A whole life policy deferred for m years is equal to a whole life policy starting now minus the first m years of coverage. }

EXAMPLE: DeMoivre’s Law (x) is subject to DeMoivre’s law for mortality with ! = 100. If 25|A 40 ?

= 0.05, what is

SOLUTION: Arch MLC, Spring 2009

c Yufeng Guo

69

Chapter 2 We saw in an earlier example that A 1

= 0.2378 for this scenario, and we also

40:25

know that 25|A 40

= A 40

A1

40:25

.

Since we saw that A 40 = 0.3167, we have that 25|A 40

= 0.3167

0.2378 = 0.0789.

Again, don’t try to memorize the formulas, try to think logically why it must be true that a whole life policy starting today is equal to a 25-year term policy plus a 25-year deferred whole life policy. } Those are the basic level benefit life insurance and endowment products covered in the chapter. Now for the more esoteric stu↵.

4.2.4 Varying Benefit Insurance • Option A reference: Actuarial Mathematics Chapter 4.2.4 • Option B reference: Models for Quantifying Risk Chapter 9.4 Name: Annually Increasing Whole Life Insurance Description: Pays 1 if you die in year 1, 2 in year 2, 3 in year 3, . . . . the more you receive upon death. bt = bt + 1c t vt = v t t

The later you die,

0 where b c is the greatest integer function.

0

Z = bt + 1cv t T

0

The symbol for the APV of this insurance is (IA)x and (IA)x = E(Z) =

Z 1 0

bt + 1cv t t px µx (t)dt .

Because of the discontinuous nature of the greatest integer function, this generally would be a pretty nasty integral to have to integrate. To get you used to some more notation, we might as well introduce some insurance that increases in value more often, say m times per year. Name: m-thly Increasing Whole Life Insurance 1 Description: Pays m at death in first m-th of a year, Pays

2 m

Arch MLC, Spring 2009

at death in second m-th of a year, etc. c Yufeng Guo

70

Chapter 2 Note that an “m-th” can be any period of time – not necessarily a month, which naturally comes to mind. It’s useful to think of it as a month while learning the concept, but just realize it can be any period of time less than a year. Also, don’t be intimidated by this idea – it’s really the same idea as annually increasing whole life, except that the increases occur more frequently than annually. ⇣

⌘

The symbol for the actuarial present value of this insurance is I (m)A . Letting m ! 1 x in the above insurance results in an insurance that pays t at time t. That is: bt = t, vt = v t Z = T vT . This is called continuously increasing insurance and the symbol for its APV is E(Z) = (IA)x . Just one more type of continuous insurance before we are ready to put it all together with some questions. Name: Annually decreasing n-year term insurance Description: During the n-year period from issue, provides a benefit at the end of the year of death equal to n k where k is the number of years completed since issue. For example, if death occurs during the first year, the benefit is n. If death occurs during the second year, benefit is n 1, etc. bt =

(

n 0

btc t  n t > n (It’s n-year term insurance.)

vt = v t t Z=

(

v T (n 0

0 btc) T  n T >n

The APV is (DA)1

x:n

= E(Z) =

Z n

v T (n

0

btc) t px µx (t)dt

Table 4.2.1 in the text is not very useful except for one very important fact that you won’t notice if you don’t read the footnotes. It tells you exactly when the rule of moments applies. It never applies for increasing or decreasing insurances - only for insurance where the benefit is always either 0 or 1. EXAMPLE: (This is mostly for review so these are not too hard. Read the directions carefully and don’t look at the next page!) Each of the following expressions is associated with a certain type of insurance or endowment. For each symbol, Arch MLC, Spring 2009

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71

Chapter 2 (A) Name the type of insurance associated.

(B) State what the benefit would be under this insurance to a person dying at t = 6.5

1.

2. A x:8

8|A x

6. (IA)1

x:8

11. (IA)1

x:4

12. A x:4

3. A 1

4. A

x:8

7. (IA)x

⇣

8. I (12)A

⌘

5. (DA)1

1 x: 8

x

x:8

⇣

9. I (3)A

⌘

x

10. A x

(I know we haven’t seen this one but try anyway.)

13. A 1

Arch MLC, Spring 2009

x:4

14. A

1 x: 4

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72

Chapter 2 SOLUTION: 1. A) 8-year deferred whole life insurance; B) 0 2. A) 8-year endowment; B) 1 3. A) 8-year term insurance; B) 1 4. A) 8-year pure endowment; B) 0 5. A) 8-year annually decreasing term; B) 2 6. A) 8-year annually increasing term; B) 7 7. A) Continuously increasing whole life; B) 6.5 8. A) Monthly increasing whole life; B) 79 12 OK, I looked at the definition of benefit for that one! 9. A) 3-thly increasing whole life; B)

20 3

10. A) Whole life; B) 1 11. A) Annually increasing 4-year term; B) 0 (the term has expired) 12. A) 4-year endowment; B) 0 (you were paid 1 at t = 4) 13. A) 4-year term; B) 0 14. A) 4-year pure endowment; B) 0 (you were paid 1 at t = 4)

}

4.3 Insurances Payable at the End of the Year of Death • Option A reference: Actuarial Mathematics Chapter 4.3 • Option B reference: Models for Quantifying Risk Chapter 9.1 These are the curtate versions of the continuous functions we studied above. For continuous insurance, benefits and discounts depended on t the time since issue. In the curtate setting, benefits and discounts will only depend on k, the number of full years that have passed since the policy was issued. This takes a little getting used to, but once you begin thinking about them in the right way, curtate versions of insurance look just like their continuous counterparts except that all the integrals are replaced by sums. Name: n-year Term Life Insurance paying unit amount at end of the year of death Description: Pays benefit at the end of year of death if death occurs within an n-yr period from issue. bk+1 =

(

1 k = 0, 1, 2, ..., n 0 k n

1

Note that we use the subscript k + 1 so that b1 refers to the benefit available at the end of the first year of coverage. If the insured dies in the first year of coverage, k = 0 since no full years have been lived. Therefore, b1 = b0+1 = bk+1 . vk+1 = v (k+1) (discount benefit to the end of year of death) Arch MLC, Spring 2009

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73

Chapter 2 Z=

(

v K+1 K = 0, 1, 2, ..., n 0 K n

1

The symbol for the actuarial present value of this insurance is A1

x:n

E(Z) = A1

x:n

=

n X1

and

v k+1 k px qx+k

0

Recall that qx+k really signifies 1 qx+k — the 1 is dropped by convention. Just as we examined the integral formula for A 1 it is worth looking at the stu↵ inside the sum here more carefully. x:n

A1

x:n

=

n X1

interest discount z }| {

v

0

k+1

·

p

k x |{z}

survival from 0 to k

prob of death between times k and k + 1 z }| { qx+k ·

For each year k after the year of issue, we are taking the present value of one dollar paid at the end of that year and multiplying it times the probability of surviving to year k (k px ) times the probability of dying in year k (qx+k ). After all, to get paid at the end of the k-th year of coverage, you have to survive to that year and then die before the year is over. Here’s the detailed formula for A 1 once again so you can see how closely connected the curtate x:n

and continuous forms of insurance are.

A1

x:n

=

discount Z n interest z}|{ vt

0

prob of death between times t and t + dt ·

t px

|{z}

survival from 0 to t

z }| {

µx (t)dt

·

Remember, with curtate, we take sums instead of integrals. Also, the present value of a benefit for death at time t = 1.1 after issue is the same as for death at time t = 1.9 after issue since both benefits would be paid at the same time. Var[Z] = A1 2

x:n

where, as before 2 A1

x:n

✓

A1

x:n

◆2

,

is just E[Z] with the force of interest doubled. In other words, 2

A1

x:n

=

n X1

v 2(k+1) k px qx+k .

0

EXAMPLE: Write A1

x:n

and A

1 x: n

in terms of p, q, and v for n = 1, n = 2.

SOLUTION: Arch MLC, Spring 2009

c Yufeng Guo

74

Chapter 2 A1

x:1

A1

x:2

= vqx and A

1 x: 1

= vpx

= vqx + v 2 px qx+1 and A

1 x: 2

= v 2 2 px

You should probably do at least up to n = 4 to make sure you have the pattern down. You want to be able to reproduce this type of thing quickly on exam day. } Let’s concentrate on A1

x:2

for a minute. Rearranging things slightly, we can see that A1

x:2

=

vqx + vpx (vqx+1 ). The first term is just one year of term insurance starting at age x and the second term is vpx times one year of term insurance starting at age (x + 1). This is an important way of looking at things. The equation says that a 2-year term insurance provides one year of term insurance plus, if you survive, you get one year of term insurance starting at age x + 1. The vpx term is discounting the second year of term insurance for interest (v) and survivorship (px ). You have to live one year to get the second year of insurance. More generally, we have our first important recursion relation. A1

x:n

= vqx + vpx A

1 x+1:n 1

You can learn recursions like this in two ways. First, know how to derive the recursive formula rigorously (refer to the textbook). After knowing how to derive the formula, explain the meaning of the formula in plain English. This is what the formula means: an n-year term insurance on x is equal to a 1-year term insurance plus, if the insured is still alive one year later, a brand new (n 1)-year term insurance on (x + 1). This makes lot of sense. However, remember to discount the “if you survive” part for both interest and survivorship. Name: Whole Life Description: Pays benefit at the end of year of death at any time during the future (over the whole life of the insured) bk+1 = 1 k = 0, 1, 2, ... vk+1 = v (k+1) k = 0, 1, 2, ... Z = v K+1 K = 0, 1, 2, ... E(Z) = Ax =

1 X

v (k+1) k px qx+k

0

After our careful recursion work above, the next formula should make sense: Ax = vqx + vpx · Ax+1 Arch MLC, Spring 2009

c Yufeng Guo

75

Chapter 2 It says that a whole life policy for (x) is the same thing as a 1-year term policy, plus, if (x) survives one year, a whole life policy starting at age x + 1. Hopefully, the pattern of conversion from continuous (payable at moment of death) to curtate (payable at end of year of death) insurances can be seen from these two examples. For the rest, we’ll just list the expression for the APV and a formula. Note that Equation 4.3.13 in the text has a typo, there should be no bar over the A. It should be Ax:n =

n X1

v k+1 k px qx+k + v n n px .

0

More insurances follow: Name: Increasing n-year term

APV (IA)1

Decreasing n-year term

(DA)1

Increasing Whole Life

(IA)x

m-year deferred n-year term

m|n Ax

x:n

x:n

Table 4.3.1 is useful for learning these types of insurance. Remember, it’s the same idea as for the payable at moment of death products. Except that now time is indicated by an integer (k) rather than a continuous measure (t). The list of recursion equations that begins on Page 119 of the text is useful and likely to be tested one way or another. It is really unfortunate that the authors decided to confuse matters by using (y x) instead of n for the number of years under the brackets! We have already discussed recursions like (a), (b), and (c), so we will look at (d), (f), and(g). I would rather write them as m|n Ax

(DA)1

x:n

= 0 + vpx

(m 1)|n Ax+1

= nvqx + vpx (DA)

1 x+1:n 1

(IA)x = Ax + vpx (IA)x+1

(d) (e) (f )

For each of these, you should convince yourself why they are logically correct. Recognize the following similarity among the formulas. In each case, PV of benefit at age x = b1 vqx + vpx · (PV of benefits at age x + 1). EXAMPLE: You are given: A1

x:30

= 0.1, Ax+30 = 0.5, and Ax:30 = 0.7. Find the value of Ax .

SOLUTION: Arch MLC, Spring 2009

c Yufeng Guo

76

Chapter 2 A whole life policy is equal to a 30-year term policy plus (if you survive) a whole life policy starting at age x + 30, so Ax = A 1

x:30

+ v 30

30 px Ax+30 .

We have everything we need in this equation except for v 30 A 1 . But x: 30

Ax:30 = A1

x:30

so A

1 x: 30

+A

30 px ,

which is just

1 , x: 30

= 0.6. Ax = 0.1 + (0.6)(0.5) = 0.4.

}

EXAMPLE: You are given: • i = 0.06

• px = 0.95

• Ax = 0.6. Find the value of Ax+1 . SOLUTION: Ax = A 1

x:1

v=

+ v px Ax+1 = vqx + v px Ax+1 .

1 = 0.9434, and qx = 0.05 1.06

0.6 = (0.9434)(0.05) + (0.9434)(0.95)Ax+1 ) Ax+1 = 0.617

Note that Ax+1 > Ax , as we would expect.

}

EXAMPLE: DeMoivre’s Law Mortality is according to DeMoivre’s Law with ! = 80. (A) Assuming i = 0.06, find A30:10 ; (B) Assuming i = 0, find (IA)30 . SOLUTION: Arch MLC, Spring 2009

c Yufeng Guo

77

Chapter 2 (A) Ax:n = A1

x:n

+A

=

1 x: n

k px

qx+k =

!

=

1 x

!

⌘

v k+1 k px qx+k + v n n px .

0

Under DeMoivre’s law,

and

n X1 ⇣

x !

k

=

k x

1 . # of yrs left to !

So the sum reduces to Ax:n =

n X1 ✓

v

!

0

A30:10 =

1

k+1

✓

9 1 X 1 50 0 1.06

= 0.02

x

◆

◆k+1

◆ 10 ✓ X 1 k

1.06

1

+ vn

+

✓

!

x !

1 1.06

n x

◆10 ✓

40 50

◆

+ 0.4467.

Now we use a fact for geometric series (this could come up on the test!) that says, for r 6= 1, n X r rn+1 (rk ) = . 1 r 1 )

◆ 10 ✓ X 1 k 1

1.06

=

(0.9434) (0.9434)11 = 7.36. 1 (0.9434)

) A30:10 = 0.02(7.36) + 0.4467 = 0.5939.

(B) This one is a little simpler than Part (A) because of the i = 0 assumption. (IA)30 =

49 X

(k + 1)v k k px qx+k

0

Since i = 0, the v’s drop out and k px qx+k

=

1 !

x

as in part (A). So (IA)30 =

49 50 1 X 1 X (50)(51) (k + 1) = k= = 25.5 50 0 50 1 2(50)

Whenever you are reading a problem and you see the assumption i = 0, it could mean that they are simplifying some really messy computations for you. } Arch MLC, Spring 2009

c Yufeng Guo

78

Chapter 2

4.4 Relationships between Insurances Payable at the Moment of death and the End of the Year of Death • Option A reference: Actuarial Mathematics Chapter 4.4 • Option B reference: Models for Quantifying Risk Chapter 9.5 The lesson of this section is that there is NOT much of a relationship between continuous insurance and curtate insurance unless you assume the Uniform Distribution of Deaths (UDD) holds in the year of death. We saw UDD back in Chapter 3 as “linear interpolation.” On the exam, SOA tends to describe this assumption as UDD. Under UDD, the following key identity holds: i A x = Ax . To remember that the multiplier is i , not i , remember that i > (the annual interest rate should be greater than the instant interest rate)1 and that A x > Ax (in A x the death benefit is paid sooner). It is also true under UDD that A(m) = x

i i(m)

Ax

(m)

where Ax is the APV of a whole life insurance that pays the death benefit and the end of (12) the m-th of a year of death. (So Ax is the APV of a whole life insurance that pays the benefit at the end of the month of death.) For completeness, we’ll list some others: i = A1

A1

x:n

x:n

i = (IA)1

(IA)1

x:n

Note that we have never used the symbol A

x:n

1. x: n

This is because a benefit of 1, payable upon

survival for n years has the same value whether things are continuous or curtate. Therefore A 1 = A 1 . This creates a problem since it means that x: n

x: n

i A x:n 6= Ax:n . Instead, we use A x:n = A 1

x:n

+A

1 x: n

i = A1

x:n

+A

1 x: n

.

EXAMPLE: You are given: 1

1+i=e =1+ +

1 2 2

Arch MLC, Spring 2009

+

1 3 3

+ .... This gives us i =

+

c Yufeng Guo

1 2 2

+

1 3 3

+ .... So i > .

79

Chapter 2 • v 20 = 0.5 •

20 px

= 0.8

• Ax = 0.4

• Ax+20 = 0.7

• Deaths are uniformly distributed over the year of death. Find A) A x , and B) A x:20 . SOLUTION: (A) i A x = Ax . v 20 = 0.5 ) i = 0.0353 ) = 0.0347. 0.0353 Ax = (0.4) = 0.407. 0.0347 (B) This one is a little harder. A x:20 = A 1

x:20

+A

1 x: 20

i = A1

x:20

= (1.0173)A1

x:20

To find A1

x:20

+ v 20 20 px

+ 0.4.

, we use the fact that Ax = A 1

x:20

) A1

x:20

= 0.4

+ v 20 20 px Ax+20 . (0.4)(0.7) = 0.12.

) A x:20 = (1.0173)(0.12) + 0.4 = 0.522.

}

EXAMPLE: Step 1: Go to Page 123 of the text. Step 2: Cover up the solution to Example 4.4.1! Step 3: Work the example. This is a very realistic example of the type of question that might show up on exam day. It requires you to use the rule of moments as well as the UDD connection between curtate and continuous functions.

Chapter 4 Suggested Problems: 6, 7, 10, 11ac, 14a, 16, 23, 26a (Solutions are available at archactuarial.com) Arch MLC, Spring 2009

c Yufeng Guo

80

Chapter 2

CHAPTER 4 Formula Summary CONTINUOUS INSURANCES n-year term:

E(Z) = A 1

=

x:n

Z n 0

✓

Var(Z) = 2A 1

A1

x:n

A x = E(Z) =

Whole Life:

Z 1 0

Ax =

Ax

µ µ+

Ax =

2

A

1 x: n

µ µ + (2 )

= E(Z) = v n n px

Var[Z] = 2 A

Endowment:

A x:n = A 1

+A

x:n

A x:n =

Z n 0

=

✓

1 x: n

A

1 x: n

◆2

1 x: n

v t t px µx (t)dt + v n n px

m-year deferred whole life:

m|A x

⌘2

are constant, then:

Pure Endowment:

If µ and

x:n

◆2

v t t px µx (t)dt

⇣

Var(Z) = 2A x

If µ and

v t t px µx (t)dt

m|A x = E(Z) =

Z 1 m

v t t px µx (t)dt

m|A x

= Ax

A1

t

µdt =

µ e µ+

x:m

are constant, then:

Z 1 m

v t px µx (t)dt = t

Arch MLC, Spring 2009

Z 1 m

e

e

µt

c Yufeng Guo

m(µ+ )

81

Chapter 2 Increasing and Decreasing Insurances: (IA)x =

Z 1

(DA)1

= E(Z) =

x:n

0

⇣

bt + 1cv t px µx (t)dt t

Z n

v T (n

IA

⌘

x

=

Z 1 0

tv t t px µx (t)dt

btc) t px µx (t)dt

0

DISCRETE INSURANCES (The usual formulas hold for Var(z).) n-year term:

E(Z) = A1

x:n

2

=

=

A1

E(Z) = Ax =

v k+1 k px qx+k

0 n X1

x:n

Whole Life:

n X1

v 2(k+1) k px qx+k

0

1 X

v (k+1) k px qx+k

0

Endowment: Ax:n =

n X1

v k+1 k px qx+k + v n n px

0

Various Recursions: (Try to understand the logic behind each one.) A1

x:n

= vqx + vpx A

(DA)1

x:n

m|n Ax

= nvqx + vpx (DA)

= 0 + vpx

x:n

(IA)x = Ax + vpx (IA)x+1

1 x+1:n 1

(m 1)|n Ax+1

Under UDD: i A x = Ax (IA)1

Ax = vqx + vpx · Ax+1

1 x+1:n 1

A(m) = x

i i(m)

Ax

A1

x:n

i = A1

x:n

i = (IA)1

x:n

BUT: A x:n 6= i Ax:n ;

Arch MLC, Spring 2009

Instead, we use A x:n = A 1

x:n

c Yufeng Guo

+A

1 x: n

i = A1

x:n

+A

1 x: n

82

Chapter 2

Past SOA/CAS Exam Questions: 1. An investment fund is established to provide benefits on 400 independent lives age x. (i) On January 1, 2001, each life is issued a 10-year deferred whole life insurance of 1000, payable at the moment of death. (ii) Each life is subject to a constant force of mortality of 0.05. (iii) 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

Solution: µ E(Z) = 1000 e µ+ Var(Z) = (1000)

2

10(µ+ )

µ e µ+2

= 1000 ·

10(µ+2 )

✓

✓

5 12

5 12

◆

◆2

·e

e

1.2

2.4

!

= 125.5 = 23, 610.16

E(S) = 400 · E(Z) = 50, 200

Var(S) = 400 · Var(Z) = 9, 444, 064 0.95 = P r

Key: A

Arch MLC, Spring 2009

E(S) k 50, 200 p p 9, 444, 064 Var(S)

S

) k = 1.645 ⇥

!

p

9, 444, 064 + 50, 200 = 55, 255

c Yufeng Guo

83

Chapter 2 2. A risky investment with a constant rate of default will pay: (i) principal and accumulated interest at 16% compounded annually at the end of 20 years if it does not default; and (ii) 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 actuarial 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

Solution: The expected values at the end of 20 years are (1.1)20 and (1.16)20 e present values at time zero are (1.1)20 ⌫ 20 and (1.16)20 e So 1.1 = 1.16e

µ

20µ .

Actuarial

20µ ⌫ 20 .

and µ = log 1.16 1.1 = 0.0531.

S(x0.5 ) = e

µx0.5

= 0.5 ) x0.5 =

log 0.5 0.6931 = = 13.05 µ 0.0531

Key: E

3. A decreasing term life insurance on (80) pays (20 if (80)dies in year k + 1, for k = 0, 1, 2, . . . , 19.

k) at the end of the year of death

(i) i = 0.06 (ii) For a certain mortality table with q80 = 0.2, the single benefit premium for this insurance is 13. (iii) For this same mortality table, except that q80 = 0.1, the single benefit premium for this insurance is P . Calculate P . (NOTE: “Single Benefit Premium” means the same thing as “Actuarial Present Value”) (A) 11.1

(B) 11.4

Arch MLC, Spring 2009

(C) 11.7

(D) 12.0 c Yufeng Guo

(E) 12.3 84

Chapter 2 Solution:

✓

(DA) 1

= 20v q80 + v p80 (DA) 1

80:20

81:19

◆

For the case when q80 = 0.2, 13 =

20(0.2) 0.8 + (DA) 1 1.06 1.06 81:19

) (DA) 1

81:19

=

13(1.06) 0.8

4

= 12.225

And for the case when q80 = 0.1, (DA) 1

= 20v(0.1) + v(0.9)(12.225)

80:20

=

2 + 0.9(12.225) = 12.267 1.06

Key: E

4. For a special 3-year term insurance on (x) , you are given: (i) Z is the present-value random variable for the death benefits. (ii) qx+k = 0.02(k + 1) k = 0, 1, 2 (iii) The following death benefits, payable at the end of the year of death: k 0 1 2

bk+1 300,000 350,000 400,000

(iv) i = 0.06 Calculate E(Z). (A) 36,800

(B) 39,100

(C) 41,400

(D) 43,700

(E) 46,000

SOLUTION: AP V (x’s benefit) =

2 X

v k+1 bk+1 k px qx+k

k=0

h

= 36,829

Arch MLC, Spring 2009

c Yufeng Guo

i

= 1000 300v(0.02) + 350v 2 (0.98)(0.04) + 400v 3 (0.98)(0.96)(0.06) Key: A

85

Chapter 2 5. 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 benefit premiums using: (i) mortality based on the Illustrative Life Table, (ii) i = 0.05 The company calculates contract premiums as 112% of benefit premiums. The single contract 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 contract premium at age 65. (A) 0.030

(B) 0.035

(C) 0.040

(D) 0.045

(E) 0.050

SOLUTION: 10,000 A63 (1.12) = 5233 A63 = 0.4672 Ax+1 =

Ax (1 + i) px

qx

A64 =

(0.4672)(1.05) 0.01788 = 0.4813 1 0.01788

A65 =

(0.4813)(1.05) 0.01952 = 0.4955 1 0.01952

Single contract premium at 65 = (1.12)(10,000)(0.4955) = 5550 5550 (1 + i) = 5233 2

=) i =

r

5550 5233

1 = 0.02984

Key A

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86

Chapter 2 6. For a whole life insurance of 1 on (41) with death benefit payable at the end of year of death, you are given: (i) i = 0.05 (ii) p40 = 0.9972 (iii) A41 (iv)

2

A41

A40 = 0.00822 2

A40 = 0.00433

(v) Z is the present-value random variable for this insurance. Calculate Var(Z). (A) 0.023

(B) 0.024

(C) 0.025

(D) 0.026

SOLUTION: Var(Z) = 2 A41 A41

A40

0.00822 = A41

= A41

✓

(A41 )2 (vq40 + vp40 A41 )

0.0028 0.9972 + A41 1.05 1.05

) A41 = 0.21650 2

A41

2

A40 = 0.00433 = 2 A41 = A41 2

✓

(E) 0.027

⇣

◆

v 2 q40 + v 2 p40 2 A41

0.0028 0.9972 2 + · A41 1.052 1.052

◆

⌘

) 2 A41 = 0.07193 Var(Z) = 0.07193

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(0.2165)2 = 0.02544

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87

Chapter 2 7. For an increasing 10-year term insurance, you are given: (i) bk+1 = 100, 000(1 + k), k = 0, 1, . . . , 9 (ii) Benefits are payable at the end of the year of death. (iii) Mortality follows the Illustrative Life Table. (iv) i = 0.06 (v) The single benefit premium for this insurance on (41) is 16,736. Calculate the single benefit premium for this insurance on (40). (A) 12,700 (B) 13,600 (C) 14,500 (D) 15,500 (E) 16,300

SOLUTION: 100, 000(IA)1

40:10|

ment]

= 100, 000⌫p40 [(IA)1

41:10|

= 15, 513.



Where A1

= A40

=

100, 000 0.99722 1.06

40:10|

0.16736

10

8,950,901 9,287,264 1.0610

10⌫ 10 9 p41 q50 ] + A1

40:10|

(100, 000) [see com-

⇥ (0.00592) + (0.02766 ⇥ 100, 000)

10 E40 A50

= 0.16132 (0.53667)(0.24905) = 0.02766. Comment: the first line comes from comparing the benefits of the two insurances. At each of age 40, 41, 42, . . . , 49 (IA)1 provides a death benefit 1 greater than (IA)1 . Hence the A1

40:10|

40:10|

term. But (IA)1

41:10|

41:10|

provides a death benefit at 50 of 10, while

(IA)1 provides 0. Hence a term involving 9| q41 = 9 p41 q50 . The various ⌫’s and p’s 40:10| just get all actuarial present values at age 40. Key: D

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88

Chapter 2 8. 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 single benefit premium for a special 2-year term insurance, with: (i) Benefits: k bk+1 0 1000 1 500 (ii) Mortality follows the Illustrative Life Table. (iii) i = 0.06 The actual experience of the fund is as follows: k 0 1

Interest Rate Earned 0.070 0.065

Number of Deaths 1 1

Calculate the di↵erence, 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 (B) 870 (C) 900 (D) 930 (E) 960

SOLUTION: A1

30:2|

= 1000⌫q30 + 500⌫ 2 1| q30 ⇣

⌘

⇣

⌘2

1 1 = 1000 1.06 (0.00153) + 500 1.06 (0.99847)(0.00161) = 2.15875 Initial fund = 2.15875 ⇥ 1000 participants = 2158.75 Let Fn denote the size of Fund 1 at the end of year n. F1 = 2158.75(1.07) 1000 = 1309.86 F2 = 1309.86(1.065) 500 = 895.00 Expected size of Fund 2 at end of year 2 = 0 (since the amount paid was the single benefit premium). Di↵erence is 895. Key: C

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89

Chapter 2 9. For a special whole life insurance on (x), payable at the moment of death: (i) µx (t) = 0.05, t > 0 (ii)

= 0.08

(iii) The death benefit at time t is bt = e0.06t , t > 0. (iv) Z is the present value random variable for this insurance at issue. Calculate V ar(Z). (A) 0.038 (B) 0.041 (C) 0.043 (D) 0.045 (E) 0.048

SOLUTION: R

E[Z] ⇣= 01 b v t ·t px µ(x + t)dt = ⌘⇥ t ⇤1 1 100 = 20 e 0.07t 0 = 57 7 R

2

R 1 0.06t 0.08t 0.05t 1 e e 0 e 20 dt

E[Z 2⇣ ] = ⌘01 bt v t ·t px µ(x + t)dt = ⇥ 0.09t ⇤1 1 100 5 = 20 e 0 = 9 9 V ar[Z] = Key: D

5 9

⇣ ⌘2 5 7

Arch MLC, Spring 2009

R 1 0.12t 0.16t 0.05t 1 1 R1 0.09t dt e e dt = 0 e 20 20 0 e

= 0.04535.

c Yufeng Guo

90

Chapter 2 10. For a group of individuals all age x, you are given: (i) 25% are smokers (s); 75% are nonsmokers (ns). (ii) s ns k qx+k qx+k 0 0.10 0.05 1 0.20 0.10 2 0.30 0.15

(iii) i = 0.02. Calculate 10, 000A1x:2e for an individual chosen at random from this group. (A) 1690 (B) 1710 (C) 1730 (D) 1750 (E) 1770

SOLUTION: Let ns = nonsmoker and s = smoker (ns)

(ns)

(s)

(s)

k = qx+k px+k qx+k px+k 0 .05 0.95 0.10 0.90 1 .10 0.90 0.20 0.80 2 .15 0.85 0.30 0.70 1(ns)

(ns)

(ns) (ns)

Ax:2e = ⌫qx + ⌫ 2 px qx+1 1 1 = 1.02 (0.05) + 1.02 2 0.95 ⇥ 0.10 = 0.1403 1(s)

(s)

(s) (s)

Ax:2e = ⌫qx + ⌫ 2 px qx+1 1 1 = 1.02 (0.10) + 1.02 2 0.90 ⇥ 0.20 = 0.2710 1 Ax:2 = weighted average = (0.75)(0.1403) + (0.25)(0.2710) = 0.1730 Key: C

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91

Chapter 2 11. Z is the present value random variable for a 15-year pure endowment of 1 on (x): (i) The force of mortality is constant over the 15-year period. (ii) ⌫ = 0.9 (iii) Var(Z) = 0.065E[Z] Calculate qx . (A) 0.020 (B) 0.025 (C) 0.030 (D) 0.035 (E) 0.040

SOLUTION: Variance = ⌫ 30 ⇥15 px ⇥15 qx

Expected value = ⌫ 15 ⇥15 px

⌫ 30 ⇥15 px ⇥15 qx = 0.065⌫ 15 ⇥15 px

⌫ 15 ⇥15 qx = 0.065 )15 qx = 0.3157

Since µ is constant ⇣

15 qx = 1

(px )15

(px )15 = 0.6843 px = 0.975,

⌘

qx = 0.025

Key: B

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Chapter 2

Problems from Pre-2000 SOA-CAS exams 1. For a special whole life insurance on (x), you are given: • µx (t) = µ, t 0 • 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 single benefit 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 greater than or equal to 0.07 but less than 0.08 greater than or equal to 0.08 but less than 0.09 greater than or equal to 0.09 but less than 0.10 greater than or equal to 0.10

2. You are given: • • • •

A35:1 = 0.9434 A35 = 0.1300 p35 = 0.9964 (IA)35 = 3.71

Calculate (IA)36 . (A) 3.81

(B) 3.88

(C) 3.94

(D) 4.01

(E) 4.08

3. You are given: • • • •

qx = 0.10 qx+1 = 0.20 i = 0.12 Deaths are uniformly distributed over each year of age.

Calculate 2A 1 . x:2

(A) 0.206

(B) 0.209

Arch MLC, Spring 2009

(C) 0.218

(D) 0.224

c Yufeng Guo

(E) 0.232

93

Chapter 2 4. You are given: • Deaths are uniformly distributed over each year of age. • i = 0.10

• qx = 0.05

• qx+1 = 0.08 Calculate A 1

x:2

(A) 0.103

(B) 0.108

(C) 0.111

(D) 0.114

(E) 0.119

5. You are given: • s(40) = 0.500 • s(41) = 0.475 • i = 0.06

• A 41 = 0.54

• Deaths are uniformly distributed over each year of age. Calculate A40 . (A) 0.483

(B) 0.517

Arch MLC, Spring 2009

(C) 0.523

(D) 0.531

c Yufeng Guo

(E) 0.645

94

Chapter 2

Solutions 1. Key: C (E[Z])2

Var(Z) = E[Z 2 ]

E[Z] = A x:10 + 0.510|A x = A x =

µ µ+µ =

e

10µ

1 2

10 Ex

· A x+10 + 0.5 

µ + 0.5 e µ+µ

e

10µ

·

10µ

1 e 4

20µ

= 0.3324 ) µ = 0.02 =

e

h

10µ

10| ExA x+10

·

µ µ+µ

i

For E[Z 2 ], we can’t use the rule of moments since the benefit is not always 1 or zero. E[Z ] = 2

Z 10 0

= (0.02)

Z 10

e

0.06t

0

dt +

Z 1 1 2t v t px µ dt + v t px µ dt 2t

10

Z 0.02 1

4

=

e

0.06t

dt =

10

1 3

3 e 12

) Var[Z] = 0.1961

0.6

4

1h e 3

0.06t

i10 0

1 h e 12

0.06t

i1 10

= 0.1961

(0.3324)2 = 0.0856

2. Key: A (IA)35 = vq35 + vp35 (A36 + (IA)36 ) A35 = vq35 + vp35 A36 ) (IA)35

A35 = vp35 (IA)36

A35:1 = A 1

35:1

[(IA)35 A35 ] vp35

+ vp35 = vq35 + vp35 = v = 0.9434

(IA)36 =

Arch MLC, Spring 2009

) (IA)36 =

3.71 0.13 = 3.808 (0.9434)(0.9964)

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95

Chapter 2 3. Key: C = ln 1.12 = 0.1133287

) 2 = 0.2266574

The i corresponding to 2 is i = e2 2

A1

x:2

2

A1

x:2

=

1 = 0.2544

0.10 0.9(0.20) + = 0.194113 1.2544 (1.2544)2

0.2544 · (0.194113) = 0.217872 0.2266574

=

4. Key: D A1

x:2

= ln 1.1 = 0.0953 i

A1

x:2

=

h

=

ih

i = A1

x:2

and v =

1 = 0.9091 1.1

vqx + v 2 px qx+1

i

i 0.1 0.9091(0.05) + (0.9091)2 (0.95)(0.08) = 0.114 0.0953

5. Key: B i A 41 = A41 ln 1.06 A41 = A 41 = · (0.54) = 0.5244202 i 0.06 s(41) 475 p40 = = = 0.95 ) q40 = 0.05 s(40) 500 Ax = vqx + vpx Ax+1 A40 = vq40 + vpx A41 =

Arch MLC, Spring 2009

1 (0.05) + 1.06

✓

1 1.06

c Yufeng Guo

◆

· 0.95 · (0.5244202) = 0.517

96

Chapter 3

ACTUARIAL MATHEMATICS: CHAPTER 5 LIFE ANNUITIES • Option A reference: Actuarial Mathematics Chapter 5 • Option B reference: Models for Quantifying Risk Chapter 10 The previous chapter discussed life insurances which provide payment contingent upon death. We now turn to a lighter topic – annuities. Annuities provide benefit payments contingent upon survival of the insured. There are several di↵erent annuity variations to learn in this chapter. Again, the critical items from this chapter are to understand the idea of an annuity and to learn the notation. Usually when we think of a life annuity, we think of giving an insurance company a pile of money, then receiving payments for some specified period of time. That is a classic annuity you purchase from a financial institution. We will be thinking of annuities in additional ways. For example: 1. Consider a 30 year fixed mortgage. In a sense, the mortgage company has purchased from the homeowner an annuity with a monthly payment interval. 2. Consider the premiums you pay after being issued a life insurance policy. The insured is paying an annuity with annual, semi-annual, or monthly payments to the insurance company. The insured is exchanging an annuity of premium payments for insurance coverage.

5.2 Continuous Life Annuities • Option A reference: Actuarial Mathematics Chapter 5.2 • Option B reference: Models for Qualifying Risk Chapter 10.1.3 In all of these examples, the payment is a unit benefit of 1 per year. It will be easy to generalize to di↵erent benefit amounts once we have the basic concepts. Name: Whole Life Annuity 97

Chapter 3 Description: Provides a payment continuously throughout the life of the annuitant. (Even though the payment is “continuous,” the rate of payment is 1 per year.) Let Y = (Present value of future payments) = a T . Then the actuarial present value for this annuity is represented by ax , and E(Y ) = ax =

Z 1 0

a t t px µ(x + t)dt .

This is the messier formula for ax and we’re going to simplify it. But first, let’s look at the integrand and convince ourselves that this is the right expression. The first part a t represents the present value of this annuity for a person that lives exactly t years. The second part t px µ(x + t) represents the probability of living exactly t years and then dying. Integrating over all possible times t produces the expected value. The text slips in some new notation at this point. Recall the actuarial present value symbol for an n-yr pure endowment, A 1 . That is the notation for a pure endowment in an “insurance” x: n

context. The notation for this same idea in an annuity context is n Ex . You need to be equally comfortable with either symbol. Now we have what we need to simplify the expression for ax . The book shows how to use the chain rule to prove ax =

Z 1 0

t

v t px dt

✓

=

Z 1 0

t Ex dt

◆

This is the form we find easiest to remember (actually the middle expression is the best). For once the book does a good job describing how to think of this integral so I will just quote the book’s explanation of it (page 135 of Bowers): This integral can be considered as involving a momentary payment of 1dt made at time t, discounted at interest back to time 0 by multiplying by v t and further multiplied by t px to reflect the probability that a payment is made at time t. Basically, as long as you live (t px ) you will continue to receive a payment of 1dt and for APV, we have to discount the payment back to time 0. Must have been a di↵erent person writing that part of the book! In general, the APV for a continuous annuity is Z 1 0

v t · Pr[payments are being made at time t] · [payment rate at time t]dt .

Just as with insurance, there is an natural recursion relationship with annuities: ax = ax:1 + vpx ax+1 . In this expression ax:1 represents the APV of a continuous annuity that pays at the rate of 1 per year for up to a year, or until you die, whichever comes first. It is called an n-year temporary life annuity (with n = 1 in this case!). The equation says that a lifetime continuous annuity on (x) is equal to a 1-year temporary life annuity plus, if you survive, another life annuity starting at age x + 1. Note that the annuity starting at age x + 1 is discounted for both interest and survivorship. Arch MLC, Spring 2009

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98

Chapter 3 EXAMPLE: You are given: •

10 p50

= 0.9;

• a60 = 8;

= 0.05 a50 = 10

Find a50:10 . SOLUTION: a50 = a50:10 + 10 E50 a60 . 10 = a50:10 + e

10(0.05)

(0.9)(8)

) a50:10 = 5.63.

}

EXAMPLE: DeMoivre’s Law Mortality is subject to DeMoivre’s Law with ! = 80 and

= 0.05. Find a30 .

SOLUTION: We’ll do this with integration by parts. a30 =

Z ! x 0

= R

Integration by Parts:

v t px dt = t

1 50

Z 50

v=

t

20e

t

e

! !

0

0.05t

x

(50

t x

dt

t) dt.

0

u dv = uv

u = 50

e

Z 50

R

v du du =

dt

dv = e

0.05t

0.05t dt

So the integral equals 1 50



20e

=

20 h e 50

=

0.05t

(50

t)

20

Z 50

e

0.05t

dt

0

0.05t

(t

20 h 0 + 20e 50

2.5

50) + 20e + 50

0.05t

i 50 0

i

20 = 12.66

A quick check is to see that the annuity is definitely less than 1/ , the value of a perpetuity. } Arch MLC, Spring 2009

c Yufeng Guo

99

Chapter 3

The most important equation so far(!) The following relation is key to life contingencies and if you are comfortable with it, you will go far on the exam! It ties together insurance and annuities. 1 = ax + A x

(5.2.8)

The book derives this using a “relation familiar from interest theory.” It says that whatever the future lifetime T of (x), 1 = aT + vT . You might remember this formula in the form aT =

1

vT

from Interest Theory. Along with this formula, two very important facts are used to derive equation (5.2.8): 1. E[v T ] = A x 2. E[a T ] = ax We should put each of these equations into words so you won’t forget them. Be familiar with these and ready to use them without prompting! 1. “The expected value of 1 discounted over the total future lifetime of (x) is equal to the APV of a benefit of 1 paid at the death of (x).” 2. To me this is just the definition of ax . Di↵erent forms of equation (5.2.8) will definitely be required on the exam so it’s a good one to know backwards, forwards, and all rearranged. The most important form might be ax =

1

Ax

(Remember This One!)

.

You might find this form the easiest to remember because it looks just like aT =

1

vT

from Interest Theory (And not by coincidence!). We know that E[a T ] = ax and we can use our new equations to find Var[a T ]. Var[a T ] = Var

"

1

vT

#

=

1 2

h

Var 1

And we know from Chapter 4 that Var[v T ] = 2A x Var[a T ] = Arch MLC, Spring 2009

2A

x

i

vT =

Var[v T ] 2

(A x )2 , so (A x )2 2

c Yufeng Guo

100

Chapter 3 EXAMPLE: Constant Force of Mortality 1. Assuming constant force of mortality, find an expression for ax in terms of µ and . 2. Assuming constant force of mortality, find an expression for Var[a T ]. ⇣

⌘

3. Find Pr a T > ax . (These are the questions of Example 5.2.1 in the book, I hope our approach is simpler for you than the text’s.)

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101

Chapter 3 SOLUTION: 1. ax = 2.

2A

Var[a T ] =

1

Ax

(A x )2

x

2

µ µ+

1

=

=

"

1 2

=

1 µ+ ✓

µ µ+2

µ µ+

◆2 #

3. We’ll use the book’s solution for this one but try to explain what they are doing. This kind of question makes an appearance now and then because it separates prepared students from unprepared students. It is the kind of question that, if you are not ready for it, can leave you kind of flustered. Let’s start by just rewriting the only part of the expression with a variable in it. ! ⇣ ⌘ 1 vT Pr a T > ax = Pr > ax Although there are lots of constants, there is still only one variable in this expression. We will get this variable by itself on one side of the inequality. ⇣

= Pr v < 1

✓

⌘

ax = Pr e

T

= Pr



T < log



= Pr T >

1

✓

log

t0 ] is just the probability that (x) lives past time t0 . We know this probability as t0 px and we know from Chapter 3, that in the presence of constant mortality, t0 px = e µt0 , so ⇣

⌘

Pr a T > ax = e

µt0

=

✓

µ µ+

◆µ

(Be sure you can work out the last step.) I realize that this problem was long with lots of messy algebra, but it is worth going over since many of the questions of this type are similar. First, isolate the random variable in the inequality, then look at what you have and interpret it in terms of t px . } Useful notes to remember: Under constant forces of interest and mortality, ax = Arch MLC, Spring 2009

1 +µ

Ax =

µ µ+

c Yufeng Guo

2A

x

=

µ 2 +µ

102

Chapter 3 EXAMPLE: Constant Force of Mortality You are given: µ = 0.025,

= 0.06.

Find ax:10 . SOLUTION: ax = ax:10 + 10 Ex ax+10 Since we have constant force of mortality, both ax and ax+10 are equal to 1 µ+

=

1 = 11.76. 0.085

Also 10 Ex

=e

0.085(10)

) ax:10 = ax

= 0.427.

10 Ex ax+10

= 6.738.

}

Name: n-yr Temporary Life Annuity Description: Payment of benefits until the earlier of death or end of n-year period. Again, Y is the present value of future benefits. Y = E(Y ) =

Z n 0

(

aT an

0T a ¯x ). (A) 0.40

(B) 0.44

Arch MLC, Spring 2009

(C) 0.46

(D) 0.48 c Yufeng Guo

(E) 0.50 118

Chapter 3 Solution:

⇣

⌘

Pr a T > ax = Pr 

= Pr T >

1

=e

=

Key: C

✓

µ µ+

◆µ

⌫T

✓

µ log µ+ 1

= t0 px where t0 = µt0

1

=

✓

> ax ◆

µ log µ+

✓

6 10

◆3 2

!

◆

= 0.4648

3. For a whole life annuity-due of 1 on (x), payable annually: (i) qx = 0.01 (ii) qx+1 = 0.05 (iii) i = 0.05 (iv) a ¨x+1 = 6.951 Calculate the change in the actuarial present value of this annuity-due if px+1 is increased by 0.03. (A) 0.16

(B) 0.17

(C) 0.18

(D) 0.19

(E) 0.20

Solution: a ¨x = 1 + ⌫px + ⌫ 2 px px+1 a ¨x+2 Let y denote the change in px+1 . a ¨(with increase) = 1 + ⌫px + ⌫ 2 px (px+1 + y)¨ ax+2 = a ¨(without) + y⌫ 2 px a ¨x+2 y⌫ 2 px a ¨x+2 = change in actuarial present value a ¨x+1 = 6.951 = 1 + ⌫px+1 a ¨x+2 = 1 +

1 (1 1.05

0.05)¨ ax+2

Therefore, a ¨x+2 = 6.577. ✓

1 0.03 1.05

◆2

0.99 ⇥ 6.577 = 0.177

Key: C Arch MLC, Spring 2009

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119

Chapter 3 4. A person age 40 wins 10,000 in the actuarial lottery. Rather than receiving the money at once, the winner is o↵ered the actuarially equivalent option of receiving an annual payment of K (at the beginning of each year) guaranteed for 10 years and continuing thereafter for life. You are given: (i) i = 0.04 (ii) A40 = 0.30 (iii) A50 = 0.35 (iv) A140:10 = 0.09 Calculate K. (A) 538

(B) 541

(C) 545

(D) 548

(E) 551

Solution: The person receives K per year guaranteed for 10 years. Therefore, K¨ a 10 = 8.4353K. The person received K per years alive starting 10 years from now. So the relevant APV is 10| a ¨40 K. Hence we have 10,000 = (8.4353 + 10 E40 a ¨50 )K. Derive

10 E40

: A40 = A140:10 + (10 E40 )A50 10 E40

=

A40

a ¨50 =

A140:10 A50

1

=

A50 1 = d

0.30 0.09 = 0.60 0.35 0.35 0.04 1.04

= 16.90

Plug in values: 10,000 = (8.4353 + (0.60)(16.90))K = 18.5753K K = 538.35 Key: A

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Chapter 3 5. Y is the present-value random variable for a special 3-year temporary life annuity-due on (x). You are given: (i)

t px

= 0.9t , t

0

(ii) K is the curtate-future-lifetime random variable for (x). (iii) Y =

8 > < 1.00, K = 0

1.87, K = 1

> : 2.72, K = 2, 3, . . .

Calculate Var(Y ). (A) 0.19

(B) 0.30

(C) 0.37

(D) 0.46

(E) 0.55

Solution: Pr(K = 0) = 1 Pr(K = 1) = 1 px

2 px

px = 0.1 = 0.9

0.81 = 0.09

Pr(K > 1) = 2 px = 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 = 6.407 Var(Y ) = 6.407

2.47152 = 0.299

Key: B

6. 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: (i)

= 0.06 (ii) A¯x = 0.40 (iii) 2 A¯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

c Yufeng Guo

(E) 11.10

121

Chapter 3 Solution: E[YAGG ] = 100 E[Y ] = 100(10,000)ax = 100(10,000)

Y

=

q

Var[Y ] =

r

(10,000)2

0.9 = Pr

1 ⇣2 Ax 2 p

AGG

=

F

E[YAGG ]



100

Y

2

⌘

Ax =

1

Ax

!

(10,000) q (0.25)

= 10,000,000 (0.16) = 50,000

= 10(50,000) = 500,000

>0

) 1.282 =

AGG

F = 1.282

AGG

F

E[YAGG ] AGG

+ E[YAGG ]

F = 1.282(500,000) + 10,000,000 = 10,641,000 Key: D

7. For a special 30-year deferred annual whole life annuity-due of 1 on (35): (i) If death occurs during the deferral period, the single benefit premium is refunded without interest at the end of the year of death. (ii) a ¨65 = 9.90 (iii) A35:30 = 0.21 (iv) A135:30 = 0.07 Calculate the single benefit premium for this special deferred annuity. (A) 1.3

(B) 1.4

(C) 1.5

(D) 1.6

(E) 1.7

Solution: Let ⇡ denote the single benefit premium. ⇡ = 30| a ¨35 + ⇡A 1

35:30

⇡=

¨35 30| a 1

=

A1

=

35:30

(0.21 1

✓

A35:30 1

A1

35:30

A1

◆

a ¨65

35:30

0.07)9.9 1.386 = = 1.49 0.07 0.93

Key: C

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122

Chapter 3 8. A government creates a fund to pay this year’s lottery winners. You are given: (i) There are 100 winners each age 40. (ii) Each winner receives payments of 10 per year for life, payable annually, beginning immediately. (iii) Mortality follows the Illustrative Life Table. (iv) The lifetimes are independent. (v) i = 0.06 (vi) 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

SOLUTION: Let Y = present value random variable for payments on one life. S=

P

Y = present value random variable for all payments. E[Y ] = 10¨ a40 = 148.166

Var[Y ] = 102

2A

A240

40

d2

= 100(0.04863

0.161322 )(1.06/0.06)2 = 705.55

E[S] = 100E[Y ] = 14,816.6 Var[S] = 100 Var[Y ] = 70,555 By normal approximation, need

) Std Dev [S] =

p

70555 = 265.62

E[S] + 1.645 Standard Deviations = 14,816.6 + (1.645)(265.62) = 15,254 Key: E

9. For a continuous whole life annuity of 1 on (x): (i) The force of interest and force of mortality are constant and equal. (ii) ax = 12.50 Calculate the standard deviation of a T (x) . (A) 1.67

(B) 2.50

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(C) 2.89

(D) 6.25

c Yufeng Guo

(E) 7.22 123

Chapter 3 SOLUTION: 1 µ+

12.50 = ax =

Ax =

µ µ+

⇣

Var a T

) µ + = 0.08 ) µ =

= 0.5

⌘

=

2A

Ax =

2 2

Ax

x 2

S.D. =

p

=

1 3

= 0.04

µ 1 = µ+2 3 1 4

0.0016

= 52.083

52.083 = 7.217

Key: E

10. For a group of individuals all age x, you are given: (i) 30% are smokers and 70% are non-smokers. (ii) The constant force of mortality for smokers is 0.06. (iii) The constant force of mortality for non-smokers is 0.03. (iv)

= 0.08 ⇣

Calculate Var a T (x) (A) 13.0

⌘

for an individual chosen at random from this group.

(B) 13.3

(C)13.8

(D) 14.1

(E) 14.6

SOLUTION: h

i

h

i

h

i

A x = E v T (x) = E v T (x) |N S ⇥ Prob(N S) + E v T (x) |S ⇥ Prob(S) =

✓

0.03 0.03 + 0.08

Similarly, 2A x = ⇣

✓

Arch MLC, Spring 2009

⇥ 0.70 +

0.03 0.03 + 0.16

Var a T (x) Key: D

◆

⌘

=

2A

◆

✓

0.6 0.06 + 0.08

⇥ 0.70 + 2

Ax

x 2

=

✓

◆

⇥ 0.30 = 0.3195

0.06 0.06 + 0.16

◆

⇥ 0.30 = 0.1923

0.1923 0.31952 = 14.1 0.082

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124

Chapter 3 11. You are given: (i) Ax = 0.28 (ii) Ax+20 = 0.40 (iii) A

1 x: 20

= 0.25

(iv) i = 0.05 Calculate ax:20 . (A) 11.0

(B) 11.2

(C) 11.7

(D) 12.0

(E) 12.3

SOLUTION: (i) ax:20 = a ¨x:20 1

(ii) a ¨x:20 =

Ax:20

(iii) Ax:20 = A1

d

x:20

(iv) Ax = A1

x:20

0.28 = A1

1 + 20 Ex

+A

1 x: 20

+ 20 Ex Ax+20

+ (0.25)(0.40) ) A1

x:20

x:20

= 0.18

Now plug into (iii): Ax:20 = 0.18 + 0.25 = 0.43 1 0.43 ⌘ = 11.97 Now plug into (ii): a ¨x:20 = ⇣ 0.05 1.05

Now plug into (i): ax:20 = 11.97

1 + 0.25 = 11.22

Key: B

12. For an annuity payable semiannually, you are given: (i) Deaths are uniformly distributed over each year of age. (ii) q69 = 0.03 (iii) i = 0.06 (iv) 1000A 70 = 530 (2)

Calculate a ¨69 (A) 8.35

(B) 8.47

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(C) 8.59

(D) 8.72

c Yufeng Guo

(E) 8.85 125

Chapter 3 SOLUTION:

ln(1.06) A70 = A 70 = (0.53) = 0.5147 i 0.06 1 A70 1 0.5147 a ¨70 = = = 8.5736 0.06 d 1.06 a ¨69 = 1 + vp69 a ¨70 = 1 +

(2)

a ¨69 = ↵(2)¨ a69

✓

◆

0.97 (8.5736) = 8.8457 1.06

(2) = (1.00021)(8.8457)

0.25739 = 8.5902

(m)

Note that the approximation a ¨x ⇡ a ¨x (m2m1) works well (closer to the key than to any other option and is o↵ from the correct answer by less than 0.01). Since m = 2, this estimate gives 1 8.8457 = 8.5957 4

13. Your company currently o↵ers 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

(C) 100,000

(D) 150,000

(E) 200,000

SOLUTION: Annuity Benefit: Death Benefit: New Benefit:

K+1

Z1 = 12,000 1 vd Z2 = Bv K+1 K+1 Z = Z1 + Z2⇣= 12,000 1 v⌘d + Bv K+1 = 12,000 + B 12,000 v K+1 d

K = 0, 1, 2, . . . K = 0, 1, 2, . . .

d

✓

◆

⇣ ⌘ 12, 000 2 Var v K+1 d 12,000 Var(Z) = 0 when B = = 150,000 0.08

Var(Z) = B

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126

Chapter 3 14. You are given: (i) µx (t) = 0.03, t (ii)

0

= 0.05

(iii) T (x) is the future lifetime random variable. (iv) g is the standard deviation of a T (x) . ⇣

Calculate P r a T (x) > ax (A) 0.53

⌘

g .

(B) 0.56 (C) 0.63 (D) 0.68 (E) 0.79

SOLUTION: R

a ¯x = R01 e 0.08t dt = 12.5 A¯x = 0R1 e 0.08t (0.03)dt = 3/8 = 0.375 2A ¯x = 1 e 0.13t (0.03)dt = 3/13 = 0.23077 0q q ⇥ ⇤ p (¯ a ) = V ar[¯ a ] = 12 2 A¯x (A¯x )2 = 400 [0.23077 hT |

Pr a ¯T | > a ¯x = Pr

h

⌫T

1 h 0.05

T|

i

h

(¯ aT | ) = P r a ¯T | > 12.5 i

h

> 6.4952 = P r 0.67524 > e i

0.67524 = P r T > ln0.05 = P r [T > 7.85374] 0.03⇥7.85374 =e = 0.79 Key: E

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i

(0.375)2 ] = 6.0048

6.0048

0.05T

i

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127

Chapter 3 15. 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: (i) For typical annuitants of all ages, mortality follows De Moivre’s Law with the same limiting age, !. (ii) i = 0.06 Calculate the annual benefit that your company can o↵er to this individual. (A) (B) (C) (D) (E)

38, 000 41, 000 46, 000 49, 000 52, 000

SOLUTION: For De Moivre’s Law: ex = ! 2 x 1 k| qx = ! x Ax =

! P x 1 k=b

⌫ k+1 k| qx =

a

1 ! x

! P x 1

⌫ k+1

k=b

Ax = !! xx a ¨x = 1 dAx e50 = 25 ) ! = 100 for typical annuitants ey = 15 ) y = Assumed age = 70 a 30 A70 = 30 = 0.45883 a ¨70 = 9.5607 500000 = b¨ a70 ) b = 52, 297 Key: E

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128

Chapter 3 16. At interest rate i: (i) a ¨x = 5.6 (ii) The actuarial present value of a 2-year certain and life annuity-due of 1 on (x) is a ¨x:2 = 5.6459. (iii) ex = 8.83 (iv) ex+1 = 8.29 Calculate i. (A) 0.077 (B) 0.079 (C) 0.081 (D) 0.083 (E) 0.084

SOLUTION: ex = px + px ex+1 ) px =

ex 1+ex+1

=

8.83 9.29

= 0.95048

a ¨x = 1 + ⌫px + ⌫ 2 ·2 px + . . . a ¨x:2 = 1 + ⌫ + ⌫ 2 ·2 px + . . . a ¨x:2 a ¨x = ⌫qx = 5.6459 5.60 = 0.0459 ⌫(1 0.95048) = 0.0459 ⌫ = 0.9269 i = ⌫1 1 = 0.0789 Key: B

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c Yufeng Guo

129

Chapter 3 17. For a special 3-year temporary life annuity-due on (x), you are given:

(i)

t 0 1 2

Annuity Payment 15 20 25

px+t 0.95 0.90 0.85

(ii) i = 0.06 Calculate the variance of the present value random variable for this annuity. (A) 91 (B) 102 (C) 114 (D) 127 (E) 139

SOLUTION: Event x=0 x=1 x=2

P rob (0.05) (0.95)(0.10) = 0.095 (0.95)(0.90) = 0.855

Present Value 15 15 + 20/1.06 = 33.87 15 + 20/1.06 + 25/1.062 = 56.12

E[X] = (0.05)(15) + (0.095)(33.87) + (0.855)(56.12) = 51.95 E[X 2 ] = (0.05)(15)2 + (0.095)(33.87)2 + (0.855)(56.12)2 = 2813.01 V ar[X] = E(X 2 ) E(X)2 = 2813.01 (51.95)2 = 114.2 Key: C

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130

Chapter 3

Problems from Pre-2000 SOA-CAS exams 1. For a 3-year temporary life annuity-due on (30), you are given: • s(x) = 1

0  x  80

x 80 ,

• i = 0.05 ( a ¨ K+1 , • Y = a ¨3 ,

K = 0, 1, 2 K = 3, 4, 5, ...

Calculate Var(Y ). (A) 0.08

(B) 0.29

(C) 0.36

(D) 0.60

(E) 0.93

2. You are given: (4)

• a ¨ 1 = 17.287 • Ax = 0.1025

• Deaths are uniformly distributed over each year of age. (4)

Calculate a ¨x . (A) 15.48

(B) 15.51

(C) 15.75

(D) 15.82

(E) 15.86

3. You are given: • T (x) is the random variable for the future lifetime of (x). • µx+t = µ, t •

0 is the constant force of mortality.

=µ

⇣

⌘

Calculate Var a T . 1 (A) 12µ

(B) µ +

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1 (C) 6µ

(D)

1 2

c Yufeng Guo

✓

µ 4 +1

◆

(E)

✓

1 6µ

◆2

131

Chapter 3 Use the following information for the rest of the questions: For (x), you are given: • T (x) is the random variable for the future lifetime of (x). • µx+t = 0.04, t •

= 0.06

4. Determine (A)

0

d dn (n Ex )

0.1 n e

(D) 0.1 e

(B)

0.1n

0.1 e

(E) 0.1 n e

0.1n

(C)

0.1n

0.1 e

0.1

0.1n

5. Calculate the standard deviation of a T . (A) 5.0

(B) 6.0

(C) 6.5

(D) 25.0

(B) 10 1

e

(E) 42.3

6. Determine ax:n . (A) 10 1

e

(D) 10 1

0.67 e

7. Determine (A)

2e

d dx

0.1n

0.1 0.1n

0.1n

(C) 10 1 + e

(E) 10 1 + 0.67 e

0.1n

0.1n

ax:n . (B)

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0.33 e

0.1n

(C) 0

c Yufeng Guo

(D) 0.33 e

0.1n

(E) 2 e

0.1n

132

Chapter 3

Solutions to Pre-2000 Problems: Chapter 5 1. Key: A This is a Demoivre’s Law problem. Death is uniformly distributed evenly over the next 50 years. Let’s do this the way we might do it in a Statistics course. Here is the distribution for Y P (Y = y) 0.02 (only if (30) dies in first year) 0.02 (only if (30) dies in 2nd year) 0.96

y a ¨1 = 1 a ¨2 = 1 + v = 1.952 a ¨3 = 1 + v + v 2 = 2.859

E[Y ] = (0.02) · 1 + (0.02)(1.952) + (0.96)(2.859) = 2.804

E[Y 2 ] = (0.02) · 12 + (0.02)(1.952)2 + (0.96)(2.859)2 = 7.943 Var[Y ] = 7.943

2. Key: A (4)

a ¨1 =

(2.804)2 = 0.0806

1 = 17.287 d(4) 1

1+i=

d(4) 4

1

!4

) d(4) = 0.05785 = 1.06

Under UDD,

i

A(4) x = h

i(4) = 4 · (1 + i)0.25 a ¨(4) x

i

Ax

i(4)

1 = 0.0587

=

1

) i = 0.06

) A(4) x = 0.10477

(4)

Ax

= 15.476

d(4)

3. Key: A Var[a T ] =

2A

(A x )2

x

2

This is constant force, so Ax =

µ µ+

=

1 2

Var[a T ] =

Arch MLC, Spring 2009

and 1 3

1 4 2

Ax =

2

=

µ 1 = µ+2 3

1 1 = 2 12 12 µ

c Yufeng Guo

133

Chapter 3 4. Key: B n Ex

= v n n px = e

n

d n Ex = dn

5. Key: A Ax =

µ µ+

Var[a T ] =

= 0.4 2A

x

µn

e

0.1e

Ax =

2

(A x )2 2

=

=e

0.1n

0.1n

µ = 0.25 µ+2

0.25 0.16 = 25 0.062

So the standard deviation is 5.

6. Key: B ax:n = ax

n Ex ax+n

=

1 µ+

e

0.1n

1 µ+

= 10

10e

0.1n

7. Key: C

d d ax:n = (10 10e 0.1n ) = 0 dx dx There is no x in the expression for ax:n . You could also answer this one intuitively. Since constant force applies, the value of this annuity has to be the same at every age. Therefore the derivative with respect to age must be zero.

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134

Chapter 4

ACTUARIAL MATHEMATICS: CHAPTER 6 BENEFIT PREMIUMS • Option A reference: Actuarial Mathematics Chapter 6 • Option B reference: Models for Quantifying Risk Chapter 11 Consider a simple market with yourself as an insurer. You would like to sell benefits to customers like the death benefits and endowments we learned about in Chapter 4. For o↵ering this coverage, you will receive a stream of premiums. This stream of premiums will be exactly like one of the annuities that we studied in Chapter 5. Also, issues of profit and expenses are not dealt with in this chapter. We will be assuming that the APV of the benefit exactly equals the APV of the annuity of contract premiums used to purchase the benefit. Even if the expected values of the benefit and the contract premiums are the same, you as the insurer are still facing the risk of a loss. Suppose, for example, you sell a discrete death benefit of 1000 in return for annual premium payments of 100 per year. Although this may be extremely profitable, the insured may die one month after issue leaving you with a huge loss. To analyze this situation, we will look the ‘Loss function’: Loss = PV of Benefits

PV of Premiums.

In this expression, the loss and PV’s are random variables (not expected values) whose values will depend on when death actually occurs. Suppose you as the insurer want your expected inflows to equal your expected outflows, so that: E(Loss) = E(PV of Benefits) E(PV of Premiums) = 0. In other words, you want E(PV of Benefits) = E(PV of Payments). Given a set of benefits, the premium that satisfies this equation is called the benefit premium. The equation itself 135

Chapter 4 is called the equivalence principle. If you keep the equivalence principle in mind, finding a benefit premium can be pretty straightforward.

6.2 Fully Continuous Premiums • Option A reference: Actuarial Mathematics Chapter 6.2 • Option B reference: Models for Quantifying Risk Chapter 11.1.2, 11.1.3, 11.1.4 Consider a whole life policy with a benefit of 1 payable immediately upon death. For any continuously paid premium (explained below), P , we have the loss function: L = vT

P aT

This loss equals the present value of the benefit (bt vt with bt = 1) minus the present value of the stream of premiums. In chapter 5, we considered only annuities with annual payments of size 1. P a T represents the expected value of a continuous stream of payments of P per year for life beginning at age x. The premium is simply a dollar amount – just like what you send your auto insurer. If the Equivalence Principle (E[L] = 0) is used to determine the continuous premiums for this whole life policy, the premium is denoted ⇣

P Ax

⌘

and is called the equivalence premium. To find the equivalence premium for this insurance, we would take expected values of the equation for L above to get ⇣

⌘

P Ax =

Ax . ax

E(L) = A x Rearranging algebraically,

⇣

P A x ax = 0. ⌘

So, the benefit premium based on the equivalence principle is simply the ratio of expected present value of benefits to the expected value of a corresponding annuity. Remember that and you basically have the most important concept in this chapter. As an insurer, you will be interested not only in the expected value of the loss on a product but also in the variance of the Loss Function. ⇣

Var(L) = Var v "

= Var v Arch MLC, Spring 2009

T

1+

t

P

!

"

P aT

⌘

= Var v

P

#

P

=

1+

c Yufeng Guo

t

P !2

1

vT

⇣

# ⌘

Var v T . 136

Chapter 4 But v T is just the present value of 1 dollar received at the moment of death (i.e., continuous whole life insurance), so we can simplify this expression further using the fact from Chapter 4 that ⇣ ⌘ ⇣ ⌘2 Var v T = 2A x Ax . ) Var(L) =

1+

P

!2 

⇣

2

Ax

Ax

⌘2

.

EXAMPLE: Constant Force of Mortality If the force of interest and force of mortality are constants, µ and , find an expression for the continuous benefit premium P for a continuous whole life policy paying a benefit of 1 per year. SOLUTION: Under constant interest and mortality assumptions, Ax =

µ µ+

and ax =

1 . µ+

The equivalence principle tells us that the APV of future premiums must equal the APV of future benefits, so: ⇣

⌘

P A x ax = A x , which reduces, after some algebra to

P (A x ) = µ This tells us that the annual premium (paid continuously!) to fund this whole life policy is equal to µ per year. This assumes that premiums are paid during the entire period that the insurance is in force. } This way of finding the benefit premium works for all types of insurance you will encounter on this exam. The key is to match up the Insurance benefit with the “corresponding annuity”. For example, the annuity stream of premium payments for ten-year term insurance is di↵erent than for whole life insurance. EXAMPLE: Constant Force of Mortality If, in the previous example, µ = 0.01, and SOLUTION: Var(L) = Arch MLC, Spring 2009

1+

P

= 0.04, find Var(L).

!2 

2

Ax

c Yufeng Guo

⇣

Ax

⌘2

137

Chapter 4 We need A x = 0.2,

ax = 20, P (A x ) = 0.01, µ 2 Ax = = 0.11. µ+2

) Var(L) = [0.11

✓

0.04] 1 +

0.01 0.04

◆2

= (0.07)(1.56) = 0.109.

}

EXAMPLE: Illustrative Life Table You are given: • Deaths are uniformly distributed over the year of death.

• Mortality and interest are as described in the illustrative life table in the text. Find the continuous annual premium for a whole life policy on (40) that pays 10,000 at the moment of death.

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138

Chapter 4 SOLUTION: We are looking for 10,000P (A 40 ) = 10,000

A 40 . a40

The illustrative life table only provides values of Ax , but since we have UDD, we can use i 0.06 A 40 = A40 = (0.163242) = 0.1661. log (1 + 0.06) a40 =

1

) 10,000P (A 40 ) =

A 40

= 14.31,

(10,000)(0.1661) = 116.06 . 14.31

}

EXAMPLE: Illustrative Life Table Find the variance of the loss function for the insurance and premium in the previous example. SOLUTION: First we find the variance as if the benefit were 1. Var(L) = ✓

0.01161 = 1+ 0.05827

1+ ◆2 "

P

!2 

2

Ax

2i + i2 2 · A40 2

= 1.438 [(1.0606)(0.048633)

⇣

Ax

⌘2 2

(0.1661)

#

(0.0276)] = 0.0345.

Since, our policy has death benefit and premium multiplied by 10,000, the variance will be (10,000)2 Var(L). So the variance we want is (10,000)2 (0.0345) = 3,450,000.

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}

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Chapter 4 Table 6.2.1 is worth some discussion. (Part of Table 6.2.1, Page 173 of Bowers et al.) Fully Continuous Benefit Premiums Loss Components

Premium Formula P =

E[bT vT ] E[Y ]

Plan Whole life insurance

bT vT

Y

1 vT

aT

n-Year term insurance

1 vT 0

aT , T  n an , T > n

P (A x:n ) =

n-Year endowment insurance

1 vT 1 vn

aT , T  n an , T > n

P (A x:n ) =

h-Payment whole life insurance

1 vT 1 vT

aT , T  h ah , T > h

h P (A x )

h-Payment, n-year endowment insurance

1 vT 1 vT 1 vn

aT , T  h ah , h < T  n ah , T > n

h P (A x:n )

n-Year pure endowment

0 1 vn

aT , T  n an , T > n

P (A

P (A x ) =

A ax

1

=

1) x: n

A1

x:n

ax:n

A x:n ax:n

Ax ax:h

=

=

A x:n ax:h

A

1 x: n

ax:n

Column 1 is pretty self-explanatory: the name of the type of insurance and payment plan (more on the payment plans in a minute). Column 2 shows the present value of the benefit that will be paid, and Column 3 shows the present value of the premiums that will pay for the benefits. Finally, the most important column is Column 4, which shows the formula and notation for the benefit premium. You want to be able to cover up Column 4 and reproduce it quickly looking only at Column 1. Notice that all of the premiums follow the same general scheme; only the forms of the annuity and the death benefit change. The most important payment scheme shown is the one having h payments. For these plans, the insured pays premiums for the first h years that the insurance is in force, at which point the insurance is “paid up” and continues in force for the rest of the coverage period without further premiums.

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Chapter 4 EXAMPLE: Fred is 50 years old today and is interested in buying some continuously payable 10-pay whole life insurance. That is, he pays his premium continuously for the next 10 years or until he dies, whichever comes first. At the moment of death, he will receive 1000. Using the following assumptions, find the benefit premium for this insurance. • a50 = 13.3 , a60 = 11.1 • A 50 = 0.249 • •

10 p50

= 0.92

= 0.04

SOLUTION: It looks like there might be too much information here and it is not obvious from the start how everything is going to fit together, so we should begin by writing down the equivalence principle and then search for the quantities we need. Having carefully studied Table 6.2.1 above we are able to correctly use the notation to write: ⇣ ⌘ h P A x ax:h = 1000A x , 10 P

We know that

⇣

⌘

A 50 a50:10 = 1000A 50 .

a50:10 = a50

10 E50 a60 ,

and 10 E50

Therefore,

) a50:10 = 13.3 ⇣

= e(

0.04)(10)

10 p50 .

(0.67032)(0.92)(11.1) = 6.455. ⌘

249 = 38.58. 6.455 So the premium for this insurance is 38.58 per year, payable continuously. 10 P A 50 =

}

What does it mean to pay 38.58 per year continuously? Suppose the annual premium is 38.58. This idea of “continuously paid” can be thought of if the insurance company has a direct tap into your checking account. It deducts money at a continuous rate such that over a year’s time, 38.58 has been removed from your account. This is pretty absurd but it fits nicely into the mathematical theory of actuarial mathematics. Furthermore, some insurances where premiums are collected weekly approach this idea. So far we have only looked at premiums paid continuously. Most people are more comfortable with the idea of paying their premiums on a discrete basis, so you can guess what’s coming next ....

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Chapter 4

6.3 Fully Discrete Premiums • Option A reference: Actuarial Mathematics Chapter 6.3 • Option B reference: Models for Quantifying Risk Chapter 11.1.1, 11.1.2 These are the discrete analogs of the continuous cases. Consider a unit whole life policy with benefit payable at the end of the year of death. (See figure)

Benefit payable at time (K+1)

x

...

x+1 x+2

x+(K+1)

x+K x+T

We have the following loss function: L = v K+1

Px a ¨ K+1 .

Note: There is no bar over the P . The symbol Px denotes a whole life premium paid annually. a ¨ K+1 is an annuity with payments due at the beginning of each year starting at policy issue and including the year of death. (See figure)

Px x

Px

Px

x+1 x+2

Px

Px

... x+(K+1)

x+K x+T

Taking the expected value gives us ⇣

E(L) = E v K+1

⌘

⇣

⌘

Px E a ¨ K+1 .

Just as with continuous premiums, we use the Equivalence Relation, E(L) = 0, to obtain 0 = Ax

Px a ¨x

So, we come back to the basic premium structure: Px = Arch MLC, Spring 2009

Ax . a ¨x

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Chapter 4 Once again, the benefit premium is just the death benefit divided by the appropriate annuity of premiums. We also have (as before) ✓

P Var(L) = 1 + d

◆2 h

2

Ax

i

(Ax )2 .

There is another form of the variance that might be necessary on the exam. Var(L) =

2A

(Ax )2 (d¨ ax )2

x

It is not obvious that these two expressions are the same, but you should be able to prove it using Px a ¨x = Ax , and a ¨x = 1 dAx . Either form of the variance might well be needed on the exam, and I would memorize both of them if you have time. If you don’t have time, just memorize the first one and practice the derivation. This structure can be extended to a variety of discrete insurance products. Table 6.3.1 is an excellent source for these formulas. (Part of Table 6.3.1, Page 173 of Bowers et al.) Fully Discrete Benefit Premiums Loss Components

Premium Formula

bK+1 vK+1

Y

1 v K+1

a ¨ K+1 , K = 0, 1, 2, · · ·

n-Year term insurance

1 v K+1 0

a ¨ K+1 , K = 0, 1, · · · , n a ¨ n , K = n, n + 1, · · ·

1

n-Year endowment insurance

1 v K+1 1 vn

a ¨ K+1 , K = 0, 1, · · · , n a ¨ n , K = n, n + 1, · · ·

1

h-Payment whole life insurance

1 v K+1 1 v K+1

a ¨ K+1 , K = 0, 1, · · · , h a ¨ h , K = h, h + 1, · · ·

1

h-Payment, n-year endowment insurance

1 v K+1 1 v K+1 1 vn

a ¨ K+1 , K = 0, 1, · · · , h a ¨ h , K = h, · · · , n 1 a ¨ h , K = n, n + 1, · · ·

1

n-Year pure endowment

0 1 vn

a ¨ K+1 , K = 0, 1, · · · , n a ¨ n , K + n, n + 1, · · ·

1

Arch MLC, Spring 2009

c Yufeng Guo

E[bK+1 vK+1 ] E[Y ]

P =

Plan Whole life insurance

Px =

P1

x:n

Ax a ¨x

=

Px:n =

h Px

=

h Px:n

P

1 x: n

A1

x:n

a ¨x:n

Ax:n a ¨x:n

Ax a ¨x:h

=

=

Ax:n a ¨x:h

A

1 x: n

a ¨x:n

143

Chapter 4 EXAMPLE: A professional cli↵ diver, age 30, wants to buy a 2-year term life policy with a benefit of 100,000 payable at the end of year of death. Interest is i = 0.08, and the mortality for the diver is shown in the table: x px 30 0.95 31 0.93 32 0.90 1. Find the equivalence premium for this insurance. 2. Find the standard deviation of the loss function. SOLUTION: ⇣

1. P30 a ¨30:2

⌘

= A1

x:2

and P (1 + v p30 ) = v q30 + v 2 p30 q31 .

)P =

0.0463 + 0.057 = 0.0550 1.8796

) 100,000P30 = 5,500.

2. Since the special variance formulas only hold for endowment insurance and whole life, we have to use elementary probabilities and the fact that Var(L) = E[L2 ]

(E[L])2 .

This will work well especially since we already know that E[L] = 0, since that is how P was chosen. Event dies 1st yr dies 2nd yr lives 2 yrs

L 100,000v 5,500 = 87,093 (100,000)v 2 5,500(1 + v) = 75,141 5,500(1 + v) = 10,593

Pr 0.05 0.0665 0.8835

–note: 0.0665 = (0.95)(0.07), and 0.8835 = (0.95)(0.93). Var(L) = E[L2 ] = (0.05)(87,093)2 + (0.0665)(75,141)2 + (0.8835)( 10,593)2 = 845,400,000. So the standard deviation of the loss function is

p

Var(L) = 29,076.

}

Semi-Continuous Premiums So far in this chapter, we have seen two types of insurance plans 1. Death benefits are paid at the moment of death and premiums are paid continuously. 2. Death benefits are paid at the end of the year of death and premiums are paid annually at the beginning of the year. Arch MLC, Spring 2009

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Chapter 4 The setup that comes closest to reality is a combination of these two: Death benefits payable at the moment of death while premiums are payable at the beginning of the year. To capture this ⇣ idea, ⌘ semicontinuous premiums were developed. Semicontinuous premiums are denoted P A x which indicates discrete premiums and continuous benefits. As you might expect ⇣

⌘

P Ax =

Ax . a ¨x

You are given:

EXAMPLE:

• Deaths are uniformly distributed over the year of death

• Interest and mortality are as given by the Illustrative Life Table. ⇣

⌘

Find P A 65 . SOLUTION:

⇣

⌘

A 65 . a ¨65 0.06 = (0.4398) = 0.4529 . log 1.06

P A 65 = UDD ) A 65 =

i

A65

⇣

⌘

) P A 65 =

0.4529 = 0.04575. 9.90

}

The 3-premium Principle: The next few examples deal with a premium concept that frequently makes an appearance on the exam. EXAMPLE: (These are just warm-ups to the main point.) 1. What premium, payable annually, is required to provide 25 years of term insurance to (30) with a death benefit of 1 ? 2. What premium, payable annually, is required to both provide 25 years of term coverage to (30) and provide (30) with a benefit of 1 upon attainment of age 55? 3. What premium, payable annually, is required to provide (30) with a benefit of w upon attainment of age 55? 4. What premium, payable annually for 25 years, will provide (30) with a whole life policy? SOLUTION: 1. P 1

30:25

2. P30:25

Arch MLC, Spring 2009

3. w P

1 30: 25

4.

25 P30

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}

145

Chapter 4 EXAMPLE:(more warmup) 1. Rank the following in order of size from largest to smallest and justify your answer in words (no mathematical proofs, please!). h Px ,

Px:h

P1

x:h

2. Xavier, Yuri, and Zelda each made a deal with an insurance company. • Xavier agreed to pay P 1

30:25

each year for 25 years and received the

corresponding coverage from the company. • Yuri agreed to pay P30:25 each year for 25 years and received the corresponding coverage from the company. • Zelda agreed to pay 25 P30 each year for 25 years and received the corresponding coverage from the company. If each of these three survived to age 55, describe what remained of each policy at that age. SOLUTION: 1. Px:h > h Px > P1 . To see why, note that all three premiums buy the same x:h

benefit for the first h years and then provide $1, Ax+h , and $0, respectively, upon attainment of age x + h. 2.

• Xavier will have nothing at age 55 (except the satisfaction of knowing that had he died in the previous 25 years, his beneficiary would be somewhat richer). • Yuri will have $1. • Zelda will have a whole life policy whose value is A55 . }

EXAMPLE: (This one is the main point!) Find an expression for

Px:n P

n Px

.

1 x: n

SOLUTION: We’ll approach this one conceptually, which is just how you should approach it on the exam. Recall that Px:n is the premium required to provide n years of $1 term coverage followed by a payment of $1 at age x + n should the insured survive to that point. Similarly, paying a premium of n Px for n years will provide n years of $1 term coverage and then will provide the insured with a whole life policy with value Ax+n should the insured survive to age x + n. Arch MLC, Spring 2009

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Chapter 4 Therefore Px:n 1 = 0 dollars of n Px is the premium required to provide 1 term coverage for n years followed by a payment of 1 Ax+n upon attainment of age x + n. But this just describes an n-year pure endowment with a payment of 1 Ax+n , and we know the premium for this is (1 Ax+n ) · P 1 . Therefore, x: n

Px:n and

= (1

n Px

Px:n P

n Px

Ax+n ) P

=1

1, x: n

Ax+n .

1 x: n

Reread this argument a couple of times if necessary and then you should be able to do the next couple of examples quickly! } EXAMPLE: 1. What is the value of

Px:n P

2. What is the value of

n Px

P

P1

x:n

?

1 x: n

P1

x:n

?

1 x: n

SOLUTION: 1. 1

2. Ax+n

}

6.4 True m-thly Payment Premiums • Option A reference: Actuarial Mathematics Chapter 6.4 • Option B reference: Models for Quantifying Risk Chapter 11.4 Sometimes premiums are paid more frequently than annually (say 12, 4, or 2 times per year). (m) Let Px denote a “true level annual benefit premium” payable in m-thly installments at the beginning of each m-thly period for a unit of whole life insurance payable at the end of the (m) year of death. Note that only one m-th of Px is paid at each installment. So if you are (12) buying a whole-life policy with a death benefit of 100, 000 and Px = 120, then you will be paying 10 each month. Table 6.4.1 shows notation and formulas for both discrete and continuous cases. Don’t be intimidated! You already know these functions and the basic premium structure. Just learn the new notation for m-thly payment and you’ll be set. It is always (m)

P# Arch MLC, Spring 2009

=

A# (m)

a ¨#

.

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Chapter 4 There is really nothing new here, the premium is the benefit divided by the annuity of payments. EXAMPLE: Illustrative Life Table You are given: ↵(2) = 1.00021

and

✓

(2) = 0.2574.

Use the Illustrative Life Table to find P (2) A 1 SOLUTION: P

A1

40:20

= A40

✓

(2)

A1

20 E40 A60

= 0.16132

40:20

◆

.

A1

40:20 (2) a ¨40:20

= 0.16132

.

1 l60 · (0.36913) 20 (1.06) l40

(0.2741)(0.36913) = 0.06013.

(2)

(2)

(2) ¨60 , 20 E40 a

a ¨40:20 = a ¨40 where

=

40:20

◆

(2)

(2) = (1.00021)(14.817)

0.2574 = 14.56,

(2)

(2) = (1.00021)(11.145)

0.2574 = 10.89.

a ¨40 = ↵(2)¨ a40 a ¨60 = ↵(2)¨ a60 (2)

)a ¨40:20 = (14.56) )P

(2)

✓

A1

40:20

◆

(0.2741)(10.89) = 11.57. =

0.06013 = 0.005197 11.57

So, for example, the biannual payment for this insurance with a benefit of 10,000 would be (0.5)(51.97) = 25.99. } On the test, you may be required to relate an m-thly premium to the corresponding annually payable premium. The connection is always through the insurance provided. EXAMPLE: Write h P (m) (A x ) as a multiple of h P (A x ). SOLUTION: Use the fact that both premiums are used to provide A x so hP

(m)

(m)

(A x ) a ¨x:h = A x = h P (A x ) a ¨x:h .

) h P (m) (A x ) = Arch MLC, Spring 2009

a ¨x:h

P (A x ). (m) h

a ¨x:h

c Yufeng Guo

} 148

Chapter 4 Percentile Premiums: This is one more topic that shows up several times in Chapter 6 and has made appearances on exams in the past. Definition: L(⇡) is the loss function for a given insurance, assuming that the premium charged is ⇡, (not necessarily the equivalence premium). So for a discrete, whole life policy, L(⇡) = v K+1

⇡a ¨ K+1 .

Before doing the following two examples, it is a good idea to start by doing example 6.3.3 b) in the text. EXAMPLE: Consider a fully continuous whole life insurance on (30). Mortality and interest are constants µ = 0.02, = 0.05, respectively. Find the smallest premium ⇡ such that the probability of loss is less than 0.5. SOLUTION: How big the loss L is depends only on how long you live. The longer you live the better it is for the insurance company (smaller loss). The way to solve this problem is to first find the age that the insured has probability 0.5 of living beyond. Then find the premium that makes the loss for that age equal to zero. The resulting premium will produce a loss half of the time. L = vT Assume 0.5 = t px = e

µt .

⇡ aT

This implies t=

log 0.5 = 34.66. 0.02

The premium that produces a loss of 0 when t = 34.66 is the one for which 0=v

34.66

⇡(a 34.66 ) = e

0.05(34.66)

) 0 = 0.1768

⇡

1

e

0.05(34.66)

0.05

!

⇡ (16.46)

) ⇡ = 0.0107

}

EXAMPLE: Consider a fully discrete whole life insurance on (30) where mortality and interest are as given by the Illustrative Life Table. Find the smallest premium ⇡ such that the probability of a positive loss on the insurance is less than 0.25.

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Chapter 4 SOLUTION: Remember, the longer you live, the lower the loss is. So we want to find t so that 75% of the time, the insured lives longer than t. We need to solve 0.75  k p30 . Since l30 = 95,014, we need to find the age a such that la = (0.75)(95014) = 71261. The correct age is somewhere between 67 and 68. Since losses are larger for earlier deaths, we play it safe by going with 67, so K = 37. Now the loss function is given by ✓ ◆ 1 38 K+1 L(⇡) = v ⇡a ¨ K+1 = ⇡a ¨ 38 . 1.06 So we want to solve

0 = 0.1092 ) ⇡ = 0.1092

⇡

1

v 38 , d

0.0566 = 0.00694 0.8908

}

One last note on this chapter. In some later chapters and in many exam questions, the phrase single benefit premium is used in the place of actuarial present value. If you pay for your insurance all in one premium at the beginning, you should expect to pay the APV (if no profit is involved!). Chapter 6 Suggested Problems: 4, 10(long but good practice! Do the 1st and last rows only), 13, 16, 17, 27, 29, 31, 32 (Solutions at archactuarial.com)

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Chapter 4

CHAPTER 6 Formula Summary Fully Continuous Equivalence Premium: ⇣

P Ax

⌘

Ax = ax

Var(L) =

1+

P

!2 

⇣

2

Ax

Ax

⌘2

Var(L) =

2A

x

⇣

Ax

( ax )2

⌘2

Under constant interest and mortality assumptions, Ax =

µ µ+

1 µ+

ax =

P (A x ) = µ

Fully discrete: Px =

Ax a ¨x

✓

Var(L) = 1 +

◆2 h

P d

Semicontinous:

2

⇣

(Ax )2

Ax

⌘

P Ax =

i

Var(L) =

2A

(Ax )2 (d¨ ax )2

x

Ax a ¨x

Sample 3-premium relation: Px:n P

n Px

=1

Ax+n

1 x: n

Be able to reason out this relation and the similar ones. m-thly premiums: (m)

P#

=

A# (m)

a ¨#

.

h-payment insurance: h P (A x )

=

Ax ax:h

h P (A x:n )

h Px

=

Ax a ¨x:h

h Px:n

=

=

A x:n ax:h

Ax:n a ¨x:h

Be familiar with the two big tables in this chapter summarizing the benefits and premiums for various types of continuous and discrete insurance. Arch MLC, Spring 2009

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Chapter 4

Past SOA/CAS Exam Questions: 1. The pricing actuary at Company XYZ sets the premium for a fully continuous whole life insurance of 1000 on (80) using the equivalence principle and the following assumptions: (i) The force of mortality is 0.15. (ii) i = 0.06 The pricing actuary’s supervisor believes that the Illustrative Life Table with deaths uniformly distributed over each year of age is a better mortality assumption. Calculate the insurer’s expected loss at issue if the premium is not changed and the supervisor is right. (A)

124

(B)

26

Solution: ⇡=

(C) 0

1000

Expected loss = 1000A80

(D) 37

R1

µe ( +µ)t dt R 10 ( +µ)t dt 0 e

(E) 220

=

µ 1000 µ+ 1 µ+

= 1000µ = 150

⇡a80 .

i 1000A80 = 1000 A80 = (1.0297)(665.75) = 685.53 a80 = Expected loss = 685.53

1

0.68553

150(5.3969) =

= 5.3969

124.

Key: A

2. For a fully discrete whole life insurance of 10,000 on (30): (i) ⇡ denotes the annual premium and L(⇡) denotes the loss-at-issue random variable for this insurance. (ii) Mortality follows the Illustrative Life Table (iii) i = 0.06 Calculate the lowest premium, ⇡ 0 , such that the probability is less than 0.5 that the loss L(⇡ 0 ) is positive. (A) 34.6

(B) 36.6

Arch MLC, Spring 2009

(C) 36.8

(D) 39.0

c Yufeng Guo

(E) 39.1

152

Chapter 4 Solution: Pr[L(⇡ 0 ) > 0] < 0.5 From Illustrative Life Table,

47 p30

) Pr[10,000v K+1

⇡0a ¨ K+1 > 0] < 0.5

= 0.50816 and

= 0.47681

48 p30

Since L is a decreasing function of K, to have Pr [L(⇡ 0 ) > 0] < 0.5 means we must have L(⇡ 0 )  0 for K 47. Highest value of L(⇡ 0 ) for K

47 is at K = 47.

L(⇡ 0 )[at K = 47] = 10,000v 47+1 L(⇡ 0 )  0 =) (609.98

⇡0 a ¨ 47+1 = 609.98

16.589⇡ 0 )  0 =) ⇡ 0 >

16.589⇡ 0

609.98 = 36.77 16.589

Key: C

3. Company ABC issued a fully discrete three-year term insurance of 1000 on Pat whose stated age at issue was 30. You are given:

(i)

x 30 31 32 33

qx 0.01 0.02 0.03 0.04

(ii) i = 0.04 (iii) Premiums are determined using the equivalence principle. During year 3, Company ABC discovers that Pat was really age 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

Solution: 0.01 0.99(0.020) (0.99)(0.98)(0.03) + + = 0.053796725 1.04 1.042 1.043

A130:3 =

a ¨30:3 = 1 +

0.99 (0.99)(0.98) + = 2.848927515 1.04 1.042

1 1000P30:3 = 1000

A131:3 = Arch MLC, Spring 2009

0.053796725 = 18.88315 2.848927515

0.02 0.98(0.03) (0.98)(0.97)(0.04) + + = 0.080215919 1.04 1.042 1.043 c Yufeng Guo

153

Chapter 4

0.98 (0.98)(0.97) + = 2.82119 1.04 1.042

a ¨31:3 = 1 +

1 1000P30:3 ⇥a ¨31:3 = BA131:3

53.2792954 = B(0.08215919)

) B = 664

Key: B

4. For a fully continuous whole life insurance of 1 on (x): (i) ⇡ is the benefit premium. (ii) L is the loss-at-issue random variable with the premium equal to ⇡. (iii) L⇤ is the loss-at-issue random variable with the premium equal to 1.25 ⇡. (iv) a ¯x = 5.0 (v)

= 0.08

(vi) Var(L) = 0.5625 Calculate the sum of the expected value and the standard deviation of L⇤ . (A) 0.59

(B) 0.71

(C) 0.86

Solution: In general Var(L) = 1 +

p 2

⇣

2A

(D) 0.89

2

Ax .

x

P (A x ) = ✓

⌘

1 ax

=

0.12 Var(L) = 1 + 0.08 ⇤

Var(L ) =

) Var(L⇤ ) = ⇣ 0.15ax = 1 E[L⇤ ] +

Arch MLC, Spring 2009

◆2 ⇣

1 5 Ax

0.08

1+

15 8

1+

12 8

0.08 = 0.12.

2

5 4 (0.12)

1+ ⇣

E[L⇤ ] = A x

(E) 1.01

⌘2

!2

2

⌘

A x = 0.5625, ⇣

2

Ax

⌘

⌘2 (0.5625) = 0.744

ax ( + 0.15) = 1

q

2

Ax .

Var(L⇤ ) = 0.7125

c Yufeng Guo

5(0.23) =

0.15 Key: B

154

Chapter 4 5. On January 1, 2002, Pat, age 40, purchases a 5-payment, 10-year term insurance of 100,000: • Death benefits are payable at the moment of death.

• Contract 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

Solution: 0L

= 100,000v 2.5

4,000¨ a 3 at i = 0.05

= 77,079

Key: A

6. For a special fully discrete 35-payment whole life insurance on (30): (i) The death benefit is 1 for the first 20 years and is 5 thereafter. (ii) The initial benefit premium paid during the each of the first 20 years is one fifth of the benefit premium paid during each of the 15 subsequent years. (iii) Mortality follows the Illustrative Life Table. (iv) i = 0.06 (v) A30:20 = 0.32307 (vi) a ¨30:35 = 14.835 Calculate the initial annual benefit premium. (A) 0.010

(B) 0.015

(C) 0.020

(D) 0.025

SOLUTION: Initial Benefit Prem = =

5(0.10248) 5(14.835)

30:20

Arch MLC, Spring 2009

=

5¨ a30:35

✓

4 A1

30:20

◆

4¨ a30:20

4(0.02933) 0.5124 0.11732 0.39508 = = = 0.015 4(11.959) 74.175 47.836 26.339

where A1

5A30

(E) 0.030

A30:20

A

1 30: 20

!

= 0.32307

c Yufeng Guo

0.29374 = 0.02933 155

Chapter 4 and a ¨30:20 =

1

A30:20 d

=

1

⇣

0.32307 0.06 1.06

⌘

= 11.959

Comment: The numerator could equally have been calculated as A30 + 4 · 20 E30 A50 = 0.10248 + (4)(0.29374)(0.24905) = 0.39510

Key B

7. For a special fully continuous whole life insurance on (x): (i) The level premium is determined using the equivalence principle. (ii) Death benefits are given by bt = (1 + i)t where i is the interest rate. (iii) L is the loss random variable at t = 0 for the insurance. (iv) 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 ) (v T A x ) (C) (1 + A x ) (D) (v T A x )(1 (v T + A x ) (E) (1 + A x )

A x)

SOLUTION: Let ⇡ denote the premium. L = bT v T

⇡a T = (1 + i)T ⇤ v T =1

E[L] = 1 )L=1 =

vT

⇡a T

⇡ax = 0 ! ⇡ =

⇡a T = 1 (1 ax

⇡a T

aT ax ax )

=

=

ax

1 ax (1 v T ) ax

vT A x 1 Ax

Key A Arch MLC, Spring 2009

c Yufeng Guo

156

Chapter 4 8. For a fully discrete 2-year term insurance of 1 on (x): (i) 0.95 is the lowest premium such that there is a 0% chance of loss in year 1. (ii) px = 0.75 (iii) px+1 = 0.80 (iv) 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

SOLUTION: v is the lowest premium to ensure a zero % chance of loss in year 1 (The present value of the payment upon death is v, so you must collect at least v to avoid a loss should death occur). Thus v = 0.95.

⇣

E(Z) = vqx + v 2 px qx+1 = 0.95 ⇥ 0.25 + (0.95)2 ⇥ 0.75 ⇥ 0.2 = 0.3729 ⌘

E Z 2 = v 2 qx + v 4 px qx+1 = (0.95)2 ⇥ 0.25 + (0.95)4 ⇥ 0.75 ⇥ 0.2 = 0.3478 ⇣

Var(Z) = E Z 2

⌘

[E(Z)]2 = 0.3478

(0.3729)2 = 0.21

Key: D

9. For a special 2-payment whole life insurance on (80): (i) Premiums of ⇡ are paid at the beginning of years 1 and 3. (ii) The death benefit is paid at the end of the year of death. (iii) There is a partial refund of premium feature: If (80) dies in either year 1 or year 3, the death benefit is 1000 + Otherwise, the death benefit is 1000.

⇡ 2.

(iv) Mortality follows the Illustrative Life Table. (v) i = 0.06 Calculate ⇡, using the equivalence principle. (A) 369

(B) 381

Arch MLC, Spring 2009

(C) 397

(D) 409

c Yufeng Guo

(E) 425

157

Chapter 4 SOLUTION: ⇣

⌘

⇡vq80 ⇡v 3 2 p80 q82 + 2 2 ✓ ◆ ✓ ◆ 0.83910 0.08030 0.83910 ⇥ 0.09561 ⇡ 1+ = 665.75 + ⇡ + 1.062 2(1.06) 2(1.06)3 ⇡ 1 + 2 p80 v 2 = 1000A80 +

⇡(1.74680) = 665.75 + ⇡(0.07156) ) ⇡ = 397.41

3,284,542 = 0.83910 3,914,365 = (1 0.08030)(1 0.08764) = 0.83910

Where 2 p80 = Or 2 p80

Key: C

10. For a special 3-year deferred whole life annuity-due on (x): (i) (ii) (iii) (iv) (v)

i = 0.04 The first annual payment is 1000. Payments in the following years increase by 4% per year. There is no death benefit during the three year deferral period. Level benefit premiums are payable at the beginning of each of the first three years. (vi) ex = 11.05 is the curtate expectation of life for (x). (vii) k k px 1 0.99 2 0.98 3 0.97 Calculate the annual benefit premium. (A) 2625

(B) 2825

(C) 3025

(D) 3225

(E) 3425

SOLUTION: ex = px + 2 px + 3 px + . . . = 11.05 Annuity = v 3 3 px 1000 + v 4 4 px ⇥ 1000 ⇥ (1.04) + . . . =

1 X

1000(1.04)k

3

v k k px = 1000v 3

k=3

= 1000v 3 (ex

0.99

0.98) = 1000

⇣

✓

⌘

1 1.04

◆3

1 X

k px

k=3

⇥ 9.08 = 8072

Let ⇡ = benefit premium: ⇡ 1 + 0.99v + 0.98v 2 = 8072 ) 2.8580⇡ = 8072 ) ⇡ = 2824 Arch MLC, Spring 2009

c Yufeng Guo

Key: B 158

Chapter 4 11. For a special fully discrete 10-payment whole life insurance on (30) with level annual benefit premium ⇡: (i) The death benefit is equal to 1000 plus the refund, without interest, of the benefit premiums paid. (ii) A30 = 0.102 (iii)

10| A30

= 0.088

(iv) (IA) 1

30:10

= 0.078

(v) a ¨30:10 = 7.747 Calculate ⇡. (A) 14.9

(B) 15.0

(C) 15.1

(D) 15.2

SOLUTION: ⇡¨ a30:10 = 1000A30 + ⇡(IA) 1

30:10

⇡=

=

(E) 15.3

+ (10⇡)

1000A30 a ¨30:10

(IA) 1

30:10

10

⇣

⇣

10| A30

10| A30

⌘

⌘

1000(0.102) 102 = = 15.024 7.747 0.078 10(0.088) 6.789

Key: B

12. (We’ll see more or this normal approximation idea in later chapters. The idea is to find the mean and variance for all of the lives and then treat the problem as if it is a normal distribution with that mean and variance.) 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: (i) µ = 0.04 (ii)

= 0.06

(iii) 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

Arch MLC, Spring 2009

(C) 500

(D) 610

c Yufeng Guo

(E) 720

159

Chapter 4 SOLUTION: Let Z be the present value random variable for one life. Let S be the present value random variable for the 100 lives. E(Z) = 10

Z 1

e t eµt µ dt = 10

5

⇣

E Z

2

⌘

= 10

2

✓

◆

µ e 2 +µ ⇣

Var(Z) = E Z 2

⌘

(2 +µ)5

µ e +µ

= 10

2

✓

( +µ)5

◆ 0.04 ⇣ e 0.16

[E(Z)]2 = 11.233

= 2.426 0.8

⌘

= 11.233

2.4262 = 5.348

E(S) = 100 E(Z) = 242.6 Var(S) = 100 Var(Z) = 534.8 F 242.6 p = 1.645 ! F = 281 534.8

Key: A

13. For a whole life insurance of 1000 on (x) with benefits payable at the moment of death: (i) t

=

(

(ii) µx (t) =

0.04, 0.05,

(

0 < t  10 10 < t

0.06, 0.07,

0 < t  10 10 < t

Calculate the single benefit premium for this insurance. (A) 379

(B) 411

(C) 444

(D) 519

(E) 594

SOLUTION: Since this is CFM, the solution is faster if we use the facts that Ax =

µ , µ+

A1

x:10

µ (1 µ+

=

10 Ex )

and 10 Ex

Then the solution is

=e 

1000A x = 1000 A 1 Arch MLC, Spring 2009

10(µ+ )

x:10

+ 10 Ex · A x+10

c Yufeng Guo

160

Chapter 4 = 1000

✓

◆⇣ 0.06 1 0.06 + 0.04 h

10(0.06+0.04)

e ⇣

= 1000 (0.6) 1

e

1

⌘

⌘

+e

+ 0.5833e

10(0.06+0.04)

1

i

✓

0.07 0.07 + 0.05

◆

= 593.86

14. Two actuaries use the same mortality table to price a fully discrete 2-year endowment insurance of 1000 on (x). (i) Kevin calculates non-level benefit premiums of 608 for the first year and 350 for the second year. (ii) Kira calculates level annual benefit premiums of ⇡. (iii) d = 0.05 Calculate ⇡. (A) 482

(B) 489

(C) 497

(D) 508

(E) 517

SOLUTION: d = 0.05 ) v = 0.95 From Kevin’s work:

608 + 350vpx = 1000vqx + 1000v 2 px (px+1 + qx+1 ) 608 + 350(0.95)px = 1000(0.95)(1 608 + 332.5px = 950(1 px =

px ) + 1000(0.9025)px (1) px ) + 902.5px

342 = 0.9 380

Now to arrive at Kira’s value for 1000Px:2 : 608 + 350(0.95)(0.9) = 1000Px:2 [1 + (0.95)(0.9)] 1000Px:2 =

Arch MLC, Spring 2009

299.25 + 608 = 489.08 1.855

c Yufeng Guo

161

Chapter 4 15. 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: (i)

= 0.04.

(ii) µx (t) = 0.02, t

0

(iii) The single benefit premium for this insurance is equal to Var(Z). Calculate b. (A) 2.75 (B) 3.00 (C) 3.25 (D) 3.50 (E) 3.75

SOLUTION: E[Z] = bA¯x since constant force A¯x = µ/(µ + ) bµ E(Z) = µ+ = b(0.02) (0.06) = b/3 T 2 T 2 2 ¯ V ar[Z] ✓ = V ar[b⌫ ] =◆b V ar[⌫ ] = b ( Ax = b2

h

µ µ+2

i

⇣

µ µ+

2 1 2 = b2 10 9 =b V ar(Z) ⇣ ⌘ = E(Z) 4 2 b 45 = 3b

⇣

⌘

4 b 45 = Key: E

1 3

⌘2

⇣

4 45

A¯2x )

⌘

) b = 3.75

Arch MLC, Spring 2009

c Yufeng Guo

162

Chapter 4 16. For a special 3-year term insurance on (30), you are given: (i) Premiums are payable semiannually. (ii) Premiums are payable only in the first year. (iii) Benefits, payable at the end of the year of death, are: k bk+1 0 1000 1 500 2 250 (iv) Mortality follows the Illustrative Life Table. (v) Deaths are uniformly distributed within each year of age. (vi) i = 0.06 Calculate the amount of each semiannual benefit premium for this insurance. (A) 1.3 (B) 1.4 (C) 1.5 (D) 1.6 (E) 1.7

SOLUTION: A1

30:3|

= 1000⌫q30 + 500⌫ 2 1| q30 + 250⌫ 3 2| q30 ⇣

⌘⇣

⌘

⇣

⌘2

1 1.53 1 = 1000 1.06 (0.99847) 1000 +500 1.06 = 1.4434 + 0.71535⇣ + 0.35572 = 2.51447 ⌘ (2) 30:1|

a ¨

= 1/2 + 1/2

1 1.06

1/2

(1

⇣

1.61 1000

⌘

+250

⇣

1 1.06

⌘3

(0.99847) (0.99839)

(1/2)q30 ) = 1/2 + 1/2 (0.97129) (1

⇣

1.70 1000

(1/2)(0.00153))

= 1/2 + (1/2)(0.97129)(0.999235) = 0.985273. 2.51447 Annualized premium = 0.985273 = 2.552 2.552 Each semiannual premium = 2 = 1.28. Key: A

Arch MLC, Spring 2009

c Yufeng Guo

163

⌘

Chapter 4 17. Company ABC sets the contract premium for a continuous life annuity of 1 per year on (x) equal to the single benefit premium calculated using: (i)

= 0.03

(ii) µx (t) = 0.02, t

0

However, a revised mortality assumption reflects future mortality improvement and is given by µx (t) =

(

0.02 for t  10 0.01 for t > 10

Calculate the expected loss at issue for ABC (using the revised mortality assumption) as a percentage of the contract premium. (A) 2% (B) 8% (C) 15% (D) 20% (E) 23%

SOLUTION: Let P denote the contract premium. R R P =a ¯x = 01 e t e µt dt = 01 e 0.05t dt = 20 P E[L] = a ¯RIM P x R 10 IM P 0.03t a ¯x = 0 e e 0.02t dt + e 0.03(10) e 0.02(10) 01 e 0.5 e 0.5 = 1 0.05 + e0.04 = 23 E[L] = 23 20 = 3 E[L] 3 P = 20 = 15% Key: C

Arch MLC, Spring 2009

c Yufeng Guo

0.03t e 0.01t dt

164

Chapter 4 18. For a fully continuous whole life insurance of 1 on (x), you are given: (i) The forces of mortality and interest are constant. (ii) 2 A¯x = 0.20 (iii) P¯ (A¯x ) = 0.03 (iv) 0 L is the loss-at-issue random variable based on the benefit premium. Calculate Var(0 L). (A) 0.20 (B) 0.21 (C) 0.22 (D) 0.23 (E) 0.24

SOLUTION: P¯ (A x ) = µ = 0.03 2A = 0.20 = µ = x 2 +µ

0.03 ) 2 +0.03 1 2 2A 2 0.20 ( ) (A x ) V ar(0 L) = x( a) = ( 0.06 )32 = 2 0.09 µ 1 where A = µ+ = 0.03 0.09 = 3 1 1 a ¯ = µ+ = 0.09

= 0.06 0.20

Key: A

19. For a fully discrete whole life insurance of 100,000 on each of 10,000 lives age 60, you are given: (i) The future lifetimes are independent. (ii) Mortality follows the Illustrative Life Table. (iii) i = 0.06 (iv) ⇡ is the premium for each insurance of 100,000. Using the normal approximation, calculate ⇡, such that the probability of a positive total loss is 1%. (A) 3340 (B) 3360 (C) 3380 (D) 3390 (E) 3400 Arch MLC, Spring 2009

c Yufeng Guo

165

Chapter 4 SOLUTION: A60 = 0.36913 d = 0.05660 2A = 0.17741 60q and 2 A60 A260 = 0.202862. Expected Loss on one policy is E(L(⇡)) = 100, 000 + ⇡d A60 2 2 Variance on one policy is V ar(L(⇡)) = 100, 000 + ⇡d A60 On the 10000 lives,

⇡ d A260

E[S] = 10, 000E[L(⇡)] and V ar[S] = 10, 000V ar[L(⇡)] The ⇡ is such that 0 E[S]/V ar[S] = 2.326 since (2.326) = 0.99

10, 000

⇡ d

100, 000 +

100 100, 000 + 100

⇡ d

q

2A

⇡ d

60

A60 A260

= 2.326

100, 000 + ⇡d 0.36913 = 2.326 100, 000 + ⇡d 0.202862

⇡ d

63.087 ⇡d 3691300 = 0.4719 100, 000 + ⇡d ⇡ ⇡ 63.087 3691300 = 47190 + 0.4719 d d ⇡ 3691300 + 47190 = = 59705.885 d 63.087 0.4719 ⇡ = 59706d = 3379 Key: C

Arch MLC, Spring 2009

c Yufeng Guo

166

Chapter 4 20. For a special fully discrete 3-year term insurance on (x): (i) bk+1 =

(

0 1, 000(11

f or k = 0 k) f or k = 1, 2

(ii) k qx+k 0 0.200 1 0.100 2 0.097 (iii) i = 0.06 Calculate the level annual benefit premium for this insurance. (A) 518 (B) 549 (C) 638 (D) 732 (E) 799

SOLUTION: Actuarial present value Benefits equals =

(0.8)(0.1)(10, 000) (0.8)(0.9)(0.097)(9, 000) + = 1,239.75 1.062 1.063 ✓

(0.8) (0.8)(0.9) 1,239.75 = P 1 + + 2 1.06 1.062 Key: A

Arch MLC, Spring 2009

P = 517.53 ) 518

c Yufeng Guo

◆

= P (2.3955)

167

Chapter 4

Problems from Pre-2000 SOA-CAS exams 1. 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. • L is the loss-at-issue random variable.

• Premiums are determined using the equivalence principle. •

Var(Z) = 0.36 Var(L)

• ax = 10

⇣

⌘

Calculate P A x . (A) 0.03

(B) 0.04

Arch MLC, Spring 2009

(C) 0.05

(D) 0.06

c Yufeng Guo

(E) 0.07

168

Chapter 4 2. You are an actuary reviewing a tentative pricing calculation for a fully discrete whole life insurance of 1000 on (60). The tentative benefit premium seems surprisingly high. You discover that an incorrect value for q60 was used. You are given: • i = 0.06

• The incorrect value q60 = 0.10 was used.

• Based on that incorrect value, 1000 A60 = 405.54. • The correct value is q60 = 0.01.

• For all other ages, the correct mortality rates were used.

• L⇤ denotes the insurer’s loss-at-issue random variable, assuming the tentative benefit premium is not corrected. Calculate: E [L⇤ ] (A) -110

(B) -105

(C) -100

(D) -95

(E) -90

3. For a 5-year endowment insurance of 1 on (35), you are given: • The death benefit is payable at the moment of death.

• Premiums, payable continuously, are determined using the equivalence principle. • µ35 (t) = 0.05, t •

0

= 0.04

• t L is the prospective loss at time t. ⇣

⌘

Calculate 1000 P A 35:5 . (A) 151

(B) 173

Arch MLC, Spring 2009

(C) 198

(D) 208

c Yufeng Guo

(E) 226

169

Chapter 4

Solutions to Pre-2000 Exam Questions: Chapter 6 1. Key: B Var[Z] = 2A x Var[L] =

1+ )

P (A x ) P

1+

!2

!2

⇣

(A x )2 (A x )

2

2

Ax

1 = 0.36

)

) P (A x ) = Also P (A x ) =

Ax 1 ax = ax ax

)

P (A x ) =

1

=

1+

1+ P

=

!

=

P

!2

Var[Z]

5 3

2 3

2 ax = 1 3

ax ax

⌘

ax

) ax =

3 5

0.4 = 0.04 10

2. Key: E E[L⇤ ] = 1000A60

Pa ¨60

To calculate A and a ¨, we use the corrected q. To get P , we use the incorrect q. First P: A60 A60 0.40554 P = 1000 = 1000 1 A60 = 1000 = 38.615 a ¨60 10.502127 d A61 is the same before and after the correction: 0.40554 = v(0.1) + v(0.9) A61 ) A61 =

(1.06)(0.40554) 0.9

0.1

= 0.3665

Now for the corrected A and a ¨.

A60 = vq60 + vp60 A61 =

1 1 (0.01) + (0.99)(0.3665) = 0.3517 1.06 1.06

)a ¨60 =

1

A60 = 11.45 d

E[L⇤ ] = 1000A60

Pa ¨60

Where A and a ¨ are from the correct q; and P is from the incorrect q. E[L⇤ ] = 1000(0.35175)

Arch MLC, Spring 2009

38.615(11.45) ⇡

c Yufeng Guo

90

170

Chapter 4 3. Key: D

⇣

1000P A 35:5

⌘

1000A 35:5

=

a35:5 = a35

a35:5 5 E35 a40

Since this is constant force, this equals 1 µ+

e

0.45

1 µ+

= 11.11

A 35:5 = 1 ⇣

) 1000P A 35:5

Arch MLC, Spring 2009

e

0.45

(11.11) = 4.026

a35:5 = 0.8390 ⌘

=

c Yufeng Guo

839.0 ⇡ 208 4.026

171

Chapter 4

Arch MLC, Spring 2009

c Yufeng Guo

172

Chapter 5

ACTUARIAL MATHEMATICS: CHAPTER 7 BENEFIT RESERVES • Option A reference: Actuarial Mathematics Chapter 7 • Option B reference: Models for Quantifying Risk Chapter 12 As we calculated premiums in the previous chapter, we depended upon the equivalence principle. That is, at the time of policy issue, E (PV of Future Benefits) = E (PV of Future Premiums) . This is true at issue. But consider a person who is now 95 years old with a 100,000 whole life policy with annual premiums of 300 that she has been paying every year since age 35. The expected value of the benefits and premiums were equal at issue, but now, E (PV of Future Benefits) >>>> E (PV of Future Premiums) . Intuitively, we know that 100,000 will be paid at some point in the next 10 years or so. However, there are only 3,000 worth of premiums coming in over that time. Hopefully, the insurer has been saving up some of the premiums received over the years to cover this upcoming benefit. In fact, the insurer is legally required to do so! This accumulation is called a “reserve” and is one reason for the existence of the actuarial profession. In the rest of this chapter, you can think of the reserve, at time t after issue, as the amount of money the insurance company must have saved up to be able to provide for the future benefits of the policy. It makes sense that this amount should equal the APV of future benefits minus the APV of future premiums. This chapter covers the calculation of benefit reserves. A benefit reserve is the di↵erence between the expected value of future benefits and the expected value of future premiums: tV

= E(PV of future benefits) 173

E(PV of future premiums)

Chapter 5 Read t V as “the benefit reserve at time t”. Question: Why must it be true that the benefit reserve is 0 at time t = 0?

7.2 Fully Continuous Benefit Reserves • Option A reference: Actuarial Mathematics Chapter 7.2 • Option B reference: Models for Quantifying Risk Chapter 12.3.2, 12.3.3 Consider a unit whole ⇣ life⌘ policy on a fully continuous basis with an annual continuous benefit premium rate of P A [x] . Let the Prospective Loss at time t be: tL

= v T (x)

t

⇣

⌘

P A [x] a T (x)

t

This is a familiar idea in a new setting. In plain English, the loss function at time t, t L, equals the value of future benefits, v T (x) t minus the present value of future premi⇣ present ⌘ ums, P A [x] a T (x) t . (Note again the importance of understanding the notation. The loss formula above is for a whole life policy – you will need to be able to transform this formula for other types of insurance.) The Benefit Reserve, t V is the expected value of t L: tV

= E[t L].

This reserve takes into account mortality probabilities as well as interest. tV

⇣

⌘

A [x] = A [x]+t

⇣

⌘

P A [x] a[x]+t .

Remember, the [x] signifies a person “selected at age x.” If no special selection occurred, simply drop the “[ ]” notation. In fact, you might find the look of the benefit reserve more natural without the [x] notation ⇣

⌘

t V A x = A x+t

EXAMPLE:

⇣

⌘

P A x ax+t .

You are given • lx = 100

x, for 0 < x < 100.

• i = 0.08 Find

20 V 40 .

SOLUTION: Arch MLC, Spring 2009

c Yufeng Guo

174

Chapter 5 This is just a DeMoivre’s Law question in disguise (with ! = 100). At the end of the example, we’ll talk about other ways to recognize DeMoivre. 20 V 40

⇣

= A 60

⌘

P A 40 a60

(I know this is getting repetitive, but once again, this is just PVFB - PVFP for a person now age 60, who started the policy at age 40.) First we’ll find A 60 , and this will give us a60 as well using the fundamental identity. A 60 =

Z ! 60

v t t px µx+t dt.

0

At this point you can remember that for DeMoivre, f (x + t) = t px µx+t is uniform over (60, 100), or you can work it out using t px

=

!

x !

t x

and µx+t =

!

1 x

t

.

Either way the integral equals 1 40 =

Z 40

dt =

t

e

0

h 1 e (0.07696)(40)

0.07696t

i40

=

0

a60 = Similarly, A 40 =

1

1

1 40

A 60

1

=

Z 40

e

0.07696t

dt

0

e 40(0.07696) = 0.3099 = A 60 . 3.0784

0.6901 = 8.97 0.07696

e 60(0.07696) = 0.2144 60(0.07696)

Finally, we can use the fundamental identity to say that ⇣

⌘

P A 40 = =

A 40 A 40 = a40 1 A 40

(0.07696)(0.2144) = 0.02099 . 0.786

Putting it all together, we have 20 V 40

= 0.3099

(0.02099)(8.97) = 0.122

}

As you have just seen, they can test lots of chapters with a single reserve question. Often the SOA will slip you a DeMoivre’s Law question without telling you it is DeMoivre (as in this example). You can recognize a DeMoivre question if you are given any of the following as being true for all x between 0 and !. (You might want to bookmark this page.) • lx = !

x (The ! can be replaced by any number.)

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Chapter 5 • s(x) = 1 • µ(x) =

• F (x) =

x !

1 ! x,

=

! x !

or µx (t) =

1 ! x t

x !

Table 7.2.1 is a good one to understand. Make sure you can recognize the di↵erent types of insurance presented. Read the subscripts of the A’s and a’s carefully! Fully Continuous Benefit Reserves: Age at Issue x, Duration t, Unit Benefit Actuarial Notation

Plan Whole life insurance n-Year term insurance

tV

tV

n-Year endowment insurance

tV

h-Payment years, whole life insurance

hV t

h-Payment years, n-year endowment insurance

n-Year pure endowment n-Year Deferred whole life annuity

hV t

tV

(A x ) 1

(A x:n )

Prospective Formula A x+t

P (A x )ax+t

8 < A 1 x+t:n t :

0

(A x:n )

(

A x+t:n 1

(A x )

(

A x+t A x+t

(A x:n )

8 > < A x+t:n t

(A

8 < A

1) x: n

t V (n| ax )

t

(

P A1

x:n

◆

ax+t:n

P (A x:n )ax+t:n

t

t

h P (A x )ax+t:h t

A x+t:n > : 1 : 1

✓

n t| ax+t

h P (A x:n )ax+t:h t

P (A

1 )ax+t:n t x: n

P (n| ax )ax+t:n

th

t

1 x+t: n t

t : K(x)+1

X

⇡j v j

K(x) = h, h + 1, . . .

h

j=h

Note that h L is zero for K(x) < h because that means (x) died in the first h years. Since we are considering future losses while standing at time t = h, they are zero in this case since (x) has already died and we have already taken our loss! As before, the reserve formula is the expected value of the loss function, conditional on K(x) > h. hV

=

1 X

= E [h L | K(x)

bh+j+1 v j+1 j px+h qx+h+j

j=0

h] 1 X

⇡h+j v j j px+h

j=0

This is the Prospective Formula for a benefit reserve. It is very messy but is easier to reproduce if you keep two ideas in mind. 1. Recall that the Reserve equals the APV of benefits (benefits discounted for interest and mortality) minus the APV of premiums (again, discounted for interest and mortality). The format of the reserve above fits this definition. 2. All of the p’s and q’s start at x + h because we are looking at the situation h years after policy issue. Therefore, the insured, who was (x) at policy issue, is now (x + h). Similarly, we are only concerned with b’s and ⇡’s for time h and beyond.

EXAMPLE: A fully discrete whole life insurance on (40) features premiums that will be level in years 1-10 and then will double and remain fixed. The death benefit for all years will be 10,000. You are given: • A40 = 0.15,

A50 = 0.24,

• i = 0.07, •

10 p40

Find:

= 0.96.

1) ⇡0 , and 2)

10 V

for this insurance.

SOLUTION: Arch MLC, Spring 2009

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Chapter 6 1. The APV of premiums has to equal the APV of the benefits. ⇡0 a ¨40 + 10 E40 · ⇡0 a ¨50 = 10,000A40 . The left-hand side is correct since we can think of the premium structure as paying ⇡0 forever and then paying an extra ⇡0 starting 10 years from now. a ¨40 = a ¨50 =

1

A40 0.85 = = 12.99, d 0.0654

1

A50 0.76 = = 11.62, d 0.0654

10 E40 =

✓

1 1.07

◆10

(0.96) = 0.488.

⇡0 (12.99) + (0.488)⇡0 (11.62) = 10,000(0.15) ) ⇡0 = 80.38 2.

equals PVFB - PVFP:

10 V

10 V

= 10,000 A50

2⇡0 a ¨50 = 2400

2(80.38)(11.62) = 532.

Note that if we made the premium jump too big at year 10 (say a multiple of 4), we could even achieve a negative reserve. } EXAMPLE: Illustrative Life Table A fully discrete whole life policy is issued to (50) with some special features. The death benefit is 1000 for all years. The premium for the first year will be set equal to the APV of the first year’s benefit and the remaining premiums will all be level. Mortality and interest are according to the ILT. Find (i) ⇡0 , (ii) ⇡1 , (iii)

1V

.

SOLUTION: (i) ⇡0 = 1000 A 1

50:1

= 1000 vq50 ✓

= (943.4) 1 Arch MLC, Spring 2009

✓

1 = 1000 1 1.06

88,979 89,509

c Yufeng Guo

◆

l51 l50

◆

= 5.59 199

Chapter 6 (ii) We can set up the equivalence equation to find ⇡1 since all the premiums after the first one will equal ⇡1 . 5.59 + 1 E50 · ⇡1 a ¨51 = 1000A50 ) 5.59 + v p50 ⇡1 (13.08) = 249.05 ) ⇡1 = 19.85.

(iii) Since it is just one year, I will go with the retrospective approach: ¨50:1 1 V = 5.59 s

1000 1 k50 = 5.59

= 5.59 = 5.59

a ¨50:1 vp50

1 (0.9434)(0.994)

1000

a ¨50:1

1000

1 Ex

A1

50:1

1 Ex

A1

50:1

vp50

5.59 = 0. (0.9434)(0.994)

This may be surprising but it makes sense. If the first year of premium exactly covers expected deaths in the first year, there is nothing left over to provide a reserve for the survivors. (For a more general example of this type of problem see Example 8.2.1 in the text.) } Now, let’s look at the continuous analog of the discrete case. Consider a fully continuous insurance on (x). What does ‘fully continuous’ imply with respect to premium and benefit payment? Let bt = benefit payable at the moment of death. ⇡t = annual rate of benefit premiums payable continuously at t. As in the discrete case, the benefits and premiums can change over the life of the contract. So bs is not necessarily equal to bt and ⇡s is not necessarily equal to ⇡t when s 6= t. For the continuous case, the loss function at time t is given by tL

=

8 < 0

: bT (x) v T (x) t

Z T (x) t

T (x)  t

⇡u v u

t du

T (x) > t

where, as in the discrete case, t L = 0 for T (x)  t since that means (x) died prior to the time at which future losses are being considered. The reserve at time t can be expressed as tV

=

Z 1 0

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bt+u v

u

u px+t

µx (t + u) du

c Yufeng Guo

Z 1 0

⇡t+r v r r px+t dr. 200

Chapter 6 Note the similarities and di↵erences between the discrete and continuous cases. In both cases, tL

= P V (future benefits)

P V (future premiums) ,

and tV

= AP V (future benefits)

AP V (future premiums) .

And in both cases, t L is a random variable and t V = E[t L|T (x) > t], but there are some di↵erences: Discrete ! Continuous X

!

Z

K(x) ! T (x) Use of p, q ! Use of p, µ. Make sure you recognize the significance of the each of these di↵erences.

EXAMPLE: For a continuous whole life policy of 1 on (x), premiums will remain constant over the life of the contract but benefits will grow at a rate of 5 percent per year (compounded continuously). If mortality and interest are constants µ = 0.025 and = 0.06, respectively, find the benefit premium for this insurance. SOLUTION: The death benefit at time t is bt = e0.05t so the equivalence equation becomes ⇡ax =

Z 1 0

v bt t px µdt = t

)⇡

1 µ+

) ⇡ · (11.76) =

Z 1

e

0.06t

0

= (0.025)

· e0.05t · e

Z 1

e

0.035t

0.025t

(0.025)dt

dt

0

0.025 h e 0.035

0.035t

i1 0

= ( 0.714)( 1)

) ⇡ = 0.061. The insured has to live quite a while for the company to break even on this insurance! }

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Chapter 6

8.3 Recursion Relations for Fully Discrete Benefit Reserves • Option A reference: Actuarial Mathematics Chapter 8.3 • Option B reference: Models for Qualifying Risk Chapter 12.2.2 Luckily, the first reserve recursion is intuitive! Let’s consider what h V represents. It is the amount of money the insurance company must have saved at time h to have enough (along with future premiums) to provide future benefits. Looking only one year ahead, we need to start the year with the quantity hV

+ ⇡h

to have enough to pay a death benefit if the insured dies plus provide a reserve at time h + 1 if the insured lives. In other words hV

+ ⇡h = v bh+1 qx+h + v · h+1 V · px+h .

Make sure this equation in some form is completely familiar and logical to you. You will almost certainly find it useful on the exam in some form. Remember - “Those who die get the benefit, those who live get the reserve”. Multiplying both sides of this equation by (1 + i) gives (h V + ⇡h )(1 + i) = bh+1 qx+h + h+1 V px+h . This says that if we start the year with (h V + ⇡h ) and let it accumulate interest for a year, we will have enough to pay o↵ those who died and have a reserve in place for those who live. This is very intuitive! I don’t care which form you learn, but learn this equation!

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Chapter 6 Some Terminology from the text: The policy year from time t = h to time t = (h + 1) is called “policy year h + 1”. This seems screwy at first but it is accurate since the year from t = 0 to t = 1 ought to be policy year 1. For policy year h + 1, the following terminology applies: hV

+ ⇡h hV

h+1 V

! “initial benefit reserve for policy year h + 1” ! “terminal benefit reserve for year h” ! “terminal benefit reserve for year h + 1”

These names are logical if you think about them and you need to be able to instantly write down the correct corresponding expression whether a question uses ‘initial’ reserves or ‘terminal’ reserves. The last idea from this section on reserves is that of Net Amount at Risk. This refers to the amount of money the insurance company could have to pay the insured above and beyond the reserve accumulated. In other words, the net amount at risk for policy year h + 1 is (bh+1 h+1 V ) . In the insurance business, this is an important quantity. It is the amount of money the company will have to produce from sources other than the insured’s benefit reserve if the insured dies in policy year h + 1. OK – that is a lot of theory, here are a couple of examples. EXAMPLE: Mortality and interest are as given in the Illustrative Life Table. For a fully discrete whole life policy on (40) with a benefit of 1000, find (A) the terminal benefit reserve for policy year 10, (B) the initial benefit reserve for policy year 11, (C) the terminal net amount at risk for policy year 10. SOLUTION: (A) This is just

10 V40

= A50

P40 a ¨50 . According to the ILT,

A50 = 0.249,

a ¨50 = 13.27,

A40 0.161 = = 0.0109. a ¨40 14.82 = 0.249 (0.0109)(13.27) = 0.104.

P40 = 10 V40

So the reserve for this policy is 100010 V40 = 104. (B) This is 1000 10 V40 + 1000P40 = 114.9. (C) b10

10 V40

= 1000

Arch MLC, Spring 2009

104 = 896. c Yufeng Guo

} 203

Chapter 6 EXAMPLE: The insured (x) has purchased a 20-year term policy which pays a flat benefit of 100,000 at the end of the year of death plus, if the insured lives for 20 years, all premiums will be returned (without interest). You are given: •

20 Ex

= 0.286,

• a ¨x:20 = 11.88, • Ax = 0.129

10 Ex

= 0.543,

a ¨x:10 = 7.73, Ax+10 = 0.201

Ax+20 = 0.305

Find (A) the benefit premium, (B) the terminal reserve for the 10th policy year. SOLUTION: (A) We should start by setting the APV of premiums equal to the APV of benefits. Since this is a non-standard insurance, I will use ⇡ for the premium. ⇡a ¨x:20 = 100,000A1

x:20

A1

x:20

= Ax

20 Ex Ax+20

+ (20⇡) 20 Ex

= 0.129

(0.286)(0.305) = 0.042

) ⇡ (11.88) = 4200 + (5.72)⇡ ) ⇡ = 681.82 (B) Since the premiums are constant but the benefits are complicated, the retrospective method might work best: 100,000 · 10 Vx = ⇡ s¨x:10 = (681.82) = (681.82)

7.73 0.543

a ¨x:10 10 Ex

100,000

100,000

0.129

100,000 Ax

A1

x:10

10 Ex

10 Ex Ax+10 10 Ex

(0.543)(0.201) = 6049. 0.543

Note that since we carried the premium through at its full value, we also had to carry the death benefit through. You have to be careful when doing these to see that all parts of the formula are assuming the same death benefit. (Leaving out the 100,000 would have been assuming a death benefit of 1 for past benefits, while the premium assumes a death benefit of 100,000.) } Arch MLC, Spring 2009

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Chapter 6

8.4 Benefit Reserves at Fractional Durations • Option A reference: Actuarial Mathematics Chapter 8.4 • Option B reference: Models for Qualifying Risk Chapter 12.6 Consider a fully discrete insurance with benefits bj+1 payable at the end of the year. This insurance is purchased by annual benefit premiums of ⇡j , where j = 0, 1, 2, . . .. Note that by this time, saying “end of year benefit” and “annual premiums” should seem redundant once we’ve said we are working with fully discrete insurance. In this section, we pursue the “Interim Benefit Reserve,” which sounds like what it is – the reserve at some point in time between policy anniversaries. The notation for this reserve is: h+s V for h = 0, 1, 2, . . . and 0 < s < 1. So, s represents a fraction of a year.

h

h+3

h+2

h+1

h+s

For fractional year reserves (reserves at time h + s where 0 < s < 1), we basically want to be able to do two things. 1. Write and understand the messy formula that gets us from 2. Write and understand a very simple formula that relates

h+s V

h+s V

to

h+1 V

.

to both h V and

h+1 V

First recall the recursive formula for h V : hV

+ ⇡h = vbh+1 qx+h + v

h+1 V

px+h .

For h+s V , we are standing at time (h + s) so ⇡h is in the past and will not be in the formula. In addition all discounting will be for (1 s) years instead of 1 year. Finally, any survival functions will start at age (x + h + s) and involve survival for (1 s) years. The result is h+s V

= v1

s

bh+1

1 s qx+h+s

+ v1

s

h+1 V 1 s px+h+s .

This is a pretty messy formula, but if you think about what it means, it is possible to remember. In plain English - “The reserve at time h + s must be enough to provide the benefit bh+1 for those who die in the next (1 s) years plus provide h+1 V at the end of the year for those who survive the next (1 s) years.” By the way, the text uses the name interim benefit reserve for the fractional year reserve above, so that is the name to expect to see on the exam. It would be nice if the interim benefit reserve (at time h + s) were the linear interpolation of the beginning-of-year reserve and end-of-year reserve for the same year. Then it would be true that s) (h V + ⇡h ) + (s) (h+1 V ) (UDD) h+s V = (1 Arch MLC, Spring 2009

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Chapter 6 In fact this is only true under Uniform Distribution of Deaths - as you may have guessed from the UDD written next to the equation! It is a formula worth remembering since the exam writers often give you UDD as a simplifying assumption, and the formula is nearly correct even when UDD does not apply. There is a useful way of rewriting the UDD linear interpolation: h+s V

= (1

(h V )

s)

+(s)

| {z }

terminal reserve at h

(h+1 V ) | {z }

terminal reserve at h + 1

+

(1

|

s)⇡h

{z

}

unearned premium

So, in the UDD case, we have linear interpolation between the two “end point” terminal reserves, plus the unearned benefit premium. The unearned benefit premium can be thought of as follows: A premium was paid at the beginning of the year to cover the whole year. At time s, where (0 < s < 1), only a part of the premium has been “earned” by the insurance company. If the policy were canceled, a pro rata portion of the premium might be returned to the policyholder since it has not been earned. In other words, a premium is fully earned only when the time-period for which it was intended to provide coverage has passed. (1-s).(ann prem) is earned after h+s.

s.(ann prem) is earned before h+s.

h+s

h

h+1

EXAMPLE: Suppose a premium of 100 was paid on a fully discrete insurance policy. As of April 27th of the year (not a leap year), how much premium has been earned? How much is unearned? In this case, what are ⇡h and s? (Work it out before looking at the solution!) SOLUTION: ⇡h = 100 Jan + Feb + March + (27 days of April ) = 31 + 28 + 31 + 27 = 117 days. 117 = 0.3205, 365 0.3205) ⇤ 100 = 67.95 of unearned premium. s=

(1

s)⇡h = (1

Then there must be 100 67.95 = 32.05 of earned premium. Unearned premium is a simple concept that might seem a little difficult at first, but is easier to understand if you think about it logically. } Arch MLC, Spring 2009

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Chapter 6 EXAMPLE: A non-standard insurance was bought on (x) 10.5 years ago and you are asked to estimate the reserve as of today. You are given • b11 = 10 • 11 V = 43 • = 0.06 • qx+10 = 0.1 (A) Find 10.5 V assuming constant force of mortality. (B) Find 10.5 V assuming uniform density of deaths in the year of death. SOLUTION: This one is definitely hard. We have to remember some stu↵ from all the way back in Chapter 3. For both (A) and (B), the reserve we need at t = 10.5 is enough to pay a death benefit at the end of the year for those who die, plus enough to fund 11 V = 43 for those who live. (Or you can memorize that messy formula!) 10.5 V

= v 0.5 b11

0.5 qx+10.5

+ v 0.5

11 V 0.5 px+10.5

(This is messy, think about why each part is right.) (A) Assuming constant force of mortality, y qx+t

=1

pyx ) 0.5 qx+10.5 = 1

0.5 px+10.5

10.5 V

=e

(0.06)(0.5)

=1

0.5 qx+10.5

(10)(0.0513) + e

(0.9)0.5 = 0.0513 = 0.9487 (0.06)(0.5)

(43)(0.9487)

= 0.498 + 39.59 = 40.09. (B) Assuming UDD, y qx+t

=

yqx 1 tqx

0.05 (0.5)qx+10 = = 0.0526 1 (0.5)qx+10 0.95 Note: I used the Chapter 3 formula here but I tend to go right to the last step by saying, the probability of dying in the last half of the year is the number that die in the last half (0.05 since UDD and 0.1 die over entire year) divided by the number alive at midyear (0.95 out of 1 alive at beginning of year). ) 0.5 qx+10.5 =

0.5 px+10.5

10.5 V

=e

(0.06)(0.5)

=1

0.5 qx+10.5

(10)(0.0526) + e

= 0.9474 (0.06)(0.5)

= 0.510 + 39.53 = 40.04 Arch MLC, Spring 2009

c Yufeng Guo

(43)(0.9474) } 207

Chapter 6 EXAMPLE: For a non-standard insurance policy, you are given: • Deaths are uniformly distributed over the year of death, • ⇡h = 25 for all values of h, • 20 V = 250, 21 V = 265. Find

20.25 V

.

SOLUTION: We are almost always happy to see UDD and this question is no exception: The reserve at time t = 10.25 will be the weighted average of the beginning of year reserve and the end of year reserve. We should be sure that we put the heavier weight (0.75 in this case) on the reserve that is closer in time. 20.25 V

= (0.75)(20 V + ⇡20 ) + (0.25) 21 V

= (0.75)(275) + (0.25)(265) = 272.50.

}

8.5 The Hattendorf Theorem • Option A reference: Study Note MLC-25-05 Section 8.5 from the second printing of Actuarial Mathematics, Second Edition • Option B reference: Models for Qualifying Risk Chapter 12.2.3 Recall from Section 8.2 that the loss function h L is the present value at time h of all future losses.

hL

=

8 0 > > <

K(x) = 0, 1, . . . h

1

K(x)

> b v K(x)+1 h > : K(x)+1

X

⇡j v j

h

K(x) = h, h + 1, . . .

j=h

In this section, we are interested in the present value of losses that occur in the next year only. That is, we are standing at time t = h and we are interested in the present value of losses that occur in the next year. How do we measure those “losses”? They are the present value of any benefits paid plus any increase in reserve minus any premiums paid. The symbol for the present value at time h of losses incurred between times h and h + 1 is ⇤h . (That is a Greek capital L. I guess L was already taken by h L and I have no idea why Bowers feels the h belongs after the symbol rather than before it as with h L. We will give the formula for ⇤h and then explain why each of the terms is correct. ⇤h =

8 > < 0

K(x) = 0, 1, . . . h 1 v bh+1 ⇡h + ( Vh ) K(x) = h > : ( ⇡h ) + (v h+1 V h V ) K(x) = h + 1, h + 2, . . .

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Chapter 6 • First, if K(x) < h then the insured person is already dead and no loss will be incurred on their policy next year (seems clear enough!) so Vh = 0. • Second, if the insured dies during the next year (i.e, if K(x) = h), then a loss equal to the death benefit will be su↵ered but it will o↵set by the release of the reserve. Since the person has died by year-end, we no longer need to hold a reserve for that policy. • Third, if the insured is still alive at the end of next year, then the loss amount is equal to any increase in the reserve minus premium paid at the beginning of the year. If we assume that the insured dies after time t = h (otherwise the losses for next year are not very interesting), we have the random variable [⇤h |K(x)

h]

which has the following probability distribution: v bh+1 ⇡h + v

Pr [⇤h = |K(x) qx+h px+h

⇡h Vh h+1 V hV

h]

Using this, we can quickly calculate that E [⇤h |K(x)

h] = (v bh+1

⇡h

Vh ) · qx+h + ( ⇡h + v

= vbh+1 qx+h + v h+1 V px+h

h+1 V

hV

) · px+h

(⇡h + h V )

Now you can use the standard recursion relation for reserves (earlier in this chapter) to show that this equals zero. Therefore, E [⇤h |K(x)

h] = 0

The expected loss for next year equals zero given survival to today. Similar techniques (elementary statistics) can be used to show that Var [⇤h |K(x)

h] = v 2 (bh+1

h+1 V

)2 px+h qx+h

You are required to be able to use this formula for the exam. Earlier in this chapter, we saw that E[h L|K(x) h] = h V . The point to the Hattendorf Theorem is that the variance of the loss function h L can be calculated using a recursion that involves the variance of ⇤h . Sparing you the derivation (see Bowers pages 244-245), we have Var[h L|K(x)

h] = Var[⇤h |K(x)

h] + v 2 px+h Var [h+1 L|K(x)

h + 1]

This is a messy formula but it is a little easier to remember to think about what it says. The left-hand-side is the variance in present value of all future losses and the equation says that that quantity is equal to the variance in present value of next years losses plus the variance in present value for years after next year if you live till next year. Like the formula above, you want to remember this one for the exam. Arch MLC, Spring 2009

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Chapter 6 EXAMPLE: A fully discrete 5-year term insurance on (50) with a face amount of 1000, you have calculated the following: •

h 0 1 2 3 4

q50+h 0.005890 0.006459 0.007017 0.007633 0.008309

hV 0.00 1.04 1.64 1.73 1.21

• i = 0.06 For h = 4, 3, 2, find Var [h L|K(50)

h]

SOLUTION: • h = 4: Since the policy only lasts for 5 years, it is clear that the variance of the loss function for the rest of the lifetime of the policy at time t = 4 (4 L) is equal to the variance in the loss function for next year only ( ⇤4 ). Therefore Var[4 L|K(x)

4] = Var[⇤4 |K(x)

4] = v 2 (b5

5V

)2 p54 · q54

b5 = 1000 and since the policy expires at t = 5, the necessary reserve at that time (after benefits are paid) is 5 V = 0. Finally, v=

1 = 0.9434 1.06

So we have Var[⇤4 |K(x)

4] = (0.9434)2 (1000

0)2 (0.991691) (0.008309) = 7334

• h = 3: Var[3 L|K(x)

3] = Var[⇤3 |K(x)

= Var[⇤3 |K(x)

4]

3] + (0.9434)2 (0.992367)(7334)

= v 2 (b4

= (0.890)(1000

3] + v 2 p53 Var [4 L|K(x)

4V

)2 p53 · q53 + 6477

1.21)2 (0.992367)(0.007633) + 6477 = 13,202

• h = 2: Var[2 L|K(x) = v 2 (b3

2] = Var[⇤2 |K(x) 3V

2] + v 2 p52 Var [3 L|K(x)

)2 p52 · q52 + v 2 p52 (13,202) = 17,847

3] }

Two final notes: Arch MLC, Spring 2009

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Chapter 6 • Be sure to do Problem 18 in the suggested exercises, it is good practice with topics from earlier chapters as well as with Hattendorf’s Theorem. • Don’t ignore this Hattendorf material! It’s on the syllabus. Chapter 8 Suggested Problems: 14, 30, 35ab (Solutions at archactuarial.com)

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Chapter 6

CHAPTER 8 Formula Summary hL

=

8 0 > > <

K(x) = 0, 1, . . . h

1

K(x)

> b v K(x)+1 h > : K(x)+1

h V = E [h L | K(x)

h] =

1 X

X

⇡j v j

K(x) = h, h + 1, . . .

h

j=h

1 X

bh+j+1 v j+1 j px+h qx+h+j

j=0

⇡h+j v j j px+h

j=0

For the continuous case, the loss function at time t is given by tL =

tV =

tL tV

8 < 0

: bT (x) v T (x) t

Z 1 0

Z T (x) t

T (x)  t

⇡u v u

Z 1

bt+u v u u px+t µx (t + u) du

= P V (future benefits)

= AP V (future benefits)

t du

0

T (x) > t

⇡t+r v r r px+t dr.

P V (future premiums) , AP V (future premiums) .

It is essential to understand and remember the next two! hV

+ ⇡h = v bh+1 qx+h + v

h+1 V

px+h .

(h V + ⇡h )(1 + i) = bh+1 qx+h + h+1 V px+h .

Net Amount at Risk for policy year h + 1: (bh+1

h+s V

= v1

s

bh+1

h+s V

= (1

1 s qx+h+s

+ v1

s

h+1 V

)

h+1 V 1 s px+h+s .

s) (h V + ⇡h ) + (s) (h+1 V )

(UDD)

There is a useful way of rewriting the UDD linear interpolation: h+s V

= (1

(h V )

s)

| {z }

terminal reserve at h Arch MLC, Spring 2009

+(s)

(h+1 V ) | {z }

terminal reserve at h + 1 c Yufeng Guo

+

(1

|

s)⇡h

{z

}

unearned premium 212

Chapter 6

Chapter 8 More Formulas At time t = h, the variance in the present value of losses to be su↵ered in the next year equals Var [⇤h |K(x)

h] = v 2 (bh+1

h+1 V

)2 px+h qx+h

At time t = h, the variance in the present value of losses to be su↵ered in all future years equals Var[h L|K(x)

Arch MLC, Spring 2009

h] = Var[⇤h |K(x)

h] + v 2 px+h Var [h+1 L|K(x)

c Yufeng Guo

h + 1]

213

Chapter 6

Past SOA/CAS Exam Questions: 1. For a fully discrete two-year term insurance of 400 on (x): (i) i = 0.1 1 = 74.33 (ii) 400 Px:2 1 = 16.58 (iii) 400 1Vx:2

(iv) The contract premium equals the benefit premium. Calculate the variance of the loss at issue. (A) 21,615

(B) 23,125

(C) 27,450

(D) 31,175

(E) 34,150

Solution: Need to determine qx and qx+1 . Formula 7.23:

16.58 = 1V =

Formula 8.39

0V

400qx+1 1.1

= 0 = 400⌫qx qx =

74.33 ) qx+1 = 0.25

⇡ + ⌫ 1V px

⇡ ⇥ (1.1) 1V = 0.17 400 1V

Loss (Example 8.51): 0 1 2

400 400

⇣ ⇣

1 1.1 1 1.1

⌘

74.33 = 289.30

⌘2

⇣

74.33 1 +

Since E[L] = 0,

⇣

74.33 1 + 1 1.1

⌘

=

1 1.1

⌘

= 188.68

141.90

Var = (289.30)2 (0.17)+(188.68)2 (0.25)(1 0.17)+(141.90)2 (1 0.17)(1 0.25) = 34, 150 Key: E

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214

Chapter 6 2. For a fully discrete whole life insurance with non-level benefits on (70): (i) The level benefit premium for this insurance is equal to P50 . (ii) q70+k = q50+k + 0.01, k = 0, 1, . . . , 19 (iii) q60 = 0.01368 (iv)

kV

= k V50 , k = 0, 1, . . . , 19

(v)

11 V50

= 0.16637

Calculate b11 , the death benefit in year 11. (A) 0.482

(B) 0.624

(C) 0.636

(D) 0.648

(E) 0.834

Solution: = (10 V + P50 ) (1 + i)

11 V

(b11

11 V

) q80

(1)

But by the traditional formula: 11 V50

and since

10 V

= 10 V50 and 11 V

(1)

= (10 V50 + P50 ) (1 + i)

(1

11 V50 ) q60

= 11 V50 ,

11 V

= (10 V + P50 ) (1 + i) (2) ) (b11 ) b11 =

=

(1

11 V

(1

(1

) q80 = (1

11 V

) q60

q80

11 V

) q60

11 V

(2)

) q60

+ 11 V

0.16637)(0.01368) + 0.16637 = 0.64796 ⇡ 0.648 0.02368

Key: D

3. For a fully discrete 3-year endowment insurance of 1000 on (x): i) qx = qx+1 = 0.20, Calculate 1000 (A) 320

⇣

2 Vx:3

(B) 325

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ii) i = 0.06, 1 Vx:3

⌘

iii) 1000Px:3 = 373.63

.

(C) 330

(D) 335

(E) 340

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215

Chapter 6 Solution:

2⇣

10001 Vx:3 = 4 =



10002 Vx:3 = 4 =

Key: B

px

⌘

1000Px:3 + 1 Vx:3 (1 + i)

⇣

3 5

(1000qx+1 )

px+1

(373.63 + 245.06)(1.06) 0.8

) 1000

(1000qx )

(373.63)(1.06) (1000)(0.2) = 245.06 0.8 2⇣



⌘

1000Px:3 (1 + i)

2 Vx:3

1 Vx:3

⌘

(1000)(0.2)

= 569.76

3 5

= 569.76

245.06 = 324.70

4. For a fully discrete 20-year endowment insurance on (55): (i) Death benefits in year k are given by bk = (21

k), k = 1, 2, . . . , 20.

(ii) The maturity benefit is 1. (iii) Annual benefit premiums are level. (iv)

kV

denotes the benefit reserve at the end of year k, k = 1, 2, . . . , 20.

(v)

10 V

= 5.0

(vi)

19 V

= 0.6

(vii) q65 = 0.10 (viii) i = 0.08 Calculate (A) 4.5

11 V

.

(B) 4.6

(C) 4.8

(D) 5.1

(E) 5.3

SOLUTION: ⇡ denotes benefit premium. 19 V

= APV Future benefits - APV future premiums 1 ⇡ =) ⇡ = 0.326 1.08 (10 V + ⇡)(1.08) (q65 )(10) 11 V = p65 0.6 =

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c Yufeng Guo

216

Chapter 6 =

(5.0 + 0.326)(1.08) (0.10)(10) = 5.28 1 0.10

Key E

5. For a 5-year fully continuous term insurance on (x): (i)

= 0.10

(ii) All the graphs below are to the same scale. (iii) All the graphs show µx (t) on the vertical axis and t on the horizontal axis. Which of the following mortality assumptions would produce the highest benefit reserve at the end of year 2? (A) 0.08 0.06 0.04 0.02 0

(B)

0

1

2

3

4

5

(C) 0.08 0.06 0.04 0.02 0

0.08 0.06 0.04 0.02 0

0

1

2

3

4

5

0

1

2

3

4

5

(D)

0

1

2

3

4

5

0

1

2

3

4

5

0.08 0.06 0.04 0.02 0

(E) 0.08 0.06 0.04 0.02 0

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c Yufeng Guo

217

Chapter 6 Solution: Comparing B to D: Prospectively at time 2, they have the same future benefits. At issue, B has the lower benefit premium. Thus, by formula 7.2.2, B has the higher reserve. Comparing A to B: Use formula 7.3.5. At issue, B has the higher benefit premium. Until time 2, they have had the same benefits, so B has the higher reserve. Comparing B to C: Visualize a graph C* that matches graph B on one side of t=2 and matches graph C on the other side. By using the logic of the two preceding paragraphs, C’s reserve is lower than C*’s which is lower than B’s. Comparing B to E: Reserve on E are constant at 0. Key: B

6. For a fully discrete three-year endowment insurance of 10,000 on (50), you are given: (i) (ii) (iii) (iv) (v) (vi)

i = 0.03 1000q50 = 8.32 1000q51 = 9.11 10,000 1 V50:3 = 3209 10,000 2 V50:3 = 6539 0 L is the prospective loss random variable at issue, based on the benefit premium.

Calculate the variance of 0 L. (A) 277,000

(B) 303,000

Solution: ⇡ = 10,000v 0L

=

(C) 357,000

(D) 403,000

= 9708.74

6539 = 3169.74

s V50:3

8 > < 10,000v

⇡¨ a 1 = 6539 ⇡¨ a 2 = 3178.80 ⇡¨ a 3 = 83.52

10,000v 2

> : 10,000v 3

(E) 454,000

for K = 0 for K = 1 for K > 1

Pr(K = 0) = q50 = 0.00832

Pr(K = 1) = p50 q51 = (0.99168)(0.00911) = 0.0090342 Pr(K > 1) = 1

Pr(K = 0)

Var(0 L) = E since ⇡ is the benefit premium,

h

2

0L

i

Pr(K = 1) = 0.98265

(E [0 l])2 = E

h

2 0L

i

= 0.00832 ⇥ (6539)2 + 0.00903 ⇥ (3178.80)2 + 0.98265 ⇥ ( 83.52)2 = 453,895 Key: E

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218

Chapter 6 7. For a special fully discrete 3-year term insurance on (x): (i) Level benefit premiums are paid at the beginning of each year. (ii) k 0 1 2

bk+1 200,000 150,000 100,000

qx+k 0.03 0.06 0.09

(iii) i = 0.06 Calculate the initial benefit reserve for year 2. (A) 6,500

(B) 7,500

(C) 8,100

(D) 9,400

(E) 10,300

SOLUTION: Let ⇡ = benefit premium. Actuarial Present Value of benefits = = (0.03)(200,000)v + (0.97)(0.06)(150,000)v 2 + (0.97)(0.94)(0.09)(100,000)v 3 = 5660.38 + 7769.67 + 6890.08 = 20,320.13 Actuarial Present Value of benefit premiums h

i

=a ¨x:3 ⇡ = 1 + 0.97v + (0.97)(0.94)v 2 ⇡ = 2.7266⇡ 20,320.13 = 7452.55 2.7266 (7452.55)(1.06) (200,000)(0.03) 1V = 1 0.03 = 1958.46 ⇡=

Initial reserve, year two = 1 V + ⇡ = 1958.56 + 7452.55 = 9411.01 Key:D

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c Yufeng Guo

219

Chapter 6 8. Lottery Life issues a special fully discrete whole life insurance on (25): (i) At the end of the year of death there is a random drawing. With probability 0.2, the death benefit is 1000. With probability 0.8, the death benefit is 0. (ii) At the start of each year, including the first, while (25) is alive, there is a random drawing. With probability 0.8, the level premium ⇡ is paid. With probability 0.2, no premium is paid. (iii) The random drawings are independent. (iv) Mortality follows the Illustrative Life Table. (v) i = 0.06 (vi) ⇡ is determined using the equivalence principle. Calculate the benefit reserve at the end of year 10. (A) 10.25

(B) 20.50

(C) 30.75

At age x:

SOLUTION:

APV of future benefits =

⇣

1 5 Ax

⌘

1000;

(D) 41.00

(E) 51.25

APV of future premiums =

1000 4 A25 = ⇡¨ a25 by equivalence principle 5 5

⇣

4 ¨x 5a

⌘

⇡

1000 A25 1 81.65 =⇡)⇡= ⇤ = 1.258 4 a ¨25 4 16.2242 10 V

= APV(Future Benefits) - APV(Future Premiums) 1000 4 A35 ⇡¨ a35 5 5 1 4 = (128.72) (1.258)(15.3926) = 10.25 5 5 =

Key: A

9. For a special fully continuous whole life insurance on (65): (i) The death benefit at time t is bt = 1000e0.04t , t

0.

(ii) Level benefit premiums are payable for life. (iii) µ65 (t) = 0.02, t (iv)

0

= 0.04

Calculate 2 V , the benefit reserve at the end of year 2. (A) 0

(B) 29

Arch MLC, Spring 2009

(C) 37

(D) 61

(E) 83

c Yufeng Guo

220

Chapter 6 SOLUTION: At issue, actuarial present value (APV) of benefits =

Z 1 0

bt v t t p65 µ65 (t) dt = = 1000

Z 1 0

Z 1 0

t p65 µ65 (t) dt

✓

1 APV of premiums= ⇡a65 = ⇡ 0.04 + 0.02 1000 Benefit premium ⇡ = = 60 16.667 2V

=

Z 1

=

Z 1 0

⇣

1000 e0.04t

◆

⌘⇣

e

0.04t

1000e0.04(2+u) e = 1000e

0.08

0.04u

Z 1 0

= 1083.29 1 q67

t p65 µ65 (t) dt

= 1000 1 q65 = 1000

= 16.667⇡

b2+u v u u p67 µ65 (2 + u) du

0

⌘

u p67 µ65 (2

u p67 µ65 (2

⇡a67

+ u) du

+ u) du

1000 = 1083.29

(60)(16.667) 1000

1000 = 83.29

Key: E

10. For a special fully discrete whole life insurance on (x): (i) The death benefit is 0 in the first year and 5000 thereafter. (ii) Level benefit premiums are payable for life. (iii) qx = 0.05 (iv) v = 0.90 (v) a ¨x = 5.00 (vi)

10 Vx

(vii)

10 V

Calculate (A) 795

= 0.20 is the benefit reserve at the end of year 10 for this insurance. 10 V

.

(B) 1000

Arch MLC, Spring 2009

(C) 1090

(D) 1180

c Yufeng Guo

(E) 1225

221

Chapter 6 SOLUTION: v = 0.90 ) d = 0.10 ) Ax = 1 Benefit premium ⇡ = 10 Vx

5000Ax

5000vqx a ¨x

= 5000Ax+10

(0.10)(5) = 0.5

(5000)(0.5)

a ¨x+10 ) 0.2 = 1 a ¨x

=1

Ax+10 = 1 10 V

=

d¨ ax = 1

5000(0.90)(0.05) = 455 5

a ¨x+10 )a ¨x+10 = 4 5

d¨ ax+10 = 1

(0.10)(4) = 0.6

⇡¨ ax+10 = (5000)(0.6)

(455)(4) = 1180

Key: D

11. For a special fully discrete whole life insurance of 1000 on (40): (i) The level benefit premium for the first 20 years is ⇡. (ii) The benefit premium payable thereafter at age x is 1000v qx , x = 60, 61, . . . (iii) Mortality follows the Illustrative Life Table. (iv) i = 0.06 Calculate ⇡. (A) 4.79

(B) 5.11

(C) 5.34

(D) 5.75

(E) 6.07

SOLUTION: AP V Benefits = 1000A 1

40:20

1 X

+

k E40 1000v q40+k

k=20

AP V Premiums = ⇡¨ a40:20 +

1 X

k E40 1000v q40+k

k=20

The equivalence principle gives us 1000A 1

40:20

+

1 X

k E40 1000v q40+k

= ⇡¨ a40:20 +

k=20

Arch MLC, Spring 2009

k E40 1000v q40+k

k=20

) ⇡ = 1000 =

1 X

161.32 14.8166

A1

40:20

a ¨40:20

(0.27414)(369.13) = 5.11 (0.27414)(11.1454)

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222

Chapter 6 12. For a special fully discrete whole life insurance on (40): (i) The death benefit is 1000 for the first 20 years; 5000 for the next 5 years; 1000 thereafter. (ii) The annual benefit premium is 1000P40 for the first 20 years; 5000P40 for the next 5 years; ⇡ thereafter. (iii) Mortality follows the Illustrative Life Table. (iv) i = 0.06 Calculate (A) 255

21 V

, the benefit reserve at the end of year 21 for this insurance.

(B) 259

(C) 263

(D) 267

SOLUTION: 1000P40 = ✓

100020 V40 = 1000 1 21 V

=

=

(E) 271

A40 161.32 = = 10.89 a ¨40 14.8166

a ¨60 a ¨40

◆

✓

= 1000 1

(20 V + 5000P40 ) (1 + i) p60

11.1454 14.8166

◆

= 247.78

5000q60

(247.78 + 5(10.89)) (1.06) 5000(0.01376) = 255 1 0.01376

Note: For this insurance, a usual whole life policy.

20 V

= 100020 V40 because retrospectively, this is identical to

13. Michel, age 45, is expected to experience higher than standard mortality only at age 64. For a special fully discrete whole life insurance of 1 on Michel, you are given: (i) The benefit premiums are not level (ii) The benefit premium for year 20, ⇡19 , exceeds P45 for a standard risk by 0.010. (iii) Benefit reserves on his insurance are the same as benefit reserves for a fully discrete whole life insurance of 1 on (45) with standard mortality and level benefit premiums. (iv) i = 0.03 (v)

20 V45

= 0.427

Calculate the excess q64 for Michel over the standard q64 . (A) 0.012

(B) 0.014

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(C) 0.016

(D) 0.018

c Yufeng Guo

(E) 0.020 223

Chapter 6 SOLUTION: Let q64 for Michel equal the standard q64 plus c. We need to solve for c. The recursion formula for standard insurance can be written as (19 V45 + P45 ) (1.03) = q64 + (1

q64 ) · 20 V45

The recursion formula for Michel’s insurance is the same except that 0.01 is added to the premium and c is added to q64 : (19 V45 + P45 + 0.01) (1.03) = (q64 + c) + (1

q64

c) · 20 V45

The values of 19 V45 and 20 V45 are the same in the two equations since we are told that Michel’s benefit reserves are the same as for standard insurance. Subtracting the second equation from the first and rearranging terms gives: 0= )c=

(1.03)(0.01) + c (1

20 V45 )

(1.03)(0.01) 0.0103 = 0.018 = 1 20 V45 1 0.427

14. For a fully discrete 10-payment whole life insurance of 100,000 on (x), you are given: (i) i = 0.05 (ii) qx+9 = 0.011 (iii) qx+10 = 0.012 (iv) qx+11 = 0.014 (v) The level annual benefit premium is 2078. (vi) The benefit reserve at the end of year 9 is 32,535. Calculate 100,000Ax+11 . (A) 34,100

(B) 34,300

Arch MLC, Spring 2009

(C) 35,500

(D) 36,500

c Yufeng Guo

(E) 36,700

224

Chapter 6 SOLUTION: Because no premiums are paid after year 10 for (x), we know that 11 V

= Ax+11

The usual reserve recursion formula is: (h V + ⇡h )(1 + i) = bh+1 · qx+h + h+1 V · px+h Rearranging the terms we have h+1 V

=

(h V + ⇡h )(1 + i) px+h

bh+1 qx+h

(32,535+2078)(1.05) 100,000(0.011) = 35,635.642 0.989 (35,635.642+0)(1.05) 100,000(0.012) ) 11 V = = 36,657.31 = Ax+11 0.988 ) 10 V =

15. For a fully discrete whole life insurance of b on (x), you are given: (i) qx+9 = 0.02904 (ii) i = 0.03 (iii) The initial benefit reserve for policy year 10 is 343. (iv) The net amount at risk for policy year 10 is 872. (v) a ¨x = 14.65976 Calculate the terminal benefit reserve for policy year 9. (A) 280 (B) 288 (C) 296 (D) 304 (E) 312

SOLUTION: (9 V + P )(1.03) = qx+9 b + (1 qx+9 )10 V = qx+9 (b 10 V ) +10 V (343)(1.03) = 0.02904(872) +10 V )10 V = 327.97 b = (b 10 V ) +10 V = 872 + 327.97 = 1199.97 Arch MLC, Spring 2009

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225

Chapter 6 ⇣

⌘

⇣

⌘

1 0.03 P = b a¨1x d = 1200 14.65976 1.03 = 46.92 P = 343 46.92 = 296.08 9 V = initial reserve Key: C

16. For a special fully discrete 2-year endowment insurance of 1000 on (x), you are given: (i) The first year benefit premium is 668. (ii) The second year benefit premium is 258. (iii) d = 0.06 Calculate the level annual premium using the equivalence principle. (A) 469 (B) 479 (C) 489 (D) 499 (E) 509

SOLUTION: d = 0.06 ) V = 0.94 Step 1 Determine px 668 + 258⌫px = 1000⌫qx + 1000⌫ 2 px (px+1 + qx+1 ) 668 + 258(0.94)px = 1000(0.94)(1 px ) + 1000(0.8836)px (1) 668 + 242.52px = 940(1 px ) + 883.6px px = 272/298.92 = 0.91 Step 2 Determine 1000Px:2| 668 + 258(0.94)(0.91) = 1000Px:2| [1 + (0.94)(0.91)] 1000Px:2| = Key: B

[220.69+668] 1.8554

Arch MLC, Spring 2009

= 479

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226

Chapter 6 17. For a fully discrete 5-payment 10-year decreasing term insurance on (60), you are given: (i) bk+1 = 1000(10

k), k = 0, 1, 2, . . . , 9

(ii) Level benefit premiums are payable for five years and equal 218.15 each. (iii) q60+k = 0.02 + 0.001k, k = 0, 1, 2, . . . , 9 (iv) i = 0.06 Calculate 2 V , the benefit reserve at the end of year 2. (A) 70 (B) 72 (C) 74 (D) 76 (E) 78

SOLUTION: 1V

=

2V

=

218.15(1.06) 10,000(0.02) = 31.88 1 0.02 (31.88+218.15)(1.06) (9,000)(0.021) 1 0.021

= 77.66

18. For a fully discrete 3-year endowment insurance of 1000 on (x): (i) i = 0.05 (ii) px = px+1 = 0.7 Calculate the second year terminal benefit reserve. (A) 526 (B) 632 (C) 739 (D) 845 (E) 952

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Chapter 6 SOLUTION: t 0 1 2 3

px+t 0.7 0.7 -

t px

1 0.7 0.49 -

⌫t 1 0.95238 0.90703 -

⌫tt px 1 0.6667 0.4444 -

2 P ⌫ t t px = 2.1111 t=0 ✓ ◆ ⇣ a ¨

From above a ¨x:3 =

10002 Vx:3 = 1000 1

x+2:1

a ¨x:3

= 1000 1

1 2.1111

⌘

= 526

Alternatively, Px:3 = a¨ 1 d x:3

10002 Vx:3 = 1000(⌫ Px:3 ) = 1000(0.95238 0.4261) = 526 You could also calculate Ax:3 and use it to calculate Px:3 . Key: A

19. For a special fully discrete 5-year deferred whole life insurance of 100,000 on (40), you are given: (i) The death benefit during the 5-year deferral period is return of benefit premiums paid without interest. (ii) Annual benefit premiums are payable only during the deferral period. (iii) Mortality follows the Illustrative Life Table. (iv) i = 0.06 (v) (IA) 1

40:5

= 0.04042

Calculate the annual benefit premiums. (A) 3300 (B) 3320 (C) 3340 (D) 3360 (E) 3380 Arch MLC, Spring 2009

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228

Chapter 6 SOLUTION: a ¨40:5 = a ¨40 5 E40 a ¨40 = 14.8166 (0.73529)(14.1121) = 4.4401 ⇡¨ a40:5 = 100, 000A45 ⌫ 5 5 p40 + +⇡(IA)140:5 ⇣

⌘

⇡ = 100, 000A45 ⇥ 5 E40 / a ¨40:5 (IA)140:5 = 100, 000(0.20120)(0.73529)/(4.4401 0.04042) = 3363 Key: D

20. For a special fully discrete whole life insurance of 1000 on (42): (i) The contract premium for the first 4 years is equal to the level benefit premium for a fully discrete whole life insurance of 1000 on (40). (ii) The contract premium after the fourth year is equal to the level benefit premium for a fully discrete whole life insurance of 1000 on (42). (iii) Mortality follows the Illustrative Life Table. (iv) i = 0.06 (v) 3 L is the prospective loss random variable at time 3, based on the contract premium. (vi) K(42) is the curtate future lifetime of (42). Calculate E[3 L|K(42)

3].

(A) 27 (B) 31 (C) 44 (D) 48 (E) 52

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Chapter 6 SOLUTION: P40 = A40 /¨ a40 = 0.16132/14.8166 = 0.0108878 P42 = A42 /¨ a42 = 0.17636/14.5510 = 0.0121201 a45 = a ¨45 1 = 13.1121 E[3 L|K(42) 3] = 1000A45 1000P40 1000P42 a45 = 201.20 10.89 (12.12)(13.1121) = 31.39 Key: B Many similar formulas would work equally well. One possibility would be 1000 · 3 V42 + 1000P42 1000P40 , because prospectively after duration 3, this di↵ers from the normal benefit reserve in that in the next year you collect 1000P40 instead of 1000P42 .

21. For a special fully discrete 3-year endowment insurance on (75), you are given: (i) The maturity value is 1000. (ii) The death benefit is 1000 plus the benefit reserve at the end of the year of death. (iii) Mortality follows the Illustrative Life Table. (iv) i = 0.05 Calculate the level benefit premium for this insurance. (A) 321 (B) 339 (C) 356 (D) 364 (E) 373

SOLUTION: 1V

= (0 V + ⇡)(1 + i) = 1.05⇡ 1000q75 .

(1000 +1 V

Similarly, 2 V = (1 V + ⇡) ⇥ 1.05 3 V = (2 V + ⇡) ⇥ 1.05

1000q76 . 1000q77 .

1

V ) ⇥ q75

1000 = 3 V = (1.053 ⇡ + 1.052 ⇡ + 1.05⇡) Arch MLC, Spring 2009

(1000)1.052 ⇥ q75

c Yufeng Guo

(1000)1.05q76

⇤ (1000)q77

230

Chapter 6

⇡= =

1000 + 1000(1.052 q75 + 1.05q76 + q77 ) (1.05)3 + (1.05)2 + 1.05

1000(1 + 1.052 (0.05169) + 1.05(0.05647) + 0.06168) 3.310125 1000 ⇥ 1.17796 = = 355.87 3.310125

Key: C ⇤ This equation is algebraic manipulation of the three equations in three unknowns (1 V,2 V, ⇡). One method - usually e↵ective in problems where benefit = stated amount plus reserve, is to multiply the 1 V equation by 1.052 , the 2 V equation by 1.05, and add those two to the 3 V equation: in the result, you can cancel out the 1 V , and 2 V terms. Or you can substitute the 1 V equation into the 2 V equation, giving 2 V in terms of ⇡, and then substitute that into the 3 V equation.

22. For a deferred whole life annuity-due on (25) with annual payment of 1 commencing at age 60, you are given: (i) Level benefit premiums are payable at the beginning of each year during the deferral period. (ii) During the deferral period, a death benefit equal to the benefit reserve is payable at the end of the year of death. Which of the following is a correct expression for the benefit reserve at the end of the 20th year? (A) (¨ a60 /¨ s 35 )¨ s 20 (B) (¨ a60 /¨ s 20 )¨ s 35 (C) (¨ s 20 /¨ a60 )¨ s 35 (D) (¨ s 35 /¨ a60 )¨ s 20 (E) (¨ a60 /¨ s 35 )

SOLUTION: Let ⇡ be the benefit premium Let k V denote the benefit reserve a the end of year k. For any n, (n V + ⇡)(1 + i) = (q25+n ⇥n+1 V + p25+n ⇥n+1 V ) =n+1 V Thus 1 V = (0 V + ⇡)(1 + i) s2 2 V = (1 V + ⇡)(1 + i) = (⇡(1 + i) + ⇡)(1 + i) = ⇡¨ Arch MLC, Spring 2009

c Yufeng Guo

231

Chapter 6 3V

= (2 V + ⇡)(1 + i) = (⇡¨ s 2 + ⇡)(1 + i) = ⇡¨ s3

By induction (proof omitted) sn n V = ⇡¨ For n = 35, n V = a ¨60 (actuarial present value of future benefits; there are no future premiums) a ¨60 = ⇡¨ s 35 a ¨60 ⇡ = s¨ 35

For n = 20,

s 20 = 20 V = ⇡¨

✓

a ¨60 s¨35

◆

s¨20

The technique, for situations where the death benefit is a specified amount (here, 0) plus the benefit reserve is discussed in section 8.3 of Bowers. This specific problem is Example 8.3.1. Key: A

23. You are given: (i) k V A is the benefit reserve at the end of year k for type A insurance, which is a fully discrete 10-payment whole life insurance of 1000 on (x). (ii) k V B is the benefit reserve at the end of year k for type B insurance, which is a fully discrete whole life insurance of 1000 on (x). (iii) qx+10 = 0.004 (iv) The annual benefit premium for type B is 8.36. (v)

10 V

A

10

V B = 101.35

(vi) i = 0.06 Calculate

11 V

A

11

V B.

(A) 91 (B) 93 (C) 95 (D) 97 (E) 99 Arch MLC, Spring 2009

c Yufeng Guo

232

Chapter 6 SOLUTION: 11 V

11 V

=

⇣

=

⇣

A

B

10 V

10 V

A

B

+0

⌘ (1 + i)

qx+10 ⇥ 1000 px+10

px+10

+ ⇡B

⌘ (1 + i)

px+10

(1)

qx+10 ⇥ 1000 px+10

(2)

Taking Equation (1) minus Equation (2) implies that 11 V

A

11

VB =

= (101.35

⇣

10 V

A

8.36)

10

VB

⇡B

⌘ (1 + i)

px+10

(1.06) = 98.97 1 0.004

Key: E

Arch MLC, Spring 2009

c Yufeng Guo

233

Chapter 6

Problems from Pre-2000 SOA-CAS exams 1. For a special fully discrete whole life insurance on (40), you are given: • The net premium for this insurance is equal to P20 . • k V = k V20 , k = 0, 1, . . . , 19 •

11 V

= 11 V20 = 0.08154

• q40+k = q20+k + 0.01, k = 0, 1, . . . , 19 • q30 = 0.008427

Calculate b11 , the death benefit in year 11. (A) 0.457

(B) 0.468

(C) 0.480

(D) 0.491

(E) 0.502

2. For a deferred temporary life annuity on (57), you are given: • µ = 0.04 •

= 0.06

• The premiums are payable continuously for the first two years at the rate of P . • Annuity benefits are paid at the beginning of the year. • The following annuity payment schedule Year Annuity Benefit

1 2 3 4 5 6 7 8 9+ 0 0 0 10 8 6 4 2 0

Calculate the reserve at the end of year 3. (A) 23.95

(B) 24.95

(C) 25.45

(D) 25.95

(E) 26.45

3. For a fully discrete 10-payment whole life insurance of 1000 on (x), you are given: • i = 0.06

• qx+9 = 0.01262

• The annual benefit premium is 32.88.

• The benefit reserve at the end of year 9 is 322.87. Calculate 1000 Px+10 . (A) 31.52

(B) 31.92

Arch MLC, Spring 2009

(C) 32.32

(D) 32.72

c Yufeng Guo

(E) 33.12

234

Chapter 6 Use the following information for the next 3 questions (note – we have revised this question to make interest assumptions consistent with the current version of the exam): For a special fully discrete 20-year endowment insurance on (68), you are given: • Mortality follows the Illustrative Life Table. • i = 0.06

• Net premiums, for k 6= 5, are given by ⇡k = 1000P68 ,

k = 0, 1, 2, 3, 4, 6, 7, . . . , 19

• ⇡5 > 0

• Death benefits, for k + 1 6= 16, are given by bk+1 = 1000,

k + 1 = 1, 2, . . . , 14, 15, 17, 18, 19, 20

• b16 > 0

• The endowment benefit is 1000.

• The following values, based on the Illustrative Life Table at 6% interest: 1000P68 = 53.21 1000A68:20 = 500.05

4. Calculate (A) 489

17 V

a ¨68:20 = 8.8325 a ¨85:3 = 2.501

.

(B) 509

(C) 587

(D) 715

(E) 725

5. Calculate 2 V . (A) 59

(B) 67

(C) 80

(D) 91

(E) 99

6. For this question only, you are given additional information: 5V

= 148.00

and

6V

= 282.00

Calculate ⇡5 . (A) 147

(B) 152

Arch MLC, Spring 2009

(C) 157

(D) 162

(E) 167

c Yufeng Guo

235

Chapter 6 Use the following information for the next 4 questions: The random variable k L is the prospective loss at time k for a fully discrete 3-year endowment insurance of 3 on (x). You are given: • i = 0.10 • qx = 0.009 • The premium is 3Px:3 = 0.834 •

3 · k Vx:3 0.898 1.893 3.000

k 1 2 3

7. Calculate qx+1 . (A) 0.007

(B) 0.011

(C) 0.015

(D) 0.019

(E) 0.023

8. Calculate 0 L, given that (x) dies in the first year. (A) 0.08

(B) 0.23

(C) 0.63

(D) 1.25

(E) 1.89

9. Calculate Var [1 L]. (A) 0.011

(B) 0.016

(C) 0.021

(D) 0.026

(E) 0.031

(C) 0.033

(D) 0.042

(E) 0.054

10. Calculate Var [⇤0 ]. (A) 0.016

(B) 0.024

11. For a fully discrete whole life insurance of 1 on (30), you are given: • Var [10 L|K(x) 10] = y + v 2 p40 Var [11 L|K(x) • Mortality follows the illustrative live table. • i = 0.06

11]

Calculate y. (A) 0.0022

(B) 0.0023

Arch MLC, Spring 2009

(C) 0.0024

(D) 0.0025

c Yufeng Guo

(E) 0.0026

236

Chapter 6

Solutions to Pre-2000 Problems: Chapter 8 1. Key: E 11 V 11 V20

= (10 V + P20 ) (1 + i)

(b11

= (10 V20 + P20 ) (1 + i)

11 V

(1

) q50

(1)

11 V20 ) q30

Also, 10 V

= 10 V20

and

11 V

= 11 V20

So, 11 V

= (10 V + P20 ) (1 + i)

(1

11 V

) q30

(2)

Subtracting equations (2) from equation (1) gives (b11

11 V

b11 = =

(1

) q50 = (1

(1

11 V

11 V

) q30

q50

) q30

+ 11 V

0.08154)(0.008427) + 0.08154 = 0.502 0.018427

2. Key: E Reserve = PV[Future Benefits] - PV[Future Premiums] There are no future premiums at time 3, so the reserve equals 10 + 8vp60 + 6v 2 2 p60 + 4v 3 3 p60 + 2v 4 4 p60 = 10 + 8e

0.06

e

0.04

⇣

+6 e

0.06

= 10 + 8e

e

0.1

0.04

⌘2

+ 6e

⇣

+4 e

0.2

+ 4e

0.06

e

0.04

0.3

+ 2e

⌘3

⇣

+2 e

0.4

0.06

e

0.04

⌘4

= 10 + 7.239 + 4.912 + 2.963 + 1.341 = 26.45

Arch MLC, Spring 2009

c Yufeng Guo

237

Chapter 6 3. Key: E 10 V

= 1000 Ax+10 = =

(9 V + P )(1.06) qx+9 (1000) px+9

(322.87 + 32.88)(1.06) 12.62 = 369.13 0.98738 1 0.36913 a ¨x+10 = = 11.14537 0.06 1.06

1000 Px+10 =

1000 Ax+10 369.13 = = 33.1196 a ¨x+10 11.14537

4. Key: E 17 V

A85:3 = 1

h

= 1000 A85:3

da ¨85:3 = 1

) 17 V = 858.44

P68 a ¨85:3

i

(0.0566)(2.501) = 0.85844 53.21(2.501) = 725.36

5. Key: A Since the premiums and benefits for the first 2 years are the same as for a whole life insurance on (68), the reserve at the end of year should be the same. 2V

✓

= 1000 2 V68 = 1000 1

a ¨70 a ¨68

◆

From the illustrative life table, this equals ✓

= 1000 1

8.5693 9.1066

◆

= 59.00

6. Key: A (5 V + ⇡5 ) (1 + i) = p73 (6 V ) + 1000 · q73 ) ⇡5 = =

Arch MLC, Spring 2009

p73 6 V + 1000 · q73 1+i

(0.9567)(282) + 1000(0.0433) 1.06

c Yufeng Guo

5V

148 = 147.37

238

Chapter 6 7. Key: B 1V

+ P = 3v · qx+1 + 2 V · v · px+1

(0.898 + 0.834) =

3 1.893 · qx+1 + (1 1.1 1.1

qx+1 )

1.893 3 1.893 = · qx+1 1.1 1.1

1.732

qx+1 = 0.0110

8. Key: E 0L

=

= 3v 1

3 1.1

Px:3 a ¨1

0.834 = 1.89

9. Key: A Var[1 L|K(x)

1] = Var[⇤1 |K(x)

1] + v 2 px+1 Var [2 L|K(x)

2]

But since this is endowment insurance, a person alive at time t = 2 will receive the same payment at the end of the year whether they live or die. Therefore, Var [2 L|K(x)

2] = 0

and Var[1 L|K(x)

1] = Var[⇤1 |K(x)

1] = v 2 (b2

2V

)2 px+1 qx+1

Two problems earlier, we saw that qx+1 = 0.011 so that px+1 = 0.989 and Var[1 L|K(x)

1] =

✓

1 1.1

◆2

(3

1.893)2 (0.989)(0.011) = 0.011

10. Key: C Since survival to time zero is guaranteed: Var[⇤0 ] = Var[⇤0 |K(x) =

Arch MLC, Spring 2009

✓

1 1.1

◆2

(3

0] = v 2 (b1

1V

)2 px qx

0.898)2 (0.991)(0.009) = 0.033

c Yufeng Guo

239

Chapter 6 11. Key: C Var[10 L|K(x)

10] = Var[⇤10 |K(x)

So

y = Var [⇤10 |K(x) 11 V30

y=

Arch MLC, Spring 2009



1 (1 1.06

10] + v 2 p40 Var [11 L|K(x)

10] = v 2 (b11

p40 q40

a ¨41 = 0.0885 a ¨30

=1

0.0885)

2 11 V30 )

11]

2

(0.99722)(0.00278) = 0.0024

c Yufeng Guo

240

Chapter 7

ACTUARIAL MATHEMATICS: CHAPTER 9 MULTIPLE LIFE FUNCTIONS • Option A reference: Actuarial Mathematics Chapter 9 • Option B reference: Models for Quantifying Risk Chapter 13 So far, we have developed the basic ideas of actuarial mathematics in the context of just one insured life. We now expand that theory to cover multiple lives. In this chapter there are no new actuarial concepts to learn. However, you have to understand the single life material covered in earlier chapters in order to grasp the multiple life functions. A lot of the multiple life material parallels the single life material very closely, so it shouldn’t be too strenuous a chapter, though it is long! An important idea to remember as we go through Chapter 9 is that of independence among lives. For the most part, this chapter considers multiple lives to be independent. While this is not likely to be true for a group of people buying a policy together, it has historically been assumed for the sake of convenience. As a student, be thankful for this assumption! It saves you tons of headaches. Instead of discussing a “life,” we now consider a “status.” Think of two people on a desert island. We can define two types of common status. First, a “joint life status” survives as long as both people survive on the island. As soon as one person dies (or is voted o↵ the island), the joint life status terminates. Alternatively, a “last survivor status” survives as long as at least one of the people remains alive on the island. We will come back to this idea, but be familiar with the thought of a status continuing or terminating, as opposed to a life continuing or terminating.

9.2 Joint Distributions of Future Lifetimes • Option A reference: Actuarial Mathematics Chapter 9.2 241

Chapter 7 • Option B reference: Models for Quantifying Risk Chapter 13.1 We are going to bypass the theory behind joint distributions and just work a couple of examples. The solutions will involve enough explanation so that you should have the idea after working through them. One important concept from this section is that of the joint survival function. Recall that, in the single life case, the survival function is sT (x) (t) = s(t) = Pr [T (x) > t] = t px . Similarly, joint survival function is sT (x)T (y) (s, t) = Pr [T (x) > s and T (y) > t]. Most of the time, when we encounter this function, we will actually be interested in the probability that both (x), and (y) survive t years, that is t pxy

= sT (x)T (y) (t, t) = Pr [T (x) > t and T (y) > t].

EXAMPLE: (x) and (y) are independent lives. The force of mortality for each is µx (t) = 501 t for 0 < t < 50. Find the joint survival function for T (x) and T (y): sT (x)T (y) (t, t). SOLUTION: First find sT (x) (t) and sT (y) (t) and then we can just multiply since the lives are R µ(t)dt independent. We know from Chapter 3 that s(t) = e , ) sT (x) (t) = e

Rt

1 0 50 ⌧

d⌧

= e[ln(50

⌧ )]t0

= eln[

) sT (x)T (y) (t, t) = sT (x) (t) · sT (y) (t) =

✓

50 t 50

] = 50 t 50

50 t 50

◆2

.

}

EXAMPLE: Lives (x) and (x + 3) are independent. If 3 qx = 0.08 and 6 qx = 0.2, find the value of 3 pxy . SOLUTION: Since the lives are independent, 3 pxy

= Pr [T (x) > t and T (y) > t] = Pr[T (x) > t] · Pr[T (y) > t] = 3 px 3 px+3 . 3 px

=1

3 qx

= 0.92 and 6 px = 0.8.

We also need 3 px+3 . To get it we notice that 6 px

= 3 px · 3 px+3 .

So 3 px+3 = 0.87, and 3 pxy = (0.92)(0.87) = 0.80. Arch MLC, Spring 2009

c Yufeng Guo

} 242

Chapter 7

9.3 Joint Life Status • Option A reference: Actuarial Mathematics Chapter 9.3 • Option B reference: Models for Quantifying Risk Chapter 13.1 A joint life status survives as long as all members of a set survive, then terminates upon the first death. The notation that follows parallels the single life case: for (x), we have t px , while for (xy), a two life status, we have t pxy . The quantity t pxy represents the probability that the joint-life status (x, y) will survive for at least t years. In less mathematical and more intuitive terms, t pxy is the probability that both (x) and (y) will survive for at least the next t years. Joint-Life: Cumulative Distribution Function The distribution function for T , the time to failure of the joint-life status, is FT (t) = Pr {min [T (x), T (y)]  t} = t qxy = P r (T  t)

=1

sT (x)T (y) (t, t).

When T (x) and T (y) are independent, (and on the exam, this will almost always be the case), we can express the above relations in terms of single life functions: FT (t) = t qxy = 1

t px t py

= t qx + t qy

t qx t qy

This tells us that when (x) and (y) are independent, t qxy

= t qx + t qy

t qx t qy .

This says that the probability that at least one of (x) and (y) will die in the next t years is equal to the probability that (x) will die plus the probability that (y) will die minus the probability that both will die. (We have to subtract out the probability that both will die because it has been counted in both t qx and t qy .)

The following is important! So we have established that if (x) and (y) are independent lives then t pxy

= t px t py and t qxy = t qx + t qy

t qx t qy .

Both of these relations are extremely important and you must be ready to produce them at any moment! Because the one for p’s is simpler, you could just remember that one and then get the other if you need it using t qxy

=1

t pxy

=1

t px t py

=1

(1

t qx )(1

t qy )

= t qx + t qy

t qx t qy .

If that doesn’t seem quick and easy to you, definitely remember both relations. They can be intuitive if you think about them. Arch MLC, Spring 2009

c Yufeng Guo

243

Chapter 7 Note the basic idea we have discovered so far: GENERAL (POSSIBLY DEPENDENT) CASE INDEPENDENT CASE

=) =)

DIFFICULT! MUCH EASIER!

Fortunately, questions on the exam tend to be of the easier variety. EXAMPLE: x and y are independent lives with px = 0.9, and qxy = 0.15. Find qy . SOLUTION: t qxy

= t qx + t qy

=) 0.15 = 0.1 + t qy

t qx t qy

(0.1) t qy

) t qy = 0.056.

}

EXAMPLE: x and y are independent lives:

n px

= 0.75,

n py

= 0.60.

1. What is the probability that exactly one of (x) and (y) will be alive at the end of n years? 2. What is the probability that at least one of (x) and (y) will die in the next n years? 3. What is the probability that (x) will survive the next n years and y will not? 4. What is the probability that both lives will die in the next n years? SOLUTION: 1. Either x lives and y dies or y lives and x dies. The probability of this is n px n q y

2.

n qxy

3.

n px

4.

n qx

= n qx + n qy

+ n qx n py = (0.75)(0.4) + (0.25)(0.6) = 0.45. n qx n qy

= (0.25) + (0.4)

(0.25)(0.4) = 0.55.

· n qy = (0.75)(0.4) = 0.3.

· n qy = (0.25)(0.4) = 0.1.

}

Useful Formula: Just as ex , the expected future lifetime of (x), is given by ex =

Z 1 0

t px dt,

the complete expectation for the joint-life status, exy , is given by exy = Arch MLC, Spring 2009

Z 1 0

t pxy

dt.

c Yufeng Guo

244

Chapter 7 EXAMPLE: x and y are independent lives that have constant forces of mortality equal to µx = 0.03 and µy = 0.05, respectively. Find exy . SOLUTION:

Z 1

exy =

0

t pxy

dt

Since the lives are independent, this equals Z 1 0

t px t py dt =

Z 1

e

(0.03)t

e

(0.05)t

dt =

0

Z 1

e

(0.08)t

dt =

0

1 = 12.5 0.08

This means that the expected future time before at least one of x and y dies is 12.5 years. } EXAMPLE: (50) and (60) are independent lives that are both subject to DeMoivre’s Law for mortality with ! = 100. Find e50:60 . SOLUTION: exy = For DeMoivre’s Law, t px

Z 1 0

=

)e50:60 =

· t px dt

t px

!

x !

Z 1 0

t x

t p50

.

· t p60 dt.

Since they both can stay alive for at most 40 years (! = 100) we can put 40 for the upper limit of integration. Along with substituting for t px and t py , this gives e50:60 =

Z 40 ✓ 50

50

0

t

◆✓

40 t 40

◆

dt

There is a tendency to shy away from a problem like this once you see it has an integral, but most of the integrals they give on the exam are polynomials or a simple exponential. We definitely want to do these! =

1 2000

Z 40 ⇣

2000

0

= Arch MLC, Spring 2009

⌘

90t + t2 dt =

1 [80,000 2000



1 2000t 2000

1 45t2 + t3 3

72,000 + 21,333] = 14.66. c Yufeng Guo

40 0

} 245

Chapter 7 Joint-Life: Probability Distribution Function and Force of Failure INDEPENDENT CASE: Recall from Chapter 3 that, in the single life case, fT (x) (t) = f (t) = t px µx (t) = t px µ(x + t). Similarly, fT (xy) (t) = t pxy µxy (t) = t px t py µxy (t). It just so happens, that when (x) and (y) are independent, µxy (t) = µ(x + t) + µ(y + t)

(Remember This!).

This says that the Force of Failure for the joint-life status (x, y) is equal to the sum of the forces of mortality for (x) and (y). Therefore, fT (xy) (t) = t px t py (µ(x + t) + µ(y + t)) . That takes care of the PDF and force of failure in the independent case. In the general case, the force of failure is given by GENERAL CASE: The PDF is very messy! (I.e., Forgetaboutit!) Force of Failure – see below. µxy (t) =

fT (xy) (t) 1 FT (xy) (t)

This is just like µx (t) =

f (t) t px

from Chapter 3. This form doesn’t simplify any further so a question of this type for joint life distributions would have to be somewhat straightforward. EXAMPLE: (50) and (60) are independent lives that are both subject to DeMoivre’s Law for mortality with ! = 100. Find µxy (t). SOLUTION: µxy (t) = µ(x + t) + µ(y + t) Now we can either remember for DeMoivre’s law that µ(x + t) = or we can derive it using t px = In any case, µxy (t) = Arch MLC, Spring 2009

1 50

t

!

+

x !

t x

1 40

!

t

=

1 x

t d dt t px

and µ(x + t) =

(50

90 2t t)(40

c Yufeng Guo

t)

t px

.

. } 246

Chapter 7 EXAMPLE: The lives (x) and (y) have constant forces of mortality. You are given that ex = 20 and ey = 40. Find exy . SOLUTION: 20 =ex =

1 (since constant force of mortality) µx ) µx = 0.05.

Similarly, µy = 0.025. Now, the joint status (xy) also has a constant force of mortality equal to µxy = 0.05 + 0.025 = 0.075 )exy =

1 = 13.33. 0.075

This approach is a lot faster than using integrals (see the constant force of mortality question above, which you ought to be able to more quickly now!) } Curtate Life Functions The text just includes a few blurbs about curtate life functions and there doesn’t seem to be anything new that you don’t already know as long as you remember that for independent lives, k pxy = k px k py , k qxy

= k qx + k qy

k qx k qy .

The probability that a joint-life status terminates during the period (k, k+1) can be expressed as follows: Pr(k < T  k + 1) = Pr(T  k + 1) Pr(T  k) = k pxy

k+1 pxy

= k pxy qx+k:y+k . (This is nothing new, and don’t let the colon in x + k : y + k disturb you! The authors should really have included it in the xy notation as well. You can ignore it.) As always, keep in mind the English translation of each line. The second line of the equation says that for the first death to occur in the period (k, k + 1), both (x) and (y) first have to survive k years (k pxy ) but they cannot both survive k + 1 years. The third line can be interpreted similarly: “For the first death to occur in (k, k + 1) both must survive k years and then at least one of the two must die in the k + 1st year.” Again, the independent assumption is very nice in that qx+k:y+k = qx+k + qy+k Arch MLC, Spring 2009

c Yufeng Guo

qx+k qy+k . 247

Chapter 7 Recall that K is the number of complete years completed prior to failure for the status. The probability function (pf ) is as follows: Pr(K = k) = k pxy qx+k:y+k = k| qxy . And finally, the curtate complete expectation of life ex has the expected formula based on that from Chapter 3 for single lives: exy = E[K(xy)] =

1 X

k+1 pxy .

k=0

The text gives very little attention to the curtate functions, so understanding these formulas should be all you need to know as long as you remember what you already know for single life functions.

9.4 Last Survivor Status • Option A reference: Actuarial Mathematics Chapter 9.4 • Option B reference: Models for Quantifying Risk Chapter 13.2 This type of status exists as long as at least one member of the set is alive. The status terminates upon the final death of the set of members. The Following is Important! Here is a key point to this chapter that is new to the joint life functions: suppose you have two individual lives (x) and (y) represented by remaining life functions T (x) and T (y). Then T (xy) and T (xy) represent both lives. The quantity T (xy) represents the remaining time until the first death, while T (xy) represents the remaining time until the last death. No matter how things turn out for these two lives, T (xy) will be the time until one of the deaths and T (xy) will be the time until the other death. However, the exact same statement can be made of T (x) and T (y)! Therefore, it must be true that T (xy) + T (xy) = T (x) + T (y)

((A) A key Relation!).

All of the following must be true for exactly the same reasons: T (xy) · T (xy) = T (x) · T (y),

(B)

cT (xy) + cT (xy) = cT (x) + cT (y) for c > 0,

(C)

FT (xy) (t) + FT (xy) (t) = FT (x) (t) + FT (y) (t),

(D)

fT (xy) (t) + fT (xy) (t) = fT (x) (t) + fT (y) (t),

(E)

t pxy

Arch MLC, Spring 2009

+ t pxy = t px + t py .

(F )

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248

Chapter 7 EXAMPLE: You are given: • (40) and (50) are independent lives

• Mortality is as given in the Illustrative Life Table. Find

20 p40:50 .

SOLUTION: t pxy

= t px + t py

20 p40:50

20 p40

=

t pxy

= t px + t p y

= 20 p40 + 20 p50

(20 p40 ) (20 p50 )

l60 81,881 = = 0.879, l40 93,132

) 20 p40:50 = 0.879 + 0.739

Arch MLC, Spring 2009

t p x t py

similarly,

20 p50

= 0.739.

(0.879)(0.739) = 0.968.

c Yufeng Guo

}

249

Chapter 7 EXAMPLE: 1. Assuming the future lifetimes of (x) and (y) are independent, write k pxy in terms of k px and k py . 2. Assuming the future lifetimes of (x) and (y) are independent, write k qxy in terms of k qx and k qy . SOLUTION: 1. The probability that at least one of (x) and (y) will be alive after k years is the probability that (x) will be alive plus the probability that (y) will be alive minus the probability that both will be alive. Or, k pxy

= k p x + k py

k p x k py .

2. Both have to die, t qxy

= k qx k qy .

}

9.5 More Probabilities and Expectations • Option A reference: Actuarial Mathematics Chapter 9.5 • Option B reference: Models for Quantifying Risk Chapter 13.1.7, 13.2.2 In this section, the text introduces a few new formulas. No big new ideas here – just some good formulas that the SOA could put on the exam (Remember from Chapter 3 that ex = E[T (x)]): Continuous exy =

exy =

Z 1

t pxy

Z 1

t pxy

0

0

Curtate exy =

dt

1 X

k pxy

1 X

k pxy

1

exy =

dt

1

exy =ex + ey

exy

exy = ex + ey

exy

Formulas for the Variance of the future lifetime (in the continuous case) are Var[T (xy)] = E[T (xy)2 ]

E[T (xy)]2 = 2

Z 1 0

Var[T (xy)] = E[T (xy) ] 2

E[T (xy)] = 2 2

Z 1 0

t t pxy dt

(exy )2 ,

t t pxy dt

(exy )2 .

EXAMPLE: Mr. and Mrs. Moneybags Mr. and Mrs. Moneybags have promised to donate their entire fortune to the national home for unwanted cats after they both have died. They are ages 50 and Arch MLC, Spring 2009

c Yufeng Guo

250

Chapter 7 60 respectively, and their mortality is given by Demoivre’s Law with ! = 100. How long can the home for unwanted cats expect to wait before receiving their donation? (Hint: we saw in an earlier problem that, for this pair, e50:60 = 14.66.) SOLUTION: We are looking for e50:60 , the expected future lifetime of the status (xy). We know that e50:60 =e50 + e60 e50:60 . e50 =

Z 50 0

t p50 dt =

Z 50 ! 0

50 t dt = ! 50

Z 50 50

50

0

(Or we can use the fact that under DeMoivre, ex =

t

! x 2 .)

dt = 25

Similarly,

e60 = 20. )e50:60 = 20 + 25

14.66 = 30.34.

}

EXAMPLE: Mr. and Mrs. Moneybags For Mr. and Mrs. Moneybags, write an integral expression for Var[T (xy)]. The integrand should be a function of t only. SOLUTION: Var[T (xy)] = 2

Z 1 0

t t pxy dt

(exy )2 = 2

Z 40 ✓ 50

t

0

t

50

◆✓

40 t 40

◆

dt

(14.66)2

That will do it - note that there is no reason we are unable to do this integral, it is just a polynomial if we multiply it out. } Later in the chapter it will be useful to notice that many of these formulas are exactly the same for a single-life status (Chapters 3 and 4), joint-life status, and last survivor status, except that the status (x, xy, xy) changes. It is useful to recognize that, in general, a status can be denoted u. Then, write whatever the function is (Ax , Var[T (x)], etc.) with u as the ‘single-life status’. Then depending on which case you’re using, plug in either u = xy or u = xy into the formula. So, if Var[T (u)] = 2

Z 1

t t pu dt

Z 1

t t pxy dt

0

then Var[T (xy)] = 2

0

We will see more of this later. Arch MLC, Spring 2009

c Yufeng Guo

⇣

eu

⌘2

,

(exy )2 .

251

Chapter 7 A sneaky question the SOA could pose deals with the covariance of T (xy) and T (xy). Here is what you need to know: ⇣

⌘

Cov[T (xy), T (xy)] = Cov[T (x), T (y)] + E[T (x)] = Cov[T (x), T (y)] + ( ex

E[T (xy)]

exy )( ey

·

⇣

E[T (y)]

⌘

E[T (xy)]

exy ).

This formula applies to both dependent lives and independent lives. Hopefully, any exam question will be in the case where (x) and (y) are independent lives, in which case the covariance term on the right-hand-side is 0. The formula for this is: Cov[T (xy), T (xy)] = (ex

exy )(ey

exy ).

A question using this formula is more likely than the one above.

EXAMPLE: Mr. and Mrs. Moneybags For Mr. and Mrs. Moneybags, let T1 be the random variable representing the time of the first death, and let T2 be the random variable representing the time of the second death. Find Cov [T1 , T2 ]. SOLUTION: We want Cov[T (xy), T (xy)] = (ex

exy )(ey

exy ) = (25

14.66)(20

14.66) = 55.22

Note that this means T (xy) and T (xy) are positively correlated. This makes sense, if T (xy) is small (meaning the first death occurred quickly), it increases the chance that T (xy) will be small. }

EXAMPLE: Mr. and Mrs. Moneybags Mr. and Mrs. Moneybags, have amended their will such that a donation will be made to the national home for cats only if the first death occurs between 10 and 20 years from now. What is the probability that the national home for cats will get a donation from the Moneybags? SOLUTION: We want the probability that the first death does not occur in the next ten years but does occur in the next twenty years. This is 10 pu

Arch MLC, Spring 2009

20 pu ,

c Yufeng Guo

252

Chapter 7 where u = xy. So the answer is 10 p50:60

20 p50:60

= 10 p50 10 p60

By recalling that for DeMoivre, t px = 10 p50:60

20 p50:60

! x t ! x ,

20 p50 20 p60 .

we get that

= (0.8)(0.75)

(0.6)(0.5) = 0.3.

Note that we used the fact that the lives were independent. That’s ok because each person’s mortality is given by DeMoivre regardless of when the other dies. So the lives are independent. } EXAMPLE: Mr. and Mrs. Moneybags Mr. and Mrs. Moneybags, have amended their will yet again such that a donation will be made to the national home for cats only if the second death occurs between 10 and 20 years from now. What is the probability that the national home for cats will get a donation from the Moneybags? SOLUTION: We want the probability that the second death does not occur in the next 10 years but does occur in the next 20 years. Again this is 10 pu

But this time, u = xy. So we want = (10 p50 + 10 p60 = [0.8 + 0.75

20 pu .

10 p50:60

10 p50 10 p60 )

(0.8)(0.75)]

10 p50:60

(20 p50 + 20 p60 [0.6 + 0.5

20 p50 20 p60 )

(0.6)(0.5)] = 0.15

}

9.6 Dependent Lifetime Models 9.6.1 Common Shock –(Non-Theoretical Version) • Option A reference: Actuarial Mathematics Chapter 9.6 • Option B reference: Models for Quantifying Risk Chapter 13.6 We’ll skip the theoretical aspects of this section and concentrate on what you are likely to need for the exam. Arch MLC, Spring 2009

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253

Chapter 7

Symbols

Meaning

T ⇤ (x), T ⇤ (y)

What the future lifetimes of (x) and (y) would be if there were no common peril. These two are independent. What the future lifetime of both would be if the only way to die was the common peril

Z

T (x), T (y)

The actual future lifetimes of (x) and (y) with both the common peril and everything else. These two are not independent of each other.

T ⇤ (x) and T ⇤ (y) refer to what the future lifetimes of (x) and (y) would be if they were completely independent and there were no common perils (i.e., just like everything we have looked at before this section!) The symbol Z represents what either of their lifetimes would be if the only peril they faced in life was the common peril. Since each of (x) and (y) must in fact face the common peril and everything else in life that could kill them, the resulting survival function is the product of the survival functions for 1) the common peril, and 2) everything else. The result is sT (x) (t) = sT ⇤ (x) (t) · sZ (t) = sT ⇤ (x) (t) · e

t

,

sT (y) (t) = sT ⇤ (y) (t) · sZ (t) = sT ⇤ (y) (t) · e

t

,

sT (x)T (y) = sT ⇤ (x) (t) sT ⇤ (y) (t) sZ (t) = sT ⇤ (x) (t) sT ⇤ (y) (t) e

t

.

“But wait a minute!”, you may say, “I know full well that sT (x)T (y) = sT (x) (t) · sT (y) (t) since that is just t pxy = t px t py !!” Not Quite! That is true only when (x) and (y) are independent lives. The common shock assumption makes (x) and (y) have dependent future lifetimes because they are exposed to a “common” peril. EXAMPLE: In the presence of a common shock peril, write sT (xy) (t) in terms of sT ⇤ (x) (t), sT ⇤ (y) (t), and sZ (t). SOLUTION: Arch MLC, Spring 2009

c Yufeng Guo

254

Chapter 7 We know that sT (xy) (t) = sT (x) (t) + sT (y) (t) ) sT (xy) (t) = sT ⇤ (x) (t) e

t

+ sT ⇤ (y) (t) e

= [sT ⇤ (x) (t) + sT ⇤ (y) (t)

sT (xy) (t). t

sT ⇤ (x)T ⇤ (y) (t) e

sT ⇤ (x)T ⇤ (y) (t)]e

t

.

t

}

Recall from Section 9.4 that if T (x) and T (y) are independent, then µxy (t) = µ(x + t) + µ(y + t). Since, in the context of common shock, T ⇤ (x), T ⇤ (y), and Z are all independent, it makes sense that µxy (t) = µ(x + t) + µ(y + t) + . (Note that = µZ (t).) So at each moment, the joint life status is threatened by 3 di↵erent forces of mortality. Notational Comment: Make sure you recognize that, throughout this section, the ‘normal’ random variables – i.e., with no common shock associated – are designated with a ‘*’. This seems a little backwards, since in previous chapters there was no ‘*’ in use. EXAMPLE: (x) and (y) are subject to constant forces of mortality µx = 0.02 and µy = 0.03 respectively. In addition, both lives are subject to a common shock peril with = 0.01. Find

1. 8 pxy , and

2. exy .

SOLUTION: 1. µxy (t) = 0.02 + 0.03 + 0.01 = 0.06, a constant. Therefore, 8 pxy

2. Since µxy is a constant, exy =

=e 1 0.06

0.06(8)

= 0.619.

= 16.67

}

EXAMPLE: (x) and (y) are subject to constant forces of mortality µx = 0.02 and µy = 0.03 respectively. In addition, both lives are subject to a common shock peril with = 0.01. Find the probability that the common peril will end the pair of lives. SOLUTION: For the common peril to cause the failure of the joint life status at time t. The pair must survive all perils up to time t and then get hit by the force of mortality . The probability that this will happen at some time t is Z 1 0

t pxy

Arch MLC, Spring 2009

dt =

Z 1

e

(µx +µy + )

dt =

0

c Yufeng Guo

µx + µy +

= 0.167.

} 255

Chapter 7

9.7 Insurance and Annuity Benefits 9.7.1 Survival Statuses • Option A reference: Actuarial Mathematics Chapter 9.7.1 • Option B reference: Models for Quantifying Risk Chapter 13.6 In this section, the text basically repeats many formulas from chapters 4 and 5, so most explanations will be brief. However, don’t underestimate the importance of this section! In many ways, it is some of the most testable stu↵ for multiple lives. CURTATE INSURANCE AND ANNUITY FUNCTIONS: Consider a whole life insurance with unit benefit payable at the end of year of ‘death’ of the survival status u (u can be x, xy, or xy) As before, let K be the “curtate-future-lifetime” of u. So if u = x, then K is the number of future complete years lived by x, whereas if u = xy, then K is the number of complete years lived by both x and y. (Get the idea? If so, go ahead and say to yourself what K is if u = xy.) Now, using what you know for single-life functions, you can write down most simple actuarial functions for the multiple-life case. For example, we know that Au =

1 X

v k+1 k pu qu+k ,

k=0

so Axy =

1 X

v k+1 k pxy qx+k:y+k .

k=0

When doing this translation process, you have to be careful dealing with the last survivor status. It happened with the first-to-die status that we could write k pxy qx+k:y+k by essentially just replacing u with xy. With last survivor status, you have to be a lot more careful. As we will see in the next EXAMPLE below. The key way to think about it is that Au =

1 X

v k+1 Pr[K = k],

k=0

and then you have to think to figure out logically what is the probability that the last survivor status fails in year k + 1. I.e., what is Pr[K = k]? If desired, you can go further with this translation when (x) and (y) are independent lives by translating the entire sum into one containing only single life functions, using the single life translations of k pxy and qx+k:y+k . For the joint-life status, the messy result would be Axy =

1 X

v k+1 k px k py (qx+k + qy+k

qx+k qy+k ).

k=0

Just as in Chapter 4, the random variable Z = v K+1 represents the present value at the time the policy is issued of the benefit payment. Also, Au = E[Z] Arch MLC, Spring 2009

and

Var[Z] = 2 Au

c Yufeng Guo

(Au )2 . 256

Chapter 7 Therefore, if u = xy, Var[Z] = 2 Axy

(Axy )2 .

Nothing really new here; just making sure you know what you know! EXAMPLE: (x) and (y) bought a last survivor curtate life insurance policy together with a benefit of 1,000,000 payable on the last death of the two. Write expressions for expected value of the benefit and the variance of the present value of the death benefit. SOLUTION: 1. 1,000,000 Axy Ok, that was a pretty cheap answer. You probably wrote something more complex, like Axy =

1 X

v k+1 Pr[K = k].

k=0

Then you might have gotten muddled over what to do with Pr[K = k]. The problem is that k pxy qx+k:y+k is not the right expression since qx+k:y+k is the probability of the status failing in year k + 1 if both are alive at the beginning of the year, but that might not be the case even if the last survivor status survives k years (i.e., it may be that only one of x and y is alive at the start of the last year). This is a hard one! You have to recognize that Pr[K = k] = fK(xy) (k) and use the relation fK(xy) (k) = fK(x) (k) + fK(y) (k) = k px qx+k + k py qy+k So Axy =

1 X

k=0

v k+1

⇣

fK(xy) (k)

k pxy qx+k:y+k .

k px qx+k + k py qy+k

⌘

k pxy qx+k:y+k .

and our answer is 1,000,000 times that. This is a messy answer to a messy example, but it contains lots of useful review – for me, the hardest part is remembering that Pr[K = k] = fK(xy) (k)! Review this example. 2. If the benefit were 1, then Var[Z] = 2 Axy

(Axy )2 .

Since we want Var[1,000,000 Z] the variance of the present value of the benefit is h i (1,000,000)2 2 Axy (Axy )2 . We’ll leave it at that! Arch MLC, Spring 2009

}

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257

Chapter 7 EXAMPLE: Given the following information, what is the actuarial present value of a continuous life annuity that pays 25,000 per year until the last death of (65) and (70)? • a65 = 10.2

• a65:5 = 4.6

• a65:70 = 6.2

• 5 E65 = 0.82 SOLUTION: We want to find a65:70 and we know that a65:70 = a65 + a70

a65:70 .

We have all of those quantities except for a70 , for that we need to use a65 = a65:5 + 5 p65 v ) a70 =

5

a70 .

10.2 4.6 = 6.83, 0.82

and a65:70 = 10.2 + 6.83

6.2 = 10.83.

So our answer is (25,000)(10.83).

}

It is worth noting in this example that a65:70 < a70 < a65 < a65:70 . You should reason out why this ought to be the case. CONTINUOUS INSURANCE AND ANNUITY FUNCTIONS: Since the last example had to do with continuous multiple-life functions, this is a good time to mention that the same process that we use to get from single-life to multiple-life cases for curtate insurance and annuities also works for continuous functions. Since we gave a specific life insurance example in the Curtate case, we will look at an annuity example for the continuous case. Recall from Chapter 5 that for a continuous annuity of 1 per year until failure of the survival status u, the present value of future benefits is given by the random variable Y = aT , au = and

Z 1

Var[Y ] = Arch MLC, Spring 2009

0

2A

v t t pu dt,

u

⇣

Au

2

c Yufeng Guo

⌘2

. 258

Chapter 7 EXAMPLE: Write out the corresponding formula for axy and Var[Y ] where Y is the present value random variable for a joint-life continuous life annuity. SOLUTION: axy =

Z 1

Var[Y ] =

0

2A

v t t pxy dt,

xy

⇣

A xy

2

⌘2

.

Again, we could take this further and get everything in terms of single life functions if we are told that (x) and (y) are independent lives. } EXAMPLE: (x) and (y) are subject to constant forces of mortality µx = 0.03 and µy = 0.01, respectively. The force of interest is = 0.06. The pair have purchased a whole life policy that pays 1000 upon the first death of the two. (A) Find the actuarial present value of this insurance. (B) Find the standard deviation of the present value of this insurance. SOLUTION: (A) We are looking for A 50:60 . The force of mortality for this joint life status will be µxy = µx + µy = 0.04. Since this is a constant, we know the APV of the benefit is just 1000 · A 50:60 = 1000 · (B)

h

µ50:60 µ50:60 +

= 1000 ·

i

⇣

Var v T = 2A 50:60 A 50:60 =

2

h

A 50:60

0.04 = 400. 0.1

⌘2

µ50:60 0.04 = = 0.25 µ50:60 + 2 0.16 i

) Var v T = 0.25

(0.4)2 = 0.09.

The variance of our insurance is h

i

Var 1000v T = (1000)2 (0.09) = 90,000. The standard deviation of the present value is = Arch MLC, Spring 2009

p

90,000 = 300.

c Yufeng Guo

} 259

Chapter 7 EXAMPLE: Assuming (40) and (50) are independent lives and using the Illustrative Life Table, find values for (A) A40:50 , (B) Var[¨ a K+1 ], where K is the number of complete years that pass before the first death between the two. (C) A40:50 .

Arch MLC, Spring 2009

c Yufeng Guo

260

Chapter 7 SOLUTION: (A) We want Ax:x+10 where x = 40, which equals 0.294. (B)

h

i

Var a ¨ K+1 =

2A

(A40:50 )2

40:50

d2

0.124 (0.294)2 = 11.73. (0.0566)2

=

where we used the fact that d=1 (C) A40:50 = A40 + A50

1 =1 1+i

1 = 0.0566. 1.06

A40:50 = 0.161 + 0.249

0.294 = 0.116.

}

Some final formulas specific to multiple life functions - we will only show them in the continuous case, but they hold for the curtate case as well (we have seen a couple of these already): T (xy) + T (xy) = T (x) + T (y) v T (xy) + v T (xy) = v T (x) + v T (y) v T (xy)+T (xy) = v T (x)+T (y) =) v T (xy) v T (xy) = v T (x) v T (y) axy + axy = ax + ay A xy + A xy = A x + A y . In addition, if T (x) and T (y) are independent, then Cov [T (xy), T (xy)] =

⇣

ex

exy

⌘⇣

ey

⌘

exy ,

(I included this one even though you have already seen it because the book messed up the notation.) h i ⇣ ⌘⇣ ⌘ Cov v T (xy) , v T (xy) = A x A xy A y A xy . EXAMPLE:

h

i

(x) and (y) are independent lives. Find an expression for Cov aT (xy) , aT (xy) in terms of APV life formulas. SOLUTION: h

i

Cov aT (xy) , aT (xy) = Cov "

#

"

1

v T (xy) 1 ,

v T (xy)

#

h i v T (xy) v T (xy) 1 = Cov , = 2 Cov v T (xy) , v T (xy)

= Arch MLC, Spring 2009

⇣

Ax

A xy

⌘⇣

Ay

2

c Yufeng Guo

A xy

⌘

.

} 261

Chapter 7

9.7.2 Special Two-Life Annuities • Option A reference: Actuarial Mathematics Chapter 9.7.2 • Option B reference: Models for Quantifying Risk Chapter 13.4.2 In many pension plans, upon retirement of a married worker, an annuity is set up so that the stream of payments will decrease after the death of one of the partners. This section gives some insight into that sort of annuity. The whole section is an example, so we’ll walk through one similar to the book’s. You should also work through the example in the book: Example 9.7.2 – (Page 284 of the text). EXAMPLE: Mom and Pop want to buy a lifetime continuous two-life annuity to take care of their retirement needs. They want the annuity to pay 1 as long as both are alive, but they figure either spouse can get by on 0.6 once the other is deceased. So they have decided to buy an annuity that pays: (i) 1 per year while both (x) and (y) are alive; and (ii) 0.6 per year while only one of (x) and (y) is alive. (A) Derive an expression for the annuity’s present value random variable: The most obvious way to write the APV is:

⇣

h

1 · a T (xy) + (0.6) a T (xy) ⌘

i

a T (xy) .

Note that a T (xy) a T (xy) indicates that we will pay 1 per year as long as either of (x) or (y) is alive but we take that payment back while both are alive. This present value random variable can be expressed as : Z = (0.6)a T (xy) + (0.4)a T (xy) . (B) Derive an expression for the annuity’s actuarial present value in terms of ax , ay , axy . The APV is the expected value of the equation in the solution to (A): E [Z] = (0.6)axy + (0.4)axy . To get to ax , ay , and axy , we can use axy + axy = ax + ay to rewrite the APV as (0.6)ax + (0.6)ay

(0.2)axy .

(C) Derive an expression for the variance of the random variable in (A) under the assumption that Mom and Pop have independent lives. ⇣

⌘

Var[Z] = Var (0.6) a T (xy) + (0.4) a T (xy) . Arch MLC, Spring 2009

c Yufeng Guo

262

Chapter 7 Hopefully you remember from Statistics that Var(aS + bT ) = a2 Var(S) + b2 Var(T ) + 2ab Cov(S, T ). ⇣

⌘

⇣

⌘

⇣

⌘

) Var[Z] = (0.36) Var a T (xy) + (0.16) Var a T (xy) + (0.48) Cov a T (xy) , a T (xy) . We know that ⇣

Var a T (xy)

⌘

=

2A

xy

⇣

A xy

2

⌘2

⇣

, Var a T (xy)

⌘

=

2A

⇣

A xy

xy

2

⌘2

.

Now for the covariance term: ⇣

Cov a T (xy) , a T (xy)

⌘

=

⇣

Ax

A xy

⌘⇣

Ay

A xy

2

⌘

,

as we saw earlier. Therefore, Var[Z] is equal to 1 2

⇢

h

(0.36) 2A xy

⇣

A xy

⌘2 i

h

+ (0.16) 2A xy

⇣

A xy

⌘2 i

⇣

+ (0.48) A x

A xy

⌘⇣

Ay

A xy

⌘

9.7.3 Reversionary Annuities • Option A reference: Actuarial Mathematics Chapter 9.7.3 • Option B reference: Models for Quantifying Risk Chapter 13.4.4 Reversionary annuities are really nothing new except for the name and some notation. A reversionary annuity is payable during the existence of one status (u), but only if another status (v) has failed. For example, an annuity may be purchased to provide lifetime income to Sue, but the income only kicks in once her husband Bob has passed away. In this case, Sue would be (u) and Bob would be (v). Consider an annuity of 1 per year payable continuously to (y) after the death of (x). The present value at time 0 is Z = a T (y) a T (xy) . This is somewhat intuitive. It says Z is the present value of an annuity that pays as long as (y) is alive minus the present value of an annuity that pays as long as both (x) and (y) are alive. It is as if (y) is getting a life annuity but has agreed to give it back as long as both (x) and (y) are alive. The symbol for the APV of this insurance is ax|y and in the subscript, the person that has to die comes before the ‘|’. ax|y = E[Z] = ay Arch MLC, Spring 2009

c Yufeng Guo

axy . 263

.

Chapter 7 EXAMPLE: Find an expression for the actuarial present value of a single continuous life annuity that pays as follows: • 1 per year while both (x) and (y) are alive, • 0.6 per year if (x) is alive and (y) is not, • 0.5 per year if (y) is alive and (x) is not. SOLUTION: The way this question is written, it is clear that this is the sum of 3 separate annuities. The first is just axy . The second is (0.6)ay|x = (0.6) (ax The last one is (0.5)ax|y = (0.5) (ay

axy ) . axy ) .

Adding these up gives an APV equal to (0.5)ay + (0.6)ax

(0.1)axy .

}

EXAMPLE: Pat and Kelly are independent lives ages 30 and 35, respectively. They want to buy a two-life discrete annuity that will pay 1000 at the beginning of any year in which either of them is alive and over 60. (So the maximum payment in any year is 1000.) Find an expression for the APV of this annuity. SOLUTION: This one is tricky since the annuity will start in 25 years if Kelly is alive then, but won’t start until later if Kelly dies before then. The key idea is to think of this as an annuity that • pays 1000 per year to Pat if he is alive and over 60, • pays 1000 per year to Kelly if she is alive and over 60, and • takes back 1000 per year if both are alive and over 60. Now it is easier to write the APV: 1000 30| a ¨30 + 1000 25| a ¨35

1000 30| a ¨30:35 .

Note that the joint-life piece is deferred 30 years because there is no way both can be over 60 before that time. } One last example that I didn’t manage to fit in earlier in the Chapter: this one doesn’t teach a lot but is the kind of quirky question that shows up sometimes. I find this type very hard to get correct unless you have seen one or two like it. QUIRKY EXAMPLE: Assume that all lives are independent, Find

20 p40:60

= 0.4, and

10 p40

= 0.8.

30 p50 .

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c Yufeng Guo

264

Chapter 7 SOLUTION: With no idea how to get started, let’s start by writing down the two-life function in terms of single-life functions. 0.4 = 20 p40 20 p60 . If we are lucky, we will notice that this expression is also equal to 0.4 = 40 p40 . Now we can use the fact that 40 p40

= 10 p40 · 30 p50

to see that 0.4 = (0.8)30 p50 ) 30 p50 = 0.5.

When you get questions like this, you might feel like the exam writer is testing your cleverness as much as your knowledge of the exam material. }

9.9 Simple Contingent Functions • Option A reference: Actuarial Mathematics Chapter 9.7.2 • Option B reference: Models for Quantifying Risk Chapter 13.3 The last topic in this chapter is insurances that depend upon the order of death within a status. For example, a policy might pay on the first death of a couple if and only if the husband is the first to die. We have a little bit of new notation: n q1 xy

where 1 over the x denotes “(x) dies before (y)” and the usual meaning for n – that the death occurs within n years. So n q1 gives the probability that (x) will die in the next n years and xy

(y) will be alive at the moment x dies. (I.e., (y) is alive when (x) kicks the bucket but (y) does not necessarily live the full n years.) If we aren’t told that (x) and (y) are independent, then we have an awkward formula: n q1 xy

=

Z n 0

Pr[T (y) > t | T (x) = t] t px µ(x + t)dt.

If we’re lucky and the SOA gives a case where (x) and (y) are independent (this will almost surely be the case), we can use the formula: n q1 = xy

Z n 0

t py t px µ(x

+ t) dt.

Let’s go over this formula in English: First, t is the time variable that we’re allowing to run from 0 to n. For all times t, both (x) and (y) survive to t, but then (x) is hit by the force Arch MLC, Spring 2009

c Yufeng Guo

265

Chapter 7 of mortality in the interval (t, t + dt). Thus, in all cases analyzed here, (y) survives longer than (x). nq 2 xy

is the symbol representing the probability that (y) dies after (x), but within an n-yr

period from now. Note that above, we knew (y) died after (x), but there is no upper limit on when (y) dies. Here there is an upper limit on when (y) dies – n years. In the independent case, nq 2 xy

=

Z n 0

t q x t py

µ(y + t)dt

The integral makes sense, it integrates over all times t, the probability that (x) is dead and (y) is alive, times the probability that (y) gets hit by the force of mortality at that time. nq 2 xy

This equation says that n q

2 xy

= n qy

nq 1, xy

equals the probability that (y) dies in the next n years minus

the probability that (y) dies in the next n years and dies first. Note that the book has the 1 in the wrong place in this equation (Equation 9.9.4 in the book). The non-independent case is messy and is unlikely (but possible) to be tested. EXAMPLE: (x) and (y) are subject to the constant forces of mortality µx = 0.03 and µy = 0.05, respectively. (A) Find the probability that (x) dies before (y). (B) Find the probability that (x) dies before (y) and in the next 4 years. (C) Find the probability that both (x) and (y) will die in the next 4 years with (x) dying first.

SOLUTION: (A) This is 1 q1 x:y

=

Z 1

t pxy

0.08t

(0.03) dt =

0

µx (t) dt =

Z 1

0.03t

e

e

0.05t

0

3 (0.03) dt = . 8

(B) 4 q1 x:y

=

Z 4

e

0

(C) This is 4 q

2. x:y

4q 2 x:y

0.03 h e 0.08

0.08t

i4 0

3 = (1 8

0.726) = 0.103.

We can find this one by calculating the integral =

Arch MLC, Spring 2009

Z 4 0

t q x t py

µy (t) dt =

Z 4

(1

0

c Yufeng Guo

e

0.03t

)(e

0.05t

)µy (t) dt. 266

Chapter 7 Or we could save some trouble by noticing that 4q

2 x:y

= 4 qy

If you think about it, n q1 and n q xy

⇣

4 q1 = 1 x:y

2 xy

e

0.05(4)

⌘

0.103 = 0.078.

}

are the same except that, with n q 2 , we know that (y) xy

did not survive the full n years. That is, in both cases, (x) dies within n years and before (y). Therefore, n q 2 = n q1 n py n q x . xy

xy

This can be thought of as the case in which (y) outlives (x), who died within the n-year period. However, we’re taking away the possibility that (y) survives the full n-year period. EXAMPLE: A1

x:y

represents the actuarial present value of an insurance that pays 1 at the

moment of death of (x) if (y) is alive at the moment (x) dies. Assuming (x) and (y) are independent lives, find an integral expression for A 1 . x:y

SOLUTION: A1 = x:y

Z 1 0

v t pxy µ(x + t)dt = T

Z 1 0

v t t px t py µ(x + t)dt.

}

EXAMPLE: Assuming (x) and (y) are independent, write the integral expression for A SOLUTION: A

2 x:y

=

Z 1 0

v t t qx t py µ(y + t)dt.

2. x:y

}

Chapter 9 Suggested Problems: 4(assume the lives are independent), 10, 12, 13, 22, 24, 25, 26, 28 (Solutions at archactuarial.com)

Arch MLC, Spring 2009

c Yufeng Guo

267

Chapter 7

CHAPTER 9 Formula Summary If (x) and (y) are independent lives, t pxy

= t px t py

and

= t qx + t qy

t qxy

t qx t qy

µxy (t) = µ(x + t) + µ(y + t) fT (xy) (t) = t px t py (µ(x + t) + µ(y + t)) Force of Failure for non-independent lives: µxy (t) =

fT (xy) (t) 1 FT (xy) (t)

The probability that a joint-life status terminates during the period (k, k + 1) is k pxy

k+1 pxy

=

k pxy qx+k:y+k .

The “Laws of addition”: T (xy) + T (xy) = T (x) + T (y)

T (xy) · T (xy) = T (x) · T (y)

v T (xy) + v T (xy) = v T (x) + v T (y)

v T (xy) v T (xy) = v T (x) v T (y)

A xy + A xy = A x + A y

axy + axy = ax + ay

FT (xy) (t) + FT (xy) (t) = FT (x) (t) + FT (y) (t)

fT (xy) (t) + fT (xy) (t) = fT (x) (t) + fT (y) (t)

t pxy

+ t pxy = t px + t py

Continuous exy =

exy =

Curtate

Z 1

t pxy dt

Z 1

t pxy dt

0

0

exy =

1 X

k pxy

1 X

k pxy

1

exy =ex + ey

exy =

1

exy = ex + ey

exy

Var[T (xy)] = E[T (xy)2 ]

E[T (xy)]2 = 2

Z 1 0

Var[T (xy)] = E[T (xy)2 ]

E[T (xy)]2 = 2

Z 1 0

exy

t t pxy dt

(exy )2

t t pxy dt

(exy )2

Cov[T (xy), T (xy)] = Cov[T (x), T (y)] + ( ex

exy )( ey

Cov[T (xy), T (xy)] = (ex

exy ).

Arch MLC, Spring 2009

exy )(ey

c Yufeng Guo

exy )

268

Chapter 7

Common Shock: sT (x) (t) = sT ⇤ (x) (t) · sZ (t) = sT ⇤ (x) (t) · e

t

,

sT (y) (t) = sT ⇤ (y) (t) · sZ (t) = sT ⇤ (y) (t) · e

t

,

sT (x)T (y) = sT ⇤ (x) (t) sT ⇤ (y) (t) sZ (t) = sT ⇤ (x) (t) sT ⇤ (y) (t) e

t

.

Insurances based on multiple lives: Axy =

1 X

v k+1 k px k py (qx+k + qy+k

qx+k qy+k )

k=0

Axy =

1 X

k=0

v k+1

⇣

k px qx+k

+ k py qy+k

k pxy qx+k:y+k

⌘

If T (x) and T (y) are independent, then h

i

⇣

Cov v T (xy) , v T (xy) = A x

Arch MLC, Spring 2009

A xy

c Yufeng Guo

⌘⇣

Ay

A xy

⌘

269

Chapter 7

Past SOA/CAS Exam Questions: 1. For a last-survivor insurance of 10, 000 on independent lives (70) and (80), you are given: (i) The benefit, payable at the end of the year of death, is paid only if the second death occurs during year 5. (ii) Mortality follows the Illustrative Life Table. (iii) i = 0.03 Calculate the actuarial present value of this insurance. (A) 235

(B) 245

(C) 255

SOLUTION: AP V = ⌫ 5 4 p70

=

=

= 4 p70 + 4 p80

l75 53, 960.81 = = 0.815592 l70 66, 161.55 5 p70:80

= 5 p70 + 5 p80

AP V = ⌫ 5 (0.953912

(E) 275

5 p70:80

l74 56, 640.51 = = 0.856094, l70 66, 161.55 4 p70:80

5 p70

4 p70:80

(D) 265

=

4 p80 4 p70

⇥ 4 p80 = 0.953912

5 p80 5 p70

0.926690) =

l84 26, 607.34 = = 0.679736 l80 39, 143.65

=

l85 23, 582.46 = = 0.602459 l80 39, 143.65

⇥ 5 p80 = 0.926690

0.027222 = 0.02348 1.035

Key: A

2. For a last-survivor whole life insurance of 1 on (x) and (y): (i) The death benefit is payable at the moment of the second death. (ii) The independent random variables T ⇤ (x), T ⇤ (y), and Z are the components of a common shock model. T ⇤ (x)

(iii) T ⇤ (x) has an exponential distribution with µx

T ⇤ (y)

(iv) T ⇤ (y) has an exponential distribution with µy

(t) = 0.03, t

0.

(t) = 0.05, t

0.

(v) Z, the common shock random variable, has an exponential distribution with µZ (t) = 0.02, t 0. (vi)

= 0.06

Calculate the actuarial present value of this insurance. (A) 0.216

(B) 0.271

Arch MLC, Spring 2009

(C) 0.326

(D) 0.368

c Yufeng Guo

(E) 0.423 270

Chapter 7 Solution: µTx (x) = µTx

⇤ (x)

+ µZ = 0.03 + 0.02 = 0.05

µTy (y) = µTy

⇤ (y)

+ µZ = 0.05 + 0.02 = 0.07

⇤ (y)

+ µZ = 0.03 + 0.05 + 0.02 = 0.10

µxy = µTx

⇤ (x)

+ µTy

T (x)

Ax =

µx

T (x) µx

+

=

0.05 = 0.4545 0.11

=

0.07 = 0.5385 0.13

=

0.1 = 0.6250 0.16

T (y)

Ay =

µy

T (y) µy

Axy = Axy = Ax + Ay

+

µxy µxy +

Axy = 0.4545 + 0.5385 +

0.6250 = 0.3680

Key: D

3. You have calculated the actuarial present value of a last-survivor whole life insurance of 1 on (x) and (y). You assumed: (i) The death benefit is payable at the moment of death. (ii) The future lifetimes of (x) and (y) are independent, and each life has a constant force of mortality with µ = 0.06. (iii)

= 0.05

Your supervisor points out that these are not independent future lifetimes. Each mortality assumption is correct, but each includes a common shock component with constant force 0.02. Calculate the increase in the actuarial present value over what you originally calculated. (A) 0.020

(B) 0.039

(C) 0.093

(D) 0.109

(E) 0.163

SOLUTION: Original calculation (assuming independence): µx = 0.06,

µy = 0.06

µxy = 0.06 + 0.06 = 0.12

Arch MLC, Spring 2009

Ax =

µx µx +

=

0.06 = 0.54545 0.06 + 0.05

Ay =

µy µy +

=

0.06 = 0.54545 0.06 + 0.05

c Yufeng Guo

271

Chapter 7 A xy = A xy = A x + A y

µxy µxy +

=

0.12 = 0.70588 0.12 + 0.05

A xy = 0.54545 + 0.54545

0.70588 = 0.38502

Revised calculation (common shock model): µx = 0.06, µTx ⇤(x) = 0.04 µy = 0.06, µTy ⇤(y) = 0.04 µxy = µTx ⇤(x) + µTy ⇤(y) + µZ = 0.04 + 0.04 + 0.02 = 0.10 Ax =

µx µx +

=

0.06 = 0.54545 0.06 + 0.05

Ay =

µy µy +

=

0.06 = 0.54545 0.06 + 0.05

A xy =

µxy µxy +

A xy = A x + A y

=

0.10 = 0.66667 0.10 + 0.05

A xy = 0.54545 + 0.54545

Di↵erence = 0.42423

0.66667 = 0.42423

0.38502 = 0.03921

Key: B

4. For independent lives (x) and (y): (i) qx = 0.05 (ii) qy = 0.10 (iii) Deaths are uniformly distributed over each year of age. Calculate

0.75 qxy .

(A) 0.1088

(B) 0.1097

(C) 0.1106

(D) 0.1116

(D) 0.1125

Solution: 0.75 px 0.75 py 0.75 qxy

Arch MLC, Spring 2009

=1

=1

(0.75)(0.05) = 0.9625

=1

(0.75)(0.10) = 0.925

0.75 pxy

=1

(0.9625)(0.925) = 0.1097

c Yufeng Guo

Key: B

272

Chapter 7 5. For independent lives (50) and (60): µ(x) =

1 , 0  x < 100 100 x

Calculate e50:60 . (A) 30

(B) 31

(C) 32

(D) 33

(E) 34

Solution: t px

= exp



Z t 0

ds 100 x

0

0

e50:60

"

t2 2

#50

= 25

"

t2 2

#40

= 20

1 dt = 50t 50 50 t

1 dt = 40t 40 40 t

Z 40 50 0

i

s)|t0 =

x

e50:60 = e50 + e60

Z 40 40

e60 =

e50:60 =

s

Z 50 50

e50 =

h

= exp ln(100

t 40 t · dt = 50 40

1 = 2000t 2000

Z 40 0

e50:60 = 25 + 20

0

0

1 ⇣ 2000 2000

t3 45t + |40 3 0 2

!

100 x t 100 x

⌘

90t + t2 dt

= 14.67

14.67 = 30.33

Key: A

6. A continuous two-life annuity pays: • 100 while both (30) and (40) are alive;

• 70 while (30) is alive but (40) is dead; and • 50 while (40) is alive but (30) is dead.

The actuarial present value of this annuity is 1180. Continuous single life annuities paying 100 per year are available for (30) and (40) with actuarial present values of 1200 and 1000, respectively. Calculate the actuarial present value of a two-life continuous annuity that pays 100 while at least one of them is alive. (A) 1400

(B) 1500

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(C) 1600

(D) 1700

c Yufeng Guo

(E) 1800 273

Chapter 7 SOLUTION: 1180 = 70a30 + 50a40 1180 = (70)(12) + (50)(10) a30:40 = a30 + a40

20a30:40

20a30:40

) a30:40 = 8

a30:40 = 12 + 10

8 = 14

100a30:40 = 1400 Key: A

7. (x) and (y) are two lives with identical expected mortality. • Px = Py = 0.1

• Pxy = 0.06, where Pxy is the annual benefit premium for a fully discrete insurance of 1 on (xy). • d = 0.06 Calculate the premium Pxy , the annual benefit premium for a full discrete insurance of 1 on (xy). (A) 0.14

(B) 0.16

(C) 0.18

(D) 0.20

(E) 0.22

SOLUTION: Ps =

1 a ¨s

d, where s can stand for any of the statuses under consideration. a ¨s =

1 Ps + d

)a ¨x = a ¨y =

a ¨xy =

1 = 6.25 0.1 + 0.06

1 = 8.333 0.06 + 0.06

a ¨xy + a ¨xy = a ¨x + a ¨y a ¨xy = 6.25 + 6.25 Pxy =

1 4.167

8.333 = 4.167 0.06 = 0.18

Key C

Arch MLC, Spring 2009

c Yufeng Guo

274

Chapter 7 8. For a select-and-ultimate table with a 2-year select period: x 48 49 50 51

p[x] 0.9865 0.9858 0.9849 0.9838

px+2 x + 2 0.9713 50 0.9698 51 0.9682 52 0.9664 53

p[x]+1 0.9841 0.9831 0.9819 0.9803

Keith and Clive are independent lives, both age 50. Keith was selected at age 45 and Clive was selected at age 50. Calculate the probability that exactly one will be alive at the end of three years. (A) Less than 0.115 (B) At least 0.115, but less than 0.125 (C) At least 0.125, but less than 0.135 (D) At least 0.135, but less than 0.145 (E) At least 0.145

SOLUTION: Prob{only 1 survives} = 1 - Prob{both survive} - Prob{neither survives} =1 =1

3 p50

⇥ 3 p[50]

(1

3 p50 )

1

3 p[50]

⌘

(0.9713)(0.9698)(0.9682) (0.9849)(0.9819)(0.9682) |

{z

=0.91202

(1

}|

0.912012)(1

= 0.140461

Arch MLC, Spring 2009

⇣

c Yufeng Guo

{z

0.936320

0.93632)

}

Key: D

275

Chapter 7 9. For two independent lives now age 30 and 34, you are given: x 30 31 32 33 34 35 36 37

qx 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

Calculate the probability that the last death of these two lives will occur during the 3rd year from now (i.e. 2| q30:34 ) (A) 0.01

(B) 0.03

(C) 0.14

(D) 0.18

(D) 0.24

SOLUTION: 2| q30:34

= 2 p30:34

3 p30:34

Thanks to independence the following two expressions are true: 2 p30:34

h

i h

i

= 2 p30 · 2 p34 = (0.9)(0.8) · (0.5)(0.4) = 0.144 h

i h

i

3 p30:34

= 3 p30 · 3 p34 = (0.72)(0.7) · (0.2)(0.3) = 0.03024

2 p30:34

= 2 p30 + 2 p34

Now 3 p30:34

= 3 p30 + 3 p34

2 p30:34

3 p30:34

2| q30:34

= 0.776

= 0.72 + 0.2

= 0.504 + 0.06

0.144 = 0.776 0.03024 = 0.53376

0.53376 = 0.24224

10. You are given: (i) Mortality follows DeMoivre’s law with ! = 105. (ii) (45) and (65) have independent future lifetimes. Calculate e45:65 (A) 33

(B) 34

Arch MLC, Spring 2009

(C) 35

(D) 36

(E) 37

c Yufeng Guo

276

Chapter 7 SOLUTION: e45:65 =e45 + e65 Under DeMoivre’s Law, ex =

! x 2 ,

so we have e45 = 30,

e45:65 =

Z 40 0

t p45:65 dt =

=

Z 40 0

Z 40 "

e45:65

e65 = 20

t p45 · t p65 dt =

1

0

#

Z 40 ✓

1

0

t 60

◆✓

1

t 40

◆

dt

t t2 + dt = 15.56 24 2400

)e45:65 = 30 + 20

15.56 = 34.4

11. For independent lives (35) and (45): (i) 5 p35 = 0.90 (ii) 5 p45 = 0.80 (iii) q40 = 0.03 (iv) q50 = 0.05 Calculate the probability that the last death of (35) and (45) occurs in the 6th year. (A) 0.0095 (B) 0.0105 (C) 0.0115 (D) 0.0125 (E) 0.0135

SOLUTION: =5| q35 +5| q45 5| q35:45 5 p35 q40 + 5 p45 q50 5 p35:45 q40:50 p40:50 ) 5 p35 q40 + 5 p45 q50 5 p35 ⇥5 p45 (1 p40 p50 ) 5 p35 q40 + 5 p45 q50 5 p35 ⇥5 p45 (1 = (0.9)(0.3) + (0.8)(0.05) (0.9)(0.8)[1 (0.97)(0.95)] = 0.01048. 5| q35:45

Alternatively, 6 p35 =5 p35 ⇥ p40 = (0.90)(1 Arch MLC, Spring 2009

0.03) = 0.873 c Yufeng Guo

277

Chapter 7 =5 p50 ⇥ p50 = (0.80)(1 0.05) = 0.76 q 5| 35:45 =5 p35:45 6 p35:45 = (5 p35 +5 p45 5 p35:45 ) (6 p35 +6 p45 6 p35:45 ) = (5 p35 +5 p45 5 p35 ⇥5 p45 ) (6 p35 +6 p45 6 p35 ⇥6 p45 ) 6 p45

= (0.90 + 0.80 0.90 ⇥ 0.80) = 0.98 0.96952 = 0.01048 Key: B

(0.873 + 0.76

0.873 ⇥ 0.76)

12. You are given: (i) T (x) and T (y) are not independent. (ii) qx+k = qy+k = 0.05, k = 0, 1, 2, . . . (iii) k pxy = 1.02k px · k py , k = 1, 2, 3, . . . Into which of the following ranges does ex:y , the curtate expectation of life of the last survivor status, fall? (A) ex:y  25.7

(B) 25.7 < ex:y  26.7

(C) 26.7 < ex:y  27.7

(D) 27.7 < ex:y  28.7 (E) 28.7 < ex:y

SOLUTION: ex = ey =

1 P

k=1

t px

= 0.95 + 0.952 + . . .

= 1 0.95 0.95 = 19 exy = pxy + 2 pxy + . . . = 1.02(0.95)(0.95) + 1.02(0.95)2 (0.95)2 + . . . ⇥ ⇤ 2 = 1.02 0.952 + 0.954 + . . . = 1.02(0.95) = 9.44152 1 0.952 exy = ex + ey exy = 28.56 Key: D

13. You are pricing a special 3-year annuity-due on two independent lives, both age 80. The annuity pays 30,000 if both persons are alive and 20,000 if only one person is alive. You are given: Arch MLC, Spring 2009

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Chapter 7 k k p80 1 0.91 (i) 2 0.82 3 0.72 (ii) i = 0.05 Calculate the actuarial present value of this annuity. (A) 78,300 (B) 80,400 (C) 82,500 (D) 84,700 (E) 86,800

SOLUTION: Calculate the probability that both are alive or both are dead. P(both alive) = k pxy = k px · k py P(both dead) = k qxy = k qx · k qy P(exactly one alive) = 1 k pxy k qxy Only have to do two year’s worth so have table Pr(both alive) 1 (0.91)(0.91) = 0.8281 (0.82)(0.82) = 0.6724

Pr(both dead) 0 (0.09)(0.09) = 0.0081 (0.18)(0.18) = 0.0324

AP V Annuity = 30, 000 80, 431

⇣

1 1.050

+

0.8281 1.051

+

0.6724 1.052

Pr(only one alive) 0 0.1638 0.2952

⌘

+ 20, 000

⇣

0 1.050

+

0.1638 1.051

+

0.2952 1.052

⌘

=

Alternatively, 0.6724 a ¨xy = 1 + 0.8281 1.05 + 1.052 = 2.3986 0.91 0.82 a ¨x = a ¨y = 1 + 1.05 + 1.05 2 = 2.6104 AP V = 20, 000¨ ax + 20, 000¨ ay 10, 000¨ axy (it pays 20,000 if x alive and 20,000 if y alive, but 10,000 less than that if both are alive) = (20, 000)(2.6104) + (20, 000)(2.6104) (10, 000)2.3986 = 80, 430 Other alternatives also work.Key: B

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Chapter 7 14. For a special fully continuous last survivor insurance of 1 on (x) and (y), you are given: (i) T (x) and T (y) are independent. (ii) µx (t) = 0.08, t > 0 (iii) µy (t) = 0.04, t > 0 (iv)

= 0.06

(v) ⇡ is the annual benefit premium payable until the first of (x) and (y) dies. Calculate ⇡. (A) 0.055 (B) 0.080 (C) 0.105 (D) 0.120 (E) 0.150

SOLUTION: µxy (t) = µx (t) + µy (t) = 0.08 + 0.04 = 0.12 A¯x = µx (t)/(µx (t) + ) = 0.5714 A¯y = µy (t)/(µy (t) + ) = 0.4 A¯xy = µxy (t)/(µxy (t) + ) = 0.6667 a ¯xy = 1/(µxy (t) + ) = 5.556 ¯ ¯ A¯xy A¯xy = 0.5714 + 0.4 0.6667 = 0.3047 ¯ = Ax + Ay P remium = 0.304762/5.556 = 0.0549 Key: A

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Chapter 7 15. The mortality of (x) and (y) follows a common shock model with components T ⇤ (x), T ⇤ (y) and Z. (i) T ⇤ (x), T ⇤ (y) and Z are independent and have exponential distributions with respective forces µ1 , µ2 and . (ii) The probability that (x) survives 1 year is 0.96. (iii) The probability that (y) survives 1 year is 0.97. (iv)

= 0.01

Calculate the probability that both (x) and (y) survive 5 years. (A) 0.65 (B) 0.67 (C) 0.70 (D) 0.72 (E) 0.74

SOLUTION: 0.96 = e (µ1 + ) µ1 + = ln(0.96) = 0.04082 µ1 = 0.04082 = 0.04082 0.01 = 0.03082. Similarly µ2 = ln(0.97) = 0.03046 0.01 = 0.02046 µxy = µ1 + µ2 + = 0.03082 + 0.02046 + 0.01 = 0.06128. (5)(0.06128) = e 0.3064 = 0.736. 5 pxy = e Key: E

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Chapter 7 16. For (80) and (84), whose future lifetimes are independent: x 80 81 82 83 84 85 86

px 0.50 0.40 0.60 0.25 0.20 0.15 0.10

Calculate the change in the value 2| q80:84 if p82 is decreased from 0.60 to 0.30. (A) 0.03

(B) 0.06

(C) 0.10

(D) 0.16

(E) 0.19

SOLUTION: =2| q80 +2| q84 2| q80:84 = 0.5 ⇥ 0.4 ⇥ (1 0.6) + 0.2 ⇥ 0.15 ⇥ (1

0.1) = 0.10136

Using new p82 value of 0.3 = 0.5 ⇥ 0.4 ⇥ (1 0.3) + 0.2 ⇥ 0.15 ⇥ (1

0.1) = 0.16118

2| q80:84

Change = 0.16118

0.10136 = 0.06

Alternatively, 2 p80 = 0.5 ⇥ 0.4 = 0.20 3 p80 =2 p80 ⇥ 0.6 = 0.12 2 p84 = 0.20 ⇥ 0.15 = 0.03 3 p84 =2 p84 ⇥ 0.10 = 0.003 2 p80:84 =2 p80 +2 p84 2 p80 · 2 p84 since independent = 0.20 + 0.03 (0.20)(0.03) = 0.224 3 p80:84 =3 p80 +3 p84 3 p80 · 3 p84 = 0.12 + 0.003 (0.12)(0.003) = 0.12264 2 q80:84 =2 p80:84 3 p80:84 = 0.224 0.12264 = 0.10136 Revised 3 p80 = 0.20 ⇥ 0.30 = 0.06 (0.06)(0.003) = 0.06282 3 p80:84 = 0.06 + 0.003 0.06282 = 0.16118 2| q80:84 = 0.224 change = 0.16118 0.10136 = 0.06. Key: B Arch MLC, Spring 2009

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Chapter 7

Problems from Pre-2000 SOA-CAS exams 1. Individual A is subject to the following law of mortality: µx = 0.04,

for all x

Calculate e35:40 , assuming (35) and (40) are subject to the same law of mortality as individual A. (A) 11.5

(B) 12.0

(C) 12.5

(D) 13.0

(E) 13.5

2. For a fully continuous last-survivor whole life insurance of 1 on (x) and (y), you are given: • T (x) and T (y) are independent. • µx (t) = µy (t) = 0.06, •

t>0

= 0.04

• Premiums are payable until the first death. Calculate the annual benefit premium. (A) 0.033

(B) 0.042

(C) 0.055

(D) 0.072

(E) 0.120

3. (40) and (50) are independent lives. Mortality follows De Moivre’s law with ! = 100. Calculate the probability that at least one of (40) and (50) will die within 10 years. (A) 1/30

(B) 3/10

(C) 1/3

(D) 2/3

(E) 7/10

4. (Same setup as previous problem:) (40) and (50) are independent lives. Mortality follows De Moivre’s law with ! = 100. Calculate the probability that the second death occurs between times t = 10 and t = 20. (A) 1/10

(B) 1/5

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(C) 4/15

(D) 1/3

(E) 2/5

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Chapter 7 Use the following information for the next 4 questions: You are given: • (30) and (50) are independent lives, each subject to constant force of mortality with µ = 0.05. •

= 0.03

5. Calculate (A) 0.155

10 q30:50 .

(B) 0.368

(C) 0.424

(D) 0.632

(E) 0.845

6. Calculate e30:50 . (A) 10

(B) 20

(C) 30

(D) 40

(E) 50

7. Calculate Var[T (30 : 50)]. (A) 50

(B) 100 ⇥

(C) 150

(D) 200 ⇤

(E) 400

8. Calculate Cov T (30 : 50), T 30 : 50 . (A) 10

(B) 25

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(C) 50

(D) 100

(E) 200

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Chapter 7

Solutions to Pre-2000 Exam Questions: Chapter 9 1. Key: C Since the lives are independent, the joint-life force of mortality is µ35:40 = µ35 + µ40 = 0.08 So the joint-life, force of mortality is also constant. e35:40 =

1 = 12.5 0.08

2. Key: D Constant Force of Mortality: Ax = Ay = µxy = µx + µy

µ µ+

A xy = 0.6 + 0.6

1 µ+µ+

P =

0.06 = 0.6 0.1 0.12 = 0.75 0.12 + 0.04

) A xy =

A xy = A x + A y axy =

=

=

0.75 = 0.45

1 = 6.25 0.16

A xy 0.45 = = 0.072 axy 6.25

3. Key: C 10 q40:50

=1

=1

10 p40:50

10 p40

Since this is De Moivre’s law: 10 p40

=

5 6

and

) 10 q40:50 = 1

✓

· 10 p50

10 p50

=

4 5

◆

=

1 3

5 4 · 6 5

4. Key: A 10|10 q40:50

= 10|10 q40 + 10|10 q50

= 10 p40 · 10 q50 + 10 p50 · 10 q60 =

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5 6

✓ ◆

1 5

+

4 5

✓ ◆

1 4

5 6

✓ ◆

4 5

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10|10 q40:50

10 p40:50

4 5

· 10 q50:60

✓ ◆

3 4

=

1 10

285

Chapter 7 5. Key: A 10 q30:50

10(0.05)

)

10 p30:50

= 10 p30 + 10 p50

10 p30:50

e

=1

+e

10(0.05)

10 q30:50

=1

e

10 p30:50

20(0.05)

= 0.845

0.845 = 0.155

6. Key: C Constant Force of mortality: µxy = µx + µy = 0.1 exy =

1 = 10 µxy

ex =ey =

1 = 20 0.05

exy =ex + ey

exy = 30

7. Key: B µxy = 0.1, a constant. So, Var[T ] =

8. Key: D

⇣

Cov = ex

Arch MLC, Spring 2009

exy

⌘⇣

ey

1 2

(µxy )

=

⌘

1 = 100 0.01

exy = (20

c Yufeng Guo

10)(20

10) = 100

286

Chapter 8

ACTUARIAL MATHEMATICS: CHAPTER 10 MULTIPLE DECREMENT MODELS • Option A reference: Actuarial Mathematics Chapter 10 • Option B reference: Models for Quantifying Risk Chapter 15 For most of this text, we’ve encountered models with one life and one decrement (death). We also looked at models with multiple lives and one decrement in Chapter 9. Now we are back to models with one life, but these lives face multiple decrements. For example, consider a group insurance plan provided by an employer. At time t = 0, there is some number of employees covered (the radix ! l0 ). However, over time, this number will decrease due to death, retirement, disability, or changing jobs (these are all di↵erent decrements). To accurately model costs of such plans, along with retirement plans or disability insurance, we need models that take into account both time of termination and cause of termination.

10.2 Two Random Variables • Option A reference: Actuarial Mathematics Chapter 10.2 • Option B reference: Models for Quantifying Risk Chapter 15.1 In the models we will be developing in this chapter, there are two random variables. You are already familiar with time of decrement, which is denoted by T (x). Now, we add a discrete random variable for cause of decrement, J(x). For example, if death is decrement 1 and withdrawal is decrement 2, then J(x) = 1 corresponds to decrement by death, and J(x) = 2 corresponds to decrement by withdrawal. (j)

There are a few new notational concepts in this chapter. For example, t qx denotes the (⌧ ) probability of decrement in the next t years due to cause j. Also, t qx denotes the probability of decrement due to all causes before time t. 287

Chapter 8 Suppose the only decrements to a retirement plan are (1) death and (2) retirement. (1) (2) For example, if q63 = 0.03 and q63 = 0.3 , then (⌧ )

(1)

(2)

q63 = q63 + q63 = 0.33. And

(⌧ )

(⌧ )

p63 = 1

q63 = 0.67,

(⌧ )

where p63 is the probability of surviving all causes of decrement for 1 year. It is always true that (⌧ ) (⌧ ) t px = 1 t qx .

(⌧ )

(j)

The symbol µx represents the force of decrement due to all causes combined, whereas µx represents the force of decrement due only to decrement j. As you might guess, m X

=

) µ(⌧ x

µ(j) x ,

j=1

where m is the number of di↵erent decrements trying to get at (x). Just as for single decrement functions, it is also true that ) µ(⌧ x =

d dt

⇣

(⌧ )

t px

(⌧ ) t px

and (⌧ ) t px = e

⌘

Rt 0

=

d dt

(⌧ )

⇣

(⌧ )

t qx

(⌧ ) t px

µx (s) ds

⌘

,

.

EXAMPLE: (1)

(2)

(⌧ )

If µx (t) = 0.03 and µx (t) = 0.1t for all t, find a formula for t px in terms of t. SOLUTION: (⌧ ) t px = e

Rt 0

(⌧ )

µx (s) ds

Rt

=e

0

(0.1s+0.03) ds

=e

(0.05)t2 (0.03)t

.

}

Compare the following two integral expressions: (j) t qx =

(⌧ ) t qx =

Z t 0

Z t 0

(⌧ ) (j) s px µx (s) ds,

(⌧ ) (⌧ ) s px µx (s) ds.

The first integral adds up over all times s, the probability of surviving to s times the probability of ‘decrementing’ in the next instant due to cause j. The second integral is the same Arch MLC, Spring 2009

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Chapter 8 (⌧ )

except that all decrements are taken into account. Note that in both integrals, s px gives the probability of surviving to time s, this is because no matter how you ‘decrement’ at time s, you must survive all decrements up to time s before decrementing at time s. These integrals could be used to show the very intuitive fact that (⌧ ) t qx =

m X

(j) t qx .

j=1

Probability density functions: Since there are several decrements now, the p.d.f. has two variables. fT,J (t, j) is the probability of dying at time t from cause j. The discussion above means that ) (j) fT,J (t, j) = t p(⌧ x µx (t).

A nice summary of the p.d.f’s you need to have in your bag of tricks is given below. Questions on these are not plentiful on past tests but they do occur. It is important that you keep track of the name of each one, since a question may ask for one by name without giving the functional expression. (⌧ )

(j)

fT,J (t, j) = t px µx (t)

Joint PDF Marginal PDF of J Marginal PDF of T

(j)

fJ (j) = 1 qx (⌧ )

(⌧ )

fT (t) = t px µx (t).

Note that the Marginal PDF of J represents the probability of decrementing due to cause j at some in the future. Similarly, the Marginal PDF of T represents the probability of decrementing at time t due to any cause. There is one other important PDF to know. It is a conditional probability that represents the probability of decrementing due to cause j, given that you decrement at a specified time t. The symbol is fJ|T (j|t) (more nearly indecipherable notation!), and (j)

The Conditional PDF

fJ|T (j|t) =

µx (t) (⌧ )

µx (t)

.

EXAMPLE: You are given: (1)

2 25

(2)

7 25

• µx (t) = • µx (t) =

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Chapter 8 Find the p.d.f.’s for (A) the joint distribution, (B) the marginal distributions, and (C) the probability that (x) decrements due to cause 2 if decrement occurs at t = 1.

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Chapter 8 SOLUTION: (A) Joint Distribution: ) (j) fT,J (t, j) = t p(⌧ x µx (t)

We know that ) (1) (2) µ(⌧ x (t) = µx (t) + µx (t) =

Therefore,

Rt

) ) t p(⌧ x =e

fT,J (t, j) = e

(0.36)t

0

9/25 dt

µ(j) x (t) =

(B) Marginal Distributions: fJ (j) = 1 qx(j) = =

=e

9 . 25

(0.36)t

8 (0.36)t · 2 > < e 25 > : e (0.36)t · 7 25

Z 1 0

for j = 1 for j = 2

(⌧ ) (j) t px µ (t) dt

8 R1 (0.36)t 2 dt > < 0 e 25

for j = 1

> : R 1 e (0.36)t 7 dt for j = 2 0 25 8 2 > < 9 for j = 1

=

> : 7

for j = 2

9

Note that we could also have gotten these using fJ (j) =

Z 1 0

fT,J (t, j)dt.

Now for the marginal distribution of T : ) (⌧ ) fT (t) = t p(⌧ x µx (t) = e

(0.36)t

9 . 25

(C) Bypassing the ugly notation of the conditional distribution, this probability is given by (j) µx (t) , (⌧ ) µx (t) where j = 2 and t = 1. So the probability we want is 7 25 9 25

=

7 9

The conditional distribution turned out to be a constant since both µ’s were constant, but that will not always be the case. } Arch MLC, Spring 2009

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Chapter 8 For a (much) more complex example of the same ideas, see Example 10.2.1 in the text. EXAMPLE: In a 2-decrement environment, you are given: (1)

• µx (t) = 0.3t, t •

(⌧ ) 2 px

0

= 0.3

Find e

R2

(⌧ ) t px

=e

0

SOLUTION: For t = 2, we have (⌧ ) 2 px

=e

R2 0

(⌧ )

µx (t) dt

=e ) )e

R 2h 0

(⌧ ) 2 px

R2 0

(1)

(2)

µx (t) dt

Rt 0

.

(⌧ )

µx (s) ds

i

(2)

µx (t)+µx (t) dt

=e (2)

(0.6)

µx (t) dt

=

·e

=e

R2 0

(2)

R2 0

(1)

µx (t) dt

·e

R2 0

(2)

µx (t) dt

µx (t) dt

0.3 = 0.547. 0.549

}

10.3 Random Survivorship Group 10.4 Deterministic Survivorship Group • Option A reference: Actuarial Mathematics Chapter 10.3, 10.4 • Option B reference: Models for Quantifying Risk Chapter 15.1.1 Sections 10.3 and 10.4 are really just two di↵erent ways at looking at the same thing so we will consider them together. Suppose we have a group of people all at some age a at a given point in time. The number of (⌧ ) people in this group at time 0 is denoted by la . (We’ll get to what the ⌧ represents here.) We assume that each member of the group has a joint pdf for time until decrement and cause of decrement given by ) (j) fT,J (t, j) = t p(⌧ a µa (t) for t

0, j = 1, 2, . . . , m.

(j)

Let n dx represent the expected number of lives to leave the group during the age-interval (x, x + n) due to cause j. This is looking at the number of people that die in the next n years after reaching age x as a random variable. If we were looking at things deterministically, we would be saying that we know ahead of time that the number of people that will decrement Arch MLC, Spring 2009

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Chapter 8 (j)

in the next n years due to cause j will definitely be n dx . In that case we don’t have to talk about it as an expected value. For the purposes of the exam, this distinction is not very important. Note that even though we are interested in who leaves the group between ages x and (x + n), (2) we are calculating this number while everyone is currently age a. So 3 dx refers to the number (⌧ ) of people among the original la that live to age x and then decrement due to cause 2 before reaching age (x + 3). EXAMPLE: (j)

(⌧ )

(j)

1. Find an expression for n dx in terms of la , p, and n qx . (j)

(⌧ )

2. Find an integral expression for n dx in terms of la

(⌧ )

(j)

, t pa , and µa (t).

SOLUTION: 1. This one is not too bad. The number that decrement from cause j among (⌧ ) the la people that start out should just be (j) n dx

= la(⌧ ) · x

(⌧ ) (j) a pa n q x

2. We want to integrate over all times t between t = (x a) and t = (x a + n). The probability of surviving all decrements for t years and then decrementing in the next instant due to cause j: (j) n dx

=

la(⌧ )

Z x a+n x a

(⌧ ) (j) t pa µa (t) dt.

}

Recall that ⌧ indicates all causes, so we have (⌧ ) n dx =

m X

(j) n dx .

j=1 (⌧ )

Similarly, the la symbol represents the set of lives in the group that will eventually decrement (j) due to some cause (i.e. everyone!). If we write la , we are referring to the people in the group who eventually decrement due to cause j. Clearly, la(⌧ ) =

m X

la(j) .

j=1

For example, say that we begin with 100 people. And suppose that we expect that over time 40 will decrement due to retirement, 40 will leave to pursue other jobs, 10 will leave due to disability, and 10 will leave due to death. Our total is the sum of these various “decrement pieces.” Algebraically, la(⌧ ) = la(1) + la(2) + la(3) + la(4) = 40 + 40 + 10 + 10. Arch MLC, Spring 2009

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Chapter 8 We now have the formulas we need to construct a multiple decrement table. We saw the idea of a single decrement (death) mortality table in Chapter 3. This table is simply an extension to allow for multiple decrements. It allows the user to track the number of terminations due to specific causes. Read Example 10.3.1 in the text and then try to complete the multiple decrement mortality (⌧ ) table begun below (the solution is on the next page.) Assume l40 = 100. (1)

(2)

x qx qx 40 0.01 0.05

(⌧ )

px

(⌧ )

lx

(1)

dx

(2)

dx

41 0.02 0.06 42 0.03 0.07 43 0.04 0.08

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Chapter 8 Completed Multiple Decrement Table x qx qx 40 0.01 0.05

(1)

(2)

px 0.94

(⌧ )

lx 100.0

(⌧ )

dx 1.00

(1)

dx 5.00

(2)

41 0.02 0.06

0.92

94.0

1.88

5.64

42 0.03 0.07

0.90

86.5

2.59

6.06

43 0.04 0.08

0.88

77.9

3.12

6.23

EXAMPLE: Using the table above, determine each of the following. (1) 2 q42

(⌧ ) 2 q42

(2) 1|2 q40

SOLUTION: The first one is the number of people who decrement due to cause 1 at ages 42 and 43 divided by the number alive at age 42. (1)

(1)

2 q42 =

(1)

d42 + d43 5.71 = = 0.066 l42 86.5

We could also have done this from earlier principles using p’s and q’s: (1) 2 q42

Next –

(⌧ )

(⌧ ) (1)

(⌧ )

d42 + d43 8.65 + 9.35 = = 0.208. l42 86.5

(⌧ )

2 q42 =

(2)

(1)

= q42 + p42 q43 .

(2)

1|2 q40 =

(2)

d41 + d42 5.64 + 6.06 = = 0.117. l40 100

}

EXAMPLE: Using the table above, determine each of the following. (⌧ ) 2 p41

SOLUTION: (⌧ ) 2 p41 (⌧ ) 2 d41

(⌧ )

=

(⌧ ) 2 d41

l43 77.9 = = 0.829, l41 94

(⌧ )

= d41 + d42 = (1.88 + 5.64) + (2.59 + 6.06) = 16.17.

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} 295

Chapter 8 EXAMPLE: (j)

(j)

(⌧ )

Write qx in terms of dx and lx . SOLUTION: (j)

qx(j) =

dx

(⌧ )

lx

}

.

10.5 Associated Single Decrement Tables • Option A reference: Actuarial Mathematics Chapter 10.5 • Option B reference: Models for Quantifying Risk Chapter 15.4.1 In Section 10.5 the text introduces an idea that has several names – “net probabilities of decrement”, “independent rate of decrement”, and the SOA-approved term absolute rate 0(j) (j) of decrement. The idea also has a new symbol t qx as opposed to the t qx we’ve seen 0(j) previously. The symbol t qx represents the probability of decrement from cause j only, ignoring the other possible causes. In other words, is the probability that (x) would decrement from cause j in the next t years if there were no other causes of decrement. A not-so-obvious, but important fact is the following: (j) t qx

 t qx0(j) .

(Understand why this is true!!)

This is true because if the other decrements are not around for the next t years, the probability that decrement j will get you is higher. For example suppose that every year, both death and retirement have a 40 percent chance of acting on (x). If you take away the possibility of retirement, the probability that (x) will decrement by dying is much higher – we lose all those cases in which (x) might have retired before dying. Understanding this concept should help your intuition when trying to answer multiple decrement types of questions. 0(j)

Since t qx takes into account the force of decrement for cause j only (forsaking all others), 0(j) it then follows that t px also ignores all other possible causes of decrement: 1

0(j) 0(j) t qx = t px = exp



Z t 0

µ(j) x (s)ds .

EXAMPLE: Jake Jake has an old car, age x, that faces two forces of decrement (as far as Jake is concerned). They are (1) breakdown and (2) sale. Both forces of decrement are (1) (2) constants: µx = 0.03 and µx = 0.1. 0 (1)

Find qx

(1)

and qx .

Arch MLC, Spring 2009

c Yufeng Guo

296

Chapter 8 SOLUTION: 0

qx(1) =

Z 1 0

0 (1) (1) t px µx (t) dt

Notice that the second force of decrement doesn’t show up at all in the calculation 0 (1) 0 (1) of qx . It doesn’t exist if Jake agrees not to sell his car. We need t px , so 0 (1)

t px 0

qx(1) =

Z 1

(1)

µx (s) ds

=e

(0.03)dt = 1

e

=e

0.03t

e

Rt 0

0.03t

.

0.03

= 0.0296.

0

0 (1)

(We could also have used qx

=1

qx(1) =

0 (1)

px .) Z 1 0

(⌧ ) (1) t px µx (t) dt

Now there are two decrements so we have to survive all decrements up to the time the first decrements hit. )

qx(1)

0 (1)

=

Z 1

e

(0.13)t

(0.03) dt =

0

0.03 ⇣ 1 0.13

e

0.13

⌘

= 0.0281.

(1)

Note that qx > qx . If Jake agrees not to sell his car next year, there is a greater probability that he will experience a breakdown. }

10.5.1 Basic Relationships • Option A reference: Actuarial Mathematics Chapter 10.5.1 • Option B reference: Models for Quantifying Risk Chapter 15.1-15.4 For m causes of decrement, (⌧ ) t px = exp

⇢

Z th 0

i

(2) (m) µ(1) x (s) + µx (s) + · · · + µx (s) ds .

So, using a little algebra for exponential functions, we can see that (⌧ ) t px =

m Y

0(i) t px

i=1 0(j)

(⌧ ) t px

It follows from this relation that t px

0(j)

and we have already seen that qx

(j)

qx .

EXAMPLE: (I hope by now you are working all of the exercises before looking at the solutions!) You are given: (1)

(2)

0 (1)

• qx = 0.2, qx = 0.1, and qx

= 0.21.

0 (2)

Find qx . Arch MLC, Spring 2009

c Yufeng Guo

297

Chapter 8 SOLUTION: Often with associated decrement problems, it is easiest to work through the p’s. ) p(⌧ x =1 0

qx(2) = 0.7

qx(1) ⇣

0

) (1) (2) p(⌧ = 1 x = p x px

0

qx(1) ⇣

0

1

0

) 0.7 = (0.79) 1 0

⌘⇣

qx(2)

) qx(2) = 0.114

qx(2) ⌘

⌘

}

10.5.4 Uniform Distribution Assumption for Multiple Decrements • Option A reference: Actuarial Mathematics Chapter 10.5.4 • Option B reference: Models for Quantifying Risk Chapter 15.4.1 Assuming a Uniform Distribution of Decrement over the interval (x, x + 1) implies that (j) t qx

= t · qx(j) and t qx(⌧ ) = t · qx(⌧ ) .

Also, at any time t between 0 and 1, (⌧ ) (j) t px µx (t)

= qx(j) .

(U DD)

This tells us that the pdf over each year is a constant. Finally, it also gives us a handy (j) formula for µx (t): (j) (j) qx qx µ(j) (t) = = . (U DD) x (⌧ ) (⌧ ) p 1 t q x x t EXAMPLE: You are given: (1)

(2)

• qx = 0.05 and qx = 0.08,

• All decrements are uniform in the multiple decrement context over the interval (x, x + 1). (1)

(1)

Find µx (0.25) and µx (0.75). SOLUTION: (1)

µ(1) x (0.25) =

qx 1

(⌧ ) (0.25) qx

=

0.05 = 0.052 0.97

=

0.05 = 0.055 0.90

(1)

µ(1) x (0.75) Arch MLC, Spring 2009

=

qx 1

(⌧ ) (0.75) qx

c Yufeng Guo

298

Chapter 8 Note that if decrements are uniformly distributed, then the force of decrement must increase over time. This could help you see that some answers you get are reasonable compared to each other. } Letting 0  s  1, we can derive that 0(j) s px

=

(⌧ ) s px

=

⇣

) p(⌧ x

When s = 1, this tells us that p0(j) x

(j)

⇣

x ⌘ q(⌧ ) qx

(U DD)

.

(j)

x ⌘ q(⌧ ) qx

(U DD)

.

This formula is the same as the one in the constant force of decrement case if s = 1. I.e. they are the same when you are only considering full-year decrements. Once again, note that (j) (⌧ ) this formula cannot be used in the event that px = 0 or px = 0. More about this in Section 10.6. IMPORTANT NOTE: These messy formulas result when we assume that the decrements (j) are uniform in the multiple decrement context, i.e., when we assume the t qx are uniformly distributed. A more common assumption that we will see in Section 10.6 is to assume that the individual decrements are uniformly distributed in the associated single decre0 (j) ment table, i.e., we assume the t qx are uniformly distributed. This leads to relatively simple formulas. This is a distinction that you have to keep straight when reading questions of this type or you may waste lots of time and still get the wrong answer. END OF IMPORTANT NOTE

EXAMPLE: Jake 2 Jake has the same old car, age x, that faces two forces of decrement. They are (1) breakdown and (2) sale. But now Jake believes that the decrements are uniform (1) over the interval (x, x + 1) in the multiple decrement context, with qx = 0.2 and (2) qx = 0.1. Find the probability that the car will break down before half a year has passed if Jake agrees ahead of time NOT to sell the car next year. SOLUTION: 0 (1)

(1)

We are looking for qx , and this one is trickier than it looks. Although t qx 0 (1) uniformly distributed on (x, x + 1), t qx is not. We can use 0(j) s px

=

With s = 0.5 and j = 1, this becomes 0(1) 0.5 px

Arch MLC, Spring 2009

=

⇣

⇣

(j)

(⌧ ) s px

(1)

(⌧ ) 0.5 px

x ⌘ q(⌧ ) qx

is

x ⌘ q(⌧ )

⇣

qx

= 1

c Yufeng Guo

.

(0.5)qx(⌧ )

⌘2 3

= 299

Chapter 8 = (0.85)(0.66667) = 0.897 0

) 0.5 qx(1) = 1

0.897 = 0.103.

}

10.6 Construction of a Multiple Decrement Table • Option A reference: Actuarial Mathematics Chapter 10 • Option B reference: Models for Quantifying Risk Chapter 15.1, 15.2 We have already done some Decrement Table development earlier in this chapter. The di↵erence here is that we are including and using ideas from the Associated Single Decrement Table (i.e. the “primes”). It turns out that life is often a lot easier if we are told the “primes” are uniform, than if we are told the “non-primes” are uniform.

CASE I : Two decrements that are uniformly distributed in the associated single decrement table 0 (j)

0 (j)

In this case t qx = tqx , the distribution of each decrement would be uniform if the other decrement were removed. When this is true, there are two key formulas: qx(1)

0 (1)

= qx

✓

1

1 0 (2) q 2 x

◆

qx(2)

0 (2)

= qx

✓

1

1 0 (1) q 2 x

◆

Read these equations carefully and think about them! One way to remember them is to keep 0 (1) (1) in mind that qx should be a little less than qx .

CASE II : Three decrements that are uniformly distributed in the associated single decrement table No need to memorize the following difficult formula. You can always derive it from the scratch: ✓ ◆ 1 0 (2) 1 0 (3) 1 0 (2) 0 (3) 0 qx(1) = qx(1) 1 qx qx + qx qx 2 2 3 (2)

(3)

Similar formulas hold for qx and qx .

CASE III : Multiple decrements – some are uniformly distributed in the associated single decrement table and some are not. These have to be done on a case-by-case basis. We’ll will see them in the examples. EXAMPLE: Jake 3 Jake has the same old car, age x, that faces two forces of decrement. They are (1) breakdown and (2) sale. But now Jake believes that the decrements are uniform over the interval (x, x+1) (1) (2) in the associated single decrement table. He also assumes that qx = 0.2 and qx = 0.1 Arch MLC, Spring 2009

c Yufeng Guo

300

Chapter 8 Find the probability that the car will break down before a year has passed if Jake were to agree ahead of time NOT to sell the car next year.

Arch MLC, Spring 2009

c Yufeng Guo

301

Chapter 8 SOLUTION: 0 (1) We are looking for qx . We know that 0 (1)

0.2 = qx

0 (2)

0.1 = qx

⇣ ⇣

⌘

0 1 (2) 2 qx ⌘ 0 1 (1) 2 qx

1 1

0 (2)

Now using the second equation to solve for qx , and substituting into the first equation, we get ! 0.05 0 (1) 0.2 = qx 1 0 (1) 1 (0.5)qx Which leads to the following quadratic equation: 0.2

0

0

⇣

(0.1)qx(1) = qx(1) 0.95

or

⇣

0

(0.5) qx(1)

⌘2

0

0.5qx(1)

⌘

0

(1.05)qx(1) + 0.2 = 0

Using the quadratic formula that we learned in (junior?) high school, we have 0

qx(1) = 0.212 0 (1)

We can tell there is a good chance we did things right because the qx (1) bigger than qx .

that we got is a little }

EXAMPLE: Jake 4 Jake has the same old car, age x, that faces two forces of decrement but the model has changed. The decrements are still (1) breakdown and (2) sale. But now Jake believes that the breakdown decrement is uniform over the interval (x, x + 1) in the associated single decrement table. The sale decrement can only take place at year-end. You are given: 0 (1)

= 0.25,

0 (2)

= 0.5.

• qx • qx

(1)

(2)

(⌧ )

Find qx , qx , and qx . SOLUTION: To do this one we should note that during the year, while the first decrement is acting on the car, the second decrement is in fact absent. After all, if Jake sells, he won’t sell until year-end. Therefore 0 qx(1) = qx(1) = 0.25. For the second decrement, we note that Jake can only sell the car if it is still around after a whole year of being ‘hit’ by the first decrement. So qx(2) = (1 Arch MLC, Spring 2009

0

0

qx(1) )qx(2) = (0.75)(0.5) = 0.375 c Yufeng Guo

302

Chapter 8 0 (1)

Note that there was no 12 before qx in the expression above because the first decrement operated for a whole year before the sale decrement acted on the car. (⌧ )

Finally, qx = 0.375 + 0.25 = 0.625. We could have gotten this quantity directly from the given information using 0

⇣

0

0

) (1) (2) p(⌧ = 1 x = px px (⌧ )

Then qx = 1

qx(1)

⌘⇣

⌘

0

qx(2) = 0.375

0.375 = 0.625.

}

EXAMPLE: Jake 5 Jake has the same old car, age x, that faces two forces of decrement but the model has changed. The decrements are still (1) breakdown and (2) sale. But now Jake believes that the breakdown decrement is uniform over the interval (x, x + 1) in the associated single decrement table. The sale decrement can only take place at mid-year. You are given: 0 (1)

= 0.25,

0 (2)

= 0.5.

• qx • qx

(1)

(2)

Find qx and qx . SOLUTION: If you can do this one confidently, you are probably ready for what the SOA can throw at you! To do this one right, we really need to break the year into 3 parts: 1st half, mid-year, 2nd-half. During the 1st half of the year, the only decrement acting on the car is (1). So the probability of the car decrementing due to cause (1) during the 1st half of the year is 1 0 (1) q = 0.125. 2 x We used the fact that the first decrement is uniform over (x, x + 1) so that 0 (1) t qx

0

= tqx(1) .

At mid-year, in order for the car to decrement due to cause (2) it must have survived the 1st half of the year, so the probability of decrement due to cause (2) is ✓

qx(2) = 1

◆

1 0 (1) 0 (2) q qx = (0.875)(0.5) = 0.4375. 2 x

Now, during the second half of the year, the car can decrement due to cause (1) only if it survived both the 1st half of the year and survived mid-year. So the probability of the car decrementing due to cause (1) during the 2nd half of the year is (1 Arch MLC, Spring 2009

0.125

0

0.4375) @

1

1

0 1 (1) 2 qx A 1 0 (1) q x 2

c Yufeng Guo

= 0.0625 303

Chapter 8 (1)

(2)

Therefore qx = 0.125 + 0.0625 = 0.1875 and qx = 0.4375. We can check our work by (⌧ ) (1) (2) (⌧ ) checking that px =⌘ 1 qx qx comes very close (equal except for rounding) to px = ⇣ 0 (1) ⌘ ⇣ 0 (2) 1 qx 1 qx . On the exam you won’t have to check since you will have one of the 5 answer choices! } EXAMPLE: Jake 6 – Last Jake Problem Jake has decided to refine his model by adding a 3rd decrement, collision. It is assumed that decrements (1) and (3), breakdown and collision, are both uniform over (x, x + 1) in the associated single decrement table. It is also assumed that the car may be sold only at year-end. You are given: 0 (1)

= 0.2,

0 (2)

= 0.5,

0 (3)

= 0.4 (Jake is a poor driver).

• qx • qx • qx

Find the probabilities of decrement for each of the three causes. SOLUTION: For the entire year until the last moment, decrements (1) and (3) are operating in the presence of each other and both are uniform in the associated single decrement table so qx(1) qx(3)

0 (1)

= qx

0 (3)

= qx

✓ ✓

◆

1

1 0 (3) q = (0.2)(0.8) = 0.16. 2 x

1

1 0 (1) q = (0.4)(0.9) = 0.36. 2 x

◆

Finally, decrement (2) operates only at year-end, so ⇣

qx(2) = 1

qx(1)

⌘

0

qx(3) qx(2) = (0.48)(0.5) = 0.24.

}

As a last comment to this section, let’s consider Examples 10.6.1 and 10.6.2 in the text. (1) (2) (3) The whole point to these examples is that in 10.6.1, qx , qx , qx were calculated assuming a constant force of mortality whereas in Example 10.6.2, it was assumed that the decrements were uniformly distributed in each year of age in the associated single-life decrement table. You should be able to duplicate the first couple of entries for either table. Finally, Example 10.6.3 gives a more complicated version of our examples above. Chapter 10 Suggested Problems: 1, 2, 4, 5, 7, 8, 10, 14, 17, 28, 29 (Solutions at archactuarial.com)

Arch MLC, Spring 2009

c Yufeng Guo

304

Chapter 8

CHAPTER 10 Formula Summary ) µ(⌧ x

=

m X

µ(j) x ,

j=1

Rt

(⌧ ) t px = e

(j) t qx = (⌧ ) = t qx

Z t 0

Z t 0

0

(⌧ )

µx (s) ds

(⌧ ) (j) s px µx (s) ds,

(⌧ ) (⌧ ) s px µx (s) ds =

m X

(j) t qx

j=1

(⌧ )

(j)

Joint PDF

fT,J (t, j) = t px µx (t)

Marginal PDF of J

fJ (j) = 1 qx

Marginal PDF of T

fT (t) = t px µx (t).

(j) (⌧ ) n dx = la · x

(j)

(⌧ )

(⌧ ) (j) = la(⌧ ) · a pa n q x (⌧ ) = n dx

Z x a+n x a

m X

(j) n dx

m X

la(j)

(⌧ )

(⌧ ) (j) t pa µa (t) dt

j=1

la(⌧ ) =

j=1 (j)

qx(j) = (j) t qx

1

Arch MLC, Spring 2009

0(j) t qx

=

0(j) t px

dx

(⌧ )

lx

 t qx0(j)

= exp



Z t

c Yufeng Guo

0

µ(j) x (s)ds

305

Chapter 8 For m causes of decrement, (⌧ ) t px = exp

⇢

Z th 0

i

(2) (m) µ(1) x (s) + µx (s) + · · · + µx (s) ds (⌧ ) t px

=

m Y

0(i) t px

i=1 0(j) t px

(⌧ ) t px

UDD in the Multiple Decrement Table over the interval (x, x + 1): (j) t qx

= t · qx(j) and t qx(⌧ ) = t · qx(⌧ ) (⌧ ) (j) t px µx (t)

= qx(j)

(j)

µ(j) x (t) =

0(j) s px

qx

(⌧ ) t px

=

⇣

(j)

=

qx

(⌧ )

1

t qx (j)

(⌧ ) s px

x ⌘ q(⌧ ) qx

(j)

(⌧ )

Note that this formula cannot be used in the event that px = 0 or px = 0. If two decrements are uniformly distributed in the associated single decrement table: qx(1)

0 (1)

= qx

✓

1

1 0 (2) q 2 x

◆

qx(2)

0 (2)

= qx

✓

1

1 0 (1) q 2 x

◆

Read these equations carefully and think about them! One way to remember them is to keep 0 (1) (1) in mind that qx should be a little less than qx . If three decrements are uniformly distributed in the associated single decrement table: qx(1)

0 (1)

= qx

(2)

✓

1

1 0 (2) q 2 x

1 0 (3) 1 0 (2) 0 (3) q + qx qx 2 x 3

◆

(3)

Similar formulas hold for qx and qx .

Arch MLC, Spring 2009

c Yufeng Guo

306

Chapter 8

Past SOA/CAS Exam Questions: 1. In a double decrement table: (⌧ )

(i) l30 = 1000,

(⌧ )

l32 = 472

0(1)

(ii) q30 = 0.100, (iii)

(1) 1| q30

0(2)

q30 = 0.300

= 0.075 (2)

Calculate q31 . (A) 0.11

(B) 0.13

Solution:

(C) 0.14

(⌧ )

(D) 0.15

(1)

(E) 0.17

(2)

p30 = p0 30 p0 30 = (0.9)(0.7) = 0.63 (⌧ )

(⌧ ) (⌧ )

l31 = p30 l30 = 630 (1)

(1) (⌧ )

d31 = 1| q30 l30 = 75 (⌧ )

(⌧ )

(⌧ )

d31 = l31 (2)

l32 = 630

(⌧ )

(1)

d31 = d31

(2)

q31 =

d31 = 158

(2)

d31

(⌧ ) l31

472 = 158

=

75 = 83

83 = 0.1317 ⇡ 0.13 630

Key: B

(⌧ )

2. For a double decrement table with l40 = 2000: x 40 41

(1)

qx 0.24 --

(2)

qx 0.10 --

q 0 (1) x 0.25 0.20

q 0 (2) x y 2y

(⌧ )

Calculate l42 . (A) 800

(B) 820

Arch MLC, Spring 2009

(C) 840

(D) 860

(E) 880

c Yufeng Guo

307

Chapter 8 Solution:

(⌧ )

(1)

(2)

q40 = q40 + q40 = 0.34 (1)

0.34 = 1 (2)

(2)

p0 40 p0 40

=1

(2)

0.75p0 40 (2)

p0 40 = 0.88

) q 0 40 = 0.12 = y

(2)

q 0 41 = 2y = 0.24 (⌧ )

q41 = 1

(0.8)(1

(⌧ )

l42 = 2000(1

0.24) = 0.392

0.34)(1

0.392) = 803

Key: A

3. For a multiple decrement table, you are given: (i) Decrement (1) is death, decrement (2) is disability, and decrement (3) is withdrawal. (1)

(ii) q 0 60 = 0.010 (2)

(iii) q 0 60 = 0.050 (3)

(iv) q 0 60 = 0.100 (v) Withdrawals occur only at the end of the year. (vi) Mortality and disability are uniformly distributed over each year of age in the associated single decrement tables. (3)

Calculate q60 . (A) 0.088

(B) 0.091

(C) 0.094

(D) 0.097

(E) 0.100

Solution: Since only decrements (1) and (2) occur during the year, probability of reaching the end of the year is (1)

(2)

p0 60 ⇥ p0 60 = (1

0.01)(1

0.05) = 0.9405

Probability of remaining through the year is (1)

(2)

(3)

p0 60 ⇥ p0 60 ⇥ p0 60 = (1 Arch MLC, Spring 2009

0.01)(1

c Yufeng Guo

0.05)(1

0.10) = 0.84645 308

Chapter 8 Probability of exiting at the end of the year is (3)

q60 = 0.9405

0.84645 = 0.09405

Key: C

4. For students entering a three-year law school, you are given: (i) The following double decrement table:

Academic Year 1 2 3

For a student at the beginning of that academic year, probability of Withdrawal for Survival Academic All Other Through Failure Reasons Academic Year 0.40 0.20 – – 0.30 – – – 0.60

(ii) Ten times as many students survive year 2 as fail during year 3. (iii) The number of students who fail during year 2 is 40% of the number of students who survive year 2. Calculate the probability that a student entering the school will withdraw for reasons other than academic failure before graduation. (A) Less than 0.35 (B) At least 0.35, but less than 0.40 (C) At least 0.40, but less than 0.45 (D) At least 0.45, but less than 0.50 (E) At least 0.50

Solution: (⌧ )

Let l0 = number of students entering in year 1. Superscript (f ) denotes academic failure, superscript (w) denotes withdrawal, subscript is “age” at start of year; equals year - 1. (⌧ )

p0 = 1 (⌧ )

ls

0.40 (⌧ ) (f )

= 10l2 q2

0.20 = 0.40 (f )

) q2

Arch MLC, Spring 2009

= 0.1 c Yufeng Guo

309

Chapter 8

(w)

q2

(⌧ )

(f )

= q2

(⌧ ) (f )

l1 q1

q2

(f )

(w) 3 q0

⇣

= 0.4 1 (f )

(⌧ )

(⌧ )

= 0.4 l1

q1

p1 = 1

= (1.0

h

q1

=

(f )

q1

(w)

= q0

0.6)

0.1 = 0.3

(f )

1

(w)

q1 (f )

q1 ⌘

0.3

q1

0.28 = 0.2 1.4

(w)

q1

⇣

=1

(⌧ ) (w)

+ p0 q 1

0.2

⌘i

0.3 = 0.5

(⌧ ) (⌧ ) (w)

+ p 0 p1 q 2

= 0.2 + (0.4)(0.3) + (0.4)(0.5)(0.3) = 0.38 Key: B

5. For a multiple decrement model on (60): (1)

(i) µ60 (t), t (ii)

(⌧ ) µ60 (t)

Calculate (A) 0.03

0, follows the Illustrative Life Table. (1)

= 2µ60 (t), t

0,

(⌧ ) 10| q60

(B) 0.04

(C) 0.05

(D) 0.06

(E) 0.07

SOLUTION: (⌧ ) k px = e

Rk 0

(⌧ )

µx (t)dt

o=e

Rk 0

(1)

2µx (t)dt

✓

= e

Rk 0

(1)

µx (t)dt

◆2

= (k px )2 where k px is from the ILT, since µ(1) follows ILT. 10 p60

=

6,616,155 = 0.80802, 8,188,074

(⌧ ) 10| q60

(⌧ )

= 10 p60

(⌧ ) 11 p60

= 0.808022

11 p60

= (10 p60 )2

=

6,396,609 = 0.78121 8,188,074

(11 p60 )2 from ILT.

0.781212 = 0.0426

Key B

Arch MLC, Spring 2009

c Yufeng Guo

310

Chapter 8 6. For students entering a college, you are given the following from a multiple decrement model: (i) 1000 students enter the college at t = 0. (ii) Students leave the college for failure (1) or all other reasons (2). (iii) µ(1) (t) = µ 0  t  4 µ(2) (t) = 0.04 0  t < 4

(iv) 48 students are expected to leave the college during their first year due to all causes. Calculate the expected number of students who will leave because of failure during their fourth year. (A) 8

(B) 10

(C) 24

(D) 34

(E) 41

SOLUTION: (⌧ )

d0 = 1000

Z 1

e

(µ+0.04)t

⇣

(µ + 0.04)dt = 1000 1

0

e µ + 0.04 =

(µ+0.04)

(µ+0.04)

⌘

= 48

= 0.952

ln(0.952) = 0.049 =) µ = 0.009

(1)

d3 = 1000 = 1000

e

0.009 ⇣ e 0.049

Z 4

e

0.049t

(0.009)dt

3

(0.049)(3)

e

(0.049)(4)

⌘

= 7.6

Key A

7. Don, age 50, is an actuarial science professor. His career is subject to two decrements: (i) Decrement 1 is mortality. The associated single decrement table follows De Moivre’s law with ! = 100. (ii) Decrement 2 is leaving academic employment, with (2)

µ50 (t) = 0.05, t

0

Calculate the probability that Don remains an actuarial science professor for at least five but less than ten years. (A) 0.22

(B) 0.25

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(C) 0.28

(D) 0.31

c Yufeng Guo

(E) 0.34 311

Chapter 8 SOLUTION: 0 (1)

(⌧ )

5 p50 = 5 p50

0 (2)

5 p50 =

Similarly (⌧ )

10 p50 =

✓

(⌧ ) 5|5 q50

✓

100 100 ◆

100 100

60 e 50 (⌧ )

= 5 p50

◆

55 e 50

(0.05)(5)

(0.05)(10)

(⌧ ) 10 p50

= (0.9)(0.7788) = 0.7009

= (0.8)(0.6065) = 0.4852

= 0.7009

0.4852 = 0.2157

Key: A

8. For a double decrement model: (i) In the single decrement table associated with cause (1), 0 (1)

q40 = 0.100 and decrements are uniformly distributed over the year. (ii) In the single decrement table associated with cause (2), 0 (2)

q40 = 0.125 and all decrements occur at time 0.7. (2)

Calculate q40 . (A) 0.114

(B) 0.115

(C) 0.116

(D) 0.117

(E) 0.118

SOLUTION: Only decrement 1 operates before t = 0.7 0 (1) 0.7 q40

0 (1)

= (0.7)q40 = (0.7)(0.10) = 0.07 since UDD

Probability of reaching t = 0.7 is 1

0.07 = 0.93

Decrement 2 operates only at t = 0.7, eliminating 0.125 of those who reached 0.7 (2)

q40 = (0.93)(0.125) = 0.11625 Key: C

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312

Chapter 8 9. For a double-decrement model: 0(1)

t 60 ,

0(2)

t 40 ,

(i) t p40 = 1 (ii) t p40 = 1

0  t  60. 0  t  40.

(⌧ )

Calculate µ40 (20). (A) 0.025

(B) 0.038

(C) 0.050

(D) 0.063

(E) 0.075

SOLUTION: Note that above 40, decrement 1 is DeMoivre with omega = 100; decrement 2 is DeMoivre with omega = 80. (1) (2) That means µ40 (20) = 1/40 = 0.025; µ40 (20) = 1/20 = 0.05 (⌧ ) µ40 (20) = 0.025 + 0.05 = 0.075. Or from basic definition of µ, (⌧ ) 60 t 40 t 2400 100t+t2 t p40 = 60 ⇥ 40 = 2400 (⌧ )

d(t p40 )/dt = ( 100 + 2t)/2400 at t = 20 gives 60/2400 = 0.025 (⌧ ) 20 p40 = (2/3) ⇤ (1/2) = 1/3 (⌧ ) (⌧ ) (⌧ ) µ40 (20) = [ d(t p40 )/dt]/20 p40 = 0.025/(1/3) = 0.075

Key: E

10. A fully discrete 3-year term insurance of 10,000 on (40) is based on a double-decrement model, death and withdrawal: (i) Decrement 1 is death. (1)

(ii) µ40 (t) = 0.02, t

0

(iii) Decrement 2 is withdrawal, which occurs at the end of the year. 0(2)

(iv) q40+k (t) = 0.04, k = 0, 1, 2 (v) ⌫ = 0.95. Calculate the actuarial present value of the death benefits for this insurance. (A) 487

(B) 497

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(C) 507

(D) 517

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(E) 527

313

Chapter 8 SOLUTION: 0(1)

At any age, px = e 0.02 = 0.9802 0(1) (1) qx = 1 0.9802 = 0.0198, which is also qx , since decrement 2 occurs only at the end of the year. Actuarial present value (AP V ) at the start of each year for that year’s death benefits = 10, 000 ⇤ 0.0198⌫ = 188.1 (⌧ ) px = 0.9802 ⇤ 0.96 = 0.9410 (⌧ ) Ex = px ⌫ = 0.941⌫ = 0.941 ⇤ 0.95 = 0.8940 AP V of death benefit for 3 years 188.1 + E40 ⇤ 188.1 + E40 ⇤ E41 ⇤ 188.1 = 506.60 Key: C

11. (50) is an employee of XY Z Corporation. Future employment with XY Z follows a double decrement model: (i) Decrement 1 is retirement. (ii)

(1) µ50 (t)

=

(

0.00 0  t  5 0.02 5t

(iii) Decrement 2 is leaving employment with XY Z for all other causes. (iv)

(2) µ50 (t)

=

(

0.05 0  t  5 0.03 5t

(v) If (50) leaves employment with XY Z, he will never rejoin XY Z. Calculate the probability that (50) will retire from XY Z before age 60. (A) 0.069

(B) 0.074

(C) 0.079

(D) 0.084

(E) 0.089

SOLUTION: (⌧ ) 5 p50 (1) 5 q55

= e (0.05)(5) = e 0.25 = 0.7788 R (1) = 05 5 µ55 (t) ⇥ e (0.03+0.02)t dt = ( 0.02/0.05)e = 0.4(1 e 0.25 ) = 0.0885 (⌧ ) (1) Probability of retiring before 60 ⇥ 5 p50 5 q55 = 0.7788 ⇤ 0.0885 = 0.0689 Key: A

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0.05t |5 0

314

Chapter 8 12. For a double decrement table, you are given: 0(1)

= 0.2.

0(2)

= 0.3.

(i) qx (ii) qx

(iii) Each decrement is uniformly distributed over each year of age in the double decrement table. Calculate

(1) 0.3 qx+0.1

.

(A) 0.020 (B) 0.031 (C) 0.042 (D) 0.053 (E) 0.064

SOLUTION: 0(1) 0(2)

(⌧ )

px = px p x (1) qx

=

h

=



0(1)

ln(px ) 0(⌧ ) ln(px )

ln(0.8) ln(0.56)

(1) 0.3 qx+0.1

i

=

= 0.8(0.7) = 0.56. (⌧ )

qx

since U DD in double decrement table

0.44 = 0.1693 (1)

0.3qx (⌧ ) 1 0.1qx

= 0.053

To elaborate on the last step: Number dying from cause 1 between x + 0.1 and x + 0.4 (1) 0.3 qx+0.1 = Number alive at x+0.1 Since U DD in double decrement, =

(⌧ ) (1) lx (0.3)qx (⌧ ) (⌧ ) lx (1 0.1qx )

!

Key: D

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315

Chapter 8 13. For a special fully discrete 3-year term insurance on (55), whose mortality follows a double decrement model: (i) Decrement 1 is accidental death; decrement 2 is all other causes of death. (ii) (1)

(2)

x qx qx 55 0.002 0.020 56 0.005 0.040 57 0.008 0.060 (iii) i = 0.06 (iv) The death benefit is 2000 for accidental deaths and 1000 for deaths from all other causes. (v) The level annual contract premium is 50. (vi) 1 L is the prospective loss random variable at time 1, based on the contract premium. (vii) K(55) is the curtate future lifetime of (55). Calculate E[1 L|K(55)

1].

(A) 5 (B) 9 (C) 13 (D) 17 (E) 20

SOLUTION: Actuarial present value (AP V ) of future benefits = = [(0.005)2000 + 0.04(1000)]/1.06 + (1 0.005 0.04)(0.008 · 2000 + 0.06 · 1000)/1.062 = 47.17 + 64.60 = 111.77 AP V of future premiums = [1 + (1 0.005 = (1.9009)(50) = 95.05 E[1 L|K(55) 1] = 111.77 95.05 = 16.72. Key: D

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0.04)/1.06]50

c Yufeng Guo

316

Chapter 8

Problems from Pre-2000 SOA-CAS Exams 1. You’re given the following double-decrement model for students writing actuarial exams: •

x 21 22 23

(1)

qx 0.008 0.015 0.025

(2)

qx 0.15 0.20 0.25

(1)

• qx denotes the probability of decrement due to attaining fellowship. (2)

• qx denotes the probability of decrement due to all other causes. • Decrements occur at the end of the year. You are also given: • After attaining fellowship, the only decrement is mortality. • After attaining fellowship, µ = 0.04.

Calculate the probability that a student writing exams now age 21, will be living and a Fellow three years later. (A) 0.031

(B) 0.036

(C) 0.038

(D) 0.043

(E) 0.046

2. For a triple-decrement model, you are given: • Decrement 1 is uniformly distributed over each year of age. • Decrement 2 occurs only at the end of the year.

• Decrement 3 occurs only at the beginning of the year. •

x 60 61 62

(⌧ )

lx 100,000

0 (1)

qx 0.14

0 (2)

qx 0.1 0.1

0 (3)

qx 0.1 0.2

45,516

(1)

Calculate q61 . (A) 0.070

(B) 0.074

Arch MLC, Spring 2009

(C) 0.078

(D) 0.082

c Yufeng Guo

(E) 0.086

317

Chapter 8 3. For a double-decrement model, you are given: 1

(1)

• µ10 (t) =

30

(⌧ )

• µ10 (t) =

t 50

600

,

0  t < 30

2t , 50t + t2

0  t < 20

Calculate the probability that (10) will terminate from cause 2 during the 6th year. (A) 0.0225

(B) 0.0242

(C) 0.0392

(D) 0.0408

(E) 0.0650

Use the following information for the next 4 questions. For a double-decrement model, you are given: 2 , t 0 60 + t 3 (2) • µ40 (t) = , t 0 60 + t • T is the time-until-decrement random variable for (40). (1)

• µ40 (t) =

• J is the cause-of-decrement random variable for (40).

4. Calculate fT,J (20, 2). (A) 0.0059

(B) 0.0066

(C) 0.0076

(D) 0.0089

(E) 0.0099

(C) 0.0148

(D) 0.0159

(E) 0.0165

5. Calculate fT (20). (A) 0.0127

(B) 0.0136

6. Calculate fJ (2). (A) 0.33

(B) 0.40

(C) 0.50

(D) 0.60

(E) 0.67

(C) 0.50

(D) 0.60

(E) 0.67

7. Calculate fJ|T (1|10). (A) 0.33

(B) 0.40

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318

Chapter 8 8. From a double-decrement table, you are given: • l30 = 1000 0 (1)

• q30 = 0.1 0 (2)

• q30 = 0.2 •

(1) 1| q30

= 0.05

(⌧ )

• l32 = 562 (2)

Calculate q31 . (A) 0.125

(B) 0.130

(C) 0.145

(D) 0.150

(E) 0.170

9. You are given the following extract from a triple-decrement table: x 60 61 62 63

(1)

qx 0.010

(2)

qx 0.050

(3)

qx 0.020

(⌧ )

qx

(⌧ )

lx 10,000

0 (1)

0 (2)

0 (3)

qx

qx

px

0.023

0.033

0.990

0.076 0.098 (⌧ )

Calculate l63 . (A) 7172

(B) 7175

Arch MLC, Spring 2009

(C) 7951

(D) 7954

c Yufeng Guo

(E) 8031

319

Chapter 8

Solutions to Pre-2000 Problems: Chapter 10 1. Key: B (1)

x 21 22 23

(2)

qx 0.008 0.015 0.025

(⌧ )

qx 0.15 0.20 0.25

qx 0.158 0.215 0.275

(⌧ )

px 0.842 0.785 0.725

(⌧ ) n px

0.842 0.661 0.479

n 1 2 3

The probability we are looking for is given by (1)

(⌧ )

(1)

(⌧ )

(⌧ )

(1)

q21 · p22 · p23 + p21 · q22 · p23 + p21 · p22 · q23 = (0.008)e

0.04

0.04

·e

+ (0.842)(0.015)e

0.04

+ (0.842)(0.785)(0.025)

= 0.007385 + 0.012135 + 0.016524 = 0.036

2. Key: B (⌧ )

First, we need l61

(⌧ )

0 (1)

0 (3)

0 (2)

p60 = p60 · p60 · p60 = (0.86)(0.9)(0.9) = 0.6966 (⌧ )

l61 = 69,660 (1)

(1)

To get to q61 , we need d61 . For the 69,660 people alive at age 61, the 3rd decrement strikes first and leaves 69,660(0.8) = 55,728 lives. Next, the 55,728 remaining lives are hit by the 1st decrement, but to figure out how (1) many decrement due to cause 1, we need q61 . We use the fact that 0 (1)

0 (2)

0 (3)

p61 · p61 · p61 = 0 (1)

) p61 =

45,516 = 0.6534 69,660

0.6534 = 0.9075 (0.9)(0.8)

0 (1)

) q61 = 1

0.9075 = 0.0925

(1)

) d61 = 55,728(0.0925) = 5154.84 (1)

) q61 =

Arch MLC, Spring 2009

5154.84 = 0.074 69,660

c Yufeng Guo

320

Chapter 8 3. Key: D (2) 5| q10

=

Z 6

(⌧ ) t p10

5

(⌧ )

(2)

· µ10 (t) dt

(1)

(2)

µ10 (t) = µ10 (t) + µ10 (t) 50

(2)

) µ10 (t) =

2t 50t + t2

600

1 = 30t 600

t2 2

5

=

t

0 (2)

= t p10 · t p10 =

) #6

30

0 (1)

(⌧ ) t p10

"

1

30 t (30 t)(20

t)

=

1 20

t

30 t 20 t · 30 20

Z 6 30 5

t 20 t 1 · · dt 30 20 20 t 

1 = 180 600

18

150 +

25 24.5 = = 0.0408 2 600

4. Key: D (⌧ ) (j)

fT,J (t, j) = t p40 µ40 (t) (⌧ )

(1)

(1)

µ40 (t) = µ40 (t) + µ40 (t) = (⌧ ) t p40

=e

Rt

5 ds 0 60+s

=e

) fT,J (20, 2) =

✓

5 ln(60+s)|t0

60 80

◆5 ✓

3 80

◆

5 60 + t =

✓

60 60 + t

◆5

= 0.0089

5. Key: C ) (⌧ ) fT (t) = t p(⌧ x · µx (t)

fT (20) =

Arch MLC, Spring 2009

✓

60 80

◆5 ✓

5 80

c Yufeng Guo

◆

= 0.0148

321

Chapter 8 6. Key: D FJ (j) = 1 qx(j) =

Z 1 0

(⌧ ) (j) t p40 µ40 (t) dt

From work two problems ago, we know this equals ◆5 ✓ Z 1✓ 60

60 + t

0

= (60) (3) 5

Z 1

(60 + t)

◆

3 60 + t

dt "

# 5 1

(60 + t) ds = (60) (3) 5

6

5

0

= (60)5 (3) ·

(60) 5

5

0

= 0.6

7. Key: B fJ|T (j|t) =

(j) µ40 (t) (⌧ ) µ40 (t)

) fJ|T (1|10) =

⇣

(1) µ40 (10) (⌧ ) µ40 (10)

2 70

⌘

= ⇣ ⌘ = 0.4 5 70

8. Key: D (2)

(⌧ )

We need to know d31 . Then dividing by l31 will give the answer we are looking for. (⌧ )

(⌧ )

(⌧ )

l31 = p30 · l30 = (0.9)(0.8)(1000) = 720 (⌧ )

(⌧ )

(⌧ )

d31 = l31

l32 = 720

(1)

(⌧ )

(1)

562 = 158

d31 = 1| q30 · l30 = (0.05)(1000) = 50 (2)

) d31 = 158 (2)

q31 =

50 = 108

108 = 0.15 720

9. Key: C (⌧ )

(⌧ )

(⌧ )

l61 = l60 · p60 = 10,000(1 (⌧ )

(⌧ )

0.01

(⌧ )

l62 = l61 · p61 = 9200(1 (⌧ )

(⌧ )

(⌧ )

⇣

0 (1)

l63 = l62 · p62 = 8501 p62

0.05

0.02) = 9200

0.076) = 8501 ⌘⇣

0 (2)

p62

⌘⇣

0 (3)

p62

= 8501 · (0.977)(0.967)(0.99) = 7951 Arch MLC, Spring 2009

c Yufeng Guo

⌘

322

Chapter 9

ACTUARIAL MATHEMATICS: CHAPTER 11 APPLICATIONS OF MULTIPLE DECREMENT THEORY • Option A reference: Actuarial Mathematics Chapter 11 • Option B reference: Models for Quantifying Risk Chapter 15.6

11.2 Actuarial Present Values and Their Numerical Estimation Multiple decrement theory comes into play when the benefits to be paid vary based on which decrement eventually gets the insured. For example, consider double indemnity provisions in many life insurance policies. The benefit paid might be X in the event of a “natural” death, but will be 2X if the death is due to an accident. In general, what we have to do to find the actuarial present value of this type of insurance is to find the APV separately for all possible benefits, and then sum them: A=

m Z 1 X

j=1

0

(j)

) (j) Bx+t v t t p(⌧ x µx (t)dt ,

where there are m decrements and B (j) refers to the benefit paid for decrement j. (1)

Note that if there is only one decrement (m = 1) and the benefit is 1 (Bx+t = 1), then the formula above reduces to the standard actuarial present value for a whole life policy of 1 with immediate payment of claims. It is often easier to think of a policy with a double indemnity provision as two separate policies - one that pays a unit upon death, regardless of cause, and a second that pays a unit if the death is accidental. For example there are two ways to write down A for an n-year 323

Chapter 9 term policy with a double indemnity provision in which B (1) = 2 and B (2) = 1 (accidental death is decrement 1 and natural death is decrement 2). The hard way is A=2

Z n 0

v

t

(⌧ ) (1) t px µx (t)

We can think of this as

dt +

Z n 0

(1)

) (2) v t t p(⌧ x µx (t) dt.

(2)

A = 2A 1

+A1 .

x:n

x:n

The more useful way to write this that is more often called for on the exam is as (1)

A = A1

+A1 ,

x:n

where A 1

x:n

x:n

is a standard n-year term policy that pays a unit death benefit regardless of the

cause of death.

11.3 Benefit Premiums and Reserves This section (like the last one) doesn’t really introduce anything new. We are just taking unusual benefit patterns and breaking them down into pieces to calculate APV (in the previous section) and benefits and reserves (in this section). Consider a fully discrete whole life policy issued to a person age x with a double indemnity provision. It is common for the double indemnity provision to be applicable only prior to age 65. The benefit is 1 for non-accidental death and 2 for accidental death. Again, we will break this policy into its two parts to calculate premiums. First, we calculate the premium for the benefit of 1 paid regardless of cause of death:

Px(⌧ ) =

1 X

(⌧ )

) v k+1 k p(⌧ x qx+k

k=0 1 X

. v

k

(⌧ ) k px

k=0

Second, we calculate the premium due to pay for the double indemnity rider. Note that this benefit only applies to age 65, so premiums are paid only to age 64 (since this is a fully discrete policy):

(2) 65 x Px

=

64 Xx

(2)

) v k+1 k p(⌧ x qx+k

k=0 64 Xx

, v

k

(⌧ ) k px

k=0

where (2) signifies accidental death. Arch MLC, Spring 2009

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324

Chapter 9 Each of the above premium calculations is of the form: Benefit Premium =

Exp PV of Benefits Exp PV of Corresponding Annuity

This is the pattern you are already familiar with from earlier sections. The benefit reserve is calculated in the usual way General Case: Reserve = (APV future benefits) - (APV of future premiums) In this case, it is the reserve for a usual death benefit of 1 plus a reserve for a benefit of 1 for accidental death before age 65. (2) k V = k V x + k Vx =

"1 X

v

h+1

(⌧ ) h px+k

(⌧ ) qx+k+h

Px(⌧ )

h=0

+

"64 x k X

v

h+1

1 X

v

h

(⌧ ) h px+k

h=0 (⌧ ) h px+k

(2) qx+k+h

(2) 65 x Px

h=0

64X x k

v

h

#

(⌧ ) h px+k

h=0

#

.

Before we get to an example, let me say that what you really need to get from this section is that we can describe some pretty messy insurances by breaking a complex policy down into its fairly easy parts, and then summing them back to get the whole. EXAMPLE: A certain 40-year old wishes to purchase a 3-year term life insurance policy that pays 100,000 at the end of year of death in the event of a natural death (cause 1) and pays double indemnity in the event of an accidental death (cause 2). The associated single-life decrement table is shown below. i = 0.06. x 40 41 42

0 (1)

0 (2)

qx qx 0.02 0.005 0.03 0.005 0.04 0.005

Each of the decrements is uniformly distributed in the year of death in the associated single-life decrement table. 1. Complete the multiple decrement table; 2. Find the actuarial present value of this insurance policy; 3. Find the level benefit premium for this insurance; 4. Find the benefit reserve at the end of the second year. SOLUTION: Arch MLC, Spring 2009

c Yufeng Guo

325

Chapter 9 1. For each year we use the fact that qx(1)

0 (1)

= qx

✓

◆

1 0 (2) q , 2 x

1

(2)

and the corresponding formula for qx . 0 (1)

x 40 41 42

0 (2)

(1)

(2)

(⌧ )

qx qx qx qx qx 0.02 0.005 0.01995 0.00495 0.0249 0.03 0.005 0.02993 0.00493 0.0349 0.04 0.005 0.03990 0.00490 0.0448

2. We can view this policy as a 3-year term policy on both causes plus a 3-year term policy on cause 2 only. The value of the 3-year term policy that pays regardless of cause of death is (⌧ )

(⌧ )

A1

40:3

= (1.06

1

(⌧ ) (⌧ )

(⌧ ) (⌧ ) (⌧ )

= vq40 + v 2 p40 q41 + v 3 p40 p41 q42

)(0.0249) + (1.06

2

)(0.975)(0.0349) + (1.06

3

)(0.941)(0.0448)

= 0.08913 Similarly, the APV of the extra accidental death policy (a 3-year term policy on cause 2) is given by (2)

A1

(2)

40:3

= (1.06

1

(⌧ ) (2)

(⌧ ) (⌧ ) (2)

= vq40 + v 2 p40 q41 + v 3 p40 p41 q42

)(0.00495) + (1.06

2

)(0.975)(0.004925) + (1.06

3

)(0.941)(0.0049)

= 0.01282 So the actuarial present value of the entire policy is 100,000(0.08913 + 0.01282) = 10,195.05 3. The benefit premium is (⌧ )

⇡ = 100, 000 (⌧ )

(2)

A1

(⌧ ) (⌧ )

40:3

a ¨x:3 = 1 + vp40 + v 2 p40 p41 = 1 + (1.06

+A1

40:3

a ¨x:3 1

)0.9751 + (1.06

2

)0.9411 = 2.757

10, 195.05 = 3, 697.21 2.757 PVFP), which equals ⇡=

4. The reserve is (PVFB

⇣

(⌧ )

(2)

100,000v q42 + q42 = 100,000(1.06

1

)(0.0448 + 0.0049)

⌘

⇡ 3, 697.21 = 991.47

}

Chapter 11 Suggested Problems: 1,2

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326

Chapter 9

CHAPTER 11 Formula Summary A=

m Z 1 X

j=1

0

Px(⌧ ) =

(j)

) (j) Bx+t v t t p(⌧ x µx (t)dt

1 X

(⌧ )

) v k+1 k p(⌧ x qx+k

k=0 1 X

) v k k p(⌧ x

k=0

ARCH Sample Exam Problem 1. Kyle enters a company pension plan at age 25. The only two decrements that might a↵ect Kyle at work are death and retirement. If Kyle survives to age 65, the retirement plan at the company will begin making annual payments of $400 times the number of years Kyle has worked at the company. If Kyle dies before retiring, he will receive a flat, end-of-year death benefit of $100,000. Mortality and interest follow the Illustrative Life Table with i = 6%. Calculate the actuarial present value of these benefits on the day Kyle begins work. (A) 11,000

(B) 13,000

(C) 15,000

(D) 17,000

(E) 19,000

Solution: 1. 100,000A1

25:40

✓

= 100,000 A25

1 (1.06)40 ✓

◆

40 p25 · A65 + 16,000 ·

= 100,000 0.08165 Arch MLC, Spring 2009

+ 40 E25 (400)(40)¨ a65 1 (1.06)40

40 p25

·a ¨65

◆

75,340 0.09722 · (0.43980) 95,650

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327

Chapter 9 +16,000(0.09722)

75,340 (9.8969) 95,650

= 4797 + 12,126 = 16,923 Key: D

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328

Chapter 9

Past SOA/CAS Exam Questions: 1. A special whole life insurance is issued on (x). The basic death benefit is 1 during policy year one and 2 thereafter. An additional benefit of 2 is provided if death is accidental. You are given • Benefits are payable at the moment of death.

• The force of mortality due to accidental death is µ(ad) (x + t) = 0.005, t 0. • µ(⌧ ) (x + t) = 0.040, t •

0.

= 0.06

Calculate the Actuarial Present Value for this insurance. (A) 0.777

(B) 0.812

(C) 0.827

(D) 0.844

(E) 0.862

Solution: APV =

Z 1 0

=

Z 1

e

v

t

(⌧ ) (⌧ ) t px µ dt

0.06t

e

0.04t

+

) vp(⌧ x

Z 1

v

t

0

(0.04)dt + e

0.06

(⌧ ) (⌧ ) t px+1 µ dt

e

0.04

0

Z 1

e

+

Z 1 0

0.06t

e

) (ad) 2v t t p(⌧ dt x µ

0.04t

(0.04)dt

0

+

Z 1

2e

0.06t

Z 1

e

e

0.04t

(0.005)dt

0

= (0.0862)

0.1t

dt = 0.862 Key:E

0

2. For a triple-decrement model, you are given: • Decrement 1 is uniformly distributed over each year of age. • Decrement 2 occurs only at the end of the year.

• Decrement 3 occurs only at the beginning of the year. •

x 60 61 62

(⌧ )

lx 100,000

0 (1)

qx 0.14

0 (2)

qx 0.1 0.1

0 (3)

qx 0.1 0.2

45,516

(1)

Calculate q61 . (A) 0.070

(B) 0.074

Arch MLC, Spring 2009

(C) 0.078

(D) 0.082

c Yufeng Guo

(E) 0.086 329

Chapter 9 Solution:

(⌧ )

(1) (2) (3)

p60 = p60 p60 p60 = (0.86)(0.9)(0.9) = 0.6966 (⌧ )

) l61 = 69, 660

At age 61, after decrement 3 occurs, there are 69,660(0.8) = 55,728 lives remaining. So 0 (1) 55,728·q61 lives leave due to Decrement 1. ⇣

1

0 (1)

q61

⌘⇣

⇣

) 1 (1)

0 (2)

p61

⌘⇣

0 (1)

q61

0 (3)

p61

⌘

⌘

=

45,516 = 0.6534 69,660

(0.9)(0.8) = 0.6534.

0 (1)

) q61 = 0.0925 0 (1)

d61 = 55,728 · q61 = 5154.84 (1)

q61 = 0.074 Key: B

3. For a special whole life insurance of 100,000 on (x), you are given: (i)

= 0.06.

(ii) The death benefit is payable at the moment of death. (iii) If death occurs by accident during the first 30 years, the death benefit is doubled. (⌧ )

(iv) µx (t) = 0.008, t (v)

(1) µx (t)

= 0.001, t accident.

0 (1)

0, where µx

is the force of decrement due to death by

Calculate the single benefit premium for this insurance. (A) 11,765

(B) 12,195

(C) 12,622

(D) 13,044

(D) 13,235

Solution:

APV of regular death benefit =

Z 1

(100,000)(e

t

)(0.008)(e

µt

)dt

0

=

Z 1

(100,000)(e

0.06t

)(0.008)(e

0.008t

)dt

0

0.008 0.06 + 0.008 = 11,764.71 = 100,000

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330

Chapter 9

Z 30

APV of accidental death benefit =

(100,000)(e

t

)(0.001)(e

µt

)dt

0

Z 30

=

(100,000)(e

0.06t

)(0.001)(e

0.008t

)dt

0

1

e 2.04 0.068 = 1,279.37 = 100

Total APV = 11,765 + 1279 = 13,044

Key: D

4. A whole life policy provides that upon accidental death as a passenger on an airplane a benefit of 1,000,000 will be paid. If death occurs from other accidental causes, a death benefit of 500,000 will be paid. If death occurs from a cause other than an accident, a death benefit of 250,000 will be paid. (i) Death benefits are payable at the moment of death. 1 (ii) µ(1) = where (1) indicates accidental death as a passenger on an air2,000,000 plane. 1 (iii) µ(2) = 250,000 1 (iv) µ(3) = 10,000 (v) = 0.06 Calculate the single benefit premium for this insurance. (A) 450

(B) 460

(C) 470

(D) 480

(E) 490

SOLUTION: ) (1) (2) (3) µ(⌧ x = µx + µx + µx = 0.0001045 (⌧ ) t px

=e

0.0001045t

The present value of future benefits equals 1,000,000

Z 1

e

t

0

=

1,000,000 2,000,000

Z 1

(⌧ ) (1) t px µx dt+500,000

e

0

Arch MLC, Spring 2009

0.0601045t

dt+

Z 1

t

e

0

500,000 250,000

Z 1

(⌧ ) (2) t px µx dt+200,000

e

0.0601045t

0

= 27.5(16.6377) = 457.54 c Yufeng Guo

dt+

Z 1

250,000 10,000

t

e

0

Z 1

e

(⌧ ) (3) t px µx dt

0.0601045t

dt

0

Key: B 331

Chapter 9 5. For a special whole life insurance: (i) The benefit for accidental death is 50,000 in all years. (ii) The benefit for non-accidental death during the first 2 years is return of the single benefit premium without interest. (iii) The benefit for non-accidental death after the first 2 years is 50,000. (iv) Benefits are payable at the moment of death. (1)

(v) Force of mortality for accidental death: µx (t) = 0.01, t (vi) Force of mortality for non-accidental death: (vii)

(2) µx (t)

0

= 2.29, t

0

= 0.10

Calculate the single benefit premium for this insurance. (A) 1,000 (B) 4,000 (C) 7,000 (D) 11,000 (E) 15,000

SOLUTION:

P =P

Z 2

) (1) (2) µ(⌧ x (t) = µx (t) + µx (t) = 0.01 + 2.29 = 2.30

⌫

) (2) ·t p(⌧ x µx (t)dt + 50,000

t

0

P =P

Z 2 0

e "

0.1t

P 1

e

2.3t

2.29 ·

·2.29dt+50,000 1

#

Z 2

⌫

t

0

Z 2 0

e

) (1) ·t p(⌧ x µx (t)dt + 50,000 0.1t

e

2.3t

"

Z 1

·0.01dt+50,000

e 2(2.4) 1 = 50000 0.01 · 2.4

2

) (⌧ ) ⌫ t ·t p(⌧ x µx (t)dt

Z 1

e

0.1t

e

2

e 2(2.4) e 2(2.4) + 2.3 · 2.4 2.4

#

2.3t

·2.3dt

P = 11, 194 Key: D

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332

Chapter 9 6. A special whole life insurance on (x) pays 10 times salary if the cause of death is an accident and 500, 000 for all other causes of death. You are given: (⌧ )

(i) µx (t) = 0.01, t (accident)

(ii) µx

0

(t) = 0.001, t

0

(iii) Benefits are payable at the moment of death. (iv)

= 0.05

(v) Salary of (x) at time t is 50, 000e0.04t , t

0.

Calculate the actuarial present value of the benefits at issue. (A) 78,000 (B) 83,000 (C) 92,000 (D) 100,000 (E) 108,000

SOLUTION: µ(accid) = 0.001 µ(total) = 0.01 µ(other) = 0.01 0.001 = 0.009 Actuarial present value =

Z 1

500,000e

0.05t

e

0.01t

(0.009)dt + 10

0

Arch MLC, Spring 2009

500,000e0.04t e

0.05t

e

0.01t

(0.001)dt

0

= 500,000 Key: D

Z 1



0.009 0.001 + = 100,000 0.06 0.02

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333

Chapter 9

Arch MLC, Spring 2009

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334

Chapter 10

ACTUARIAL MATHEMATICS: CHAPTER 15 INSURANCE MODELS INCLUDING EXPENSES • Option A reference: Actuarial Mathematics Chapter 15 • Option B reference: Models for Quantifying Risk Chapter 15.6-15.7 There are two important topics in this section. The first is Expenses which are not too difficult. The main point of the expenses section is to finally deal with the fact that all of the premiums and reserves that we have talked about so far in this course have only been sufficient to cover benefits. So far, they’ve not covered any other cash demands an insurance company faces (like the salary and study materials for actuarial students). The second topic is Asset Shares, which will require some new notation and explanation. One important point to asset shares is to have a method for assigning company assets to di↵erent policyholders in the event the insurance company demutualizes or faces bankruptcy.

15.2 Expense Augmented Models Expenses are not too hard to deal with as long as you treat them as just another benefit that the insurance company must provide. Then when calculating the Expense-loaded Premium G, (G is for Gross Premium as opposed to Net Benefit Premium P ) you just need to include the expenses in the equivalence relation. A couple of examples should make this clear. EXAMPLE: You are given: • Mortality follows the Illustrative Life Table. 335

Chapter 10 • i = 6%

• A person aged (40) will purchase a fully discrete whole life insurance with death benefit equal to 1000. • Expenses include 10% of gross premium plus 5 per year, payable at the beginning of each year. 1. Find the gross premium necessary to fund all benefits and expenses. 2. Find the premium amount necessary to cover expenses. (This is called the Expense Premium and is defined to be the gross premium minus the net benefit premium.) 3. Find the benefit reserve and expense reserve (the portion of the total reserve necessary to pay future expenses) at time t = 10.

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Chapter 10 SOLUTION: 1. The equivalence relation to fund both benefits and expenses is G¨ a40 = (0.1)G¨ a40 + 5¨ a40 + 1000A40 5¨ a40 + 1000A40 0.9¨ a40 From the illustrative life table we obtain )G=

5(14.82) + 161.32 = 17.65 0.9(14.82)

G= 2. The net benefit premium is

1000A40 161.32 = = 10.885 a ¨40 14.82 So the expense premium is 17.65

10.885 = 6.77

3. There are lots of ways to calculate the benefit reserve, we’ll take a shortcut with the annuity formula: ✓

1000 1

a ¨50 a ¨40

◆

✓

= 1000 1

13.27 14.82

◆

= 104.59

The expense reserve is the actuarial present value of Future Expenses minus Future Expense Premiums (0.1)(17.65)¨ a50 + 5¨ a50

6.77¨ a50 = 0

For this example, the level expense premium exactly funds that year’s expenses and the expense reserve is exactly zero in all years. }

EXAMPLE: You are given: • Mortality follows the Illustrative Life Table. • i = 6%

• A person aged (40) will purchase a fully discrete whole life insurance with death benefit equal to F . • Expenses include 10% of gross premium, payable at the beginning of each year. In addition there is an initial expense of 20 at policy issue and a final expense of 80 at the time the death benefit is paid. • The level gross premium is equal to 30. 1. Find the face amount F of the policy. 2. Find the expense reserve time t = 10. Arch MLC, Spring 2009

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337

Chapter 10 SOLUTION: 1. G¨ a40 = (0.1)G¨ a40 + 20 + 80A40 + F · A40

Since G = 30, we have

(30)¨ a40 = (0.1)(30)¨ a40 + 20 + 80A40 + F · A40 Using values from the illustrative life table and solving for F , we arrive at a face amount equal to F = 2276 2. The net benefit premium is 2276A40 367.23 = = 24.78 a ¨40 14.82 So the level expense premium is 30.00

24.78 = 5.22

The expense reserve is (0.1)(30)¨ a50 + 80A50 = (3)(13.27) + 80(0.249)

5.21¨ a50

5.22(13.27) =

9.54

The expense reserve is negative! This is due to the fact that the initial charge was large in present value relative to the final charge at time of issue. This phenomenon will reverse in later years. For example, the expense reserve at time t = 30 is (0.1)(30)¨ a80 + 80A80 5.22¨ a80 = (3)(5.91) + 80(0.666)

5.22(5.91) = 40.16

}

15.4 More Expenses This section allows for more variation in types and level of expense than we saw in 15.2. We will include one example here to complement Example 15.4.1 in the text but Example 15.4.1 gives excellent coverage of this material. Anyone who takes the time to understand it in all its details should be ready for any questions on this material. This section also includes a classification scheme for di↵erent types of expenses. They are broken into Investment Expenses and Insurance Expenses • Investment Expenses - Costs of analyzing, buying, selling, and servicing the investments used to back insurance company reserves. Investment expenses are usually accounted for by slightly reducing the interest rate assumed during the life of the policy.

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Chapter 10 • Insurance Expenses 1. Acquisition Expense - Costs associated with acquiring the business. These can include commissions, underwriting costs, advertising, and setting up new policy records. 2. Maintenance Expense - regularly occurring expenses such as premium collection, policy changes, policyholder correspondence. 3. General Expense - Expenses that aren’t easy to allocate to particular policies. Examples include research, actuarial and legal services, accounting, legal compliance. 4. Settlement Expenses - Expenses associated with validating and paying a claim: claim investigation, legal defense, administrative costs of paying benefits. EXAMPLE: For a fully discrete 10-year endowment insurance on (50), you are given the following: • Percent of premium expenses consist of commissions equal to 50% of the gross premium in the first year followed by 5% of gross premium in years 2 through 10. • Expenses include acquisition expense of 20, settlement expenses equal to 10 per 1000 of face amount plus 80, and annual maintenance expenses equal to 5 plus 2 per 1000 of face amount. • Acquisition expense is due at time of policy issue, settlement expenses are due when the benefit is paid, and annual maintenance expenses are due at the time annual premium is paid. If the face amount of the policy is b, find the gross premium G in terms of b and the usual actuarial symbols for insurances and annuities. SOLUTION: G·a ¨50:10 = 0.45G + 0.05G¨ a50:10 + |

{z

}

Percent of Premium

+

20 |{z}

Acquisition

10 · b 2·b A50:10 + 80A50:10 + 5¨ a50:10 + a ¨ +b · A50:10 1000 50:10} |1000 {z } | {z Maintenance Settlement

Note that the right-hand side of the equation has been broken into the four types of expenses plus the death benefit. Rearranging gives us: G·¨ a50:10

✓

◆

✓

◆

10b 2b = b+ ·A50:10 +0.45G+ 0.05G + a ¨ +20+80A50:10 +5¨ a50:10 1000 1000 50:10 )G=

Arch MLC, Spring 2009

(1.01b + 80)A50:10 + (5 + 0.002b)¨ a50:10 + 20 0.95¨ a50:10 c Yufeng Guo

0.45 339

Chapter 10 We can separate this premium into a component that varies directly with the face amount plus a component that is independent of the face amount: )G=b

"

1.01A50:10 + 0.002¨ a50:10 0.95¨ a50:10

0.45

#

+

"

80A50:10 + 5¨ a50:10 + 20 0.95¨ a50:10

0.45

#

}

NOTES: • Insurance companies occasionally determine a price per unit of coverage for insurance policies. So the number they are interested in is the per unit gross premium which equals G/b. For the example just above, an expression for the per-unit premium would be " # " # 1.01A50:10 + 0.002¨ a50:10 a50:10 + 20 G 1 80A50:10 + 5¨ = + b 0.95¨ a50:10 0.45 b 0.95¨ a50:10 0.45 • The per unit gross premium is awkward since it has a portion that depends on b. This can be taken care of by calling the portion of G that is independent of b the policy fee or the expense policy fee. The policy fee is the portion of each premium that goes to pay per policy expenses (those expenses that are independent of the face amount of the policy). In the example above, the policy fee is 80A50:10 + 5¨ a50:10 + 20 0.95¨ a50:10

0.45

EXAMPLE: For a fully discrete whole life insurance on (30), you are given: • Mortality follows the illustrative life table. • i = 0.06

• Percent of premium expenses are 15% in each year

• Per policy expenses include 100 at policy issue, 30 each year the policy is in force, and 50 payable at the end of year of death. Calculate the expense policy fee for this policy.

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340

Chapter 10 SOLUTION: We need to keep in mind that percent-of-premium charges apply to the policy fee as well as the rest of the gross premium. Therefore, the policy fee g must satisfy g¨ a30 = 0.15g¨ a30 + 100 + 30¨ a30 + 50A30 100 + 30(15.86) + 50(0.1025) 0.85 36.62 )g= = 43.08 0.85

g(15.86) =

}

15.6.1 Asset Shares Asset shares are a tool used to project the accumulation of assets backing a block of insurance policies and to assign those assets to each policy. First some terminology: • G - the level annual contract premium; • k AS - the asset share assigned to the policy at time t = k; • ck - the fraction of the contract premium paid for expenses at time k (in other words c · G is the expense premium); • ek - expenses paid per policy at time t = k; (1)

• qx+k - the probability of death before age x + k + 1 for an insured who is now x + k; (2)

• qx+k - the probability of decrement by withdrawal before age x + k + 1 for an insured who is now x + k; • k CV - the cash amount due to the policyholder as a withdrawal benefit in the event they decrement by withdrawal. The CV stands for cash value. • bk - death benefit due at time t = k for a death in the k-th policy year. The definition of the k-th Asset Share k AS is the actuarially accumulated value at time k of all past premiums, minus expenses, death benefits, and withdrawal benefits. If the initial asset share is 0 AS (which may or may not be equal to zero), then 1 AS must satisfy [0 AS + G(1

c0 )

) e0 ] (1 + i) = qx(1) · b1 + qx(2) · 1 CV + p(⌧ x · 1 AS

This should remind you a lot of the reserve recursion from Chapter 8 of Bowers. The di↵erences are • We are using contract premiums, expenses, and withdrawal benefits, Arch MLC, Spring 2009

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341

Chapter 10 • The initial asset share may be non-zero (whereas the initial benefit reserve is zero by definition), • The final asset share may also be non-zero.

The general recursion formula is [k AS + G(1

ck )

(1)

(2)

(⌧ )

ek ] (1 + i) = qx+k · bk+1 + qx+k · k+1 CV + px+k · k+1 AS

EXAMPLE: For fully discrete whole life insurance on (80), you are given the following: • • • • •

The face amount is 1000, and the annual contract premium is 200. Expenses include 10% of premium plus 20 in all years. The cash value during any year is 50% of all premiums paid to that point. i = 6%. Mortality and withdrawal probabilities are given by the following table: t 0 1 2 3

(d)

qx+t 0.10 0.15 0.20 0.25

(w)

qx+t 0.30 0.25 0.10 0.10

Assuming that the initial asset share for the policy is 15, find the asset share amounts at times t = 1, 2, 3. SOLUTION: [0 AS + G(1 [15 + 200(0.9)

c0 )

) e0 ] (1 + i) = qx(1) · b1 + qx(2) · 1 CV + p(⌧ x · 1 AS

20] (1 + 0.06) = (0.1) · 1000 + (0.3)(100) + (0.6) · 1 AS ) 1 AS = 92.5

Similarly, we can recursively produce the following: t 0 1 2 3

t AS

15 92.5 112.8 84.5 }

Chapter 15 Suggested Problems: 5,6,10 Arch MLC, Spring 2009

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342

Chapter 10

CHAPTER 15 Formula Summary There are not a lot of formulas for the expense-augmented premium sections of this chapter. Remember to treat expenses as a benefit that has to be paid each year.

Asset Shares [k AS + G(1

ck )

(1)

(2)

(⌧ )

ek ] (1 + i) = qk · bk+1 + qk · k+1 CV + px+k · k+1 AS

Past SOA/CAS Exam Questions: 1. For a special fully discrete whole life insurance on (x), you are given: • The net single premium is 450.

• The level annual expense loaded premium determined by the equivalence principle is 85. • Death is the only decrement.

• Expenses, which occur at the beginning of the policy year, are as follows: First Year Renewal Years Percentage of Premium 80% 10% Per policy 25 25 Calculate ax . (A) 8.06

(B) 8.89

Arch MLC, Spring 2009

(C) 9.06

(D) 9.89

c Yufeng Guo

(E) 10.06

343

Chapter 10 2. For a fully discrete 3-year endowment insurance of 1000 on (x), you are given: • i = 0.10 • Expenses, which occur at the beginning of the policy year, are as follows: Percentage of Premium Per policy

First Year Renewal Years 20% 6% 8 2

• The gross annual premium is equal to 314. • The following double-decrement table: (⌧ )

(d)

(w)

k px+k qx+k qx+k 0 0.54 0.08 0.38 1 0.62 0.09 0.29 2 0.50 0.50 0.00 • The following table of cash values and asset shares: k 0 1

k+1 CV

247 571

k AS

0 173

Calculate 2 AS. (A) 257

(B) 326

(C) 415

(D) 423

(E) 452

3. For a fully discrete whole life insurance of 1000 on (40), you are given: • • • •

Death and withdrawal are the only decrements. Mortality follows the Illustrative Life Table. i = 0.06 The probabilities of withdrawal are: (w) q40+k

=

(

0.2, 0,

k=0 k>0

• Withdrawals occur only at the end of the year. The following expenses are payable at the beginning of the year: Percent of Per 1000 Premium Insurance All Years 10% 1.50 • k CV 40 =

• 2 AS = 24

1000k · k V40 , k  3 3

Calculate the gross premium, G. (A) 15.33

(B) 15.79

Arch MLC, Spring 2009

(C) 16.25

(D) 16.77

c Yufeng Guo

(E) 17.02 344

Chapter 10 Use the following information for numbers 4 through 7. For a semi-continuous 20-year endowment insurance of 25,000 on (x), you are given: • The following expenses are payable at the beginning of the year: Percent of Per 1000 Premium Insurance Per Policy First Year 25% 2.00 15.00 Renewal 5% 0.50 3.00 • Deaths are uniformly distributed over each year of age. • A x:20 = 0.4058 • A

1 x: 20

= 0.3195

• a ¨x:20 = 12.522 • i = 0.05

• Premiums are determined using the equivalence principle.

4. Calculate the expense-loaded first-year premium including policy fee assuming that per-policy expenses are matched separately by first-year and renewal policy fees. (A) 884

(B) 899

(C) 904

(D) 909

(E) 924

5. Calculate the expense-loaded renewal premiums including policy fee assuming that perpolicy expenses are matched separately by first-year and renewal policy fees. (A) 884

(B) 887

(C) 899

(D) 909

(E) 912

6. Calculate the level annual policy fee to be paid each year. (A) 3.00

(B) 3.60

(C) 3.82

(D) 4.24

(E) 4.51

7. Calculate the level annual expense-loaded premium. (A) 884

(B) 888

Arch MLC, Spring 2009

(C) 893

(D) 909

(E) 913

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345

Chapter 10 8. For a fully discrete whole life insurance of 100, 000 on (35) you are given: (i) Percent of premium expenses are 10% per year. (ii) Per policy expenses are 25 per year. (iii) Per thousand expenses are 2.50 per year. (iv) All expenses are paid at the beginning of the year. (v) 1000P35 = 8.36 Calculate the level annual expense-loaded premium using the equivalence principle. (A) 930

(B) 1041

(C) 1142

(D) 1234

(E) 1352

9. For a fully discrete whole life insurance of 1000 on (x): (i) Death is the only decrement. (ii) The annual benefit premium is 80. (iii) The annual contract premium is 100. (iv) Expenses in year 1, payable at the start of the year, are 40% of contract premiums. (v) i = 0.10 (vi) 1000 · 1 Vx = 40 Calculate the asset share at the end of the first year. (A) 17

(B) 18

(C) 19

(D) 20

(E) 21

10. For a fully discrete whole life insurance of 1000 on (50), you are given: (i) The annual per policy expense is 1. (ii) There is an additional first year expense of 15. (iii) The claim settlement expense of 50 is payable when the claim is paid. (iv) All expenses, except the claim settlement expense, are paid at the beginning of the year. (v) Mortality follows De Moivre’s law with ! = 100. (vi) i = 0.05 Calculate the level expense-loaded premium using the equivalence principle. (A) 27

(B) 28

Arch MLC, Spring 2009

(C) 29

(D) 30

(E) 31

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Chapter 10

SOLUTIONS to Past SOA-CAS Exam Problems: 1. 85¨ ax = 450 + 0.7(85) + 0.1(85)¨ ax + 25¨ ax a ¨x =

85

450 + 0.7(85) = 9.89 (0.1)(85) 25

ax = a ¨x

1 = 8.89

The Key is B

2. 2 AS

=

[1 AS + G(1

c1 )

(d)

e1 ] (1 + i)

(w) 2 CV qx+1

1000qx+1

(⌧ )

px+1

[173 + 314(1

0.06)

2] (1.1) 1000(0.09) 0.62 = 414.82

571(0.29)

The Key is C

3. NOTE: Although it is not a strictly necessary assumption in actuarial science, this problem requires you to assume that 0 AS = 0. Therefore, on the exam, it is a good idea to assume the initial asset share is equal to zero unless otherwise stated. k+1 AS

(⌧ )

· px+k = [k AS + G(1

ck )

ek ] (1 + i)

(1)

1000qx+k

(2)

qx+k · k+1 CV

The mortality comes from the single life table (even though there are two decrements), while withdrawal is given in the problem: (1)

q40 = 0.00278,

(2)

q40 = 0.2

And we can use the illustrative life table to calculate: 1 CV40

✓

1000 1000 = · 1 V40 = 1 3 3

) 1 AS(0.99722)(0.8) = [0 + G(0.9)

a ¨41 a ¨40

◆

✓

1000 = 1 3

1.50] (1.06)

) 1 AS(0.79778) = G(0.954) ) 1 AS = 1.196G

2 AS(0.99702)

Arch MLC, Spring 2009

= [1 AS + G(0.9)

◆

1000(0.00278)

= 2.947

0.2(2.947)

4.96

6.22

1.50] (1.06)

c Yufeng Guo

14.686 14.817

1000(0.00298) 347

Chapter 10 ) (24)(0.99702) = [2.096G

7.72](1.06)

2.98

) G = 15.79

Key: B

4. Since the policy fee is to be split between first-year and renewal premiums with the fees allocated accordingly, we first want to calculate the level expense-loaded premium necessary to pay the benefit and expenses not covered by the policy fee. We will call this premium K. K¨ ax:20 = 25,000A x:20 + 25,000(0.0015 + 0.0005¨ ax:20 ) + 0.2K + 0.5K¨ ax:20 Note that 5 of the per premium expense for the first year was carved out to go with the renewal amounts to make the annuities we have to deal with more simple. The same was done with the expenses that depend on face amount. )K=

25[405.80 + 1.5 + 0.5(12.522)] 0.95(12.522) 0.2 ) K = 883.99

They sneakily made this one of the options, but we need to add the expenses loaded first year policy fee which must pay the first year per-policy expense (15) and must take premium loading into account. G1 = 883.99 +

15 = 903.99 1 0.25

Key: C

5. For renewal years, the same base premium K is required to cover all but the policy fee (3) plus we need 3/(0.95) to cover the policy fee. G2 = 883.99 +

1

3 = 887.15 0.05

Key: B

6. Now we assume instead that a level policy fee g is collected each year: g¨ ax:20 = 12 + 3¨ ax:20 + 0.2g + 0.05g¨ ax:20 g=

12 + 3(12.522) = 4.24 (0.95)(12.522) 0.2

Key: D

7. G = K + g = 883.99 + 4.24 = 888.23 Arch MLC, Spring 2009

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Key: B 348

Chapter 10 8. Key: D Let G be the expense-loaded premium. Actuarial present value (APV) of benefits = 100, 000A35 APV of premiums = G¨ a35 APV of expenses = [0.1G + 25 + (2.50)(100)]¨ a35 Equivalence principle: G¨ a35 = 100, 000A35 + (0.1G + 25 + 250)¨ a35 A35 G = 100, 000 a¨35 + (0.1G + 275) 0.9G = 100, 000P35 + 275 G = (100)(8.36)+275 = 1234 0.9 9. Key: A 10001 Vx = ⇡(1 + i) qx (1000 40 = 80(1.1) qx (1000 40) qx = 8896040 = 0.05 (G expenses)(i+1) 1000qx 1 AS = px = =

10001 Vx )

(100 (0.4)(100))(1.1) (1000)(0.05) 1 0.05 60(1.1) 50 = 16.8 0.95

10. Key: E Let G be the expense-loaded premium. Actuarial present value (AP V ) of benefits = 1000A50 . AP V of expenses, except claim expense = 15 + 1 ⇥ a ¨50 AP V of claim expense = 50A50 (50 is paid when the claim is paid) AP V of premiums = G¨ a50 Equivalence principle: G¨ a50 = 1000A50 + 15 + 1 ⇥ a ¨50 + 50A50 50 +15 G 1 = 1050A a ¨50 For De Moivre’s with ! = 100, x = 50 A50 = a ¨50 = 1 dA50 = 13.33248 Solving for G, G = 30.88

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a 50 50

= 0.36512

349

Chapter 10

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Chapter 11

DANIEL CHAPTER 1 MULTI-STATE TRANSITION MODELS FOR ACTUARIAL APPLICATIONS Once you think about it, and get used to the terminology, a Multi-State Transition Model is very well named. For starters, it’s a model which has more than one state in it – such as living vs. dead, or healthy vs. partially impaired vs. totally disabled. Furthermore, the model contains the likelihood of moving back and forth (or “making transitions”) among these states. (For the living vs. dead model, you are already familiar with qx , the probability of “transition” from the living state to the dead state.) The living versus dead model is not a perfect example of a transition model because the transitions are usually regarded as being only one way (for insurance purposes). Transition models are really designed to handle more complex situations when it is possible to move back and forth between various states. Daniel’s study note does a good job of introducing the concept (Chapter 1), layering in costs associated with the movements among states (Chapter 2), and tying it all together through examples throughout the note.

1.1 Introduction Examples 1-6 in the study note outline several multi-state models, moving for simple to more complex. As you read through these, it should be useful to sketch a diagram of each model discussed. For all the models, the process of moving between the states over time is called a Markov process. We will introduce the concepts of homogeneous and non-homogeneous Markov Chains through an example and then make the definitions and terminology more precise afterward. EXAMPLE: NEWCO Auto Insurance 1 The NEWCO Auto Insurance company classifies its policyholders into two groups: Hi Risk and Low Risk. Each year a driver is assigned to one of the two categories 351

Chapter 11 based on his status for the previous year and his driving experience for the previous year. Thus, for each policyholder there are two states: State 0: State 1:

Low Risk High Risk

Suppose that each year, for a person in State 0, the probability of remaining in State 0 is 0.85 and the probability of moving to State 1 is 0.15. These are referred to as the transition probabilities for State 0 and using the symbols in the study note, we have: Q(0,0) = 0.85, Q(0,1) = 0.15 For drivers currently in State 1, we will assume that Q(1,0) = 0.3,

Q(1,1) = 0.7

Here is a transition diagram that represents the process: 0.15

Low Risk

0.30

High Risk

It is handy to be able to represent all of these transition probabilities at once in matrix form, the result is the transition probability matrix: Q=

0.85 0.15 0.30 0.70

We will see later that the transition probability matrix is extremely useful for modeling the transition process over time. Two Important Notes • The probability that a policyholder will be in State 1 next year depends only on what state they are in this year. It does not matter what state they were in last year or any other year. This makes the process a memoryless process and is what defines a Markov process. • In this example we have assumed that the probability of moving from State 0 to State 1 is the same in every year. If the transition probability matrix remains the same in every period, the process is called a homogeneous Markov Chain (or just Markov chain). In reality, the transition probability matrix is likely to change over time. For instance, a policyholder with the company for 20 years might have a lower probability of moving from Low Risk to High Risk than a brand new policyholder. If the transition probability matrix changes over time, we say the process is a non-homogeneous Markov chain and we use the symbol Q(i,j) n to represent the probability that a policyholder in State i in period n will move to State j for period n + 1. (See next example.) } Arch MLC, Spring 2009

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Chapter 11

EXAMPLE: NEWCO Auto Insurance 2 For a young policy holder (x), NEWCO models the risk class of the policyholder using a non-homogeneous Markov Chain with 2 states, Low Risk and High Risk. The transition probability matrices for the first 5 years (time periods 0 to 4)are: Q0 =

0.7 0.3 0.2 0.8 Q3 =

,

Q1 =

0.8 0.2 0.4 0.6

0.8 0.2 0.2 0.8

,

Q4 =

,

Q2 =

0.8 0.2 0.3 0.7

0.9 0.1 0.5 0.5

1. (Easy) If the policyholder is in State 0 during Period 2, what is the probability they will be in State 1 during Period 3? 2. If the policy holder is in State 0 during Period 3, what is the probability they will be in State 1 during Period 5? SOLUTION: 1. This is (0,1)

Q2

= 0.2

2. The symbol for what we are looking for here is (0,1) 2 Q3

The superscript (0, 1) refers to the transition we are interested in (from State 0 to State 1) as usual; the 3 subscript refers to the time period in which we assume we start in State 0; and the 2 subscript before the Q refers to how many time periods are involved in the transition. There are two ways to approach this problem. First, we could note that to get from state 0 to state 1 in 2 periods there are two possibilities. The policyholder must pass through one of the following sequences of states at periods 3, 4, and 5. 0 !0 !1 0 !1 !1 Therefore, using the matrices above, the probability of being in State 1 in period 5 is (0.8)(0.1) + (0.2)(0.5) = 0.18 Arch MLC, Spring 2009

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Chapter 11 Be sure you recognize where I got each of the numbers in the line above. In the notation we have been using, the quantity we want is (0,1) 2 Q3

(0,0)

= Q3

(0,1)

· Q4

(0,1)

+ Q3

(1,1)

· Q4

This notation reveals the second approach that can be used to solve this problem. It turns out that you can form the 2-step transition matrix 2 Q3

by matrix-multiplying Q3 and Q4 . That is 2 Q3

= Q3 · Q4 =

0.8 0.2 0.4 0.6

·

0.9 0.1 0.5 0.5

=

0.82 0.18 0.66 0.34 (0,1)

Now we can read o↵ the probability we want, which is 2 Q3

= 0.18.

Important Note: The fact that we can obtain a k-step transition matrix by multiplying k 1-year transition matrices is what makes Markov Chains useful. In the most general notation: k Qn

= Qn Qn+1 · · · Qn+k

1

Reread these two examples if necessary. You should be ready now for the more careful mathematical introduction in the next section.

1.2 Non-homogeneous Markov Chains From the text, we have the following definition: M is a non-homogeneous Markov Chain when M is an infinite sequence of random variables M0 , M1 , . . . with the following properties: • Mn denotes the State number at time n. • Each Mn is a discrete-type random variable over r values. • The transition probabilities are history independent: Q(i,j) = P r[Mn+1 = j|Mn = i] n

(i,j)

The transition probability Qn should be interpreted as the probability of moving to state j at time n + 1 if you are in state i at time n. Also, history independent means that it doesn’t matter which states have been visited in the past. All that matters is where you are at time n.

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Chapter 11 EXAMPLE: Mortality as a Non-homogeneous Markov Chain Human mortality can be considered to be a non-homogeneous Markov chain with two states, Alive and Dead. Suppose q65 = 0.05. This number gives us the probability of moving from the Alive state of living at age 65 to the state of dead at age 66. In the language of Markov chains, we would say (alive,dead)

q65 = Q0

= 0.05

If we were to define State 0 to be the Alive state and State 1 to be the Dead state, we would have the more familiar form: (alive,dead)

q65 = Q0

(0,1)

= Q0

= 0.05

The transition probability matrix for this example is Q0 =

0.95 0.05 0.00 1.00

and in general for a basic survival model, Qn =

px+n qx+n 0.00 1.00

Note that State 1 is special. You can enter, but you can never leave. More Notation: (i)

(i,i)

• Pn = Qn is the probability of staying in state i over the next unit of time. So for (0) (1) the mortality example above, Pn = px+n for all values of n and Pn = 1 for all values of n. • The probability that a subject in state i at time n remains in that state through time n + k is (i) k Pn

(i)

(i)

= P r[Mn+1 = Mn+1 = · · · = Mn+k = i|Mn = i] = Pn(i) Pn+1 · · · Pn+k

1

• If we’re interested in transition probabilities across more than a single time period (and we are), we need (i,j) = P r[Mn+k = j|Mn = i] k Qn We use k Qn to denote the entire k-step r-by-r transition probability matrix in which entry (i, j) represents the probability of being in State j at time n + k given that we (i,i) are in State i at time n. Note that k Qn represents the probability of being in State i after k time periods given that we are in State i at time n. This is very di↵erent from (i) (i,i) allows for the possibility of leaving the state and re-entering it later, k Pn since k Qn (i) whereas k Pn allows only the case where the subject remains in State i throughout the k periods. Arch MLC, Spring 2009

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Chapter 11 You should be able to reason why it must be true in all cases that (i) k Pn

 k Q(i,i) n

As we saw in the NEWCO Auto Insurance 2 example above, the following provides a handy way to evaluate k-step probabilities for Markov Chains: Theorem 1.18 k Qn

= Qn · Qn+1 · · · Qn+k

1

Or, in the case of a homogeneous Markov chain, k Qn

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= Qk

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356

Chapter 11 EXAMPLE: Let Q be the transition probability matrix for a homogeneous Markov chain.

Q=

0.9 0.1 0 0.4 0.5 0.1 0.3 0.2 0.5

(A) How many di↵erent states does the Markov chain represented by this matrix have? (B) The first time that the process enters State 2, what is the probability it will next be in State 0? (C) The fifth time that the process enters State 2, what is the probability it will next be in State 0? (D) If the process is currently in State 1, what is the probability that the process will be in State 1 two periods from now? (E) If the process is currently in State 1, what is the probability that the process will stay in State 1 throughout the next two periods? SOLUTION: (A) 3; usually we would call these States 0,1, and 2 with State 0 being the one in which the process begins. (B) Q(2,0) = 0.3 (C) 0.3 - the answer has to be the same as the one in (B) because this is a Markov Chain. (D) We could just multiply Q times itself and find the(1,1) entry. A faster method that works for the kinds of questions that end up on the exam is the following: Let (pn , qn , rn ) be a vector that represents the probabilities of being in states 0, 1, and 2, respectively at time n. So we start out with (p0 , q0 , r0 ) = (0, 1, 0) since we are certain that we are in State 1 at time t = 0. Now (p1 , q1 , r1 ) = (0, 1, 0) ·

0.9 0.1 0 0.4 0.5 0.1 0.3 0.2 0.5

= (0.4, 0.5, 0.1)

The vector on the right shows the probabilities of being in States 0, 1, and 2 at time t = 1. To get the probabilities of being in each state at time t = 2: (p2 , q2 , r2 ) = (0.4, 0.5, 0.1) · Arch MLC, Spring 2009

0.9 0.1 0 0.4 0.5 0.1 0.3 0.2 0.5

c Yufeng Guo

= (0.59, 0.31, 0.1) 357

Chapter 11 Now we can read o↵ 0.31 as the probability of being in State 1 at time t = 2. This method is especially handy if you need to find two of the probabilities for a couple of time periods after start. Also, note that you can check your work by making sure the 3 probabilities in the state probability vector add to 1 at each step. (E) This is just

h

Q(1,1)

i2

= (0.5)2 = 0.25

}

EXAMPLE: For a certain Markov chain, the transition matrix given by P=

0.9 0.1 0.4 0.6

If the process is in State 0 at time t = 4, what is the probability that the process will be in State 1 at time t = 7, three periods later? SOLUTION: At t = 4, the probability vector is (1,0): (1, 0) ·

0.9 0.1 0.4 0.6

= (0.9, 0.1);

(0.85, 0.15) ·

(0.9, 0.1) · 0.9 0.1 0.4 0.6

0.9 0.1 0.4 0.6

= (0.85, 0.15)

= (0.825, 0.175)

From this vector, we can see that the probability of being in State 1 at time t = 7 is 0.175. }

Chapter 1 Suggested Problems: 1-8 (Solutions at archactuarial.com) Also, we strongly recommend that you read all of Daniel’s examples. Focus especially on the CCRC example since that has been a favorite with exam makers in the past.

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Chapter 11

CHAPTER 2 – CASH FLOWS AND THEIR ACTUARIAL PRESENT VALUES Section 2.1 Introduction Chapter 2 shows how the concepts of Chapter 1 can be fit into an actuarial framework involving present values of future contingent cash flows. There are two basic types of cash flows we need to be able to deal with: • Cash Flows While in States - This is the most natural type of cash flow and the easiest to deal with as far as calculating actuarial present values (APVs). Two Examples: 1. Premium payments for Hi Risk and Low Risk drivers. To figure out the APV of a policyholders future premium payments, you need to figure out how many periods that policyholder expects to spend in each state, calculate the premium due at each time, and discount for the appropriate number of time periods. 2. Disability insurance payments payable to an insured during each time period that they are in the disabled state. • Cash Flows upon Transitions - These are cash flows that are payable specifically upon transition from one state to another. For example, life insurance makes a one-time payment upon transition from the Living state to the Dead state. Calculating the APV of Cash Flows upon Transitions is a bit more complicated than for Cash Flows while in States. We will deviate from the study note author’s approach and start with the easier case of Cash Flows while in states. Also, the concepts of benefit premiums and benefit reserves will be introduced throughout the chapter rather than separately at the end.

Cash Flows while in states Notation: Let l C (i) denote the cash flow at time l if the subject is in State i at time l. For example, if State 1 is the state of being a Low-risk driver, then 5 C (1) could be used to represent the premium due from a policyholder at time t = 5 if they are in the Low-risk state at that time. Suppose there are two possible states, Low-risk (State 1) and High-risk (State 2), and we are now at time t = 0 with a policy holder in State 2. Then the premium we expect to receive at time t = 5 is h

(2,1)

5 Q0

i h

·

5C

(1)

i

+

h

(2,2)

5 Q0

i h

·

5C

(2)

i

That messy formula is just the payment for each state times the probability of being in that state at time t = 5. To calculate the APV of this single payment at time t = 0, we need to discount the payment for 5 years, or AP V0 =

h

(2,1)

5 Q0

i h

·

5C

(1)

i

v5 +

h

(2,2)

5 Q0

i h

·

5C

(2)

i

v5

That, in a nutshell, is all you need to know for this section. The goal is to find the APV of all future payments for di↵erent states given that we start in a certain State i at time t = n. Arch MLC, Spring 2009

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Chapter 11 Before writing down the one big formula for this section, it will be helpful to collect all of the formal notation into one place. • l C (i) - Cash flow due at time l if the subject is in State i.

• k Q(s,i) - The probability of being in State i at time t = n + k if in State s at time t = n. n • AP Vs@n (C (i) ) - The Actuarial Present Value at time t = n of all future payments to be made while in State i, given that the subject is in State s at time t = n. • k vn - The present value of 1 paid k years after time t = n. Theorem 2.12 APV of cash flows while in states Let l C (i) denote the cash flow at time l if the subject is in state i at time l. Suppose that the subject is in state s at time n. Then the APV of these cash flows, as seen from time n, is AP Vs@n (C (i) ) =

1 X

k=0

(i) [k Q(s,i) n ] · [n+k C ] · [k vn ]

To interpret this formula, it is helpful to recall a very similar formula from Chapter 4 of the Bowers text. interest discount prob of death between times k and k + 1 z }| { 1 X z }| { k+1 Ax = v · · qx+k k px |{z}

0

survival from 0 to k Note that we’re summing, over all possible times of occurrence, the present value of a cash flow (benefit of 1). In this study note, we are doing the same type of calculation but the cash flow is we’re calculating the APV of a C (i) instead of 1. In similar terms, our formula is: AP Vs@n (C ) = (i)

1 int. discount X z}|{ k vn

0

cash flow amount ·

(s,i) k Qn

| {z }

prob of moving from s to i in k steps

·

z }| { n+k C

(i)

EXAMPLE: NEWCO 3 (Work this one before looking at the solution!) For the NEWCO Auto Insurance 2 example given above in section 1.1, recall that we had a non-homogeneous Markov Chain and we were provided the first 5 transition probability matrices, Q0 , Q1 , ..., Q4 . We were told the Policyholder begins in State 0. We are now given the additional information that premiums are equal to 500 per year for those in the Low Risk state and 800 per year for those in the High Risk state. Premiums are payable at the beginning of each year and the interest rate for all years is i = 8%. Find the actuarial present value of future premiums for this policyholder for the first 5 years of the policy. Arch MLC, Spring 2009

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Chapter 11 SOLUTION: We could do this exactly according to the formula in Theorem 2.12, but it will be easier on exam day to apply probability state vectors. In Year 1, we are in State 0, so the premium is exactly 500. In Year 2, the probability state vector is (1, 0) · Q0 = (1, 0) ·

0.7 0.3 0.2 0.8

= (0.7, 0.3)

So the expected premium to be collected in Year 2 is (0.7)(500) + (0.3)(800) = 590 In Year 3, the state vector is (0.7, 0.3) · Q1 = (0.62, 0.38) So the expected premium for Year 3 is (0.62)(500) + (0.38)(800) = 614 Similar calculations result in the following table: Year 1 2 3 4 5

Premium 500 590 614 617 606.8

Now the APV of all future premiums is 500 + 590v + 614v 2 + 617v 3 + 606.8v 4 = 500 + 590(0.926) + 614(0.857) + 617(0.794) + 606.8(0.735) = 2509

}

Two comments on this example: • For most people (certainly for me) it is easier to reason through the APV as in the calculation above rather than to memorize a messy formula like the one in Theorem 2.12. • The APV of future premiums would have been di↵erent if the policyholder had begun in State 1. For practice, it might be a good idea for you to repeat the entire calculation in the example for that case.

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Chapter 11 The next example will be used to introduce the idea of benefit premiums. Also, we will consider for the first time, interest rates that can vary by year. EXAMPLE: Long Term Care 1 CARECO-LTC is a long term insurance company that sells 4-year Long-term care insurance policies to 75-year olds. CARECO-LTC models the status of a policyholder using a homogeneous Markov chain with 3 states: • State 1: Active

• State 2: Disabled • State 3: Dead

The transition probability matrix for the process is Q=

0.7 0.2 0.1 0.3 0.5 0.2 0.0 0.0 1.0

Each time period represents one year, benefits are payable at the beginning of the year and are equal to 20,000 per year spent in the disabled status. Interest rates are assumed to be 4% for the first year, 6% for the second, and 8% thereafter. (Try to work out each of the following without looking at the solution. Do this example twice if necessary. All parts of it involve essential skills for the exam.) 1. Premiums of P will be paid at the beginning of each year spent in the Active State. If a policyholder always begins in the Active state, find the APV of future premiums. 2. Again, assuming policyholders always begin in the Active state, find the APV of future benefits. 3. Find the benefit premium for this insurance.

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Chapter 11 SOLUTION: 1. It will be helpful to have the probability state vectors for all years. Let ⇡n be the probability state vector for period n, then ⇡0 = (1, 0, 0) 0.7 0.2 0.1 0.3 0.5 0.2 0.0 0.0 1.0

⇡1 = (1, 0, 0) ·

0.7 0.2 0.1 0.3 0.5 0.2 0.0 0.0 1.0

⇡2 = (0.7, 0.2, 0.1) ·

= (0.7, 0.2, 0.1)

= (0.55, 0.24, 0.21)

And similarly we can calculate ⇡3 = (0.457, 0.23, 0.313) The result is the following probability vectors: Year 0 1 2 3

⇡n (1, 0, 0) (0.70, 0.20, 0.10) (0.55, 0.24, 0.21) (0.46, 0.23, 0.31)

So we have the following table: Time t 0 1 2 No. of active people 1 0.7 0.55 No. of disabled people 0 0.2 0.24 No. of the dead people 0 0.1 0.21 The active people pay the premium. The The APV of all future premiums is (1)P +(0.7)

✓

◆

✓

3 0.46 0.23 0.31 disabled people get the benefits. ◆

✓

◆

1 1 1 1 1 1 P +(0.55) · P +(0.46) · · P 1.04 1.04 1.06 1.04 1.06 1.08 = 2.558P

2. The APV of all benefit payments is h

✓

◆ ✓ ◆ ✓ ◆i 1 1 1 1 1 1 0+(0.2) +(0.24) · +(0.23) · · (20,000) 1.04 1.04 1.06 1.04 1.06 1.08

= 12,064

3. The benefit premium is the premium such that the APV of future premiums equals the APV of future benefits, or 2.558P = 12,064 ) P = 4716 Arch MLC, Spring 2009

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Chapter 11

Cash Flows upon transitions So far, all of the cash flows we have evaluated have involved payments made due to being in a certain state for a particular time period. We also want to be able to value cash flows that are made upon transition from one state to another. For example, life insurance benefits aren’t paid for every period you are dead, just for the single period in which you move from the Alive state to the Dead state. Notation 2.5 – Cash flows at transition Let l+1 C (i,j) denote the cash flow at time l + 1 if the subject is in state i at time l and state j at time l + 1.

Actuarial Present Values The formula for calculating the APV of a cash flow C (i,j) if the person moves from state i to state j at any time in the future is very similar to the one for C (i) . The only di↵erence is in the term that refers to probability payment, we need to recognize that for the payment C (i,j) to occur at time l + 1, two things must happen. We have to be in State i at time l and we have to be in State j at time l + 1. Assuming we start out in some specific state s, the probability of this occurring is (s,i) (i,j) l Qn Qn+l The resulting formula for the APV of a cash flow to be paid upon transition from State i to State j is

AP Vs@n (C (i,j) ) =

1 int. discount X z }| { k+1 vn

0

cash flow amount ·

(i,j)

(s,i) k Qn Qn+k

|

{z

}

prob of moving from s to i, then i to j

·

z

}|

n+k+1 C

{

(i,j)

This formula is identical to the one for C (i) , except for the probability term mentioned above and the fact that interest is discounted to the time of payment a year later. Rearranging of the terms gives us Theorem 2.7: Let l+1 C (i,j) denote the cash flow at time l + 1 if the subject is in state i at time l and state j at time l + 1. Suppose that the subject is now in state s at time n. Then the APV of these cash flows, as seen from time n, is AP Vs@n (C (i,j) ) =

1 X 0

Arch MLC, Spring 2009

(i,j)

(i,j) [k Q(s,i) ] · [k+1 vn ] n Qn+k ] · [n+k+1 C

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Chapter 11 EXAMPLE: Long Term Care 2 CARECO-LTC has decided to add a death benefit to their product. If the policyholder dies while active, they will receive 50,000 for the period in which they enter the Dead state. They will receive 25,000 if they die while disabled. Find the additional premium necessary to fund the death benefit. SOLUTION: Recall that the state probability vectors for each time period were given by Year 0 1 2 3

⇡n (1, 0, 0) (0.70, 0.20, 0.10) (0.55, 0.24, 0.21) (0.46, 0.23, 0.31)

Death benefit payable in Period 2 can only occur as a transition from alive to dead and occurs with probability 0.1. The APV of 2nd-period death benefit is ✓

◆

1 (0.1)(50,000) = 4808 1.04

The present value of both types of death benefits for entering the dead state in Period 3 is ✓

◆h i 1 1 (0.7)(0.1)(50,000) + (0.2)(0.2)(25,000) = 4082 1.04 1.06

The present value of 4th period death benefits is ✓

◆h i 1 1 1 (0.55)(0.1)(50,000) + (0.24)(0.2)(25,000) = 3318 1.04 1.06 1.08

The present value of death benefit for dying upon transition from the 4th period to the 5th is ✓

◆h i 1 1 1 1 (0.46)(0.1)(50,000) + (0.23)(0.2)(25,000) = 2683 1.04 1.06 1.08 1.08

(Note that it makes sense that we need to calculate the APV of 4 di↵erent possible times of death since 4 premiums are paid.) So the total actuarial present value of all death benefits is 4808 + 4082 + 3318 + 2683 = 14,891

2.558P = 14,891 Arch MLC, Spring 2009

) P = 5821

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365

Chapter 11 Reserves could be calculated for cash flows upon transition just as they were for cash flows while in states. EXAMPLE: Long Term Care 3 CARECO-LTC is a long term insurance company that sells 3-year Long-term care insurance policies to 75-year olds. CARECO-LTC models the status of a policyholder using a homogeneous Markov chain with 3 states: • State 1: Active

• State 2: Disabled • State 3: Dead

The transition probability matrix for the process is Q=

0.7 0.2 0.1 0.3 0.5 0.2 0.0 0.0 1.0

Each time period represents one year. Premiums are paid by the active in the beginning of each year. 5,000 disability benefit is paid to the disabled at the end of each year. 10,000 death benefit is paid to the dead at the end of the year. Interest rate is 5%. 1. Find the benefit premium. 2. Using the retrospective method, find the benefit reserve 1 V for a policyholder that is Active at t = 1. 3. Using the prospective method, find the benefit reserve 1 V for a policyholder that is Active at t = 1. 4. Using the retrospective method, find the benefit reserve 2 V for a policyholder that is Active at t = 2. 5. Using the prospective method, find the benefit reserve 2 V for a policyholder that is Active at t = 2.

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Chapter 11 SOLUTION: 1. We previously produced the following table: Time t 0 1 2 3 No. of the active 1 0.7 0.55 0.457 No. of the disabled 0 0.2 0.24 0.230 No. of the dead 0 0.1 0.21 0.313 Total: active+disabled+dead 1 1 1 1 PV of future premiums is P (1 + 0.7v + 0.55v 2 ) PV of future disability benefits is 5, 000(0.2v + 0.24v 2 + 0.23v 3 ) PV of future death benefits is 10, 000[0.1v+(0.21 0.1)v 2 +(0.313 0.21)v 3 ] = 10, 000(0.1v+0.11v 2 +0.103v 3 ) Please note that PV of future death benefits is not 10, 000(0.1v + 0.21v 2 + 0.313v 3 ) The number of deaths in Row 4 is the accumulative deaths and the people who died in the previous year still stay in the population (“once dead, always dead“). This can be seen from the fact that the total population is always one at Row 5 even though death occurs each year. Since one death gets 10,000 only once, we need to calculate the incremental deaths per year. There’s 0.1 death in Year 1; 0.21 death in Year 1 and Year 2; and 0.313 death in Year 1, Year 2, and Year 3. As a result, there’s 0.1 death in Year 1; 0.11 death in Year 2; and 0.103 death in Year 3. Using Equivalence Principle, we get: P (1+0.7v+0.55v 2 ) = 5, 000(0.2v+0.24v 2 +0.23v 3 )+10, 000(0.1v+0.11v 2 +0.103v 3 ) P = 2712.54 2. Reserve at t = 1 using the retrospective method: 2712.54(1.05)

5, 000(0.2) 0.7

10, 000(0.1)

= 1211.67

3. Reserve at t = 1 using the prospective method: 5, 000(0.24v + 0.23v 2 ) + 10, 000(0.11v + 0.103v 2 ) 0.7 = 1211.67

2712.54(0.7 + 0.55v)

4. Reserve at t = 2 using the retrospective method: 2712.54(1.052 + 0.7 ⇥ 1.05)

5, 000(0.2 ⇥ 1.05 + 0.24) 0.55 = 1062.35

10, 000(0.1 ⇥ 1.05 + 0.11)

5. Reserve at t = 1 using the prospective method: 5, 000(0.23v) + 10, 000(0.103v) 0.55 Arch MLC, Spring 2009

c Yufeng Guo

2712.54(0.55)

= 1062.35

367

Chapter 11 This is a good time to point out that the Daniel note has a lot of examples worth reading. Note that the material in Chapter 3 is there only to have some matrices for reference by some examples in Chapter 1 and 2. We suggest that you read each of the examples carefully and work out the numbers he obtains. Useful problems are indicated below. Chapter 2 Suggested Problems: Page 15, #1-5, Page 18, #1-6, Page 20, #1-6 (Complete Solutions at archactuarial.com)

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Chapter 11

ARCH Warm-up Questions: 1. The Valley U. football team plays 1 game per week all year and has game results that follow a Markov process. If the team wins this week then the probability of winning next week is 0.8. If the team loses this week, then the probability of winning next week is 0.3. If Valley U. loses this week what is the probability that they will win two weeks from now? (A) 0.24

(B) 0.30

(C) 0.45

(D) 0.55

(E) 0.64

Use the following information for the next two questions: Janet is a world class sprinter. As time passes, it becomes more and more likely that she’ll drop from being a world class sprinter (state 0) to a decent sprinter (state 1). It is possible, however, with lots of training and hard work, to move from being a decent sprinter back to world class. For t = 1, . . . , 10, you are given: Qt =

1 0.5

(0,1)

Qt (0.1)t (1,0) (0.05)t 1 Qt

2. If Janet is a world class sprinter at time t = 4, what is the probability that she’ll still be a world class sprinter 4 periods later? (A) 0.086

(B) 0.114

(C) 0.162

(D) 0.192

(E) 0.213

3. What is the probability that she’ll remain a world class sprinter throughout the same time period? (A) 0.014

(B) 0.036

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(C) 0.051

(D) 0.074

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(E) 0.086

369

Chapter 11 Use the following information for the next two questions: An actuarial student has the following probabilities of changing titles (and salaries!) each year. The titles and salaries (paid in a lump sum at the beginning of each year) are as follows: • Actuarial Student (State 0), $40, 000 • Actuarial Analyst (State 1), $60, 000

• Associate Actuary (State 2), $80, 000 The probability matrix for career progression is as follows: Qt =

0.35 0.60 0.05 0 0.50 0.50 0 0 1

4. If Luke is just starting his career in State 0, what is the probability that he will be an Associate Actuary at the end of 3 years? (A) 0.42

(B) 0.49

(C) 0.56

(D) 0.63

(E) 0.70

5. Suppose that i = 0.07 and that Luke receives his full annual salary at the beginning of each year. What is the present value of Luke’s salary over the next three years? (A) 107,000

(B) 117,000

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(C) 127,000

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(D) 137,000

(E) 147,000

370

Chapter 11 Use the following information for the next two questions: You own a special two-year disability income policy. If you are healthy (state 0), you pay a full premium at the beginning of the year. If you are partially disabled (state 1), you pay a half premium. If you are fully disabled (state 2), you pay no premium. All transitions take place at year end. At the moment you move to the partially disabled state, you receive a benefit of 10,000. If you move to the fully disabled state, you receive a benefit of 40,000. If you are healthy, you receive no benefit. The probability matrix is as follows: Qn =

0.7 0.2 0.1 0.5 0.3 0.2 0.1 0.3 0.6

6. Ignoring the two year constraint on the policy, what is 3 Q(1,0) ? (A) 0.1

(B) 0.3

(C) 0.5

(D) 0.6

(E) 0.7

7. If the insured begins in the healthy state, what is the benefit premium for this two-year policy (i = 0.06)? (A) 7,800

(B) 8,700

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(C) 10,600

(D) 11,500

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(E) 12,400

371

Chapter 11

Solutions 1. The transition probability matrix for Valley U. is given by P =

0.8 0.2 0.3 0.7

Where State 0 represents a win this week. (0, 1) ·

0.8 0.2 0.3 0.7

= (0.3, 0.7)

(0.3, 0.7) ·

0.8 0.2 0.3 0.7

= (0.45, 0.55)

So the probability of winning two weeks after a loss is 0.45. Key: C (0,0)

2. SOLUTION: We’re looking for 4 Q4 signifying starting and ending in State 0, beginning at time t = 4, ending at time t = 8. 1 0 , so the probability of being in

The state probability vector for time t = 4 is state zero four periods later is: 1 0

·

0.60 0.40 0.30 0.70

·

0.50 0.50 0.25 0.75

·

1 0

·

0.60 0.40 0.30 0.70

=

0.6 0.4

·

0.5 0.5 0.25 0.75

0.4 0.6

·

0.4 0.6 0.2 0.8

0.28 0.72

·

0.3 0.7 0.15 0.85

0.40 0.60 0.20 0.80

=

=

=

·

0.30 0.70 0.15 0.85

0.6 0.4

0.4 0.6

0.28 0.72

0.192 0.808

So the answer is 0.192 – D. 3. SOLUTION: (0,0)

We’re looking for 4 P4 signifying staying in State 0, beginning at t = 4, ending at time t = 8. We can multiply: (0.6)(0.5)(0.4)(0.3) = 0.036 Arch MLC, Spring 2009

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372

Chapter 11 So the answer is B. 4. SOLUTION: 1 0 0

0.35 0.60 0.05 0 0.50 0.50 0 0 1

·

0.35 0.60 0.05

0.35 0.60 0.05 0 0.50 0.50 0 0 1

·

0.1225 0.5100 0.3675

·

=

=

0.35 0.60 0.05 0 0.50 0.50 0 0 1

0.35 0.60 0.05

0.1225 0.5100 0.3675

=

0.0429 0.3285 0.6286

so the answer is 0.6286 – solution D. 5. SOLUTION: He gets 40 (salary in 1000s) immediately. The APV of the payment at the beginning of Year 2 is: (0.35)(40) + (0.60)(60) + (0.05)(80) = 50.467 1.07 The present value for potential salaries at the beginning of year 3 is: (0.1225)(40) + (0.51)(60) + (0.3675)(80) = 56.686 1.072 So the combined APV for all 3 years is 40 + 50.467 + 56.686 = 147.153 The answer is E. 6. SOLUTION: We’re looking for the probability of being partially disabled today and being in the healthy state 3 periods later: 0 1 0

0.5 0.3 0.2 Arch MLC, Spring 2009

0.7 0.2 0.1 0.5 0.3 0.2 0.1 0.3 0.6

·

·

0.7 0.2 0.1 0.5 0.3 0.2 0.1 0.3 0.6 c Yufeng Guo

=

=

0.5 0.3 0.2

0.52 0.25 0.23 373

Chapter 11

0.52 0.25 0.23

·

0.7 0.2 0.1 0.5 0.3 0.2 0.1 0.3 0.6

=

0.512 0.248 0.240

The answer is C. 7. SOLUTION: The state probability vectors for times t=1 and t=2 are as follows: 1 0 0

0.7 0.2 0.1 0.5 0.3 0.2 0.1 0.3 0.6

·

0.7 0.2 0.1

·

0.7 0.2 0.1 0.5 0.3 0.2 0.1 0.3 0.6

=

=

0.7 0.2 0.1

0.60 0.23 0.17

The equivalence relation is: 



P (0.7) (0.2)(0.5 · P ) (10000)(0.2) + (40000)(0.1) (0.23)(10000) + (0.17)(40000) P+ + = + 1.06 1.06 1.06 1.062 (1.754717)P = 13759.34 P = 7841 The answer is A.

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Chapter 11

Past SOA/CAS Exam Questions: 1. For Shoestring Swim Club, with three possible financial states at the end of each year: (i) State 0 means cash of 1500. If in state 0, aggregate member charges for the next year are set equal to operating expenses. (ii) State 1 means cash of 500. If in state 1, aggregate member charges for the next year are set equal to operating expenses plus 1000, hoping to return the club to state 0. (iii) State 2 means cash less than 0. If in state 2, the club is bankrupt and remains in state 2. (iv) The club is subject to four risks each year. These risks are independent. Each of the four risks occurs at most once per year, but may recur in a subsequent year. (v) Three of the four risks each have a cost of 1000 and a probability of occurrence 0.25 per year. (vi) The fourth risk has a cost of 2000 and a probability of occurrence 0.10 per year. (vii) Aggregate member charges are received at the beginning of the year. (viii) i = 0 Calculate the probability that the club is in state 2 at the end of three years, given that it is in state 0 at time 0. (A) 0.24

(B) 0.27

(C) 0.30

(D) 0.37

(E) 0.56

Solution: P02 = P12 , and therefore it is not necessary to track states 0 and 1 separately. P12 = Pr [1 big or (no big and 2 or more smalls)] h

i

= 0.1 + (0.9) 3(0.25)2 (0.75) + (0.25)3 = 0.1 + 0.14 = 0.24 If you enter state 2, you stay there. Probability of not entering in 3 years = (1

0.24)3 = 0.44.

Probability of being in 2 after 3 years = 1

0.44 = 0.56.

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Key: E

375

Chapter 11 Use the following information for the next two questions: The Simple Insurance Company starts at time t = 0 with a surplus of S = 3. At the beginning of every year, it collects a premium of P = 2. Every year, it pays a random claim amount: Claim Amount Probability of Claim Amount 0 0.15 1 0.25 2 0.50 4 0.1 Claim amounts are mutually independent. If, at the end of the year, Simple’s surplus is more than 3, it pays a dividend equal to the amount of surplus in excess of 3. If Simple is unable to pay its claims, or if its surplus drops to 0, it goes out of business. Simple has no administrative expenses and its interest income is 0.

2. Determine the probability that Simple will ultimately go out of business. (A) 0.00

(B) 0.01

(C) 0.44

(D) 0.56

(E) 1.00

3. Calculate the expected dividend at the end of the third year. (A) 0.115

(B) 0.350

(C) 0.414

(D) 0.458

(E) 0.550

SOLUTION to 2: Simple’s surplus at the end of each year follows a Markov process with four states: State 0: out of business State 1: ending surplus 1 State 2: ending surplus 2 State 3: ending surplus 3 (after dividend, if any) State 0 is absorbing (recurrent). All other states are transient states. Thus, eventually Simple must reach state 0. Key: E SOLUTION to 3: (See solution to problem 2 for definition of states) t = 0: h

0 0 0 1

Arch MLC, Spring 2009

2

i6 6 6 4

1.0 0.1 0.1 0.0

0.0 0.5 0.0 0.1

0.00 0.25 0.50 0.00

0.00 0.15 0.40 0.90

3

7 h i 7 7 = 0.0 0.1 0.0 0.9 at t = 1 5

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Chapter 11 t = 1: h

0.0 0.1 0.0 0.9

2

i6 6 6 4

1.0 0.1 0.1 0.0

0.0 0.5 0.0 0.1

0.00 0.25 0.50 0.00

0.00 0.15 0.40 0.90

3

7 h i 7 7 = 0.01 0.14 0.025 0.825 at t = 2 5

Expected dividend at the end of the third year = 3 X

k=0

( probability in state k at t = 2) ⇤ (expected dividend if in state k)

0.01⇤0+0.14⇤0+0.025⇤(0⇤0.85+1⇤0.15)+0.825⇤(0⇤0.6+1⇤0.25+2⇤0.15) = 0.4575 Key D

4. A machine is in one of four states (F, G, H, I) and migrates annually among them according to a Markov process with transition matrix: F G H I

F 0.20 0.50 0.75 1.00

G 0.80 0.00 0.00 0.00

H 0.00 0.50 0.00 0.00

I 0.00 0.00 0.25 0.00

At time 0, the machine is in State F. A salvage company will pay 500 at the end of 3 years if the machine is in State F. Assuming v = 0.90, calculate the actuarial present value at time 0 of this payment. (A) 150

(B) 155

(C) 160

(D) 165

(E) 170

Solution: M = Initial state matrix = [ 1 0 0 0 ] 2 6 6 4

T = One year transition matrix = 6

0.20 0.80 0 0 0.50 0 0.50 0 0.75 0 0 0.25 1.00 0 0 0

3 7 7 7 5

M ⇥ T = [ 0.20 0.80 0 0 ]

(M ⇥ T ) ⇥ T = [ 0.44 0.16 0.40 0 ]

((M ⇥ T ) ⇥ T ) ⇥ T = [ 0.468 0.352 0.08 0.10 ] Arch MLC, Spring 2009

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Chapter 11 Probability of being in State F after three years = 0.468. Actuarial present value = 0.468v 3 (500) = 171 Key: E Notes: • Only the first entry of the last matrix need be calculated (verifying that the four sum to 1 is useful “quality control.”) • It would be valid to calculate T3 here, but advancing M one year at a time seems easier.

5. For a perpetuity-immediate with annual payments of 1: (i) The sequence of annual discount factors follows a Markov chain with the following three states: State number Annual discount factor, v

0 0.95

1 0.94

2 0.93

(ii) The transition matrix for the annual discount factors is: 2

3

0.0 1.0 0.0 6 7 4 0.9 0.0 0.1 5 0.0 1.0 0.0

Y is the present value of the perpetuity payments when the initial state is 1. Calculate E[Y ]. (A) 15.67

(B) 15.71

(C) 15.75

(D) 16.82

(E) 16.86

SOLUTION: The process begins in State 1 and is guaranteed to return to State 1 two steps later with the same prospective value then as we have now. Let E0 be the event that the first transition is to State 0 and let E2 be the event that the first transition is to State 2. h

) E[Y ] = E[Y |E0 ] · Pr[E0 ] + E[Y |E2 ] · Pr[E2 ] i

h

i

= 0.94 · 1 + 0.95 · (1 + E[Y ]) · 0.9 + 0.94 · 1 + 0.93 · (1 + E[Y ]) · 0.1 = 1.6497 + 0.8037 E[Y ] + 0.814 + 0.0874 E[Y ] 1.6497 + 0.1814 E[Y ] = = 16.82 1 0.8037 0.0874

Key: D

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Chapter 11 6. For a Markov model with three states, Healthy (0), Disabled (1), and Dead (2): (i) The annual transition matrix is given by 0 1 2

0 1 2 3 0.70 0.20 0.10 6 7 4 0.10 0.65 0.25 5 0 0 1 2

(ii) There are 100 lives at the start, all Healthy. Their future states are independent. Calculate the variance of the number of the original 100 lives who die within the first two years. (A) 11

(B) 14

(C) 17

(D) 20

(E) 23

SOLUTION: Ways to go 0 ! 2 in 2 years 0 0 2; p = (0.7)(0.1) = 0.07 0 1 2; p = (0.2)(0.25) = 0.05 0 2 2; p = (0.1)(1) = 0.1 Total = 0.22 Binomial m = 100 q = 0.22 V ar = (100)(0.22)(0.78) = 17 Key: C

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Chapter 11 7. An insurance company issues a special 3-year insurance to a high risk individual. You are given the following homogenous Markov chain model: (i) State 1: active State 2: disabled State 3: withdrawn State 4: dead Transition probability matrix: 1 2 3 4

2 1 6 6 6 4

2 3 4 0.4 0.2 0.3 0.1 0.2 0.5 0 0.3 0 0 1 0 0 0 0 1

(ii) Changes in state occur at the end of the year.

3 7 7 7 5

(iii) The death benefit is 1000, payable at the end of the year of death. (iv) i = 0.05 (v) The insured is disabled at the end of year 1. Calculate the actuarial present value of the prospective death benefits at the beginning of year 2. (A) 440

(B) 528

(C) 634

(D) 712

(E) 803

SOLUTION: A For death occurring in year 2 AP V = 0.3⇥1000 = 285.71 1.05 For death occurring in year 3, two cases: (1) State 2 ! State 1 ! State 4 : (0.2 ⇥ 0.1) = 0.02 (2) State 2 ! State 2 ! State 4 : (0.5 ⇥ 0.3) = 0.15 AP V = Total 0.17 Total AP V = 285.71 + 154.20 = 439.91. Key: A

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0.17⇥1000 1.052

= 154.20

380

Chapter 11 8. For a Markov model for an insured population: (i) Annual transition probabilities between health states of individuals are as follows: Healthy Sick Terminated

Healthy 0.7 0.3 0.0

Sick 0.1 0.6 0.0

Terminated 0.2 0.1 1.0

(ii) The mean annual health care cost each year for each health state is: Healthy Sick Terminated

Mean 500 3000 0

(iii) Transitions occur at the end of the year. (iv) i = 0. Calculate the expected future health care costs (including the current year) for an insured individual whose current state is healthy. Recall: a11 a12 a21 a22 where d = a11 a22

! 1

=

a22 /d a21 /d

a12 /d a11 /d

!

a12 a21 .

(A) 5100 (B) 5600 (C) 6100 (D) 6600 (E) 7100

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Chapter 11 SOLUTION: Non absorbing matrix T = which is an absorbing state. ! 0.3 0.1 I T = 0.3 0.4 (I

T)

1

=

0.4 0.09 0.3 0.09

0.1 0.09 0.3 0.09

!

=

0.7 0.1 0.3 0.6

!

, the sub-matrix excluding ”Terminated”,

4.44¯ 1.1¯1 3.3¯3 3.3¯3

!

Future costs for a healthy = 4.4¯ 4 ⇥ 500 + 1.1¯1 ⇥ 3000 = 5555.

Key: B

9. For a Markov model for an insured population: (i) Annual transition probabilities between health states of individuals are as follows:

Healthy Sick Terminated

Healthy 0.7 0.3 0.0

Sick 0.1 0.6 0.0

Terminated 0.2 0.1 1.0

(ii) The mean annual healthcare cost each year for each health state is:

Healthy Sick Terminated

Mean 500 3000 0

(iii) Transitions occur at the end of the year. (iv) i = 0. A contract premium of 800 is paid each year by an insured not in the terminated state. Calculate the expected value of contract premiums less healthcare costs over the first 3 years for a new healthy insured. (A)

390

(B)

200

(C)

20

(D) 160 (E) 340 Arch MLC, Spring 2009

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382

Chapter 11 SOLUTION: 0 1 0 1 0.7 0.1 0.2 0.52 0.13 0.35 B C B C T = @ 0.3 0.6 0.1 A T 2 = @ 0.39 0.39 0.22 A 0 0 1 0 0 1

Actuarial present value (A.P.V.) prem = 800(1 + (0.7 + 0.1) + (0.52 + 0.13)) = 1, 960. A.P.V. claim = 500(1 + 0.7 + 0.52) + 3000(0 + 0.1 + 0.13) = 1800. Di↵erence = 160 Key: D

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Chapter 11

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Chapter 12

DANIEL STUDY NOTE ON POISSON PROCESS 5.3 The Poisson Process First, go to the SOA website and download Jim Daniel’s study note on Poisson process. Jim’s study note is an easy read. Not to repeat what Daniel said, I’m going to explain Poisson process using Chapter 5 of the textbook Probability Models by Ross. However, you don’t need to have Ross’s textbook Probability Models to follow along. In the life contingencies book, the two most important mortality rules to keep track of are the constant force of mortality and DeMoivre’s Law. In the same way, the Poisson process is definitely the most important stochastic process to know and understand well in the probability book. (Markov is probably second!)

5.3.1 Counting Processes This section formalizes a few ideas that are fairly simple. A stochastic process is also a counting process if N (t) represents the total number of occurrences of some event up to time t. For example, if N (t) is the number of people who have entered a store at or before time t, then N (t), t 0, is a counting process. Some quick facts about a counting process N (t): 1. N (t)

0.

2. N (t) 2 {0, 1, 2, . . .}.

3. If s < t, then N (s)  N (t). 4. For s < t, N (t) interval (s, t].

N (s) equals the number of events that have occurred in the time

A counting process has independent increments if the number of events occurring in di↵erent time periods are unrelated. For example, if you believe the number of car accidents in July does not depend on the number in May, you would say the increments are independent. 385

Chapter 12 A counting process has stationary increments if the number of events in an interval depend only on the length of the interval. This is a stronger statement than that of independence. For the car accident example, a stationary increments assumption would imply that you expect July and May to have the same number of accidents since they have the same number of days.

5.3.2 Definition of the Poisson Process The most important counting process for this exam is the Poisson process. A Poisson Process is one for which P {N (s + t)

N (s) = k} = e

t(

t)k k!

k = 0, 1, . . . .

This is true for any time s and any length of time t. The constant parameter of the process.

is called the rate or

Notice that in the right hand side of the formula, s doesn’t show up at all. This tells us that no matter when we start a period of time of length t, the distribution for the number of occurrences is the same. So a Poisson Process has stationary increments. EXAMPLE: Car Counting You have been assigned to watch cars pass at a rural intersection and to model the number of cars passing as a Poisson process, N (t), with parameter 3 where t is measured in minutes. 1. What is the probability that 0 cars will pass by in the next 2 minutes? 2. Same question for 1, 2, 3, 4 cars respectively. SOLUTION: 1. P {N (2) = 0} = e 2.

t(

t)0 =e 0!

3·2 (3

· 2)0 =e 0!

P {N (2) = 1} = e

6

.

1 6 (6)

= 6e 6 1! (6)2 P {N (2) = 2} = e 6 = 18e 6 2! (6)3 P {N (2) = 3} = e 6 = 36e 6 3! (6)4 P {N (2) = 4} = e 6 = 54e 6 . 4!

}

We could use a bunch of calculus to prove the following important fact that is true for any Poisson Process: E[N (t)] = t and Var[N (t)] = t. So, in the example above, the expected number of cars to pass p by each minute is 3 and the standard deviation in the number that pass each minute is 3. Arch MLC, Spring 2009

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Chapter 12

5.3.3 Interarrival and Waiting Time Distributions Continuing with the cars at the intersection in the last example, there is another important random variable closely related to the Poisson process for the number of cars that have passed. It is called the interarrival time between occurrences – the waiting time between events. So for the intersection, the interarrival time random variable Tn is concerned with the time we have to wait after one car passes for the next one to arrive. It turns out that if the random variable for number of cars passing is Poisson with rate , then the random variable for the interarrival time is a continuous random variable and is exponentially distributed with mean 1 . In the car example, the waiting time is exponentially distributed with mean 13 (minutes). The expected waiting time between cars is 20 seconds, which makes sense if we expect to see 3 per minute. More formally: Let T1 be the first time an event in a Poisson process occurs. For n > 1, let Tn be the time since occurrence Tn 1 . Given a Poisson process, Tn , n = 1, 2, . . . are independent random variables with an exponential distribution with mean 1 . Let Sn be the time of the n-th occurrence of the event that is being modeled as a Poisson process. It is clear that Sn =

n X

Ti for n

1.

i=1

It turns out that Sn has a gamma distribution with parameters n and . This is less important than the fact that the interarrival time is exponentially distributed but you still should know it for the exam. The probability density function for Sn is fSn (t) = e FSn (t) =

t

1 X

e

( t)n 1 . (n 1)! t(

j=n

E[Sn ] =

n

,

t)j . j!

Var[Sn ] =

n 2

.

Note: For the exam, you will be given the probability density functions for Poisson, Exponential, and Gamma distributions. So, it is up to you whether to memorize the formulas or not – but it certainly cannot hurt to know these by memory (in addition to the mean and variance for each) so they will be automatic when you need them. I would probably memorize the ones for Poisson and Exponential, and use the formula sheet for Gamma. In any case, it is essential that you become very familiar with the relations between the Poisson process, the interarrival time, and the waiting time distributions and that you are able to work your way between them comfortably. Note that if you memorize the Gamma distribution stu↵, then you already have the exponential distribution stu↵ by letting n = 1 in the formulas for the Gamma Distribution. Arch MLC, Spring 2009

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Chapter 12 EXAMPLE: The following questions refer to the car counting question above. 1. If a car has just passed, what is the probability of not seeing another car for 60 seconds? 2. If no car has passed for the last 20 seconds, what is the probability of not seeing another car for 60 seconds? 3. What is the expected waiting time for the arrival of the next car? 4. What is the variance of the waiting time until the arrival of the next car? SOLUTION: 1. For an exponential distribution with mean equal to 1 , F (t) = 1 implies that the probability of seeing no cars for 1 minute is 1

F (1) = e

(1)

= e(

3)(1)

e

t.

This

= 0.0498.

2. This is a memoryless process, so the answer must be the same as in (1) ! 0.0498. 3. This is the mean =

1

= 13 , or 20 seconds.

4. The variance for an exponential distribution with mean ✓ ◆2 1

1

is

1 = , or 6.67 seconds. 9

}

EXAMPLE: The following questions also refer to the car counting question. 1. What is the expected waiting time until the 10th car passes? 2. What is the variance in the time until the 10th car passes? 3. What is the probability that S4  2? SOLUTION: 1. E[S10 ] =

10

=

10 3

minutes.

2. Var[S10 ] = Arch MLC, Spring 2009

10 2

=

c Yufeng Guo

10 . 9 388

Chapter 12 3. You can use the Gamma distribution, or you can note that this is the same as the probability that, for the Poisson Process, N (2) 4. (That is, the 4th car pass in two minutes or less if and only if the number of cars passing in the next two minutes is 4 or more. Pr[N (2) =1

4] = 1

Pr[N (2) < 4]

(Pr[N (2) = 0] + Pr[N (2) = 1] + Pr[N (2) = 2] + Pr[N (2) = 3])

These are exactly the numbers we calculated in the original car counting example. So our answer is 1

e

6

=1

6e

6

61e

18e 6

6

36e

6

= 0.849

}

5.3.4 Further Properties of Poisson Processes Suppose that customers arrive at a certain convenience store at a Poisson rate of 20 per hour. So N (t), the number of people arriving at the store before time t, is a Poisson process with parameter = 20. Furthermore, suppose that each person that enters the station is male with probability p and is female with probability 1 p. Finally, let N1 (t) and N2 (t) represent the number of men and women, respectively, that enter the store before time t. Then N1 (t) and N2 (t) are both Poisson processes with parameters p and (1 p) , respectively. In addition, although it is somewhat surprising, the two random variables N1 (t) and N2 (t) are independent as well. Even if 50 men enter the store in the next hour, the expected number of women that will enter the store is still (1 p) . EXAMPLE: An emergency veterinary clinic cares for only cats and dogs. Animals arrive at a poisson rate of 5 per hour and each arrival is a dog with probability 0.6. 1. What is the expected time until the eighth dog arrives? 2. What is the probability that no cats will be treated in the next hour? 3. If the vet wants to have treatment capacity to treat 1 standard deviation above the expected number of cats to arrive each hour and 2 standard deviations more than the expected number of dogs to arrive each hour, the clinic must be equipped to handle the arrival of how many cats and dogs in a single hour? SOLUTION: 1. Let Nd (t) be the random variable for the number of dogs that arrive each hour. Then Nd (t) is Poisson with rate = (0.6)(5) = 3. This means that the expected waiting time until the 8th canine arrival has a gamma distribution with mean 83 = 2.67. So the expected waiting time until the eighth arrival is 2.67 hours. Arch MLC, Spring 2009

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Chapter 12 2. Cat arrivals follow a Poisson distribution with parameter probability of no arrivals in the next hour is P [N (1) = 0] = e

t(

t)n =e n!

2·1 (2

= (0.4)(5) = 2, so the

· 1)0 =e 0!

2

.

3. The variances in the number of cats and dogs that arrive each hour p are 2 and 3, respectively, so the clinic must be ready to handle the arrival of (2 + 2) cats and p (3 + 2 3) dogs in a single one-hour period. } EXAMPLE: Car Care A very specialized car care center accepts only Fords and Chevys. It is assumed that the number of Fords that arrive at the center each day is given by a Poisson distribution with parameter F = 6. The number of Chevys that arrive at the center each day is given by a Poisson distribution with parameter C = 8. (A) What is the expected waiting time until the 4th car arrives? (B) What is the variance in the number of cars that arrives in a 5-day week? SOLUTION: (A) Since NF and NC are both Poisson distributions, N = NF + NC is also Poisson and has parameter = 14. Therefore the waiting time until the 4th car arrives has a Gamma distribution with mean 4 E[S4 ] = = 0.286. 14 (B) The number of cars arriving in a 5 day period has a Poisson distribution with 5 · 14 = 70, so the variance in the number arriving is 70.

= }

The next application is to determine the probability that n events occur in one Poisson process before m events occur in a di↵erent, independent Poisson process. For example, if we were counting the number of cars passing at two di↵erent intersections, we might be interested in the probability that 20 cars pass one intersection before 10 cars pass at the other. Let N1 (t) and N2 (t) be independent Poisson processes with rates 1 and 2 , respectively. Let 2 be the time of the m-th event of N (t). Sn1 be the time of the n-th event of N1 (t) and Sm 2 So some facts that we already know (and want to remember for the exam!) are 1. S11 and S12 are exponentially distributed with means

1

2 have gamma distributions with means 2. Sn1 and Sm

and

Arch MLC, Spring 2009

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n 1

1

and m 2

1 2

, respectively.

, respectively. 390

Chapter 12 Back to the point of this application – we are interested in finding 2 P {Sn1 < Sm },

which is the probability that event n of the N1 process occurs before event m the N2 process. To make a long story a little shorter, P {Sn1

<

2 Sm }

=

n+m X1 k=n

n+m k

1

!✓

1 1

+

2

◆k ✓

2 1

+

2

◆n+m 1 k

.

This formula is ugly and as we’ll see in the example below, this type of problem can be approached more simply. At a minimum, though, be comfortable with the two special cases below: ✓ ◆n 1 1 P {Sn1 < S12 } = and P {S11 < S12 } = . + + 1 2 1 2 EXAMPLE: Car Care For the Car Care center described in the previous example: (A) What is the probability that 3 Chevys arrive before the first Ford arrives? (B) What is the probability that 3 Chevys arrive before 2 Fords arrive? SOLUTION: (A)

✓

◆3

✓

◆

8 3 = 0.187. 14 F + C Note that although a total of four cars are involved, we only need to see the first three cars to know if three Chevys will arrive before the first Ford (B) As in the last problem, we will know if 3 Chevys will arrive before 2 Fords by looking at the next 4 cars that arrive. We need to find the probability that 3 or more Chevys are among the next 4 cars. The probability that all 4 cars are Chevys is ✓ ◆ 4 ✓ ◆0 8 6 14 14 while the probability that there will be exactly 3 Chevys among the next four cars is !✓ ◆ ✓ ◆ 8 3 6 1 4 1 14 14 P {S3C < S1F } =

C

=

!

4 Note that we had to include the term to account for the fact that the 1 Ford could have been any of the next four cars. All this information give us the following: P {S3C < S2F } = 4

✓

8 14

◆3 ✓

6 14

◆

+1

✓

= 0.320 + 0.107 = 0.427. Arch MLC, Spring 2009

c Yufeng Guo

8 14

◆4 ✓

6 14

◆0

} 391

Chapter 12

5.3.5 Conditional Distribution of the Arrival Times Although this section contains a lot of interesting math and examples, it is not easy to make much of it relevant to the exam. The following is a short explanation of some of the points and one testable example. Concept: In a Poisson process, if we know that exactly one event occurred between time 0 and time t, then the distribution of the time at which the event occurred is uniform over the interval [0, t]. For the cars passing the intersection example: if we know that 1 car passed by in the last 30 seconds, it is equally likely to have passed by at any moment during that 30 seconds. The x probability that it passed by in the first x seconds of that interval is 30 . Concept: In the cats and dogs example, it was assumed that the probability an arriving animal was a dog with probability 0.6 no matter what time the animal arrived. It is possible to let this probability change with time - for example, it could be 0.6 for the first two hours (0 < t < 2), 0.5 for the next two hours, etc. If the probability P1 that the arriving animal is a dog is kept constant, and if N1 is the random variable for the number of dogs arriving, then E[Ni (t)] = Pi t. If the probability varies with time Pi = Pi (t), then E[Ni (t)] =

Z t 0

Pi (s)ds.

EXAMPLE: An emergency veterinary clinic cares for only cats and dogs. Animals arrive at a Poisson rate of 5 per hour and the clinic is open for 6 hours each day. The likelihood that the arriving animal is a dog, given that it arrives at time t during the day, is given in the following table. t 0t A > C and the answer is B!

32. SOLUTION: 1 L is the loss random variable of a 2-year endowment insurance on (x + 1) given that (x + 1) dies in Year 1. Time t Attained age Premium collected Death benefit paid 1L

= 1000v

0 (x + 1) 1000P30:3

1 (x + 2) 1000

1000P30:3

(P30:3 + d)¨ ax:3 = 1 ! P30:3 = 0.42744 1L

= 1000(v

P30:3 ) = 1000(1.05

1

0.42744) = 524.941

The answer is D.

33. SOLUTION: In Bowers, formula 5.2.9 gives Var(a T ) =

⇣

2A

x

⌘

(A x )2 2

As shown in example 5.2.1, Ax = Arch MLC, Spring 2009

µ 0.08 = = 0.6154 +µ 0.05 + 0.08 c Yufeng Guo

444

Chapter 13 Ax =

2

So, Var(a T ) =

0.4444 0.61542 0.052

µ 0.08 = = 0.4444 2 +µ 0.10 + 0.08

= 26.27. The answer is C.

34. SOLUTION: V ar(S) = E(N ) V ar(X) + E 2 (X) V ar(N ) With binomial distribution, E(N ) = np and V ar(N ) = npq In this problem, E(N ) = n(0.25) and V ar(N ) = n(0.25)(0.75) With this claim distribution, E(X) = 0(0.3) + 20(0.3) + · · · = 19, E(X 2 ) = 02 (0.3) + 202 (0.3) + · · · = 550, so V ar(X) = 550

192 = 189

So, V ar(S) = 919.5 = [n(0.25)] [189] + [192 ] [n(0.25)(0.75)] which leads to n = 8, the answer is B.

35. SOLUTION: 10 V

= 5000Ax+10

⇡¨ ax+10

First, calculate ⇡ 5000Ax+10 4000vqx x+10 = 5000A a ¨x a ¨x 5000( 14 0.1) 4000 0.9⇥0.02 = 4

⇡= =

4000vqx a ¨x

= 5000( a¨1x

d)

4000vqx a ¨x

732

Next, calculate Ax+10 and a ¨x+10 10 Vx

=1

a ¨x+10 a ¨x

a ¨x+10 = (1

ax 10 Vx )¨

Ax+10 = 1

d¨ ax+10 = 1

Finally,

10 V

= (1

= 5000Ax+10

0.6)0.4 = 1.6 0.1 ⇥ 1.6 = 0.84 ⇡¨ ax+10 = 5000 ⇥ 0.84

732 ⇥ 1.6 = 3, 028.8

36. SOLUTION: A single premium is equal to the APV at issue. It is critical to recognize that AP V = a ¨x = bx + (bx+1 )(v)(px ) + (bx+3 )(v 2 )(px )(px+1 )(¨ ax+2 ) Arch MLC, Spring 2009

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445

Chapter 13 ) AP VA = 1 + (2)(0.9524)(0.99) + (3)(0.9524)2 (0.99)(0.95)(6.951) = 20.675 AP VB = 3 + (2)(0.9524)(0.99) + (1)(0.9524)2 (0.99)(0.95)(6.951) = 10.815

Then, 20.675

10.815 = 9.86 – the answer is D!

Note: in both calculations of APV, you use the number (2)(0.9524)(0.99) = 1.8857. As you work an exam problem, try to recognize when you need numbers you’ve already calculated – this can be a time saver!!

37. SOLUTION: E(S) = (30)(20) = 600 V ar(S) = (30)(4)2 + (2.5)2 (20)2 = 2980 Normal Approximation: ✓

548 600 P r(S < 548) = P r Z < 54.589 1

(0.95) = 1

◆

= P r(Z <

0.95)

0.83 = 0.17

The answer is A.

38. SOLUTION: A1

x:5

=

If he doesn’t go: Z 5 0

v t fT (nogo) (t)dt =

Z 5

1 1 e 0.05(5) dt = = 0.0885 50 50(0.05)

0.05t

e

0

If he does go: A1

x:5

=

Z 1 0

Z 5

v t fT (go) (t)dt +

1

v t 1 px · t

1 px+1

· µno

go (x

+ t)dt

From age x + 1 on, the probability distribution for the future lifetime is uniform with maximum 49 years to live so t 1 px+1

This gives us =

Z 1

e

0.05t

0

1 = 10 1 = 10 Arch MLC, Spring 2009

e

e

· µno

1 dt + 10 0.05t

0.05 1

0.05

0.05

go (x

!

!1 0

Z 5

+ t) =

e

1 49

0.05t

(0.9)

e

0.05t

1

0.9 + 49

0.9 + 49

e

c Yufeng Guo

0.05 0.05(5)

1 dt 49

!5

e 0.05

1 0.05(1)

!

446

Chapter 13 = 0.0975 + 0.0633 = 0.161 So the di↵erence is APV(Go) - APV(No Go) = 0.161

0.0885 = 0.0725. The key is C.

39. SOLUTION: N | Poisson but binomial.

is gamma. Then the unconditional distribution of N is negative

First, find the parameters of N . N | is Poisson. Then E(N | ) = V ar(N | ) = Double expectation:

E(N ) = E [E(N | )] = E ( ) = E( ) = 5 Variance formula: V ar(N ) = E [V ar(N | )] + V ar [E(N | )] = E( ) + V ar( ) = 5 + 25 = 30 Next, look up the mean and variance formula from Exam M Table. For negative binomial random variable N with parameters and r: E(N ) = r V ar(N ) = r (1 + ) P (N = k) =

r(r + 1)(r + 2)...(r + k k!(1 + )r+k

1)

k

The two parameters of negative binomial random variable N is: =

V ar(N ) E(N )

r= P (N = 1) =

E(N )

1=5 =1

r 5 = 2 = 0.1389 r+1 (1 + ) 6

Please note that in a given day, the parameter is constant. Consequently, the probability of finding one coin in the third minute is the same of probability of finding one coin in one minute, no matter it’s the first minute, second minute, or any other one minute. Only the length of the time interval matters; the starting point doesn’t matter. The answer is B.

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447

Chapter 13 40. SOLUTION: 2| q40:44

= 3 q40:44

3 q40:44

= (3 q40 )(3 q44 ) = (1

3 p40 )(1

3 p43 )

= (1

p40 p41 p42 )(1

2 q40:44

= (2 q40 )(2 q44 ) = (1

2 p40 )(1

2 p43 )

= (1

p40 p41 )(1

2 q40:44

p43 p44 p45 )

p43 p44 )

Table 1: 3 q40:44

= (1

0.9 ⇥ 0.8 ⇥ 0.7)(1

2 q40:44

= (1

2| q40:44

= 0.46624

0.9 ⇥ 0.8)(1

0.5 ⇥ 0.4 ⇥ 0.3) = 0.46624

0.5 ⇥ 0.4) = 0.22400

0.22400 = 0.24224

Table 2: 3 q40:44

= (1

0.8 ⇥ 0.7 ⇥ 0.6)(1

2 q40:44

= (1

2| q40:44

= 0.648064

0.8 ⇥ 0.7)(1

0.4 ⇥ 0.3 ⇥ 0.2) = 0.648064

0.4 ⇥ 0.3) = 0.387200

0.387200 = 0.260864

Absolute di↵erence: 0.260864

0.24224 = 0.018624

The answer is A.

Arch MLC, Spring 2009

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448

Chapter 14

SOLUTION TO MAY 2007 MLC

449

www.archactuarial.com

www.guo.coursehost.com

450

Solution to May 2007 MLC Problem 1 The only tricky part is the notation of the force of mortality. The force of mortality is expressed as a function of the attained age = ( ) instead of a function of future life time ( + ). Most candidates are familiar with ( + ) and less familiar with . Make sure you understand how and ( + ) are related. Consider ( ), whose future life time is ( ). If we treat the attained age as the random variable, then the force of mortality at age is 0 ( ) ( ) = ( )= = 1 ( ) ( ) 0 ( ) Using the basic calculus rule [ln ( )] = , we have: ( ) ( )= Change Z = ln

0

+

( )

=

( + ) = ( ) = exp

Set

=

[ln ( )]

( ) = ( ) to :

( ) Z

µ Z

=

[ln ( )]

+

[ln ( )] = ln ( + )

¶

+

( )

. Then

ln ( )

becomes the future life time ( ). We have: Z Z + ( ) = ( + )

(1)

0

= exp

µ Z

+

( )

¶

= exp

µ Z

( + )

0

¶

(2)

Remember Equation 1 and 2. Now we are ready to tackle the problem. Z 75 ( ) = 0 107 We are given µ71 Z 71+4 ¶ ( ) = 0 107 4 71 = exp 5 70

=

70 (4 71 )

71 3 70

=

2 71

(4

71 )

=

0 95 0 96

0 107

= 0 889 17

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451

Problem 2 This is a simple problem. 2 ³ ´ 2 = 2 ( )| 1 1+ 1 0 3443 = 1+ =

+

=

= 1 904 4

1 1 = = = = 0 207 95 +2 1 + 2( ) 1 + 2 (1 904 4) ´ 0 207 95 0 34432 ³ = = 13 970 ( )| 0 082

2

Problem 3 5 [60]

[60]+5

=1

[60]

The above formula holds for a fully discrete whole life regardless of mortality. The selection period is 3 years. Hence [60]+5 = 65 5 [60]

65

=1

[60]

= 9 8969 1 1 0 359 [65] = 11 324 3 = [60] = 1 1 06 1 9 8969 = 0 126 5 [60] = 1 11 324 3 10005 [60] = 126 65

Problem 4 2

( )=

2

=

0 0143

0 06532

= 0 01 148 7 2 (1 ) (1 0 0653) (p ) = 0 01 148 7 = 0 1072 150000 ( ) = (150000) 0 1072 = 16080 2

p

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452

Problem 5 Exponentially distributed inter-arrival time m Number of arrivals is a Poisson distribution To specify the parameter of the Poisson distribution, use common sense. If the inter-arrival time is months, then every months there’s one more event. So events occur at the average rate of 1 per month or 2 per 2-month. No events by 1/31/ 2007 doesn’t have any impact on the distribution of future events (the Poisson arrival process always starts afresh and doesn’t remember what happened in the past). Let represent the number of events that have occurred during 1/31/2007 and 3/31/2007. is a Poisson random variable with an average arrival rate of = 2 = 2 per 2-month. ( = ) = 22! (

3) = 1 =1

[ (³ = 0) + ´( = 1) + 2 2 1 + 2 + 22! = 0 323

(

= 2)]

Problem 6 The key to solving this problem is precisely specifying the model. Total donation at the end of 7 days: = 1+ 2+ + is the donation by the -th person who donates. ’s are independent identically distributed with a common mean and common variance: ( ) = 15, ( ) = 75 is the total number of people who donate. is Poisson with parameter of an average rate = 10 (7) (0 8) = 56 arrivals per 7-day. ( )= ( ) = 56 and

are independent

Total withdrawals at the end of 7 days: = 1+ 2+ + is the withdrawal by the -th person who draws food. ’s are independent identically distributed with a common mean and common variance: ( ) = 40, ( ) = 533 is the total number of people who withdraw food is Poisson with parameter of an average rate = 10 (7) (0 2) = 14 arrivals per 7-day ( )= ( ) = 14 and are independent Net amount at the end of 7 days: = and are independent and approximately normal. Hence is approximately normal (linear combination of independent normal random variables is also normal) Net amount at = 0 is zero. We are asked µ to find ¶( 600 ( ) ( 600) = 1 ( 600) = 1

600) at the end of 7 days.

( )

( )= ( ) ( )= ( )

(

(

) ( )= ) = 15 (56) = 840

(

)+

c °Yufeng Guo

(

)

452

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453

)= ( ) ( )+ 2( ) ( ) = 75 (56) + 152 (56) = 16 800 ( ) = ( ) ( ) = 40 (14) = 560 ( )= ( ) ( )+ 2( ) ( ) = 533 (14) + 402 (14) = 29 862 ( ) = 840 560 = 280 ( ) = 16 800 + 29³ 862 = 46 ´ 662 600 280 ( 600) = 1 =1 (1 48) = 0 069 4 46 662 (

Problem 7 Since we know the benefit premium for the first 3 years, we can use the retrospective method. Reserve at = 3 is , where =accumulating benefit premiums for the first 3 years to = 3 =accumulating death benefits for the first 3 years to = 3 is the same regardless of the benefit premium pattern Choice B,C, and D have same benefit premium for the first 3 years. They have the same reserve at = 3. B,C, and D can be eliminated. Of A and E, the accumulated value of the benefit premiums for the first 3 years is higher for E than for A. E produces higher reserve than for A (and also higher than B,C, and D). Problem 8 This is a di cult problem. If you can’t solve it, don’t feel too bad. Define being called by the parent is death. Then we have two lives Kevin and Kira. Set the unit of time is an hour (instead of one year in a typical life contingency problem). The constant force of mortality is =07 =06 Kira’s½ total score is 0 if = 100 000 ( ) if ( )= [ = 100 000 = 100 000

() [ [

[ () To see why, notice ( )= ( ( )= (

() ()

() () ( )] 0 + [ ( )] [ ( )]

( )] = )= )=

() ()

=

+

¡

¡

07

06

¢

() ( )]

( )]

[100 000

( )]

07 0 7+0 6

¢

=07 07 =06 06

So the future lives of Kevin and Kira are both exponentially distributed. The joint life distribution is

(

)=

(

)=

Z

Z

0

0

( )

07

( )=07 (

)

=

Z

0

06

Z

07

06

07

06

06

0

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=

Z

06 0

06

µZ

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07

¶

07

0

=

454

07 0 7+0 6

You can save time if you know the following rule about two independent exponential random variables competing: Two independent exponential random variables 1 and 2 : 1 2 1 2 1 ( ) = 2 ( ) = Then the probability that 1 arrives before 2 is: 1 ( 1 2) = 1+ 2 1 You can use common sense to memorize why ( 1 2 ) = 1 + 2 . In this example, the Kevin’s death arrives at the rate of 0.7 death per hour and Kira 0.6 death per hour. This is similar to: Car A arrives at 0.7 car per hour. Car B arrives at 0.6 car per hour. So on average we see a total of 1.3 car per hour. If we see one car, then there’s 0.7/1.3=7/13 chance that this car is Car A (i.e. there’s 7/13 chance that Car A arrives before Car B) and 6/13 chance that this car is Car B (i.e. there’s 6/13 chance that Car B arrives before Car A). Let’s continue. = 016 ( ) = 100 000³ [ ( )] ´() 07 1 = 100 000 0 7+0 6 0 6 = 89 743 59

Problem 9 The typical formula is Z (2) ( ) (2) ( ) 25 25 = 0

=

Z

(2)

0

( )×

0 (1) 25

×

0

(2) 25

Cause (2) works only at = 1 5 and = 3 5. During [0 1 5) (1 5 3 5), and (3 5 1], only cause

1 is at work. Hence the above formula needs to be modified as 0 0 (2) (1) (1) 25 =1 5 25 × [(2)’s death rate at = 1 5] +3 5 25 × [(2)’s death rate at = 3 5] time 0

(1) 25

Cause (2) decrement rate # of people hit by cause (2)

0 1

1 5 1 051 = 0 98 3 4 (0 12) = 0 09 (0 98) 0 09 = 0 088 2

3 5 1 3(05 1) = 0 94 1 4 (0 12) = 0 03 0 94 (0 03) = 0 028 2

During [0 1 5), only cause 1 is at work (so we are in the world of a single decrement table). There will be 0.98 survivor immediately before = 1 5. Then at the next instant, cause (2) hits (0 98) 0 09 = 0 088 2 people. Similarly, the interval (1 5 3 5) is in the world of single decrement table. Immediately before = 3 5, 0.94 people survive cause (1). In the next instant, cause (2) hits 0 94 (0 03) = 0 028 2 people. So the total number of people hit by cause (2) in the first year is (2) 25 = 0 088 2 + 0 028 2 = 0 116 4

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455

Problem 10 This is a clever problem testing the recursive formula of whole life insurance. However, wording is a little vague. This is what SOA means: On 12/31/2007, two insureds (65) and (66) each buy a whole life insurance with $1000 death benefit. The interest rate is 10% for the first policy year and 6% thereafter. 65 = 0 01 and 66 = 0 012 . In addition, 1000 66 = 300. Calculate 1000 65 . Use the recursive formula = + +1 . Remember that the interest rate may change. 1 (0 01) + 1 1 1 (0 99) 6% 65 = 1 1 66 6% 1 (0 012) + 1 06 1 (0 988) 6% 66 = 1 06 67 To find 6% 67 : 1 = 1 1 (0 012) + 1 1 1 (0 988) 66 0 3 = 1 1 1 (0 012) + 1 1 1 (0 988) 6% 67 = 0 321 86

6% 67 6% 67

Final answer: 6% 1 (0 012) + 1 06 1 (0 988) 0 321 86 = 0 311 32 66 = 1 06 1 (0 01) + 1 1 1 (0 99) 0 311 32 = 0 289 279 65 = 1 1 1000 65 = 289 3 Problem 11 time 0 1+ % of premium expense (for 10 years) 9% + 20% 9% fixed expense (for 20 years) 5+5 5 = 1000 + 0 2 + 5 + 0 09 + 5 1 40:10| 40:10| 40:20| 40:20|

illustrative life table 14 8166 13 2668 11 1454

40:10|

= =

1

40:20| 40:20|

1000 161 32

|

40 50 60

=

40 40 40

100010 536 67

100020 274 14

369 13 10 20 20

= 14 8166 0 53667 × 13 2668 = 7 696 7 0 27414 × 0 36913 = 0 060 13 60 = 0 16132

40 50 40

40 60

= 14 8166

0 27414 × 11 1454 = 11 761 2

7 696 7 = 1000 (0 060 13) + 0 2 + 5 + 0 09 (7 696 7) + 5 (11 761 2) = 18 215

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Problem 12 56 57

(1)

(2)

( )

0 005 0 008

0 04 0 06

= 0 054 0 068

(1)

+

(2)

( )

=1 0 955 0 932

( )

(55) 1 means that the insured is still alive at age 56. Now we are standing at the insured is aged 65 and still alive. 1 | (55) 1 is distributed as follows: death time 1 2 1 2 2 3 2 3 3 Total (1) ( ) 1| 56 = 56 (2) ( ) 1| 56 = 56 ( ) ( ) 2 56 = 56

cause accident other accident other any (1) 57 (2) 57 ( ) 57

| (55) 1 ¡ ¢ 2000 1 06 1 50 = 1836 79 ¡ ¢ 1 1000 1 06 50 = 893 40 ¡ ¢ ¡ ¢ 2 2000 1 06 50 1 + 1 06 1 = 1682 82 ¡ ¢ ¡ ¢ 1000 1 06 2 50 1 + 1 06 1 = 792 83 ¡ ¢ 50 1 + 1 06 1 = 97 17 1

= 1, where

Probability (1) 56 = 0 005 (2) 56 = 0 04 (1) 1| 56 = 0 007 64 (2) 1| 56 = 0 05 73 ( ) 2 56 = 0 890 06 1

= 0 955 × 0 008 = 0 007 64 = 0 955 × 0 06 = 0 057 3 = 0 955 × 0 932 = 0 890 06

Check total probability: 0 005 + 0 04 + 0 007 64 + 0 05 73 + 0 890 06 = 1

OK

If your total probability is not one, you made some errors. Let’s reorganize the above table: [1 = | (55) 1] 97 17 0 890 06 792 83 0 05 73 893 40 0 04 1682 82 0 007 64 1836 79 0 005 Total 1

[1 | (55) 1] 0 890 06 0 890 06 + 0 05 73 = 0 947 36 0 947 36 + 0 04 = 0 987 36 0 987 36 + 0 007 64 = 0 995 0 995 + 0 005 = 1 0

So the smallest

(55)

such

[1

|

1]

0 95 is

SOA’s answer 893 is problematic because [1 893| (55) 1] = [1 = 97 17| (55) 1] + [1 = 792 83| = 0 890 06 + 0 05 73 = 0 947 36

(55)

= 893 40

1]

Of the SOA’s 5 choices, (E) 943 is the smallest given by SOA that satisfies [1 | (55) 1] 0 95 [1 943| (55) 1] = [1 = 97 17| (55) 1] + [1 = 792 83| (55) 1] + [1 = 893 40| = 0 890 06 + 0 05 73 + 0 04 = 0 987 36

(55)

1]

If you chose (E) and your exam score is 5, you might consider writing a letter to SOA justifying your answer. c °Yufeng Guo

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457

Problem 13 SOA wants you to use the Hattendorf theorem (Actuarial Mathematics Section 8.5 page 244): 2 2 [ | ( ) ] = [ ( +1 )] [ +1 | ( ) + 1] +1 + + + + If you are one of the lucky few who have memorized this di cult formula, congratulations. Since the policy ends at = 3, [3 | ( ) 3] = 0 [2 | ( ) 2] 2 2 [3 | ( ) 3] = [ ( 3 3 )] +2 +2 + +2 2 = [ ( 3 3 )] +2 +2 £ ¤2 = 1 1 1 (1000 0) (0 5) (0 5) = 206611 57

[1 | ( ) 1] 2 2 = [ ( 2 2 )] [2 | ( ) 2] +1 +1 + +1 £ ¤ 2 = 1 1 1 (1000 120 833) (0 6) (0 4) + 1 1 2 (0 6) (206611 57) = 255761 36 If you are one of many who don’t bother memorizing the Hattendorf theorem, congratulations. You can solve this problem using basic principles. To simplify calculation, assume the death benefit is $1 (instead of $1000). Calculate the benefit premium: time people alive deaths

0 1

1 07 1 07=03

2 0 7 (0 6) = 0 42 0 7 0 42 = 0 28

Equivalence principle: ¡ ¢ 1 + 0 7 + 0 42 2 = 0 3 + 0 28 2 + 0 21 0 3 × 1 1 1 + 0 28 × 1 1 2 + 0 21 × 1 1 = 1 + 0 7 × 1 1 1 + 0 42 × 1 1 2

3 3

= 0 333 71

Find the distribution of the random variable 1 | 1 is distributed as follows: death time 1 2 2 3 3

|

1 2

3 0 7 (0 6) (0 5) = 0 21 0 42 0 21 = 0 21

( )

1. We are standing at = 1.

( ) 1 = 1 1 1 0 333 71 = 0 575 38 ¢ ¡ (1 + ) = 1 1 2 ¡ 0 333 71 ¢1 + 1 1 1 = 0 189 36 (1 + ) = 0 333 71 1 + 1 1 1 = 0 637 08

1

|

(55)

Probability 04 0 6 (0 5) = 0 3 1 04 03=03

£[1 |2 ( ) 1] = ¤ 0 575 38 (02 4) + 0 189 36 (0 23) 0 637 08 (0 3) =20 095 836 | ( ) 1 = 0 575 38 (0 4) + 0 189 36 (0 3) + ( 0 637 08) (0 3) 1 = 0 264 943 298 56 [1 | ( ) 1] = 0 264 943 298 56 0 095 836 2 = 0 255 759 Since the death benefit is 1000, the variance is 10002 × 0 255 759 = 255 759 You can use BA II Plus or BA II Professional calculator 1-V Statistics Worksheet to quickly calculate the variance. To 1-V Statistics Worksheet, we need to scale the probability to an integer. So 0.4, 0.3, and 0.3 are scaled to 4, 4, and 3. Enter c °Yufeng Guo

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01 = 0 575 38 02 = 0 189 36 03 = 0 637 08

458

01 = 4 02 = 3 03 = 3

You should get: = 0 095836 = 0 50572597 Then 2 = 0 505725972 = 0 255 759

10002 × 0 255 759 = 255 759

You can use 1-V statistics worksheet to calculate the mean and variance of a discrete random variable as long as you scale the probabilities to integers. By the way, the reserve given to you 0.95833, 0.120833, and 0 at = 1 2 3 is really not necessary because you can calculate the reserve yourself. The reserve at = 1 under the retrospective method is: (1 + ) 0 333 71 (1 1) 0 3 = = 0 095 83 07 The reserve at = ¡ 2 under ¢ the prospective method is: =05 11 1 0 333 71 = 0 120 835 +2 SOA purposely gives you the reserves because it wants you to use the Hattendorf theorem. Problem 14 To have 0

(30)

(35)

5, we let two things happen:

• Let (35) die at time or (35) = . Since the probability that a continuous random variable takes on a single value is zero (i.e. [ (35) = ] = 0), to produce probability we have to express (35) in terms of a range. So we write (35) + . This means that (35) dies in the tiny interval [ + ], which is the statistical statement that (35) dies at time . The probability is (35) ( ) • Let (30) die during the interval [ + 5]. The probability is [ (30) + 5] = ( 30 ) ( +5 30 )

= =

Z [0

(30) [

Z0

(35)

(35)

[

] ( [ (30)

+

]

[ (30)

]

[ (30) Z

]

0

Z

(30)

5+ ]=

[ (30)

]

5]

+

(35)

[

+ 5])

[

(35)

+

]

[ (30)

+ 5]

0

[

(35)

+

]

[ (30)

]=

[ (30)

(35)] = 1

0

So the correct answer should have the term 1 answer. Let’s finish up the problem. + 5] = [ (30) 5] [ (30) + 5 Z [ (30) [

0Z

=

0

= (5

£

30 )

(35)

Z

(35)

()

+

]

+ 5] Z ¤ (5 30 ) ( 35 ) = (5 30 )

35

( )(

35 )] (

| (30)

5] = (5

30 ) (

is the right

35 )

[ (30)

0

[

. This immediately tells us that

35 )

£

(35)

()

¤

(

35 )

0

c °Yufeng Guo

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www.archactuarial.com The first

35

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in the integration is the survival function of someone who is initially 35. The second

35 in the integration is the survival function of someone who was initially 30 years old 5 years ago and who becomes age 35 now. By the phrase "The future lifetimes of (30) and (35) are independent and identically distributed," SOA wants to say that two 35 terms are identical. In other words, SOA wants to say that two people with the same attained age have the same survival function.

Then

Z

[

35

( )(

35 )] (

35 )

refers to the following probability: "There are two independent

0

lives each aged 35. Their future lifetimes are identically distributed. What’s the chance that one (35) will die before the other (35)?" Clearly, the probability is 1/2 because each (35) has 50% chance of dying before Zthe other. (5

30 )

[

35

( )(

35 )] (

35 )

= 0 5 (5

30 )

=05

0

Then the final answer is 1

05 .

Please note that the phrase "The future lifetimes of (30) and (35) are independent and identically distributed" is problematic. It means that (30) and (35) are independent identically distributed (instead of meaning 35 is the same regardless of you are initially 35 or you are initially younger but grow up to age 35, as SOA wants us to interpret).

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460

Problem 15 A probability tree diagram will enable you to quickly solve the problem. L=Independent Living. C=Health Center. D=Dead. Next, draw the following diagram. =0

=1

=2 0 4L

0 6L

0 4C

=3 02 0 1C 07 0 01 09

02

0

1L

0 3C

04

0 01 09

06

01

For example, 1 Independent Living at = 0 will become [0 6 0 3 0 1 ] at = 1. Similarly, 1 at = 1 becomes [0 4 0 4 0 2 ] at = 2. Finally, after = 3, each state remains unchanged (i.e. each state becomes an absorbing state). So nobody collects benefits at 3 time 0 1 2 3 # of people in 1 0 6 0 6 (0 4) = 0 24 # of people 0 3 0 6 (0 4) = 0 24 0 6 (0 4) (0 1) = 0 024 To calculate how many people collect benefits at = 1 2 3, pay attention to the bold numbers in the diagram. From = 0 to = 1, 1 From = 1 to = 2, 0 6

0 3 . Hence 0 3 collects 1000 benefits at = 1. 0 4 . Hence 0 6 (0 4) = 0 24 collects 1000 benefits at = 2. c °Yufeng Guo

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From = 2 to = 3, 0 4

0 1 . Hence 0 6 (0 4) (0 1) = 0 024

461 collects 1000 benefits at = 3.

The equivalence principle: ¡

1 + 0 6 + 0 24

= 1000

2

¢

¡ = 1000 0 3 + 0 24

2

+ 0 024

3

0 3 × 1 25 1 + 0 24 × 1 25 2 + 0 024 × 1 25 1 + 0 6 × 1 25 1 + 0 24 × 1 25 2

¢

c °Yufeng Guo

3

= 248 46

461

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462

Problem 16 First, label the 0 1 0 02 05 1 01 06 2 01 05

transition matrix so we can quickly find the probability. 2 03 03 04

For example, from the above table, we easily see that the probability of "1 coin today and 1 coin tomorrow" is (1 1) = 0 6. Next, draw the following diagram: =0 =1 =2 0, Pr0.2 0, Pr 0.1 1, Pr0.5 2, Pr0.3

1, Pr 1

1, Pr 0.6

0, Pr 0.1 1, Pr 0.6 2, Pr 0.3

2, Pr 0.3

0, Pr 0.1 1, Pr 0.5 2, Pr 0.4

=# of coins found at = 1 =# of coins found at = 2 We need to find ( + 3) Possible combinations of (

):

(

=1 = 2) =1 = 2) = 0 6 (0 3) = 0 18 ( =2 = 1 or 2) ( =2 = 1 or 2) = 0 3 (0 5 + 0 4) = 0 27 (

(

+

3) = 0 18 + 0 27 = 0 45

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463

Problem 17 N=No accident. Y=Yes, there’s at least one accident. Draw the following diagram: =0 =1 =2 =3 N0 8 N0 8 02 N0 8 N0 7 02 03 1 N0 8 N0 7 02 02 N0 7 03 03

At = 0, there’s no accident with potability 1 (so 1 at = 0). Then at = 1, there’s 0.8 chance of no accident ( 0 8) and 0.2 chance of accident ( 0 2). So on and so forth. Payout #1: Pay 100 for each node at = 1 2 3 time Pr ( ) 1 08 2 0 82 + 0 2 (0 7) = 0 78 3 0 83 + 0 8 (0 2) (0 7) + 0 2 (0 7) (0 8) + 0 2 (0 3) (0 7) = 0 778

Payout #2: Pay at = 3 if the pattern is ( ) = 0 83 Equivalence: ¡ 100 0 8 + 0 78

2

+ 0 778

3

¢

= 0 83

at = 1 2 3

3

¢ ¡ 100 0 8 + 0 78 2 + 0 778 3 = ´ ³ 0 83 3 3 = 100 × 0 8 0 8 (1 + )2 + 0 78 (1 + ) + 0 778 ¢ ¡ = 100 × 0 8 3 0 8 × 1 042 + 0 78 × 1 04 + 0 778 = 479 39

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464

Problem 18 A logic approach is [67]+1

=

67+2

69

We are given: To simplify, set 3

[67] [67]+1

3 [ ]+1 = 4 +1=

[67]

[ +1]

4

+2

=5

[ +1]+1

=4 [ ] 4 +1 = µ 5 [ ]+1 ¶ 4 4 4 7700 1 = 0 03 67+1 = 68 = [67]+1 = 5 5 5 8000 µ 3 3 3 4 3 4 = = = = × × 1 67 [67] [67 1]+1 4 [66]+1 4 5 4 5 ¡ 4 ¢¡ ¢ 1 [67] 1 [67] [67]+1 = 69 0 03) (1 0 01 463) = 7700 [67] (1 7700 = 8056 [67] = (1 0 03) (1 0 01 463) [

1]+1

8000 8200

¶

= 0 01 463

Problem 19 10 40

50

=1

13 40

53

=1

40

=13 =1+

10 40

40

50

50

Set

40

= 53 = 10 + (2 50 ) 2 + (3 50 ) 3 50

= 51 = 52 = =1+ + 2 2+

53

50 50

10 = 1 + + 10 = 1 + 1 06

2 2 1

+ 10 3 + 2 1 06

3 3

53

3 2

+ 10 3 1 06

3

Try out the 5 answers. The correct answer is

= 0 954

c °Yufeng Guo

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465

Problem 20 Year 1 Person

D only 0 001 0 0002 = 0 000 8

D&A 0 0002

A only 0 001 0 0002 = 0 000 8

Probability that both D and A survive the first year 0 0008 0 0002 0 0008 = 0 998 2 Year 2 Person

D only 0 001

D&A 0

A only 0 001

Probability that both D and A survive the second year 0 001 0 0 001 = 0 998 Probability that both D and A survive 2 years: 0 998 2 × 0 998 = 0 996 2 Problem 21 Old model (DeMoivre model): = New model (general DeMoivre model):

=(

)

The new and the old have the same . To solve this problem quickly, you’ll need memorize the following formula: under the general DeMoivre model This is why. = Z =

+

(

(

) "µ

0

=

+1

Set 2

(

=

µ ) = 1 ) µ Z = 1 0 # ¶

=

¶

+1

1

+1 ¶

=

+1

0

= 1. Then the general DeMoivre model becomes the familiar DeMoivre model and

=

.

We are given:

30

=

4 3

30 4 30 = × +1 3 1+1 60

60 = +1

=

70

=

90

70 2

30

1 2 = +1 3

60 = 20 15

+1=15 = 90

= 10 c °Yufeng Guo

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466

Problem 22 Let ½= (40) 1000 0 1 = 2500 0 1

if if

10 10 1 ln 0 7 = 3 57 01

1000

01

700

01

07

2500

01

700

01

700 2500

3 57 if 12 73 if

10 10

½

700 = 0

... 700

| DeMoivre (

3 57 |

1 700 ln = 12 73 0 1 2500

10 |

... 700

1 1 1 ()= = = 100 40 60 Z 3 57 Z 12 73 700) = () + () 0

=

10

12 73 |

40 = 60

3 57 + 2 73 = 0 105 60

Problem 23 (2)

(2) (2)

In a single decrement table, 1|1 = 6= +1 . In a double decrement table, 1|1 +1 ; there’s (2) (2) ( ) (2) no such a symbol as . The correct formula is 1|1 = . To be hit by cause (2) in Year 2, +1 the person needs to survive all causes in Year 1 and then be hit by cause (2) in Year 2. The first sentence of the problem says that force of decrement for each cause is constant in each (2) (1) year of age. So = 0 2 means that cause (2) has a constant mortality 0.2 in Year 1. +1 = 0 15 means that cause (1) has a constant mortality 0.15 in Year 2. ( )

=

(2) +1

=

( ) +1

=

exp

³

(2) +1

0

(1)

(2)

³ = 1

0

(1)

´

exp

³

(2)

´

= (1

0 1) exp ( 0 2) = 0 736 9

(2) +1 ( ) +1 ( ) +1 0

(1) +1

( ) +1

=

0

´

( ) +1

0

(2) +1

= exp

=

( ) +1 (1) +1

0 287 7 (2) (1 +1 = 0 437 7 (2) ( ) (2) = 1|1 +1

³

(1) +1

´³ 1

(2) +1

´

( ) +1

= 0 645 5 = 0 437 7

0

= exp ( 0 15) (1 =

0 25) = 0 645 5

ln 0 645 5 = 0 437 7

0 15 = 0 287 7

0 645 5) = 0 233 = 0 736 9 (0 233) = 0 171 7 c °Yufeng Guo

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Problem 24 :3|

=1+

2

+2

By Equation µ 2,Z 75+ ( ) 75 = exp 75Z Z 75+ 75 ( ) ( ) 0

¶

µZ

= exp

75+

( )

0

= 0 01 (75 + )

Z

75

0

12

( )

¶¸

12

0 01 (75)

0

h i = exp 0 01 (75)1 2 0 01 (75 + )1 2 h i 12 12 0 01 (75 + 1) = 0 971 9 75 = exp 0 01 (75) h i 12 12 0 01 (75 + 2) = 0 944 5 2 75 = exp 0 01 (75) 75

:3|

100

¡ = 1 + 0 971 9 1 11 :3|

1

¢

¡ + 0 944 5 1 11

= 100 (2 642 ) = 264 2

2

¢

= 2 642

Problem 25 The that arrive Zduring 7am to 7:25am is: Z 25 average number Z 10 of trains Z 20 25 () = 0 05 + 0 1 = 0 5 + 0 75 + 0 5 = 1 75 + 0 0 10 200 20 (

= 4) =

1 75 1

754 = 0 0679 4!

0 07

Problem 26 Mistakes occur at the rate of 4 per day. So on average there’s one mistake every 1 4 day. The inter-arrival time (i.e. the time between two failures) is an exponential random variable with mean 1 4 day. = time (in days) when the 289th mistake occurs 1 = time (in days) when the 1st mistake occurs 2 = time (in days) between the occurrence of the 1st mistake and the 2nd mistake ... 289 = time (in days) between the occurrence of the 288th mistake and the 289th mistake = 1 + 2 + + 289 where ’s are independent identically distributed exponential random 2 variables with a common mean 1 4 and a common variance (1 4) . Under the central limit theorem, is approximately normal. ( ) = 289 (1 4) = 72 25 p (

( )= 68) = 1

( ) = 289 (1 4)

2

289 (1 4) = 4 25 (

68) = 1

µ

68

72 25 4 25

¶

= 0 841 3

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Problem 27 R ( )= 0

=

= 0 04

=

¡

2

¢

=

2 12 +

R

0

=

¡

R

0 06

=

0

¢2

=

R

1 06 +

468

= 0 03636

2 12

0

0 04 = 0 01 851 9 2 12 + 0 04

( ) = 0 01 851 9

0 036362 = 0 017 2

Problem 28 Don’t be scared by the cash value. Just treat it as another expense. Asset share is similar to reserve EXCEPT 1. Reserve considers death only, ignoring lapse; asset share considers both death and lapse, 2. Reserve uses the benefit premium; asset share uses the contract (gross) premium. 3. Reserve considers the death benefits only, ignore expenses; asset share consider death benefit and expenses. Consider the time interval Time 0 Age 40 Gross premium Renewal expense Cash value Asset share

[15 16] 15 45 90 0 05 (90) = 4 54

16 46

1150

1320

The source of funding at = 15 is: 90 4 45 + 1150 = 1235 55 This accumulates with interest to = 16: 1235 55 (1 08) = 1334 394 The fund pays the following 3 items at = 16: • Death benefit for each death. 10000

( 55

• Lapse expense (i.e. CV) for each lapse.

)

= 10000 (0 004) = 40 ( 55

³ • Asset share for each policy remaining. 1320 1 1248 72

)

= 0 05

( 55

)

( 55

)

´

= 1320 (1

0 004

0 05) =

Funding=Spending 1334 394 = 40 + 0 05 + 1248 72 = 913 48 You don’t need to memorize any complex recursive formula about the asset share. Just remember that the funding is equal to the spending.

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469

Problem 29 = 1

30:30| 30

30

60

+ (30

1

30:30|

=

(30

30 )

30 ) (10

50 )

30

= (20

= 11 1454

1

30:30|

30 ) 200 60

30

= 0 10248

60

= (20

30 ) (10

50 )

= 0 10248

60

= 0 29374 × 0 51081 = 0 150 05

= 0 36913

(0 150 05) 0 36913 = 0 047 09

= 0 047 09 + 200 × 0 150 05 × 11 1454 Problem 30 R

( )

=3

0

( )

R

=

(1)

+

(2)

0 06

0 06 +

( )

(1)

0

+

= 0 02 + 0 04 = 0 06

0

=

R

= 351

0 06 0 06 +

=3 +

R

0

0 02

0 04 0 06 +

0 06

+

R

( )

(2)

0

R

0

0 04

0 06

=01

c °Yufeng Guo

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