May 1, 2017 | Author: Jose Martinez-Gracida | Category: N/A
Deeper Understanding, Faster Calculation: MLC, Spring 2011 Yufeng Guo November 28, 2010
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Contents 0 INTRODUCTION The origin of this study guide . . . . . . . Two sections each with a separate table of Wanting for more practice problems . . . MLC syllabus change in Spring 2012 . . . About Yufeng Guo . . . . . . . . . . . . .
. . . . . contents . . . . . . . . . . . . . . .
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 . . . . . . . . . . . . . . . . . . . . . . . . .
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Past SOA/CAS Exam Questions: . . . . . . . . . . . . . . . . . . . . . . . . . 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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97 97 100 106 112 116 116 117 118 131 133
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 Difference 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
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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 . . . . . . . . . . . . . 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 . . . . . . . . . . . . . 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 . . . . . . .
BENEFIT 197 . . . . . . 197 . . . . . . 202 . . . . . . 205 . . . . . . 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 Guo MLC, Spring 2011
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CASE III : Multiple decrements – some are uniformly distributed the associated single decrement table and some are not. . . . . . . CHAPTER 10 Formula Summary . . . . . . . . . . . . . . . . . . . . . . . Past SOA/CAS Exam Questions: . . . . . . . . . . . . . . . . . . . . . . . Problems from Pre-2000 SOA-CAS Exams . . . . . . . . . . . . . . . . . . Solutions to Pre-2000 Problems: Chapter 10 . . . . . . . . . . . . . . . . .
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300 305 307 317 320
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 . . . . . . . . . Guo MLC, Spring 2011
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5.4.1 Nonhomogeneous Poisson Process . . . 5.4.2 Compound Poisson Process . . . . . . . 5.4.3 Conditional or Mixed Poisson Processes: CHAPTER 5 Formula Summary . . . . . . . . . . . ARCH Warm-up Problems: . . . . . . . . . . . . . . Solutions: . . . . . . . . . . . . . . . . . . . . . . . . Past SOA/CAS Exam Questions: . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . Gamma-Poisson Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
393 395 398 401 402 403 404
13 Practice Exam 417 Answer Key for Practice Exam . . . . . . . . . . . . . . . . . . . . . . . . . . 430 14 SOLUTION TO MAY 2007 MLC
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INTRODUCTION The origin of this study guide Most of the study guide is from the original Arch manual written by by two gifted actuaries, Nathan Hardiman and Robin Cunningham. My great thanks to them for passing the Arch manual to me. In 2006 I ”inherited” the Arch manual when Nathan and Robin, who, each having a busy day job, no longer had time to maintain or expand the Arch manual. From 2006 to the Fall 2010, I had two study guides on MLC: the Arch and my Deeper Understanding manual. Managing two separate manuals by the same author on the same exam can be confusing to readers, who have difficulty deciding which one to buy. To overcome this confusion, I decided to discontinue the Arch manual starting from Spring 2011 and transfer the Arch manual content to my Deeper Understanding manual for MLC. Thanks to the Arch manual I finally learned LaTeX. Many of my Deeper Understanding manuals were written in Microsoft Word with the MathType plugin. Though ”Word+MathType” gets one started fast and easy, documents produced this way are a nightmare to maintain. Wanting to break away from ”Word+MathType,” I tried several times to learn LaTeX but always went nowhere. After taking over the Arch manual, I had no choice but to learn LaTeX (because the Arch was written in LaTeX). After struggling several days, finally I ”got it” and became proficient in LaTeX.
Two sections each with a separate table of contents There are two sections that don’t appear in the overall table of contents. These two sections have separate tables of contents. One section is about Actuarial Mathematics Chapter 9 Multiple Life Functions; the other about the Poisson process. These two sections are toward the end of this book (so scroll toward the end of this document to read these two sections).
Wanting for more practice problems The strength of this study guide is that it explains the core concepts well. It gets you up and running quickly toward learning the fundamentals of the core MLC knowledge. 9 9 of 674
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The weakness of this study guide is that it doesn’t have a great number of practice problems. If you use this book to learn the core concepts but need additional practice problems, you can download the past course 3 and Exam M problems from the SOA website http: //www.soa.org/ and the CAS website http://www.casact.org/admissions/studytools/.
MLC syllabus change in Spring 2012 SOA already announced that the MLC syllabus is to be changed in Spring 2012. SOA stated: Changes are coming to the learning objectives and required readings for Exam MLC effective with the Spring 2012 exam administration. What is changing and when? The learning objectives for Exam MLC are being revised to reflect current theory and practice. A draft of the revised learning objectives has been prepared to give advance notice for this change. Information regarding textbooks and study materials for the exam will be released by the end of the second quarter of 2011. The first Exam MLC administration using the new learning objectives will be held in the Spring of 2012. Why Spring 2012? Currently, the life contingencies topic is taught as a twosemester fall/spring course sequence at most universities. As such the end of the spring semester is an ideal time to offer the first exam administration with the revised syllabus. A textbook covering all the learning objectives is not currently available. However, it is anticipated that a suitable textbook will be available in early 2011.
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: 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: Guo MLC, Spring 2011
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• 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://actuary88.com. If you have questions, you can email Mr. Guo at yufeng
[email protected]. 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.
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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 different 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 13 of 674
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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 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) − F (x) = [1 − s(z)] − [1 − s(x)] = s(x) − s(z) s(x)
F(x)
O
x
the pdf y=f(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 Guo MLC, Spring 2011
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inconsistent in their treatment of F and s! If two different random variables, say X and Y , referred to the future lifetimes of two different 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.
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F (40) − F (17) F (40) − F (10)
or
s(17) − s(40) s(10) − s(40)
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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 ≥ 0
t px
= 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, ω). (ω 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-offs 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.) Guo MLC, Spring 2011
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Useful formulas: t px
=
t qx
t|u qx
x+t p0 x p0
=1−
=
s(x + t) s(x)
s(x + t) s(x)
s(x + t) − 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 =
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. Guo MLC, Spring 2011
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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+1 px
=
k px
∗ qx+k
=
k| qx
(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 = =
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s(37) − s(38) s(18)
63 − 62 1 = . 82 82
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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. µ(x) =
f (x) −s0 (x) = 1 − F (x) s(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)
s(x)
f (x)
µ(x)
f (u) du
−
1−e
f (u) du
e−
F (x)
F (x)
1 − s(x)
Rx
s(x)
1 − F (x)
s(x)
R∞
f (x)
F 0 (x)
−s0 (x)
f (x)
µ(x)
F 0 (x) 1−F (x)
−s0 (x) s(x)
f (x) s(x)
0
x
Rx 0
Rx
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−
Rx 0
µ dt
= e−µx .
2. t px
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=
s(x + t) = e−µt . s(x)
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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 ) ω = limiting age at which probability of survival = 0 (s(x) = 0 for all x ≥ ω) 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 n x x − lx+n
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 Guo MLC, Spring 2011
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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 = 1 − 92,872.6 97,706.6 = 0.04947. = 92,595.7 93,131.6 (0.0032) = 0.003182.
3.
38 q3
= 1 − 38 p3 = 1 −
4.
2| q40
= 2 p40 · q42
l41 l3
♦
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
∞
∞
Z
f (t)dt = 1.
t px µ(x + t)dt = 0
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)] =
Z
∞ t px dt
0
Z
Var[T (x)] = 2 0
∞
◦2
t · t px dt− 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)] = Guo MLC, Spring 2011
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Chapter 1
or
s[x + m(x)] 1 = s(x) 2
or m px
1 = . 2
The curtate-expectation-of-life is E[K(x)] and is denoted ex (no circle). To remember ◦ the difference 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)] =
∞ X
k px
1 ∞ X
Var[K(x)] =
(2k − 1) · k px − e2x
1
EXAMPLE: Constant Force of Mortality ◦
◦
Find e0 and e50 if the force of mortality is a constant µ. SOLUTION: ◦
e0 =
Z
∞
◦
Z
e−µt dt =
0
0
e50 =
∞
Z t p0 dt =
∞
∞
Z t p50 dt
=
e−µt dt =
0
0
−1 −µt e µ
∞
−1 −µt e µ
= 0
1 µ
∞
= 0
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) =
50−x 50
0 < x < 50 Otherwise
0
for all x between 0 and 50, find e0 and e45 . SOLUTION: e0 =
50 X 1
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t p0
=
50 X 50 − t 1
50
= 50 −
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Chapter 1
= 50 −
e45 =
5 X
1 (50)(51) = 24.5 50 2
t p45 =
5 X s(45 + t)
1
s(45)
1
=
=
5 X 5−t 1
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 . 1
Z
lx+t dt
Lx = 0
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
Z n Lx
lx+t dt
= 0
n mx
lx − lx+n n Lx
=
The remaining of Section 3.5.1 has obscure symbols Tx and α(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). Z
Tx =
∞
lx+t dt
(3.5.16)
0
Note from the definitions that you can think of Tx as
∞ Lx .
The final symbol is α(x). It’s the expected death time given x dies next year. Guo MLC, Spring 2011
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Chapter 1
R 1 lx+t R1 R1 t· l tf (t)dt tt px µ(x+t)dt 0 0 x α (x) = E [T |T < 1] = R 1 = R1 = R0 1 lx+t 0
f (t)dt
0
t px µ(x+t)dt
0
lx
µ(x+t)dt µ(x+t)dt
R1 t·lx+t µ(x+t)dt = R0 1 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 α(x) = R 1 = R01 = 01 tdt = 12 . 0
cdt
0
dt
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
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Chapter 1
SOLUTION: (A) 1
Z
1
Z
l5+t dt =
L5 =
t p5 l5 dt. 0
0
Since l5 = l0 e−µ·5 = 1000e−0.5 = 606.5, we have Z
1
L5 = 606.5
h
e−0.1t dt = 606.5 −10e−0.1t
0
h
i1 0
i
= 606.5 10(1 − e−0.1 ) = 577.16. (B) m5 =
l5 − l6 606.5 − 548.8 = 0.10 = L5 577.16
This approximates the rate at which people were dying between the 5th and 6th years. (C) ∞
Z
−0.1(5+t)
1000e
T5 =
Z
dt = 606.5
∞
e−0.1t dt = 6065
0
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 t px dt
0
= n-year temporary complete life expectancy of (x) ◦
= ex:n (p.71)
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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 different 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 eω and eω and working backward. For ex , using the recursion once will produce eω−1 , the second iteration will produce eω−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 ω. The formulas are: for ex , u(x) = ex c(x) = px d(x) = px Starting Value = eω = u(ω) = 0 So to start, set x + 1 = ω and the recursion will produce u(x) = u(ω − 1). ◦
For ex ,
◦
u(x) =ex Z
c(x) =
1 s px ds
0
d(x) = px ◦
Starting Value =eω = u(ω) = 0
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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
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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: Z
1
Z
1
dt = f (t)
f (t)dt = f (t)
qx =
0
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 − t)qx . Later I’ll show you how to derive other formulas in Balducci assumptions.
Function t qx
Uniform Distribution tqx
Constant Force 1 − ptx
Hyperbolic
µ(x + t)
qx 1−tqx
− log px
qx 1−(1−t)qx
1−t qx+t
(1−t)qx 1−tqx
1 − p1−t x
(1 − t)qx
y qx+t
yqx 1−tqx
1 − pyx
yqx 1−(1−y−t)qx
t px
1 − tqx
ptx
px 1−(1−t)qx
qx
−ptx log px
qx p x [1−(1−t)qx ]2
t px µ(x
+ t)
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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: f (t) = qx = constant t
Z t qx
t
Z
= 1 − t qx = 1 − tqx
µ(x + t) = µx (t) = t px µ(x y px+t
=
dt = tqx 0
0
0 t px
f (t) qx = 1 − tqx t px
+ t) = f (t) = qx
s(x)t+y px s(x + t + y) = = s(x + t) s(x)t px y qx+t
t
qx dt = qx
f (t)dt =
=
Z
= 1 − y px+t
1−t qx+t
=
t+y px
t px yqx = 1 − tqx
=
1 − (t + y)qx 1 − tqx
(1 − t)qx 1 − tqx
Under constant force of mortality, for 0 ≤ t ≤ 1: µx (t) = µ t px
=e
−
Rt 0
µdt
= e−µt
px = e−µ µ = − ln px t px
=e
y px+t =
−µt
t+y px t px
t
= (e−µ ) = (px )t =
(px )t (t+y)
(px )
= (px )y
Finally, let’s derive log s(x + t) = (1 − t) log s(x) + t · log s(x + 1) s(x + t) = s(x)e−µt ⇒ log s(x + t) = log s(x) − µt s(x + 1) = s(x)e−µ ⇒ log s(x + 1) = s(x) − µ ⇒ (1 − t) log s(x) + t log s(x + 1) = (1 − t) log s(x) + t log s(x) − µt = log s(x) − µt ⇒ log s(x + t) = (1 − t) log s(x) + t log s(x + 1) Guo MLC, Spring 2011
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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 px s(x + t) = 1−(1−t)q x
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
However, t px =
s(x+t) s(x)
1−t qx+t
= (1 − t)qx .
s(x + 1) px =1− = (1 − t)qx s(x + t) s(x + t) px s(x + t) = 1 − (1 − t)qx
=1−
= s(x + t). This gives us: t px =
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 d px qx px = t px = − dt dt 1 − (1 − t)qx [1 − (1 − t)qx ]2
Derive µ(x + t): µ(x + t) =
qx f (t) = 1 − (1 − t)qx t px
Derive y qx+t : y qx+t
=1−
s(x + t + y) s(x + t)
px Use the formula: s(x + t) = 1−(1−t)q x Replace t with t + y, assuming 0 ≤ t + y ≤ 1:
s(x + t + y) = Guo MLC, Spring 2011
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y qx+t
Chapter 1
=1−
Finally, let’s derive
px s(x + t + y) 1 − (1 − t)qx = =1− s(x + t) 1 − (1 − t − y)qx 1 − (1 − t − y)qx
1 s(x+t)
=
1−t s(x)
+
t s(x+1)
s(x + t) =
px 1 − (1 − t)qx
1 1 − (1 − t)(1 − px ) (1 − t)px + t t 1 − (1 − t)qx = = = (1 − t) + = s(x + t) px px px px Since s(x) = 1 and s(x + 1) = px , we have: 1−t t t + =1−t+ s(x) s(x + 1) px ⇒
1−t t 1 = + 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: 0.5 qx
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= (0.5)(0.1) = 0.05
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Chapter 1
• CF: 0.5 qx
= 1 − (0.9)0.5 = 0.0513
• Balducci: 0.5 qx
(B)
=
(0.5)(0.1) = 0.0526. 1 − (0.5)(0.1)
• UDD: 0.5 qx+0.5
=
(0.5)(0.1) = 0.0526 1 − (0.5)(0.1)
• CF: 0.5 qx+0.5
= 1 − (0.9)0.5 = 0.0513
• Balducci: 0.5 qx+0.5
= (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) = (ω − x)−1 and s(x) = 1 −
x , ω
where 0 ≤ x < ω
Gompertz’ Law: µ(x) = Bcx and s(x) = exp[−m(cx − 1)] where B > 0, c > 1, m =
B , x≥0 log c
Makeham’s Law: µ(x) = A + Bcx and s(x) = exp[−Ax − m(cx − 1)] where B > 0, A ≥ −B, c > 1, m =
B , x≥0 log c
Weibull’s Law: µ(x) = kxn and s(x) = exp(−uxn+1 ) Guo MLC, Spring 2011
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Chapter 1
where k > 0, n > 0, u =
k , x≥0 (n + 1)
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 ω. 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 effort to understand the formulas and what they mean. EXAMPLE: Under DeMoivre’s law, show that 1. t px
=
ω−x−t ω−x
2. t qx
3.
= ◦
t ω−x
e0 = 4. e0 = 5.
◦
ex = 6. ex =
Guo MLC, Spring 2011
ω 2
ω−1 2 ω−x 2
ω−x−1 2
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Chapter 1
SOLUTION: 1. t px
=
ω−x−t s(x + t) = s(x) ω−x
2. t qx
= 1 − t px =
t ω−x
3. ◦
e0 =
Z
"
0
4. e0 =
ω X
k p0
ω X ω−k
=
1
1
5. ◦
ex =
#
(ω − t)2 ω−t ω dt = − = ω 2ω 2
ω
ω−x
Z 0
ω
=ω−
ω ω−1 1X k= ω 1 2
ω−x−t ω−x dt = ω−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 ω−x−1 1 ◦ ex =ex − = 2 2 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) = ω−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 Guo MLC, Spring 2011
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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 different 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 effect” wears off 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 Guo MLC, Spring 2011
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Chapter 1
for health to deteriorate since she was examined at age x − 1 (one year ago) rather than at age x. EXAMPLE: Select and Ultimate Life Table Using the select and ultimate life table shown above find the value of 1000
Guo MLC, Spring 2011
3 q32 − 3 q[32] .
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Chapter 1
SOLUTION: 3 q32
deals only with the ultimate table so I am only interested in the values of lx + 2 in Column (6). 3 q32
= 1 − 3 p32 = 1 −
9,888 l35 =1− = 0.001313 l32 9,901
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 off 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.)
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Chapter 1
CHAPTER 3 Formula Summary s(x) = 1 − F (X) = 1 − Pr(X ≤ x) Pr(x < X ≤ z) = F (z) − F (x) = s(x) − s(z) Pr(x < X ≤ z|X > x) =
h R t t px
=e
t|u qx
−
0
µ(x+s)ds
i t px
= t px ∗ u qx+t
t|u qx
=
[F (z) − F (x)] [s(x) − s(z)] = [1 − F (x)] [s(x)]
s(x + t) s(x)
t qx
= t+u qx − t qx
= 1 − t px
t|u qx
= t px − t+u px
Pr(K(x) = k) =
k px
− k+1 px
=
k px
∗ qx+k
=
k| qx
Life Tables: lx = l0 ∗ s(x) qx =
n qx
lx − lx+1 lx
=
lx − lx+n lx
n dx
= lx − lx+n
px =
n px
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−µx ◦
ex =
Guo MLC, Spring 2011
1 µ
t px
= e−µt
Var[T ] =
c
Yufeng Guo
1 µ2
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Chapter 1
Expected Future Lifetime ◦
ex = E[T (x)] =
Z
∞
Z t px dt
Var[T (x)] = 2
0
ex = E[K(x)] =
0
∞ X
k px
Var[K(x)] =
∞ X
∞
◦2
t · t px dt− ex
(2k − 1) · k px − e2x
1
1
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! 1
Z
Lx =
Z
lx+t dt
0
0
mx =
(lx − lx+1 ) Lx
Z
n
lx+t dt
n Lx =
n mx
=
lx − lx+n n Lx
∞
lx+t dt
Tx = 0
Tx ◦ = ex lx
n Lx
lx
n
Z
=
t px dt
◦
=ex:n
0
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
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
R
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. Guo MLC, Spring 2011
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Chapter 1
De Moivre’s Law: µ(x) = (ω − x)−1 and s(x) = 1 −
x , ω
where 0 ≤ x < ω
Gompertz’ Law: µ(x) = Bcx and s(x) = exp[−m(cx − 1)] where B > 0, c > 1, m =
B , x≥0 log c
Makeham’s Law: µ(x) = A + Bcx and s(x) = exp[−Ax − m(cx − 1)] B , x≥0 log c
where B > 0, A ≥ −B, c > 1, m =
Weibull’s Law: µ(x) = kxn and s(x) = exp(−uxn+1 ) where k > 0, n > 0, u =
k , x≥0 (n + 1)
DeMoivre’s Law and Modified DeMoivre’s Law: If x is subject to DeMoivre’s Law with maximum age ω, 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 1 µ(x) = ω−x
Modified DeMoivre c µ(x) = ω−x ω−x c ω ω−x c ω
s(x) =
ω−x ω
s(x) =
lx = l0 ·
ω−x ω
l0
◦
ex = E[T ] = Var[T ] = t px
=
(ω−x)2 12
ω−x−t ω−x
µx (t) =
◦
ω−x 2
1 ω−x−t
ω−x c+1
ex = Var[T ] = t px
=
(ω−x)2 c (c+1)2 (c+2) ω−x−t ω−x
µx (t) =
c
c ω−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 different algebraic form. Guo MLC, Spring 2011
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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)
s(x)
F (x)
F (x)
1 − s(x)
Rx
s(x)
1 − F (x)
s(x)
R∞
f (x)
F 0 (x)
−s0 (x)
f (x)
µ(x)
F 0 (x) 1−F (x)
−s0 (x) s(x)
f (x) s(x)
Assumptions for fractional ages. Uniform Function Distribution tqx t qx
f (x) 0
µ(x)
f (u) du
1 − e−
f (u) du
e−
x
Rx 0
µ(x) e−
Constant Force 1 − ptx
Hyperbolic
− log px
qx 1−(1−t)qx
1−t qx+t
(1−t)qx 1−tqx
1 − p1−t x
(1 − t)qx
y qx+t
yqx 1−tqx
1 − pyx
yqx 1−(1−y−t)qx
t px
1 − tqx
ptx
px 1−(1−t)qx
qx
−ptx log px
qx px [1−(1−t)qx ]2
Guo MLC, Spring 2011
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Yufeng Guo
µ(t) dt
µ(t) dt
Rx 0
µ(t) dt
tqx 1−(1−t)qx
qx 1−tqx
+ t)
0
µ(x)
µ(x + t)
t px µ(x
Rx
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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 = ω − x, 0 ≤ x ≤ ω (iii) T (x) is the future lifetime random variable. Calculate Var[T (10)]. (A) 65
(B)93
(C) 133
Solution: Z
◦
e0 =
ω
e10 = 40
Var [T (x)] = 2 0
t ω2 ω dt = ω − = = 25 ⇒ ω = 50 ω 2ω 2
40
Z 0
Z
(E) 333
1−
0 ◦
(D) 178
t 402 1− dt = 40 − = 20 40 (2)(40)
"
t t2 t3 t 1− dt − (20)2 = 2 − 40 2 3 · 40
#40
− (20)2 = 133 0
Key: C
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
Guo MLC, Spring 2011
(C) 0.0790
(D) 0.0796
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Yufeng Guo
(E) 0.0800
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Chapter 1
Solution: 0.0408 = µ(81.5) =
q81 ⇒ q81 = 0.0400 1 − (1/2)q81
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 ω = 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
Solution: STANDARD: ◦
Z
e25:11 =
11
(1 −
0
MODIFIED: p25 = e− ◦
e25:11
t t2 )dt = t − |11 = 10.1933 75 2 × 75 0
R1 0
0.1ds
= e−0.1 = 0.90484
1
10
t )dt 74 0 0 Z 1 Z 10 t = e−0.1t dt + e−0.1 (1 − )dt 74 0 0 ! −0.1 2 t 1−e −0.1 = +e t− |10 0.1 2 × 74 0 Z
Z
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 difference is 0.8047. Key: D
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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 ≥ 0 (ii)
0.4 p0
= 0.50
Calculate F .
(A) -0.20
(B) -0.09
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(C) 0.00
(D) 0.09
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Yufeng Guo
(E) 0.20
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Chapter 1
Solution: 0.4 p0
=e
= 0.5 = e
−0.4F −
h
e2x 2
−
R 0.4 0
i0.4 0
⇒ 0.5 = e−0.4F −0.6128
(F +e2x )dx
=e
−0.4F −
e0.8 −1 2
⇒ ln(0.5) = −0.4F − 0.6128
⇒ −0.6931 = −0.4F − 0.6128
⇒ F = 0.20
Key: E
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
Guo MLC, Spring 2011
(C) 276
(D) 279
(E) 282
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Chapter 1
Solution: From UDD,
Likewise, from
l96.5 =
l96 + l97 . 2
480 =
600 + l97 −→ l97 = 360 2
l97 = 360 and l97.5 = 288, we get l98 = 216. 216 l98 = = 0.6 l97 360
e−µ =
For constant force, 0.5 px
1
= e−0.5µ = (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 enter◦ ing the second year, e1:1.5 . (A) 1.25
(B) 1.3
(C) 1.35
(D) 1.4
(E) 1.45
Solution: Z
◦
e1:1.5 =
1.5 t p1 dt
0
1
Z
=
Z
x p2 dx
0
Z
=
0.5
t p1 dt + 1 p1
1
(1 − 0.1t)dt + 0.9
0
Z
0 0.5
(1 − 0.05x)dx
0
"
0.1t2 = t− 2
#1
"
0.05x2 + 0.9 x − 2 0
= 0.95 + 0.444 = 1.394
Guo MLC, Spring 2011
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Yufeng Guo
#0.5 0
Key: D
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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 ω = 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 =
ω − 30 . 2 ◦
ω = 100 ⇒e30 =
Prior to medical breakthrough ◦0
(E)108
100−30 2
= 35.
◦
After medical breakthrough e30 =e30 +4 = 39. ◦0
⇒e30 = 39 =
ω 0 − 30 ⇒ ω 0 = 108. 2
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
Guo MLC, Spring 2011
(B) 0.43 ≤ P < 0.45
(C) 0.45 ≤ P < 0.47
(E) 0.49 ≤ P ≤ 1.00
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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: (
0.04, 0.05,
µ(x) =
0 < x < 40 x > 40
◦
Calculate e25:25 . (A) 14.0
(B) 14.4
(C) 14.8
Solution: ◦
e25:25 = Z
=
15
Z
−0.04t
e
(D) 15.2
(D) 15.6
15
Z t p25 dt
10 t p40 dt
+ 15 p25 0
0
dt + e
−
R 15 0
0.04 ds
0
Z
10
e−0.05t dt
0
1 1 = 1 − e−0.60 + e−0.60 1 − e−0.50 0.04 0.05
= 11.2797 + 4.3187 = 15.60 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 ω = 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
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(C) 0.88
(D) 0.92
c
Yufeng Guo
(E) 0.96 49 49 of 674
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Chapter 1
SOLUTION: From the Illustrative Life Table: l50 = 0.8951 l0 Z
∞
Z
50
Z
1.2 f2 (t) dt 50
0
0
∞
kf1 (t) dt +
fT (t) dt =
1=
l40 = 0.9313 l0
= kF1 (50) + 1.2 (F2 (∞) − F2 (50)) = k (1 − 50 p0 ) + 1.2(1 − 0.5) = k(1 − 0.8951) + 0.6 ⇒k= For x ≤ 50, Z
1 − 0.6 = 3.813 1 − 08951
x
3.813f1 (t) dt = 3.813F1 (x)
FT (x) = 0
This gives us the following two results: l40 FT (40) = 3.813 1 − l0
l50 FT (50) = 3.813 1 − l0
10 p40
=
= 0.262 = 0.400
1 − FT (50) 1 − 0.400 = = 0.813 1 − FT (40) 1 − 0.262
12. For a double decrement table, you are given: (1)
(τ )
(i) µx (t) = 0.2µx (t), t > 0 (τ )
(ii) µx (t) = kt2 , t > 0 0 (1)
(iii) qx
= 0.04 (2)
Calculate 2 qx . (A) 0.45
(B) 0.53
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(C) 0.58
(D) 0.64
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(E) 0.73
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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
= 1−e−
R1 0
(τ )
0.2µx (t) dt
= 1−e−
R1 0
0.2kt2 dt
= 1−e−0.2k/3
⇒ e−0.2k/3 = 0.96 (τ ) µ(1) x (t) = 0.2µx (t)
(2) 2 qx
Z
= 0
2
(τ ) t px
·
µ(2) x (t)dt
(τ ) ⇒ µ(2) x (t) = 0.8µx (t)
Z
= 0
2
(τ ) t px
) (τ ) · (0.8)µ(τ x (t)dt = 0.82 qx
(τ )
To get 2 qx , we use (τ ) 2 qx
− ) = 1 − 2 p(τ x =1−e
R2 0
(τ )
µx (t) dt
= 1 − e−0.2k/3
40
= 1 − e−
R2 0
kt2 dt
= 1 − e−8k/3
= 1 − (0.96)40
⇒ 2 qx(τ ) = 0.8046 ⇒ 2 qx(2) = (0.8)(0.8046) = 0.644
13. You are given: ◦ (i) e30:40 = 27.692 (ii) s(x) = 1 − ωx , x ≤ ω (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 ω−30−t ω−30 dt Guo MLC, Spring 2011
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Chapter 1
t2 2(ω−30) 800 − ω−30
= t−
i
|40 0
= 40 = 27.692 ⇒ ω = 95 t p30
=
65−t 65
Now, realize (after getting ω = 95) that T (30) is uniformly on (0, 65), its variance is just the variance of a continuous uniform random variable: 2 = 352.08 V ar = (65−0) 12 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 Guo MLC, Spring 2011
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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 0 (1 − t qx )dt = 0 (1 − tqx )dt
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−1 = 0.5(1 qx + 1 px + 1 px ) + 1 px × ex+1:n−1 = ◦ 0.51 qx + 1 px (1 + ex+1:n−1 ). Set n = ∞, we have: e˙ x = 21 qx + px (1 + e˙ x ) e˙ [80] = 21 q[80] + p[80] (1 + e˙ 81 ) 8.5 = 21 (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] = 21 q[81] + p[81] (1 + e˙ 82 ) = 21 (1 − 0.912) + (0.912)(1 + 8.043) = 8.206 Key: C
15. You are given: (
µ(x) =
0.05 50 ≤ x < 60 0.04 60 ≤ x < 70
Calculate
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
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Chapter 1
4|14 q50 = 4 p50 − 18 p50 = 0.8187 − 0.4404 = 0.3783 Key: A
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Chapter 1
16. For a population which contains equal numbers of males and females at birth: (i) For males, µm (x) = 0.10, x ≥ 0 (ii) For females, µf (x) = 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−(0.1)(61) + 0.5e−(0.08)(61) = 0.00492 q60 = 1 − Key: B
0.00492 0.005354
= 0.081
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Chapter 1
Problems from Pre-2000 SOA-CAS exams s
1. You are given: µ(x) =
1 , 0 ≤ x < 80. 80 − x
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)
e−4 2
(B)
e−2 2
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(C)
ln 2 2
(D)
ln 4 2
c
Yufeng Guo
(E) 1
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Chapter 1
6. Calculate mx , the central-death-rate at age x. (A)
e−2 2
(B) e−2
(C) 2e−2
7. You are given: x α , 0 ≤ x < ω, s(x) = 1 − ω
(D) 1
(E) 2
where α > 0 is a constant.
◦
Calculate µx · ex . (A)
α α+1
(B)
αω α+1
(C)
α2 α+1
(D)
α2 ω−x
(E)
α(ω − x) (α + 1)ω
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
(B) 85
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(C) 94
(D) 178
(E) 213
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Chapter 1
Solutions to Chapter 3 1. Key: A
We need to find t such that t p20 = 0.5 ⇒ 0.5 = e−
Rt 0
µ(20+s) ds
= e−
Rt 0
(60−s)−0.5 ds
√
t
√
0.5 = e2[(60−s) ]0 = e2[ 60−t− 60] h√ √ i ln 0.5 = 2 60 − t − 60 √ ⇒ 60 − t = 7.40 ⇒ t = 5.25
2. Key: D
∞
Z
lx+1+t dt
Tx+1 = 0
lx+1 =
6.4 =8 0.8 ∞
Z
Tx+1 = 8
⇒ lx+1+t = 8(0.8)t
(0.8)t dt =
0
=
8 (0.8)t |∞ 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 0.5 = t px = e−µ·t = e−2t ⇒t=
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Chapter 1
6. Key: E mx = =
lx − lx+1 lx (1 − e−µ ) = R1 Lx 0 lx · t px dt
lx (1 − e−2 ) 1 − e−2 == 1 =2 R1 −2 lx 0 e−µt dt 2 (1 − e )
7. Key: A This is a Modified DeMoivre problem with constant equal to α. So µ(x) =
α , ω−x
◦
ex =
ω−x α+1
Multiplying these two quantities gives α α+1
8. Key: B
In both cases the probability is given by 0.5 p60
· 0.5 q60.5 + 1 p60 · 0.5 q61
(0.5)(0.3) UDD: [1 − (0.5)(0.3)] 1−(0.5)(0.3) + (0.7) · (0.5)(0.4) = 0.2900 = f
(0.5)(0.3) 0.7 Balducci: 1−(0.5)(0.3) + (0.7) · 1
(0.5)(0.4) 1−(0.5)(0.4)
= 0.2985 = g
10,000(g − f ) = 85
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MLC Problem Solving Dictionary: Multiple Life Function ONLY Yufeng Guo November 28, 2010
492 of 674
Contents 1 Integrals R 1.1 0 − . . R 1.2 0 − . . R 1.3 0 2 − . 1.4 [( ∧ ) ] . 1.4.1 If is 1.4.2 If is 2 Multiple lives 2.1 2.2 2.3 2.4
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7 42 48 57
1 + 2 = 1 :| :| :| 1 + 2 = 1 :| :| :| . . . . 1 + 2 = + 1 = 1 . . . . . .
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57 57 57 57
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57
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58
2.5.3 1 + 1 = . . . . . . . . . . . . . . . . . . . . . . −2 = 1 −1 =1 . . . . . . . . . . . . . . . . . . 1
58 58
2.6.1
:|
1
:|
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:|
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:|
1
:| 1
1
. . . .
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\:|
+
:|
:|
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1
:|
+
\:|
=
1
:|
=
58
1
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58
2.6.3 − 2 = − 1 = 1 . . . . . . . . . . . . . . Also see . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
58 66
1
:|
−
:|
:|
. . . . . .
:|
:|
1
\:|
:|
−
1
\:|
1
2.6.2
2 2
=
1
\:|
=
1
\:|
−
1
=
−
1
=
:| :|
:| :|
3 First death, last death 3.1
1 1 1 2 2 2 3
. . . .
2.5.2
2.7
:|
. . . . . .
. . . .
2.5.1
2.6
1
1 , 2 . . . . . :| :| 1 , 2 . . . . . :| :| . . . . . . . . 1 , 2 + =1 . . . 1 2 2.4.1 2.4.2 2.4.3
2.5
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67
First to die . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
67
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CONTENTS 3.2 3.3 3.4
3.5
3.6
CONTENTS
Last-to-die . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Typical problems . . . . . . . . . . . . . . . . . . . . . . . . . . () () . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.1 constant . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.2 UDD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.3 Also See . . . . . . . . . . . . . . . . . . . . . . . . . . . . mean and variance: () () () () ()| ()| 3.5.1 constant force of mortality . . . . . . . . . . . . . . . . . . 3.5.2 de Moivre’s law . . . . . . . . . . . . . . . . . . . . . . . . 3.5.3 Also see . . . . . £. . . . . . . . ¤. . . . ³. . . . . . . . . ´. . [ () ()] () () ()| ()| .
4 Joint life annuity 4.1 Formulas . . . . . . . . . . . 4.1.1 Reversionary annuity . 4.1.2 Also see . . . . . . . . 4.2 Complex annuity . . . . . . .
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68 72 88 90 96 102 103 103 106 111 111 125 125 126 127 127
5 Solution to Sample MLC (Multiple Life Function) 129 5.1 q1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 0
5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10
q31 . . . . . . . . . . . . . q46 joint life annuity immediate q73 exactly one alive . . . . . . 0 q91 . . . . . . . . . . . . . q94 () . . . . . . . . . . . . q104 . . . . . . . . . . . . . q112 joint life annuity . . . . . q123 . . . . . . . . . . . . q128 . . . . . . . . . . . .
5.11 5.12 5.13 5.14 5.15 5.16 5.17 5.18 5.19 5.20
q150 q173 q193 q261 q262 q263 q265 q266 q268 q269
0
. . . . . . . . joint life annuity . . . . . . . . . . 2 . . . . . . . . . . . . . . 1 . . . . . . . 2 1 . . . . . . . . 1 , 2 . . . . . . . . . . . . . . . . . . . . .
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129 130 130 131 131 132 132 132 133
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133 133 134 134 135 135 135 135 136 137
0
5.21 q270 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 5.22 q271 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
5.23 q272 [ ()] . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 5.24 q273 [ () ()] . . . . . . . . . . . . . . . . . . . . . . 138 www.actuary88.com
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CONTENTS 5.25 5.26 5.27 5.28 5.29
q278 q279 q280 q281 q282
CONTENTS
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Preface This is Guo’s problem solving dictionary for Exam MLC for Multiple Life Function only (Actuarial Mathematics Chapter 9). When I was studying for the actuarial exam, I often bought a primary study manual and secondary study manual (in case that the primary manual didn’t explain some topics clearly enough). Even having two study guides, however, I often found that some topics were still not explained clearly in either the manual. Then I thought how nice it would be if I could just buy a small study manual that covered only some of the most difficult topics. Multiple Life Function is one of the most difficult topics in Exam MLC. If you already have a study manual, chances are that some concepts are still murky in your head. Hence this study manual. This study guide offers crystal clear explanation to many core concepts on multiple life function and provides challenging problems. This study guide doesn’t Common Shock. Why not? Because Professor Jim Daniel has published a fine essay on Common Shock as a free download (and as an ad for his MLC seminar). You don’t need to read another essay from me. Just go to his website http://www.actuarialseminars.com/Misc/Shock.pdf and download the wonderful pdf file. By the way, consider Professor Jim Daniel’s seminar especially if your employer foots the bill. This study guide has many heavy duty integrals. To better explain how I evaluate difficult integrals, I also included a chapter on integrals. At the end of this book is my solution to Sample MLC problems about Multiple Life Function. This book is the exclusive property of Yufeng Guo. Redistribution of this book in any form without the author’s permission is prohibited. Please report any errors to
[email protected].
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2.1.
1
: |
,
CHAPTER 2. MULTIPLE LIVES
2
: |
1
:|
,
2
: |
Problem 2.7. You are given the following information about a 10-year term insurance on two independent lives and : • If either life dies within the 10 year term, 1 is immediately paid on the first death if it is the life of ; 05 is immediately paid on the first death if it is the life of • If the second death occurs within the remainder of the 10 years term, twice the amount previously paid is paid immediately upon the second death • () = 004 • () = 006 • = 008 Calculate the net single premium of this insurance policy. Solution. payment 1 upon ’s death if dies first in 10-yr term
NSP 1
2 upon ’s death if dies second in the remainder of 10-yr term
2
05 upon ’s death if dies first in 10-yr term
05
1 upon ’s death if dies second in the remainder of 10-yr term
1
:10|
=
1
:10|
0185 49 2
:10|
1
:10|
=
1
:10|
0278 23 2
2
:10|
1
:10|
2
:10|
¢ ¢ ¡ 004 ¡ 1 − −10( +) = 1 − −10(004+008) = 0232 94 + 004 + 008 ´ ³ ¢ ¡ 004 1 − −10(004+006+008) = = 1 − −10( + +) = + + 004 + 006 + 008 = 1
:10|
−
1
:10|
= 0232 94 − 0185 49 = 0047 45
¢ ¢ ¡ ¡ 006 1 − −10( +) = 1 − −10(006+008) = 0322 89 + 006 + 008 ³ ´ ¢ ¡ 006 1 − −10(004+006+008) = = 1 − −10( + +) = + + 004 + 006 + 008
:10|
= 1
:10|
Total NSP: 1 + 2 :10|
:10|
2
−
:10|
1
:10|
+ 05
= 0322 89 − 0278 23 = 0044 66
1
:10|
+
2
:10|
= 0185 49 + 2 (0044 66) +
05 (0278 23) + 0047 45 = 0461 38 www.actuary88.com
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CHAPTER 2. MULTIPLE LIVES
1
: |
,
2.1.
2
: |
1
: |
,
2
: |
Comments. = 1 , 2 = 2 . Notation. 1 :10| :10| :10| :10| Don’t be scared by the phrase "in the remainder of the 10 year term." If is the 2nd death within the 10-year term, then automatically dies in the remainder of the 10 year term (because needs to die first within 10 years.) Finally, since as:
1
:10|
+2
1
:10|
2
:10|
+
+05
2
:10|
1
:10|
= 1
+
:10|
, the total NSP can be simplified
2
:10|
= 2
2
:10|
+05
1
:10|
+ 1
:10|
= 2 (0044 66) + 05 (0278 23) + 0232 94 = 0461 38 Problem 2.8. You are given the following information about a whole life insurance on two independent lives and : • It pays 1 immediately upon ’s death if dies before • It pays 2 immediately upon ’s death if dies after • It pays 3 immediately upon ’s death if dies before • It pays 4 immediately upon ’s death if dies after • () = 002 • () = 003 • = 004 Calculate the net single premium of this insurance policy. Solution. payment 1 upon ’s death if dies first
NSP 1
2 upon ’s death if dies second
2
3 upon ’s death if dies first
3
4 upon ’s death if dies second
4
2
1
2
002 = = 0333 33 + 002 + 004 002 = = 0222 22 = + + 002 + 003 + 004 = − 1 = 0333 33 − 0222 22 = 0111 11
=
1
2
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MLC Problem Solving Dictionary: Poisson Process ONLY Yufeng Guo November 28, 2010
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Contents 1 Old SOA/CAS problems
1
2 Recent SOA problems
5
3 Original problems
7
4 Bernoulli Process (not in Daniel’s study note but good to know) 19 5 FAQ about the Daniel’s study note
25
iii
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Preface This is Guo’s problem solving dictionary for Exam MLC for Poisson process only (Daniel study note). This book is the exclusive property of Yufeng Guo. Redistribution of this book in any form without the author’s permission is prohibited. Please report any errors to
[email protected].
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CHAPTER 3. ORIGINAL PROBLEMS 2 1 3 (0561 80) + (0337 08) + (0101 12 ) = 12 ( | = 1 2 3) = 02 02 02 303 4 The expected total hunting time, given that he has been hunting for 3 hours is: 3 + 12 303 4 = 15 303 4 Problem 3.3. Customers arrive at a bank according to a Poisson process with = 2 customers per hour. Calculate the probability that 4 customers arrives between 8 a.m. and 10 a.m. and the third customer arrives between 9 : 20 a.m. and 9 : 30 a.m.. Solution. Set 8 a.m. as time zero. Let’s use minutes to measure time. Consider the interval (0 120), the time between 8 a.m. and 10 a.m.. Let represent a time point between 9 : 20 a.m. and 9 : 30 a.m.. So ∈ (80 90). 4 Customers arrive during (0 120) and the 3rd customer arrives during (80 90). This means: • 2 customers arrive during (0 ). Probability: [ () = 2] = −
()2 2!
• 1 (the third) customer arrives during ( + ). Prob: [ () = 1] = 1 () = −() ≈ −() 1! • 1 (the 4th) customer arrives during ( 120). Prob: [ (120 − ) = 1] = (120 − ) −(120−) 1! 2
() (120 − ) 4 2 (120 − ) ××−(120−) = −120 Total probability: − 2! 1! 2 1 2 = per minute (2 per hour) where = 60 30 Next, we add up the above probability for any ∈ (80 90): R 90 −120 4 2 (120 − ) 4 −120 R 90 2 = (120 − ) 80 80 2 2 4 −120 = 2 R 90 80
µ
1 30
¶4
−120
2
1 30
=
µ
1 30
¶4
−4
2
¢ R 90 ¡ (120 − ) = 80 1202 − 3 = 2
¢ 1¡ 4 90 − 804 = 2517 500 4
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c °Yufeng Guo
µ ¶90 ¢ ¡ 1 4 3 40 − = 40 903 − 803 − 4 80 15
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CHAPTER 3. ORIGINAL PROBLEMS
4 −120 R 90 2 (120 − ) = 80 2 −2 846 27 × 10
µ
1 30
¶4
−4
2
× 2517 500 =
1007 −4 = 2 648
If you want to use an hour as the unit time, then = 2 Consider the interval (0 2), the time between 8 a.m. and 10 a.m.. Let ¶ point between 9 : 20 a.m. and 9 : 30 a.m.. So µ represent a time 4 30 3 20 = 1 = . ∈ 1 60 3 60 2 µ ¶ 4 3 4 Customers arrive during (0 2) and the 3rd customer arrives during . 3 2 This means: • 2 customers arrive during (0 ). Probability: [ () = 2] = −
()2 2!
• 1 (the third) customer arrives during ( + ). Prob: [ () = 1] = 1 () = −() ≈ −() 1! • 1 (the 4th) customer arrives during ( + 2) ≈ ( 2). Prob: [ (2 − ) = 1] = (2 − ) −(2−) 1! ()2 (2 − ) 4 2 (2 − ) × × −(2−) = −2 2! 1! µ 2 ¶ 4 3 : Next, we add up the above probability for any ∈ 3 2 R 32 −2 4 2 (2 − ) 4 −2 R 32 2 24 −2(2) R 32 2 = (2 − ) = (2 − ) = 43 43 43 2 2 2 1007 −4 = 2 846 27 × 10−2 648 Total probability: −
Problem 3.4. Customers arrive at a bank according to a Poisson process with a constant . Calculate the conditional probability that given 4 customers arrives between 8 a.m. and 10 a.m., the third customer arrives between 9 : 20 a.m. and 9 : 30 a.m.. Solution. Set 8 a.m. as time zero. Let’s use an hour as the unit. Consider the interval (0 2), the time between 8 a.m. and 10 a.m.. = 4 customers arrive during the interval (0 2). ¶ µ 4 3 = 3rd customers arrive during the interval 3 2 www.actuary88.com
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