ARCH MLC

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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.

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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.

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Yufeng Guo

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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.

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0 INTRODUCTION 9

The origin of Arch Manual . . . 9

Praises of the Arch Manual originally written by Nathan and Robin . . . 10

About Yufeng Guo . . . 10

How to use this manual . . . 11

1 ACTUARIAL MATHEMATICS: CHAPTER 3 SURVIVAL DISTRIBU-TIONS AND LIFE TABLES 13 3.2.1 The Survival Function . . . 15

3.2.2 Time-until-Death for a Person Age x . . . 16

3.2.3 Curtate-Future-Lifetime . . . 19

3.2.4 Force of Mortality . . . 20

3.3-3.5 Life Tables . . . 21

3.5.2 Recursion Formulas . . . 27

3.6 Assumptions for Fractional Ages . . . 28

3.7 Some Analytical Laws of Mortality . . . 33

Modified DeMoivre’s Law . . . 35

3.8 Select and Ultimate Tables . . . 35

Conclusion . . . 38

CHAPTER 3 Formula Summary . . . 39

Past SOA/CAS Exam Questions: . . . 43

Problems from Pre-2000 SOA-CAS exams . . . 56

Solutions to Chapter 3 . . . 58

2 ACTUARIAL MATHEMATICS CHAPTER 4 – LIFE INSURANCE 61 4.2 Insurances Payable at the Moment of Death . . . 62

TYPES OF INSURANCE . . . 63

4.2.1 Level Benefit Insurance . . . 63

4.2.2 Endowment Insurance . . . 66

4.2.3 Deferred Insurance . . . 68

4.2.4 Varying Benefit Insurance . . . 70

4.3 Insurances Payable at the End of the Year of Death . . . 73

4.4 Relationships between Insurances Payable at the Moment of death and the End of the Year of Death . . . 79

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Solutions . . . 95

3 ACTUARIAL MATHEMATICS: CHAPTER 5 LIFE ANNUITIES 97 5.2 Continuous Life Annuities . . . 97

The most important equation so far(!) . . . 100

5.3 Discrete Life Annuities . . . 106

5.4 Life Annuities with m-thly Payments . . . 112

Continuous Annuities: . . . 116

Discrete annuities: . . . 117

Solutions to Pre-2000 Problems: Chapter 5 . . . 133

4 ACTUARIAL MATHEMATICS: CHAPTER 6 BENEFIT PREMIUMS 135 6.2 Fully Continuous Premiums . . . 136

6.3 Fully Discrete Premiums . . . 142

6.4 True m-thly Payment Premiums . . . 147

Solutions to Pre-2000 Exam Questions: Chapter 6 . . . 170

5 ACTUARIAL MATHEMATICS: CHAPTER 7 BENEFIT RESERVES 173 7.2 Fully Continuous Benefit Reserves . . . 174

7.3 Other Methods for Calculating the Benefit Reserve . . . 177

1) Prospective Formula . . . 177

2) Retrospective Formula . . . 177

3) Premium Di↵erence Formula . . . 180

4) Paid-Up Insurance Formula . . . 180

5) Other Reserve Formulas . . . 181

7.4 Fully Discrete Reserves . . . 183

7.5 Benefit Reserves on a Semi-Continuous Basis . . . 186

7.6 Benefit Reserves Based on True m-thly Benefit Premiums . . . 187

Continuous Reserve Formulas: . . . 188

Discrete Reserves: . . . 188

Solutions to Pre-2000 Questions: Chapter 7 . . . 196

6 ACTUARIAL MATHEMATICS: CHAPTER 8 ANALYSIS OF BENEFIT RESERVES 197 8.2 Benefit Reserves for General Insurances . . . 197

8.3 Recursion Relations for Fully Discrete Benefit Reserves . . . 202

8.4 Benefit Reserves at Fractional Durations . . . 205 c Yufeng Guo

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8.5 The Hattendorf Theorem . . . 208

Chapter 8 More Formulas . . . 213

Solutions to Pre-2000 Problems: Chapter 8 . . . 237

7 ACTUARIAL MATHEMATICS: CHAPTER 9 MULTIPLE LIFE FUNC-TIONS 241 9.2 Joint Distributions of Future Lifetimes . . . 241

9.3 Joint Life Status . . . 243

The following is important! . . . 243

9.4 Last Survivor Status . . . 248

9.5 More Probabilities and Expectations . . . 250

9.6 Dependent Lifetime Models . . . 253

9.6.1 Common Shock –(Non-Theoretical Version) . . . 253

9.7 Insurance and Annuity Benefits . . . 256

9.7.1 Survival Statuses . . . 256

9.7.2 Special Two-Life Annuities . . . 262

9.7.3 Reversionary Annuities . . . 263

9.9 Simple Contingent Functions . . . 265

Solutions to Pre-2000 Exam Questions: Chapter 9 . . . 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 asso-ciated single decrement table . . . 300

CASE II : Three decrements that are uniformly distributed in the as-sociated single decrement table . . . 300

CASE III : Multiple decrements – some are uniformly distributed in the associated single decrement table and some are not. . . 300

Problems from Pre-2000 SOA-CAS Exams . . . 317

Solutions to Pre-2000 Problems: Chapter 10 . . . 320 c Yufeng Guo

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9 ACTUARIAL MATHEMATICS: CHAPTER 11 APPLICATIONS OF

MUL-TIPLE DECREMENT THEORY 323

11.2 Actuarial Present Values and Their Numerical Estimation . . . 323

11.3 Benefit Premiums and Reserves . . . 324

ARCH Sample Exam Problem . . . 327

Solution: . . . 327

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

Asset Shares . . . 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

12 DANIEL STUDY NOTE ON POISSON PROCESS 385 5.3 The Poisson Process . . . 385

5.3.1 Counting Processes . . . 385

5.3.2 Definition of the Poisson Process . . . 386

5.3.3 Interarrival and Waiting Time Distributions . . . 387

5.3.4 Further Properties of Poisson Processes . . . 389

5.3.5 Conditional Distribution of the Arrival Times . . . 392

5.4 Generalizations of the Poisson Process . . . 393

5.4.1 Nonhomogeneous Poisson Process . . . 393

5.4.2 Compound Poisson Process . . . 395

5.4.3 Conditional or Mixed Poisson Processes: Gamma-Poisson Model 398 CHAPTER 5 Formula Summary . . . 401

ARCH Warm-up Problems: . . . 402

Solutions: . . . 403

Past SOA/CAS Exam Questions: . . . 404 c Yufeng Guo

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13 ARCH Practice Exam 417 Answer Key for Practice Exam . . . 430

14 SOLUTION TO MAY 2007 MLC 449

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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 daunt-ing challenge of plowdaunt-ing through difficult textbooks and masterdaunt-ing 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 study-ing for exams, they created their own study framework and philosophy for quickly passstudy-ing 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.

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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 recom-mending 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:

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Fall 2002 Passed Course 1 Spring 2003 Passed Course 2,3 Fall 2003 Passed Course 4 Spring 2004 Passed Course 6 Fall 2004 Passed Course 5 Spring 2005 Passed Course 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 seehttp://www.guo.coursehost.com andhttp://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 Under-standing 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

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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 prac-tice. 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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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).

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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) O _x _z F(x) s(x) the pdf y=f(x)

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

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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 prob-ability 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. F (40) F (17) F (40) F (10) or s(17) s(40) s(10) s(40) c Yufeng Guo

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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):

tqx = probability that person age x will die within t years

= Pr[T (x)  t] where t 0

tpx = probability that person age x will survive at least t years

= Pr[T (x) > t] where t 0

x x+t Age

tqx t xp

In the figure,tqxis the probability that (x)’s death will occur in the age-interval (x, x+t), and tpx 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:

• tp0 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 ontpx ortqx which you will learn with practice.

The symbol

t|uqx

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+uqx tqx tpx t+upx

(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.)

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Useful formulas: tpx = x+t p0 xp0 = s(x + t) s(x) tqx = 1 s(x + t) s(x) t|uqx = s(x + t) s(x + t + u) s(x) = s(x + t) s(x) ⇤ s(x + t) s(x + t + u) s(x + t) = tpx⇤uqx+t

This last equation makes sense. It says “The probability of (x) dying between t and t + u years from now (t|uqx) is equal to the probability that (x) will first survive t years (tpx) and

then die within u years (uqx+t).”

If you don’t remember anything else from the above, remember the following!

tpx=

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|6qx in terms of F and then in terms of p.

SOLUTIONS: 1. 25p52 2. 17q74 3. q33 4. 7|17q43 5. _5|6qx = s(x + 5) s(x + 11) s(x) = F (x + 11) F (x + 5) 1 F (x) =5px(1 6px+5) or =5px 11px. }

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

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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) = kpx k+1px

= kpx⇤ 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= s(37) s(38) s(18) = 63 62 82 = 1 82. } c Yufeng Guo

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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) 1 F (x) =

s0(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 (x) F (x) 1 s(x) R₀xf (u) du 1 e Rx 0 µ(t) dt s(x) 1 F (x) s(x) R_x1f (u) du e R x 0 µ(t) dt f (x) F0(x) s0(x) f (x) µ(x) e R x 0 µ(t) dt µ(x) _{1 F (x)}F0(x) _s(x)s0(x) f (x)_s(x) 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. tpx = e µt SOLUTION: 1. s(x) = e Rx 0 µ dt= e µx. 2. tpx = s(x + t) s(x) = e µt_. _} c Yufeng Guo

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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 !)

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

Relationships: lx = l0⇤ s(x) qx = lx lx+1 lx nqx = lx lx+n lx npx = lx+n lx ndx = lx 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 c Yufeng Guo

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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 40d2. 3. Find 38q3. 4. Find 2|q40. SOLUTION: 1. s(42) = 92,595.7_{100,000 = 0.925957.} 2. 40d2 = l2 l42= 5230.6 3. 38q3= 1 38p3 = 1 l_l41₃ = 1 92,872.6_{97,706.6 = 0.04947.} 4. 2|q40=2p40· q42= 92,595.7_{93,131.6(0.0032) = 0.003182.} }

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) =tpxµ(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 tpxµ(x + t)dt =

Z ₁

0 f (t)dt = 1.

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

e_x= E[T (x)] = Z ₁ 0 tpxdt Var[T (x)] = 2Z 1 0 t_·tpxdt e 2 x (3.5.4)

The book shows how to figure both of these formulas out with integration by parts in Sec-tion 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 thatmpx=mqx. 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)] = 1

2 c Yufeng Guo

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or s[x + m(x)] s(x) = 1 2 or mpx = 1 2.

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 1 kpx Var[K(x)] =X1 1 (2k 1) ·kpx e2x

Find e0 and e50 if the force of mortality is a constant µ.

SOLUTION: e₀= Z ₁ 0 tp0dt = Z ₁ 0 e µt_{dt =} 1 µ e µt 1 0 = 1 µ e50= Z ₁ 0 t p50dt = Z ₁ 0 e µtdt =  ₁ µ e µt 1 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

0 Otherwise

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

SOLUTION: e0 = 50 X 1 tp0 = 50 X 1 50 t 50 = 50 1 50 50 X 1 t c Yufeng Guo

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= 50 1 50 (50)(51) 2 = 24.5 e45= 5 X 1 tp45= 5 X 1 s(45 + t) s(45) = 5 X 1 5 t 5 = 4 + 3 + 2 + 1 + 0 5 = 2. }

More Life Functions:

The expression Lxdenotes 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+tdt

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:

nLx = Z n 0 lx+tdt nmx = lx lx+n nLx

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

lx+tdt (3.5.16)

Note from the definitions that you can think of Tx as1Lx.

The final symbol is ↵(x). It’s the expected death time given x dies next year. c Yufeng Guo

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↵ (x) = E [T|T < 1] = R1 0 tf (t)dt R1 0 f (t)dt = R_R01ttpxµ(x+t)dt 1 0 tpxµ(x+t)dt = R01t· lx+t lx µ(x+t)dt R1 0 lx+t lx µ(x+t)dt = R_R01t·lx+tµ(x+t)dt 1 0 lx+tµ(x+t)dt

If UDD, then f(t) = qx= c is a constant. Then ↵(x) =

R1 0 t·cdt R1 0 cdt = R 1 0 tdt R1 0 dt =R1 0 tdt = 12.

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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SOLUTION: (A) L5 = Z 1 0 l5+tdt = Z 1 0 t p5l5dt. Since l5 = l0e µ·5= 1000e 0.5= 606.5, we have L5 = 606.5 Z 1 0 e 0.1tdt = 606.5h 10e 0.1ti1 0 = 606.5h10(1 e 0.1)i= 577.16. (B) m5 = l5 l6 L5 = 606.5 548.8 577.16 = 0.10

This approximates the rate at which people were dying between the 5th and 6th years. (C) T5= Z ₁ 0 1000e 0.1(5+t)_{dt = 606.5}Z 1 0 e 0.1tdt = 6065

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

lx

= ex

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 lxsurvivors

at age x as: nLx lx = Z n 0 tpxdt nLx lx

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

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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 pxds d(x) = px

Starting Value =e!= u(!) = 0

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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 tpx 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 tpx 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. tqx = s(x) s(x + t) s(x) = t s(x) s(x + 1) s(x) = tqx f (t) = d dttqx = qx c Yufeng Guo

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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:

qx= Z 1 0 f (t)dt = f (t) Z 1 0 dt = f (t)

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 s(x + t) = 1 t s(x) + t s(x + 1) All you need to know about Balducci assumption is this:

1 tqx+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 tqx+t = (1 t)qx. Later I’ll show you how to

derive other formulas in Balducci assumptions.

Uniform Constant

Function Distribution Force Hyperbolic

tqx tqx 1 ptx 1 (1 t)qtqx x µ(x + t) qx 1 tqx log px qx 1 (1 t)qx 1 tqx+t (1 t)q_{1 tq}_xx 1 p1 tx (1 t)qx yqx+t _{1 tq}yqx_x 1 pyx 1 (1 y t)qyqx x tpx 1 tqx ptx 1 (1 t)qpx x tpxµ(x + t) qx ptxlog px _{[1 (1 t)q}qxpx _x_]2 c Yufeng Guo

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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 tqx= Z t 0 f (t)dt = Z t 0 qxdt = qx Z t 0 dt = tqx tpx = 1 tqx= 1 tqx µ(x + t) = µx(t) = f (t) tpx = qx 1 tqx tpxµ(x + t) = f (t) = qx ypx+t= s(x + t + y) s(x + t) = s(x)t+ypx s(x)tpx = t+ypx tpx = 1 (t + y)qx 1 tqx yqx+t= 1 ypx+t= yqx 1 tqx 1 tqx+t= (1 t)qx 1 tqx

Under constant force of mortality, for 0  t  1: µx(t) = µ tpx = e Rt 0µdt = e µt px= e µ µ = ln px tpx = e µt = (e µ)t= (px)t ypx+t= t+y px tpx = (px)t (px)(t+y) = (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)

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Under Balducci assumption: The starting point of Balducci assumption is

1 tqx+t= (1 t)qx

You can derive all the other formulas from this starting point. For example, let’s derive the formula fortpx.

T (x) 0 t 1

Age x x + t x + 1

Number of people alive s(x) = 1 s(x + t) = px

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

This is how to derive s(x + t) = 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 formula1 tqx+t= (1 t)qx. 1 tqx+t= 1 s(x + 1) s(x + t) = 1 px s(x + t) = (1 t)qx s(x + t) = px 1 (1 t)qx

However,tpx= s(x+t)_s(x) = s(x + t). This gives us: tpx= _{1 (1 t)q}px _x.

Next, let’s derive the formula f(t) =tpxµ(x + t) =tpxµx(t)

f (t) = d dttpx = d dt px 1 (1 t)qx = qxpx [1 (1 t)qx]2 Derive µ(x + t): µ(x + t) = f (t) tpx = qx 1 (1 t)qx Deriveyqx+t: yqx+t= 1 s(x + t + y) s(x + t) Use the formula: s(x + t) = px

1 (1 t)qx

Replace t with t + y, assuming 0  t + y  1: s(x + t + y) = px

1 (1 t y)qx

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

Finally, let’s derive 1

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

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

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

A couple of time-saving formulas that are valid under UDD only!! ex= ex+1₂

Var(T ) = Var(K) + 1 12 EXAMPLE:

You are given that qx = 0.1. Find (A)0.5qx, (B) 0.5qx+0.5 under each of

• Uniform Density of Deaths (UDD)

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

(A) • UDD:

0.5qx= (0.5)(0.1) = 0.05

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• CF: 0.5qx= 1 (0.9)0.5 = 0.0513 • Balducci: 0.5qx = ₁ (0.5)(0.1)_(0.5)(0.1) = 0.0526. (B) _{• UDD:} 0.5qx+0.5 = (0.5)(0.1) 1 (0.5)(0.1) = 0.0526 • CF: 0.5qx+0.5 = 1 (0.9)0.5 = 0.0513 • Balducci: 0.5qx+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 pro-fession, 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 1)] where B > 0, c > 1, m = B log c, x 0 Makeham’s Law: µ(x) = A + Bcx and s(x) = exp[ Ax m(cx 1)] where B > 0, A B, c > 1, m = B log c, x 0 Weibull’s Law: µ(x) = kxn and s(x) = exp( uxn+1) c Yufeng Guo

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where k > 0, n > 0, u = k

(n + 1), x 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. tpx= ! x t ! x 2. tqx= t ! x 3. e0= !₂ 4. e0 = ! 1 2 5. e_x= ! x 2 6. ex = ! x 1 2 c Yufeng Guo

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SOLUTION: 1. tpx= s(x + t) s(x) = ! x t ! x 2. tqx= 1 tpx = t ! x 3. e0= Z ! 0 ! t ! dt = " (! t)2 2! # = ! 2 4. e0 = ! X 1 kp0 = ! X 1 ! k ! = ! 1 ! ! X 1 k = ! 1 2 5. ex= Z ! x 0 ! x t ! x 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 1₂ = ! ₂x 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

µ(x) = c

! 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

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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

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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⇣3q32 3q[32]

⌘

.

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SOLUTION:

3q32 deals only with the ultimate table so I am only interested in the values of

lx+ 2 in Column (6). 3q32= 1 3p32= 1 l35 l32 = 1 9,888 9,901 = 0.001313

For3q[32], we need l[32] and l[32]+3 which is just l35 since the select period is only

2 years. So 3q[32]= 1 l35 l_[32] = 1 9,888 9,899 = 0.001111 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

5p[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.)

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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) = [F (z)_[1 F (x)] F (x)] = [s(x) s(z)] [s(x)] tpx = e h Rt 0µ(x+s)ds i tpx= s(x + t) s(x) tqx = 1 tpx t|uqx =tpx⇤uqx+t t|uqx=t+uqx tqx t|uqx=tpx t+upx Pr(K(x) = k) = kpx k+1px = kpx⇤ qx+k = k|qx Life Tables: lx = l0⇤ s(x) ndx = lx lx+n qx= lx lx+1 lx px= lx+1 lx nqx = lx lx+n lx n px= lx+n lx

Constant Force of Mortality

If the force of mortality is a constant µ for every age x, s(x) = e µx tpx = e µt ex= 1 µ Var[T ] = 1 µ2 c Yufeng Guo

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Expected Future Lifetime ex= E[T (x)] = Z 1 0 t pxdt Var[T (x)] = 2 Z 1 0 t_·tpxdt e 2 x ex= E[K(x)] = 1 X 1 kpx Var[K(x)] = 1 X 1 (2k 1) ·kpx e2x

Under UDD only,

ex= ex+1₂ Var(T ) = Var(K) +₁₂1

Median future lifetime Pr[T (x) > m(x)] = 1 2 s[x + m(x)] s(x) = 1 2 mpx= 1₂

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

Lx = Z 1 0 lx+tdt nLx= Z n 0 lx+tdt mx = (lx lx+1) Lx n mx= lx lx+n nLx Tx= Z ₁ 0 lx+tdt Tx lx = ex n Lx lx =Z n 0 t pxdt =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,

a(x) = R1 0 t· lx+tµ(x + t)dt R1 0 lx+tµ(x + t)dt = R1 0 t·tpxµ(x + t)dt R1 0 tpxµ(x + t)dt

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.

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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(c}x _1)]

where B > 0, c > 1, m = B

log c, x 0 Makeham’s Law: µ(x) = A + Bcx _{and s(x) = exp[ Ax} _m(cx _1)]

where B > 0, A B, c > 1, m = B

log c, x 0 Weibull’s Law: µ(x) = kxn _{and s(x) = exp( ux}n+1₎

where k > 0, n > 0, u = k

(n + 1), x 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 Modified DeMoivre

µ(x) = _{! x}1 µ(x) = _{! x}c s(x) = ! x_! s(x) = ! x_! c lx= l0·! x_! l0 ! x_! c ex= E[T ] = ! x₂ ex= ! x_c+1 Var[T ] = (! x)2 12 Var[T ] = (! x)2_c (c+1)2_(c+2) tpx= ! x t_{! x} tpx= ⇣ ! x t ! x ⌘c µx(t) = _{! x t}1 µx(t) = _{! x t}c

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

80 2

• · · ·

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

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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) µ(x) F (x) F (x) 1 s(x) R₀xf (u) du 1 e Rx 0 µ(t) dt s(x) 1 F (x) s(x) R_x1f (u) du e R x 0 µ(t) dt f (x) F0(x) s0(x) f (x) µ(x) e R x 0 µ(t) dt µ(x) _{1 F (x)}F0(x) _s(x)s0(x) f (x)_s(x) µ(x)

Assumptions for fractional ages.

Uniform Constant

Function Distribution Force Hyperbolic

tqx tqx 1 ptx 1 (1 t)qtqx x µ(x + t) qx 1 tqx log px qx 1 (1 t)qx 1 tqx+t (1 t)q_{1 tq}_xx 1 p1 tx (1 t)qx yqx+t _{1 tq}yqx_x 1 pyx 1 (1 y t)qyqx x tpx 1 tqx ptx 1 (1 t)qpx x tpxµ(x + t) qx ptxlog px _{[1 (1 t)q}qxpx _x_]2 c Yufeng Guo

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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 (E) 333 Solution: e0= Z ! 0 ✓ 1 t ! ◆ dt = ! ! 2 2! = ! 2 = 25 ) ! = 50 e₁₀= Z 40 0 ✓ 1 t 40 ◆ dt = 40 40 2 (2)(40) = 20 Var [T (x)] = 2Z 40 0 t ✓ 1 t 40 ◆ dt (20)2= 2 " t2 2 t3 3 · 40 #40 0 (20)2 _{= 133} 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 (C) 0.0790 (D) 0.0796 (E) 0.0800

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Solution: 0.0408 = µ(81.5) = q81 1 (1/2)q81 ) q81= 0.0400 Similarly, q80= 0.0200 and q82= 0.0600 2q80.5=1/2q80.5+1/2p80.5 h q81+ p81·1/2q82 i = 0.01 0.99 + 0.98 0.99[0.04 + 0.96(0.03)] = 0.0782 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 Solution: STANDARD: e_25:11 = Z 11 0 (1 t 75)dt = t t2 2 ⇥ 75 |110 = 10.1933 MODIFIED: p25= e R1 0 0.1ds= e 0.1= 0.90484 e_25:11 = Z 1 0 t p25dt + p25 Z 10 0 (1 t 74)dt = Z 1 0 e 0.1tdt + e 0.1 Z 10 0 (1 t 74)dt = 1 e 0.1 0.1 + e 0.1 t t2 2 ⇥ 74 ! |100 = 0.95163 + 0.90484(9.32432) = 9.3886

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

The di↵erence is 0.8047. Key: D

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

x l_[x] l_[x]+1 lx+2 x + 2

60 80,625 79,954 78,839 62 61 79,137 78,402 77,252 63 62 77,575 76,770 75,578 64

Assume that deaths are uniformly distributed between integral ages. Calculate0.9q[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 0.9q[60]+0.6= 80,222.4 79,396.5 80,222.4 = 0.0103 Key: B 5. Given: (i) µ(x) = F + e2x_{, x} ₀ (ii) 0.4p0= 0.50 Calculate F . (A) -0.20 (B) -0.09 (C) 0.00 (D) 0.09 (E) 0.20 c Yufeng Guo

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Solution: 0.4p0= 0.5 = e R0.4 0 (F +e 2x)_dx = e 0.4F h e2x 2 i0.4 0 = e 0.4F ⇣ e0.8 1 2 ⌘ ) 0.5 = e 0.4F 0.6128 ) 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= 1000kp95, 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 Survivors 95 1000 95.5 800 96 600 96.5 480 97 97.5 288 98

The actuary decides to change his assumption for mortality at fractional ages to the constant force assumption. He retains his original assumption for eachkp95.

Calculate the revised expected number of survivors at age 97.5. (A) 270 (B) 273 (C) 276 (D) 279 (E) 282

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Solution: From UDD, l96.5=

l96+ l97

2 .

480 = 600 + l97

2 ! l97= 360

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

For constant force, e µ= l98

l97 = 216 360= 0.6 0.5px= e 0.5µ= (0.6) 1 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, e_1:1.5.

(A) 1.25 (B) 1.3 (C) 1.35 (D) 1.4 (E) 1.45 Solution: e_1:1.5 = Z 1.5 0 tp1dt =Z 1 0 tp1dt +1p1 Z 0.5 0 xp2dx =Z 1 0 (1 0.1t)dt + 0.9 Z 0.5 0 (1 0.05x)dx = " t 0.1t 2 2 #1 0 + 0.9 " x 0.05x 2 2 #0.5 0 = 0.95 + 0.444 = 1.394 Key: D c Yufeng Guo

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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 (E)108

Solution: For deMoivre’s law, e30= ! ₂30.

Prior to medical breakthrough ! = 100)e30= 100 30₂ = 35.

After medical breakthrough e0₃₀=e30+4 = 39.

)e030= 39 =

!0 30

2 ) !0 = 108. Key E

9. For a select-and-ultimate mortality table with a 3-year select period: (i) x q_[x] q_[x]+1 q_[x]+2 qx+3 x + 3 60 0.09 0.11 0.13 0.15 63 61 0.10 0.12 0.14 0.16 64 62 0.11 0.13 0.15 0.17 65 63 0.12 0.14 0.16 0.18 66 64 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 (B) 0.43  P < 0.45 (C) 0.45  P < 0.47 (D) 0.47  P < 0.49 (E) 0.49  P  1.00

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Solution: 5p[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 < x < 40 0.05, x > 40 Calculate e_25:25. (A) 14.0 (B) 14.4 (C) 14.8 (D) 15.2 (D) 15.6 Solution: e_25:25= Z 15 0 t p25dt +15p25 Z 10 0 t p40dt =Z 15 0 e 0.04t_{dt +}✓_e R₀150.04 ds◆ Z 10 0 e 0.05t_dt = 1 0.04 ⇣ 1 e 0.60⌘+ e 0.60  ₁ 0.05 ⇣ 1 e 0.50⌘ = 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 ! = 100.

(iii) fT(t) = ( k f1(t), 0  t  50 1.2 f2(t), 50 < t Calculate10p40. (A) 0.81 (B) 0.85 (C) 0.88 (D) 0.92 (E) 0.96 c Yufeng Guo

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SOLUTION: From the Illustrative Life Table: l50 l0 = 0.8951 l40 l0 = 0.9313 1 =Z 1 0 fT(t) dt = Z 50 0 kf1(t) dt + Z ₁ 50 1.2 f2(t) dt = kF1(50) + 1.2 (F2(1) F2(50)) = k (1 50p0) + 1.2(1 0.5) = k(1 0.8951) + 0.6 ) k = ₁1 ₀₈₉₅₁0.6 = 3.813 For x  50, FT(x) = Z x 0 3.813f1(t) dt = 3.813F1(x)

This gives us the following two results: FT(40) = 3.813 ✓ 1 l40 l0 ◆ = 0.262 FT(50) = 3.813 ✓ 1 l50 l0 ◆ = 0.400 10p40= 1 FT(50) 1 FT(40) = 1 0.400 1 0.262 = 0.813

12. For a double decrement table, you are given: (i) µ(1) x (t) = 0.2µ(⌧ )x (t), t > 0 (ii) µ(⌧ ) x (t) = kt2, t > 0 (iii) q0(1) x = 0.04 Calculate 2qx(2). (A) 0.45 (B) 0.53 (C) 0.58 (D) 0.64 (E) 0.73 c Yufeng Guo

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SOLUTION:

We can use the exponential formulation for p0_(x)

0.04 = q0₍₁₎ x = 1 p 0₍₁₎ x = 1 e R1 0 µ (1) x (t) dt= 1 e R 1 0 0.2µ (⌧ ) x (t) dt= 1 e R 1 0 0.2kt 2_dt = 1 e 0.2k/3 ) e 0.2k/3= 0.96 µ(1)_x (t) = 0.2µ(⌧ )_x (t) ) µ(2)x (t) = 0.8µ(⌧ )x (t) 2q(2)x = Z 2 0 t p(⌧ )_x _{· µ}(2)_x (t)dt = Z 2 0 t p(⌧ )_x _{· (0.8)µ}(⌧ )_x (t)dt = 0.8₂q(⌧ )_x To get₂qx(⌧ ), we use 2qx(⌧ )= 1 2p(⌧ )x = 1 e R2 0 µ (⌧ ) x (t) dt= 1 _e R2 0 kt 2_dt = 1 e 8k/3 = 1 ⇣e 0.2k/3⌘40= 1 (0.96)40 )2qx(⌧ )= 0.8046 )2qx(2)= (0.8)(0.8046) = 0.644

13. You are given: (i) e_30:40 = 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: e_30:40 =R₀40tp30dt =R40 0 ! 30 t! 30 dt c Yufeng Guo

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=ht _{2(! 30)}t2 i_|40 0 = 40 800 ! 30 = 27.692 ) ! = 95 tp30= 65 t₆₅

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:

V ar = (65 0)₁₂ 2 = 352.08 Key: B

14. For a life table with a one-year select period, you are given: (i) 80 1000 90x l[x] d[x] lx+1 e8.5[x]

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

p_[80]= ₁₀₀₀910 = 0.91 p_[81]= 830₉₂₀ = 0.902 p81= 830₉₁₀ = 0.912

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

We have: e[80]= e80:1 +1p[80]⇥ e81

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Notice under UDD, tpx is a straight line. Since ex:1 is the area of the function tpx

bounded by t = 0 and t = 1, under UDD we have e_x:1 = 0.5(0px+1px) = 0.5(1 +1px)

You can also derive the above equation using the standard formula: e_x:1 =R1

0 tpxdt =

R1

0 (1 tqx)dt =

R1

0 (1 tqx)dt

Since qx is a constant, you can verify that:

R1

0 (1 tqx)dt = 1 0.5qx = 0.5(1 +1px)

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

0.51qx+1px(1 + ex+1:n 1).

Set n = 1, we have: ˙ex = 1₂qx+ px(1 + ˙ex)

˙e[80]= 12q[80]+ p[80](1 + ˙e81) 8.5 = 1 2(1 0.91) + (0.91) (1 + ˙e81) ˙e81= 8.291 ˙e81= 1₂q81+ p81(1 + ˙e82) ˙e82= 8.043 ˙e[81]= 12q[81]+ p[81](1 + ˙e82) = 1 2(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|14q50. (A) 0.38 (B) 0.39 (C) 0.41 (D) 0.43 (E) 0.44 SOLUTION: 4p50= e (0.05)(4)= 0.8187 10p50= e (0.05)(10)= 0.6065 8p60= e (0.04)(8)= 0.7261 18p50= (10p50)(8p60) = 0.6065 ⇥ 0.7261 = 0.4404 c Yufeng Guo

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4|14q50=4p50 18p50= 0.8187 0.4404 = 0.3783

Key: A

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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 q60for 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 (MaleNewborn) > 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 _0.0053540.00492 = 0.081

Key: B

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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:

• tpx = (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 (C) ln 2 2 (D) ln 4 2 (E) 1 c Yufeng Guo

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6. Calculate mx, the central-death-rate at age x.

(A) e 2

2 (B) e 2 (C) 2e 2 (D) 1 (E) 2

7. You are given: s(x) = ✓ 1 x ! ◆↵ , 0_{ x < !, where ↵ > 0 is a constant.} Calculate µx· ex. (A) ↵ ↵ + 1 (B) ↵! ↵ + 1 (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 (B) 85 (C) 94 (D) 178 (E) 213

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Solutions to Chapter 3

1. Key: A We need to find t such that tp20= 0.5

) 0.5 = e R0tµ(20+s) ds= e Rt 0(60 s) 0.5_ds = e2[(60 s)0.5]t 0 = e2[ p 60 t p60] ln 0.5 = 2hp60 t p60i )p60 t = 7.40 _{) t = 5.25} 2. Key: D Tx+1= Z ₁ 0 lx+1+tdt lx+1 = 6.4 0.8 = 8 ) lx+1+t= 8(0.8)t Tx+1 = 8 Z ₁ 0 (0.8)t_{dt =} 8 ln 0.8(0.8)t|10 = 8 ln 0.8 = 35.9

3. Key: A This is constant force of mortality (CFM) with µ = 2. E[T ] = 1 µ = 0.5 =ex 4. Key: A (CFM) Var[T ] = 1 µ2 = 0.25 5. Key: C 0.5 =tpx= e µ·t = e 2t ) t = ln 2₂ c Yufeng Guo

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6. Key: E mx= lx lx+1 Lx = lx(1 e µ) R1 0 lx·tpxdt = lx(1 e 2) lxR₀1e µtdt == ₁1 e 2 2(1 e 2) = 2

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

0.5p60·0.5q60.5+1p60·0.5q61 UDD: [1 (0.5)(0.3)] (0.5)(0.3) 1 (0.5)(0.3) + (0.7) · (0.5)(0.4) = 0.2900 = f Balducci: 0.7 1 (0.5)(0.3) (0.5)(0.3) 1 + (0.7) · (0.5)(0.4) 1 (0.5)(0.4) = 0.2985 = g 10,000(g f ) = 85 c Yufeng Guo

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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.