Arch MLC Spring 2010

Arch MLC: Spring 2010

Yufeng Guo

October 26, 2009

actuary88.com Chapter 0

Feature of Arch MLC Manual

• Cover everything in the SOA MLC syllabus.

• Thorough explanation of the core concepts with worked out problems

• Reference to Actuarial Mathematics and Models for Quantifying Risk. You can useeither 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 alsoinstall the PDF in your computer.

• PDF has detailed bookmarks for quick reference.

• PDF has a clickable table of contents for quick reference.

Arch MLC, Spring 2010 c©Yufeng Guo 2

Contents

0 INTRODUCTION 9The origin of Arch Manual . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9Praises of the Arch Manual originally written by Nathan and Robin . . . . . 10About Yufeng Guo . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10How to use this manual . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1 ACTUARIAL MATHEMATICS: CHAPTER 3 SURVIVAL DISTRIBU-TIONS AND LIFE TABLES 13

3.2.1 The Survival Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.2.2 Time-until-Death for a Person Age x . . . . . . . . . . . . . . . . . . . . 163.2.3 Curtate-Future-Lifetime . . . . . . . . . . . . . . . . . . . . . . . . . . . 193.2.4 Force of Mortality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.3-3.5 Life Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.5.2 Recursion Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.6 Assumptions for Fractional Ages . . . . . . . . . . . . . . . . . . . . . . . 283.7 Some Analytical Laws of Mortality . . . . . . . . . . . . . . . . . . . . . . 33Modified DeMoivre’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.8 Select and Ultimate Tables . . . . . . . . . . . . . . . . . . . . . . . . . . 35Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38CHAPTER 3 Formula Summary . . . . . . . . . . . . . . . . . . . . . . . . . 39Past SOA/CAS Exam Questions: . . . . . . . . . . . . . . . . . . . . . . . . . 43Problems from Pre-2000 SOA-CAS exams . . . . . . . . . . . . . . . . . . . . 56Solutions to Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

2 ACTUARIAL MATHEMATICS CHAPTER 4 – LIFE INSURANCE 614.2 Insurances Payable at the Moment of Death . . . . . . . . . . . . . . . . . 62TYPES OF INSURANCE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

4.2.1 Level Benefit Insurance . . . . . . . . . . . . . . . . . . . . . . . 634.2.2 Endowment Insurance . . . . . . . . . . . . . . . . . . . . . . . . 664.2.3 Deferred Insurance . . . . . . . . . . . . . . . . . . . . . . . . . . 684.2.4 Varying Benefit Insurance . . . . . . . . . . . . . . . . . . . . . . 70

4.3 Insurances Payable at the End of the Year of Death . . . . . . . . . . . . 734.4 Relationships between Insurances Payable at the Moment of death and theEnd of the Year of Death . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79CHAPTER 4 Formula Summary . . . . . . . . . . . . . . . . . . . . . . . . . 81Past SOA/CAS Exam Questions: . . . . . . . . . . . . . . . . . . . . . . . . . 83

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Problems from Pre-2000 SOA-CAS exams . . . . . . . . . . . . . . . . . . . . 93Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

3 ACTUARIAL MATHEMATICS: CHAPTER 5 LIFE ANNUITIES 975.2 Continuous Life Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . 97The most important equation so far(!) . . . . . . . . . . . . . . . . . . . . . . 1005.3 Discrete Life Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1065.4 Life Annuities with m-thly Payments . . . . . . . . . . . . . . . . . . . . . 112CHAPTER 5 Formula Summary . . . . . . . . . . . . . . . . . . . . . . . . . 116

Continuous Annuities: . . . . . . . . . . . . . . . . . . . . . . . . . . . 116Discrete annuities: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

Past SOA/CAS Exam Questions: . . . . . . . . . . . . . . . . . . . . . . . . . 118Problems from Pre-2000 SOA-CAS exams . . . . . . . . . . . . . . . . . . . . 131Solutions to Pre-2000 Problems: Chapter 5 . . . . . . . . . . . . . . . . . . . 133

4 ACTUARIAL MATHEMATICS: CHAPTER 6 BENEFIT PREMIUMS 1356.2 Fully Continuous Premiums . . . . . . . . . . . . . . . . . . . . . . . . . . 1366.3 Fully Discrete Premiums . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1426.4 True m-thly Payment Premiums . . . . . . . . . . . . . . . . . . . . . . . 147CHAPTER 6 Formula Summary . . . . . . . . . . . . . . . . . . . . . . . . . 151Past SOA/CAS Exam Questions: . . . . . . . . . . . . . . . . . . . . . . . . . 152Problems from Pre-2000 SOA-CAS exams . . . . . . . . . . . . . . . . . . . . 168Solutions to Pre-2000 Exam Questions: Chapter 6 . . . . . . . . . . . . . . . 170

5 ACTUARIAL MATHEMATICS: CHAPTER 7 BENEFIT RESERVES 1737.2 Fully Continuous Benefit Reserves . . . . . . . . . . . . . . . . . . . . . . 1747.3 Other Methods for Calculating the Benefit Reserve . . . . . . . . . . . . . 177

1) Prospective Formula . . . . . . . . . . . . . . . . . . . . . . . . . . 1772) Retrospective Formula . . . . . . . . . . . . . . . . . . . . . . . . . 1773) Premium Difference Formula . . . . . . . . . . . . . . . . . . . . . . 1804) Paid-Up Insurance Formula . . . . . . . . . . . . . . . . . . . . . . 1805) Other Reserve Formulas . . . . . . . . . . . . . . . . . . . . . . . . 181

7.4 Fully Discrete Reserves . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1837.5 Benefit Reserves on a Semi-Continuous Basis . . . . . . . . . . . . . . . . 1867.6 Benefit Reserves Based on True m-thly Benefit Premiums . . . . . . . . . 187CHAPTER 7 Formula Summary . . . . . . . . . . . . . . . . . . . . . . . . . 188Continuous Reserve Formulas: . . . . . . . . . . . . . . . . . . . . . . . . . . . 188

Discrete Reserves: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188Past SOA/CAS Exam Questions: . . . . . . . . . . . . . . . . . . . . . . . . . 190Problems from Pre-2000 SOA-CAS exams . . . . . . . . . . . . . . . . . . . . 194Solutions to Pre-2000 Questions: Chapter 7 . . . . . . . . . . . . . . . . . . . 196

6 ACTUARIAL MATHEMATICS: CHAPTER 8 ANALYSIS OF BENEFITRESERVES 197

8.2 Benefit Reserves for General Insurances . . . . . . . . . . . . . . . . . . . 1978.3 Recursion Relations for Fully Discrete Benefit Reserves . . . . . . . . . . 2028.4 Benefit Reserves at Fractional Durations . . . . . . . . . . . . . . . . . . . 205



8.5 The Hattendorf Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

CHAPTER 8 Formula Summary . . . . . . . . . . . . . . . . . . . . . . . . . 212

Chapter 8 More Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213

Past SOA/CAS Exam Questions: . . . . . . . . . . . . . . . . . . . . . . . . . 214

Problems from Pre-2000 SOA-CAS exams . . . . . . . . . . . . . . . . . . . . 234

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



Problems from Pre-2000 SOA-CAS exams . . . . . . . . . . . . . . . . . . . . 283

Solutions to Pre-2000 Exam Questions: Chapter 9 . . . . . . . . . . . . . . . 285

8 ACTUARIAL MATHEMATICS: CHAPTER 10 MULTIPLE DECREMENTMODELS 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 inthe 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



9 ACTUARIAL MATHEMATICS: CHAPTER 11 APPLICATIONS OF MUL-TIPLE DECREMENT THEORY 323

11.2 Actuarial Present Values and Their Numerical Estimation . . . . . . . . 32311.3 Benefit Premiums and Reserves . . . . . . . . . . . . . . . . . . . . . . . 324CHAPTER 11 Formula Summary . . . . . . . . . . . . . . . . . . . . . . . . . 327ARCH Sample Exam Problem . . . . . . . . . . . . . . . . . . . . . . . . . . 327Solution: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327Past SOA/CAS Exam Questions: . . . . . . . . . . . . . . . . . . . . . . . . . 329

10 ACTUARIAL MATHEMATICS: CHAPTER 15 INSURANCE MODELSINCLUDING EXPENSES 335

15.2 Expense Augmented Models . . . . . . . . . . . . . . . . . . . . . . . . . 33515.4 More Expenses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33815.6.1 Asset Shares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341CHAPTER 15 Formula Summary . . . . . . . . . . . . . . . . . . . . . . . . . 343

Asset Shares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343Past SOA/CAS Exam Questions: . . . . . . . . . . . . . . . . . . . . . . . . . 343SOLUTIONS to Past SOA-CAS Exam Problems: . . . . . . . . . . . . . . . . 347

11 DANIEL CHAPTER 1 - MULTI-STATE TRANSITION MODELS FORACTUARIAL APPLICATIONS 351

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3511.2 Non-homogeneous Markov Chains . . . . . . . . . . . . . . . . . . . . . . 354CHAPTER 2 – CASH FLOWS AND THEIR ACTUARIAL PRESENT VALUES359

Section 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . 359Cash Flows while in states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359Cash Flows upon transitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364Actuarial Present Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 364ARCH Warm-up Questions: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372Past SOA/CAS Exam Questions: . . . . . . . . . . . . . . . . . . . . . . . . . 375

12 DANIEL STUDY NOTE ON POISSON PROCESS 3855.3 The Poisson Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385

5.3.1 Counting Processes . . . . . . . . . . . . . . . . . . . . . . . . . . 3855.3.2 Definition of the Poisson Process . . . . . . . . . . . . . . . . . . 3865.3.3 Interarrival and Waiting Time Distributions . . . . . . . . . . . . 3875.3.4 Further Properties of Poisson Processes . . . . . . . . . . . . . . 3895.3.5 Conditional Distribution of the Arrival Times . . . . . . . . . . . 392

5.4 Generalizations of the Poisson Process . . . . . . . . . . . . . . . . . . . . 3935.4.1 Nonhomogeneous Poisson Process . . . . . . . . . . . . . . . . . 3935.4.2 Compound Poisson Process . . . . . . . . . . . . . . . . . . . . . 3955.4.3 Conditional or Mixed Poisson Processes: Gamma-Poisson Model 398

CHAPTER 5 Formula Summary . . . . . . . . . . . . . . . . . . . . . . . . . 401ARCH Warm-up Problems: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402Solutions: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403Past SOA/CAS Exam Questions: . . . . . . . . . . . . . . . . . . . . . . . . . 404


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

14 SOLUTION TO MAY 2007 MLC 449


actuary88.com Chapter -1


Chapter 0

INTRODUCTION

The origin of Arch Manual

The Arch Manual was originally written by two gifted actuaries, Nathan Hardiman and RobinCunningham.

In the late 90’s, Nathan Hardiman and Robin Cunningham worked full-time at the formerJefferson Pilot Financial. Nathan and Robin, like other exam candidates, faced the daunt-ing challenge of plowing through difficult textbooks and mastering the fundamentals of lifecontingency theories and probability models to pass the Course 3 exam, the most difficult ofthe 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 requiredto read encyclopedia-like textbooks such as Actuarial Mathematics and Probability Models,gain sophisticated understanding of complex concepts such as multiple decrements, MarkovChain, 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 passingCourse 3. After passing the exam in one sitting using their unique study methods, Nathanand Robin decided to jointly write a new study guide that would enable candidates to build acore body of knowledge for Course 3 quickly. They wanted their manual to use straight talkand 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 popularstudy manual for Course 3 and Exam M.

Arch’s power lies in its simplicity. While textbooks talk fancy, Arch talks simple. Whiletextbooks rigorously prove theorems, Arch explains the intuition. While textbooks demandattention to everything, Arch separates the critical from the trivial.

While Arch sells well, Nathan and Robin climbed corporate ladder higher and higher. With

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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 businessand passed on the copyright and ownership of Arch M manual to Yufeng Guo, who is theauthor of Deeper Understanding manuals for Exam P, FM, and M.

Nathan and Robin’s contribution to actuarial education was not just the Arch manual butmore importantly the Arch’s effective teaching style. Before Arch was published, manythought that learning difficult things such as Course 3 ought to be slow and painful. Arch’sstraight talk and down-to-earth examples showed the actuarial community that learningdifficult actuarial theories can indeed by fast-paced and enjoyable.

Praises of the Arch Manual originally written by Nathan andRobin

-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]. NowI think I wasted a whole set of money on the others since they always end upconfusing 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 andunderstand the material without your study guide.

-I want to personally thank both of you for the fantastic and brilliant work thatyou did on ARCH. Seeing as it’s not my first time tackling this exam, I’ve hadthe chance to use [several other manuals]; however this is by far superior to allof those products. I have and will continue to recommend it to others in mycompany.

-I would first like to say that I am very happy with your manual so far. I feelthat I am progressing through the syllabus much faster than I would have withoutit, and the depth of understanding that I am on my own giving up due to my notusing the texts themselves is more than compensated for by the excellent coverageof the important topics in your manual.

About Yufeng Guo

Yufeng Guo was born in central China. After receiving his Bachelor’s degree in physics atZhengzhou University, he attended Beijing Law School and received his Masters of law. Hewas 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 actuarialcareer and passed exams 1, 2, 3, 4, 5, 6, and 7 in rapid succession after discovering a successfulstudy strategy. Mr. Guo’s exam records are as follows:



Fall 2002 Passed Course 1

Spring 2003 Passed Course 2,3


Spring 2004 Passed Course 6


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 detailssee http://www.guo.coursehost.com and http://www.myactuaryexam.com.

If you have questions, you can email Mr. Guo at yufeng [email protected].

FAQ

I notice you have two study guides for MLC: Arch MLC and Deeper Understanding MLC.What’s the difference? Which one should I buy?

Difference: 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 onlyone guide, choose the one that better fits your need. For example, if you already have astudy guide and want to learn calculation shortcuts, buy Deeper Understanding: MLC. Ifyour 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 asefficiently as possible. The material for the course is broken down into different chapters fromthe textbooks. The chapters are presented using down-to-earth explanations. In addition, Ipoint 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 tobenefit a great deal if every time you get to an example, you cover up the solution and


http://www.guo.coursehost.com

http://www.myactuaryexam.com


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 calculatenumerical answers. Therefore, much of our explanation of the material is done by way ofnumerical examples and practice questions. Examples range from very simple ones (to makesure you know the basic concepts), to thought provoking ones (to help you think about whatyou’ve learned and really understand it), to exam questions from prior exams (to get youready for exam day).

I also suggest problems from the texts for you to work. Many of the problems in thetext 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 atwww.archactuarial.com.

A formula summary for each chapter is included. These summaries are intended to serve asa reference as you familiarize yourself with the syllabus material.

Finally, there’s a full length practice exam of new questions. This practice exam is designedto be used in conjunction with the prior Course 3 and Exam M problems in the SOA websiteat 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 forindividual 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.


http://www.actuarialbookstore.com

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Chapter 1

ACTUARIAL MATHEMATICS:CHAPTER 3 SURVIVALDISTRIBUTIONS AND LIFETABLES

• 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 understandingof this material in order to pass the exam. However, you do not need to understand themajority of the underlying theory in this text. The key points that a student must learnfrom this text are:

KEYPOINTS:

1. Notation – much of this notation is new. While it can be confusing at first, there is somelogic to it. It will help you to remember and understand the many symbolsif you regularly translate the notation into words as you read.

2. Basic ideas – for example, chapter four introduces a variety of types of insurance. Youwill want to make sure you have an understanding of these different products and theirbenefit 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 upondeath. The second part of the chapter considers the same products except that thebenefit is paid at the end of the year in which death occurs. It is helpful to realize thatyou are really learning only one set of products, with a couple of benefit options, ratherthan two sets of products. These parallels run throughout the text (e.g., continuous vscurtate 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 theformulas, you will be ready to answer questions quickly. This manual has tools to helpyou learn these formulas, so don’t feel overwhelmed!

To the text!!!



Chapter 3 is all about notation, definitions, and a few basic ideas that are essential tolife contingencies. If you can make yourself comfortable with the symbols and methods ofChapter 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 deathYou 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 occurprior to (or at) age x”. In statistics, F (x) is the cumulative distribution function for thefuture 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 greaterthan 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. Readit 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 pdfy=f(x)

The figure shows the probability distribution function f(x) for death at age x. For any valueof 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 inFX(x) ... you can ignore it. I don’t know if the authors realize it but they are being a little



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 theF and s straight for each kid. That’s all the subscript is indicating.

3.2.2 Time-until-Death for a Person Age x



Newborns are great, but if our pension and insurance companies are going to make moneywe 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 personwho 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 diebefore 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 diesbetween 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)



2. “The probability of death between ages 20 and 35, given that the newbornwill not attain age 80.” ♦

Now, let the symbol “(x)” represent a person age x and let T (x) be the future lifetime of aperson 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

t px t xq

In the figure, tqx is 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, ω). (ω representsthe 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-offson tpx or tqx 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, eachof 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 tyears but not t+ 1 years.)



Useful formulas:

tpx =x+tp0

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 + uyears from now (t|uqx) is equal to the probability that (x) will first survive t years (tpx) andthen 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, butdoesn’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 insymbols. You can also make flash cards and quiz yourself.



3.2.3 Curtate-Future-Lifetime


• 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 nodecimal, 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−x100 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. ♦



3.2.4 Force of Mortality


• 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 instantgiven survival up to that time. This is an instantaneous measure, rather than an intervalmeasure. There is good bit of theory in this section, but the most important items are thefollowing formulas and the table of relationships.

µ(x) =f(x)

1− F (x)=−s′(x)

s(x)(3.2.13)

It is very important to know the relationships and requirements given in Table 3.2.1. Thesewill probably be tested on the exam. Below is a summary of the useful information in thistable. 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)∫ x

0 f(u) du 1− e−∫ x0µ(t) dt

s(x) 1− F (x) s(x)∫∞x f(u) du e−

∫ x0µ(t) dt

f(x) F ′(x) −s′(x) f(x) µ(x) e−∫ x0µ(t) dt

µ(x) F ′(x)1−F (x)

−s′(x)s(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−∫ x0µdt = e−µx.

2.

tpx =s(x+ t)

s(x)= e−µt. ♦



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 intosystems for calculating reserves, premium rates, and the surrender cash value of an insurancepolicy. Learning Life Tables will not only help you pass Exam MLC, it also helps you whenyou 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)

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

lxndx = lx − lx+n

Illustrative Life Table: Basic FunctionsAge lx dx 1,000 qx

0 100,000.0 2,042.2 20.41 97,957.8 131.6 1.42 97,826.3 119.7 1.23 97,706.6 109.8 1.1...

......

...40 93,131.6 259.0 2.841 92,872.6 276.9 3.042 92,595.7 296.5 3.243 92,299.2 317.8 3.4



EXAMPLE: Life Table Mortality

Above is an excerpt from the Illustrative Life Table in the book. The followingquestions are all based on this excerpt.

1. Find s(42).

2. Find 40d2.

3. Find 38q3.

4. Find 2|q40.

SOLUTION:

1. s(42) = 92,595.7100,000 = 0.925957.

2. 40d2 = l2 − l42 = 5230.6

3. 38q3 = 1− 38p3 = 1− l41l3

= 1− 92,872.697,706.6 = 0.04947.

4. 2|q40 = 2p40 · q42 = 92,595.793,131.6(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 densityfunction 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 tyears and then be hit at that instant by the force of mortality.Among other things, this tells us that∫ ∞

0tpxµ(x+ t)dt =

∫ ∞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

◦ex= E[T (x)] =

∫ ∞0

tpxdt

Var[T (x)] = 2

∫ ∞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 suchthat mpx = mqx. In other words, it is the number of years that (x) is equally likely to surviveor not survive. It can be found by solving any of the following:

Pr[T (x) > m(x)] =1

2



ors[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 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)] =∞∑1

kpx

Var[K(x)] =∞∑1

(2k − 1) · kpx − e2x


Find◦e0 and

◦e50 if the force of mortality is a constant µ.

SOLUTION:

◦e0=

∫ ∞0

tp0 dt =

∫ ∞0

e−µtdt =

[−1

µe−µt

]∞0

=1

µ

◦e50=

∫ ∞0

tp50 dt =

∫ ∞0

e−µtdt =

[−1

µe−µt

]∞0

=1

µ

If the force of mortality is constant, your future expected lifetime is the samewhether 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 < 500 Otherwise

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

SOLUTION:

e0 =50∑1

tp0 =50∑1

50− t50

= 50− 1

50

50∑1

t



= 50− 1

50

(50)(51)

2= 24.5

e45 =5∑1

tp45 =5∑1

s(45 + t)

s(45)=

5∑1

5− t5

=4 + 3 + 2 + 1 + 0

5= 2. ♦

More Life Functions:

The expression Lx denotes the total expected number of years, full or fractional, lived betweenages x and x+ 1 by survivors of the initial group of l0 lives. Those who survive to x+ 1 willlive one year between x and x + 1, contributing one full year to Lx. Those who die duringthe year will contribute a fraction of a year to Lx.

Lx =

∫ 1

0lx+tdt

The expression mx is the central death rate over the interval x to x+ 1. Make sure not toconfuse 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 =

∫ n

0lx+tdt

nmx =lx − lx+n

nLx

The remaining of Section 3.5.1 has obscure symbols Tx and α(x). They rarely show up inthe 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 l0initial members (i.e. the lx people still alive at age x). Be careful with notation. This is notT (x), the future lifetime of (x).

Tx =

∫ ∞0

lx+tdt (3.5.16)

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.



α (x) = E [T |T < 1] =

∫ 1

0tf(t)dt∫ 1

0f(t)dt

=

∫ 1

0ttpx µ(x+t)dt∫ 1

0tpxµ(x+t)dt

=

∫ 1

0t· lx+t

lxµ(x+t)dt∫ 1

0

lx+tlx

µ(x+t)dt=

∫ 1

0t·lx+t µ(x+t)dt∫ 1

0lx+t µ(x+t)dt

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

∫ 1

0t·cdt∫ 1

0cdt

=

∫ 1

0tdt∫ 1

0dt

=∫ 1

0 tdt = 12 .


If l0 = 1000 and the force of mortality is a constant µ = 0.1, find

(A) L5

(B) m5

(C) T5



SOLUTION:

(A)

L5 =

∫ 1

0l5+t dt =

∫ 1

0tp5 l5dt.

Since

l5 = l0e−µ·5 = 1000e−0.5 = 606.5,

we have

L5 = 606.5

∫ 1

0e−0.1t dt = 606.5

[−10e−0.1t

]10

= 606.5[10(1− e−0.1)

]= 577.16.

(B)

m5 =l5 − l6L5

=606.5− 548.8

577.16= 0.10

This approximates the rate at which people were dying between the 5th and6th years.

(C)

T5 =

∫ ∞0

1000e−0.1(5+t) dt = 606.5

∫ ∞0

e−0.1t dt = 6065

So if we add up all of the time lived by each of the people alive at t = 5, weexpect to get a total of 6065 years, or 10 years per person. ♦

Relationship:Txlx

=◦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 lx survivorsat age x as:

nLxlx

=

∫ n

0tpxdt

nLxlx

= n-year temporary complete life expectancy of (x)

=◦ex:n (p.71)



3.5.2 Recursion Formulas


• Option B reference: Models for Quantifying Risk Chapter 6

These are basically ways to avoid working integrals. They are based on the Trapezoid Rulefor 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 formulasyourself to convince yourself of their equivalence. Then, learn whichever form you find morestraightforward.

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 iterationwill produce eω−2, etc. until you get all the way back to e0, when you will have produced alist 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

c(x) =

∫ 1

0spxds

d(x) = px

Starting Value =◦eω= u(ω) = 0



3.6 Assumptions for Fractional Ages

• Option A reference: Actuarial Mathematics Chapter 3.6


(OK, you can start paying attention again ....)

The random variable T is a continuous measure of remaining lifetime. The life table has beendeveloped as an approximation of T , using a curtate variable K. As we’ve discussed, K isonly defined at integers. So, we need some way to measure between two integer ages. Threepopular methods were developed.

For all of the methods that follow, let x be an integer and let 0 ≤ t ≤ 1. Suppose that weknow the value of s(x) for the two integers x and x + 1 and we want to approximate s atvalues 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 thatthe 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 ofthe 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. Forexample, if s(x + 0.5) = 0.9, we say that for each unit of people at age x, we have 0.9 unitof people at age x + 0.5, with one unit being one billion, one million, or any other positiveconstant.

tqx =s(x)− s(x+ t)

s(x)= t

s(x)− s(x+ 1)

s(x)= tqx

f(t) =d

dttqx = qx



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

qx =

∫ 1

0f(t)dt = f(t)

∫ 1

0dt = 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 for0 ≤ t ≤ 1.

Method 3: Harmonic Interpolation: This method is more commonly called the “Balducciassumption” or the “Hyperbolic assumption.”

Forget about the complex formula

1

s(x+ t)=

1− ts(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 chanceof 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 toderive other formulas in Balducci assumptions.

Uniform ConstantFunction Distribution Force Hyperbolic

tqx tqx 1− ptxtqx

1−(1−t)qx

µ(x+ t) qx1−tqx − log px

qx1−(1−t)qx

1−tqx+t(1−t)qx1−tqx 1− p1−t

x (1− t)qx

yqx+tyqx

1−tqx 1− pyxyqx

1−(1−y−t)qx

tpx 1− tqx ptxpx

1−(1−t)qx

tpxµ(x+ t) qx −ptx log pxqxpx

[1−(1−t)qx]2



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 =

∫ t

0f(t)dt =

∫ t

0qxdt = qx

∫ t

0dt = tqx

tpx = 1− tqx = 1− tqx

µ(x+ t) = µx(t) =f(t)

tpx=

qx1− tqx

tpxµ(x+ t) = f(t) = qx

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

s(x+ t)=s(x)t+ypxs(x)tpx

=t+ypx

tpx=

1− (t+ y)qx1− tqx

yqx+t = 1− ypx+t =yqx

1− tqx

1−tqx+t =(1− t)qx1− tqx

Under constant force of mortality, for 0 ≤ t ≤ 1:

µx(t) = µ

tpx = e−∫ t0µdt = e−µt

px = e−µ

µ = − ln px

tpx = e−µt = (e−µ)t

= (px)t

ypx+t =t+ypx

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)



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 derivethe formula for tpx.

T (x) 0 t 1

Age x x+ t x+ 1

Number of people alive s(x) = 1 s(x+ t) = px1−(1−t)qx s(x+ 1) = px

This is how to derive s(x+ t) = px1−(1−t)qx .

First, we set the starting population at s(x) = 1 for convenience. You can set it to anypositive 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−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 = px

1−(1−t)qx .

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

f(t) = − d

dttpx = − d

dt

px1− (1− t)qx

=qxpx

[1− (1− t)qx]2

Derive µ(x+ t):

µ(x+ t) =f(t)

tpx=

qx1− (1− t)qx

Derive yqx+t:

yqx+t = 1− s(x+ t+ y)

s(x+ t)

Use the formula: s(x+ t) = px1−(1−t)qx

Replace t with t+ y, assuming 0 ≤ t+ y ≤ 1:

s(x+ t+ y) =px

1− (1− t− y)qx



yqx+t = 1− s(x+ t+ y)

s(x+ t)= 1− 1− (1− t)qx

1− (1− t− y)qx=

px1− (1− t− y)qx

Finally, let’s derive 1s(x+t) = 1−t

s(x) + ts(x+1)

s(x+ t) =px

1− (1− t)qx

1

s(x+ t)=

1− (1− t)qxpx

=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− ts(x)

+t

s(x+ 1)= 1− t+

t

px

⇒ 1

s(x+ t)=

1− ts(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 +

1

2

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



• CF:

0.5qx = 1− (0.9)0.5 = 0.0513

• Balducci:

0.5qx =(0.5)(0.1)

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



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 theexam. The book describes four basic analytical laws/formulas. Of these 4 laws, De Moivre’sLaw is frequently tested in the exam. The other 3 laws are rarely tested in the exam (I wouldskip 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

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)



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 shouldbecome intimately familiar with De Moivre’s Law. DeMoivre’s Law says that at age x, youare 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’tread the proofs, still make an effort to understand the formulas and what they mean.

EXAMPLE:

Under DeMoivre’s law, show that

1.

tpx =ω − x− tω − x

2.

tqx =t

ω − x3.

◦e0=

ω

2

4.

e0 =ω − 1

2

5.◦ex=

ω − x2

6.

ex =ω − x− 1

2



SOLUTION:

1.

tpx =s(x+ t)

s(x)=ω − x− tω − x

2.

tqx = 1− tpx =t

ω − x

3.◦e0=

∫ ω

0

ω − tω

dt =

[−(ω − t)2

2ω

]=ω

2

4.

e0 =ω∑1

kp0 =ω∑1

ω − kω

= ω − 1

ω

ω∑1

k =ω − 1

2

5.◦ex=

∫ ω−x

0

ω − x− tω − x

dt =ω − x

2

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

ex =◦ex −

1

2=ω − x− 1

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

µ(x) =c

ω − xwhere c is a positive constant. This gives rise to a set of formulas for each of the quantitiesfound for standard DeMoivre’s Law (c = 1) in the example above. All of the formulas forModified DeMoivre’s Law are listed in the formula summary at the end of this chapter.

3.8 Select and Ultimate Tables





Suppose you are trying to issue life insurance policies and two 45 year-old women apply forpolicies. You want to make sure you charge appropriate premiums for each one to cover thecost of insuring them over time. One of the women is simply picked from the populationat large. The second women was picked from a group of women who recently passed acomprehensive physical exam with flying colors – significantly healthier than the generalpopulation.

Is it equitable to charge the same premium to the two women? No – because you haveadditional information about the second woman that would cause you to expect her to havebetter “mortality experience” than the general population. Thus, to her premiums, you mightapply a “select” mortality table that reflects better the mortality experience of very healthy45 year olds. However, after 15 years, research might show that being very healthy at 45does not indicate much of anything about health at age 60. So, you might want to go backto using standard mortality rates at age 60 regardless of status at age 45. After all, 15 yearsis 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 periodof time, you expect mortality to be different than that for the general population – the “selectperiod”. However, at some point, you’re just not sure of this special status anymore, so thosefolks 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 thatfor 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 3231 0.234 0.352 0.459 9,902.89 9,900.58 9,897.09 3332 0.250 0.377 0.500 9,898.75 9,896.28 9,892.55 3433 0.269 0.407 0.545 9,894.29 9,891.63 9,887.60 3534 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, recentlypassing 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



for health to deteriorate since she was examined at age x − 1 (one year ago) rather than atage x.

EXAMPLE: Select and Ultimate Life Table

Using the select and ultimate life table shown above find the value of

1000(

3q32 − 3q[32]

).



SOLUTION:

3q32 deals only with the ultimate table so I am only interested in the values oflx + 2 in Column (6).

3q32 = 1− 3p32 = 1− l35

l32= 1− 9,888

9,901= 0.001313

For 3q[32], we need l[32] and l[32]+3 which is just l35 since the select period is only2 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 aselect group. People you are sure are healthy should be less likely to die than someonedrawn from the general population.

• To follow the people alive from the 9898.75 ‘selected’ at age 32, you first follow thenumbers to the right until you hit the ultimate column and then proceed down theultimate column. This is useful! You can quickly evaluate that

5p[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 thenotation 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.)



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)− F (x)]

[1− F (x)]=

[s(x)− s(z)][s(x)]

tpx = e

[−∫ t0µ(x+s)ds

]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

lxpx =

lx+1

lx

nqx =lx − lx+n

lxnpx =

lx+n

lx

Constant Force of MortalityIf the force of mortality is a constant µ for every age x,

s(x) = e−µx tpx = e−µt

◦ex=

1

µVar[T ] =

1

µ2



Expected Future Lifetime

◦ex= E[T (x)] =

∫ ∞0

tpxdt Var[T (x)] = 2

∫ ∞0

t · tpxdt−◦e

2

x

ex = E[K(x)] =∞∑1

kpx Var[K(x)] =∞∑1

(2k − 1) · kpx − e2x

Under UDD only,

◦ex= ex +

1

2Var(T ) = Var(K) +

1

12

Median future lifetime

Pr[T (x) > m(x)] =1

2

s[x+m(x)]

s(x)=

1

2mpx =

1

2

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

Lx =

∫ 1

0lx+tdt nLx =

∫ n

0lx+tdt

mx =(lx − lx+1)

Lxnmx =

lx − lx+n

nLx

Tx =

∫ ∞0

lx+tdt

Txlx

=◦ex

nLxlx

=

∫ n

0tpxdt =

◦ex:n

a(x) is the average number of years lived between age x and age x + 1 by those of thesurvivorship group who die between age x and age x + 1. With the assumption of uniformdistribution of deaths over the interval (x, x+ 1),

a(x) = 1/2.

Without this assumption,

a(x) =

∫ 10 t · lx+t µ(x+ t)dt∫ 1

0 lx+t µ(x+ t)dt=

∫ 10 t · tpx µ(x+ t)dt∫ 1

0 tpx µ(x+ t)dt

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



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

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)

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 ω, then all of the relations on the leftbelow are true. The relations on the right are for Modified DeMoivre’s Law with c > 0.

DeMoivre Modified DeMoivre

µ(x) = 1ω−x µ(x) = c

ω−x

s(x) = ω−xω s(x) =

(ω−xω

)clx = l0 · ω−xω l0

(ω−xω

)c◦ex= E[T ] = ω−x

2

◦ex= ω−x

c+1

Var[T ] = (ω−x)2

12 Var[T ] = (ω−x)2 c(c+1)2(c+2)

tpx = ω−x−tω−x tpx =

(ω−x−tω−x

)cµx(t) = 1

ω−x−t µx(t) = cω−x−t

Be careful on the exam - Modified DeMoivre problems are often disguised in a question thatstarts 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 writtenin a different algebraic form.



Note that in this table, the function listed at left are given in terms of the functions acrossthe top row!

F (x) s(x) f(x) µ(x)

F (x) F (x) 1− s(x)∫ x

0 f(u) du 1− e−∫ x0µ(t) dt

s(x) 1− F (x) s(x)∫∞x f(u) du e−

∫ x0µ(t) dt

f(x) F ′(x) −s′(x) f(x) µ(x) e−∫ x0µ(t) dt

µ(x) F ′(x)1−F (x)

−s′(x)s(x)

f(x)s(x) µ(x)

Assumptions for fractional ages.Uniform Constant

Function Distribution Force Hyperbolic

tqx tqx 1− ptxtqx

1−(1−t)qx

µ(x+ t) qx1−tqx − log px

qx1−(1−t)qx

1−tqx+t(1−t)qx1−tqx 1− p1−t

x (1− t)qx

yqx+tyqx

1−tqx 1− pyxyqx

1−(1−y−t)qx

tpx 1− tqx ptxpx

1−(1−t)qx

tpxµ(x+ t) qx −ptx log pxqxpx

[1−(1−t)qx]2



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 ofactuarial 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 (D) 178 (E) 333

Solution:◦e0=

∫ ω

0

(1− t

ω

)dt = ω − ω2

2ω=ω

2= 25⇒ ω = 50

◦e10=

∫ 40

0

(1− t

40

)dt = 40− 402

(2)(40)= 20

Var [T (x)] = 2

∫ 40

0t

(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



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

[q81 + p81 · 1/2q82

]=

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 ω = 100. If she takes up hotair ballooning for the coming year, her assumed mortality will be adjusted so that forthe 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 ifshe takes up hot air ballooning.

(A) 0.10 (B) 0.35 (C) 0.60 (D) 0.80 (E) 1.00

Solution: STANDARD:

◦e25:11 =

∫ 11

0(1− t

75)dt = t− t2

2× 75|110 = 10.1933

MODIFIED:

p25 = e−∫ 1

00.1ds = e−0.1 = 0.90484

◦e25:11 =

∫ 1

0tp25dt+ p25

∫ 10

0(1− t

74)dt

=

∫ 1

0e−0.1tdt+ e−0.1

∫ 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 difference is 0.8047. Key: D



4. You are given the following extract from a select-and-ultimate mortality table with a2-year select period:

x l[x] l[x]+1 lx+2 x+ 2

60 80,625 79,954 78,839 6261 79,137 78,402 77,252 6362 77,575 76,770 75,578 64

Assume that deaths are uniformly distributed between integral ages.

Calculate 0.9q[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



Solution:

0.4p0 = 0.5 = e−∫ 0.4

0(F+e2x)dx

= e−0.4F−

[e2x

2

]0.40 = 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 thenext three years.

The actuary starts by calculating the expected number of survivors at each integral ageby

l95+k = 1000 kp95, k = 1, 2, 3

The actuary subsequently calculates the expected number of survivors at the middle ofeach year using the assumption that deaths are uniformly distributed over each year ofage.

This is the result of the actuary’s model:

Age Survivors

95 100095.5 80096 60096.5 48097 −−97.5 28898 −−

The actuary decides to change his assumption for mortality at fractional ages to theconstant force assumption. He retains his original assumption for each kp95.

Calculate the revised expected number of survivors at age 97.5.

(A) 270 (B) 273 (C) 276 (D) 279 (E) 282



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)12 = 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:◦e1:1.5 =

∫ 1.5

0tp1dt

=

∫ 1

0tp1dt+ 1p1

∫ 0.5

0xp2dx

=

∫ 1

0(1− 0.1t)dt+ 0.9

∫ 0.5

0(1− 0.05x)dx

=

[t− 0.1t2

2

]1

0

+ 0.9

[x− 0.05x2

2

]0.5

0

= 0.95 + 0.444 = 1.394 Key: D



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 thelimiting age.

Assuming de Moivres law still applies after the medical breakthrough, calculate thenew limiting age.

(A) 104 (B) 105 (C) 106 (D) 107 (E)108

Solution: For deMoivre’s law,◦e30=

ω − 30

2.

Prior to medical breakthrough ω = 100⇒◦e30= 100−302 = 35.

After medical breakthrough◦e′30=

◦e30 +4 = 39.

⇒◦e′30= 39 =

ω′ − 30

2⇒ ω′ = 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 6361 0.10 0.12 0.14 0.16 6462 0.11 0.13 0.15 0.17 6563 0.12 0.14 0.16 0.18 6664 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



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 < 400.05, x > 40

Calculate◦e25:25 .

(A) 14.0 (B) 14.4 (C) 14.8 (D) 15.2 (D) 15.6

Solution:◦e25:25 =

∫ 15

0tp25 dt+ 15p25

∫ 10

0tp40 dt

=

∫ 15

0e−0.04t dt+

(e−∫ 15

00.04 ds

)∫ 10

0e−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 ω = 100.

(iii)

fT (t) =

{k f1(t), 0 ≤ t ≤ 501.2 f2(t), 50 < t

Calculate 10p40.

(A) 0.81 (B) 0.85 (C) 0.88 (D) 0.92 (E) 0.96



SOLUTION: From the Illustrative Life Table:

l50

l0= 0.8951

l40

l0= 0.9313

1 =

∫ ∞0

fT (t) dt =

∫ 50

0kf1(t) dt+

∫ ∞50

1.2 f2(t) dt

= kF1(50) + 1.2 (F2(∞)− F2(50))

= k (1− 50p0) + 1.2(1− 0.5)

= k(1− 0.8951) + 0.6

⇒ k =1− 0.6

1− 08951= 3.813

For x ≤ 50,

FT (x) =

∫ x

03.813f1(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

(τ)x (t), t > 0

(ii) µ(τ)x (t) = kt2, t > 0

(iii) q′(1)x = 0.04

Calculate 2q(2)x .

(A) 0.45 (B) 0.53 (C) 0.58 (D) 0.64 (E) 0.73



SOLUTION:

We can use the exponential formulation for p′(x)

0.04 = q′(1)x = 1−p′(1)

x = 1−e−∫ 1

0µ(1)x (t) dt = 1−e−

∫ 1

00.2µ

(τ)x (t) dt = 1−e−

∫ 1

00.2kt2 dt = 1−e−0.2k/3

⇒ e−0.2k/3 = 0.96

µ(1)x (t) = 0.2µ(τ)

x (t) ⇒ µ(2)x (t) = 0.8µ(τ)

x (t)

2q(2)x =

∫ 2

0tp

(τ)x · µ(2)

x (t)dt =

∫ 2

0tp

(τ)x · (0.8)µ(τ)

x (t)dt = 0.82q(τ)x

To get 2q(τ)x , we use

2q(τ)x = 1− 2p

(τ)x = 1− e−

∫ 2

0µ(τ)x (t) dt = 1− e−

∫ 2

0kt2 dt = 1− e−8k/3

= 1−(e−0.2k/3

)40= 1− (0.96)40

⇒ 2q(τ)x = 0.8046

⇒ 2q(2)x = (0.8)(0.8046) = 0.644

13. You are given:

(i)◦e30:40 = 27.692

(ii) s(x) = 1− xω , 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:◦e30:40 =

∫ 400 tp30dt

=∫ 40

0ω−30−tω−30 dt



=[t− t2

2(ω−30)

]|400

= 40− 800ω−30

= 27.692⇒ ω = 95

tp30 = 65−t65

Now, realize (after getting ω = 95) that T (30) is uniformly on (0, 65), its variance isjust the variance of a continuous uniform random variable:

V ar = (65−0)2

12 = 352.08Key: B

14. For a life table with a one-year select period, you are given:

(i)x l[x] d[x] lx+1

◦e[x]

80 1000 90 − 8.581 920 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] = 910l82 = l[81] − d[81] = 830

p[80] = 9101000 = 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



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◦ex:1 = 0.5(0px + 1px) = 0.5(1 + 1px)

You can also derive the above equation using the standard formula:◦ex:1 =

∫ 10 tpxdt =∫ 1

0 (1− tqx)dt =∫ 1

0 (1− tqx)dt

Since qx is a constant, you can verify that:∫ 10 (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 =∞, we have: ex = 12qx + px (1 + ex)

e[80] = 12q[80] + p[80] (1 + e81)

8.5 = 12(1− 0.91) + (0.91) (1 + e81)

e81 = 8.291e81 = 1

2q81 + p81 (1 + e82)e82 = 8.043e[81] = 1

2q[81] + p[81] (1 + e82)

= 12(1− 0.912) + (0.912)(1 + 8.043)

= 8.206Key: C

15. You are given:

µ(x) =

{0.05 50 ≤ x < 600.04 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



4|14q50 = 4p50 − 18p50 = 0.8187− 0.4404 = 0.3783Key: A



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 (MaleNewborn)∗P [T (MaleNewborn) > 60]+P (Femalenewborn)∗P [T (FemaleNewborn) > 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− 0.004920.005354 = 0.081

Key: B



Problems from Pre-2000 SOA-CAS exams

1. You are given: µ(x) =

√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



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

ω

)α, 0 ≤ x < ω, 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 uniformdistribution of deaths assumption.

• g is the probability that (60) will die between ages 60.5 and 61.5 under the Balducciassumption.

Calculate 10,000(g − f).

(A) 0 (B) 85 (C) 94 (D) 178 (E) 213



Solutions to Chapter 3

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

⇒ 0.5 = e−∫ t0µ(20+s) ds = e−

∫ t0

(60−s)−0.5 ds

= e2[(60−s)0.5]t

0 = e2[√

60−t−√

60]

ln 0.5 = 2[√

60− t−√

60]

⇒√

60− t = 7.40 ⇒ t = 5.25

2. Key: D

Tx+1 =

∫ ∞0

lx+1+t dt

lx+1 =6.4

0.8= 8 ⇒ lx+1+t = 8(0.8)t

Tx+1 = 8

∫ ∞0

(0.8)t dt =8

ln 0.8(0.8)t|∞0

=−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 ] = 1µ2

= 0.25

5. Key: C0.5 = tpx = e−µ·t = e−2t

⇒ t =ln 2

2



6. Key: E

mx =lx − lx+1

Lx=lx(1− e−µ)∫ 10 lx · tpx dt

=lx(1− e−2)

lx∫ 1

0 e−µt dt

==1− e−2

12(1− e−2)

= 2

7. Key: A This is a Modified DeMoivre problem with constant equal to α. So

µ(x) =α

ω − x,

◦ex=

ω − xα+ 1

Multiplying these two quantities gives

α

α+ 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: 0.71−(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


Arch MLC Spring 2010

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