ACTEX 2010 SOA Exam MLC /CAS Exam 3L

Introduction

ACTEX Study Manual for SOA Exam MLC, CAS Exam 3L This study manual will cover the major topics in the required readings for SOA Exam MLC, the Life Contingencies segment of the Actuarial Models exam, and the life contingencies portion of CAS Exam 3L, and will show you how to solve exam problems that have been given on prior life contingencies exams and are relevant to the current syllabus. It is focused on providing the understanding that is required to solve practical problems in an exam setting. Theoretical concepts are introduced through motivating examples to make learning as direct as possible. Every new explanation of a concept or skill will be followed by a computational exercise for you to do immediately to test your understanding and reinforce your learning. Actuarial exam problems will be included not only at the end of chapters (which we call modules), but will also be included within the text. At the beginning of each module is a list of the CAS Exam 3L and SOA Exam MLC learning objectives that are covered by the module. ▪

For the references and readings “AM” and “Option 1” refers to Actuarial Mathematics, Second Edition, 1997, by Bowers, N.L., Gerber, H.U., Hickman, J.C., Jones, D.A. and Nesbitt, C.J.

▪

“MQR” and “Option 2” refers to Models for Quantifying Risk, Third Edition, 2009, by Cunningham, R., Herzog, T. and London, R.L.

The material that is not required for CAS Exam 3L is explicitly noted. Also note that, per the list of Learning Objectives for CAS Exam 3L, 33-37% of CAS Exam 3L is based on statistical topics that are not covered in this study manual. There are 13 modules in this study manual. The first two are review, and cover prerequisite interest theory and probability theory that you will need to know for the exam. We suggest that you work through the review rapidly, and use the review units as a resource that you can return to later if you discover that you are having trouble applying one of the prerequisite skills. The content directly related to the learning objectives for Exam MLC begins in Module 3. We have a recommended study procedure based on our experience helping actuarial students prepare for the life contingencies actuarial exam. 1) Work through all the modules in order and do not skip the review. Module 1 on interest theory will go relatively quickly. Module 2 on probability may introduce concepts that are new to you.

©ACTEX 2010 Luckner, Hassett, Stewart, & Steeby

SOA Exam MLC /CAS Exam 3L

2) As you work through the modules, immediate reinforcement exercises will appear in the text. Don’t skip them. They are essential basic calculations, and if you cannot do them you probably did not grasp what you just read. 3) Beginning in Module 3, every module will have a set of basic problems and a final section of prior exam problems with worked solutions. Try to work each problem before consulting the solution. Do not skip the exam problems. Some of them introduce additional problem-solving techniques that will help with the exam. 4) At the end of the manual there will be another set of prior exam problems to serve as a final exam. These will be in random order so that you can work on problem type identification as well as basic skills. 5) After the problem set in 4) above, there are original practice exams. Use these exams to give yourself exam-taking experience by taking them under exam conditions. When you do this should depend on when you finish the modules. Completing a module a week is a reasonable goal, but that is for initial understanding and a few weeks of review are usually helpful after that before taking the practice exams. 6) Following the original practice exams is the SOA Spring 2007 MLC Exam, the only MLC exam publicly released. We have given the questions and solutions separately so that you can take this exam under exam conditions if you wish. We have provided white space so that you can write in notes, expand calculations or identify questions that need further thought. Students are all different, and develop their own personal approaches to problems. Don’t be afraid to write yours down – in pencil for possible revision. We encourage you to make use of the ACTEX www.actexmadriver.com. Three areas there can be helpful:

web

site

at

1) Customer feedback. You can use this to let us know about possible typographical errors or topics that you believe to be missing. That will result in postings to the next two areas. 2) Errata and Updates. We will post error corrections here as soon as we are aware of any errors. 3) Free Downloads. We will provide a download of pages on any topic which is added to the syllabus after this manual is printed, and will provide supplements on material that needs additional emphasis. Let us know what you think is needed. We advise every student to also regularly check the SOA web site for announcements. This manual is self-contained, but we refer you to the required textbooks noted above, and study notes for deeper understanding.



The required study notes are Poisson Processes (and mixture distributions), Daniel 2008 Multi-state transition models with actuarial applications, Daniel We also cite two other books for reference for basic probability and statistics. Probability for Risk Management (Second Edition), Hassett, Stewart. ACTEX. 2006. Loss Models (Third Edition), Klugman, Panjer and Wilmot. Wiley, 2008. Loss Models is required for Exam C. For basic statistics, we refer you to Introductory Statistics (Eighth Edition), Weiss. Pearson. 2008 On the following pages you will find a map of the chapters and a table of contents with a detailed topic list. That table of contents has also been sorted into an alphabetical content index at the end of volume II. We have included all of these to make your study more efficient. There is a large amount of material in Exam MLC, and it is easy to get topics confused or simply lose track of where they are. We encourage you to take a few minutes to review the topic map, the table of contents, and the index. We apologize in advance for errors, typographical or otherwise, that you might find. We would greatly appreciate it if you forward them to ACTEX so that we can correct them. Any errors that are found will be posted to an errata file at the ACTEX website, www.actexmadriver.com. We hope you find this study manual helpful as you prepare for the MLC exam, and wish the best of luck on the exam. October, 2009

Warren Luckner, FSA, CFA, MAAA Matt Hassett, PhD Donald Stewart, PhD Amy C. Steeby, MBA



Table of Contents Module 1

Interest Theory Review Section 1.1 1.2

Title Preliminary Discussion Detailed Review

Topic

Time value of Money Present and Future Value Interest Rate vs Discount Rate Notation for Interest Theory Investment Growth Functions Force of Interest Nominal Rates of Interest 1.3

Annuities Unit Annuities Immediate or Due Geometric Series Annuity Immediate Calculations Perpetuities Annuity Due Calculations Continuous Annuities Relating Discount, Force of Interest and Interest Rate Calculator Time Value of Money Annuities with Varying Payments Increasing Annuities with Terms in Arithmetic Progression Decreasing Annuities with Terms in Arithmetic Progression Annuities with Terms in Geometric Progression Equations of Value Deferred Annuity Identity Variable Annuities

1.4

Module 2

Formula Sheet Module 1

Page 1 1 1 1 2 3 3 5 6 9 9 9 10 11 12 15 17 18 18 19 21 23 25 27 28 29

Probability Review Section 2.1 2.2

Title Preliminary discussion Discrete Random Variables

Topic

Random variable definition Discrete Random Variable -Probability Function Discrete Random Variable -Mean and Variance Standard Deviation Mean and Variance of Y = aX+b Expectation of g(X) Moment Generating Functions -Discrete Cumulative Distribution F(x) -Discrete 2.3

2.4

2.5

Continuous Random Variables Density Function f(x) Cumulative Distribution F(x) -Continuous Continous Random Variable -Mean, Variance, Moments Moment Generating Functions -Continuous Moments -raw and central Coefficient of Variation Median and Percentiles Mixed Distributions Commonly Used Discrete Random Variables Uniform (Discrete) Binomial Random Variable Geometric Random Variable Negative Binomial Random Variable Poisson Random Variable Commonly Used Continuous Random Variables Uniform (Continuous) Normal Random Variable Standard Normal Exponential Random Variable Gamma Function Gamma Random Variable

Page 1 1 1 1 2 2 3 4 4 6 8 8 9 10 11 12 12 13 14 15 15 17 20 22 25 26 26 28 28 30 32 32

©ACTEX 2010 Luckner, Hassett, Stewart, Steeby Page TOC- 1

SOA Exam MLC / CAS Exam 3L

Table of Contents 2.6

2.7

2.7

Module 3

Multivariate Probability Bivariate Probability -Discrete Marginal Distribution -Discrete Conditional Distribution -Discrete Tree Diagrams Bayes Theorem Double Expectation Theorem Bivariate Probability -Continuous Marginal Distribution -Continous Conditional Distribution -Continous Independent Random Variables Covariance Sums of Independent Random Variables Individual Model Collective Model Moments of S Convolutions Normal Approximation Central Limit Theorem Moment Generating Functions for Sums of Independent Random Variables Sum of Independent Normal random Variables Sum of Independent Poisson Random Variables Sum of Independent Negative Binomial (Geometric) Random Variables Sum of Independent Gamma (Exponenetial) Random Variables Waiting Time Compound Distributions -Collective Model Compound Poisson Collective Model: Mean and Variance Collective Model: Normal Approximation Compound Distributions -Moment and Probability Generating Function Formula Sheet Module 2

35 35 35 36 38 38 42 43 44 45 46 47 48 48 48 48 49 51 51 52 53 55 56 57 57 58 59 59 60 62 63

The Poisson Process Section 3.1

Title The Poisson Distribution

Topic Poisson probabilities, mean and variance, probability generating fctn. Poisson approximation to the binomial Sum of indpendent Poisson random variables Decomposition of Poisson random variable

3.2

Creating New Counting Models Compound Frequency Models -Definition Compound Frequency Models -Normal Approximation

3.3

Counting Processes Counting Process definition Poisson Process definition Poisson Process decomposition Wating time: Poisson Poisson process: nonhomogeous

3.4

Mixture Models Discrete Mixtures Mean and Variance of a Mixture Distribution Continuous Mixtures

3.5 3.6 3.7 3.8 3.9

Module 4

Formula Sheet Module 3 Practice Problems Module 3 Practice Problem Solutions Module 3 Exam problems Module 3 Exam problem solutions Module 3

Page 2 2 2 3 4 7 7 7 10 10 10 11 13 15 17 18 19 21 25 27 29 32 39

Continuous Survival Models Section 4.1

Title Preliminary discussion

Topic Survival Model

Page 2 2




Age at Failure Random Variable

2 3 3 3 6 7 8 8 10 11 13 13 14 15 15 15 16 17 17 17 18 19 23 23 23 26 27 29 34 38

Age at Failure Random Variable Age at Death Random Variable qx-px notation for F(x) and S(x) Probabilities for an Age-to-Failure Random Variable Hazard Rate Force of Mortality Constant Force Model Cumulative Hazard Function Complete Expectation of Life 4.3

4.4

4.5

4.6 4.7 4.8 4.9 4.10

Module 5

Important Survival Models Survival Model -Uniform (De Moivre) Survival Model -Exponential Memoryless Property - Exponential Survival Model Gompertz Distribution Makeham Distribution Weibull Distribution Random Variable for Time Until Death for a Life Age x Survival Function -Life Age x T(x) -Survival Function T(x) - Cumulative Distribution and Density T(x) -Mean and Variance The Central Rate of Failure Central rate of Failure Central rate of Death Formula Sheet Module 4 Practice Problems Module 4 Practice Problem Solutions Module 4 Exam problems Module 4 Exam problem solutions Module 4

The Life Table Approach Section 5.1

5.2

5.3 5.4

5.5 5.7 5.8 5.9 5.10 5.11

Module 6

Title What is a Life Table?

Topic

Life Table Survival Models in Life Table Language More Functions in Life Table Form Force of Mortality in Life Table Notation Survival Model -Density, Mean and Variance in Life Table Language Exposure T0 and Tx Remaing Lifetime at age x -conditional analysis Remaing Lifetime at age x -expected value over next n years only Survival Function -shifts in force of mortality nLx -Total life years lived over next n-years only Central Rate in Life Table Notation Curtate Future Lifetime Curtate Future Lifetime Interpolation Between Integral Ages Linear Interpolation between Integral Ages UDD (Uniform Distribution of Deaths) Constant Force Interpolation Balducci Interpolation Select Mortality Select-and-Ultimate Mortality Tables Formula Sheet Module 5 Practice Problems Module 5 Practice Problem Solutions Module 5 Exam problems Module 5 Exam problem solutions Module 5

Page 3 3 5 7 7 8 9 9 13 16 18 19 21 23 23 25 25 26 29 29 35 37 38 40 42 47 53

Contingent Payment Models for Life Insurance Section 6.1

Title

Topic

Introduction Actuarial Present Value Intorduced

Page 2 2




6.3

6.4

6.5

6.6 6.7

6.8

6.9

6.10

6.11 6.12 6.13 6.14 6.15

Module 7

Whole Life Insurance Payable at the Moment of Death Whole Life Insurance Payable at the Moment of Death Level Benefit Insurance Whole Life Insurance Payable at the Moment of Death -Expected Value Whole Life Insurance Payable at the Moment of Death -Variance DeMoivre -Whole Life Expected Value and Variance Constant Force -Whole Life Expected Value and Variance Level Benefits other than 1 Term Insurance Payable at the Moment of Death Term Insurance Payable at the Moment of Death Term Ins. Payable at Moment of Death- Expected Value and Variance DeMoivre -Term Insurance Expected Value and Variance Constant Force - Term Insurance Expected Value and Variance Endowment Insurance Pure endowment Pure endowment -Expected Value and Variance Endowment Insurance Endowment Insurance -Expected Value Endowment Insurance -Variance Deferred Insurance Payable at Time of Death Deferred Insurance Payable at the Moment of Death Deferred Insurance Payable at the Moment of Death -Expected Value Relating Whole Life, Term and Deferred Insurance Whole Life Insurance-Sum of Term and Deferred Insurances Insurance Payable at the End of the Year of Death Insurance Payable at the End of the Year of Death Whole Life Ins. Payable at the End of the Year of Death-Expected Value Term Ins. Payable at the End of the Year of Death -Expected Value Deferred Ins. Payable at the End of the Year of Death -Expected Value Endowment Ins. Payable at the End of the Year of Death -Exp. Value Whole Life Ins. Payable at the End of the Year of Death -Variance Recursion Relations for Insurance Recursion -Illustrative Life Table Variable Benefit Insurance Variable Benefit Insurance Annually Increasing or Decreasing Insurance Annually Increasing or Decreasing Insurance -Recursion Relating Discrete and Continuous Insurances Relating Discrete and Continuous Insurances m-thly adjustment for whole life (UDD) Whole Life Insurance Applications Normal Approximation Risk Assessments Force of Interest and Mortality Shifts Percentiles for Loss Random Variable Z Formula Sheet Module 6 Practice Problems Module 6 Practice Problem Solutions Module 6 Exam problems Module 6 Exam problem solutions Module 6

4 4 4 5 5 6 8 10 11 11 12 12 13 15 15 15 17 17 18 19 19 19 23 23 25 25 25 26 26 27 27 29 31 34 34 36 40 43 43 46 47 47 49 51 53 56 58 63 75

Life Annuities Section 7.1

7.2

7.3

7.4 7.5

Title Continuous Annuities

Topic

Whole Life Annuities- Continuous Whole Life Annuities- Continuous -Expected Value Whole Life Annuities- Continuous -Expected Value -Constant Force Whole Life Annuities- Continuous -Variance Whole Life Annuities- Continuous -Distribution of Random Variable Continuous Temporary Life Annuities Temporary Life Annuity -Continuous Temporary Life Annuity -Continuous -Expected Value Temporary Life Annuity -Continuous -Variance Deferred Life Annuities Deferred Life Annuity-Continuous Deferred Life Annuity-Continuous-Expected Value Relating Life, temporary and Deferred Annuities Relating Life, temporary and Deferred Annuities The Certain-and-Life Annuity Certain-and-Life Annuity -Continous

Page 2 2 3 4 5 7 9 9 9 10 11 11 11 13 13 14 14



Table of Contents 7.6 7.7

7.8

7.9 7.10 7.11 7.12 7.13

Module 8

Actuarial Accumulated Future Value Actuarial Accumulated Future Value -Continous Discrete Life Annuities Whole Life Annuity Due -Random Variable Whole Life Annuity Due -Expected Value Whole Life Annuity Due -Variance Temporary Life Annuity Due -Random Variable Temporary Life Annuity Due -Expected Value Deferred Life Annuity Due -Expected Value Certain-and-Life Annuity Due Actuarial Accumulated Future Value -Discrete Life Annuity Immediate Life Annuities with m-thly payments m-thly Interest and Discount review m-thly Life Annuity Due -timeline m-thly Life Annuity Due -UDD assumption m-thly Life Annuity Due - Simple Approximation m-thly Life Annuity Due - α-β formula Formula Sheet Module 7 Practice Problems Module 7 Practice Problem Solutions Module 7 Exam problems Module 7 Exam problem solutions Module 7

16 16 18 18 19 20 20 21 22 23 26 27 29 29 30 30 31 32 33 35 37 42 51

Premiums Section 8.1

8.2

8.3

8.4

8.5

8.6 8.7 8.8

Title

Topic

Introduction Benefit Premium Net Premium Single Benefit Premium Fully Continuous Benefit and Premium Loss Function -Continuous Whole Life and Annuity Premium -Continuous Whole Life and Annuity Loss Function -Continuous Whole Life and Annuity -Variance Premium Identities -Continuous Whole Life and Annuity Premium -Continuous n-year Term Premium -Continuous n-year Endowment Premium -Continuous h-payment Whole Life Premium -Continuous h-payment n-year Endowment Premium -Continuous Payment pure Endowment Premium -Continuous Variable Benefit Example Fully Discrete Benefit and Premium Loss Function -Discrete Whole Life and Annuity Premium -Discrete Whole Life and Annuity Loss Function -Discrete Whole Life and Annuity -Variance Premium -Fully Discrete n-year Term Premium -Fully Discrete n-year Endowment Premium -Fully Discrete h-payment Whole Life Premium -Fully Discrete h-payment n-year Endowment Premium -Fully Discrete n-year Pure Endowment Premium -Geometric Series Problems Semi-Continuous Benefit Premiums Premium -Semicontinuous Premium -Semi-continous n-year Term Premium -Semi-continous n-year Endowment Premium -Semi-continous h-payment Whole Life Premium -Semi-continous h-payment n-year Endowment m-thly Benefit Premiums Premium -mthly payment Premium -mthly payment n-year Term Premium -mthly payment n-year Endowment Premium -mthly payment h-payment Whole Life Premium -mthly payment h-payment n-year Endowment Applying the Equivalence Principle to Variable Benefits Premium -Variable Benefit Examples The Loss Function and Percentile Premiums Percentile Premiums The Loss Function L when Premiums May Not Satisfy the Equivalence Principle Loss Function -Variance -General

Page 2 2 2 2 3 3 5 6 7 11 12 13 14 15 17 18 18 19 19 20 20 20 20 20 21 23 23 23 23 23 23 25 25 28 28 28 28 29 30 32 32 37 37




Another Useful Identity

39 39 40 45 48 54 69

Life Annuity -Premium Formula 8.10 8.11 8.12 8.13 8.14

Module 9


Benefit Reserves Section 9.1 9.2

9.3

9.4

9.5

9.6

9.7 9.8

9.9 9.10 9.11 9.12 9.13 9.14

Module 10

Title Topic Introduction Prospective and Retrospective Reserves Prospective Method for Reserves -Introduction Retrospective Method for Reserves -Introduction Reserves in the Continuous Case Reserves -Fully Continuous Whole Life -Prospective Loss Function -Fully Continuous Whole Life -Variance at time t Reserves -Premium Difference Formula -Continuous Reserves -Paid-up Insurance Formula -Continuous Reserves -Annuity Reserve Formula -Continuous Reserves -Benefit Formula -Continuous Reserves -Fully Continuous Whole Life -Retrospective -Continuous Reserves -Fully Continuous n-year Term -Prospective Reserves -Fully Continuous n-year Endowment -Prospective Reserves -Fully Continuous h-payment Whole Life -Prospective Reserves -Fully Continuous h-payment n-year Endowment -Prospective Fully Discrete Level Benefit Reserves Initial Reserve Terminal Reserve Reserves -Fully Discrete Whole Life -Prospective Loss Function -Fully Discrete Whole Life -Variance at time t Reserves -Premium Difference Formula -Discrete Reserves -Paid-up Insurance Formula -Discrete Reserves -Annuity Reserve Formula -Discrete Reserves -Benefit Formula -Discrete Reserves -Retrospective -Discrete Reserves -Fully Discrete n-year Term -Prospective Reserves -Fully Discrete n-year Endowment -Prospective Reserves -Fully Discrete h-payment Whole Life -Prospective Reserves -Fully Discrete h-payment n-year Endowment -Prospective Semi-continuous or m-thly Payment Benefit Premiums Reserves -Semi-continous Whole Life with Annual Premiums Reserves -Semi-continous Whole Life with m-thly Premiums General Benefit Reserves -the Discrete Case Reserves -Variable Benefit and/or Premium Reserves -Recursion Reserves -Problem Examples Using the Distribution of the Loss Random Variable Directly to Find the Mean and Variance Loss Function -Distribution and Mean-Variance Calculation Hattendorf's Theorem Hattendorf's Theorem Net Amount at Risk Interpolating Reserves for Fractional Periods Formula Sheet Module 9 Practice Problems Module 9 Practice Problem Solutions Module 9 Exam problems Module 9 Exam problem solutions Module 9

Page 2 4 4 5 6 6 7 8 9 11 12 12 16 16 16 17 18 18 18 19 20 21 21 21 21 21 22 22 22 23 24 24 24 26 26 28 31 37 37 41 41 42 45 47 49 51 57 69

Multiple Life Random Variables Section 10.1

Title

Topic

Introduction Joint Life Status Introduced Last Survivor Status Introduced

Page 3 3 3




The Joint Life Status Survival Model Joint Life Status -Random Variable Joint Life Status -Probabilities Joint Life Status -Force of Mortality Joint Life Status -Constant Force Results Premium Benefits for the Joint Life Status Joint Life Status -Insurance Calculations The Last Survivor Status Last Survivor Status -Random Variable Last Survivor Status -Probabilities Last Survivor Status + Joint Life Status Identities Common Shock Common Shock -Random Variable and Calculations Contingent Probability Functions Formula Sheet Module 10 Practice Problems Module 10 Practice Problem Solutions Module 10 Exam problems Module 10 Exam problem solutions Module 10

10.3 10.4

10.5 10.5 10.7 10.8 10.9 10.10 10.11

Module 11

4 4 4 6 7 9 9 11 11 11 13 15 15 18 21 22 24 29 38

Multiple Decrements Section 11.1

Title A Two-Decrement Example

11.3

Relating t q x ( i ) and t q x ¢ ( i ) tt

11.5 11.6 11.7 11.8 11.9 11.10

Module 12

Page 3 3 3 4 5 6 9 9 9 9 9 12

Decrement introduced Forces of Decrement -2 decrements Force of Decrement -Total -2 decrements Associated Single Decrement -2 decrements Decrement -Density Function -2 decrements The General Multiple Decrement Model Forces of Decrement -n decrements Force of Decrement -Total -n decrements Associated Single Decrement -n decrements Decrement -Density Function -n decrements Additional Examples

11.2

11.4

Topic



14 14 16 18 18 21 22 25 29 39

Associated Single Decrements UDD Decrements and Total Decrement UDD Life tables for Multiple Decrements Decrement -Life Table Notation and Problems Formula Sheet Module 11 Practice Problems Module 11 Practice Problem Solutions Module 11 Exam problems Module 11 Exam problem solutions Module 11 q

x

x

Insurance Models Including Expense Section 12.1

Title A Basic Identity

12.2

Timing of Expenses

12.3

Constant Annual Expense

Topic Life Annuity Immediate / Annuity Due Identity Expense: First Year and Continuing Equivalence Principle Including Expense Gross Premium Reserves -with Constant Expense

12.4

Variable Annual Expense Gross Premium with different First Year and Continuing Expense Reserves -with Different First Year and Continuing Expense

12.5

Expense Reserves Level Expense Premium Expense Reserve

12.6 12.7 12.8 12.9

Expense Example The Expenses of an Insurance Organization Expenses at Death Expense at Death Example Including Costs per 1000 of Insurance Expenses: Cost per 1000 Example

Page 2 2 3 3 4 4 4 6 8 8 9 10 10 10 12 14 15 15 16 16




Asset Share Calculations

18 18 18 18 22 23 25 30 37

Cash Value Withdrawal Asset Share Calculations 12.11 12.12 12.13 12.14 12.15

Module 13


Multi-State Transition Models Section 13.1 13.2

13.3 13.4 13.5

13.6 13.7 13.8 13.9 13.10 13.11 13.12 13.13

Title Introduction Markov Chains by Example

Topic

Transition Probabilty Transition Matrix Markov Chain Markov Chain -Finite Markov Chain -Homogeneous Absorbing State Markov Chain - non-homogeneous Longer Term Transition Probabilities Transition Matrix Multiplication for Longer Term Probablities The Probability Distribution at Time n Actuarial Present Value of Cash Flows Upon Transition Benefit Upon Transition Cash Flow upon Transition Markov Chaim -Actuarial Present Value of Cash Flows Upon Transition Actuarial Present Value of Cash Flows While in States Markov Chain ; Actuarial Present Value of Cash Flows While in States Benefit Premiums and Reserves Markov Chain: Benefit Premiums and Reserves Notation from the Study Note Formula Sheet Module 13 Practice Problems Module 13 Practice Problem Solutions Module 13 Exam problems Module 13 Exam problem solutions Module 13

Page 2 3 3 3 3 3 4 4 5 6 7 10 11 11 11 11 14 14 16 16 18 19 20 22 27 35

Final Problem Set Practice Exams 1-3 and Solutions Spring 2007 Exam MLC and Solutions Tables Index



Page M4- 1

Module 4 – Continuous Survival Models

Continuous Survival Models MLC/3L LEARNING OBJECTIVES AND REFERENCES LEARNING OBJECTIVES: Survival models 1. Define survival-time random variables a. for one life, in the single- decrement model; 2. Calculate the expected values, variances, probabilities, and percentiles for survival-time random variables. REFERENCES: AM Ch 3: 3.1 thru 3.3.1 MQR Ch 3

CAS EXAM 3L LEARNING OBJECTIVES – SURVIVAL MODELS

KNOWLEDGE STATEMENTS

1. For continuous univariate probability distributions for failure time random variables, develop expressions in terms of the life table functions, lx, qx, px, nqx, npx, and m|nqx, for the cumulative distribution function, the survival function, the probability density function and the hazard function (force of mortality), and be able to:

a. Failure time random variables

•

Establish relations between the different functions

•

Develop expressions, including recursion relations, in terms of the functions for probabilities and moments associated with functions of failure time random variables, and calculate such quantities using simple failure time distributions

•

Express the effect of explanatory variables on a failure time distribution in terms of proportional hazards and accelerated failure time models

b. SEE MODULE 5 c. Cumulative distribution functions d. Survival functions e. Probability density functions f. Hazard functions g. Relationships between failure time random variables in the functions above

The distributions may be left-truncated, rightcensored, both, or neither. READINGS Option 1: Chapter 3 (excluding 3.6 and 3.8); Option 2: Chapters 3.1-3.4, 4.1-4.4



Page M4-2


Section 4.1 - Preliminary Discussion In previous sections, we used selected random variables to represent the frequency and severity of claims. Another quantity of great importance in actuarial science is the length of a person’s life. When we use a particular random variable X to represent an individual’s age at death we refer to it as a survival model. Engineers also study survival models, but instead of time until death of a person they study things like the time until failure of a device. I was once an expert witness in a lawsuit involving the premature failure of a medical device, and used basic survival model mathematics in my analysis. In this module we will study the use of various continuous random variables in survival analysis. Some of the random variables looked at here (such as the uniform and exponential) will have already been used as severity random variables in previous modules. However, we will introduce some new concepts and notations since actuaries have their own special notation for some survival analyses, and knowledge of it is essential for the exams. In Module 5 we will look at discrete random variables for survival, and introduce life tables and their analysis.



Page M4- 3


Section 4.2 – Age at Failure Random Variable Let X represent the age at which an object of interest fails. This failure could be a death or a device failure: I replaced a light bulb ten minutes before writing this. Thus we could refer to X as an age at failure random variable or an age at death random variable. These terms are used interchangeably in practice. It is common to use X for the full age at death and T to denote the random variable for the remaining lifetime of an individual who has already reached a certain age x. This leads to the relation X = x + T. Entire Lifetime

0

X

0

x

X Future Lifetime

The letters X and T are not always used this way, and you must be careful not to make assumptions. Textbook writers tend to use notation any way they want to! We will use the X, T notation consistently here, since it is the convention used in AM and MQR. For problems involving the future lifetime T of a person aged x, it is common to use the notation (x) to represent a person of that age. In an exam problem, I would be referred to as (64), not Matt, age 64.

New Notation for F(x) and S(x) In actuarial notation the probability that a person born at time 0 dies at or before time x is denoted by x q0 . This is a new notation for the already familiar CDF F(x), since (4.1)

x

q0 = P( X £ x ) = F (x )

The probability that a person born at time 0 dies after age x is denoted by x p0 . This is a new notation for the already familiar survival function, S(x), since



Page M4-4


(4.2)

x

p0 = P( X > x ) = 1 - F ( x ) = S( X )

Important Note!

The reason to use the name survival function for S(x) is now obvious: in a survival model context it represents the probability that an individual survives to age x or beyond.

Caution!

AM uses a small s in the notation S(x). This use of s(x) instead of the familiar S(x) is just another notational variation you should know.

We have already encountered the uniform and exponential random variables in previous modules. These two random variables are not good models for human lifetimes, but they can be used for other failure analyses. We will use them as examples for basic concepts because the computations involved are simpler than computations for more useful random variables. This relative simplicity of computation also means that you are likely to see the uniform and the exponential on Exam MLC/3L. Example (4.3)

Let age at failure X be uniform on [0,100] (DeMoivre) (We have already found F(x) for the uniform in Module 2.) Then x x q0 = P ( X £ x ) = F ( x ) = 100 x

p0 = P( X > x ) = 1 - F (x ) = 1 -

x 100 - x = 100 100

Example (4.4)

Let the age at failure X be exponential with density function f (x ) = .05e -.05 x for x > 0. (We have already discussed F(x) for the exponential in Module 2.) Then -0.05 x x q0 = P ( X £ x ) = F ( x ) = 1 - e x

p0 = P( X > x ) = 1 - F ( x ) = S( x) = e -0.05 x



Page M4- 5


Note that in the survival model context the 1 -x is replaced by the notation f (x ) = e θ

θ

Keys to Success

notation f (x) = λe -λ x . The choice of θ as parameter is useful in severity analysis because θ is the mean. The choice of λ of is useful in survival problems because λ represents the failure rate – which we shall demonstrate in a few pages.

The graph of a survival function for an age at failure random variable has a pattern which we can see in the graphs of the functions S(x) in the last two examples.

Uniform and Exponential Survival Functions 1.00 0.90 0.80 0.70 S(X)

0.60 Uniform

0.50

Exponential

0.40 0.30 0.20 0.10 0.00 0

20

40

60

80

100

Age x

Note: The initial value S(0) is always equal to 1. (4.5)

lim S( x ) = 0 x ¥

The formula above says that the limit of the survival probability as age increases is zero. In other words, as age approaches ∞, the probability of survival is zero. We are born, and ultimately we die.



Page M4-6


Probabilities for an Age-to-Failure Random Variable We know that for an age at failure random variable with density f(x) x

F (x) =

ò f (t ) dt 0

and probabilities are found by P (a £ X £ b) =

b

ò f (x)dx = F (b) - F (a) = S(a) - S(b) . a

We have also observed that you can save time on probability calculations by using F(x) or S(x) when available to avoid integration. Example (4.6)

Let the age at failure X be exponential with density function f (x ) = 0.05e -0.05 x for x > 0. Find P(1< X <2). Solution: P(1 < X < 2) = S(1) - S(2) = e -.05(1) - e -.05(2) = 0.0464

Exercise 1:

Let the age at failure random variable X be uniform on [0, 100]. Find P(30 < X < 60) Answer:

0.30

Note that we encounter the usual paradoxical result for continuous distributions that the “area under the curve” exactly at x = a is equal to zero (this is the area of a line segment). Symbolically, this is: P ( X = a) =

a

ò f (x)dx = 0 . a



Page M4- 7


f(x) is not the probability of death at age x; in fact, f(x) could be greater than 1. f(x) is a function which is used to calculate the probability of death in an interval.

Caution!

Because of its interval properties, we can say that for a small value of dx, f(x)dx approximates the probability of death in the interval [ x, x + dx ] since f(x)dx approximates the area under the curve y=f(x) between x and x+dx. Example (4.7)

Let the age at failure X be exponential with density function f (x ) = 0.05e -0.05 x for x > 0.

Find P(1< X <1.01) exactly and using the approximation f(x)dx. Solution: P(1 < X < 1.01) = S(1) - S(1.01) = e -0.05(1) - e -0.05(1.01) = 0.000475 Using the approximation with x = 1 and dx = 0.01 P(1 < X < 1.01) » f (1)(0.01) = 0.05e -0.05(1) (0.01) = 0.000476

The Failure Rate (or Hazard Rate) Function for X As humans age we observe that deaths in an age group seem to be occurring at an increasing rate. Naturally, we are interested in the rate of death or failure. The failure rate (also known as the hazard rate) for an age at failure random variable is denoted by μ(x) or λ(x). The text AM primarily uses the μ notation, but MQR and other texts use λ. These two forms are interchangeable, and are defined by (4.8)

μ (x) = λ (x) =

f (x) S ¢(x ) =S (x) S( x)

The final expression results from the fact that S ¢(x ) = -F ¢(x ) = -f (x ) .



Page M4-8


It is essential to know the definition of the failure rate for Exam MLC/3L and be able to use it.

Keys to Success

It helps to try to understand what the failure rate does. For small dx, λ(x)dx (or, in different notation, μ (x)dx ) represents the probability that death occurs in the interval (x,x+dx] given that survival to age x has occurred. This means it is a conditional tool.

We can illuminate this interpretation by writing (4.9)

S( x)λ ( x ) dx = S(x )μ (x ) dx = f (x )dx

What (4.9) says is: (Probability of survival to age x)(Prob. of death between ages x and x+dx given survival to x) =P(death between ages x and x+dx)

For an exponential random variable with density function f (x ) = μ e -μ x , the failure rate is (4.10)

μ (x ) =

f (x ) μ e -μ x = -μ x = μ S (x) e

This has a natural interpretation. For our example in which μ = 0.05, the failure rate is always 5%. Given that you have survived to age x, the conditional probability that you will die before age x + dx is approximately 0.05dx. Actuaries also refer to μ (x ) as the force of mortality or hazard rate at age x. As we saw in (4.10), the force of mortality never changes for the exponential random variable. Because of this, it is also referred to as a constant force model. The failure rate for a uniform random variable is not constant, as we shall see in the next example.



Page M4- 9


Example (4.11)

Let age at failure X be uniform on [0,100]. Find λ ( x ) =

f (x) S (x)

(DeMoivre) Solution: 1 f (x) 1 , 0≤ x <100. λ (x) = = 100 = S ( x ) 100 - x 100 - x 100 Exercise 2:

Let the age at failure random variable X have survival function 2

S( x) =

(10 - x ) 100

Find λ(x ) .

,

0 £ x < 10 .

2

Answer:

10 - x

The uniform does show a greater hazard rate for older individuals, as we see in the graph below. Hazard rate for uniform random variable 1.00 0.90 0.80 hazard rate

0.70 0.60 0.50 0.40 0.30 0.20 0.10 0.00 0

20

40

60

80

100

Age x

The term force of mortality might remind you of the force of interest reviewed in Module 1. The two are similar:



Page M4-10


The force of interest is the instantaneous rate of growth for continuous compounding, and The force of mortality is the instantaneous rate of death of a continuous age of failure random variable. The force of mortality also has useful properties that parallel properties of the force of interest. One extremely important property enables us to start with λ(x) and find S(x) and thus f(x). The property involves integrating λ(x). If we define the cumulative hazard function by (4.12)

Λ (x) =

x

ò λ (u) du 0

It can be shown that (4.13)

S (x) = e

-Λ( x )

Equation (4.13) is analogous to (1.15) in Module 1. We will give a derivation of (4.13), because understanding this is important. But first let’s see an example. Example (4.14)

Find f(x) if you are given

λ (x) = Solution:

Λ (x) =

2x for 0≤ x ≤2. 4 - x2 x

ò 0

æ 4 ö÷ x 2u du = - ln(4 - u 2 ) = - ln(4 - x 2 ) + ln(4) = ln çç 2 0 çè 4 - x 2 ÷÷ø 4-u

Then for 0≤ x ≤2

S (x) = e

- Λ( x )

=

4 - x2 x2 = 14 4

x2 4 x f (x ) = 2 F (x ) =



Page M4-11


There is an intuitive interpretation of this. Λ(x) continuously adds up all the forces of mortality that have been applied on [0, x]. You can think of Λ(x) as the total amount of force of mortality that has been applied up to age x. -Λ x Then e ( ) reduces the survivorship by the total amount of accumulated mortality. To derive (4.13) we first note that since S ¢( x) = -F ¢(x ) = -f ( x ) ,

μ (x) = λ ( x ) =

-S ¢(x ) f (x) d ln(S(x )) = =S (x) S(x ) dx

It follows that

Λ (x) =

x

ò λ (u) du = - ln(S(u))

x 0

= - ln S ( x ) + ln S(0) = - ln S( x )

0

In the last step we used the fact that S(0) = 1. It follows immediately that (4.15)

e

- Λ( x )

= e ln S( x ) = S( x )

Exercise 3:

Find S(x) if λ(x ) =

3 , 10 - x

0 £ x < 10 .

Answer:

(10- x )3 1000

The Complete Expectation of Life at Birth We have already seen how to find moments of random variables. If X is an age at death random variable, the actuarial terminology for the mean (first moment) is the complete expectation of life at birth. The actuarial notation for the mean is (4.16)



e 0 = E( X ) =

ò

¥

0

xf (x ) dx

This is just a new name for a familiar concept. We still use the same defining calculation for the mean, as indicated by the last integral in (4.16). For age



Page M4-12


at failure random variables, there is another way to calculate E(X), as shown in the next theorem: If X is a continuous random variable defined on [0,∞) and E(X) is defined, then (4.17)



e 0 = E( X ) =

ò

¥

0

S( x) dx

The result in (4.17) can be proven using integration by parts. The proof can be found in Probability for Risk Management, page 262. This result will help us to understand alternate versions of some of the later formulas for expected lifetimes. In the next example we illustrate (4.17) for a uniform distribution. Example (4.18)

Let age at failure X be uniform on [0,100]. We already know that E(X) = 50. Check this using(4.17). Solution: We have already shown that S( x) = ¥

100

ò S(x)dx = ò 0

0


100 - x x = 1. Thus 100 100

2 ö æ æ ö çç1 - x ÷÷dx = çç x - x ÷÷ èç 100 ø÷ 200 ÷ø÷ èç

100

= 50 0


Page M4-13


Section 4.3 – Important Survival Models In this section, we will look at five age-at-death random variables which are prominent in exam study material. The first two, the uniform and the exponential, are familiar to us. They are prime candidates for Exam MLC/3L questions because they are simple and can be more completely understood, and because calculations for them can be done more rapidly in an exam setting. The last three, the Gompertz, the Makeham and the Weibull are new to us. They have hazard rates that are more appropriate for human populations and have been used as models for human survival.

The Uniform Distribution

We have already looked at the uniform age at death random variable X on [0,100] . The general uniform age-at-failure random variable X is defined on

[0, ω ]

Recall from Module 2 that the uniform distribution on the interval [0, ω ]

is often referred to as a DeMoivre distribution in actuarial literature. The general results for it follow exactly as in the example for [0,100] . For 0 £ x £ω : (4.19)

f (x) =

(4.20)

F (x) =

(4.21)

(4.22)

1

ω x

ω

S( x ) =

ω-x ω

λ (x) =

1 ω-x

The mean and variance were seen in Module 2: (4.23)



E( X ) = e 0 =

ω 2

V (X ) =

ω2 12

These formulas are important for Exam MLC/3L, and you should know them!

The following example is adapted from an old SOA Exam 3 problem which we’ll see in Module 5 after some new notation is introduced. ©ACTEX 2010 Luckner, Hassett, Stewart, & Steeby


Page M4-14


Example (4.24)

You are given 

(i)

e 0 = 25

(ii)

X is uniform on [0, ω ] .

Find V(X). Solution: 

Since e 0 =

ω 2

= 25 , ω = 50. Then V (X ) =

ω2 12

=

2500 1 = 208 12 3

Exercise 4: 

If X is uniform on [0, ω ] and e 0 = 45 , Find V(X).

Answer:

675

The uniform survival model is useful, but not appropriate for human survival analysis. For the uniform on [0,100] ten percent of the deaths occur in [0,10] and ten percent of the deaths occur in [90,100]. This is not the way human populations work.

The Exponential Distribution

We have already looked at the exponential age at death random variable X with rate parameter λ = 0.05. For the general rate parameter λ we have for x ≥ 0, (4.25)

f ( x ) = λe - λ x

(4.26)

F ( x ) = 1 - e -λ x

(4.27)

S( x ) = e -λ x



Page M4-15


λ(x ) = λ

(4.28)

The mean and variance for this form were given in Module 2: (4.29)

E( X ) =

1

V (X ) =

λ

1

λ2

These formulas are important for Exam MLC/3L, and you should know them!

The exponential has the memoryless property (discussed in Module 3). The mathematical expression of this is (4.30)

P ( X > s + t | X > s) = P( X > t )

In words, this says that once you have survived s years, you have the same probability of surviving t more years as a newborn. Some electrical components behave in this fashion, but humans don’t. You do not need to know the derivation for Exam MLC/3L. We include it for completeness. For an exponential with parameter λ, P ( X > s + t | X > s) = =

P( X > s + t & X > s) P( X > s) P( X > s + t ) P( X > s) - λ s +t )

e ( e -λs = e -λt =

= P( X > t )

The Gompertz and Makeham Distributions The Gompertz and Makeham distributions were suggested as models of human survival in the nineteenth century. Both distributions are defined by their hazard rates, hence we can use the following equation to fully describe the distributions.



Page M4-16


(4.31)

S (x) = e

-Λ( x )

The Gompertz distribution is defined by

λ ( x ) = Bc x ,

(4.32)

where x > 0, B > 0 and c > 1 This failure rate clearly increases with age, as desired for human survival. One can find S(x) using S (x) = e

- Λ( x )

x ù é B æ çç1-c ö÷÷ú ê ç ê ø÷úû

= e ëln c è

.

The Makeham distribution generalizes the Gompertz. It is defined by

λ ( x ) = A + Bc x ,

(4.33)

where x > 0, B > 0, c > 1, and A > -B This failure rate also increases with age. One can find S(x) using S (x) = e

- Λ( x )

=e

x é B æ ù ç1-c ö÷÷- Ax ú ê ÷ø êë ln c ççè ûú

.

Note that the illustrative life table in AM was constructed using a Makeham law for age 13 and greater. The law was given on page 78 of AM as x

1,000μ ( x ) = 0.7 + 0.05 (10.04 ) .

The Weibull Distribution

The Weibull distribution can be constructed from an increasing failure rate of the simple form

λ ( x ) = kx n ,

(4.34)

x ≥ 0, k > 0, n > -1 This leads to a survival function of the form S( X ) = e


-

kx n +1 n+1


Page M4-17


Section 4.4 - Random Variable for Time Until Death for a Life-Age-x In the preceding sections we have analyzed the random variable X for a newborn, where time starts at X=0. We will now study the random variable T for the remaining time until death for a person who has already reached age x. Recall that a person aged x is denoted (x). T is also written as T(x) to include information about the age x. Note: Some authors may write Tx instead of T (x) , but this causes some confusion, because there is another actuarial notation meaning for Tx .We will discuss the other meaning later.

Caution!

Finding the Survival Function for T(x) The survival function for T(x) is denoted by t px : (4.35)

t

px = ST (t ) = P( X > x + t | X > x)

Above we used a subscript to write ST (t ) because we will also be looking at S X (x ) in the next few pages. Usually the context is clear, and we will just use S(t ) . We can find S(t ) using information about the original survival function for X. Since we are given that survival to age x has occurred, S(t ) is a conditional probability: t

px = ST (t ) = P ( X > x + t | X > x ) P( X > x + t & X > x) P( X > x) P( X > x + t ) = P( X > x )

=

=


SX (x + t ) SX (x)


Page M4-18


It is crucial to remember that (4.36) t

px =

SX (x + t ) SX (x)

=

p0 p x 0

x +t

You should also be aware of the following property of the uniform distribution. Let X be uniform on [0, ω]. Then, the survival function for T(x) is given by

t

px =

SX (x + t ) SX (x)

ω - (x + t ) ω for 0 ≤ t ≤ ω - x = ω-x ω ω x) - t ( = ω-x This shows that T(x) is uniform on [0, ω - x ] . For example, if ω = 100 and

x=40, the remaining time until death is uniform on [0, 60]. We already noted this property in Module 2. It is very useful to know for Exam MLC/3L. Note that we have already found the distribution of T(x) for the exponential distribution. Due to the memoryless property, if X is exponential with parameter λ then for any value of x, the remaining lifetime T(x) has the same distribution.

The Cumulative Distribution and Density for T(x)

Once we have found the survival function S(t ) for T(x), it is a simple matter to find the cumulative distribution function and the density function. However, we must introduce more actuarial notation to go with this process. The actuarial notation for F (t ) is t q x . (4.37) t

q x = P (T ( x ) £ t ) = FT (t ) = 1 -

SX (x + t ) SX (x)

To find the density, differentiate the above expression with respect to t, and remember that x is a constant. (4.38)

fT (t ) =

d FT (t ) = dt


-

d SX (x + t ) f X (x + t ) dt = SX (x) SX (x)


Page M4-19


(4.37) and (4.38) are easy to remember. To get the survival function and density function for T(x), evaluate the original survival and density for X at x+t and divide by S(x). There is an unfortunate amount of competing notation here. It is most important to remember the principle of dividing by S(x) to get results for T(x).

Formula (4.38) can be restated in actuarial notation. Recall that we can write f (x ) = S(x )λ ( x ) Then we can rewrite (4.38) as (4.39)

fT (t ) =

f X (x + t ) SX (x)

=

S X ( x + t ) λ(x + t ) SX (x)

= ST (t ) λ( x + t )

This is written in actuarial notation as

fT (t ) = t px μ x +t

(4.40)

Equations (4.39) and (4.40) are also important to remember and fairly intuitive: to die at t years after age x you must first survive to age x, live t more years, then be hit by the force of mortality at age x + t. Remain calm in the face of this large array of theory. Much of what you do in exam problems will deal with the uniform and the exponential, and we have already seen that T(x) is pretty simple for them.

The Notations px and qx

This is important enough to deserve a separate section. When the value of t is 1, we drop the leading 1 subscript: (4.41)

1

px = px and 1 qx = q x

The Mean and Variance of T(x)

Once we know distribution of T(x), we can find the mean and variance as we would with any random variable. There is more actuarial notation to 

introduce. The expected value of T(x) is denoted by e x .



Page M4-20


(4.42)



e x = E éëT ( x )ùû

Example (4.43)

Let X be uniform on [0, 100]. 

Find e 40 = E éëT (40)ùû and V éëT (40)ùû .

Solution: T(40) is uniform on [0, 60].  0 + 60 e 40 = = 30 . 2 (60 - 0)2 V éëT (40)ùû = 12 Exercise 5:

If X is uniform on [0,80] , 

Find e 50 and V (T50 ) . Answer:

15, 75

Example (4.44) 

Let X be exponential with rate parameter λ = 0.05. Find e 40 .

Solution: T(40) is also exponential with rate parameter λ = 0.05. Then  1 e 40 = E éëT ( x )ùû = E ( X ) = = 20 . λ Exercise 6:

If X is exponential with rate parameter λ = 0.02 , 

Find e 20 . Answer:

50

In (4.17), we showed that for an age-at-birth random variable 

e 0 = E( X ) =


ò

¥

0

S( x) dx .


Page M4-21




A similar result holds for e x . (4.45)



e x = E éëT ( x )ùû =

ò

¥

0

ST (t ) dt =

ò

¥

0

S X (x + t ) dt = S X (x)

ò

¥

0

t

px dt

The next example reviews the computations for T(x) for a simple density function. Example (4.46)

Let X be an age at death random variable with density x , 0 ≤ x ≤ 10. 50

f X (x) =

Consider the random variable T(2) for the remaining lifetime past 

X=2. Find t p2 , fT(t) [which is the density function of T(2)], and e 2 . Solution: We first need information about X. For 0 ≤ x ≤ 10: x

FX ( x ) =

ò 0

x

u u2 x2 , du = = 50 100 0 100

SX (x) = 1 -

x2 100 - x 2 = 100 100

x f (x ) 2x 50 λX ( x ) = μ x = = = 2 S( x) 100 - x 100 - x 2 100 Now we can find t p2 : 2

t

p2 =

S X (2 + t ) S X (2)

fT (t ) = t p2μ t +2



e2 =

ò

8

0

1=

(2 + t ) 100 0.96

2

=

100 - (2 + t ) 96

for 0 ≤ t ≤ 8

2 100 - (2 + t ) æç 2 (2 + t ) ö÷÷ 2 + t ÷= = ççç 2÷ 96 48 èç100 - (2 + t ) ø÷÷

æ 2 + t ö÷ 44 t çç dt = çè 48 ÷÷ø 9



Page M4-22


Exercise 7:

Let age at death random variable X have density function f (x ) =

10 - x , 0 £ x < 10 . 50

If T (3) is the random variable for remaining lifetime after age 3, 

Find t p3 , fT (t ) , and e 3 . 2

Answer:


t

p3 =

(7 - t ) 49

,

fT (t ) =

(

2 7-t 49

)



, and e

3

=

7 3


Page M4-23


Section 4.5 – The Central Rate of Failure The central rate of failure or central rate of death is just a weighted average of the failure rate over an interval. First, we need a review of weighted averages. In the familiar discrete case, we have a set of weights w i where the weights add up to one; symbolically, that is: å w i = 1 . The weighted average of numbers x1 ,  , x n is:

w1 x1 + w2 x2 +  + w n x n =

åw x i

i

.

In the continuous case, we have a weighting function w(y ) on the interval

[ a, b] ,

where the weights add up to one; symbolically, that is

ò

The continuous weighted average of λ(y ) on the interval

b a

w(y ) dy = 1 .

[ a, b]

is the

continuous sum

ò

b a

w(y )λ(y ) dy

The central rate of death on [ x, x + 1] is a weighted average with w(y ) =

S X (y )

ò

x +1 x

S X (y ) dy

We are weighting at y according to the probability of survival at y. The weight at y is the ratio of the survival function at y to the integral of the survival function over the interval. The central rate, mx is the weighted average mx =

ò

x +1 x

w(y )λX (y ) dy .

This is equal to (4.47)

x +1

mx =

ò

S X (y ) λX (y ) dy

x

x +1

ò

S X (y ) dy

x



Page M4-24


Example (4.48)

Let X be uniform on [0, 10]. Find m1 . Solution: We first need the survival and hazard rate functions. (You should know these for the uniform.)

For 0 ≤ x ≤ 10, 10 - x 10 1 λ (x) = 10 - x 1 S (x) λ (x) = 10 S (x) =

Note that the function to be integrated in the numerator of (4.47) is just the density function. In general, S ( x ) λ ( x ) = f (x ) . Now we apply (4.47): 2

1

ò 10 dy

m1 =

1

2

ò 1

æ ö çç1 - y ÷÷ dy èç 10 ÷ø

æ 1 ö÷ çç ÷ çè10 ÷ø = æ17 ÷ö çç ÷ çè 20 ÷ø =

2 » 0.1176 17

To check for plausibility, look at the values of the failure rate at the endpoints of the interval [1, 2].

λ (1) =

1 1 » 0.1111 and λ (2) = = 0.1250 9 8

The simple average of these two numbers is 0.1181, quite close to the weighted average m1.



Page M4-25


Exercise 8:

Let X be uniform on [0, 10]. Find m2 . Answer:

0.1333

As we noted in the example above, the central rate can be rewritten as (4.49)

x +1

mx =

ò

f X (y ) dy

ò

S X (y ) dy

x x +1

x

This saves steps in computation if you know the density function f(x). This averaging process can be applied to interval of any length. For the interval [x, x + n], we define the central rate of death to be (4.50)

x +n

n

mx =

ò


x

x +n

ò

S X (y ) dy

x

As usual, there is an actuarial notation form for this. It can be obtained by dividing the numerator and denominator of (4.50) by the constant SX(X). (This is the division that converts the unconditional survival function into the conditional t px .) The result is (4.51)

n

n

mx =

ò

t

px μ x +t dt

0

n

ò

t

px dt

0

Note that when the remaining lifetime random variable is exponential (constant force) with rate λ, n mx º λ . This is easy to remember. When you average a constant value λ , the resulting average equals λ .



Page M4-26


Section 4.6 - Module 4 Formula Sheet q0 = P( X £ x ) = F (x )

x

λ (x) =

f (x) S (x)



e 0 = E( X ) =

x

Λ (x) =

p0 = P( X > x ) = 1 - F ( x) = S( X ) x

ò λ (u) du

S (x) = e

-Λ( x )

0

ò

¥

0



xf (x ) dx

e 0 = E( X ) =

ò

¥

0

S( x) dx

The Uniform Distribution on [0,ω]

f (x) =

1

F (x) =

ω 

E( X ) = e 0 =

ω 2

x

S( x) =

ω

V (X ) =

ω-x ω

λ (x) =

1 ω-x

ω2 12

The Exponential Distribution with rate parameter λ

f ( x ) = λe - λ x F ( x ) = 1 - e -λ x 1 1 E( X ) = V (X ) = 2

λ

λ(x) = λ

S( x) = e -λ x

λ

Memoryless Property

P ( X > s + t | X > s) = P ( X > t )

The remaining lifetime for (x) random variable T(x)=T

t

px = ST (t ) = P ( X > x + t | X > x ) =

t

q x = P (T (x ) £ t ) = FT (t ) = 1 -

e x = E éëT ( x )ùû

SX (x)

SX (x + t ) SX (x) x +1



S X ( x + t ))

mx =

ò

fT (t ) = t px μ x +t


x

x +1

ò

S X (y ) dy

x



Page M4-27


Section 4.7 - Module 4 Practice Problems Problem Set A 1.

4 , 50 - x

For a new life, λ ( x ) =

0 £ x < 50 .



Find e 0 .

2.

If λ ( x ) =

2 , 100 - x

0 £ x < 100 , find P (20 £ X £ 50) .

3. If X is DeMoivre on [0, ω ) and V ( X ) = 468.75 , what is ω ? 4. 

X is exponential and e 30 = 40 . Find μ .

5. 2

If f ( x ) =

(30 - x ) 9,000

,

0 £ x < 30 , find t p5 .

6.

If f ( x ) =

20 - x , 200

0 £ x < 20 , find μ (10) .



Page M4-28


Problem Set B 1.

The lifetime of a machine component is modeled by α

æ xö S ( x ) = çç1 - ÷÷÷ , çè ωø

0 £ x <ω

and

α > 0.

If μ (40) = 2μ (20) , find ω . 2. 2 1 are nonsmokers and are smokers. The 3 3 remaining lifetime of a nonsmoker is uniform over [0, 80) and for a smoker it is uniform over [0,50) . For a person selected at random from those who survive to age 40, what is the probability that he will survive 5 more years?

For a population of 20 year olds,

3. 2

æ x ÷ö Given S ( x ) = çç1 ÷ , çè 100 ÷ø

0 £ x < 100 , find the probability that a person

aged 20 will die between the ages of 50 and 60. 4.

If f ( x ) =

20 - x , 200



0 £ x < 20 , find e 5 .

5.

If μ ( x ) =

2x , 400 - x 2

0 £ x < 20 , find V ( X ) .

6.

The lifetime, X, of a machine is modeled as follows: For 0 £ x < 30 , it follows the DeMoivre model on [0, 100) . For x ³ 30 , it follows the constant force model with μ = 0.02 . Find E ( X ) .



Page M4-29


Section 4.8 Module 4 Practice Problem Solutions Problem Set A 1.

First, find the cumulative hazard function Λ ( x ) :

Λ (x) =

ò

x

0

λ (u) du =

ò

x

4 = -4ln (50 - u) 50 - u 0

x

0

4

æ x ö÷ = -4 ëéln (50 - x ) - ln (50)ùû = - ln çç1 ÷ çè 50 ø÷ S (x) = e

-Λ( x )

4 æ xö lnçç1- ÷÷÷ çè 50 ÷ø

=e 4

æ x ÷ö = çç1 ÷ çè 50 ÷ø 

e0 =

ò

50

0

S ( x ) dx =

ò

50

0

4

5

æ ö æ ö çç1 - x ÷÷ dx = - 50 çç1 - x ÷÷ ÷ ÷ çè ç 50 ø 5 è 50 ø

50

0

= 10 2.

The cumulative hazard function is

Λ (x) =

ò

S (x) = e

x

0

2

x æ 2 x ö÷ du = -2ln (100 - u) = - ln çç1 ÷ çè 100 - u 100 ÷ø 0

- Λ( x )

2

æ x ÷ö = çç1 ÷ çè 100 ÷ø

2

2

æ æ 20 ö÷ 50 ö÷ - çç1 P [20 £ X £ 50] = S (20) - S (50) = çç1 ÷ ÷ ÷ èç èç 100 ø 100 ø÷ 2

2

= (0.8) - (0.5) = 0.39 3.

For X uniform on [0, ω ) , V ( X ) =

ω2 12

ω2 12

.

= 468.75 , ω 2 = 5,625 .

ω = 75 .



Page M4-30


4. 



For an exponential distribution, e n = e 0 due to the memoryless property of the exponential. 

e 0 = E (X ) = 1

= 40

μ



1

for the exponential distribution.

μ

μ = 0.025 .

5.

F (x) =

ò

x

f (t ) dt =

0

ò

2

(30 - t )

x

9,000

0

3

dt = -

(30 - t )

27,000

x

0

3

= 1-

(30 - x )

0 £ x < 30

,

27,000 3

S (x) =

t

p5 =

(30 - x )

27,000 3

S (t + 5)

(30 - t - 5) = 3 S (5) (30 - 5)

3

æ t ö÷ = çç1 ÷ çè 25 ÷ø

6.

F (x) =

ò

x

0

f (t ) dt =

ò

x

0

2

(20 - t ) 20 - t dt = 200 400

x

0

2

= 1-

(20 - x ) 400 2

S (x) =

μ (x) =

μ (10) =

(20 - x ) 400 f (x) S (x)

=

æ ö çç 20 - x ÷÷ çè 200 ø÷ é ê (20 - x ) ê êë 400

2

ù ú ú úû

=

2 20 - x

1 5



Page M4-31


Problem Set B 1.

μ (x) =

-S ¢ ( x ) S (x)

α

æ xö S ( x ) = çç1 - ÷÷ , çè ω ÷ø

,

α -1

æ 1 öæ xö S ¢ ( x ) = α çç- ÷÷ çç1 - ÷÷ çè ω ÷øèç ω ÷ø

α -1

μ (x) =

α æç xö ç1 - ÷÷÷ ç ωè ωø

=

α

æ ö çç1 - x ÷÷ çè ω ÷ø

μ (40) = 2μ (20)



ω - 20 = 2ω - 80

α

ω-x α

=

ω - 40  ω = 60

2α ω - 20

2.

20 3 = . Their remaining lifetime at age 40 is 80 4 20 3 s = 1= . Their remaining lifetime linear over [0, 60) . For smokers, 20 p20 50 5 at age 40 is linear over [0, 30) . For nonsmokers,

20

n p20 = 1-

From a population of l20 at age 20, there are smokers.

At

age

40,

there

are

2 1 l20 nonsmokers and l20 3 3

3 æç 2 ö÷ ç ÷ l20 = 0.5 l 20 4 çè 3 ø÷

nonsmokers

and

5 2 3 æç 1 ö÷ are nonsmokers and çç ÷÷ l 20 = 0.2 l20 smokers. Among 40 year olds, 7 7 5 è3ø are smokers.

5 11 = . 60 12 5 5 = 1= 30 6

n For nonsmokers, 5 p40 = 1s For smokers, 5 p40

For the total population, 5

æ 5 öæ11 ö æ 2 öæ 5 ö p40 = çç ÷÷÷ çç ÷÷÷ + çç ÷÷÷ çç ÷÷÷ èç 7 øèç12 ø èç 7 øèç 6 ø 75 84 25 = 28 =



Page M4-32


3.

The probability that a 20 year old will die between the ages of 50 and 60 is 30 p20 - 40 p20 . 2

p20 =

t

S (t + 20) S (20)

=

æ ö çç1 - t + 20 ÷÷ èç 100 ø÷ 2

æ ö çç1 - 20 ÷÷ çè 100 ø÷

2

30

p20

æ ö çç1 - 50 ÷÷ ÷ çè 0.25 25 100 ø = = = 2 0.64 64 æ ö çç1 - 20 ÷÷ ÷ çè 100 ø 2

40

p20

30

p20

æ ö çç1 - 60 ÷÷ ÷ çè 16 100 ø = = 0.64 64 9 - 40 p20 = 64

4.

F (x) =

ò

x

0

2

(20 - t ) 20 - t dt = 200 400

x

2

= 1-

(20 - x )

0

400

2

(20 - x )

S (x) =

400 2

t

p5 =

S (t + 5) S (5)

(20 - t - 5) =

2

æ t ÷ö = çç1 ÷ , çè 15 ÷ø

400 2 (20 - 5)

0 £ t < 15

400 

e5 =

ò

15

0

t

p5 dt =

ò

15

0

2

3

æ 15 çæ t ö÷ t ö÷ ÷ dt = ÷ ççç1 çç1 ÷ è 15 ø 3 è 15 ÷ø

15

0

=5



Page M4-33


5.

2x , 400 - x 2 x 2u du Λ (x) = ò 0 400 - u 2

μ (x) =

æ 400 - x 2 ö÷ x = - ln (400 - u 2 ) = - ln ççç ÷÷ 0 è 400 ø÷ S (x) = e

- Λ( x )

=e

2 çæ 400- x ÷ö÷ lnçç ÷ çè 400 ÷÷ø

f ( x ) = -S ¢ ( x ) =

400 - x 2 400

=

x 200

x2 203 40 dx = = ò0 ò0 200 600 3 4 20 20 x 3 20 E ( X 2 ) = ò x 2 f ( x ) dx = ò dx = = 200 0 0 200 800 2 æ 40 ö÷ ç V ( X ) = 200 - ç ÷÷ = 22.22 çè 3 ø 20

E (X ) =

x f ( x ) dx =

20

6.

x . S (30) = 0.7 . 100 If X is constant force with μ = 0.02 , S ( x ) = e -0.02t . For X DeMoivre on [0, 100) , S(x ) = 1 -

For the lifetime of the machine ì x ï ï , 0 £ x < 30 ï1 100 S (x) = í ï -0.02( x -30) ï ï , 30 £ x ï î0.7e

E (X ) =

ò

¥

0

S ( x ) dx =

ò

30

0

¥ æ x ö÷ -0.02( x -30) dx ÷÷ dx + 0.7ò e ççç1 30 è 100 ø

(In the second integral let u = x - 30) æ x 2 ö÷ ÷ E ( X ) = ççç x 200 ÷ø÷ è

30

+ 0.7ò 0

¥

0

e -0.02u du

= 25.5 + 0.7 (50) = 60.5



Page M4-34


Section 4.9 - Module 4 Exam Problems 1. Fall 2000 #36 SOA

Given: (i) (ii)

μ (x) = F + e 2 x , 0.4

x ³0

p0 = 0.50

Calculate F. (A)

-0.20

(B)

-0.09

(C)

0.00

(D)

0.09

(E)

0.20

2. Fall 2004 #7 CAS

Which of the following formulas could serve as a force of mortality? 1. 2. 3.

μ x = BC x , μ x = a(b + x )-1 , μ x = (1 + x )-3 ,

B > 0, C > 1

a > 0, b > 0 x ³ 0,

(A) 1 only (B) 2 only (C) 3 only (D) 1 and 2 only (E) 1 and 3 only



Page M4-35


3. Fall 2002 #1 SOA

Given: The survival function s(x ) , where s(x ) = 1,

0 £ x £1 x ü ì ï ï(e )ï ï ï s(x ) = 1 - ï í ý , 1 £ x < 4.5 ï 100 ï ï ï ï ï î þ s(x ) = 0, 4.5 £ x Calculate μ (4) . (A) 0.45 (B) 0.55 (C) 0.80 (D) 1.00 (E) 1.20 4. Fall 2003 #19 CAS

For a loss distribution where x ³ 2 , you are given: i) ii)

z2 , for x ³ 2 2x A value of the distribution function F(5) = 0.84 The hazard rate function: h(x ) =

Calculate z. (A) 2 (B) 3 (C) 4 (D) 5 (E) 6



Page M4-36


5. Fall 2004 #8 CAS 1

é æ x ö÷ù 2 ú , for 0 £ x £ 100 , calculate the probability that a life Given s( x) = ê1 - çç çè100 ÷÷øú ê ë û age 36 will die between ages 51 and 64. (A)

Less than 0.15

(B)

At least 0.15, but less than 0.20

(C)


(D)


(E)

At least 0.30

6. Fall 2003 #5 CAS

i)

Mortality follows De Moivre’s Law.

ii)

e 20 = 30



Calculate q20 . (A)

1 60

(B)

1 70

(C)

1 80

(D)

1 90

(E)

1 100



Page M4-37


7. Fall 2005 #13 Exam M The actuarial department for the SharpPoint Corporation models the lifetime of pencil sharpeners from purchase using a generalized DeMoivre model with

s ( x ) = (1 − x / ω ) , for α > 0 and 0 ≤ x ≤ ω . α

A senior actuary examining mortality tables for pencil sharpeners has determined that the original value of α must change. You are given: (i)

The new complete expectation of life at purchase is half what it was previously.

(ii)

The new force of mortality for pencil sharpeners is 2.25 times the previous force of mortality for all durations.

(iii)

ω remains the same.

Calculate the original value of α . (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 8. Fall 2005 #32 Exam M For a group of lives aged 30, containing an equal number of smokers and non-smokers, you are given:

(i)

For non-smokers, μ n ( x ) = 0.08,

(ii)

For smokers, μ s ( x ) = 0.16,

x ≥ 30

x ≥ 30

Calculate q80 for a life randomly selected from those surviving to age 80. (A) 0.078 (B) 0.086 (C) 0.095 (D) 0.104 (E) 0.112 ©ACTEX 2010 Luckner, Hassett, Stewart, & Steeby


Page M4-38


Section 4.10 - Module 4 Exam Problem Solutions 1.

First note that 0.4

p0 = 0.5 = e

-

0.4

ò0

μ ( x ) dx

=e

-

0.4

2x ò0 (F +e ) dx

The integral in the exponent is 0.4 0.4 æ ö 1 -ò (F + e 2 x ) dx = - ççFx + e 2 x ÷÷÷ 0 èç ø0 2 = - (0.4) F - 1.11277 + 0.50 = - (0.4) F - 0.61277 Thus 0.50 = e -0.4F -0.61277 . ln(0.50) = -0.69315 = -(0.4)F - 0.61277 0.08038 = 0.20 0.4

F = Answer E 2.

In order that μ X be a force of mortality it must be ³ 0 (which all three are) and must satisfy ò

¥

0

μ X dx = ¥ . The reason for the second requirement is

that for a survival function lim S( x) = 0 . Thus we need: S(¥) = 0 = e x ¥

ò

¥

0

ò

¥

0

æ B ö÷ x BC dx = çç c çèln c ÷÷ø

μ x dx

.

=¥

for B > 0, C > 1

0 ¥

a = a ln ( x + b) = ¥ b+x 0 1

ò (1 + x ) 0

¥

ò0

¥

x

¥

-

3

æ 1ö -2 dx = çç- ÷÷÷ (1 + x ) çè 2 ø

for a > 0, b > 1 ¥

= 0

1 2

Only the first two are forces of mortality. Note: The hazard function BC x indicates a Gompertz distribution, and if you knew that you could have answered without doing the integration. However the problem was not difficult if done directly. Answer D ©ACTEX 2010 Luckner, Hassett, Stewart, & Steeby


Page M4-39


3.

(This problem directly uses the definition of μ(x)) μ ( x ) =

-S ¢(x ) . S(x )

Since we want μ (4) , we use the definition of S(x ) for 1 £ x < 4.5 : ex , 100

S( x ) = 1 -

-S ¢(x ) =

ex . 100

e4 e4 μ (4) = 1004 = = 1.203 . æ e ö÷ 100 - e 4 ç ÷÷ 1 - çç è100 ø÷ Answer E 4.

The hazard rate is the force of mortality. Hence, μ (x ) = h( x) =

z2 , 2x

x ³2.

Now, we will find S( x) . Recall that z 2 is an unknown constant. S(5) = e =e

-

5

ò2

z2 dx 2x

5ù é æ 2ö ê ç z ÷÷ ú ê-çç ÷÷ ln x ú ê çè 2 ÷ø ú 2 ûú ëê

é æ z 2 ÷ö ù ê-çç ÷÷(ln5-ln2)ú ê çç 2 ÷÷ ú ëê è ø ûú

=e = 1 - F (5)

= 1 - 0.84 = 0.16

Thus

e

é æ z 2 ö÷ ù ê-çç ÷÷(ln5-ln2)ú ê çç 2 ÷÷ ú êë è ø úû

-

= 0.16

z 2 æç 5 ö÷ ln ç ÷ = ln (0.16) 2 çè 2 ø÷

é- ln (0.16)ù û =4 z2 = 2 ë ln (2.5) z =2 Answer A ©ACTEX 2010 Luckner, Hassett, Stewart, & Steeby


Page M4-40


5.

The probability of death between the ages of 51 and 64 for a person aged 36 is 15 p36 -28 p36 . 1

t

p36 =

S(36 + t ) = S(36)

é ù2 ê1 - (36 + t ) ú ê 100 úûú ëê 1

é 36 ù 2 ê1 ú êë 100 úû

1

15

p36 =

é 49 ù 2 ê ú êë100 úû é 64 ù ê ú êë100 úû

1 2

=

0.7 7 = 0.8 8

1

é 36 ù 2 ê ú êë100 úû 6 = = 0.8 8

28

p36

15

p36 -28 p36 =

1 = 0.125 8

Answer A

6.

For 20 year olds, future lifetime follows a DeMoivre model with maximum future lifetime of ω - 20 .  ω - 20 e 20 = = 30, ω - 20 = 60 2 Since deaths occur uniformly over [0,60), 1 q20 = . 60 Answer A



Page M4-41


7.

α

æ xö For original model X, S( x ) = çç1 - ÷÷÷ . çè ωø

E( X ) =

ω

ò

0

S(x ) dx =

ò

ω

0

α

æ xö ççç1 - ÷÷÷ dx è ωø

α +1 ω

ω

æ xö =ççç1 - ÷÷÷ α +1è ωø

= 0

ω

α +1

α -1

μ (x ) = -

S ¢(x ) = S( x)

xö α æç çç1 - ÷÷÷ ωè ωø

=

α

æ xö ççç1 - ÷÷÷ è ωø

æ çè

α

ω çç1 -

x ÷ö ÷ ω ÷ø

=

α

ω-x

α¢

æ xö For new model X ¢ , S( x) = çç1 - ÷÷÷ çè ωø E( X ¢) =

ω

α¢ +1 α¢ μ ¢( x) = ω-x

æ1 ö E( X ¢) = çç ÷÷÷ E( X ) and μ ¢( x ) = 2.25μ ( x ) . çè 2 ø

ω

α¢ +1

=

ω ; 2 (α + 1)

We get 2α + 2 = α ¢ + 1 2.25α = α '



α¢ 2.25α = ω-x ω-x

α =4

Answer D



Page M4-42


8.

Let T be the random variable for the future lifetime of (30). Recall that for constant force of mortality, S(t ) = e -μ t , independent of x. For non-smokers N, S N (t ) = e -0.08t = P (T > t | N ) For smokers, S S (t ) = e -0.16t = P (T > t | S ) We are asked to find q80 = 1 - p80 . Note that p80 =

S (51) S (50)

, where S(t) is the

survival function for T, the future lifetime of 30. In general S (t ) = P (T > t ) = P(T > t | S)P (S ) + P(T > t | N)P (N ) = .5e -.16t + .5e -.08t p80 =

q80

S (51) S (50)

=

.5e

-.16(51)

.5e ( = 1 - .922 = .078

-.16 50)

+ .5e

-.08(51)

+ .5e

-.08(50)

= .922

Answer A



ACTEX 2010 SOA Exam MLC /CAS Exam 3L

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