MLC Actuary Manual

7/26/2019 MLC Actuary Manual

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A Reading of the Theory of Life ContingencyModels:

A Preparation for Exam MLC/3L

Marcel B. FinanArkansas Tech University

c

All Rights Reserved

Preliminary Draft


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2

To My DaughterNadia


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Contents

Preface 11

Prerequisite Material 13

Brief Review of Interest Theory 151 The Basics of Interest Theory . . . . . . . . . . . . . . . . . . . . 152 Equations of Value and Time Diagrams. . . . . . . . . . . . . . . 263 Level Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3.1 Level Annuity-Immediate . . . . . . . . . . . . . . . . . . 283.2 Level Annuity-Due . . . . . . . . . . . . . . . . . . . . . . 303.3 Level Continuous Annuity . . . . . . . . . . . . . . . . . . 32

4 Varying Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . 344.1 Varying Annuity-Immediate . . . . . . . . . . . . . . . . . 344.2 Varying Annuity-Due . . . . . . . . . . . . . . . . . . . . . 384.3 Continuous Varying Annuities . . . . . . . . . . . . . . . . 424.4 Continuously Payable Varying Annuities . . . . . . . . . . 45

5 Annuity Values on Any Date: Deferred Annuity . . . . . . . . . . 46

A Brief Review of Probability Theory 516 Basic Definitions of Probability . . . . . . . . . . . . . . . . . . . 517 Classification of Random Variables . . . . . . . . . . . . . . . . . 578 Discrete Random Variables . . . . . . . . . . . . . . . . . . . . . 599 Continuous Random Variables . . . . . . . . . . . . . . . . . . . . 69

10 Raw and Central Moments . . . . . . . . . . . . . . . . . . . . . 7711 Median, Mode, Percentiles, and Quantiles . . . . . . . . . . . . . 8212 Mixed Distributions . . . . . . . . . . . . . . . . . . . . . . . . . 8513 A List of Commonly Encountered Discrete R.V. . . . . . . . . . 88

13.1 Discrete Uniform Distribution . . . . . . . . . . . . . . . 88

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13.2 The Binomial Distribution . . . . . . . . . . . . . . . . . 90

13.3 The Negative Binomial Distribution . . . . . . . . . . . . 9113.4 The Geometric Distribution. . . . . . . . . . . . . . . . . 9213.5 The Poisson Distribution . . . . . . . . . . . . . . . . . . 93

14 A List of Commonly Encountered Continuous R.V . . . . . . . . 9414.1 Continuous Uniform Distribution. . . . . . . . . . . . . . 9414.2 Normal and Standard Normal Distributions . . . . . . . . 9614.3 Exponential Distribution . . . . . . . . . . . . . . . . . . 10214.4 Gamma Distribution . . . . . . . . . . . . . . . . . . . . 104

15 Bivariate Random Variables . . . . . . . . . . . . . . . . . . . . 10615.1 Joint CDFs. . . . . . . . . . . . . . . . . . . . . . . . . . 10615.2 Bivariate Distributions: The Discrete Case . . . . . . . . 108

15.3 Bivariate Distributions: The Continuous Case . . . . . . 11015.4 Independent Random Variables. . . . . . . . . . . . . . . 11315.5 Conditional Distributions: The Discrete Case . . . . . . . 11615.6 Conditional Distributions: The Continuous Case . . . . . 11915.7 The Expected Value ofg(X, Y). . . . . . . . . . . . . . . 12315.8 Conditional Expectation . . . . . . . . . . . . . . . . . .127

16 Sums of Independent Random Variables. . . . . . . . . . . . . . 13216.1 Moments ofS . . . . . . . . . . . . . . . . . . . . . . . . 13216.2 Distributions Closed Under Convolution . . . . . . . . . . 13316.3 Distribution ofS: Convolutions . . . . . . . . . . . . . . 13516.4 Estimating the Distribution of S : The Central Limit

Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 13817 Compound Probability Distributions . . . . . . . . . . . . . . . 140

17.1 Mean and Variance ofS . . . . . . . . . . . . . . . . . . . 14017.2 Moment Generating Function ofS . . . . . . . . . . . . . 142

Actuarial Survival Models 14318 Age-At-Death Random Variable . . . . . . . . . . . . . . . . . . 144

18.1 The Cumulative Distribution Function ofX . . . . . . . . 14418.2 The Survival Distribution Function ofX . . . . . . . . .14718.3 The Probability Density Function ofX . . . . . . . . . . 150

18.4 Force of Mortality ofX . . . . . . . . . . . . . . . . . . . 15418.5 The Mean and Variance ofX . . . . . . . . . . . . . . . . 159

19 Selected Parametric Survival Models. . . . . . . . . . . . . . . . 16319.1 The Uniform or De Moivres Model . . . . . . . . . . . .16319.2 The Exponential Model . . . . . . . . . . . . . . . . . . . 167


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CONTENTS 5

19.3 The Gompertz Model . . . . . . . . . . . . . . . . . . . . 171

19.4 The Modified Gompertz Model: The Makehams Model . 17419.5 The Weibull Model . . . . . . . . . . . . . . . . . . . . . 17720 Time-Until-Death Random Variable . . . . . . . . . . . . . . . . 180

20.1 The Survival Function ofT(x) . . . . . . . . . . . . . . . 18020.2 The Cumulative Distribution Function ofT(x) . . . . . . 18520.3 Probability Density Function ofT(x) . . . . . . . . . . . 18920.4 Force of Mortality ofT(x) . . . . . . . . . . . . . . . . . 19320.5 Mean and Variance ofT(x) . . . . . . . . . . . . . . . . . 19820.6 Curtate-Future-Lifetime. . . . . . . . . . . . . . . . . . . 202

21 Central Death Rates . . . . . . . . . . . . . . . . . . . . . . . . 208

The Life Table Format 21322 The Basic Life Table . . . . . . . . . . . . . . . . . . . . . . . . 21423 Mortality Functions in Life Table Notation . . . . . . . . . . . . 220

23.1 Force of Mortality Function. . . . . . . . . . . . . . . . . 22023.2 The Probability Density Function ofX . . . . . . . . . . 22423.3 Mean and Variance ofX . . . . . . . . . . . . . . . . . . 22623.4 Conditional Probabilities . . . . . . . . . . . . . . . . . . 23023.5 Mean and Variance ofT(x) . . . . . . . . . . . . . . . . . 23323.6 Temporary Complete Life Expectancy . . . . . . . . . . 23623.7 The Curtate Expectation of Life . . . . . . . . . . . . . . 24023.8 The nLx Notation . . . . . . . . . . . . . . . . . . . . . . 24423.9 Central Death Rate . . . . . . . . . . . . . . . . . . . . . 247

24 Fractional Age Assumptions . . . . . . . . . . . . . . . . . . . . 25124.1 Linear Interpolation: Uniform Distribution of Deaths

(UDD) . . . . . . . . . . . . . . . . . . . . . . . . . . . 25124.2 Constant Force of Mortality Assumption: Exponential

Interpolation . . . . . . . . . . . . . . . . . . . . . . . 25924.3 Harmonic (Balducci) Assumption . . . . . . . . . . . . . 264

25 Select-and-Ultimate Mortality Tables . . . . . . . . . . . . . . . 270

Life Insurance: Contingent Payment Models 277

26 Insurances Payable at the Moment of Death . . . . . . . . . . . 27926.1 Level Benefit Whole Life Insurance . . . . . . . . . . . . 27926.2 Finite Term Insurance Payable at the Moment of Death . 28726.3 Endowments . . . . . . . . . . . . . . . . . . . . . . . . . 291

26.3.1 Pure Endowments . . . . . . . . . . . . . . . . . 291


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26.3.2 Endowment Insurance . . . . . . . . . . . . . . . 295

26.4 Deferred Life Insurance . . . . . . . . . . . . . . . . . . . 29827 Insurances Payable at the End of the Year of Death . . . . . . . 30328 Recursion Relations for Life Insurance . . . . . . . . . . . . . . . 31229 Variable Insurance Benefit . . . . . . . . . . . . . . . . . . . . . 318

29.1 Non-level Payments: A Simple Example . . . . . . . . . . 31829.2 Increasing or Decreasing Insurances Payable at the Mo-

ment of Death. . . . . . . . . . . . . . . . . . . . . . . 32229.3 Increasing and Decreasing Insurances Payable at the End

of Year of Death . . . . . . . . . . . . . . . . . . . . . 32730 Expressing APVs of Continuous Models in Terms of Discrete

Ones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331

31mthly Contingent Payments . . . . . . . . . . . . . . . . . . . . . 33732 Applications of Life Insurance . . . . . . . . . . . . . . . . . . . 341

Contingent Annuity Models 34733 Continuous Whole Life Annuities . . . . . . . . . . . . . . . . .34834 Continuous Temporary Life Annuities . . . . . . . . . . . . . . . 35635 Continuous Deferred Life Annuities . . . . . . . . . . . . . . . . 36136 The Certain-and-Life Annuity . . . . . . . . . . . . . . . . . . . 36437 Discrete Life Annuities . . . . . . . . . . . . . . . . . . . . . . . 366

37.1 Whole Life Annuity Due . . . . . . . . . . . . . . . . . .36637.2 Temporary Life Annuity-Due . . . . . . . . . . . . . . . . 37437.3 Discrete Deferred Life Annuity-Due . . . . . . . . . . . .38037.4 Discrete Certain and Life Annuity-Due . . . . . . . . . . 38337.5 Life Annuity-Immediate . . . . . . . . . . . . . . . . . . . 386

38 Life Annuities with mthly Payments . . . . . . . . . . . . . . . . 39239 Non-Level Payments Annuities . . . . . . . . . . . . . . . . . . . 401

39.1 The Discrete Case . . . . . . . . . . . . . . . . . . . . . . 40139.2 The Continuous Case . . . . . . . . . . . . . . . . . . . . 405

Calculating Benefit Premiums 40940 Fully Continuous Premiums . . . . . . . . . . . . . . . . . . . . 410

40.1 Continuous Whole Life Policies . . . . . . . . . . . . . . . 41040.2 nyear Term Policies . . . . . . . . . . . . . . . . . . . . 41740.3 Continuous nyear Endowment Insurance . . . . . . . . 42140.4 Continuous nyear Pure Endowment . . . . . . . . . . .42540.5 Continuous nyear Deferred Insurance . . . . . . . . . . 428


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40.6 Continuousnyear Deferred Whole Life Annuity . . . . . 430

41 Fully Discrete Benefit Premiums . . . . . . . . . . . . . . . . . . 43441.1 Fully Discrete Whole Life Insurance . . . . . . . . . . . . 43441.2 Fully Discrete nyear Term . . . . . . . . . . . . . . . . 43941.3 Fully Discrete nyear Pure Endowment . . . . . . . . . . 44341.4 Fully Discrete nyear Endowment Insurance . . . . . . . 44641.5 Fully Discrete nyear Deferred Insurance . . . . . . . . . 45041.6 Fully Discrete nyear Deferred Annuity-Due . . . . . . . 453

42 Benefit Premiums for Semicontinuous Models. . . . . . . . . . . 45642.1 Semicontinuous Whole Life Insurance . . . . . . . . . . . 45642.2 Semicontinuous nyear Term Insurance . . . . . . . . . . 46042.3 Semicontinuous n

year Endowment Insurance . . . . . . 463

42.4 Semicontinuous nyear Deferred Insurance . . . . . . . . 46643 mthly Benefit Premiums . . . . . . . . . . . . . . . . . . . . . . . 469

43.1 mthly Payments with Benefit Paid at Moment of Death . 46943.2 mthly Payments with Benefit Paid at End of Year of Death474

44 Non-Level Benefit/Premium Payments and the Equivalence Prin-ciple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 478

45 Percentile Premium Principle. . . . . . . . . . . . . . . . . . . . 488

Benefit Reserves 49546 Fully Continuous Benefit Reserves . . . . . . . . . . . . . . . . . 497

46.1 Fully Continuous Whole Life . . . . . . . . . . . . . . . . 49746.1.1 Reserves by the Prospective Method . . . . . . . 49746.1.2 Other Special Formulas for the Prospective Re-

serve . . . . . . . . . . . . . . . . . . . . . . . 50346.1.3 Retrospective Reserve Formula . . . . . . . . . . 507

46.2 Fully Continuous nyear Term. . . . . . . . . . . . . . . 51046.3 Fully Continuous nyear Endowment Insurance . . . . . 51446.4 Fully Continuous nyear Pure Endowment . . . . . . . . 51846.5 nyear Deferred Whole Life Annuity . . . . . . . . . . . 520

47 Fully Discrete Benefit Reserves . . . . . . . . . . . . . . . . . . . 52347.1 Fully Discrete Whole Life Insurance . . . . . . . . . . . . 523

47.2 Fully Discrete nyear Term Insurance . . . . . . . . . . . 53047.3 Fully Discrete nyear Endowment . . . . . . . . . . . . . 53347.4 Fullynyear Deferred Whole Life Annuity . . . . . . . . 537

48 Semicontinuous Reserves . . . . . . . . . . . . . . . . . . . . . . 53949 Reserves Based on True mthly Premiums . . . . . . . . . . . . . 544


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Reserves for Contracts with Nonlevel Benefits and Premiums 549

50 Reserves for Fully Discrete General Insurances . . . . . . . . . . 55051 Reserves for Fully Continuous General Insurances . . . . . . . . 557

52 Recursive Formulas for Fully Discrete Benefit Reserves. . . . . . 560

53 Miscellaneous Examples. . . . . . . . . . . . . . . . . . . . . . . 569

54 Benefit Reserves at Fractional Durations . . . . . . . . . . . . . 573

55 Calculation of Variances of Loss Random Variables: The Hat-tendorfs Theorem . . . . . . . . . . . . . . . . . . . . . . . . 578

Multiple Life Models 585

56 The Joint-Life Status Model . . . . . . . . . . . . . . . . . . . . 586

56.1 The Joint Survival Function ofT(xy) . . . . . . . . . . .58656.2 The Joint CDF/PDF ofT(xy) . . . . . . . . . . . . . . . 589

56.3 The Force of Mortality ofT(xy) . . . . . . . . . . . . . . 592

56.4 Mean and Variance ofT(xy) . . . . . . . . . . . . . . . . 594

57 The Last-Survivor Status Model . . . . . . . . . . . . . . . . . . 597

58 Relationships BetweenT(xy) and T(xy). . . . . . . . . . . . . . 606

59 Contingent Probability Functions . . . . . . . . . . . . . . . . .609

60 Contingent Policies for Multiple Lives . . . . . . . . . . . . . . . 615

61 Special Two-life Annuities: Reversionary Annuities. . . . . . . . 624

62 Dependent Future Lifetimes Model: The Common Shock . . . . 629

63 Joint Distributions of Future Lifetimes . . . . . . . . . . . . . . 635

Multiple Decrement Models 639

64 The Continuous Case . . . . . . . . . . . . . . . . . . . . . . . . 640

65 Associated Single Decrement Models. . . . . . . . . . . . . . . . 647

66 Discrete Multiple-Decrement Models. . . . . . . . . . . . . . . . 652

67 Uniform Distribution of Decrements . . . . . . . . . . . . . . . . 661

68 Valuation of Multiple Decrement Benefits . . . . . . . . . . . . . 669

69 Valuation of Multiple Decrement Premiums and Reserves . . . . 677

Incorporating Expenses in Insurance Models 685

70 Expense-Augmented Premiums. . . . . . . . . . . . . . . . . . . 686

71 Types of Expenses. . . . . . . . . . . . . . . . . . . . . . . . . . 695

72 The Mathematics of Asset Share . . . . . . . . . . . . . . . . . . 702


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Multiple-State Transition Models 709

73 Introduction to Markov Chains Process . . . . . . . . . . . . . . 71074 Longer Term Transition Probabilities . . . . . . . . . . . . . . . 71575 Valuation of Cash Flows . . . . . . . . . . . . . . . . . . . . . . 721

75.1 Cash Flows Upon Transitions. . . . . . . . . . . . . . . . 72175.2 Cash Flows while in State. . . . . . . . . . . . . . . . . . 72475.3 Benefit Premiums and Reserves . . . . . . . . . . . . . . 726

Probability Models: Poisson Processes 73376 The Poisson Process. . . . . . . . . . . . . . . . . . . . . . . . . 73477 Interarrival and Waiting Time Distributions . . . . . . . . . . . 74178 Superposition and Decomposition of Poisson Process. . . . . . . 747

79 Non-Homogeneous Poisson Process . . . . . . . . . . . . . . . . 75880 Compound Poisson Process . . . . . . . . . . . . . . . . . . . . . 76481 Conditional Poisson Processes . . . . . . . . . . . . . . . . . . . 772

Answer Key 775

BIBLIOGRAPHY 875


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Preface

The objective of this book is to present the basic aspects of the theory ofinsurance, concentrating on the part of this theory related to life insurance.An understanding of the basic principles underlying this part of the subject

will form a solid foundation for further study of the theory in a more generalsetting.This is the fourth of a series of books intended to help individuals to passactuarial exams. The topics in this manuscript parallel the topics tested onCourse MLC/3L of the Society of Actuaries exam sequence. The primaryobjective of the course is to increase students understanding of the topicscovered, and a secondary objective is to prepare students for a career in ac-tuarial science.The recommended approach for using this book is to read each section, workon the embedded examples, and then try the problems. Answer keys are

provided so that you check your numerical answers against the correct ones.Problems taken from previous SOA/CAS exams will be indicated by thesymbol.A calculator, such as the one allowed on the Society of Actuaries examina-tions, will be useful in solving many of the problems here. Familiarity withthis calculator and its capabilities is an essential part of preparation for theexamination.This work has been supported by a research grant from Arkansas Tech Uni-versity.This manuscript can be used for personal use or class use, but not for commer-cial purposes. If you encounter inconsistencies or errors, I would appreciate

hearing from you: [email protected]

Marcel B. FinanRussellville, Arkansas (July 2011)

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12 PREFACE


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Prerequisite Material

Life contingency models are models that deal with the payments (orbenefits) to a policyholder that are contingent on the continued survival (ordeath) of the person. We refer to these payments as contingent payments.

The theory of insurance can be viewed as the theory of contingent payments.The insurance company makes payments to its insureds contingent upon theoccurrence of some event, such as the death of the insured, an auto accidentby an insured, and so on. The insured makes premium payments to theinsurance company contingent upon being alive, having sufficient funds, andso on.Two important ingredients in the study of contingent models:The first is to represent the contingencies mathematically and this is doneby using probability theory. Probabilistic considerations will, therefore, playan important role in the discussion that follows. An overview of probability

theory is presented in Chapter 2 of the book.The other central consideration in the theory of insurance is the time valueof money. Both claims and premium payments occur at various, possiblyrandom, points of time in the future. Since the value of a sum of moneydepends on the point in time at which the funds are available, a method ofcomparing the value of sums of money which become available at differentpoints of time is needed. This methodology is provided by the theory ofinterest. An overview of the theory of interest is presented in Chapter 1.

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14 PREREQUISITE MATERIAL


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Brief Review of Interest Theory

A typical part of most insurance contracts is that the insured pays the in-surer a fixed premium on a periodic (usually annual or semi-annual) basis.Money has time value, that is, $1 in hand today is more valuable than $1

to be received one year. A careful analysis of insurance problems must takethis effect into account. The purpose of this chapter is to examine the basicaspects of the theory of interest. Readers interested in further and thoroughdiscussions of the topics can refer to [3].Compound interest or discount will always be assumed, unless specified oth-erwise.

1 The Basics of Interest Theory

A dollar received today is worth more than a dollar received tomorrow. Thisis because a dollar received today can be invested to earn interest. Theamount of interest earned depends on the rate of return that can be earnedon the investment. The time value of money (abbreviated TVM) quanti-fies the value of a dollar through time.Compounding is the term used to define computing a future value. Dis-counting is the term used to define computing a present value. We use theDiscount Rate or Compound Rate to determine the present value orfuture value of a fixed lump sum or a stream of payments.Under compound interest, the future value of $1 invested today over tperiodsis given by the accumulation function

a(t) = (1 +i)t.

Thus, $1 invested today worths 1 + idollars a period later. We call 1 + itheaccumulation factor.

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16 BRIEF REVIEW OF INTEREST THEORY

The function A(t) = A(0)a(t) which represents the accumulation of an in-

vestment ofA(0) for t periods is called the amount function.Example 1.1Suppose that A(t) = t2 + 10. If X invested at time 0 accumulates to$500 at time 4, and to $1,000 at time 10, find the amount of the originalinvestment, X.

Solution.We have A(0) = X = 10; A(4) = 500 = 16+ 10; and A(10) = 1000 =100+ 10. Using the first equation in the second and third we obtain thefollowing system of linear equations

16+X=500100+X=1000.

Multiply the first equation by 100 and the second equation by 16 and subtractto obtain 1600+ 100X 1600 16X= 50000 16000 or 84X= 34000.Hence,X= 3400084 = $404.76

Now, let n be a positive integer. The nthperiod of time is defined to bethe period of time between t = n 1 and t= n. More precisely, the periodnormally will consist of the time interval n 1tn. We next introducethe first measure of interest which is developed using the accumulation func-

tion. Such a measure is referred to as the effective rate of interest:The effective rate of interest is the amount of money that one unit investedat the beginning of a period will earn during the period, with interest beingpaid at the end of the period.

Example 1.2You buy a house for $100,000. A year later you sell it for $80,000. What isthe effective rate of return on your investment?

Solution.The effective rate of return is

i= 80, 000 100, 000100, 000

=20%

which indicates a 20% loss of the original value of the house


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1 THE BASICS OF INTEREST THEORY 17

If in is the effective rate of interest for the nth time period then we can

writein=

a(n) a(n 1)a(n 1) =

(1 +i)n (1 +i)n1(1 +i)n1

=i.

That is, under compound interest, the effective rate of interest is constantand is equal to the parameter iappearing in the base of the exponential formofa(t).

Example 1.3It is known that $600 invested for two years will earn $264 in interest. Findthe accumulated value of $2,000 invested at the same rate of annual com-pound interest for three years.

Solution.We are told that 600(1 +i)2 = 600 + 264 = 864. Thus, (1 +i)2 = 1.44 andsolving fori we findi = 0.2.Thus, the accumulated value of investing $2,000for three years at the rate i = 20% is 2, 000(1 + 0.2)3 = $3, 456

Example 1.4At a certain rate of compound interest, 1 will increase to 2 in a years, 2 willincrease to 3 in b years, and 3 will increase to 15 in c years. If 6 will increaseto 10 in n years, find an expression for n in terms ofa,b,and c.

Solution.If the common rate is i, the hypotheses are that

1(1 +i)a =2ln 2 =a ln (1 +i)2(1 +i)b =3ln3

2=b ln (1 +i)

3(1 +i)c =15ln 5 =c ln (1 +i)6(1 +i)n =10ln5

3=n ln (1 +i)

But

ln53

= ln 5 ln 3 = ln 5 (ln 2 + ln 1.5).Hence,

n ln (1 +i) =c ln (1 +i) a ln (1 +i) b ln (1 +i) = (c a b) ln (1 +i)


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and this implies n = c a b

The accumulation function is used to find future values. In order to findpresent values of future investments, one uses the discount function de-fined by the ratio 1(1+i)t .

Example 1.5What is the present value of $8,000 to be paid at the end of three years ifthe interest rate is 11% compounded annually?

Solution.Let F V stand for the future value and P V for the present value. We want

to find P V. We have F V =P V(1 +i)3

or P V =F V(1 +i)3

. Substitutinginto this equation we find P V= 8000(1.11)3 $5, 849.53

Parallel to the concept of effective rate of interest, we define the effectiverate of discountfor the nth time period by

dn=a(n) a(n 1)

a(n) =

(1 +i)n (1 +i)n1(1 +i)n

= i

1 +i=d.

That is, under compound interest, the effective rate of discount is constant.Now, $1 a period from now invested at the rate iworths= 1

1+i today. We

call the discount factor since it discounts the value of an investment atthe end of a period to its value at the beginning of the period.

Example 1.6What is the difference between the following two situations?(1) A loan of $100 is made for one year at an effective rate of interest of 5%.(2) A loan of $100 is made for one year at an effective rate of discount of 5%.

Solution.In both cases the fee for the use of the money is the same which is $5. Thatis, the amount of discount is the same as the amount of interest. However,

in the first case the interest is paid at the end of the period so the borrowerwas able to use the full $100 for the year. He can for example invest thismoney at a higher rate of interest say 7% and make a profit of $2 at the endof the transaction. In the second case, the interest is paid at the beginningof the period so the borrower had access to only $95 for the year. So, if this


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amount is invested at 7% like the previous case then the borrower will make

a profit of $1.65. Also, note that the effective rate of interest is taken as apercentage of the balance at the beginning of the year whereas the effectiverate of discount is taken as a percentage of the balance at the end of the year

Using the definitions of andd we have the following relations among, i,d,and : i = d

1d . d = i1+i . d = i. d = 1 .

id = i

d.

Example 1.7The amount of interest earned for one year when X is invested is $108. Theamount of discount earned when an investment grows to value Xat the endof one year is $100. Find X,i, and d.

Solution.We have iX = 108, i1+i X = 100. Thus,

1081+i = 100. Solving for i we find

i= 0.08 = 8%.Hence, X= 1080.08

= 1, 350 andd= i1+i

= 227

7.41%

Nominal Rates of Interest and DiscountWhen interest is paid (i.e., reinvested) more frequently than once per period,we say it is payable (convertible, compounded) each fraction of a pe-riod, and this fractional period is called the interest conversion period.Anominal rate of interest i(m) payable m times per period, where m is apositive integer, represents m times the effective rate of compound interestused for each of the mth of a period. In this case, i

(m)

m is the effective rateof interest for each mth of a period. Thus, for a nominal rate of 12% com-pounded monthly, the effective rate of interest per month is 1% since thereare twelve months in a year.Suppose that 1 is invested at a nominal rate i(m) compounded m times per

measurement period. That is, the period is partitioned intomequal fractionsof a period. At the end of the first fraction of the period the accumulatedvalue is 1 + i

(m)

m .At the end of the second fraction of the period the accumu-

lated value is

1 + i(m)

m

2. Continuing, we find that the accumulated value


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at the end of the mth fraction of a period, which is the same as the end of

one period, is

1 + i(m)

mm

and at the end oft years the accumulated value is

a(t) =

1 +

i(m)

m

mt.

Example 1.8Find the accumulated value of $3,000 to be paid at the end of 8 years witha rate of compound interest of 5%(a) per annum;(b) convertible quarterly;(c) convertible monthly.

Solution.(a) The accumulated value is 3, 000

1 + 0.051

8 $4, 432.37.(b) The accumulated value is 3, 000

1 + 0.05

4

84 $4, 464.39.(c) The accumulated value is 3, 000

1 + 0.0512

812 $4, 471.76Next we describe the relationship between effective and nominal rates. Ifi denotes the effective rate of interest per one measurement period equiva-lent to i(m) then we can write

1 +i=

1 + i(m)m

m

since each side represents the accumulated value of a principal of 1 investedfor one year. Rearranging we have

i=

1 +

i(m)

m

m 1

andi(m) =m[(1 +i)

1m

1].

For anyt0 we have

(1 +i)t =

1 +

i(m)

m

mt.


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Example 1.9

(a) Find the annual effective interest rate i which is equivalent to a rate ofcompound interest of 8% convertible quarterly.(b) Find the compound interest rate i(2) which is equivalent to an annualeffective interest rate of 8%.(c) Find the compound interest rate i(4) which is equivalent to a rate ofcompound interest of 8% payable semi-annually.

Solution.(a) We have

1 +i=

1 +

0.08

44 i = 1 +

0.08

44 10.08243216

(b) We have

1 + 0.08 =

1 +

i(2)

2

2i(2) = 2[(1.08) 12 1]0.07846.

(c) We have

1 +i(4)

44

= 1 +i(2)

22

i(4) = 4[(1.04)

12

1]

0.0792

In the same way that we defined a nominal rate of interest, we could also de-fine a nominal rate of discount,d(m),as meaning an effective rate of discountof d

(m)

m for each of themth of a period with interest paid at the beginning ofa mth of a period.The accumulation function with the nominal rate of discount d(m) is

a(t) =

1 d

(m)

m

mt, t0.

Example 1.10Find the present value of $8,000 to be paid at the end of 5 years using anannual compound interest of 7%(a) convertible semiannually.(b) payable in advance and convertible semiannually.


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Solution.

(a) The answer is 8, 0001 + 0.072

52 $5, 671.35(b) The answer is

8000

1 0.07

2

52$5, 602.26

Ifd is the effective discount rate equivalent to d(m) then

1 d= 1 d(m)

m m

since each side of the equation gives the present value of 1 to be paid at theend of the measurement period. Rearranging, we have

d= 1

1 d(m)

m

m

and solving this last equation for d(m) we find

d(m) =m[1 (1 d) 1m ] =m(1 1m ).Example 1.11Find the present value of $1,000 to be paid at the end of six years at 6% peryear payable in advance and convertible semiannually.

Solution.The answer is

1, 000

1 0.06

2

12= $693.84

There is a close relationship between nominal rate of interest and nominalrate of discount. Since 1 d= 1

1+i, we conclude that

1 +i(m)

m m

= 1 +i= (1 d)1 = 1 d(n)

n n

. (1.1)

Ifm= n then the previous formula reduces to1 +

i(n)

n

=

1 d

(n)

n

1.


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Example 1.12

Find the nominal rate of discount convertible semiannually which is equiva-lent to a nominal rate of interest of 12% per year convertible monthly.

Solution.We have

1 d(2)

2

2=

1 +

0.12

12

12.

Solving ford(2) we findd(2) = 0.11591

Note that formula (1.1) can be used in general to find equivalent rates ofinterest or discount, either effective or nominal, converted with any desiredfrequency.

Force of InterestEffective and nominal rates of interest and discount each measures interestover some interval of time. Effective rates of interest and discount measureinterest over one full measurement period, while nominal rates of interestand discount measure interest over mths of a period.In this section we want to measure interest at any particular moment of time.This measure of interest is called the force of interest and is defined by

t = d

dtln [a(t)]

which under compound interest we have

t = ln (1 +i) =.

Example 1.13Given the nominal interest rate of 12%, compounded monthly. Find theequivalent force of interest .

Solution.

The effective annual interest rate is

i= (1 + 0.01)12 10.1268250.

Hence,= ln (1 +i) = ln (1.1268250)0.119404.


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Example 1.14

A loan of $3,000 is taken out on June 23, 1997. If the force of interest is14%, find each of the following(a) The value of the loan on June 23, 2002.(b) The value ofi.(c) The value ofi(12).

Solution.(a) 3, 000(1 +i)5 = 3, 000e5 = 3, 000e0.7 $6, 041.26.(b) i = e 1 =e0.14 10.15027.(c) We have

1 + i(12)

1212

= 1 +i= e0.14.

Solving for i(12) we findi(12) 0.14082

Since

t = d

dtln (a(t))

we can find t given a(t). What if we are given t instead, and we wish toderive a(t) from it?From the definition oft we can write

d

dr

ln a(r) =r.

Integrating both sides from 0 to twe obtain t0

d

drln a(r)dr=

t0

rdr.

Hence,

ln a(t) =

t0

rdr.

From this last equation we find

a(t) =et0 rdr.

Example 1.15A deposit of $10 is invested at time 2 years. Using a force of interest oft = 0.2 0.02t,find the accumulated value of this payment at the end of 5years.


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Solution.

The accumulated value is

A(5) = 10a(5)

a(2)= 10e

52(0.20.02t)dt = 10e[0.2t0.01t

2]5

2$14.77


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2 Equations of Value and Time Diagrams

Interest problems generally involve four quantities: principal(s), investmentperiod length(s), interest rate(s), accumulated value(s). If any three of thesequantities are known, then the fourth quantity can be determined. In thissection we introduce equations that involve all four quantities with threequantities are given and the fourth to be found.In calculations involving interest, the value of an amount of money at anygiven point in time depends upon the time elapsed since the money waspaid in the past or upon time which will elapse in the future before it ispaid. This principle is often characterized as the recognition of the timevalue of money. We assume that this principle reflects only the effect of

interest and does not include the effect of inflation. Inflation reduces thepurchasing power of money over time so investors expect a higher rate ofreturn to compensate for inflation. As pointed out, we will neglect the effectof inflation when applying the above mentioned principle.As a consequence of the above principle, various amounts of money payableat different points in time cannot be compared until all the amounts areaccumulated or discounted to a common date, called thecomparison date,is established. The equation which accumulates or discounts each paymentto the comparison date is called the equation of value.One device which is often helpful in the solution of equations of value is thetime diagram. A time diagram is a one-dimensional diagram where the

only variable is time, shown on a single coordinate axis. We may show aboveor below the coordinate of a point on the time-axis, values of money intendedto be associated with different funds. A time diagram is not a formal part ofa solution, but may be very helpful in visualizing the solution. Usually, theyare very helpful in the solution of complex probems.

Example 2.1In return for a payment of $1,200 at the end of 10 years, a lender agrees topay $200 immediately, $400 at the end of 6 years, and a final amount at theend of 15 years. Find the amount of the final payment at the end of 15 years

if the nominal rate of interest is 9% converted semiannually.

Solution.The comparison date is chosen to be t = 0. The time diagram is given inFigure 2.1.


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2 EQUATIONS OF VALUE AND TIME DIAGRAMS 27

Figure 2.1

The equation of value is

200 + 400(1 + 0.045)12 +X(1 + 0.045)30 = 1200(1 + 0.045)20.

Solving this equation for Xwe findX$231.11Example 2.2Investor A deposits 1,000 into an account paying 4% compounded quarterly.At the end of three years, he deposits an additional 1,000. InvestorBdepositsXinto an account with force of interest t =

16+t . After five years, investors

Aand B have the same amount of money. Find X.

Solution.Consider investor As account first. The initial 1,000 accumulates at 4%compounded quarterly for five years; the accumulated amount of this pieceis

1, 0001 +0.04

4 45

= 1000(1.01)20.

The second 1,000 accumulates at 4% compounded quarterly for two years,accumulating to

1, 000

1 +

0.04

4

42= 1000(1.01)8.

The value in investor As account after five years is

A= 1000(1.01)20 + 1000(1.01)8.

The accumulated amount of investor B s account after five years is given by

B = X e50

dt6+t =X eln (

116) =

11

6X.

The equation of value at time t = 5 is

11

6X= 1000(1.01)20 + 1000(1.01)8.

Solving forXwe find X$1, 256.21


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3 Level Annuities

A series of payments made at equal intervals of time is called an annuity. Anannuity where payments are guaranteed to occur for a fixed period of timeis called an annuity-certain. In what follows we review the terminologyand notation partained to annuity-certain. By a level-annuitywe mean anannuity with fixed payments.

3.1 Level Annuity-Immediate

An annuity under which payments of 1 are made at the end of each periodforn periods is called an annuityimmediateor ordinary annuity. Thecash stream represented by the annuity can be visualized on a time diagramas shown in Figure 3.1 .

Figure 3.1

The first arrow shows the beginning of the first period, at the end of which

the first payment is due under the annuity. The second arrow indicates thelast payment datejust after the payment has been made.The present value of the annuity at time 0 is given by

an =+2 + +n =1

n

i .

That is, the present value of the annuity is the sum of the present values ofeach of the n payments.

Example 3.1

Calculate the present value of an annuityimmediate of amount $100 paidannually for 5 years at the rate of interest of 9%.

Solution.The answer is 100a5 = 100

1(1.09)50.09

388.97


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3 LEVEL ANNUITIES 29

The accumulated value of an annuityimmediate right after the nth pay-ment is made is given by

sn = (1 +i)nan =

(1 +i)n 1i

.

Example 3.2Calculate the future value of an annuityimmediate of amount $100 paidannually for 5 years at the rate of interest of 9%.

Solution.The answer is 100s5 = 100

(1.09)51

0.09

$598.47

Withan andsn as defined above we have

1

an=

1

sn+i.

A perpetuity-immediate is an annuity with infinite number of paymentswith the first payment occurring at the end of the first period. Since theterm of the annuity is infinite, perpetuities do not have accumulated values.The present value of a perpetuity-immediate is given by

a = limnan =1

i .

Example 3.3Suppose a company issues a stock that pays a dividend at the end of eachyear of $10 indefinitely, and the companies cost of capital is 6%. What is thevalue of the stock at the beginning of the year?

Solution.The answer is 10 a = 10 10.06 = $166.67


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3.2 Level Annuity-Due

Anannuitydueis an annuity for which the payments are made at the be-ginning of the payment periods. The cash stream represented by the annuitycan be visualized on a time diagram as shown in Figure 3.2 .

Figure 3.2The first arrow shows the beginning of the first period at which the firstpayment is made under the annuity. The second arrow appears n periodsafter arrow 1, one period after the last payment is made.

The present value of an annuity-due is given by

an = 1 ++2 + +n1 =1

n

d .

Example 3.4Find a

8if the effective rate of discount is 10%.

Solution.Sinced= 0.10, we have = 1 d= 0.9. Hence, a

8 = 1(0.9)

8

0.1 = 5.6953279

Example 3.5What amount must you invest today at 6% interest rate compounded annu-ally so that you can withdraw $5,000 at the beginning of each year for thenext 5 years?

Solution.The answer is 5000a

5= 5000

1(1.06)50.06(1.06)1 = $22, 325.53

The accumulated value at time n of an annuity-due is given by

sn = (1 +i)nan =

(1 +i)n 1i

=(1 +i)n 1

d .


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Example 3.6

What amount will accumulate if we deposit $5,000 at the beginning of eachyear for the next 5 years? Assume an interest of 6% compounded annually.

Solution.The answer is 5000s

5= 5000 (1.06)51

0.06(1.06)1= $29, 876.59

With an and sn as defined above we have(i) an = (1 +i)an .(ii)an =an .(iii) sn = (1 +i)sn .(iv)sn =sn .

(v) 1an =

1sn +d.

A perpetuity-due is an annuity with infinite number of payments withthe first payment occurring at the beginning of the first period. Since theterm of the annuity is infinite, perpetuities do not have accumulated values.The present value of a perpetuity-due is given by

a = limnan =

1

d.

Example 3.7

What would you be willing to pay for an infinite stream of $37 annual pay-ments (cash inflows) beginning now if the interest rate is 8% per annum?

Solution.The answer is 37a =

370.08(1.08)1

= $499.50


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3.3 Level Continuous Annuity

In this section we consider annuities with a finite term and an infinte fre-quency of payments . Formulas corresponding to such annuities are usefulas approximations corresponding to annuities payable with great frequencysuch as daily.Consider an annuity in which a very small payment dt is made at time tand these small payments are payable continuously for n interest conversionperiods. Leti denote the periodic interest rate. Then the total amount paidduring each period is k

k1dt= [t]kk1= $1.

Let an denote the present value of an annuity payable continuously for ninterest conversion periods so that 1 is the total amount paid during eachinterest conversion period. Then the present value can be found as follows:

an =

n0

tdt= t

ln

n

0

=1 n

.

Withan defined above we have

an = i

an =

d

an =

1 en

.

Example 3.8

Starting four years from today, you will receive payment at the rate of $1,000per annum, payable continuously, with the payment terminating twelve yearsfrom today. Find the present value of this continuous annuity if= 5%.

Solution.The present value is P V= 10004 a

8= 1000e0.20 1e0.40

0.05 = $5, 398.38

Next, let sn denote the accumulated value at the end of the term of anannuity payable continuously for n interest conversion periods so that 1 isthe total amound paid during each interest conversion period. Then

sn

= (1 +i)nan

= n0

(1 +i)ntdt=(1 +i)n 1

.

It is easy to see that

sn =en 1

=

i

sn =

d

sn .


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Example 3.9

Find the force of interest at which the accumulated value of a continuouspayment of 1 every year for 8 years will be equal to four times the accumulatedvalue of a continuous payment of 1 every year for four years.

Solution.We have

s8

=4s4

e8 1

=4 e4 1

e8 4e4 + 3 =0

(e4 3)(e4 1) =0Ife4 = 3 then = ln 34 0.0275 = 2.75%. Ife4 = 1 then = 0,an extrane-ous solution

Withan and sn as defined above we have

1

an=

1

sn+.

The present value of a perpetuity payable continuously with total of 1 per

period is given bya = lim

nan =

1

.

Example 3.10A perpetuity paid continuously at a rate of 100 per year has a present value of800. Calculate the annual effective interest rate used to calculate the presentvalue.

Solution.The equation of value at time t = 0 is

800 =100

= 100ln (1 +i)

.

Thus,i= e

18 1 = 13.3%


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4 Varying Annuities

In the previous section we considered annuities with level series of payments,that is, payments are all equal in values. In this section we consider annuitieswith a varying series of payments. Annuities with varying payments will becalled varying annuities. In what follows, we assume that the paymentperiod and interest conversion period coincide.

4.1 Varying Annuity-Immediate

Any type of annuities can be evaluated by taking the present value or theaccumulated value of each payment seperately and adding the results. There

are, however, several types of varying annuities for which relatively simplecompact expressions are possible. The only general types that we considervary in either arithmetic progression or geometric progression.

Payments Varying in an Arithmetic ProgressionFirst, let us assume that payments vary in arithmetic progression. The firstpayment is 1 and then the payments increase by 1 thereafter, continuing fornyears as shown in the time diagram of Figure 4.1.

Figure 4.1

The present value at time 0 of such annuity is

(Ia)n =+ 22 + 33 + +nn = an n

n

i .

The accumulated value at time n is given by

(Is)n = (1 +i)

n

(Ia)n =

sn

n

i =

sn+1

(n+ 1)i .

Example 4.1The following payments are to be received: $500 at the end of the first year,$520 at the end of the second year, $540 at the end of the third year and so


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4 VARYING ANNUITIES 35

on, until the final payment is $800. Using an annual effective interest rate of

2%(a) determine the present value of these payments at time 0;(b) determine the accumulated value of these payments at the time of thelast payment.

Solution.Inn years the payment is 500 + 20(n 1).So the total number of paymentsis 16. The given payments can be regarded as the sum of a level annuityimmediate of $480 and an increasing annuity-immediate $20, $40, , $320.(a) The present value at time t = 0 is

480a16 + 20(Ia)16 = 480(13.5777) + 20(109.7065) = $8, 711.43.

(b) The accumulated value at timet= 16 is

480s16 + 20(Is)16 = 480(18.6393) + 20(150.6035) = $11, 958.93

Next, we consider a decreasing annuity-immediate with first payment n andeach payment decreases by 1 for a total ofn payments as shown in Figure4.2.

Figure 4.2

In this case, the present value one year before the first payment (i.e., at timet= 0) is given by

(Da)n =n+ (n 1)2 + +n =n an

i .

The accumulated value at time n is given by

(Ds)n = (1 +i)n(Da)n =

n(1 +i)n sni

= (n+ 1)an (Ia)n .

Example 4.2John receives $400 at the end of the first year, $350 at the end of the secondyear, $300 at the end of the third year and so on, until the final payment of$50. Using an annual effective rate of 3.5%, calculate the present value ofthese payments at time 0.


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Solution.

In year n the payment is 400 50(n 1). Since the final payment is 50 wemust have 40050(n1) = 50. Solving for n we find n = 8. Thus, thepresent value is

50(Da)8

= 50 8 a80.035

= $1, 608.63

Example 4.3Calculate the accumulated value in Example 4.2.

Solution.The answer is 50(Ds)

8 = 50 8(1.035)8s8

0.035 = $2, 118.27

Besides varying annuities immediate, it is also possible to have varyingperpetuityimmediate. Consider the perpetuity with first payment of 1 atthe end of the first period and then each successive payment increases by 1.In this case, the present value is given by

(Ia) = limn

(Ia)n =1

i +

1

i2.

Example 4.4Find the present value of a perpetuity-immediate whose successive paymentsare 1, 2, 3, 4, at an effective rate of 6%.Solution.We have (Ia) =

1i +

1i2 =

10.06 +

10.062 = $294.44

Remark 4.1The notion of perpetuity does not apply for the decreasing case.

Payments Varying in a Geometric ProgressionNext, we consider payments varying in a geometric progression. Consider anannuity-immediate with a term ofn periods where the interest rate is iperperiod, and where the first payment is 1 and successive payments increasein geometric progression with common ratio 1 + k. The present value of thisannuity is

+2(1 +k) +3(1 +k)2 + +n(1 +k)n1 = 1 1+k1+i

n

i kprovided that k= i. Ifk = i then the original sum consists of a sum ofnterms ofwhich equals ton.


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Example 4.5

The first of 30 payments of an annuity occurs in exactly one year and is equalto $500. The payments increase so that each payment is 5% greater than thepreceding payment. Find the present value of this annuity with an annualeffective rate of interest of 8%.

Solution.The present value is given by

P V= 500 1 1.051.08

300.08 0.05 = $9, 508.28

For an annuity-immediate with a term ofn periods where the interest rate

is i per period, and where the first payment is 1 and successive paymentsdecrease in geometric progression with common ratio 1k. The presentvalue of this annuity is

+2(1 k) +3(1 k)2 + +n(1 k)n1 = 1 1k1+i

ni+k

provided that k=i. Ifk= i then the original sum becomes

+2(1 i) +3(1 i)2 + +n(1 i)n1 = 12i

1

1 i1 +i

n

Finally, we consider a perpetuity with payments that form a geometric pro-gression where 0 < 1 + k < 1 + i. The present value for such a perpetuitywith the first payment at the end of the first period is

+2(1 +k) +3(1 +k)2 + = 1 (1 +k) =

1

i k .

Observe that the value for these perpetuities cannot exist if 1 + k1 +i.Example 4.6What is the present value of a stream of annual dividends, which starts at 1

at the end of the first year, and grows at the annual rate of 2%, given thatthe rate of interest is 6% ?

Solution.The present value is 1

ik = 10.060.02 = 25


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4.2 Varying Annuity-Due

In this section, we examine the case of an increasing annuity-due. Consideran annuity with the first payment is 1 at the beginning of year 1 and then thepayments increase by 1 thereafter, continuing for n years. A time diagramof this situation is given in Figure 4.3.

Figure 4.3

The present value for this annuity-due (at time t = 0) is

(Ia)n = 1 + 2+ 32 + +nn1 = an n

n

d

and the accumulated value at time n is

(Is)n = (1 +i)n(Ia)n =

sn nd

=sn+1 (n+ 1)

d .

Example 4.7Determine the present value and future value of payments of $75 at time 0,

$80 at time 1 year, $85 at time 2 years, and so on up to $175 at time 20years. The annual effective rate is 4%.

Solution.The present value is 70a21 + 5(Ia)21 = $1, 720.05 and the future value is

(1.04)21(1, 720.05) = $3919.60

In the case of a decreasing annuity-due where the first payment is n andeach successive payment decreases by 1, the present value at time 0 is

(Da)n =n+ (n

1)+

+n1 =n an

dand the accumulated value at time n is

(Ds)n = (1 +i)n(Da)n =

n(1 +i)n snd

.


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Example 4.8

Calculate the present value and the accumulated value of a series of paymentsof $100 now, $90 in 1 year, $80 in 2 years, and so on, down to $10 at time 9years using an annual effective interest rate of 3%.

Solution.The present value is 10(Da)10 = 10 108.5302030.03/1.03 = $504.63 and the accumu-lated value is (1.03)10(504.63) = $678.18

A counterpart to a varying perpetuity-immediate is a varying perpetuity-due which starts with a payment of 1 at time 0 and each successive paymentincreases by 1 forever. The present value of such perpetuity is given by

(Ia) = limn(Ia)n = 1

d2.

Example 4.9Determine the present value at time 0 of payments of $10 paid at time 0, $20paid at time 1 year, $30 paid at time 2 years, and so on, assuming an annualeffective rate of 5%.

Solution.The answer is 10(Ia) =

10d2 = 10

1.050.05

2= $4, 410.00

Next, we consider payments varying in a geometric progression. Consider

an annuity-due with a term of n periods where the interest rate is i perperiod, and where the first payment is 1 at time 0 and successive paymentsincrease in geometric progression with common ratio 1+ k.The present valueof this annuity is

1 +(1 +k) +2(1 +k)2 + +n1(1 +k)n1 = (1 +i) 1 1+k1+i

ni k

provided that k= i. Ifk = i then the original sum consists of a sum ofnterms of 1 which equals to n.

Example 4.10

An annual annuity due pays $1 at the beginning of the first year. Eachsubsequent payment is 5% greater than the preceding payment. The lastpayment is at the beginning of the 10th year. Calculate the present value at:(a) an annual effective interest rate of 4%;(b) an annual effective interest rate of 5%.


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Solution.

(a) P V = (1.04)

1

( 1.051.04)

10

(0.040.05) = $10.44.(b) Since i = k, PV =n = $10.00

For an annuity-due with n payments where the first payment is 1 at time0 and successive payments decrease in geometric progression with commonratio 1 k. The present value of this annuity is

1 +(1 k) +2(1 k)2 + +n1(1 k)n1 = (1 +i)1 1k1+i

ni+k

provided that k=i. Ifk = i then the original sum is

1 +(1 i) +2(1 i)2 + +n1(1 i)n1 = 12d

1

1 i1 +i

n.

Example 4.11Matthew makes a series of payments at the beginning of each year for 20years. The first payment is 100. Each subsequent payment through the tenthyear increases by 5% from the previous payment. After the tenth payment,each payment decreases by 5% from the previous payment. Calculate thepresent value of these payments at the time the first payment is made usingan annual effective rate of 7%.

Solution.The present value at time 0 of the first 10 payments is

100

1 1.051.07100.07 0.05

(1.07) = 919.95.

The value of the 11th payment is 100(1.05)9(0.95) = 147.38. The presentvalue of the last ten payments is

147.381 0.951.07

10

0.07 + 0.05 (1.07)(1.07)

10 = 464.71.

The total present value of the 20 payments is 919.95 + 464.71 = 1384.66

Finally, the present value of a perpetuity with first payment of 1 at time


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0 and successive payments increase in geometric progression with common

ration 1 +k is1 +(1 +k) +2(1 +k)2 + = 1

1 (1 +k) = 1 +i

i k .

Observe that the value for these perpetuities cannot exist if 1 + k1 +i.Example 4.12Perpetuity A has the following sequence of annual payments beginning onJanuary 1, 2005:

1, 3, 5, 7, Perpetuity B is a level perpetuity of 1 per year, also beginning on January

1, 2005.Perpetuity C has the following sequence of annual payments beginning onJanuary 1, 2005:

1, 1 +r, (1 +r)2, On January 1, 2005, the present value of Perpetuity A is 25 times as large asthe present value of Perpetuity B, and the present value of Perpetuity A isequal to the present value of Perpetuity C. Based on this information, findr.

Solution.The present value of Perpetuity A is 1

d+ 2(1+i)

i2 .

The present value of Perpetuity B is 1d .The present value of Perpetuity C is 1+i

ir .We are told that

1 +i

i +

2(1 +i)

i2 =

25(1 +i)

i .

This is equivalent to12i2 + 11i 1 = 0.

Solving fori we findi= 112

. Also, we are told that

1 +i

i +

2(1 +i)

i2 =

1 +i

i

ror

25(12)(1 + 1

12) =

1 + 112112

r .

Solving forr we find r= 0.08 = 8%


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4.3 Continuous Varying Annuities

In this section we look at annuities in which payments are being made con-tinuously at a varying rate.Consider an annuity forn interest conversion periods in which payments arebeing made continuously at the rate f(t) at exact momentt and the interest

rate is variable with variable force of interest t. Then f(t)e t0 rdrdt is the

present value of the payment f(t)dt made at exact moment t. Hence, thepresent value of thisnperiod continuous varying annuity is

P V =

n0

f(t)et0

rdrdt. (4.1)

Example 4.13Find an expression for the present value of a continuously increasing annuitywith a term ofn years if the force of interest is and if the rate of paymentat time t is t2 per annum.

Solution.Using integration by parts process, we find n

0

t2etdt=t2

et

n0

+2

n0

tetdt

=

n2

en

2t

2et

n

0

+ 2

2 n

0

etdt

= n2

en 2n

2en

2

3et

n0

= n2

en 2n

2en 2

3en +

2

3

=2

3 en

n2

+

2n

2 +

2

3

Under compound interest, i.e.,t = ln (1 +i), formula (4.1) becomes

P V = n

0

f(t)tdt.

Under compound interest and with f(t) = t (an increasing annuity), thepresent value is

(Ia)n =

n0

ttdt=an nn


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and the accumulated value at time n years is

(Is)n = (1 +i)n(Ia)n =

sn n

.

Example 4.14Sam receives continuous payments at an annual rate of 8t + 5 from time 0 to10 years. The continuously compounded interest rate is 9%.(a) Determine the present value at time 0.(b) Determine the accumulated value at time 10 years.

Solution.(a) The payment stream can be split into two parts so that the present value

is 8(Ia)10

+ 5a10

.

Since

i=e0.09 1 = 9.4174%a10

=1 (1.094174)10

0.09 = 6.59370

(Ia)10

=6.59370 10(1.094174)10

0.09 = 28.088592

we obtain

8(Ia)10

+ 5a10

= 8 28.088592 + 5 6.59370 = 257.68.(b) The accumulated value at time 10 years is

257.68 (1.094174)10 = 633.78For a continuous payable continuously increasing perpetuity (wheref(t) =t),the present value at time 0 is

(Ia) = limn

an nn

= limn

1(1+i)n

n(1 +i)n

= 1

2.

Example 4.15Determine the present value of a payment stream that pays a rate of 5t attime t. The payments start at time 0 and they continue indefinitely. Theannual effective interest rate is 7%.


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Solution.

The present value is

5(Ia) = 5

[ln (1.07)]2 = 1, 092.25

We conclude this section by considering the case of a continuously decreasingcontinuously payable stream in which a continuous payment is received fromtime 0 to time n years. The rate of payment at timet is f(t) = n t, andthe force of interest is constant at . The present value is

(Da)n =nan

(Ia)n

=n1 n

an n

n

=n an

and the accumulated value

(Ds)n =n(1 +i)n sn

.

Example 4.16Otto receives a payment at an annual rate of 10

t from time 0 to time

10 years. The force of interest is 6%. Determine the present value of thesepayments at time 0.

Solution.Since

i= e0.06 1 = 6.184%a10 =

1 (1.06184)100.06

= 7.5201

the present value is then

(Da)10

=10 7.52010.06

= 41.33


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4.4 Continuously Payable Varying Annuities

In this section, we consider annuities where payments are made at a con-tinuous rate but increase/decrease at discrete times. To elaborate, considerfirst an increasing annuity with payments made continuously for n conver-sion periods with a total payments of 1 at the end of the first period, a totalpayments of 2 in the second period, , a total payments of n in the nthperiod. The present value of such annuity is

I(a)n =s1 + 2s1 2 + +ns

1n =s

1(Ian ) =

an nn

.

The accumulated value of this annuity is

I(sn ) = (1 +i)nI(a)n =

sn n

.

Next, we consider a decreasing annuity with payments made continuouslyfor n conversion periods with a total payments ofn at the end of the firstperiod, a total payments ofn 1 for the second period, , a total paymentsof 1 for the nth period. The present value at time 0 of such annuity is

D(a)n =ns1 + (n 1)s1 2 + +s1 n =s1 (Dan ) =n an

.

The accumulated value at time n of this annuity is

D(s)n = (1 +i)nD(an ) =

n(1 +i)n sn

.


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5 Annuity Values on Any Date: Deferred An-

nuityEvaluating annuities thus far has always been done at the beginning of theterm( either on the date of, or one period before the first payment) or at theend of the term (either on the date of, or one period after the last payment).In this section, we shall now consider evaluating the(1) present value of an annuity more than one period before the first paymentdate,(2) accumulated value of an annuity more than one period after the last pay-ment date,(3) current value of an annuity between the first and last payment dates.

We will assume that the evaluation date is always an integral number of pe-riods from each payment date.

(1) Present values more than one period before the first paymentdateConsider the question of finding the present value of an annuityimmediatewith periodic interest ratei and m +1 periods before the first payment date.Figure 5.1 shows the time diagram for this case where ? indicates thepresent value to be found.

Figure 5.1

The present value of an nperiod annuityimmediatem + 1 periods beforethe first payment date (called a deferred annuity since payments do notbegin until some later period) is the present value at time m discounted form time periods, that is, vman . It is possible to express this answer strictlyin terms of annuity values. Indeed,

am+n

am =1 m+n

i 1

m

i =

m m+ni

=m1 n

i =man .

Such an expression is convenient for calculation, if interest tables are beingused.


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5 ANNUITY VALUES ON ANY DATE: DEFERRED ANNUITY 47

Example 5.1

Exactly 3 years from now is the first of four $200 yearly payments for anannuity-immediate, with an effective 8% rate of interest. Find the presentvalue of the annuity.

Solution.The answer is 2002a

4= 200(a

6 a

2) = 200(4.6229 1.7833) = $567.92

The deferredannuity introduced above uses annuityimmediate. It is pos-sible to work with a deferred annuitydue. In this case, one can easily seethat the present value is given by

m

an = am+n am .Example 5.2Calculate the present value of an annuitydue paying annual payments of1200 for 12 years with the first payment two years from now. The annualeffective interest rate is 6%.

Solution.The answer is 1200(1.06)2a12 = 1200(a14 a2 ) = 1200(9.8527 1.9434)9, 491.16

(2) Accumulated values more than 1 period after the last paymentdateConsider the question of finding the accumulated value of an annuityimmediatewith periodic interest ratei and m periods after the last payment date. Fig-ure 5.2 shows the time diagram for this case where ? indicates the soughtaccumulated value.

Figure 5.2

The accumulated value of an nperiod annuityimmediatem periods afterthe last payment date is the accumulated value at timen accumulated form


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time periods, that is, (1 +i)msn .Notice that

sm+n

sm =(1 +i)m+n 1i

(1 +i)m 1i

=(1 +i)m+n (1 +i)m

i = (1 +i)m

(1 +i)n 1i

= (1 +i)msn

Example 5.3For four years, an annuity pays $200 at the end of each year with an effective8% rate of interest. Find the accumulated value of the annuity 3 years afterthe last payment.

Solution.

The answer is 200(1 + 0.08)3s4 = 200(s7 s3 ) = 200(8.92283.2464) =$1135.28

It is also possible to work with annuitiesdue instead of annuitiesimmediate.The reader should verify that

(1 +i)msn = sm+n sm .Example 5.4A monthly annuitydue pays 100 per month for 12 months. Calculate theaccumulated value 24 months after the first payment using a nominal rate of

4% compounded monthly.

Solution.The answer is 100

1 + 0.0412

12s12 0.04

12= 1, 276.28

(3)Current value between the first and last payment dateNext, we consider the question of finding the present value of an nperiodannuityimmediate after the payment at the end ofmth period where 1mn. Figure 5.3 shows the time diagram for this case.

Figure 5.3


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5 ANNUITY VALUES ON ANY DATE: DEFERRED ANNUITY 49

The current value of an nperiod annuityimmediate immediately upon

themth payment date is the present value at time 0 accumulated form timeperiods which is equal to the accumulated value at time n discounted forn mtime periods, that is,

(1 +i)man =nmsn .

One has the following formula,

(1 +i)man =nmsn =sm +anm .

Example 5.5For four years, an annuity pays $200 at the end of each half

year with an

8% rate of interest convertible semiannually. Find the current value of theannuity immediately upon the 5th payment (i.e., middle of year 3).

Solution.The answer is 200(1.04)5a8 0.04= 200(s5 0.04+a3 0.04) = 200(5.4163+2.7751) =$1, 638.28

For annuitydue we have a similar formula for the current value

(1 +i)man =nmsn = sm + anm .

Example 5.6Calculate the current value at the end of 2 years of a 10 year annuity due of$100 per year using a discount rate of 6%.

Solution.We have (1+i)1 = 1d= 10.06 = 0.94 and i= 0.060.94 . Thus, 100(.94)2a10 =$870.27

Up to this point, we have assumed that the date is an integral number ofperiods. In the case the date is not an integral number of periods from each

payment date, the value of an annuity is found by finding the value on a datewhich is an integral number of periods from each payment date and thenthe value on this date is either accumulated or discounted for the fractionalperiod to the actual evaluation date. We illustrate this situation in the nextexample.


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Example 5.7

An annuityimmediate pays $1000 every six months for three years. Calcu-late the present value of this annuity two months before the first paymentusing a nominal interest rate of 12% compounded semiannually.

Solution.The present value at time t = 0 is

1000a6 0.06 = 10001 (1.06)6

0.06 = $4917.32.

Letj be the interest rate per 2-month. Then 1+j= (1+ 0.06)13 .The present

value two months before the first payment is made is

4917.32(1.06)23 = $5112.10


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A Brief Review of ProbabilityTheory

One aspect of insurance is that money is paid by the company only if someevent, which may be considered random, occurs within a specific time frame.For example, an automobile insurance policy will experience a claim only ifthere is an accident involving the insured auto. In this chapter a brief outlineof the essential material from the theory of probability is given. Almost allof the material presented here should be familiar to the reader. A morethorough discussion can be found in [2]. Probability concepts that are notusually covered in an introductory probability course will be introduced anddiscussed in futher details whenever needed.

6 Basic Definitions of ProbabilityIn probability, we consider experiments whose results cannot be predictedwith certainty. Examples of such experiments include rolling a die, flippinga coin, and choosing a card from a deck of playing cards.By an outcome or simple event we mean any result of the experiment.For example, the experiment of rolling a die yields six outcomes, namely, theoutcomes 1,2,3,4,5, and 6.The sample space of an experiment is the set of all possible outcomesfor the experiment. For example, if you roll a die one time then the exper-iment is the roll of the die. A sample space for this experiment could be

={1, 2, 3, 4, 5, 6} where each digit represents a face of the die.Anevent is a subset of the sample space. For example, the event of rollingan odd number with a die consists of three simple events{1, 3, 5}.

51


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52 A BRIEF REVIEW OF PROBABILITY THEORY

Example 6.1

Consider the random experiment of tossing a coin three times.(a) Find the sample space of this experiment.(b) Find the outcomes of the event of obtaining more than one head.

Solution.We will use T for tail and Hfor head.(a) The sample space is composed of eight simple events:

={T T T , T T H, T HT , T HH, HT T , HT H, HHT , HHH}.(b) The event of obtaining more than one head is the set

{THH,HTH,HHT,HHH}The complement of an event E, denoted by Ec, is the set of all possibleoutcomes not in E. The union of two events A and B is the event A Bwhose outcomes are either in Aor in B. The intersection of two events Aand B is the event AB whose outcomes are outcomes of both events AandB.Two eventsA and B are said to bemutually exclusiveif they haveno outcomes in common. Clearly, for any event E, the events EandEc aremutually exclusive.

Example 6.2

Consider the sample space of rolling a die. Let A be the event of rollingan even number, B the event of rolling an odd number, and C the event ofrolling a 2. Find(a) Ac, Bc andCc.(b) A B, A C, andB C.(c)A B, A C, and B C.(d) Which events are mutually exclusive?

Solution.(a) We have Ac =B, Bc =A andCc ={1, 3, 4, 5, 6}.(b) We have

A B={1, 2, 3, 4, 5, 6}A C={2, 4, 6}B C={1, 2, 3, 5}


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6 BASIC DEFINITIONS OF PROBABILITY 53

(c)

A B=A C={2}B C=

(d)Aand B are mutually exclusive as well asB and C

Remark 6.1The above definitions of intersection, union, and mutually exclusive can beextended to any number of events.

Probability AxiomsProbability is the measure of occurrence of an event. It is a function Pr( )defined on the collection of all (subsets) events of a sample space and whichsatisfiesKolmogorov axioms:Axiom 1: For any event E, 0Pr(E)1.Axiom 2: Pr() = 1.Axiom 3: For any sequence of mutually exclusive events{En}n1, that isEi Ej =fori=j, we have

Pr (n=1En) = n=1Pr(En).(Countable Additivity)

If we let E1 = , En = for n > 1 then by Axioms 2 and 3 we have1 = Pr() = Pr(n=1En) =

n=1Pr(En) = Pr() +

n=2Pr(). This

implies that Pr() = 0. Also, if{E1, E2, , En} is a finite set of mutuallyexclusive events, then by defining Ek =for k > n and Axioms 3 we find

Pr (nk=1Ek) =n

k=1

Pr(Ek).

Any function Pr that satisfies Axioms 1-3 will be called a probability mea-sure.

Example 6.3Consider the sample space ={1, 2, 3}. Suppose that Pr({1, 2}) = 0.5and Pr({2, 3}) = 0.7. Find Pr(1), Pr(2), and Pr(3). Is Pr a valid probabilitymeasure?


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Solution.

For Pr to be a probability measure we must have Pr(1) + Pr(2) + Pr(3) = 1.But Pr({1, 2}) = Pr(1) + Pr(2) = 0.5. This implies that 0.5 + Pr(3) = 1or Pr(3) = 0.5. Similarly, 1 = Pr({2, 3}) + Pr(1) = 0.7 + Pr(1) and soPr(1) = 0.3.It follows that Pr(2) = 1 Pr(1) Pr(3) = 1 0.3 0.5 = 0.2.It can be easily seen that Pr satisfies Axioms 1-3 and so Pr is a probabilitymeasure

Probability TreesFor all multistage experiments, the probability of the outcome along anypath of a tree diagram is equal to the product of all the probabilities alongthe path.

Example 6.4In a state assembly, 35% of the legislators are Democrats, and the other 65%are Republicans. 70% of the Democrats favor raising sales tax, while only40% of the Republicans favor the increase.If a legislator is selected at random from this group, what is the probabilitythat he or she favors raising sales tax?

Solution.Figure 6.1 shows a tree diagram for this problem.

Figure 6.1

The first and third branches correspond to favoring the tax. We add theirprobabilities.

P(tax) = 0.245 + 0.26 = 0.505

Conditional Probability and Bayes TheoremWe desire to know the probability of an event Aconditional on the knowledgethat another eventBhas occurred. The information the eventBhas occurred


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6 BASIC DEFINITIONS OF PROBABILITY 55

causes us to update the probabilities of other events in the sample space.

To illustrate, suppose you cast two dice; one red, and one green. Then theprobability of getting two ones is 1/36. However, if, after casting the dice,you ascertain that the green die shows a one (but know nothing about the reddie), then there is a 1/6 chance that both of them will be one. In other words,the probability of getting two ones changes if you have partial information,and we refer to this (altered) probability as conditional probability.If the occurrence of the event A depends on the occurrence of B then theconditional probability will be denoted by P(A|B), read as the probability ofAgivenB .Conditioning restricts the sample space to those outcomes whichare in the set being conditioned on (in this case B). In this case,

P(A|B) =number of outcomes corresponding to event A and Bnumber of outcomes of B .

Thus,

P(A|B) = n(A B)n(B)

=

n(AB)n(S)

n(B)n(S)

=P(A B)

P(B)

provided that P(B)>0.

Example 6.5LetA denote the event student is female and let B denote the event stu-dent is French. In a class of 100 students suppose 60 are French, and supposethat 10 of the French students are females. Find the probability that if I picka French student, it will be a female, that is, find P(A|B).

Solution.Since 10 out of 100 students are both French and female, P(A B) = 10

100=

0.1.Also, 60 out of the 100 students are French, soP(B) = 60100 = 0.6.Hence,P(A|B) = 0.1

0.6= 1

6

It is often the case that we know the probabilities of certain events con-ditional on other events, but what we would like to know is the reverse.

That is, givenP(A|B) we would like to find P(B|A).Bayes formula is a simple mathematical formula used for calculating P(B|A)givenP(A|B).We derive this formula as follows. Let AandB be two events.Then

A= A (B Bc) = (A B) (A Bc).


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Since the events A B and A Bc are mutually exclusive, we can write

P(A) =P(A B) +P(A Bc)=P(A|B)P(B) +P(A|Bc)P(Bc) (6.1)

Example 6.6The completion of a construction job may be delayed because of a strike. Theprobabilities are 0.60 that there will be a strike, 0.85 that the construction jobwill be completed on time if there is no strike, and 0.35 that the constructionwill be completed on time if there is a strike. What is the probability thatthe construction job will be completed on time?

Solution.LetAbe the event that the construction job will be completed on time and Bis the event that there will be a strike. We are givenP(B) = 0.60, P(A|Bc) =0.85,and P(A|B) = 0.35. From Equation (6.1) we findP(A) =P(B)P(A|B) + P(Bc)P(A|Bc) = (0.60)(0.35)+(0.4)(0.85) = 0.55From Equation (6.1) we can get Bayes formula:

P(B|A) = P(A B)P(A)

= P(A|B)P(B)

P(A|B)P(B) +P(A|Bc)P(Bc) . (6.2)

Example 6.7

A company has two machinesAandBfor making shoes. It has been observedthat machineAproduces 10% of the total production of shoes while machineB produces 90% of the total production of shoes. Suppose that 1% of all theshoes produced byA are defective while 5% of all the shoes produced by Bare defective. What is the probability that a shoe taken at random from adays production was made by the machine A, given that it is defective?

Solution.We are givenP(A) = 0.1, P(B) = 0.9, P(D|A) = 0.01,and P(D|B) = 0.05.We want to findP(A|D). Using Bayes formula we find

P(A|D) =P(A D)P(D)

= P(D|A)P(A)P(D|A)P(A) +P(D|B)P(B)

= (0.01)(0.1)

(0.01)(0.1) + (0.05)(0.9)0.0217


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7 CLASSIFICATION OF RANDOM VARIABLES 57

7 Classification of Random Variables

By definition, a random variable Xis a function with domain the samplespace of an experiment and range a subset of the real numbers. The rangeofX is sometimes called the support ofX. The notation X(s) =x meansthatx is the value associated with the outcome s by the random variable X.We consider three types of random variables: Discrete, continuous, and mixedrandom variables.A random variable is said to be discrete if the range of the variable as afunction is either finite or a subset of the set of non-negative integers. Adiscrete random variable can be either finite or infinite. An example of afinite discrete random variable is the variable X that represents the age of

students in your class. An example of an infinite random variable is thevariable Xthat represents the number of times you roll a die until you get a6.(If you are extremely unlucky, it might take you a million rolls before youget a 6!).A random variable Xthat takes an uncountable number of values is said tobe continuous. An example of a continuous random variable is the randomvariable X that measures the depth of a randomly selected location in aspecific lake. The range ofXis the interval [0, M] whereM is the maximumdepth of the lake.Amixed random variable is partially discrete and partially continuous. Anexample of a mixed random variable is the random variable that represents

the weight of a tumor when it is possible that there are no tumors (andconsequently the weight would be zero).We use upper-case lettersX, Y, Z,etc. to represent random variables. We usesmall letters x,y,z, etc to represent possible values that the correspondingrandom variables X,Y,Z, etc. can take. The statement X = x defines anevent consisting of all outcomes with Xmeasurement equal to x which isthe set{s :X(s) =x}.

Example 7.1State whether the random variables are discrete, continuous, or mixed.

(a) A coin is tossed ten times. The random variable Xis the number of tailsthat are noted.(b) A light bulb is burned until it burns out. The random variable Y is itslifetime in hours.(c)Zis the income of an individual.


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Solution.

(a)Xcan only take the values 0, 1, ..., 10, so X is a discrete random variable.(b)Ycan take any positive real value, so Y is a continuous random variable.(c) Z(s) > 0 for a working individual and Z(s) = 0 for a out-of-work indi-vidual so Zis a mixed random variable

Examples of actuarial related random variables that will be encountered inthe text are: the age-at-death from birth the time-until-death from insurance policy issue. the number of times an insured automobile makes a claim in a one-yearperiod.

the amount of the claim of an insured automobile, given a claim is made(or an accident occurred). the value of a specific asset of a company at some future date. the total amount of claims in an insurance portfolio.


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8 DISCRETE RANDOM VARIABLES 59

8 Discrete Random Variables

Because the value of a random variable is determined by the outcomes of theexperiment, we may assign probabilities to the possible values of the randomvariable. The set of all probability values constitutes the distribution ofthe random variable.

Probability Mass Function (PMF)For a discrete random variable X, the distribution ofX is described by theprobability distributionor theprobability mass functiongiven by theequation

p(x) = Pr(X=x).

That is, a probability mass function (pmf) gives the probability that a dis-crete random variable is exactly equal to some value. The pmf can be anequation, a table, or a graph that shows how probability is assigned to pos-sible values of the random variable.

Example 8.1Suppose a variable X can take the values 1, 2, 3, or 4. The probabilitiesassociated with each outcome are described by the following table:

x 1 2 3 4p(x) 0.1 0.3 0.4 0.2

Draw the probability histogram.

Solution.The probability histogram is shown in Figure 8.1


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Figure 8.1

Example 8.2A committee of 4 is to be selected from a group consisting of 5 men and 5women. LetXbe the random variable that represents the number of womenin the committee. Create the probability mass distribution.

Solution.

For x = 0, 1, 2, 3, 4 we have

p(x) =

5x

54 x

104

.

The probability mass function can be described by the table

x 0 1 2 3 4p(x) 5

21050210

100210

50210

5210

Note that if the range of a random variable is Support ={x1, x2, } then

p(x) 0, xSupportp(x) = 0, xSupport


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8 DISCRETE RANDOM VARIABLES 61

Moreover, xSupportp(x) = 1.

Expected ValueWith the previous sum and the definition of weighted average with the prob-abilities of Support being the weights, we define the expected value ofXto be

E(X) =

xSupportx p(x).

The expected value ofX is also known as the mean value.

Example 8.3

Suppose that an insurance company has broken down yearly automobileclaims for drivers from age 16 through 21 as shown in the following table.

Amount of claim Probability$ 0 0.80$ 2000 0.10$ 4000 0.05$ 6000 0.03$ 8000 0.01$ 10000 0.01

How much should the company charge as its average premium in order tobreak even on costs for claims?

Solution.LetXbe the random variable of the amount of claim. Finding the expectedvalue ofXwe have

E(X) = 0(.80)+2000(.10)+4000(.05)+6000(.03)+8000(.01)+10000(.01) = 760.

Since the average claim value is $760, the average automobile insurance pre-mium should be set at $760 per year for the insurance company to breakeven

Expectation of a Function of a Random VariableIf we apply a functiong() to a random variableX, the result is another ran-dom variable Y =g(X). For example, X2, log X, 1

Xare all random variables

derived from the original random variable X.


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Var(X) =2X=E[(X E(X))2]

The variance of a random variable is typically calculated using the followingformula

Var(X) =E(X2) (E(X))2.Example 8.7We toss a fair coin and let X= 1 if we get heads, X =1 if we get tails.Find the variance ofX.

Solution.Since E(X) = 1 12 1 12 = 0 and E(X2) = 1212 + (1)2 12 = 1 we findVar(X) = 1

0 = 1

A useful identity is given in the following result

Theorem 8.2IfXis a discrete random variable then for any constants a and b we have

Var(aX+b) =a2Var(X)

Example 8.8This year, Toronto Maple Leafs t

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