actuarial mathematics (lecture notes).pdf

Lecture Noteson

Actuarial Mathematics

Jerry Alan Veeh

May 9, 2003

Copyright 2003 Jerry Alan Veeh. All rights reserved.

§0. Introduction

The objective of these notes is to present the basic aspects of the theory ofinsurance, concentrating on the part of this theory related to life insurance. Anunderstanding of the basic principles underlying this part of the subject will form asolid foundation for further study of the theory in a more general setting.

Throughout these notes are various exercises and problems. The reader shouldattempt to work all of these.

A calculator, such as the one allowed on the Society of Actuaries examinations,will be useful in solving some of the problems here. The problems contained hereare not all amenable to solution usingonly this simple calculator. A computerequipped with spreadsheet software will sometimes be useful, especially for thelaboratory exercises.


§1. Overview

The central theme of these notes is embodied in the question, “What is the valuetoday of a random sum of money which will be paid at a random time in the future?”Such a random payment is called a contingent payment.

The theory of insurance can be viewed asthe theory of contingent payments.The insurance company makes payments to its insureds contingent upon the oc-currence of some event, such as the death of the insured, an auto accident by aninsured, and so on. The insured makes premium payments to the insurance companycontingent upon being alive, having sufficient funds, and so on. A natural way tomodel these contingencies mathematically is to use probability theory. Probabilisticconsiderations will, therefore, play an important role in the discussion that follows.

The other central consideration in the theory of insurance is the time value ofmoney. Both claims and premium payments occur at various, possibly random,points of time in the future. Since the value of a sum of money depends on thepoint in time at which the funds are available, a method of comparing the valueof sums of money which become available at different points of time is needed.This methodology is provided by the theory ofinterest. The theory of interest willbe studied first in a non-random setting in which all payments are assumed to besure to be made. Then the theory will bedeveloped in a random environment, andwill be seen to provide a complete framework for the understanding of contingentpayments.


§2. Elements of the Theory of Interest

A typical part of most insurance contracts is that the insured pays the insurera fixed premium on a periodic (usually annual or semi–annual) basis. Money hastime value, that is, $1 in hand today is more valuable than $1 to be received one yearhence. A careful analysis of insurance problems must take this effect into account.The purpose of this section is to examine the basic aspects of the theory of interest.A thorough understanding of the concepts discussed here is essential.

To begin, remember the way in which compound interest works. Suppose anamountA is invested at interest ratei per year and this interest is compoundedannually. Aftern years the amount will beA(1 +i)n. The factor (1 +i)n is sometimescalled theaccumulation factor. If interest is compounded daily after the samenyears the amount will beA(1 + i

365)365n. In this discussion the interest ratei is called

thenominal annual rate of interest. Theeffective rate of interest in the examplein which interest is compounded daily is (1 +i365)

365− 1. This is the rate of interestwhich compounded annually would provide the same return.

Exercise 2–1.What is the effective rate of interest corresponding to an interest rateof 5% compounded quarterly?

It is possible that two different investment schemes with two different nominalannual rates of interest may in fact beequivalent, that is, may have equal dollarvalue at any fixed date in the future. This possibility is illustrated by means of anexample.

Example 2–1. Suppose I have the opportunity to invest $1 in Bank A which pays5% interest compounded monthly. What interest rate does Bank B have to pay,compounded daily, to provide an equivalent investment? At any timet in years the

amount in the two banks is given by(1 + 0.05

12

)12tand

(1 + i

365

)365trespectively. It

is now an easy exercise to find the nominal interest ratei which makes these twofunctions equal.

Exercise 2–2.Find the interest ratei. What is the effective rate of interest?

Situations in which interest is compounded more often than annually will arisefrequently. Some notation will be needed todiscuss these situations conveniently.Denote byi(m) the nominal annual interest rate compoundedm times per year whichis equivalent to the interest ratei compounded annually. This means that(

1 +i(m)

m

)m

= 1 + i.

Exercise 2–3.Compute 0.05(12).


§2: Elements of the Theory of Interest 5

An important abstraction of the ideaof compound interest is the idea of con-tinuous compounding. If interest is compoundedn times per year the amount after

t years is given by(1 + i

n

)nt. Letting n → ∞ in this expression produceseit, and

this corresponds to the notion of instantaneous compounding of interest. In thiscontext denote byδ the rate of instantaneous compounding which is equivalentto interest ratei. Hereδ is called theforce of interest. The force of interest isextremely important from a theoretical standpoint and also provides some usefulquick approximations.

Exercise 2–4.Show thatδ = log(1 +i).

Exercise 2–5.Find the force of interest which is equivalent to 5% compoundeddaily.

The converse of the problem of finding the amount aftern years at compoundinterest is as follows. Suppose the objective is to have an amountA n years hence.If money can be invested at interest ratei, how much should be deposited todayin order to achieve this objective? It isreadily seen that the amount required isA(1 + i)−n. This quantity is called thepresent valueof A. The factor (1 +i)−1 isoften called thediscount factor and is denoted byv.

Example 2–2. Suppose the annual interest rate is 5%. What is the present valueof a payment of $2000 payable in the year 2014? The present value (in 2004) is$2000(1 + 0.05)−10 = $1227.83.

The notion of present value is used to move payments of money through time inorder to simplify the analysis of a complexsequence of payments. In the simple caseof the last example the important idea is this. Suppose you were given the followingchoice. You may either receive $1227.83 today or you may receive $2000 in theyear 2014. If you can earn 5% on your money (compounded annually) you shouldbe indifferent between these two choices. Under the assumption of an interest rateof 5%, the payment of $2000 in 2014 can be replaced by a payment of $1227.83today. Thus the payment of $2000 can be moved through time using the idea ofpresent value. A visual aid that is often used is that of a time diagram which showsthe time and amounts that are paid. Under the assumption of an interest rate of 5%,the following two diagrams are equivalent.

Two Equivalent Cash Flows

2004 2014 2004 2014

$2000 $1227.83............................................................................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................................................................

The advantage of moving amounts of money through time is that once all


amounts are paid at the same point in time, the most favorable option is readilyapparent.

Exercise 2–6.What happens in comparing thesecash flows if the interest rate is6% rather than 5%?

In an interest payment setting, the payment of interest ofi at the end of the periodis equivalent to the payment ofd at the beginning of the period. Such a paymentat the beginning of a period is called adiscount. What relationship betweeni andd must hold for a discount payment to be equivalent to the interest payment? Thetime diagram is as follows.

Equivalence of Interest and Discount

0 1 0 1

i d............................................................................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................................................................

The relationship isd = iv follows by moving the interest payment back in timeto the equivalent payment ofiv at time 0.

Exercise 2–7.Denote byd(m) the rate of discount payablem times per year that isequivalent to a nominal annual rate of interesti. What is the relationship betweend(m) andi? Betweend(m) andi(m)? Hint: Draw the time diagram illustrating the twopayments made at time 0 and 1/m.

Exercise 2–8.Treasury bills (United States debtobligations) pay discount ratherthan interest. At a recent sale the discount rate for a 3 month bill was 5%. What isthe equivalent rate of interest?

The notation and the relationships thus far are summarized in the string ofequalities

1 + i =

(1 +

i(m)

m

)m

=

(1−

d(m)

m

)−m

= v−1 = eδ.


Problems

Problem 2–1. Show that ifi > 0 then

d < d(2) < d(3) < ⋅ ⋅ ⋅ < δ < ⋅ ⋅ ⋅ < i(3) < i(2) < i.

Problem 2–2. Show that limm→∞ d(m) = limm→∞ i(m) = δ.

Problem 2–3. Calculate the nominal rate of interest convertible once every 4 yearsthat is equivalent to a nominal rateof discount convertible quarterly.

Problem 2–4. Interest rates are not always the same throughout time. In theoreticalstudies such scenarios are usually modelled by allowing the force of interest todepend on time. Consider the situation in which $1 is invested at time 0 in anaccount which pays interest at a constant force of interestδ. What is the amountA(t) in the account at timet? What is the relationship betweenA′ (t) andA(t)? Moregenerally, suppose the force of interest at timet is δ(t). Argue thatA′ (t) = δ(t)A(t),and solve this equation to find an explicit formula forA(t) in terms ofδ(t) alone.

Problem 2–5. Show thatd = 1− v. Is there a similar equation involvingd(m)?

Problem 2–6. Show thatd = iv. Is there a similar equation involvingd(m) andi(m)?


Solutions to ProblemsProblem 2–1. An analytic argument is possible directly from the formulas.For example, (1 +i(m)/m)m = 1 + i = eδ so i(m) = m(eδ/m − 1). Considerm as acontinuous variable and show that the right hand side is a decreasing functionof m for fixed i. Can you give a purely verbal argument? Hint: How doesan investment with nominal ratei(2) compounded annually compare with aninvestment at nominal ratei(2) compounded twice a year?

Problem 2–2. Sincei(m) = m((1 + i)1/m − 1) the limit can be evaluated directlyusing L’Hopitals rule, Maclaurin expansions, or the definition of derivative.

Problem 2–3. The relevant equation is(1 + 4i(1/4)

)1/4=(1− d(4)/4

)−4.

Problem 2–4. In the constant force settingA(t) = eδt andA′ (t) = δA(t). TheequationA′ (t) = δ(t)A(t) can be solved by separation of variables.

Problem 2–5. 1− d(m)/m = v1/m.

Problem 2–6. d(m)/m = v1/mi(m)/m.


Solutions to ExercisesExercise 2–1. The equation to be solved is (1 + 0.05/4)4 = 1 + i, wherei is theeffective rate of interest.

Exercise 2–2. Takingtth roots of both sides of the equation shows thatt playsno role in determiningi and leads to the equationi = 365((1+0.05/12)12/365−1) =0.04989.

Exercise 2–3. 0.05(12) = 12((1 + 0.05)1/12− 1) = 0.04888.

Exercise 2–4. The requirement for equivalence is thateδ = 1 + i.

Exercise 2–5. Hereeδ = (1 + 0.05/365)365, so thatδ = 0.4999. So as a roughapproximation when compounding daily the force of interest is the same as thenominal interest rate.

Exercise 2–6. The present value in this case is $2000(1+ 0.06)−10 = $1116.79.

Exercise 2–7. A payment ofd(m)/m made at time 0 is required to be equivalentto a payment ofi(m)/m made at time 1/m. Henced(m)/m = v1/mi(m)/m. Sincev−1/m = (1+i)1/m = 1+i(m)/m this givesd(m)/m = 1−v1/m or 1+i = (1−d(m)/m)−m.Another relation is thatd(m)/m − i(m)/m = (d(m)/m)(i(m)/m).

Exercise 2–8. The given information isd(4) = 0.05, from which i can beobtained using the formula of the previous exercise asi = (1− 0.05/4)−4 − 1 =0.0410.

§3. Annuities Certain

Many different types of financial transactions involve the payment of a fixedamount of money at regularly spaced intervals of time for a predetermined period.Such a sequence of payments is called anannuity certain or, more simply, anannuity. A common example is loan payments. It is easy to use the idea of presentvalue to evaluate the worth of such a cash stream at any point in time. Here is anexample.

Example 3–1. Suppose you have the opportunity to buy an annuity, that is, for acertain amount paid by you today you will receive monthly payments of $400, say,for the next 20 years. How much is this annuity worth to you? Suppose that thepayments are to begin one month from today. Such an annuity is called anannuityimmediate (a truly unfortunate choice of terminology). It is useful to visualize thecash stream represented by the annuity on a time diagram.

An Annuity Immediate

0 1 2 239 240

$400↓

$400↓

. . . . . . $400↓

$400↓

Clearly you would be willing to pay today no more than the present value of thetotal payments made by the annuity. Assume that you are able to earn 5% interest(nominal annual rate) compounded monthly.The present value of the payments is

240∑j=1

(1 +.0512

)−j 400.

This sum is simply the sum of the present value of each of the payments using theindicated interest rate. It is easy to findthis sum since it involves a very simplegeometric series.

Exercise 3–1.Evaluate the sum.

Since expressions of this sort occur rather often, actuaries have developed somespecial notation for this sum. Writean for the present value of an annuity whichpays $1 at theend of each period forn periods. Then

an =n∑

j=1

vj =1− vn

i

where the last equality follows from the summation formula for a geometric series.The interest rate per period is usually not included in this notation, but when such


§3: Annuities Certain 11

information is necessary the notation isan i. The present value of the annuity in theprevious example may thus be expressed as 400a240 .05/12.

A slightly different annuity is theannuity due which is an annuity in which thepayments are made starting immediately. The notation ¨an denotes the present valueof an annuity which pays $1 at thebeginning of each period forn periods. Clearly

an =n−1∑j=0

vj =1− vn

d

where again the last equality follows by summing the geometric series. Note thatnstill refers to the number of payments. If the present time is denoted by time 0, thenfor an annuity immediate the last payment is made at timen, while for an annuitydue the last payment is made at timen − 1, that is, the beginning of thenth period.It is quite evident thatan = v an , and there are many other similar relationships.

Exercise 3–2.Show thatan = v an .

The connection between an annuity due and an annuity immediate can be viewedin the following way. In an annuity due the payment for the period is made at thebeginning of the period, whereas for an annuity immediate the payment for theperiod is made at the end of the period. Clearly a payment of 1 at the end of theperiod is equivalent to the payment ofv = 1/ (1 + i) at the beginning of the period.This gives an intuitive description of the equality of the previous exercise.

Example 3–2. Suppose that the annuity is paid continuously, that is, that the annu-itant receives money at a constant rate ofσ dollars per unit time. What value ofσmakes this continuous 20 year annuity equivalent to the discrete annuity describedabove? Two annuities are said to beequivalent if they have the same present value.For the continuous annuity the present value of theσ dt dollars received in the timeinterval (t, t + dt) is e−δtσ dt. The present value of this annuity is therefore

∫ 20

0σe−δt dt.

It is now a simple matter to findσ.

Exercise 3–3.Find σ. Note that you must first findδ which is equivalent to aninterest rate of 5% compounded monthly.

A continuous annuity of the type above which is payable at rateσ = 1 for nperiods has present value which is denoted byan .

Exercise 3–4.Show thatan =1− vn

δ.


Thus far the value of an annuity has been computed at time 0. Another commontime point at which the value of an annuity consisting ofn payments of 1 iscomputed is timen. Denote bysn the value of an annuity immediate at timen, thatis, immediately after thenth payment. Thensn = (1 + i)nan from the time diagram.The valuesn is called theaccumulated value of the annuity immediate. Similarlysn is theaccumulated value of an annuity dueat timen andsn = (1 + i)nan .

Here are two examples which further develop skill in the use of these ideas.

Example 3–3. You are going to buy a house for which the purchase price is $100,000and the downpayment is $20,000. You will finance the $80,000 by borrowing thisamount from a bank at 10% interest with a 30 year term. What is your monthlypayment? Typically such a loan isamortized, that is, you will make equal monthlypayments for the life of the loan and each payment consists partially of interest andpartially of principal. Fromthe banks point of view this transaction represents thepurchase by the bank of an annuity immediate. The monthly payment,p, is thus thesolution of the equation 80000 =pa360 0.10/12. In this setting the quoted interest rateon the loan is assumed to be compounded at the same frequency as the paymentperiod unless stated otherwise.

Exercise 3–5.Find the monthly payment. What is the total amount of the paymentsmade?

Example 3–4. Long ago I bought a new car from a local dealer. Let us say the totalcost to me was $15,000. The dealer seemed very anxious that I finance the purchasethrough him, and he presented several arguments as to why I should do so. I couldborrow the entire purchase price at 11.95% ammortized over 60 months. His firstsales pitch was as follows. If I financed the car I would pay about $5000 in interest.If I paid cash I would lose about $8000 in interest that I could earn by investing mymoney in a savings account at 8.5% interest. Thus I would gain almost $3000 byfinancing the car. Is this argument correct? If not, what’s wrong with it?

Exercise 3–6.Find the monthly payment on the car if it is financed through thedealer. What is the total interest paid? Is this a relevant fact?

The dealers second argument ran as follows. Suppose that at the end of 5years the car is worth only half its present value, namely, $7500. Let’s analyze myavailable assets under the two alternatives. If I pay cash for the car, at the end of5 years I will have clear title to a $7500 asset. If I finance the car, at the end of 5years I will have clear title to the car ($7500) plus the cash I did not pay originally($15000) plus the interest on this cash ($8000) for a total of $30500. Obviouslyonly a fool would pass up this type of opportunity!

Exercise 3–7.What are the flaws, if any, in this second argument?


Problems

Problem 3–1. Show thatan < an < an . Hint: This should be obvious from thepicture.

Problem 3–2. John borrows $1,000 from Jane at an annual effective rate of interesti. He agrees to pay back $1,000 after six years and $1,366.87 after another 6 years.Three years after his first payment, John repays the outstanding balance. What isthe amount of John’s second payment?

Problem 3–3. Suppose a loan of amountA is amortized by a series ofn payments.Denote bybk the loan balance immediately after thekth payment and writeb0 = A.Find a relationship that expressesbk+1 in terms ofbk, the interest ratei, andP = A/an .

Problem 3–4. There are two common ways of analyzing loans which are amortized.In theprospective methodthe loan balance at any point in time is seen to be thepresent value of the remaining loan payments. Use the previous problem to showthat this statement is correct.

Problem 3–5. In theretrospective methodthe loan balance at any point in time isseen to be the accumulated original loan amount less the accumulated value of thepast loan payments. Show that this formula for the loan balance is correct.

Problem 3–6. An annuity immediate pays an initial benefit of one per year, in-creasing by 10.25% every four years. The annuity is payable for 40 years. If theeffective interest rate is 5% find an expression for the present value of this annuity.

Problem 3–7. You are given an annuity immediate paying $10 for 10 years, thendecreasing by $1 per year for nine years and paying $1 per year thereafter, forever.If the annual effective rate of interest is 5%, find the present value of this annuity.

Problem 3–8. Humphrey purchases a home with a $100,000 mortgage. Mortgagepayments are to be made monthly for 30 years, with the first payment to be made onemonth from now. The rate of interest is10%. After 10 years, Humphrey increasesthe amount of each monthly payment by $325 in order to repay the mortgage morequickly. What amount of interestis paid over the life of the loan?

Problem 3–9. On January 1, an insurance company has $100,000 which is due toLinden as a life insurance death benefit.He chooses to receive the benefit annuallyover a period of 15 years, with the first payment made immediately. The benefithe receives is based on an effective interest rate of 4% per annum. The insurancecompany earns interest at an effective rate of 5% per annum. Every July 1 thecompany pays $100 in expenses and taxes to maintain the policy. How muchmoney does the company have remaining after 9 years?


Solutions to ProblemsProblem 3–1. Isn’t the chain of inequalities simply expressing the fact thatgetting a given amount of money sooner makes it worth more? An analyticproof should be easy to give too.

Problem 3–2. From Jane’s point of view the equation 1000 = 1000(1 +i)−6 +1366.87(1 +i)−12 must hold. The outstanding balance at the indicated time is1366.87(1 +i)−3, which is the amount of the second payment.

Problem 3–3. bk+1 = bk + ibk − P.

Problem 3–4. The assertion of the prospective method is thatbk = Pan−k .Plug in and show that this choice satisfies the initial conditionb0 = A and therecursion of the previous problem.

Problem 3–5. Here the assertion is thatbk = A(1 + i)k − Psk . Show that thischoice satisfies the required recursion and initial condition.

Problem 3–6. Each 4 year chunk is a simple annuity immediate. Taking thepresent value of these chunks forms an annuity due with payments every 4 yearsthat are increasing.

Problem 3–7. What is the present value of an annuity immediate paying $1 peryear forever? What is the present value of such an annuity that begins paymentsk years from now? The annuity desribed here is the difference of a few of these.

Problem 3–8. The initial monthly paymentP is the solution of 100,000 =Pa360 . The balance after 10 years isPa240 so the interest paid in the first 10years is 120P − (100,000− Pa240 ). To determine the number of new monthlypayments required to repay the loan the equationPa240 = (P + 325)ax should besolved forx. Since afterx payments the loan balance is 0 the amount of interestpaid in the second stage can then be easily determined.

Problem 3–9. Since the effective rate of interest for the insurance company is5%, the rate (0.05)(2) should be used to move the insurance company’s expensesfrom July 1 to January 1.


Solutions to ExercisesExercise 3–1. The sum is the sum of the terms of a geometric series. So∑240

j=1(1+.0512 )−j 400 = 400((1+0.05/12)−1−(1+0.05/12)−241)/ (1−(1+0.05/12)−1) =

60,610.12.

Exercise 3–2. This follows from the formulas for the present value of the twoannuities and the fact thatd = iv.

Exercise 3–3. The force of interest that is equivalent to an interest rate of 5%compounded monthly isδ = ln(1+0.05/12)12. Integration then gives the presentvalue of the continuous annuity asσ(1 − e−20δ)/δ. Soσ = 400δ/ i = 4790.03will make the continuous annuity equivalent to the original discrete annuity.

Exercise 3–4. This formula follows by straightforward integration.

Exercise 3–5. Using the earlier formula givesa360 0.10/12 = 113.95 from whichp = 702.06 and the total amount of the payments is 360p = 252740.60.

Exercise 3–6. The monthly payment is 15,000/a60 .1195/12 = 333.29 so that thetotal of the payments is 19,997.27 of which 4,997.27 is interest. The total ofthe interest payments is irrelevant, since the time at which the interest paymentis made is not taken into account.

Exercise 3–7. Where did all those payments go?

§4. Laboratory 1

1. A loan of 10,000 carries an interestrate of 9% compounded quarterly. Equalloan payments are to be made monthly for 36 months. What is the size of eachpayment?

2. Anamortization table is a table which lists the principal and interest portionsof each payment for a loan which is being amortized. Construct an amortizationtable for the loan of the previous problem. The table should have four columns:the payment number, the principal part ofthat payment, the interest part of thatpayment, and the loan balance immediately after that payment is made.

3. A loan of 10,000 is to be repaidwith equal monthly payments ofp. Theinterest rate for the first year is 1.9%, while the interest rate for the remaining 2years is 10.9%. What isp? What is the balance afterthe 6th payment? After the15th payment? What are the principal and interest components of the 6th payment?Of the 15th payment?

4. A loan of 10,000 is to be repaid as follows. Payments ofp are to be made atthe end of each month for 36 months and a balloon payment of 2500 is to be madeat the end of the 36th month as well. If the interest rate is 5%, what isp? What isthe loan balance at the end of the 12th month? What part of the 15th payment isinterest? Principal?

5. The symbola(m)n denotes the present value of an annuity immediate that pays

1/m at the end of eachmth part of a year forn years. For example, ifm = 12payments of 1/12 are made at the end of each month. Find a formula fora(m)

n .

6. The symbol ¨a(m)n denotes the present value of an annuity due that pays 1/m at

the beginning of eachmth part of a year forn years. Find a formula for ¨a(m)n .

7. An increasing annuity immediatewith a term ofn periods pays 1 at the endof the first period, 2 at the end of the second period, 3 at the end of the third period,. . ., n at the end of thenth period. Find (Ia)n , the present value of such an annuity.

8. A decreasing annuity immediatewith a term ofn periods paysn at the endof the first period,n − 1 at the end of the second period,n − 2 at the end of the thirdperiod,. . ., 1 at the end of thenth period. Find (Da)n , the present value of such anannuity.


§5. Brief Review of Probability Theory

Another 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. Forexample, an automobile insurance policy will experience a claim only if there is anaccident involving the insured auto. In this section a brief outline of the essentialmaterial from the theory of probability is given. Almost all of the material presentedhere should be familiar to the reader. The concepts presented here will play a crucialrole in the rest of these notes.

The underlying object in probability theory is asample spaceS, which is simplya set. This set is sometimes thought of as the collection of all possible outcomesof a random experiment. Certain subsets of the sample space, calledevents, areassigned probabilities by theprobability measure (or probability set function)which is usually denoted byP. This function has a few defining properties.

(1) For any eventE ⊂ S, 0 ≤ P[E] ≤ 1.

(2) P[∅ ] = 0 andP[S] = 1.

(3) If E1,E2, . . .are events andEi ∩ Ej = ∅ for i≠ j thenP[⋃∞

i=1 Ei] =∑∞

i=1 P[Ei].

From these basic facts one can deduce all manner of useful computational formulas.

Exercise 5–1.Show that ifA ⊂ B are events, thenP[A] ≤ P[B].

Another of the basic concepts is that of a random variable. Arandom variableis a function whose domain is the samplespace of a random experiment and whoserange is the real numbers.

In practice, the sample space of the experiment fades into the background andone simply identifies the random variables of interest. Once a random variable hasbeen identified, one may ask about its valuesand their associated probabilities. Allof the interesting probability information is bound up in the distribution function ofthe random variable. Thedistribution function of the random variableX, denotedFX(t), is defined by the formulaFX(t) = P[X ≤ t].

Two types of random variables arequite common. A random variableX withdistribution functionFX is discrete if FX is constant except at at most countablymany jumps. A random variableX with distribution functionFX is absolutely

continuous if FX(t) =∫ t

−∞

dds

FX(s) ds holds for all real numberst.

If X is a discrete random variable, thedensity of X, denotedfX(t) is defined bythe formulafX(t) = P[X = t]. There are only countably many values oft for whichthe density of a discrete random variable is not 0. IfX is an absolutely continuous

random variable, thedensity of X is defined by the formulafX(t) =ddt

FX(t).


§5: Brief Review of Probability Theory 18

Example 5–1. A Bernoulli random variable is a random variable which takes onexactly two values, 0 and 1. Such random variables commonly arise to indicate thesuccess or failure of some operation. A Bernoulli random variable is discrete.

Exercise 5–2.Sketch the distribution function of a Bernoulli random variable withP[X = 1] = 1/3.

Example 5–2. An exponentially distributed random variable Y with parameterλ > 0 is a non-negative random variable for whichP[Y ≥ t] = e−λ t for t ≥ 0. Such arandom variable is often used to model the waiting time until a certain event occurs.An exponential random variableis absolutely continuous.

Exercise 5–3.Sketch the distribution function of an exponential random variablewith parameterλ = 1. Sketch its density function also.

Exercise 5–4.A random variableX isuniformly distributed on an interval (a,b) ifX represents the result of selecting a number at random from (a,b). Find the densityand distribution function of a random variable which is uniformly distributed on theinterval (0,1).

Exercise 5–5.Draw a picture of a distribution function of a random variable whichis neither discrete nor absolutely continuous.

Another useful tool is the indicator function. SupposeA is a set. Theindicatorfunction of the setA, denoted1A(t), is defined by the equation

1A(t) ={

1 if t ∈ A0 if t ∉ A.

Exercise 5–6.Graph the function1[0,1)(t).

Exercise 5–7.Verify that the density of a random variable which is exponentialwith parameterλ may be writtenλ e−λ x1(0,∞)(x).

Example 5–3. Random variables which are neither of the discrete nor absolutelycontinuous type will arise frequently. As an example, suppose that person has a fireinsurance policy on a house. The amount ofinsurance is $50,000 and there is a $250deductible. Suppose thatif there is a fire the amount of damage may be representedby a random variableD which has the uniform distribution on the interval (0,70000).(This assumption means that the person is underinsured.) Suppose further that in thetime period under consideration there is a probabilityp = 0.001 that a fire will occur.Let F denote the random variable which is 1 if a fire occurs and is 0 otherwise. It iseasy to see that the sizeX of the claim to the insurer in this setting is given by

X = F[(D − 250)1[250,50250](D) + 500001(50250,∞)(D)

].


This random variableX is neither discrete nor absolutely continuous.

Exercise 5–8.Verify the correctness of the formula forX. Find the distributionfunction of the random variableX.

Often only the average value of a random variable and the spread of thevalues around this average are all that are needed. Theexpectation (or mean)of a discrete random variableX is defined byE[X] =

∑t

t fX(t), while the ex-

pectation of an absolutely continuous random variableX is defined byE[X] =∫ ∞

−∞t fX(t) dt. Notice that in both cases the sum (or integral) involves terms of the

form (possible value ofX) × (probabilityX takes on that value). WhenX is neitherdiscrete nor absolutely continuous, theexpectation is defined by the Riemann–Stieltjes integralE[X] =

∫ ∞−∞ t dFX(t), which again has the same form.

Exercise 5–9.Find the mean of a Bernoulli random variableZ with P[Z = 1] = 1/3.

Exercise 5–10.Find the mean of an exponential random variable with parameterλ = 3.

Exercise 5–11.Find the mean and variance of the random variable in the fireinsurance example given above. (Thevariance of a random variableX is definedby Var(X) = E[(X − E[X])2] and is often computed using the alternate formulaVar(X) = E[X2] − (E[X])2.)

Computation of conditional probabilities will play an important role. IfA andB are events, theconditional probability of A givenB, denotedP[A|B], is definedby P[A|B] = P[A ∩ B]/P[B] as long asP[B] ≠ 0.

The eventsA andB areindependentif P[A|B] = P[A]. The intuition underlyingthe notion of independent events is that the occurance of one of the events does notalter theprobability that the other event occurs.

Similar definitions can be given in the case of random variables. Intuitively,the random variablesX andY are independent if knowledge of the value of one ofthem does not effect theprobabilities of events involving the other. Independenceof random variables is usually assumed based on this intuition. One important factis that ifX andY are independent randomvariables thenE[XY] = E[X] E[Y].


Problems

Problem 5–1. SupposeX has the uniform distribution on the interval (0,a) wherea > 0 is given. What is the mean and variance ofX?

Problem 5–2. Themoment generating functionof a random variableX, denotedMX(t), is defined by the formulaMX(t) = E[etX ]. What is the relationship betweenM′X(0) andE[X]? Find a formula for Var(X) in terms of the moment generatingfunction ofX and its derivatives att = 0.

Problem 5–3. Express the Maclaurin expansion ofMX(t) in terms of the momentsE[X], E[X2], E[X3],. . .of X. Hint: What is the Maclaurin expansion ofex?

Problem 5–4. Find the moment generating function of a Bernoulli random variableY for whichP[Y = 1] = 1/4.

Problem 5–5. Show thatddt

ln MX(t)∣∣∣∣t=0

= E[X] andd2

dt2ln MX(t)

∣∣∣∣∣t=0

= Var(X). This

is useful when the moment generating function has a certain form.

Problem 5–6. Find the moment generating function of a random variableZ whichhas the exponential distribution with parameterλ . Use the moment generatingfunction to find the mean and variance ofZ.

Problem 5–7. A double indemnity life insurance policy has been issued to a personaged 30. This policy pays $100,000 in the event of non-accidental death and$200,000 in the event of accidental death. The probability of death during the nextyear is 0.002, and if death occurs there is a70% chance that it was due to an accident.Write a random variableX which represents the size of the claim filed in the nextyear. Find the distribution function, mean, and variance ofX.

Problem 5–8. In the preceding problem suppose that if death occurs the day of theyear on which it occurs is uniformly distributed. Assume also that the claim will bepaid immediately at death and the interest rate is 5%. What is the expected presentvalue of the size of the claim during the next year?

Problem 5–9. Show that for a random variableY which is discrete and takes non–

negative integer values,E[Y] =∞∑

i=1

P[Y ≥ i]. Find a similar alternate expression for

E[Y2].

Problem 5–10. SupposeY is a non-negative, absolutely continuous random vari-

able. Show thatE[Y] =∫ ∞

0P[Y > t] dt.


Solutions to ProblemsProblem 5–1. E[X] = a/2 and Var(X) = a2/12.

Problem 5–2. M′X(0) = E[X].

Problem 5–3. Sinceex =∞∑

k=0

xk/k!,

MX(t) = E[etX]

= E[∞∑

k=0

(tX)k/k!]

=∞∑

k=0

E[Xk]tk/k!.

Thus the coefficient oftk in the Maclaurin expansion ofMX(t) is E[Xk]/k!.

Problem 5–4. MY (t) = 3/4 + et/4.

Problem 5–6. MZ(t) = λ / (λ − t) for 0 ≤ t < λ .

Problem 5–7. Let D be a random variable which is 1 if the insured dies in thenext year and 0 otherwise. LetA be a random variable which is 2 if death is dueto an accident and 1 otherwise. ThenX = 100000AD.

Problem 5–8. If U is uniformly distributed on the integers from 1 to 365 thenE[100000ADvU] is the desired expectation. Herev = 1/ (1 + 0.05(365)/365).

Problem 5–9. Hint: In the usual formula for the expectation ofY writei =∑i

j=1 1 and then interchange the order of summation.

Problem 5–10. Use a trick like that of the previous problem. Double integralsanyone?


Solutions to ExercisesExercise 5–1. Using the fact thatB = A ∪ (B \ A) and property (3) givesP[B] = P[A] + P[B \ A]. By property (1), P[B \ A] ≥ 0, so the inequalityP[B] ≥ P[A] follows.

Exercise 5–2. The distribution functionF(t) takes the value 0 fort < 0, thevalue 2/3 for 0 ≤ t < 1 and the value 1 fort ≥ 1.

Exercise 5–3. The distribution functionF(t) is 0 if t < 0 and 1− e−t for t ≥ 0.The density function is 0 fort < 0 ande−t for t ≥ 0.

Exercise 5–4. The distribution functionF(t) takes the value 0 fort < 0, thevaluet for 0 ≤ t ≤ 1 and the value 1 fort > 1. The density function takes thevalue 1 for 0< t < 1 and 0 otherwise.

Exercise 5–5. The picture should have a jump and also a smoothly increasingportion.

Exercise 5–6. This function takes the value 1 for 0≤ t < 1 and the value 0otherwise.

Exercise 5–8. The distribution functionF(t) takes the value 0 ift < 0, the value(1−0.001)+0.001×(250/70000) for 0≤ t < 250 (becauseX = 0 if either there isno fire or the loss caused by a fire is less than 250), the value0.999+0.001×t/70000for 250≤ t < 50250 and the value 1 fort ≥ 50250.

Exercise 5–9. E[Z] = 0 × (2/3) + 1× (1/3) = 1/3.

Exercise 5–10. The expectation is∫ ∞

0 tf (t) dt =∫ ∞

0 t3e−3t dt = 1/3 usingintegration by parts.

Exercise 5–11. Notice that the loss random variableX is neither discrete norabsolutely continuous. The distribution function ofX has two jumps: one att = 0 of size 0.999 + 250/70000 and another at 50250 of size 0.001− 0.001×50250/70000. SoE[X] = 0 × (0.999 + 250/70000) +

∫ 50250250 t0.001/70000dt +

50250× (0.001− 0.001× 50250/70000). The quantityE[X2] can be computedsimilarly.

§6. Laboratory 2

One iteration of an experiment is conducted as follows. A single six sided dieis rolled and the numberD of spots up is noted. ThenD coins are tossed and thenumberH of heads observed is noted.

1. What are the possible values of the random variableD? What are the possiblevalues of the random variableH?

2. For each relevant value ofh andd computeP[H = h,D = d]. This probabilityis called thejoint density of H andD and is denotedfH,D(h,d). Put your computedvalues in the form of a rectangular table with rows indexed byh and columnsindexed byd.

3. Find the density and expectation ofD. Find the density and expectation ofH.

4. For each value ofd find the conditional density ofH givenD = d. Notation-ally, fH |D(h|d) = P[H = h|D = d].

5. For each value ofd find the conditional expectation ofH given D = d.Notationally,E[H |D = d] =

∑h

h fH |D(h|d).

6. The conditional expectation ofH givenD, denotedE[H |D], is the randomvariable whose value on the eventD = d is E[H |D = d]. Find the density of therandom variableE[H |D].

7. ComputeE[E[H |D]]. The Theorem of Total Expectation states that forany two random variablesX andY, E[E[Y |X]] = E[Y]. Do your computations ofE[H] in this problem and problem 3 reflect this fact?

8. The computations above show that it is easier to computeE[H] by firstfindingE[H |D]. Sometimes probabilistic intuition can be used to find a conditionalexpectation without computation. The intuition behind the conditional expectationE[H |D] is that this should be the expected value ofH computed after taking thevalue ofD into account. Give a probabilistic argument (without computation) thatE[H |D] = D/2.


§7. Survival Distributions

An insurance policy can embody two different types of risk. For some typesof insurance (such as life insurance) the variability in the claim is only the time atwhich the claim is made, since the amount of the claim is specified by the policy.In other types of insurance (such as auto or casualty) there is variability in both thetime and amount of the claim. The problems associated with life insurance will bestudied first, since this is both an important type of insurance and also relativelysimple in some of its aspects.

The central difficulty in issuing life insurance is that of determining the lengthof the future life of the insured. Denote byX the random variable which representsthe future lifetime of a newborn. For mathematical simplicity, assume that thedistribution function ofX is absolutely continuous. Thesurvival function of X,denoted bys(x) is defined by the formula

s(x) = P[X > x] = P[X ≥ x]

where the last equality follows from the continuity assumption. The assumptionthats(0) = 1 will always be made.

Example 7–1. In the past there has been some interest in modelling survival func-tions in an analytic way. The simplest model is that due to Abraham DeMoivre. He

assumed thats(x) = 1 −xω

for 0 < x < ω whereω is the limiting age by which

all have died. TheDeMoivre law is simply the assertion thatX has the uniformdistribution on the interval (0, ω).

Life insurance is usually issued on a person who has already attained a certainagex. For notational convenience denote such alife agedx by (x), and denote thefuture lifetime of a life agedx by T(x). What is the survival function for (x)? Fromthe discussion above, the survival function for (x) is P[T(x) > t]. Some standardnotation is now introduced. Set

tpx = P[T(x) > t]

and

tqx = P[T(x) ≤ t].

When t = 1 the prefix is ommitted and one just writespx and qx respectively.Generally speaking, having observed (x) some additional information about thesurvival of (x) can be inferred. For example, (x) may have just passed a physicalexam given as a requirement for obtaining life insurance. For now this type ofpossibility is disregarded. Operating under this assumption

tpx = P[T(x) > t] = P[X > x + t |X > x] =s(x + t)

s(x).


§7: Survival Distributions 25

Exercise 7–1.Write a similar expression fortqx.

Exercise 7–2.Show that fort ≥ s, tpx = t−spx+s spx.

There is one more special symbol. Set

t |uqx = P[t < T(x) ≤ t + u]

which represents the probability that (x) survives at leastt and no more thant + uyears. Again, ifu = 1 one writest |qx. The relationst |uqx = t+uqx − tqx = tpx − t+upx

follow immediately from the definition.

Exercise 7–3.Prove these two equalitites. Show thatt |uqx = tpx uqx+t.

Exercise 7–4.Computetpx for the DeMoivre law of mortality. Conclude that underthe DeMoivre lawT(x) has the uniform distribution on the interval (0, ω − x).

Under the assumption thatX is absolutely continuous the random variableT(x)will be absolutely continuous as well. Indeed

P[T(x) ≤ t] = P[x ≤ X ≤ x + t |X > x] = 1 −s(x + t)

s(x)

so the density ofT(x) is given by

fT(x)(t) =−s′ (x + t)

s(x)=

fX(x + t)1− FX(x)

.

Intuitively this density represents the rate of death of (x) at timet.

Exercise 7–5.Use integration by parts to show thatE[T(x)] =∫ ∞

0tpx dt. This

expectation is called thecomplete expectation of lifeand is denoted by ˚ex. Show

also thatE[T(x)2] = 2∫ ∞

0t tpx dt.

Exercise 7–6.If X follows DeMoivre’s law, what is ˚ex?

It is often useful to consider then the quantity

µx =fX(x)

1− FX(x)= −

s′ (x)s(x)

which is called theforce of mortality . Intuitively the force of mortality is theinstantaneous rate of death of (x). (In component reliability theory this function isoften referred to as thehazard rate.) Integrating both sides of this equality gives theuseful relation

s(x) = exp{−∫ x

0µt dt

}.


Exercise 7–7.Derive this last expression.

Exercise 7–8.Show thattpx = e−∫ x+t

xµs ds.

Exercise 7–9.Show that the density ofT(x) can be writtenfT(x)(t) = tpx µx+t.

If the force of mortality is constant the life random variableX has an expo-nential distribution. This is directly in accord with the “memoryless” property ofexponential random variables.This memoryless property also has the interpretationthat a used article is as good as a new one. For human lives (and most manufacturedcomponents) this is a fairly poor assumption, at least over the long term. The forceof mortality usually is increasing, although this is not always so.

Exercise 7–10.Find the force of mortality for DeMoivre’s law.

Thecurtate future lifetime of (x), denoted byK(x), is defined by the relationK(x) = [T(x)]. Here [t] is the greatest integer function. Note thatK(x) is a discreterandom variable with densityP[K(x) = k] = P[k ≤ T(x) < k + 1]. The curtatelifetime, K(x), represents the number of complete future years lived by (x).

Exercise 7–11.Show thatP[K(x) = k] = kpx qx+k.

Exercise 7–12.Show that thecurtate expectation of lifeex = E[K(x)] is given bythe formulaex =

∑∞i=0 i+1px. Hint: E[Y] =

∑∞i=1 P[Y ≥ i].


Problems

Problem 7–1. Supposeµx+t = t for t ≥ 0. Calculatetpx µx+t andex.

Problem 7–2. Calculate∂∂x tpx and

ddx

ex.

Problem 7–3. A life aged (40) is subject to an extra risk for the next year only.Suppose the normal probability of death is 0.004, and that the extra risk may beexpressed by adding the function 0.03(1− t) to the normal force of mortality for thisyear. What is the probability of survival to age 41?

Problem 7–4. Supposeqx is computed using force of mortalityµx, and thatq′x iscomputed using force of mortality 2µx. What is the relationship betweenqx andq′x?

Problem 7–5. Show that the conditional distribution ofK(x) given thatK(x) ≥ k isthe same as the unconditional distribution ofK(x + k) + k.

Problem 7–6. Show that the conditional distribution ofT(x) given thatT(x) ≥ t isthe same as the unconditional distribution ofT(x + t) + t.

Problem 7–7. TheGompertz law of mortality is defined by the requirement thatµt = Act for some constantsA andc. What restrictions are there onA andc for thisto be a force of mortality? Write an expression fortpx andex under Gompertz’ law.

Problem 7–8. Makeham’s lawof mortality is defined by the requirement thatµt = A + Bct for some constantsA, B, andc. What restrictions are there onA, Bandc for this to be a force of mortality? Write an expression fortpx and ex underMakeham’s law.


Solutions to Problems

Problem 7–1. Heretpx = e−∫ t

0µx+s ds = e−t2/2 andex =

∫ ∞

0tpx dt = √2π/2.

Problem 7–2.∂∂x tpx = tpx(µx − µx+t) and

ddx

ex =∫ ∞

0

∂∂x tpx dt = µxex − 1.

Problem 7–3. If µt is the usual force of mortality thenp40 = e−∫ 1

0µ40+s+0.03(1−s) ds.

Problem 7–4. The relationp′x = (px)2 holds, which gives a relation for thedeath probability.

Problem 7–5. P[K(x) ≤ k + l|K(x) ≥ k] = P[k ≤ K(x) ≤ k + l]/P[K(x) ≥ k] =lqx+k = P[K(x + k) + k ≤ l + k].

Problem 7–6. Proceed as in the previous problem.


Solutions to ExercisesExercise 7–1. tqx = P[T(x) ≤ t] = P[X ≤ x + t |X > x] = P[x < X ≤ x + t]/P[X >x] = (s(x) − s(x + t))/s(x).

Exercise 7–2. tpx = s(x + t)/s(x) = s(x + s + (t − s))/s(x) = (s(x + s + (t − s))/s(x +s))(s(x + s)/s(x)) = t−spx+sspx. What does this mean in words?

Exercise 7–3. For the first one,t |uqx = P[t < T(x) ≤ t + u] = P[x + t < X ≤t+u+x|X > x] = (s(x+t)−s(t+u+x))/s(x) = (s(x+t)−s(x)+s(x)−s(t+u+x))/s(x) =t+uqx − tqx. The second identity follows from the fourth term by simplifying(s(x + t) − s(t + u + x))/s(x) = tpx − t+upx. For the last one,t |uqx = P[t < T(x) ≤t + u] = P[x + t < X ≤ t + u + x|X > x] = (s(x + t) − s(t + u + x))/s(x) =(s(x + t)/s(x))(s(t + x) − s(t + u + x))/s(x + t) = tpx uqx+t.

Exercise 7–4. Under the DeMoivre law,s(x) = (ω − x)/ω so that tpx =(ω − x − t)/ (ω − x) for 0 < t < ω − x. Thus the distribution function ofT(x) ist/ (ω − x) for 0 < t < ω − x, which is the distribution function of a uniformlydistributed random variable.

Exercise 7–5. E[T(x)] =∫ ∞

0 tfT(x)(t) dt = −∫ ∞

0 ts′ (x + t)/s(x) dt =∫ ∞

0 s(x +t)/s(x) dt =

∫ ∞0 tpx dt. The fact the limt→∞ s(x + t) = 0 is assumed, since everyone

eventually dies. The other expectation is computed similarly.

Exercise 7–6. Under DeMoivre’s law, ˚ex = (ω− x)/2, sinceT(x) is uniform onthe interval (0, ω− x).

Exercise 7–7.∫ x

0 µt dt =∫ x

0 −s′ (t)/s(t) dt = − ln(s(x)) + ln(s(0)) = − ln(s(x)),sinces(0) = 1. Exponentiating both sides to solve fors(x) gives the result.

Exercise 7–8. From the previous exercise,s(x + t) = e−∫ x+t

0µs ds. Using this

fact, the previous exercise, and the fact thattpx = s(x + t)/s(x) gives the formula.

Exercise 7–9. SincefT(x)(t) = −s′ (x + t)/s(x) ands′ (x + t) = −s(x + t)µx+t by theprevious exercise, the result follows.

Exercise 7–10. From the earlier expression for the survival function underDeMoivre’s law s(x) = (ω − x)/ω, so thatµx = −s′ (x)/s(x) = 1/ (ω − x), for0 < x < ω.

Exercise 7–11. P[K(x) = k] = P[k ≤ T(x) < k+1] = (s(x+k)−s(x+k+1))/s(x) =((s(x + k) − s(x + k + 1))/s(x + k))(s(x + k)/s(x)) = kpx qx+k.

Exercise 7–12. E[K(x)] =∑∞

i=1 P[K(x) ≥ i] =∑∞

i=1 P[T(x) ≥ i] =∑∞

i=1 ipx =∑∞j=0 j+1px.

§8. Life Tables

In practice the survival distribution is estimated by compiling mortality data inthe form of alife table. An example of a life table appears at the end of these notes.Here is the conceptual model behind the entries in the table. Imagine that at time0 there arel0 newborns. Herel0 is called theradix of the life table and is usuallytaken to be some large number such as 100,000. Denote bylx the number of theseoriginal newborns who are still alive at agex. Similarly ndx denotes the number ofpersons alive at agex who die before reaching agex + n. As usual, whenn = 1 it issupressed in the notation.

Exercise 8–1.Show thatndx = lx − lx+n.

The ratiolx

l0is an estimate ofs(x) based on the collected data.Assume that in

fact s(x) =lx

l0for non-negative integer values ofx. Since earlier the assumption

was made that the life random variableX is absolutely continuous, the questionarises as to how the values of the survival function will be computed at non-integervalues ofx. There are three commonly used methods of doing this, and thesemethods produce slightly different numerical results. For the remainder of thisdiscussionsupposex is fixed (and an integer) and that0 ≤ t ≤ 1.

Under the assumption of theuniform distribution of deaths in the year ofdeath, denoted UDD, the survival function is computed by the formula

s(x + t) = (1− t)s(x) + ts(x + 1).

The UDD assumption is the one most commonly made.

The assumption of aconstant force of mortality in each year of age leads tothe formula

s(x + t) = s(x) e−µt

whereµ = − log px.

TheBalducci assumptionis expressed in the formula

1s(x + t)

=1− ts(x)

+t

s(x + 1).

Under each of the assumptions an explicit expression for all of the survivorfunctions can be found.

Exercise 8–2.Find expressions fortqx andµx+t, 0 ≤ t ≤ 1, under each of the above3 assumptions.


§8: Life Tables 31

Having observed (x) may mean more than simply having seen a person agedx.It may well mean that (x) has just passed a physical exam in preparation for buyinga life insurance policy. One would expect that the survival distribution of such aperson could be different froms(x). If this is believed to be the case the survivalfunction is actually dependenton two variables: the age atselection(applicationfor insurance) and the amount of time passed after the time of selection. A lifetable which takes this effect into account is called aselect table. A family ofsurvival functions indexed by both the age at selection and time are then requiredand notation such asq[x]+i denotes the probability that a person dies between yearsx + i andx + i + 1 given that selection ocurred at agex. As one might expect itis reasonable to suppose that after a certainperiod of time the effect of selectionon mortality is negligable. The length of time until the selection effect becomesnegligable is called theselect period. The Society of Actuaries (based in Illinois)uses a 15 year select period in its mortality tables. The Institute of Actuaries inBritain uses a 2 year select period. The implication of the select period of 15 yearsin computations is that for eachj ≥ 0, l[x]+15+j = lx+15+j.

A life table in which the survival functions are tabulated for attained ages onlyis called anaggregrate table. Generally, a select life table contains a final columnwhich constitutes an aggregate table. The whole table is then referred to as aselectand ultimate table and the last column is usually called anultimate table. Withthese observations in mind it is easy to utilize select life tables in computations.

Exercise 8–3.You are given the following extract from a 3 year select and ultimatemortality table.

x l[x] l[x]+1 l[x]+2 lx+3 x + 370 7600 7371 7984 7472 8016 7592 75

Assume that the ultimate table follows DeMoivre’s law and thatd[x] = d[x]+1 =d[x]+2 for all x. Find 1000(2|2q[71]).

§8: Life Tables 32

Problems

Problem 8–1. Graphµx+t, 0 ≤ t ≤ 1, under each of the 3 assumptions for fractionalyears.

Problem 8–2. For each of the 3 assumptions for fractional years find a formula for

tpx, 0 ≤ t ≤ 1 in terms oft andpx. For each of 20 equally space values ofpx between0 and 1, make a plot oftpx for 0 ≤ t ≤ 1 under each of the 3 assumptions. For whatvalue(s) ofpx are the assumptions numerically indistinguishable?

Problem 8–3. Use the life table to compute1/2p20 under each of the 3 assumptionsfor fractional years.

Problem 8–4. Show that under the assumption of uniform distribution of deaths inthe year of death thatK(x) andT(x) −K(x) are independent and thatT(x) −K(x) hasthe uniform distribution on the interval (0,1).

Problem 8–5. Show that under UDD ˚ex = ex + 12.

§8: Life Tables 33

Solutions to ProblemsProblem 8–3. Under UDD, tpx = (1 − t) + tpx so 1/2p20 = 1/2 + 1/2p20.Under constant force,tpx = et logpx so 1/2p20 = e(1/2) logp20. Under Balducci,1/ tpx = 1− t + t/px so1/2p20 = 1/ (1/2 + 1/2p20).

Problem 8–4. For 0 ≤ t < 1, P[K(x) = k,T(x) − K(x) ≤ t] = P[k ≤ T(x) ≤k + t] = kpx tqx+k = kpx(t − tpx+k) = tP[K(x) = k].

Problem 8–5. Use the previous problem.

§8: Life Tables 34

Solutions to ExercisesExercise 8–1. Sincendx is the number alive at agex who die by agex + n, thisis simply the number alive at agex, which is lx, minus the number alive at agex + n, which islx+n.

Exercise 8–2. Under UDD,tqx = (s(x)−s(x + t))/s(x) = (ts(x)− ts(x +1))/s(x) =tqx andµx+t = −s′ (x+t)/s(x+t) = (s(x)−s(x+1))/ ((1−t)s(x)+ts(x+1)) = (1−px)/ (1−t + tpx). Under constant force,tqx = 1− (px)t while µx+t = µx. Under Balducci,s(x + t) = s(x)s(x + 1)/ ((1− t)s(x + 1) + ts(x)) so thats(x + t)/s(x) = px/ (px + tqx).Thustqx = tqx/ (px + tqx). Alsoµx+t = −s′ (x + t)/s(x + t) = (1− qx)/ (1 + tqx).

Exercise 8–3. The objective is to compute 10002|2q[71] = 1000(2p[71]−4p[71]) =1000(l[71]+2− l[71]+4)/ l[71] = 1000(l[71]+2− l75)/ l[71], where the effect of the selec-tion period has been used. To find the required entries in the table proceed asfollows. Since 8016−7592 = 424 and using the assumption about the number ofdeaths,l[72]+1 = 8016−212 = 7804 andl72+3 = 7592−212 = 7380. Since the ulti-mate table follows DeMoivre’s Law,l71+3 = (7600 + 7380)/2 = 7490. Again us-ing the assumption about the number of deaths,l[71]+2 = (7984+7490)/2 = 7737andl[71] = 7984 + 247 = 8231. So 10002|2q[71] = 1000(7737− 7380)/8231 =43.37.

§9. Laboratory 3

1. Below is a table which gives the values ofqx for ages 1 through 105. Definel1 = 1,000,000 and use the values ofqx to computelx anddx for 1 ≤ x ≤ 106.

2. Make a plot oflx for 1 ≤ x ≤ 106.

x qx x qx x qx

1 0.000637 36 0.000841 71 0.0266272 0.000430 37 0.000904 72 0.0295653 0.000357 38 0.000964 73 0.0329314 0.000278 39 0.001021 74 0.0367385 0.000255 40 0.001079 75 0.0410026 0.000244 41 0.001142 76 0.0456997 0.000234 42 0.001215 77 0.0508338 0.000216 43 0.001299 78 0.0564879 0.000209 44 0.001397 79 0.06277710 0.000212 45 0.001508 80 0.06975711 0.000219 46 0.001629 81 0.07744412 0.000228 47 0.001762 82 0.08582813 0.000240 48 0.001905 83 0.09490414 0.000254 49 0.002060 84 0.10470015 0.000269 50 0.002225 85 0.11528916 0.000284 51 0.002401 86 0.12679817 0.000301 52 0.002589 87 0.13935318 0.000316 53 0.002795 88 0.15302119 0.000331 54 0.003023 89 0.16775720 0.000345 55 0.003283 90 0.18340821 0.000357 56 0.003583 91 0.19976922 0.000366 57 0.003932 92 0.21660523 0.000373 58 0.004332 93 0.23366224 0.000376 59 0.004784 94 0.25069325 0.000376 60 0.005286 95 0.26749126 0.000378 61 0.005833 96 0.28390527 0.000382 62 0.006414 97 0.29985228 0.000393 63 0.007014 98 0.31529629 0.000412 64 0.007616 99 0.33020730 0.000444 65 0.008207 100 0.34455631 0.000499 66 0.008777 101 0.35862832 0.000562 67 0.009318 102 0.37168533 0.000631 68 0.009828 103 0.38304034 0.000702 69 0.010306 104 0.39200335 0.000773 70 0.010753 105 1.000000


§10. Status

A life insurance policy is sometimes issued which pays a benefit at a timewhich depends on the survival characteristics of two or more people. Astatusis anartificially constructed life form for which the notion of life and death can be welldefined.

Example 10–1.A common artificial life form is the status which is denotedn . Thisis the life form which survives for exactlyn time units and then dies.

Example 10–2.Another common status is thejoint life status which is constructedas follows. Given two life forms (x) and (y) the joint life status, denotedx : y, diesexactly at the time of death of the first to die of (x) and (y).

Exercise 10–1.If (x) and (y) are independent lives, what is the survival function ofthe statusx : y?

Exercise 10–2.What is survival function ofx : n ?

Occasionally, even the order in which death occurs is important. The status

x1 : n is a status which dies at the time of death of (x) if the death of (x) occurs beforetimen. Otherwise, this status never dies.

Exercise 10–3.Under what circumstances doesx : n1

die?


§10: Status 37

Problems

Problem 10–1. Find a formula for the survival function ofx1 : n in terms of thesurvival function of (x).

Problem 10–2. If the UDD assumption is valid for (x), does UDD hold forx1 : n?

Problem 10–3. Find a formula for the survival function ofx : n1

.

Problem 10–4. If the UDD assumption is valid for (x), does UDD hold forx : n1

?

Problem 10–5. If the UDD assumption is valid for (x), does UDD hold forx : n?

Problem 10–6. If the UDD assumption is valid for each of (x) and (y) and if (x)and (y) are independent lives, does UDD hold forx : y?

§10: Status 38


Problem 10–1. P[T(x1 : n ) ≥ t] = tpx for 0 ≤ t < n andP[T(x1 : n ) ≥ t] = npx

for t ≥ n.

Problem 10–2. The UDD assuption holds forx1 : n if and only if P[T(x1 : n ) ≥k + t] = (1 − t)P[T(x1 : n ) ≥ k] + tP[T(x1 : n ) ≥ k + 1] for all integersk and all0 ≤ t ≤ 1. Now use the formula for the survival function found in the previousproblem.

§10: Status 39

Solutions to ExercisesExercise 10–1. The joint life status survivest time units if and only if both (x)and (y) survivet time units. Using the independence givess(t) = tpx tpy.

Exercise 10–2. Since a constant random variable is independent of any otherrandom variable,s(t) = tpx tpn = tpx if t ≤ n and 0 ift > n, by using the previousexercise.

Exercise 10–3. The statusx : n1

dies at timen if (x) is still alive at timen,otherwise this status never dies.

§11. Valuing Contingent Payments

Earlier, the central theme of thesenotes was asserted to be embodied in thequestion, “What is the value today of arandom sum of money which will be paidat a random time in the future?” This question can now be answered. Suppose therandom amount of money is denoted byA and the random time at which it will be paidis denoted byT. The value of this payment today is computed in two steps. First, anexpression for the present value of the payment is written. This expression will be arandom variable. Here the present value isAvT . Then the expectation of this randomvariable is computed. This expectation,E[AvT ], is the value today of the randomfuture payment. The interpretation of this amount,E[AvT ], is as theaverage presentvalue of the actual payment. Averages arereasonable in the insurance context since,from the company’s point of view, there are many probabilistically similar policiesfor which the company is obliged to pay benefits. The average cost (and income) perpolicy is therefore a reasonable startingpoint from which to determine the premium.

The expected present value is usually referred to as theactuarial present valuein the insurance context. In the next few sections the actuarial present value ofcertain standard parts of insurance contracts are computed.


§12. Life Insurance

In the case of life insurance the determination of the value of the insurancedepends on the random time of death of the insured. The amount that is paid at thetime of death is usually fixed by the policy and is non-random. Assume that theforce of interest is constant and known to be equal toδ. Also simply writeT = T(x)whenever clarity does not demand the fullnotation. The actuarial present value ofan insurance which pays 1 at the time of death is then

E[vT ]

by the priciple above. Intuitively, the actuarial present value of the benefit is thesingle premium payment that an insurance company with no operating expensesand no desire for profit would charge today in order to provide the benefit payment.For this reason the actuarial present value of a benefit is also called thenet singlepremium. The net single premium would be the idealized amount an insured wouldpay as a lump sum (single premium) at the time that the policy is issued. The caseof periodic premium payments will be discussed later.

A catalog of the various standard types of life insurance policies and the standardnotation for the associated net single premium follows. In most cases the benefitamount is assumed to be $1, and in all cases the benefit is assumed to be paid at thetime of death. Keep in mind that a fixed constant force of interest is also assumedand thatv = 1/ (1 + i) = e−δ.

Insurances Payable at the Time of Death

Type Net Single Premium

n-year pure endowment Ax:n

1 = nEx = E[vn1(n,∞)(T)]

n-year term Ax1:n

= E[vT1[0,n] (T)]

whole life Ax = E[vT ]

n-year endowment Ax:n = E[vT ∧ n]

m-year deferredn-year term m|nAx = E[vT1(m,n+m](T)]

whole life increasingmthly (IA)(m)x = E[vT [Tm + 1]/m]

n-year term increasing annually (IA)x1:n

= E[vT [T + 1]1[0,n)(T)]

n-year term decreasing annually (DA)x1:n

= E[vT (n − [T])1[0,n)(T)]

Using this table it is a simple matter to compute the net single premium in thevarious cases. The bar is indicative of an insurance paid at the time of death, whilethe subscripts denote the status whose death causes the insurance to be paid. Theseinsurances are now reviewed on a case-by-case basis.


§12: Life Insurance 42

The first type of insurance isn-year pure endowment insurancewhich paysthe full benefit amount at the end of thenth year if the insured survives at leastnyears. The notation for the net single premium for a benefit amount of 1 isA

x:n1 (or

occasionally in this contextnEx). The net single premium for a pure endowment isjust theactuarial present value of a lump sum payment made at a future date. Thisdiffers from the ordinary present value simply because it also takes into account themortality characteristics of the recipient.

Exercise 12–1.Show thatnEx = vnnpx.

The second type of insurance isn-year term insurance. The net single premiumwith a benefit of 1 payable at the time of death for an insured (x) is denotedA

x1:n.

This type insurance provides for a benefit payment only if the insured dies withinnyears of policy inception.

The third type of insurance iswhole life in which the full benefit is paid nomatter when the insured dies in the future. The whole life benefit can be obtainedby taking the limit asn → ∞ in then-year term insurance setting. The notation forthe net single premium for a benefit of 1 isAx.

Exercise 12–2.Suppose thatT(x) has an exponential distribution with mean 50. Ifthe force of interest is 5%, find the net single premium for a whole life policy for(x), if the benefit of $1000 is payable at the moment of death.

Exercise 12–3.Show thatAx = Ax1:n

+ vnnpxAx+n by conditioning on the event

T(x) ≥ n and also by direct reasoning from a time diagram by looking at thedifference of two policies.

The fourth type of insurance,n-year endowment insurance, provides for thepayment of the full benefit at the time of death of the insured if this occurs beforetime n and for the payment of the full benefit at timen otherwise. The net singlepremium for a benefit of 1 is denotedAx:n .

Exercise 12–4.Show thatAx:n = Ax1:n

+ Ax:n

1 .

Exercise 12–5.Use the life table to find the net single premium for a 5 year pureendowment policy for (30) assuming an interest rate of 5%.

Them-year deferredn-year term insurancepolicy provides provides the samebenefits asn year term insurance between timesm andm + n provided the insuredlivesm years.

All of the insurances discussed thus far have a fixed constant benefit.Increasingwhole life insurance provides a benefit which increase linearly in time. Similarly,


increasinganddecreasingn-year term insurance provides for linearly increasing(decreasing) benefit over the term of the insurance.

Corresponding to the insurances payable at the time of death are the same typeof policies available with the benefit being paid at the end of the year of death. Theonly difference between these insurancesand those already described is that theseinsurances depend on the distribution of the curtate life variableK = K(x) insteadof T. The following table introduces the notation.

Insurances Payable the End of the Year of Death

Type Net Single Premium

n-year term Ax1:n

= E[vK+11[0,n)(K)]

whole life Ax = E[vK+1]

n-year endowment Ax:n = E[v(K+1)∧ n]

m-year deferredn-year term m|nAx = E[vK+11[m,n+m)(K)]

whole life increasing annually (IA)x = E[vK+1(K + 1)]

n-year term increasing annually (IA)x1:n

= E[vK+1(K + 1)1[0,n)(K)]

n-year term decreasing annually (DA)x1:n

= E[vK+1(n −K)1[0,n)(K)]

These policies have net single premiums which can be easily computed fromthe information in the life table. The primary use for these types of policies is thecomputational connection between themand the ‘continuous’ policies describedabove. To illustrate the ease of computation when using a life table observe thatfrom the definition

Ax =∞∑

k=0

vk+1kpx qx+k =

∞∑k=0

vk+1dx+k

lx.

In practice, of course, the sum is finite. Similar computational formulas are readilyobtained in the other cases.

Exercise 12–6.Show thatAx:n

1 = Ax:n

1 and interpret the result verbally. How wouldyou computeA

x:n1 using the life table?

Under the UDD assumption it is fairly easy to find formulas which relate theinsurances payable at the time of death to the corresponding insurance payable at


the end of the year of death. For example, in the case of a whole life policy

Ax = E[e−δT(x)]

= E[e−δ(T(x)−K(x)+K(x)) ]

= E[e−δ(T(x)−K(x)) ] E[e−δK(x)]

=1δ

(1− e−δ)eδ E[e−δ(K(x)+1)]

=iδ

Ax

where the third equality springs from the independence ofK(x) and T(x) − K(x)under UDD, and the fourth equality comes from the fact that under UDD therandom variableT(x) −K(x) has the uniform distribution on the interval (0,1).

Exercise 12–7.Can similar relationships be established for term and endowmentpolicies?

Exercise 12–8.Use the life table to find the net single premium for a 5 yearendowment policy for (30) assuming an interest rate of 5%.

Exercise 12–9.An insurance which pays a benefit amount of 1 at the end of themth part of the year in which death occurs has net single premium denoted byA(m)

x .Show that under UDDi(m) A(m)

x = δ Ax.

One consequence of the exercise above isthat only the net single premiums forinsurances payable at the end of the year of death need to be tabulated, if the UDDassumption is made. This leads to a certain amount of computational simplicity.


Problems

Problem 12–1. Write expressions for all of the net single premiums in terms ofeither integrals or sums. Hint: Recall the form of the density ofT(x) andK(x).

Problem 12–2. Show thatδAx1:n

= iAx1:n

, but thatδAx:n ≠ iAx:n , in general.

Problem 12–3. Use the life table and UDD assumption (if necessary) to computeA21, A21:5 , andA

211

:5.

Problem 12–4. Show thatdAx

di= −v(IA)x.

Problem 12–5. Assume that DeMoivre’s law holds withω = 100 andi = 0.10.Find A30 andA30. Which is larger? Why?

Problem 12–6. Supposeµx+t = µ and i = 0.10. ComputeAx andAx1:n

. Do youranswers depend onx? Why?

Problem 12–7. SupposeAx = 0.25,Ax+20 = 0.40, andAx:20 = 0.55. ComputeAx:20

1

andAx1:20

.

Problem 12–8. Show that

(IA)x = vqx + v[Ax+1 + (IA)x+1]px.

What assumptions (if any) did you make?

Problem 12–9. What change inAx results if for some fixedn the quantityqx+n isreplaced withqx+n + c?

Problem 12–10.What is the change inAx if δ is replaced byδ +c? If µ is replacedby µ + c?


Solutions to ProblemsProblem 12–1. The densities required arefT(x)(t) = tpx µx+t and fK(x)(k) =kpx qx+k respectively.

Problem 12–2. δAx1:n

= δ(Ax − e−δnnpxAx+n) = iAx − ivn

npxAx+n = iAx1:n

.

Problem 12–3. UseδA21 = iA21, A21:5 = A211

:5+5E21 and the previous problem.

Problem 12–4. Just differentiate under the expectation in the definition ofAx.

Problem 12–5. Clearly A30 > A30 since the insurance is paid sooner in thecontinuous case. Under DeMoivre’s law the UDD assumption is automatic andA30 = 1

70

∫ 700 e−δt dt.

Problem 12–6. The answers do not depend onx since the lifetime is exponentialand therefore ageless.

Problem 12–7. The two relationsAx = Ax1:20

+ v20npxAx+20 andAx:20 = A

x1:20

+vn

npx along with the fact thatAx:20

1 = vnnpx give two equations in the two sought

after unknowns.

Problem 12–8. Either the person dies in the first year, or doesn’t. If she doesn’tbuy an increasing annually policy for (x + 1) and a whole life policy to make upfor the increasing part the original policy would provide.

Problem 12–9. The new benefit is the old benefit plus a pure endowmentbenefit ofcv at timen.


Solutions to ExercisesExercise 12–1. Since if the benefit is paid, the benefit payment occurs at timen, nEx = E[vn1[n,∞)(T(x))] = vnP[T(x) ≥ n] = vn

npx.

Exercise 12–2. Under the assumptions given the net single premium isE[1000vT(x)] =

∫ ∞0 1000e−0.05t(1/50)e−t/50 dt = 285.71.

Exercise 12–3. For the conditioning argument, break the expectation into twopieces by writingAx = E[vT ] = E[vT1[0,n] (T)]+ E[vT1(n,∞)(T)]. The first expecta-tion is exactlyA

x1:n. For the second expectation, using the Theorem of Total Ex-

pectation givesE[vT1(n,∞)(T)] = E[E[vT1(n,∞)(T)|T ≥ n]]. Now the conditionaldistribution ofT given thatT ≥ n is the same as the unconditional distribution ofT(x+n)+n. Using this fact gives the conditional expectation asE[vT1(n,∞)(T)|T ≥n] = E[vT(x+n)+n1(n,∞)(T(x + n) + n)]1(n,∞)(T) = vnAx+n1(n,∞)(T). Taking expecta-tions gives the result. To use the time diagram, imagine that instead of buyinga whole life policy now, the insured pledges to buy ann year term policy now,and if alive aftern years, to buy a whole life policy at timen (at agex + n). Thiswill produce the same result. The premium for the term policy paid now isA

x1:n

and the premium for the whole life policy at timen is Ax+n. This latter premiumis only paid if the insured survives, so the present value of this premium is thesecond term in the solution.

Exercise 12–4. Using the definition and properties of expectation givesAx:n =E[vT1[0,n] (T) + vn1(n,∞)(T)] = E[vT1[0,n] (T)] + E[vn1(n,∞)(T)] = A

x1:n

+ Ax:n

1 .

Exercise 12–5. The net single premium for the pure endowment policy isv5

5p30 = (1.05)−5l35/ l30 = (1.05)−595808/96477 = 0.778.

Exercise 12–6. Ax:n

1 = E[vn1[n,∞)(T)] = E[vn1[n,∞)(K)] = Ax:n

1 = vnnpx =

vnlx+n/ lx.

Exercise 12–7. Since term policies can be expressed as a difference of premi-ums for whole life policies, the answer is yes.

Exercise 12–8. The net single premium for a pure endowment policy isv5

5p30 = (1.05)−5l35/ l30 = (1.05)−595808/96477 = 0.778. For the endowmentpolicy, the net single premium for a 5 year term policy must be added to thisamount. From the relation given earlier,A

301

:5= A30−v5

5p30A35. The relationship

between insurances payable at the time of death and insurances payable at theend of the year of death is used to complete the calculation.

Exercise 12–9. Notice that [mT(x)] is the number of fullmths of a yearthat (x) lives before dying. (Here [a] is the greatest integer function.) So thenumber ofmths of a year that pass until the benefit for the insurance is paidis [mT(x)] + 1, that is, the benefit is paid at time ([mT(x)] + 1)/m. From herethe derivation proceeds as above.A(m)

x = E[v([mT]+1)/m] = E[v([m(T−K+K)]+1)/m] =E[vK]E[v([m(T−K)]+1)/m]. Now T −K has the uniform distribution on the interval(0,1) under UDD, so [m(T − K)] has the uniform distribution over the integers0,. . ., m − 1. SoE[v([m(T−K)]/m] =

∑m−1j=0 vj/m × (1/m) = (1/m)(1 − v)/ (1 − v1/m)

from the geometric series formula. Substituting this in the earlier expression


givesA(m)x = Axv−1v1/m(1/m)(1−v)/ (1−v1/m) = Axδ/ i(m) sincei(m) = m(v−1/m−1).

§13. Laboratory 4

1. Show thatAx = vqx + vpxAx+1. Derive a similar formula forAx.

2. The one step recursion formulas derived in problem 1 are especially usefulfor computational purposes. The formulas are used to work backwards from largeattained ages to smaller ones, since at large attained ages everyone is dead andthe net premium for the insurance must be zero. Use the values ofqx given inLaboratory 3 andi = 5% to compute the values ofAx andAx for x = 1 to x = 106.Place the result of your computations into a nice table.


§14. Life Annuities

The basic study of life insurance concludes by developing techniques for un-derstanding what happens when premiums are paid monthly or annually instead ofjust when the insurance is issued. In the non–random setting asequence of equalpayments made at equal intervals in time was referred to as anannuity. Hereinterest centers on annuities in which the payments are made (or received) only aslong as the insured survives.

An annuity in which the payments are made for a non–random period of timeis called anannuity certain. From the earlier discussion, the present value of anannuity immediate (payments begin one period in the future) with a payment of 1in each period is

an =n∑

j=1

vj =1− vn

i

while the present value of an annuity due (payments begin immediately) with apayment of 1 in each period is

an =n−1∑j=0

vj =1− vn

1− v=

1− vn

d.

These formulas will now be adapted to the case ofcontingent annuitiesin whichpayments are made for a random time interval.

Suppose that (x) wishes to buy a life insurance policy. Then (x) will pay apremium at the beginning of each year until (x) dies. Thus the premium paymentsrepresent alife annuity due for (x). Consider the case in which the payment amountis 1. Since the premiums are only paid annually the term of this life annuity dependsonly on the curtate life of (x). There will be a total ofK(x) + 1 payments, so theactuarial present value of the payments is ¨ax = E[aK(x)+1 ] where the left member isa notational convention. This formula gives

ax = E[aK(x)+1 ] = E[1 − vK(x)+1

d] =

1− Ax

d

as the relationship between this life annuity due and the net single premium for awhole life policy. A similar analysis holds for life annuities immediate.

Exercise 14–1.Compute the actuarial present value of a life annuity immediate.What is the connection with a whole life policy?

Exercise 14–2.A life annuity due in which payments are madem times per yearand each payment is 1/m has actuarial present value denoted by ¨a(m)

x . Show thatA(m)

x + d(m) a(m)x = 1.


§14: Life Annuities 51

Example 14–1.The Mathematical Association of America offers the followingalternative to members aged 60. You can pay the annual dues and subscription rateof $90, or you can become a life member for a single fee of $675. Life membersare entitled to all the benefits of ordinarymembers, including subscriptions. Shouldone become a life member? To answer this question, assume that the interest rate is5% so that the Life Table at the end of the notes can be used. The actuarial presentvalue of a life annuity due of $90 per year is

901− A60

1− v= 90

1− 0.4121951− 1/1.05

= 1110.95.

Thus one should definitely consider becoming a life member.

Exercise 14–3.What is the probability that you will get at least your money’s worthif you become a life member? What assumptions have you made?

Pension benefits often take the form of a life annuity immediate. Sometimesone has the option of receiving a higher benefit, but only for a fixed number of yearsor until death occurs, whichever comes first. Such an annuity is called atemporarylife annuity .

Example 14–2.Suppose a life annuity immediate pays a benefit of 1 each yearfor n years or until (x) dies, whichever comes first. The symbol for the actuarialpresent value of such a policy isax:n . How does one compute the actuarial presentvalue of such a policy? Remember that for a life annuity immediate, payments aremade at the end of each year, provided the annuitant is alive. So there will be atotal of K(x) ∧ n payments, andax:n = E[

∑K(x)∧ nj=1 vj]. A similar argument applies

in the case of ann year temporary life annuity due. In this case, payments aremade at the beginning of each ofn years, provided the annuitant is alive. In thiscase ¨ax:n = E[

∑K(x)∧ (n−1)j=0 vj] = E[ 1−v(K(x)+1)∧ n

d ] where the left member of this equalityintroduces the notation.

Exercise 14–4.Show thatAx:n = 1− d ax:n . Find a similar relationship forax:n .

Especially in the case of pension benefits it is more realistic to assume that thepayments are made monthly. Suppose payments are madem times per year. In thiscase each payment is 1/m. One could begin from first principles (this makes a goodexercise), but instead the previously established facts for insurances together withthe relationships between insurances and annuities given above will be used. Using


the obvious notation gives

a(m)x =

1−A(m)x

d(m)

=1− i

i(m) Ax

d(m)

=1− i

i(m) (1− dax)

d(m)

=id

i(m)d(m)ax +

i(m) − ii(m)d(m)

where at the second equality the UDD assumption was used.

Exercise 14–5.Find a similar relationship for an annuity immediate which pays1/m m times per year.

A useful idealization of annuities payableat discrete times is an annuity payablecontinuously. Such an annuity does not exist in the ‘real world’, but it provides auseful connecting bridge between certaintypes of discrete annuities. Suppose thatthe rate at which the benefit is paid is constant and is 1 per unit time. Then duringthe time interval (t, t + dt) the amount paid isdt and the present value of this amountis e−δt dt. Thus the present value of such a continuously paid annuity over a periodof n years is

an =∫ n

0e−δt dt =

1− e−δn

δ.

A life annuity which is payable continuouslywill thus have actuarial present value

ax = E[aT(x) ] = E[1 − e−δT(x)

δ].

Exercise 14–6.Show thatAx = 1− δax. Find a similar relationship forax:n .

This point of view makes it easy to understand certain modifications of discreteannuities. When annuity benefit payments are made at discrete intervals it iscustomary to provide a final adjustment which takes into account the death of theannuitant between payment periods. One such modified annuity is called acompleteannuity immediate whose actuarial present value is denoted by ˚ax. The paymentscheme for such an annuity is $1 at theend of each of the years 1 throughK(x) − 1plus a final adjustment payment made at the time of death in order to make thisscheme actuarially equivalent to a continuous annuity. To determine the size of theadjustment payment, first determine the rate at which a continuous annuity mustbe paid in order to be equivalent to a discrete annuity immediate. The followingpicture illustrates the cash flows.


Equivalence of an Annuity Immediate and a Continuous Annuity

0 1 0 1

1 σ............................................................................................................................................................................................................................................................................................................. .............................................................................................................................................................................................................................................................................................................

Equating the present value of the two cash streams givesv =∫ 1

0σ e−δt dt from

whichσ = δ/ i in order for the streams to be equivalent. It follows that the amountof the final payment, made at timeT(x), for the complete annuity immediate mustbe

eδT(x)∫ T(x)

K(x)

δi

e−δt dt =∫ T(x)−K(x)

0

δi

eδt dt.

Also

ax = E[∫ T(x)

0

δi

e−δt dt] =δi

ax.

Exercise 14–7.When payments are made on anmthly basis (each payment being1/m) the actuarial present value of a complete annuity immediate is denoted by ˚a(m)

x .Find a formula for the adjustment payment and the actuarial present value in thiscase.

Premium payments to an insurance company often take the form of anappor-tioned annuity due. Here the payment scheme consists of $1 immediately and atthebeginning of each year through and including timeK(x) less a refund payment(from the company) at the time of death. The refund payment is made because(in the premium context) the insured payed for a full year of coverage in advance.Using techniques similar to those above shows that the refund payment, made attimeT(x), should be in the amount of

eδT(x)∫ K(x)+1

T(x)

δd

e−δt dt =∫ K(x)+1−T(x)

0

δd

e−δt dt

and that the actuarial present value of an apportioned annuity due is

a{1}x = E[∫ T(x)

0

δd

e−δt dt] =δd

ax.

Here, of course, the left most member ofthe equalities is the notational convention.

Exercise 14–8.In the case ofmthly payments (each of size 1/m) find a formula fora{m}

x as well as for the size of the refund payment.

In order to compare apportioned and complete annuities, let us see how pre-miums paid by a complete annuity immediate would operate. In such a scheme,


premiums would be paid at the end of each year, except that in the year of deatha reduced premium would be paid at the time of death. When viewed from theinsurer’s viewpoint in this way it is obvious that ¨a{1}x > ax.

There is one other idea of importance. In the annuity certain setting one may beinterested in the accumulated value of the annuity at a certain time. For an annuitydue for a period ofn years the accumulated value of the annuity at timen, denotedby sn , is given by ¨sn = (1 + i)nan = (1+i)n−1

d . The present value of ¨sn is the same asthe present value of the annuity. Thus thecash stream represented by the annuity isequivalent to the single payment of the amount ¨sn at timen. This last notion has ananalog in the case of life annuities. In the life annuity context

nEx sx:n = ax:n

wherenEx = vnnpx is the actuarial present value(net single premium) of a pure

endowment of $1 payable at timen. Thus sx:n represents the amount of pureendowment payable at timen which is actuarially equivalent to the annuity.


Problems

Problem 14–1. Show that under UDD

ax < a(2)x < a(3)

x < ⋅ ⋅ ⋅ < ax < ⋅ ⋅ ⋅ < a(3)x < a(2)

x < ax.

Give an example to show that without the UDD assumption the inequalities mayfail.

Problem 14–2. Show that

ax < a(2)x < a(3)

x < ⋅ ⋅ ⋅ < ax < ⋅ ⋅ ⋅ < a{3}x < a{2}x < a{1}x .

Problem 14–3. Show that for anym we have ä{m}x < a(m)

x and thata(m)x < a(m)

x .

Problem 14–4. True or false:Ax1:n

= 1− d ax1:n

. Hint: When doesx1 : n die?

Problem 14–5. True or false: ¨sx:n ≤ sn .

Problem 14–6. Use the life table to calculate the actuarial present value of $1000due in 30 years if (40) survives.

Problem 14–7. Use the life table to computea21 anda{4}21 .

Problem 14–8. Find a general formula form|nax and use it together with the lifetable to compute5|10a20.

Problem 14–9. Proveax:n = 1Ex ax+1:n .

Problem 14–10.Show that under UDD

a(m)x = α (m)ax − β(m).

Hereα (m) =id

i(m)d(m)andβ(m) =

i − i(m)

i(m)d(m). The functionsα (m) andβ(m) defined

here are standard actuarial functions.

Problem 14–11.Show thatδ(Ia)T + TvT = aT .

Problem 14–12.Use the previous problem to show thatδ(Ia)x + (IA)x = ax. Here(Ia)x is the actuarial present value of an annuity in which payments are made at ratet at timet. Is there a similar formula in discrete time?

Problem 14–13.Show that äx = 1 + ax and that1m + a(m)

x:n = a(m)x:n + 1

mvnnpx.

Problem 14–14.Show that äx:n = ax − vnnpx ax+n and use this to compute ä21:5 .


Solutions to ProblemsProblem 14–1. As the type of annuity varies from left to right, the annuitantreceives funds sooner and thus the present value is higher.

Problem 14–2. Sincea(m)x = δ/ i(m)ax the result follows for the earlier relation-

ship between the rates of interest. A similar argument resolves the other half ofthe inequalities.

Problem 14–3. The difference betwee the two sides of the inequalities is theamount of the refund (or extra) payment.

Problem 14–4. The status dies only if (x) dies before timen. The result is true.

Problem 14–5. False.

Problem 14–8. m|nax = mExax+m − m+nExax+m+n.

Problem 14–10. See the text above.

Problem 14–11. Use integration by parts starting with the formulaδ(Ia)T =

δ∫ T

0t e−δt dt.


Solutions to Exercises

Exercise 14–1. In this case,E[aK(x) ] = E[ 1−vK(x)

i ] = 1−v−1Axi .

Exercise 14–2. Here there are [mT]+1 payments, so using the geometric seriesformula gives ¨a(m)

x = E[∑[mT]

j=0 (1/m)vj/m] = E[(1/m)(1 − v([mT]+1)/m)/ (1 − v1/m)].Now m(1− v1/m) = d(m), which gives the result.

Exercise 14–3. To get your money’s worth, you must live long enough so thatthe present value of the annual dues you would pay if you were not a life memberwill exceed $675. This gives a condition thatK(60) must satisfy if you are toget your moneys worth.

Exercise 14–4. For the first one ¨ax:n = E[ 1−v(K+1)∧ n

d ] = E[ 1−v(K+1)∧ n

d 1[0,n−1](K)] +

E[ 1−v(K+1)∧ n

d 1[n,∞)(K)] = E[ 1−v(K+1)

d 1[0,n−1](K)] + E[ 1−vn

d 1[n,∞)(K)] = (1/d)(1 − Ax:n ).A similar argument shows thatax:n = (1/ i)(A

x1:n

+ npxvn+1).

Exercise 14–5. The argument proceeds in a similar way, beginning with the

relationa(m)x = 1−v−1/mA(m)

xi(m) .

Exercise 14–6. The first relationship follows directly from the given equationand the fact thatAx = E[e−δT(x)]. SinceT(x : n ) = T(x) ∧ n a similar argumentgivesax:n = (1/δ)(1− Ax:n ).

Exercise 14–7. In this case the payment rate for the corresponding continuousannuity isδ/ i(m), which gives ˚a(m)

x = δi(m) ax and the adjustment payment as

eδT(x)∫ T(x)

[mT(x)]/m(δ/ i(m))e−δt dt =∫ T(x)−[mT(x)]/m

0 (δ/ i(m))eδt dt.

Exercise 14–8. Here the rate isδ/d(m) for the correspondingcontinuous annuityso thata{m}

x = (δ/d(m))ax and the refund payment iseδT∫ [mT]/m+1/m

Tδ

d(m) e−δt dt =∫ [mT]/m+1/m−T)0

δd(m) e−δt dt.

§15. Laboratory 5

1. Show that äx = 1 + vpxax+1. Find formulas expressing ä(m)x anda{m}

x in termsof ax.

2. The one step recursion formulas for annuities can be used just like the onestep recursions for insurances themselves. Use theqx values from Laboratory 3 andi = 5% and compute äx, a(12)

x , anda{12}x for x = 1 tox = 106. Place the result of your

computations into a nice table.


§16. Net Premiums

The common types of insurance policies can now be realistically analyzed froman insurers point of view.

To develop the ideas consider the case of an insurer who wishes to sell a fullydiscrete whole life policy which will bepaid for by equal annual premium paymentsduring the life of the insured. The terminologyfully discrete refers to the fact thatthe benefit is to be paid at the end of the year of death and the premiums are topaid on a discrete basis as well. How should the insurer set the premium? A firstapproximation is given by thenet premium. The net premium is found by using theequivalence principle: the premium should be set so that actuarial present valueof the benefits paid is equal to the actuarial present value of the premiums received.Using the equivalence principle the net premiumP should satisfy

E[vK(x)+1] = PE[aK(x)+1 ]

orAx −Pax = 0.

From here it is easy to determine the net premium, which in this case is denotedPx.

Exercise 16–1.Use the life table to find the net premium,P30, for (30) if i = 0.05.

The notation for other net premiums for fully discrete insurances parallel thenotation for the insurance policies themselves. For example, the netannual premiumfor ann year term policy with premiums payable monthly is denotedP(12)

x1:n.

Exercise 16–2.Use the life table to findP301

:10. What isP(12)

301

:10? What isP{1}

301

:10?

Exercise 16–3.An h payment whole life policy isone in which the premiums arepaid for h years, beginnning immediately. Find a formula forhPx, the net annualpremium for anh payment whole life policy.

Example 16–1.As a more complicated example consider a recent insurance ad-vertisement which I received. For a fixed monthly premium payment (which isconstant over time) one may receive a death benefit defined as follows:

1000001[0,65)(K(x)) + 750001[65,75)(K(x))

+ 500001[75,80)(K(x)) + 250001[80,85)(K(x)).

What is the net premium for such a policy? Assume that the interest rate is 5% sothat the life table can be used for computations. Using the equivalence principle,the netannual premiumP is the solution of the equation

Pa(12)x:85−x

= 100000Ax1:65−x

+ 7500065−x|10Ax + 5000075−x|5Ax + 2500080−x|5Ax


§16: Net Premiums 60

in terms of certain term and deferred term insurances.

Exercise 16–4.Compute the actual netmonthly premium for (21).

The methodology for finding the net premium for other types of insurance isexactly the same. The notation in the other cases is now briefly discussed. The mostcommon type of insurance policy is one issued on asemi-continuousbasis. Herethe benefit is paid at the time of death, but the premiums are paid on a discrete basis.The notation for the net annual premiumin the case of a whole life policy isP(Ax).The netannual premium for a semi-continuous term policy with premiums payablemthly is P(m)(A

x1:n). The notation for other semi-continuous policies is similar.

Exercise 16–5.What type of policy has net annual premiumP{m} (Ax:n )?

Policies issued on afully continuous basis pay the benefit amount at the timeof death and collect premiums in the form of a continuous annuity. Obviously, suchpolicies are of theoretical interest only. The notation here is similar to that of thesemi-continuous case, with a bar placed over theP. ThusP(Ax) is the premium ratefor a fully continuous whole life policy.


Problems

Problem 16–1. Show that ifδ = 0 thenP(Ax) = 1/ ex.

Problem 16–2. Arrange in increasing order of magnitude:P(2)(A40:25), P(A40:25),P{4} (A40:25), P(A40:25), P{12} (A40:25).

Problem 16–3. If P(Ax) = 0.03 and if interest is at the effective rate of 5%, findP{2}

x .

Problem 16–4. If 15P45 = 0.038,P45:15 = 0.056 andA60 = 0.625 findP451

:15.

Problem 16–5. Recall that apportionable annuities differ from annuities due onlyin the fact that the apportionable annuity offers the additional benefit of a ‘pre-mium refund’. LetAPR

x denote the net single premium for this refund benefit for acontinuous whole life policy with apportioned premiums payable annually. Showthat

APRx =

P(Ax)δ

(Ax − Ax).

Problem 16–6. Refer to the previous problem and show that

P(APRx ) =

P(Ax)δ ax

(Ax −Ax).

Problem 16–7. Use the equivalence principleto find the net annual premium fora fully discrete 10 year term policy with benefit equal to $10,000plus the return,with interest, of the premiums paid. Assume that the interest rate earned on thepremiums is the same as the interest rate used in determining the premium. Use thelife table to compute the premium for this policy for (21). How does this premiumcompare with 10000P

211

:10?

Problem 16–8. A level premium whole life insurance of 1, payable at the endof the year of death, is issued to (x). A premium ofG is due at the beginningof each year provided (x) survives. SupposeL denotes the insurer’s loss whenG = Px, L∗ denotes the insurer’s loss whenG is chosen so thatE[L∗ ] = −0.20, andVar(L) = 0.30. Compute Var(L∗ ).

Problem 16–9. A policy issued to (x) has the following features.

(1) Apportionable premiums are payable annually.

(2) The first premium is twice the renewal premium.

(3) Term insurance coverage for $100,000plus the difference between the firstand second premium is provided for 10 years.


(4) An endowment equal to the first year premium is paid at the end of 10 years.

(5) Death claims are paid at the moment of death.

Use the equivalence principle to find an expression for the renewal net annualpremium.

Problem 16–10.A $1000 whole life policy is issued to (50). The premiums arepayable twice a year, and are calculated on an apportionable basis. The benefit ispayable at the moment of death. Calculate the semi-annual net premium given thatA50 = 0.3,δ = 0.07, ande−0.035 = 0.9656.

Problem 16–11.Polly, aged 25, wishes to provide cash for her son Tad, currentlyaged 5, to go to college. Polly buys a policy which will provide a benefit in theform of a temporary life annuity due (contingent on Tad’s survival) in the amount of$25,000 per year for 4 years commencing on Tad’s 18th birthday. Polly will make 10equal annual premium payments beginning today. The 10 premium payments takethe form of a temporary life annuity due (contingent on Polly’s survival). Accordingto the equivalence principle, what is the amount of each premium payment? Usethe life table and UDD assumption (if necessary).

Problem 16–12.Snow White, presently aged 21, wishes to provide for the welfareof the 7 dwarfs in the event of her premature demise. She buys a whole life policywhich will pay $7,000,000 at the moment of her death. The premium payments forthe first 5 years will be $5,000 per year. According to the equivalence principle,what should her net level annual premium payment be thereafter? Use the life tableand UDD assumption (if necessary).

Problem 16–13.The Ponce de Leon Insurance Company computes premiums forits policies under the assumptions thati = 0.05 andµx = 0.01 for all x > 0. Whatis the net annual premium for a whole life policy for (21) which pays a benefitof $100,000 at the moment of death and has level apportioned premiums payableannually?


Solutions to ProblemsProblem 16–2. This is really a question about the present value of annuities.

Problem 16–3. SinceiAx = δAx andd(2)a{2}x = δax, P{2}x = (d(2)/ i)P(Ax).

Problem 16–4. Use the two equationsP45:15 = P415:15

+ 15E45/ a45:15 and

P451

:15=

A45 − 15E45A60

a45:15= 15P45 − 15E45A60/ a45:15 with the given information.

Problem 16–7. The present value of the benefit is 10000vK+11[0,10)(K) +pvK+1sK+1 1[0,10)(K) wherep is the premium.

Problem 16–8. The loss random variable is (1 +G/d)vK+1 − G/d from whichthe mean and variance in the two cases can be computed and compared.

Problem 16–10. The annual premiump satisfiespa{2}50 = 1000A50.

Problem 16–11. The premiump satisfiespa25:10 = 2500013|4a5.

Problem 16–12. The premiump satisfies 7000000A21 = 5000a21:5 + p5E21a26.


Solutions to ExercisesExercise 16–1. From the table,P30 = A30/ a30 = 0.133866/18.189 = 0.0735.

Exercise 16–2. Now P301

:10= A

301

:10/ a30:10. Also A

301

:10= A30− v10

10p30A40 =

0.133866− (1.05)−10(94926/96477)(0.201506) = 0.01214. Similarly, ä30:10 =a30 − v10

10p30a40 = 8.060, giving the premium as 0.001506. The other twopremiums differ only in the denominator, sinceP(12)

301

:10= A

301

:10/ a(12)

30:10and

P{1}

301

:10= A

301

:10/ a{1}

30:10. Now a(12)

30:10= a(12)

30 − v1010p30a

(12)40 . Since a(12)

30 =

(id/ i(12)d(12))a30 + (i(12) − i)/ i(12)d(12) and a similar expression holds for ä(12)40 ,

the value of the annuity can be computed from the life table. A similar argumentand the fact that ä{1}x = (i/δ)ax + (δ − i)/δd allows the computation of the valueof the apportioned annuity. Note that the UDD assumption has been used here.

Exercise 16–3. hPx = Ax/ ax:h .

Exercise 16–4. The net monthly premium isP/12 whereP = (100000A211

:44+

75000v4444p21A

651

:10+ 50000v54

54p21A751

:5+ 25000v59

59p21A801

:5)/ a(12)

21:64. These

values can be computed from the life table using the techniques of an earlierexercise.

Exercise 16–5. This is the premium for a continuous endowment policy withmthly apportioned premiums.

§17. Laboratory 6

1. A 40 year term insurance policy with semi-annual premiums is issued to(30). Assume thati = 5% and mortality follows the table given in Laboratory 3.Compute the semi-annual net premium if the benefit amount is 100,000 and thebenefit is paid at the moment of death.

2. Rework the computation in problem 1 for an interest rate of 4% and 6%.How much does the change in interest rate change the semi-annual premium?

3. Suppose the benefit amount of the policy in problem 1 is 100,000 for deathin the first 30 years (until age 60), and then decreases by 5,000 per year for each ofthe remaining years. What isthe semi-annual premium?


§18. Insurance Models Including Expenses

A more realistic view of the insurance business includes provisions for expenses.The profit for the company can also be included here as an expense.

The common method used for the determination of theexpense loaded pre-mium (or thegross premium) is a modification of the equivalence principle. Ac-cording to themodified equivalence principlethe gross premiumG is set so thaton the policy issue date the actuarial present value of the benefit plus expenses isequal to the actuarial present value of the premium income. The premium is usuallyassumed to be constant. Under these assumptions it is fairly easy to write a formulato determineG. Assume that the expenses in policy yeark areek−1 and are paid attime k − 1, that is, at the beginning of the year. The actuarial present value of theexpenses is then given by

E[K(x)∑k=0

vk ek] =∞∑

k=0

vk ek kpx.

Typically expenses are dependent on the premium. Also the sales commission isusually dependent on the policy size.

Example 18–1.Suppose that the first year expenses for a $100,000 semi-continuouswhole life policy are 20% of premiums plus a sales commission equal to 0.5% ofthe policy amount, and that the expenses for subsequent years are 10% of premiumplus $5. The gross premiumG for such a policy satisfies

100000Ax + (0.20G + 500) + (0.10G + 5)ax = Gax.

An important, and realistic, feature of the above example is the large amount offirst year expense. Expenses are now examined in greater detail.

Example 18–2.Let’s look at the previous example in the case of a policy for aperson aged 21. Assume that the interest rate is 5% and that the life table applies.Then

G =100,000A21 + 495 + 5a21

0.9a21 − 0.1= $604.24.

From this gross premium the company must pay $500 in fixed expensesplus 20%of the gross premium in expenses ($120.85),plus provide term insurance coveragefor the first year, for which the net single premium is 100,000A

211

:1= $123.97. Thus

there is a severe expected cash flow strain in the first policy year! The interestedreader may wish to examine the article “Surplus Loophole” inForbes, September4, 1989, pages 44-48.


§18: Insurance Models Including Expenses 67

Expenses typically consist of two parts. The first part of the expenses can beexpressed as a fraction of gross premium. These are expenses which depend onpolicy amount, such as sales commission, taxes, licenses, and fees. The other partof expenses consist of those items which are independent of policy amount suchas data processing fees, printing of actual policy documents, clerical salaries, andmailing expenses.

Studying the gross premium as a function of the benefit provided can be useful.Denote byG(b) the gross premium for a policy with benefit amountb. The valueG(0) represents the overhead involved in providing the policy and is called thepolicy fee. Typically the policy fee is not zero. The ratioG(b)/b is called thepremium rate for a policy of benefitb and reflects (approximately) the premiumchange per dollar of benefit change when the benefit amount isb.

Exercise 18–1.In the example above findR(b), the premium rate for a policy ofbenefitb.


Problems

Problem 18–1. The expense loaded annual premium for an 35 year endowmentpolicy of $10,000 issued to (30) is computed under the assumptions that

(1) sales commission is 40% of the gross premium in the first year

(2) renewal commissions are 5% of the gross premium in year 2 through 10

(3) taxes are 2% of the gross premium each year

(4) per policy expenses are $12.50 per 1000 in the first year and $2.50 per 1000thereafter

(5) i = 0.05

Find the gross premium using the life table.

Problem 18–2. A semi-continuous whole life policy issued to (21) has the followingexpense structure. The first year expense is 0.4% of the policy amount plus $50. Theexpenses in years 2 through 10 are 0.2% of the policy amount plus $25. Expenses inthe remaining years are $25,and at the time of death there is an additional expenseof $100. Find a formula forG(b). ComputeG(1) and compare it toA21.

Problem 18–3. Your company sells supplemental retirement annuity plans. Thebenefit under such a plan takes the form ofan annuity immediate, payable monthly,beginning on the annuitant’s 65th birthday. Let the amount of the monthly benefitpayment beb. The premiums for this annuity are collected via payroll deductionat the end of each month during the annuitant’s working life. Set up expenses forsuch a plan are $100. Subsequent expensesare $5 each month during the premiumcollection period, $100 at the time of the first annuity payment, and $5 per monththereafter. FindG(b) for a person buying the plan at agex. What isR(b)?

Problem 18–4. A single premium life insurance policy with benefits payable at theend of the year of death is issued to (x). Suppose that

(1) Ax = 0.25

(2) d = 0.05

(3) Sales commission is 18% of gross premium

(4) Taxes are 2% of gross premium

(5) per policy expenses are $40 the first year and $5 per year thereafter

Calculate the policy fee that should be charged.


Solutions to ProblemsProblem 18–1. 10000A30:35 +0.35G+0.05Ga30:10+(0.02G+25)a30:35+100 =Ga30:35.

Problem 18–2. bA21 + 0.002b + 25 + 0.002ba21:10 + 25a21+ 100A21 = G(b)a21.

Problem 18–3. 12b65−x|a(12)x +5×12a(12)

x +95+10065+ 112−xEx = G(b)×12a(12)

x:65−x.

Problem 18–4. G(b) = bAx + 0.18G(b) + 0.02G(b) + 35 + 5ax.


Solutions to ExercisesExercise 18–1. Since the premium isbR(b) when the benefit isb, the modifiedequivalence principle givesbAx + (0.20bR(b) + 0.005b) + (0.20bR(b) + 5)ax =bR(b)ax from whichR(b) = (bAx + 0.005b + 5ax)/b(0.9ax − 0.20).

§19. Multiple Lives

The study of the basic aspects of life insurance is now complete. Two differentbut similar directions will now be followed in the ensuing sections. On the onehand, types of insurance in which the benefit is paid contingent on the death orsurvival of more than one life will be examined. On the other hand, the effects ofcompeting risks on the cost of insurance will be studied.

The first area of study will be insurance in which the time of the benefit paymentdepends on more than one life. Recall that astatus is an artificially constructedlife form for which there is a definition of survival and death (ordecrement). Thesimplest type of status is thesingle lifestatus. The single life status (x) dies exactlywhen (x) does. Another simple status is thecertain statusn . This status dies atthe end ofn years. Thejoint life status for the n lives (x1), . . .(xn) is the statuswhich survives until thefirst member of the group dies. This status is denoted by(x1x2 . . .xn). The last survivor status, denoted by (x1x2 . . .xn) is the status whichsurvives until thelast member of the group dies.

When discussing a given status the question naturally arises as to how onewould issue insurance to such a status. If the constituents of the status are assumedto die independently this problem can be easily solved in terms of what is alreadyknown.

Example 19–1.Consider a fully discrete whole life policy issued to the joint status(xy). The net annual premium to be paid for such a policy is computed as follows.Using the obvious notation, the premium,P, must satisfy

Axy = Paxy.

Using the definition of the joint life status gives

Axy = E[vK(x)∧ K(y)+1]

and

axy =1− Axy

dwhich are obtained as previously.

Exercise 19–1.Obtain an expression forAxy in terms that can be computed fromthe life table.

Exercise 19–2.Is Axy + Axy = Ax + Ay?

A useful technique for writing computational formulas directly is to ask thequestion “Under what conditions is a payment made at timet?” The answer willusually provide a computational formula.


§19: Multiple Lives 72

Example 19–2.What isaxy? This annuity makes a payment of 1 at timek if andonly if both (x) and (y) are alive at timek, and the probability of this iskpxy = kpx kpy.

Thusaxy =∞∑

k=0

vkkpx kpy.

If one is willing to assume an analytical law of mortality computations involvingjoint lives can be simplified. Recall that two of the common analytical laws ofmortality are the Gompertz and Makeham laws. The Gompertz law has force ofmortalityµx = Bcx. It is easily seen that the joint survival of two independent lives(x) and (y) is probabilistically identical with the survival of a single life (w) if andonly if

µ(xy)+s = µx+s + µy+s = µw+s.

When (x) and (y) have mortality which follows Gompertz’ Law this relation holds ifw satisfiescx + cy = cw. A similar observation applies to Makeham’s law for whichthe force of mortality isµx = A +Bcx. In this case, however, it is necessary to mimicthe joint life (xy) by using a joint life (ww) at equal ages. Herew is the solution of2cw = cx + cy.

Exercise 19–3.Verify these assertions.

A status can also be determined by the order in which death occurs. The idea

here is similar to that used for term insurance earlier in which the statusx1 : n failsat the time of death of (x) provided (x) dies first. As a more complicated example

the status (x : y2) dies at the time of death of (y) provided (y) is the second to die.

Hence this status lives forever if (y) dies before (x). An insurance for such a statusis a simple case of what is known as acontingent insurance. Again, if the lives areassumed to fail independently it is a simple matter to reduce computations involvingcontingent insurance to the cases already considered.


Problems

Problem 19–1. Show

tpxy = tpxy + tpx(1− tpy) + tpy(1− tpx).

Problem 19–2. Supposeµx = 1/ (110− x) for 0 ≤ x < 110. Find10p20:30, 10p20:30,ande20:30.

Problem 19–3. Find an expression for the actuarial present value of a deferredannuity of $1 payable at the end of any year as long as either (20) or (25) is livingafter age 50.

Problem 19–4. Find the actuarial present value of a 20 year annuity due whichprovides annual payments of $50,000 while both (x) and (y) survive, reducing by25,000 on the death of (x) and by 12,500 on the death of (y).

Problem 19–5. Show thatnqx1y

= nqxy2 + nqx npy.

Problem 19–6. Show thatAx1y− A

xy2= Axy − Ay.

Problem 19–7. If µx = 1/ (100− x) for 0 ≤ x < 100, calculate25q25:50

2 .

Problem 19–8. In a mortality table which follows Makeham’s Law you are givenA = 0.003 andc10 = 3. Calculate∞q

401

:50if e40:50 = 17.

Problem 19–9. If the probability that (35) will survive for 10 years isa and theprobability that (35) will die before (45) isb, what is the probability that (35) willdie within 10 years after the death of (45)? Assume the lives are independent.

Problem 19–10.Plot the survival function for Gompertz law of mortality withA = 0.001 andc = 1.06.

Problem 19–11.Suppose (20) and (30) are independent lives that follow the Gom-pertz law of mortality given in the previous problem. Plot the survival for the jointlife status 20 : 30. Is there a single age (x) whose survival function is the same asthe survival function of 20 : 30?

Problem 19–12.Plot the survival function for Makeham’s law of mortality withA = 0.003,B = 0.001, andc = 1.06.

Problem 19–13.Suppose (20) and (30) are independent lives that follow the Make-ham law of mortality given in the previous problem. Plot the survival for the jointlife status 20 : 30. Is there a single age (x) whose survival function is the same asthe survival function of 20 : 30?


Solutions to ProblemsProblem 19–1. tpxy = P[[T(x) ≥ t] ∪ [T(y) ≥ t]] = P[T(x) ≥ t,T(y) ≥t] + P[T(x) ≥ t,T(y) ≤ t] + P[T(x) ≤ t,T(y) ≥ t].

Problem 19–2. From the form of the force of mortality, DeMoivre’s Lawholds.

Problem 19–3. 30|a20 + 25|a25 − 30|a20:25.

Problem 19–4. The annuity pays 12,500 for 20 years no matter what so theactuarial present value consists of 3 layers.

Problem 19–7. DeMoivre’s Law holds.

Problem 19–8. Here e40:50 =∫ ∞

0 tp40:50dt = 17. Also, by conditioning,P[T(40) < T(50)] =

∫ ∞0 P[T(50) > t]fT(40)(t) dt =

∫ ∞0 tp40:50µ40+t dt while by a

symmetric argumentP[T(40) > T(50)] =∫ ∞

0 tp40:50µ50+t dt. Using the form ofthe force of mortality under Makeham’s Law, the fact that these two probabilitiessum to 1, and the given information completes the argument.

Problem 19–9. P[T(35) > T(45) + 10] =∫ ∞

0 P[T(35) > t + 10]tp45µ45+t dt =∫ ∞0 P[T(35)> t+10|T(35)≥ 10]P[T(35)≥ 10]tp45µ45+t dt = a

∫ ∞0 P[T(45)+10>

t + 10]tp45µ45+t dt = a∫ ∞

0 (tp45)2µ45+t dt =∫ ∞

0 −tp45ddt tp45 dt = a/2. Thus the

desired probability is 1− a/2− b.


Solutions to ExercisesExercise 19–1. Using the independence givestpxy = tpxtpy, so thatAxy =E[vK(xy)+1] =

∑∞k=0 vk+1(kpxy − k+1pxy) =

∑∞k=0 vk+1(kpxkpy − k+1pxk+1py).

Exercise 19–2. Intuitively, either (x) dies first or (y) dies first, so the equationis true. This can be verified by writing the expectations in terms of indicators.

Exercise 19–3. Under Makeham the requirement is that (A+Bcx+s)+(A+Bcy+s) =(A + Bcw+s) + (A + Bcw+s) for all s, and this holds ifcx + cy = 2cw.

§20. Laboratory 7

1. Find an expression for the net single premium for a whole life policy issued

to (x1y), a status which fails when (x) dies if T(x) < T(y). Use this expression andthe life table data of Laboratory 3 to compute the premium for a $1,000,000 policy

issued to (301

: 40). Use 6% as the interest rate.

2. Find an expression for the net single premium for a whole life policy issued to

(xy2), where the benefit is paid on the death of (y) if T(x) < T(y). Use this expression

and the life table data of Laboratory 3to compute the premium for a $1,000,000

policy issued to (30 : 402

). Use 6% as the interest rate.

3. Show that ifX andY are independent random variables and one of them isabsolutely continuous thenP[X = Y] = 0. Hence under the standard assumptions ofthis section no two people can die simultaneously.

4. One model for joint lives which allows for simultaneous death is thecommonshock model. The intuition is that the two lives behave almost independently exceptfor the possibility of death by a common cause. The model is as follows. LetT ∗ (x),T ∗ (y), andZ be independent random variables. Assume thatT ∗ (x) andT ∗ (y) havethe distribution of the remaining lifetimes of (x) and (y) as given by the life table.The random variableZ represents the time of occurrence of the common catastrophewhich will kill any survivors. The common shock model is that the true remaininglifetimes of (x) and (y) are given asT(x) = min{T ∗ (x),Z} andT(y) = min{T ∗ (y),Z}respectively. What is the probability that (x) and (y) die simultaneously in thismodel? What is the survival function for the joint life status (xy) in this model?Answer these two questions in general, and then in the special case in whichT ∗ (x), T ∗ (y), andZ have exponential distributions with parametersµx, µy, andµz

respectively.


§21. Multiple Decrement Models

In contrast to the case in which a status is defined in terms of multiple lives, theway in which a single life fails can be studied. This point of view is particularlyimportant in the context of the analysis of pension plans. In such a setting a personmay withdraw from the workforce (a ‘death’) due to accident, death, or retirement.Different benefits may be payable in each of these cases. Another common typeof insurance in which a multiple decrement model is appropriate is in the doubleindemnity life policy. Here the benefit is twice the face amount of the policy ifthe death is accidental. In actuarial parlance the termination of a status is called adecrementandmultiple decrement modelswill now be developed. These modelsalso go by the name of competing risk models in other contexts.

To analyze the new situation,introduce the random variableJ = J(x) which is adiscrete random variable denoting the cause of decrement of the status (x). Assumethat J(x) has as possible values the integers 1, . . . ,m. It is clear that all of theinformation of interest is contained in thejoint distribution of the random variablesT(x) andJ(x). Note that this joint distribution is of mixed type since (as always)T(x)is assumed to be absolutely continuous whileJ(x) is discrete. The earlier notation ismodified in a fairly obvious way to take into account the new model. For example,

tq(j)x = P[0 < T(x) ≤ t, J(x) = j].

and

tp(j)x = P[T(x) > t, J(x) = j].

Here∞q(j)x gives the marginal density ofJ(x). To discuss the probability of death

due to all causes the superscript (τ ) is used. For example,

tq(τ )x = P[T(x) ≤ t] =

m∑j=1

tq(j)x

and similar expressions for the survival probability and the force of mortality canbe obtained. Althoughtq(τ )

x + tp(τ )x = 1 a similar equation for the individual causes

of death fails unlessm = 1. For the force of mortality

µ(τ )x+t =

fT(x)(t)P[T(x) > t]

and

µ(j)x+t =

fT(x) J(x)(t, j)P[T(x) > t]

.

One must be careful in the use of these formulas. In particular, while

tp(τ )x = exp{−

∫ t

0µ(τ )

x+s ds}


§21: Multiple Decrement Models 78

it is also the case that

tp(j)x ≠ exp{−

∫ t

0µ(j)

x+s ds}

as one might not at first expect. This latter integral does have an important usewhich is explored below.

An important practical problem is that of constructing a multiple decrement lifetable. To see how such a problem arises consider the case of a double indemnitywhole life policy. Assume that the policy will pay an amount $1 at the end of theyear of death if death occurs due to non-accidental causes and an amount of $2 ifthe death is accidental. Denote the type of decrement as 1 and 2 respectively. Thepresent value of the benefit is then

vK(x)+1 1{1} (J(x)) + 2vK(x)+11{2} (J(x)) = J(x)vK(x)+1.

To compute the net premium it remains to compute the expectation of this quantity.This computation can only be completed ifp(j)

x is known. How are these probabilitiescalculated?

There are two basic methodologies used. If a large group of people for whichextensive records are maintained is available the actual survival data with the deathsin each year of age broken down by cause would also be known. It is then very easyto construct the multiple decrement table. This is seldom the case.

Example 21–1.An insurance company has a thriving business issuing life insuranceto coal miners. There are three causes of decrement (death): mining accidents, lungdisease, and other causes. From the company’s vast experience with coal miners adecrement (life) table for these three causes of decrement is available. The companynow wants to enter the life insurance business for salt miners. Here the two causes ofdecrement (death) are mining accidents and other. How can the information aboutmining accidents for coal miners be used to get useful information about miningaccidents for salt miners?

A simple-minded answer to the question raised in the example would be tosimply lift the appropriate column from the coal miners life table and use it for thesalt miners. Such an approach fails, because it does not take into account the factthat there are competing risks, that is, the accident rate for coal miners is affected bythe fact that some miners die from lung disease and thus are denied the opportunityto die from an accident. The death rate for each causein the absence of competingrisk is needed.

To see how to proceed the multiple decrement process is examined in a bit moredetail. Some auxillary quantities are introduced. Define

tp′ (j)x = exp{−∫ t

0µ(j)

x+s ds}


and

tq′ (j)x = 1− tp′ (j)x .

The “probability” tq′ (j)x is called thenet probability of decrement (or absoluterate of decrement). It is these “probabilities” that represent the death rates in theabsence of competing risks. To see why this interpretation is reasonable, note that

µ(j)x+t =

fT(x) J(x)(t, j)P[T(x) > t]

=fX J(x)(x + t, j)/s(x)

s(x + t)/s(x)

=fX J(x)(x + t, j)P[X > x + t]

.

This shows thatµ(j)x+t represents the rate of death due to causej among those surviving

up to timex + t.

These “probabilities” may be used to obtain the desired entries in a multipledecrement table as follows. First

tp(τ )x =

m∏j=1

tp′ (j)x .

This shows how one can pass from the absolute rate of decrement to total survivalprobabilities. Note that this relationship implies that the rates are generally largerthan the total survival probability. Then, under theassumption of constant force ofmortality for each decrement over each yearof age in the multiple decrement table,

q(j)x =

∫ 1

0sp

(τ )x µ(j)

x+s ds

=∫ 1

0sp

(τ )x µ(j)

x ds

=µ(j)

x

µ(τ )x

∫ 1

0sp

(τ )x µ(τ )

x ds

=µ(j)

x

µ(τ )x

q(τ )x

=logp′ (j)x

logp(τ )x

q(τ )x .

This solves the problem of computing the entries in a multiple decrement tableunder the stated assumption about the structure of the causes of decrement in thattable.

Exercise 21–1.What happens ifp(τ )x = 1?


Exercise 21–2.Show that the same formula results if one assumes instead that thetime of decrement due to each cause of decrement in the multiple decrement tablehas the uniform distribution over the year of decrement. (This assumption meansthat tq(j)

x = t ⋅ q(j)x .)

Exercise 21–3.Assume that two thirds of all deaths at any age are due to accident.What is the net single premium for (30) for a double indemnity whole life policy?How does this premium compare with that of a conventional whole life policy?

The previous computations were based on assumptions about the causes ofdecrement within the multiple decrement table. In some contexts it is more sensibleto make assumptions about the structure of theindividual causes of decrement asif each were acting independently, that is,to make assumptions about the absoluterate of decrement in the single decrement tables.

Example 21–2.Suppose we are designing a pension plan and that there are twocauses of decrement: death and retirement. In many contexts (such as teaching) itis reasonable to assume that retirements all occur at the end of a year, while deathscan occur at any time. How could we construct a multiple decrement table whichreflects this assumption?

One common assumption about a single decrement is the assumption of uniformdistribution of deaths in the year of death. In the multiple decrement context thistranslates in the statement that for 0≤ t ≤ 1

tq′ (j)x = t q′ (j)x .

Exercise 21–4.Show that under this assumption we havetp′ (j)x µ(j)x+t = q′ (j)x for 0 ≤

t ≤ 1. Hint: Computeddt tp′ (j)x in two different ways.

If this uniformity assumption is made for all causes of decrement it is then easyto construct the multiple decrement table. The computations are illustrated for thecase of 2 causes of decrement. In this setting

q(1)x =

∫ 1

0sp

(τ )x µ(1)

x+s ds

=∫ 1

0sp′ (1)

x sp′ (2)x µ(1)

x+s ds

= q′ (1)x

∫ 1

0sp′ (2)

x ds

= q′ (1)x

∫ 1

0(1− sq′ (2)

x ) ds

= q′ (1)x (1−

12

q′ (2)x )


with a similar formula forq(2)x . It is easy to see how this procedure could be modified

for different assumptions about the decrement in each single decrement table.

Exercise 21–5.Construct a multiple decrement table in which the first cause ofdecrement is uniformly distributed and the second cause has all decrements occurat the end of the year. The pension plan described in the example above illustratesthe utility of this technique.

Another approximation which is used to connect single and multiple decrementtables makes use of the functions

Lx =∫ 1

0lx+t dt

mx =lx − lx+1

Lx.

The functionmx is called thecentral death rate at agex. The central rate of deathis used in a special technique, called thecentral rate bridge, in the context ofmultiple decrement tables. This technique is now briefly described. Define

m(τ )x =

∫ 1

0tp

(τ )x µ(τ )

x+t dt∫ 1

0tp

(τ )x dt

and

m(j)x =

∫ 1

0tp

(τ )x µ(j)

x+t dt∫ 1

0tp

(τ )x dt

and

m′ (j)x =

∫ 1

0tp′ (j)x µ(j)

x+t dt∫ 1

0tp′ (j)x dt

.

The central rate bridge is based on thefollowing approximation. First, under theUDD assumption in each single decrement table

m′ (j)x =q′ (j)x

1− 12q′ (j)x

.

Second, under the UDD assumption in the multiple decrement table

m(j)x =

q(j)x

1− 12q(τ )

x.


Thirdly, under the constant force assumption in the multiple decrement table

m(j)x = µ(j)

x = m′ (j)x .

Now assume that all of these equalities are goodapproximations in any case. Thisassumption provides a way of connecting the single and multiple decrement tables.There is no guarantee of the internal consistency of the quantities computed in thisway, since, in general, the three assumptions made are not consistent with eachother. The advantage of this method is thatthe computations are usually simplerthan for any of the ‘exact’ methods.

Exercise 21–6.Show that each of the above equalities hold under the stated as-sumptions.


Problems

Problem 21–1. Assume that each decrement has a uniform distribution over eachyear of age in the multiple decrement table to construct a multiple decrement tablefrom the following data.

Age q′ (1)x q′ (2)

x q′ (3)x

62 0.020 0.030 0.20063 0.022 0.034 0.10064 0.028 0.040 0.120

Problem 21–2. Rework the preceding exercise using the central rate bridge. Howdifferent is the multiple decrement table?

Problem 21–3. In a double decrement table where cause 1 is death and cause 2 iswithdrawal it is assumed that deaths are uniformly distributed over each year of agewhile withdrawals between agesh andh + 1 occur immediately after attainment ofageh. In this table one sees thatl(τ )

50 = 1000,q(2)50 = 0.24, andd(1)

50 = 0.06d(2)50 . What

is q′ (1)50 ? How does your answer change if all withdrawals occur at midyear? At the

end of the year?

Problem 21–4. How would you construct a multiple decrement table if you weregivenq′ (1)

x , q′ (2)x , andq(3)

x ? What assumptions would you make, and what formulaswould you use? What if you were givenq′ (1)

x , q(2)x , andq(3)

x ?



Problem 21–1. First, p(τ )62 = (.98)(.97)(.80) andq(τ )

62 = 1 − p(τ )62. Also p′ (j)62 =

1 − q′ (j)62 . From the relationq(j)62 =

logp′ (j)62

logp(τ )62

q(τ )62 the first row of the multiple

decrement table can be found.

Problem 21–3. From the informationd(2)50 = 240 andd(1)

50 = 14. Since with-drawals occur at the beginning of the year there are 1000− 240 = 760 peopleunder observation of whom 14 die. Soq′ (1)

50 = 14/760. If withdrawals occur atyear end all 1000 had a chance to die soq′ (1)

50 = 14/1000.

Problem 21–4. The central rate bridge could be used. Is there an exact methodavailable?


Solutions to ExercisesExercise 21–1. What would this mean forµ (τ )

x and the derivation?

Exercise 21–2. The assumption is thattq(j)x = tq(j)

x for all j. Hencetp(τ )x =

1 − tq(τ )x andµ (j)

x+s = dds sq

(j)x / sp(τ )

x = q(j)x / sp(τ )

x . Substitution and integration gives

p′ (j)x = e−∫ 1

0µ(j)

x+s ds = (1 − q(τ )x )q(j)

x /q(τ )x . Sincep(τ )

x = 1− q(τ )x , the result follows by

substitution.

Exercise 21–3. The actuarial present value of the benefit is (1/3) × 1 × Ax +(2/3)× 2× Ax = (5/3)Ax, from which the premium is easily calculated.

Exercise 21–4. From the definition, ddt tp′ (j)x = −tp′ (j)x µ (j)

x+t, while from the

relationtp′ (j)x = 1− tq′ (j)x , ddt tp′ (j)x = −q′ (j)x .

Exercise 21–5. Since cause 1 obeys UDD,q(1)x = q′ (1)

x

∫ 10 sp′ (2)

x ds as in thederivation above. For cause 2,sp′ (2)

x = 1 for s < 1, soq(1)x = q′ (1)

x . For cause 2proceed as in the derivation above to getq(2)

x = −∫ 1

0 sp′ (1)x

dds sp′ (2)

x ds. Now sp′ (2)x

is constant except for a jump of size−q′ (2)x at s = 1. Henceq(2)

x = q′ (2)x p′ (1)

x =q′ (2)

x (1− q′ (1)x ).

Exercise 21–6. Under UDD in the single decrement tabletp′ (j)x = 1− tq′ (j)x and

tp′ (j)x µ (j)x+t = q′ (j)x som′ (j)x =

∫ 10 q′ (j)x dt/

∫ 10 (1 − tq′ (j)x ) dt = q′ (j)x / (1 − 1

2q′ (j)x ). Under

UDD in the multiple decrement tableµ (j)x+s = q(j)

x / sp(τ )x so that substitution gives

the result. Under the constant force assumption in the multiple decrement table,µ (j)

x+s = µ (j)x for all j andm(i)

x = µ (j)x = m′ (j)x by substitution.

§22. Laboratory 8

The service table at the end of these notes contains information about a groupof workers. There are 4 causes of decrement for this population. The first causeis death (d), the second cause is withdrawal (w) (termination of employment), thethird cause is incapacity (i), and the fourth cause is retirement (r).

1. Suppose a concerted effort by the company reduces the rate of on the jobinjury (incapacity) by 1/3 at all ages. Recompute the entries in the service table.

2. Suppose that in addition to the incapacity improvement an enhanced salaryand benefits plan reduces the withdrawal rate at all ages by 1/4. Recompute theentries in the service table.


§23. Insurance Company Operations

The discussion thus far has been about individual policies. In the next fewsections the operations of the company asa whole are examined. This examinationbegins with an overview of the accounting practices of an insurance company. Thisis followed by a study of the behavior of the loss characteristics of groups of similarpolicies. This last study leads to another method of setting premiums for a policy.


§24. Net Premium Reserves

A realistic model for both insurance policies and the method and amount ofpremium payment is now in hand. The nextquestion is how accounting principlesare applied to the financial operations of insurance companies.

A basic review of accounting principles is given first. There are three broadcategories of items for accounting purposes: assets, liabilities, and equity.Assetsinclude everything which is owned by the business.Liabilities include everythingwhich is owed by the business.Equity consists of the difference in the value of theassets and liabilities. Equity could be negative. In the insurance context liabilitiesare referred to asreserve and equity assurplus. When an insurance policy isissued the insurance company is accepting certain financial obligations in return forthe premium income. The basic question is how this information is reflected in theaccounting statements of the company. Some of the different accounting proceduresavailable will now be described. Keep inmind that this discussion only concernshow the insurance company prepares accounting statements reflecting transactionswhich have occurred. The method by which gross (or net) premiums are calculatedis not being changed!

Example 24–1.Suppose the following data for an insurance company is given.

Income for Year Ending December 31, 1990Premiums 341,000Investment Income 108,000Expenses 112,000Claims and Maturities 93,000Increases in Reserves —Net Income —

Balance SheetDecember 31, 1989 December 31, 1990

Assets 1,725,000 —Reserves — 1,433,000Surplus 500,000 —

The missing entries in the tables can be filled in as follows (amounts in thou-sands). Total income is 341 + 108 = 449while total expenses are 112 + 93 = 205,so net income (before reserve contributions) is 449− 205 = 244. Now the reservesat the end of 1989 are 1,725− 500 = 1,225, so the increase in reserves must be1,433− 1,225 = 208. The net income is 244− 208 = 36. Hence the 1990 surplusis 536 and the 1990 assets are 1,969.


§24: Net Premium Reserves 89

The central question in insurance accounting is “How are liabilities measured?”The answer to this question has some veryimportant consequences for the operationof the company, as well as for the financial soundness of the company. The generalequation is

Reserve at timet = Actuarial Present Value at timet of future benefits

−Actuarial Present Value at timet of future premiums.

The only accounting assumption required is one regarding the premium to be usedin this formula. Is it the net premium, gross premium, or ???

The only point of view adopted here is that liabilities are measured as thenet level premium reserves. This is the reserve computed under theaccountingassumption that the premium charged for the policy is the net level premium. To seethat this might be a reasonable approach, recall that the equivalence principle setsthe premium so that the actuarial present value of the benefit is equal to the actuarialpresent value of the premiums collected. However, it is clear that after the policyis issued the present value of the benefits and of the un-collected premiums will nolonger be equal, but will diverge in time.This is because the present value of theunpaid benefits will be increasing in time and the present value of the uncollectedpremiums will decrease in time. The discrepency between these two amounts at anytime represents an unrealized liability to the company. To avoid a negative surplus(technical bankruptcy), this liability mustbe offset in the accounting statments ofthe company by a corresponding asset. Assume (for simplicity) that this asset takesthe form of cash on hand of the insurance company at that time. How does onecompute the amount of the reserve at any timet under this accounting assumption?This computation is illustrated in the context of an example.

Example 24–2.Consider a fully discrete whole life policy issued to (x) in whichthe premium is payable annually and is equal to the net premium. What is thereserve at timek, wherek is an integer? To compute the reserve simply note thatif (x) has survived until timek then the (curtate) remaining life ofx has the samedistribution asK(x + k). The outstanding benefit has present valuevK(x+k)+1 whilethe present value of the remaining premium income is ¨aK(x+k)+1 times the annualpremium payment. Denote bykL the random variable which denotes the size of thefuture loss at timek. Then

kL = vK(x+k)+1 − PxaK(x+k)+1 .

The reserve, denoted in this case bykVx, is the expectation of this loss variable.Hence

kVx = E[kL] = Ax+k −Pxax+k.

This is called theprospective reserve formula, since it is based on a look at thefuture performance of the insurance portfolio.


A word about notation. In the example above the reserve has been computed fora discrete whole life policy. The notation for the reserves for other types of policiesparallel the notation for the premiums for the policy. ThuskVx1:n

is the reserve attime k for ann year term policy. When discussinggeneral principals the notation

kV is used to denote the reserve at timek for a general policy.

Exercise 24–1.What types of policies have reservestV(Ax1:n

), kV(Ax1:n

), andkV(ax)?

Certain timing assumptions regarding disbursements and receipts have beenmade in the previous computation. Such assumptions are always necessary, so theyare now made explicit. Assume that a premium payment which is due at timet ispaid at timet+; an endowment benefit due at timet is paid at timet+; a death benefitpayment due at timet is assumed to be paid at timet−, that is, just before timet.Interest earned for the period is received at timet−. ThustVx includes any interestearned and also the effects of any non-endowment benefit payments but excludesany premium income payable at timet and any endowment payments to be made attime t. Also assume that the premium charged is the net level premium. Thereforethe full technical description of what has been computed is thenet level premiumterminal reserve. One can also compute thenet level premium initial reservewhich is the reserve computed right at timet. This initial reserve differs from theterminal reserve by the amount of premium received at timet and the amount of theendowment benefit paid at timet. Ordinarily one is interested only in the terminalreserve.

In the remainder of this section methodsof computing the net level premiumterminal reserve are discussed. For succintness, the term ‘reserve’ is always takento mean the net level premium terminal reserve unless there is an explicit statementto the contrary.

Exercise 24–2.Show thatkVx = 1 −ax+k

ax. From this lim

k→∞kVx = 1. Why is this

reasonable?

Exercise 24–3.Use the Life Table to compute the reserve for the first five yearsafter policy issue for a fully discrete whole life policy to (20). Assume the policyamount is equal to $100,000 and the premium is the net premium.

The reserve can be viewed in a differentway. Typically an insurance companyhas many identical policies in force. One may benefit by studying the cash flowassociated with this group of policies on the average. Here is an example.

Example 24–3.Let us examine the expected cash flow associated with a whole lifepolicy issued to (x). Assume the premium is the net level premium and that thepolicy is fully discrete. In policy yeark + 1 (that is in the time interval [k, k + 1))there are the followingexpected cash flows.


Time Income Cash on Handk− (benefits just paid, interest just received)kVx

k Px kVx + Px

k + 1− −qx+k kVx + Px − qx+k

k + 1− i(kVx + Px) (1 + i)(kVx + Px) − qx+k

This final cash on hand at timek+1−must be equal to the reserve for the policiesof the survivors. Thus

px+k k+1Vx = (1 + i)(kVx + Px) − qx+k.

This provides an important formula connecting successive reserves.

Exercise 24–4.Show that1Ex+k k+1Vx = kVx + Px − vqx+k.

The analysis of the previous example illustrates a general argument connectingthe reserves at successive time points.

kV = Actuarial Present Value at timek of benefits payable in [k, k + 1)

−Actuarial Present Value at timek of premiums payable in [k, k + 1)

+ vpx+kk+1V.

Such recursive formulas for reserves are especially useful for computational pur-poses.

There are other ways to compute the reserve. First the reserve may be viewed asmaintaining the balance between income and expenses. Since at time 0 the reserve is0 (because of the equivalence principle) the reserve can also be viewed as balancingpast income and expenses. This leads to theretrospective reserve formula

kEx kVx = Pxax:k −Ax1:k.

This formula is derived as follows. Recall that

Ax = Ax1:k

+ vkkpx Ax+k

andax = ax:k + vk

kpx ax+k.

Since the reserve at time 0 is zero,

0 = Ax −Pxax =(

Ax1:k

+ vkkpx Ax+k

)−Px

(ax:k + vk

kpx ax+k

)

wherek is an arbitrary positive integer. Rearranging terms and using the prospectiveformula for the reserve given above produces the retrospective reserve formula.


Exercise 24–5.Sometimes the retrospective reserve formula is written as

hVx = Pxax:h / hEx − hkx = Px sx:h − hkx

wherehkx is called theaccumulated cost of insurance. Find an expression forhkx.How would tkx be computed?

It is relatively straightforward to writeexpressions for the reserve for any ofthe many possible types of insurance policy. Doing this is left as an exercise forthe reader. One should keep in mind that the important point here is to be able to(ultimately) write a formula for the reserve which one can compute with the dataavailable in the life table. Hence continuous and/ormthly payment schemes needto be reduced to their equivalent annual forms. Recursive formulas are also oftenused.


Problems

Problem 24–1. True or False: For 0≤ k < n, kVx:n = 1−ax+k:n−k

ax:n. What happens at

k = n?

Problem 24–2. Find a formula for the reserve at the end of 5 years for a 10 yearterm policy with benefit $1 issued to(30) on a net single premium basis.

Problem 24–3. Show that for 0≤ t ≤ n

tV(Ax:n ) =(

P(Ax+t:n−t ) − P(Ax:n ))

ax+t:n−t .

This is called thepremium difference formula for reserves. Find similar formulasfor the other types of insurance.

Problem 24–4. Show that for 0≤ t ≤ n

tV(Ax:n ) =

(1−

P(Ax:n )P(Ax+t:n−t )

)Ax+t:n−t .

This is called thepaid up insurance formula for reserves. Find similar formulasfor the other types of insurance.

Problem 24–5. Find Px1:n

if nVx = 0.080,Px = 0.024 andPx:n

1 = 0.2.

Problem 24–6. Given that10V35 = 0.150 and that20V35 = 0.354 find10V45.

Problem 24–7. Write prospective and retrospective formulas for4020V(A20), the re-

serve at time 20 for a semi-continuous40 payment whole life policy issued to(20).

Problem 24–8. For a general fully discrete insurance let us suppose that the benefitpayable if death occurs in the time interval (h − 1,h] is bh and that this benefit ispaid at timeh, that is, at the end of the year of death. Suppose also that the premiumpaid for this policy at timeh isπh. Show that for 0≤ t ≤ 1

tpx+k k+tV + v1−ttqx+k bk+1 = (1 + i)t(kV +πk).

This gives a correct way to interpolate reserves at fractional years.

Problem 24–9. In the notation of the preceding problem show that for 0≤ t ≤ 1

k+tV = v1−t (1−tqx+k+t bk+1 + 1−tpx+k+t k+1V) .

Problem 24–10.Show that under UDD,hkV(Ax:n ) = (i/δ)hkV

x1:n+ h

kVx:n

1 .


Problem 24–11.Show that under UDD,hkV (m)x:n = h

kVx:n +β(m)hP(m)x:n kVx1:n

. This givesthe reserves for a policy withmthly premium payments in terms of the reserves fora policy with annual premium payments.

Problem 24–12.Show thathkV (m)(Ax:n ) = h

kV(Ax:n ) + β(m)hP(m)(Ax:n )kVx1:nunder

UDD.

Problem 24–13.The amount at risk in year k for a discrete insurance is thedifference between the benefit payment payable at the end of yeark andkV. Findthe mean and variance of the amount at risk in year 3 of a 5 year term policy issuedto (30) which pays a benefit of 1 at the end of the year of death and has net levelpremiums.

Problem 24–14.Suppose that 1000tV(Ax) = 100, 1000P(Ax) = 10.50, andδ =0.03. Findax+t.

Problem 24–15.Calculate20V45 given thatP45 = 0.014, P45:20

1 = 0.022, and

P45:20 = 0.030.

Problem 24–16.A fully discrete life insurance issued to (35) has a death benefit of$2500 in year 10. Reserves are calculated ati = 0.10 and the net annual premiumP. Calculateq44 given that9V + P = 10V = 500.


Solutions to ProblemsProblem 24–1. Use the prospective formula andAx:n + dax:n = 1 to see theformula is true. Whenk = n the reserve is 1 by the timing assumptions.

Problem 24–2. Prospectively the reserve isA315:5

.

Problem 24–3. Use the prospective formula and the premium definitions.

Problem 24–5. From the retrospective formulanExnVx = Pxax:n − Ax1:n

. Nowdivide by ax:n .

Problem 24–6. Use the prospective formula and the relationAx + dax = 1 toobtainkVx = 1− ax+k/ ax.

Problem 24–7. The prospective and retrospective formulas are4020V(A20) =

A40 − Pa40:20 and4020V(A20) = (Pa20:20 − A

201

:20)/20E20.

Problem 24–8. The value of the reserve, given survival, plus the present valueof the benefit, given death, must equal theaccumulated value of the prior reserveand premium.

Problem 24–13. The amount at risk random variable is1{2} (K(30))− 3V301

:5.

Problem 24–14. Use the prospective reserve formula and the relationshipAx + δax = 1.

Problem 24–15. Use the retrospective formula.

Problem 24–16. By the general recursion formula1E4410V = 9V+P−2500vq44.


Solutions to ExercisesExercise 24–1. tV(A

x1:n

) is the reserve at timet for a fully continuousn year

term insurance policy,kV(Ax1:n

) is the reserve at timek for a semi-continuousnyear term policy, andkV(ax) is the reserve at timek for a life annuity.

Exercise 24–2. SinceAx + dax = 1, kVx = Ax+k − (Ax/ ax)ax+k = 1− dax+k − (1−dax)ax+k/ ax = 1− ax+k/ ax.

Exercise 24–3. The reserve amounts are easily computed using the previousexercise as 1000001V20 = 100000(1− 19.014/19.087) = 382.46, 1000002V20 =775.40, 1000003V20 = 1184.06, 1000004V20 = 1613.67, and 1000005V20 =2064.24.

Exercise 24–4. Just multiply the previous equation byv = (1 + i)−1.

Exercise 24–5. hkx = Ax1:k/ kEx, which can be easily computed from the life

table using previous identities.

§25. Laboratory 9

1. Return to Laboratory 6 and find the reserve at the end of each policy yearfor the term policy of problem 1 at an interest rate of 5%. Begin by deriving arecurrence relation between the reserves for successive years.

2. Find the reserve at the end of each policy year for the term policy with de-clining benefits given in problem 3 of Laboratory 6. Begin by deriving a recurrencerelation between the reserves for successive years.

The problems below all concern the following situation. Your company hasjust sold 3 year term insurance policies to a group of 20 persons, each aged 30. Thebenefit amount for each policy is $1,000,000 and is payable at the moment of death.Each policy has level premiums payable at the beginning of each policy year. Themortality characteristics areq30 = 0.000444,q31 = 0.000499,q32 = 0.000562, andq33 = 0.000631. The interest rate is 4%.

3. Make a table showing the reserves for this group of policies at the end ofeach policy year.

4. Your company charges a premium which is twice the net premium. Make atable showing the expected free cash flow generated by this group of policies at theend of each year. Assume that there are no expenses.

5. Estimate the probability that this group of policies will require a cashinfusion from the other business of the company at some point during the life ofthese policies. Also estimate the amount of cash required (if any). Carefully explainyour methodology and state your assumptions.


§26. The Individual Risk Model

An insurance company has a large number of policies in force at any giventime. This creates a financial risk for the company. There are two aspects to theproblem of analyzing this risk. First, one must be able to estimate the amount ofrisk. Secondly, one must be able to model the times at which claims will occur inorder to avoid cash flow difficulties. The problem of modeling the amount of riskwill be studied first.

In theindividual risk model the insurers total riskS is assumed to be expressablein the formS = X1 + . . .+ Xn whereX1, . . . ,Xn are independent random variableswith Xi representing the loss to the insurer on insured uniti. HereXi may be quitedifferent than the actual damages suffered by insured uniti. In theclosed modelthe number of insured unitsn is assumed to be known and fixed. A model in whichmigration in and out of the insurance system is allowed is called anopen model.The individual risk model is appropriate when the analysis does not require theeffect of time to be taken into account.

The first difficulty is to find at least areasonable approximation to the proba-bilistic properties of theloss random variablesXi. This can often be done using datafrom the past experience of the company.

Example 26–1.For short term disability insurance the amount paid by the insurancecompany can often be modeled asX = cY wherec is a constant representing thedaily rate of disability payments andY is the number of days a person is disabled.One then is simply interested in modelling the random variableY. Historical datacan be used to estimateP[Y > y]. In this contextP[Y > y], which was previouslycalled the survival function, is referred to as thecontinuance function. The samenotion can be used for the daily costs of a hospitalization policy.

The second difficulty is to uncover theprobabilistic properties of the randomvariableS. In theoretical discussions the idea of conditioning can be used to findan explicit formula for the distribution function of a sum of independent randomvaribles.

Example 26–2.SupposeX andY are independent random variables each havingthe exponential distribution with parameter 1. By conditioning

P[X + Y ≤ t] =∫ ∞

−∞P[X + Y ≤ t |Y = y] fY(y) dy

=∫ t

0P[X ≤ t − y] fY(y) dy

=∫ t

0

(1− e−(t−y)

)fY(y) dy

= 1− e−t − te−t


§26: The Individual Risk Model 99

for t ≥ 0.

This argument has actually shown that ifX andY are independent and absolutelycontinuous then

FX+Y(t) =∫ ∞

−∞FX(t − y) fY(y) dy.

This last integral is called theconvolution of the two distribution functions and isoften denotedFX ∗ FY .

Exercise 26–1.If X andY are absolutely continuous random variables show thatX + Y is also absolutely continuous and finda formula for the density function ofX + Y.

Exercise 26–2.Find a similar formula ifX andY are both discrete. Use this formulato find the density ofX + Y if X andY are independent Bernoulli random variableswith the same success probability.

An approach which requires less detailed computation is to appeal to the CentralLimit Theorem.

Central Limit Theorem. If X1, . . . ,Xn are independent randomvariables then the

distribution ofn∑

i=1

Xi is approximately the normal distribution with meann∑

i=1

E[Xi]

and variancen∑

i=1

Var(Xi).

The importance of this theorem lies in the fact that the approximating normaldistribution does not depend on the detailed nature of the original distribution butonly on the first two moments. The accuracy of this approximation will be exploredin the exercises and laboratory.

Example 26–3.You are a claims adjuster for the Good Driver Insurance Companyof Auburn. Based on past experience the chance of one of your 1000 insureds beinginvolved in an accident on any given day is 0.001. Your typical claim is $500. Whatis the probability that there are no claims made today? If you have $1000 cash onhand with which to pay claims, what is the probability you will be able to pay allof todays claims? How much cash should you have on hand in order to have a99% chance of being able to pay all of todays claims? What assumptions have youmade? How reasonable are they? Whatdoes this say about the solvency of yourcompany?

It is easily seen, by using the Central Limit Theorem, that if an insurancecompany sold insurance at the pure premium not only would the company onlybreak even (in the long run) but due to random fluctuations of the amount ofclaims the company would likely go bankrupt. Thus insurance companies charge


an amount greater than the pure premium. A common methodology is for thecompany to charge (1+θ) times the pure premium. When this scheme is followedθis called therelative security loadingand the amountθ × (pure premium) is calledthesecurity loading. This is a reasonable procedure since the insureds with largerexpected claims pay a proportionate share of the loading. The relative loadingθ isusually adjusted to achieve a certain measure of protection for the company.

Example 26–4.Suppose that a company is going to issue 1,000 fire insurancepolicies each having a $250 deductible, and a policy amount of $50,000. Denoteby Fi the Bernoulli random variable which is 1 if theith insured suffers a loss, andby Di the amount of damage to theith insureds property. SupposeFi has successprobability 0.001 and that the actual damageDi is uniformly distributed on theinterval (0,70000)). What is the relative loading so that the premium income willbe 95% certain to cover the claims made? Using the obvious notation, the totalamount of claims made is given by the formula

S =1000∑i=1

Fi[(Di − 250)1[250,50250](Di) + 500001(50250,∞)(Di)

]

where theF’s and theD’s are independent (why?) and for eachi the conditionaldistribution ofDi given Fi = 1 is uniform on the interval (0,70000). The relativesecurity loading is determined so that

P[S ≤ (1 +θ) E[S]] = 0.95.

This is easily accomplished by using the Central Limit Theorem.

Exercise 26–3.ComputeE[S] and Var(S) and then use the Central Limit Theoremto findθ. What is the probability of bankruptcy whenθ = 0?

Another illustration is in connection withreinsurance. It is generally not goodpractice for an insurance company to have all of its policy holders homogeneous,such as all located in one geographicalarea, or all of the same physical type. Amoments reflection on the effect of a hurricane on an insurance company with allof its property insurance business located in one geographic area makes this pointclear. An insurance company may diversify its portfolio of policies (or just protectitself from such a concentration of business) by buying or selling reinsurance. Thecompany seeking reinsurance (thecedingcompany) buys an insurance policy fromthe reinsurer which will reimbursethe company for claims above theretentionlimit . For stop lossreinsurance, the retention limit applies on a policy-by-policybasis to those policies covered by the reinsurance. The retention limit plays thesame role here as a deductible limit in a stop loss policy. Usually there is onereinsurance policy which covers an entire package of original policies. Forexcessof lossreinsurance, the retention limit is applied to the total amount of claims for thepackage of policies covered by the insurance, not the claims of individual policies.


Example 26–5.You operate a life insurance company which has insured 2,000 30year olds. These policies are issued in varying amounts: 1,000 people with $100,000policies, 500 people with $500,000 policies, and 500 people with $1,000,000 poli-cies. The probability that any one of the policy holders will die in the next year is0.001. Stop loss reinsurance may be purchased at the rate of 0.0015 per dollar ofcoverage. How should the retention limit be set in order to minimize the probabil-ity that the total expenses (claims plus reinsurance expense) exceed $1,000,000 isminimized? LetX, Y, andZ denote the number of policyholders in the 3 catagoriesdying in the next year. ThenX has the binomial distribution based on 1000 trialseach with success probability 0.001,Y has the binomial distribution based on 500trials each with success probability 0.001, andZ has the binomial distribution basedon 500 trials each with success probability 0.001. If the retention limit is set atrthen the costC of claims and reinsurance is given by

C = (100000∧ r)X + (500000∧ r)Y + (1000000∧ r)Z

+ 0.0015[1000(100000− r)+ + 500(500000− r)+ + 500(1000000− r)+] .

It is then a relatively straightforward,though tedious, task to use the central limittheorem to estimateP[C ≥ 1,000,000].

Exercise 26–4.Verify the validity of the above formula. Use the central limittheorem to estimateP[C ≥ 1,000,000] as a function ofr. Find the value(s) ofrwhich minimize this probability.


Problems

Problem 26–1. The probability of an automobile accident in a given time period is0.001. If an accident occurs the amount of damage is uniformly distibuted on theinterval (0,15000). Find the expectation and variance of the amount of damage.

Problem 26–2. Find the distribution and densityfor the sum of three independentrandom variables each uniformly distributed on the interval (0,1). Compare theexact value of the distribution function at a few selected points (say 0.25, 1, 2.25)with the approximation obtained from the central limit theorem.

Problem 26–3. Repeat the previous problem for 3independent exponential randomvariables each having mean 1. It may help to recall the gamma distribution here.

Problem 26–4. A company insures 1000 essentially identical cars. The probabilitythat any one car is in an accident in any given year is 0.001. The damage to a carthat is involved in an accident is uniformly distributed on the interval (0,15000).What relative security loadingθ should be used if the company wishes to be 99%sure that it does not lose money?


Solutions to ProblemsProblem 26–1. The amount of damage isBU whereB is a Bernoulli variablewith success probability 0.001 andU has the uniform distribution.

Problem 26–4. The loss random variable is of the form1000∑i=1

BiUi.


Solutions to ExercisesExercise 26–1. Differentiation of the general distribution function formulaabove givesfX+Y(t) =

∫ ∞−∞ fX(t − y)fY (y) dy.

Exercise 26–2. In the discrete case the same line of reasoning givesfX+Y(t) =∑y fX(t−y)fY(y). Applying this in the Bernoulli case,fX+Y(t) =

∑my=0

( nt−y

)pt−y(1−

p)n−t+y(m

y

)py(1− p)m−y = pt(1− p)n+m−t

∑my=0

( nt−y

)(my

)=(n+m

t

)pt(1− p)m+n−t.

Exercise 26–3. The loadingθ is chosen so thatθE[S]/ √Var(S) = 1.645, fromthe normal table. Whenθ = 0 the bankruptcy probability is about 1/2.

Exercise 26–4. Direct computation using properties of the binomial distribu-tion givesE[C] = (100000∧ r)×1+(500000∧ r)×(1/2)+(1000000∧ r)×(1/2)+0.0015

[1000(100000− r)+ + 500(500000− r)+ + 500(1000000− r)+

]and also

Var(C) = (100000∧ r)2×0.999+(500000∧ r)2×0.999/2+(1000000∧ r)2×0.999/2.The probability can now be investigated numerically using the Central LimitTheorem approximation.

§27. Laboratory 10

Another method of uncovering the probabilistic properties of the random vari-able S is to use simulation. Suppose that acompany insures 10,000 essentiallyidentical cars. The probability that any one car is in an accident in any given year is0.001. The damage to a car that is involved in an accident is uniformly distributedon the interval (0,20000).

1. Write the loss random variableS in this case. FindE[S] and Var(S).

2. Use the Central Limit Theorem to find the relative security loadingθ thatshould be used if the company wishes to be 99% sure that it does not lose money.

3. Consider one of the loss random variablesX that occur in the expression forS. Explain how a random number generator could be used to simulateX.

4. Use the method of problem 3 to simulate 5,000 observations onS. Makea histogram of these observations. Basedon this histogram, what relative securityloading should be used if the company wishes to be 99% sure that it does not losemoney? Compare the result here with that of problem 2.

5. Show that ifY is a random variable with increasing distribution functionFY(t) and if U is uniformly distributed on the interval (0,1), thenF−1

Y (U) has thesame distribution asY. This gives a general method to use for simulating randomvariables.


§28. The Collective Risk Model and Ruin Probabilities

Some of the consequences of thecollective risk modelwill now be examined.In the collective risk model the time at which claims are made is taken to account.Here the aggregate claims up to timet is assumed to be given by

∑N(t)k=1 Xk where

X1,X2, . . .are independent identically distributed random variables representing thesizes of the respective claims,N(t) is a stochastic process representing the numberof claims up to timet, andN and theX’s are independent. The object of interest isthe insurer’s surplus at timet, denoted byU(t), which is assumed to be of the form

U(t) = u + ct −N(t)∑k=1

Xk

whereu is the surplus at timet = 0, andc represents the rate of premium income.Special attention will be given to the problem of estimating the probability that theinsurance company has negative surplus at some timet since this would mean thatthe company is ruined.

To gain familiarity with some of the ideas involved, the simpler classical gam-bler’s ruin problem will be studied first.


§29. Stopping Times and Martingales

A discrete time version of the collective risk model will be studied and someimportant new concepts will be introduced.

Suppose that a gambler enters a casino withz dollars and plays a game of chancein which the gambler wins $1 with probabilityp and loses $1 with probabilityq = 1−p. Suppose also that the gambler will quit playing if his fortune ever reachesa > z and will be forced to quit by being ruined if his fortune reaches 0. The maininterest is in finding the probability that the gambler is ultimately ruined and alsothe expected number of the plays in the game.

In order to keep details to a minimum, the case in whichp = q = 1/2 willbe examined first. Denote byXj the amount won or lost on thejth play of thegame. These random variables are all independent and have the same underlyingdistribution function. Absent any restrictions about having to quit the game, thefortune of the gambler afterk plays of the game is

z +k∑

j=1

Xj.

Now in the actual game being played the gambler either reaches his goal or is ruined.Introduce a random variable,T, which marks the play of the game on which thisoccurs. Technically

T = inf{ k : z +∑k

j=1 Xj = 0 ora} .

Such a random variable is called arandom time. Observe that for this specificrandom variable the event [T ≤ k] depends only on the random variablesX1, . . . ,Xk.That is, in order to decide at timek whether or not the game has ended it is notnecessary to look into the future. Such special random times are calledstoppingtimes. The precise definition is as follows. IfX1,X2, . . .are random variables andT is a nonnegative integer valued random variable with the property that for eachintegerk the event [T ≤ k] depends only onX1, . . . ,Xk thenT is said to be astoppingtime (relative to the sequenceX1,X2, . . .).

The random variablez +∑T

j=1 Xj is the gambler’s fortune when he leaves thecasino, which is eithera or 0. Denote byψ(z) the probability that the gamblerleaves the casino with 0. Then by direct computationE[z +

∑Tj=1 Xj] = a(1 − ψ(z)).

A formula for the ruin probabilityψ(z) will be obtained by computing this sameexpectation in a second way.

Each of the random variablesXj takes values 1 and−1 with equal probability,soE[Xj] = 0. Hence for any integerk, E[

∑kj=1 Xj] = 0 too. So it is at least plausible

that E[∑T

j=1 Xj] = 0 as well. Using this fact,E[z +∑T

j=1 Xj] = z, and equating this


§29: Stopping Times and Martingales 108

with the expression above givesz = a(1−ψ(z)). Thusψ(z) = 1− z/a for 0 ≤ z ≤ aare the ruin probabilities.

There are two important technical ingredients behind this computation. Thefirst is the fact thatT is a stopping time. The second is the fact that the gamblinggame withp = q = 1/2 is a fair game. The notion of a fair game motivates thedefinition of a martingale. SupposeM0, M1, M2, . . .are random variables. Thesequence is said to be amartingale if E[Mk |Mk−1, . . . ,M0] = Mk−1 for all k ≥ 1. Inthe gambling context, ifMk is the gambler’s fortune afterk plays of a fair game thengivenMk−1 the expected fortune after one more play is stillMk−1.

Exercise 29–1.Show thatMk = z +∑k

j=1 Xj (with M0 = z) is a martingale.

Example 29–1.The sequenceM0 = z2 andMk =(

z +∑k

j=1 Xj

)2− k for k ≥ 1 is also

a martingale. This follows from the fact that knowingM0, . . . ,Mk−1 is the same asknowingX1, . . . ,Xk−1 and the fact that theX’s are independent.

Exercise 29–2.Fill in the details behind this example.

The important computational fact is theOptional Stopping Theorem whichstates that if{Mk} is a martingale andT is a stopping time thenE[MT ] = E[M0]. Inthe gambling context this says that no gambling strategyT can make a fair gamebiased.

Example 29–2.Using the martingaleM0 = z2 and Mk =(z +∑k

j=1 Xj

)2− k for

k ≥ 1 along with the same stopping timeT as before can provide informationabout the duration of the gambler’s stay in the casino. The random variableMT =(

z +∑T

j=1 Xj

)2−T has an expectation which is easily computed directly to beE[MT] =

a2(1 − ψ(z)) − E[T]. By the optional stopping theorem,E[MT ] = E[M0] = z2.Comparing these two expressions givesE[T] = a2(1 − ψ(z)) − z2 = az − z2 as theexpected duration of the game.

The preceding example illustrates the general method. To analyze a particularproblem identify a martingaleMk and stopping timeT. Then computeE[MT ] in twoways, directly from the definition and by using the optional stopping theorem. Theresulting equation will often reveal useful information.

Uncovering the appropriatemartingale is often the most difficult part of theprocess. One standard method is the following. IfX1,X2, . . . are independent andidentically distributed random variables define

Wk =et∑k

j=1Xj

E[et∑k

j=1Xj].


Notice that the denominator is nothing more than the moment generating functionof the sum evaluated att. For each fixedt the sequenceWk is a martingale (hereW0 = 1). This follows easily from the fact that ifX andY are independent thenE[et(X+Y)] = E[etX ] E[etY ]. This martingale is calledWald’s martingale (or theexponential martingale) for theX sequence.

Exercise 29–3.Show that{Wk : k ≥ 0} is a martingale no matter what the fixedvalue oft is.

In many important cases a non-zero value oft can be found so that the denomi-nator part of the Wald martingale is 1. Using this particular value oft then makesapplication of the optional stopping theorem neat and easy.

To illustrate the technique consider the following situation which is closerto that of the collective risk model. Suppose the insurer has initial reservezand that premium income is collected at the rate ofc per unit time. Also,Xk

denotes the claims that are payable at timek, and theX’s are independent andidentically distributed random variables. The insurers reserve at timek is thenz + ck −

∑kj=1 Xj = z +

∑kj=1(c −Xj). Denote byT the time of ruin, so that

T = min{ k : z + ck −k∑

j=1

Xj ≤ 0} .

The objective is to study the probabilityψ(z) that ruin occurs in this setting.

As a first step, notice that ifE[c − Xj] ≤ 0, ruin is guaranteed since premiumincome in each period is not adequate to balance the average amount of claims inthe period. So to continue, assume thatE[c − Xj] > 0.

Under this assumption, suppose there is a numberτ so thatE[eτ (c−Xj)] = 1.This choice ofτ in Wald’s martingale makes thedenominator 1, and shows that

Mk = eτ (z+ck−∑k

j=1 Xj) is a martingale. Computing the expectation ofMT using theOptional Stopping Theorem givesE[MT ] = E[M0] = eτz. Computing directly gives

E[MT ] = E[eτ (z+cT−∑T

j=1 Xj) |T < ∞] ψ(z). Hence

ψ(z) = eτz/E[eτ (z+cT−∑T

j=1 Xj) |T < ∞].

A problem below will show thatτ < 0, so the denominator of this fraction is largerthan 1. Henceψ(z) ≤ eτz. The ruin probability decays exponentially as the initialreserve increases.

The usual terminology defines theadjustment coefficient R = −τ . Thusψ(z) ≤ e−Rz. So a large adjustment coefficient implies that the ruin probabilitydeclines rapidly as the initial reserve increases.


Problems

Problem 29–1. By conditioning on the outcome of the first play of the game showthat in the gambler’s ruin problemψ(z) = pψ(z + 1) +qψ(z −1). Show that ifp = qthere is a solution of this equation of the formψ(z) = C1 + C2z and findC1 andC2

by using the natural definitionsψ(0) = 1 andψ(a) = 0. Show that ifp ≠ q there is asolution of the formψ(z) = C1 + C2(q/p)z and find the two constants. This providesa solution to the gambler’s ruin problemby using difference equations instead ofprobabilistic reasoning.

Problem 29–2. In the gambler’s ruin problem, show that ifp ≠ q the choicet =ln(q/p) makes the denominator of Wald’s martingale 1. Use this choice oft and theoptional stopping theorem to find the ruin probability in this case.

Problem 29–3. Supposep ≠ q in the gambler’s ruin problem. DefineM0 = z andMk = z +

∑kj=1 Xj − k(p−q) for k ≥ 1. Show that the sequenceMk is a martingale and

use it to computeE[T] in this case.

Problem 29–4. Suppose thatc > 0 is a number andX is a random variable whichtakes on only non-negative values. Suppose also thatE[c − X] > 0. Show that ifc − X takes on positive and negative values then there is a numberτ < 0 so thatE[eτ (c−X)] = 1.



Problem 29–2. ψ(z) = (q/p)a−(q/p)z

(q/p)a−1 .

Problem 29–3. E[T] = zq−p −

aq−p

1−(q/p)z

1−(q/p)a .

Problem 29–4. Define a functionf (v) = E[ev(c−X)]. Then f ′ (v) = E[(c −X)ev(c−X)] and f ′′ (v) = E[(c − X)2ev(c−X)] > 0. Thusf is a convex function andthe graph off is concave up. Nowf (0) = 1 andf ′ (0) = E[(c − X)] > 0. Thusthe graph off is above 1 to the right of 0, and below 1 (initially) to the left of0. Sincec −X takes on negative values, limv→−∞ f (v) = ∞, so there is a negativevalue ofv at whichf (v) = 1, by continuity.


Solutions to ExercisesExercise 29–1. KnowingM0, . . . ,Mk−1 is the same as knowingX1, . . . ,Xk−1. SoE[Mk |M0, . . . ,Mk−1] = E[Mk |X0, . . . ,Xk−1] = z +

∑k−1j=1 Xj + E[Xk |X0, . . . ,Xk−1] =

Mk−1 since the last expectation is 0 by independence.

Exercise 29–2. First writeMk =(

z +∑k−1

j=1 Xj + Xk

)2− k =

(z +∑k−1

j=1 Xj

)2+

2Xk(z +∑k−1

j=1 Xj) + X2k − k. Take conditional expectations using the fact thatXk

is independent of the otherX’s andE[Xk] = 0 andE[X2k ] = 1 to obtain the result.

Exercise 29–3. IndependencegivesE[et∑k

j=1Xj ] = E[et

∑k−1

j=1Xj ]×E[etXk]. Direct

computation of the conditional expectation gives the result.

§30. The Collective Risk Model Revisited

The ideas developed in connection with the gambler’s ruin problem will nowbe used to compute the ruin probability in the collective risk model. Since theprocesses are now operating in continuoustime the details are more complicatedand not every step of the arguments will be fully justified.

In this setting the claims process is∑N(t)

k=1 Xk whereX1,X2, . . . are independentidentically distributed random variables representing the sizes of the respectiveclaims,N(t) is a stochastic process representing the number of claims up to timet,andN and theX’s are assumed to be independent. Theinsurer’s surplus is given byU(t) = u + ct −

∑N(t)k=1 Xk, whereu > 0 is the surplus at timet = 0 andc > 0 is the rate

at which premium income arrives per unit time. The probability of ruin with initialsurplusu will be denoted byψ(u).

As in the discrete time setting, the Wald martingale will be used together with theOptional Stopping Theorem in order to obtain information about the ruin probability.Here the denominator of the Wald martingale isE[eνU(t)], and the first step is to find

aν ≠ 0 so thatE[eν(ct−∑N(t)

k=1 Xk)] = 1 no matter the value oft.

The new element in this analysis is the random sum∑N(t)

k=1 Xk. Now for eachfixed t, N(t) is a random variable which is independent of theX’s. The momentgenerating function of this sum can be easily computed by conditioning on the valueof the discrete random variableN(t).

E[eν∑N(t)

k=1 Xk ] = E[E[eν∑N(t)

k=1 Xk |N(t)]]

=∞∑

j=0

E[eν∑N(t)

k=1Xk |N(t) = j] P[N(t) = j]

=∞∑

j=0

E[eν∑j

k=1Xk] P[N(t) = j]

=∞∑

j=0

(E[eνX]

)jP[N(t) = j]

=∞∑

j=0

ej ln(E[eνX ]) P[N(t) = j]

= MN(t)(ln(MX(ν))).

Hence there is aν≠0 so thatE[eν(ct−∑N(t)

k=1Xk)] = 1 if and only ifeνct MN(t)(ln(MX(−ν))) =

1 for all t. Suppose for now that there is a numberR > 0 so that

e−Rct MN(t)(ln(MX(R))) = 1

for all t. This numberR is called theadjustment coefficient. The existence of anadjustment coefficient will be investigated a bit later. Using−R as the value ofν in


§30: The Collective Risk Model Revisited 114

Wald’s martingale shows that

Wt = e−R(u+ct−∑N(t)

k=1Xk)

is a martingale.

Define a stopping timeTa by Ta = inf{ s : u + cs −∑N(s)

k=1 Xk ≤ 0 or≥ a} whereais an arbitrary but fixed positive number. It is intuitively clear thatTa is a stoppingtime in an appropriate sense in the new continuous time setting. Now by the OptionalStopping Theorem,E[WTa ] = e−Ru. Direct computation gives

E[WTa] = E[e−R(u+cTa−∑N(Ta)

k=1 Xk) |u + cTa −N(Ta)∑k=1

Xk ≤ 0] P[u + cTa −N(Ta)∑k=1

Xk ≤ 0]

+ E[e−R(u+ct−∑N(t)

k=1 Xk) |u + cTa −N(Ta)∑k=1

Xk ≥ a] P[u + cTa −N(Ta)∑k=1

Xk ≥ a].

Since this equation is valid for any fixed positivea, and sinceR > 0, it is possibleto take limits asa → ∞. Since lima→∞ P[u + cTa −

∑N(Ta)k=1 Xk) ≤ 0] = ψ(u) and

lima→∞ e−Ra = 0 the following result is obtained.

Theorem. Suppose that in the collective risk model the adjustment coefficientR > 0satisfiese−Rct MN(t)(ln(MX(R))) = 1 for all t. Let T = inf{ s : u + cs −

∑N(s)k=1 Xk ≤ 0}

be the random time at which ruin occurs. Then

ψ(u) =e−Ru

E[e−R(u+cT−∑N(T)

k=1 Xk) |T < ∞]≤ e−Ru.

Exercise 30–1.Why is the last inequality true?

As in the discrete time model, the existence of an adjustment coefficient guar-antees that the ruin probability decreases exponentially as the initial surplusuincreases.

In general there is no guarantee that an adjustment coefficient will exist. Forcertain particular types of models the adjustment coefficient can explicitly be found.Moreover, a more detailed analysis of the claims process can be made in thesespecial cases.

The more restrictive discussion begins by examining the nature of the processN(t), the total number of claims up to timet. A common assumption is that thisprocess is a Poisson process with constant intensityλ > 0. What this assumptionmeans is the following. SupposeW1,W2, . . .are independent identically distributedexponential random variables with mean 1/λ and common densityλ e−λ x1(0,∞)(x).


The W ’s are the waiting times between claims. The Poisson process can thenbe viewed as the number of claims that arrive up to timet. This means thatN(t) = inf{ k :

∑k+1j=1 Wj > t} . It can be shown that for any fixedt the random variable

N(t) has the Poisson distribution with parameterλ t and that the stochastic process{N(t) : t ≥ 0} hasindependent increments, that is, whenevert1 < t2 < . . . < tn arefixed real numbers then the random variablesN(t2) − N(t1), . . . ,N(tn) − N(tn−1) areindependent. Using this, direct computation gives

E[eνN(t)] =∞∑

j=0

eνjP[N(t) = j]

=∞∑

j=0

eνje−λ t(λ t)j/ j!

= e−λ t∞∑

j=0

(eνλ t)j/ j!

= eλ t(eν −1).

This simple formula for the moment generating function ofN(t) leads to a simpleformula for the adjustment coefficient in this case. The general equation for theadjustment coefficient was earlier found to bee−Rct MN(t)(ln(MX(R))) = 1. Takinglogarithms and using the form of the moment generating function ofN(t) shows thatthe adjustment coefficient is the positive solution of the equation

λ + cR = λMX(R).

An argument similar to that given in the discrete time case can be used to show thatthere is a unique adjustment coefficient in this setting.

Exercise 30–2.Verify that the adjustment coefficient, if it exists, must satisfy thisequation.

Example 30–1.Suppose all claims are for a unit amount. ThenMX(ν) = eν so theadjustment coefficient is the positive solution ofλ + cR = λ eR. Note that there isno solution ifc ≤ λ . But in this case the ruin probability is clearly 1.

Exercise 30–3.Show that ifc ≤ λE[X] the ruin probability is 1. Show that ifc > λE[X] the adjustment coefficient always exists and hence the ruin probabilityis less than 1.

The previous exercises suggest that only the case in whichc > λE[X] is ofinterest. Henceforth writec = (1 +θ)λE[X] for someθ > 0. Hereθ is the relativesecurity loading.

Even more detailed information can be obtained whenN(t) is a Poisson process.To do this define a stopping timeTu = inf{ s : U(s) < u} to be the first time that


the surplus falls below its initial level and denote byL1 = u −U(Tu) the amount bywhich the surplus falls below its initial level. Then

P[Tu < ∞,L1 ≥ y] =1

(1 +θ)E[X]

∫ ∞

y(1− FX(x)) dx.

The proof of this fact is rather technical.

proof : Let h > 0 be small. ThenP[N(h) = 0] = e−λ h ≈ 1, P[N(h) = 1] = λ he−λ h ≈ λ h andP[N(h) ≥ 2] ≈ 0. Denote byR(u, y) the probability that with an initial surplus ofu the firsttime the surplus drops below 0, the surplus actually drops below−y. Conditioning on thevalue ofN(h) gives

R(u, y) ≈ (1− λ h)R(u + ch, y) + λ h

(∫ u

0R(u − x, y)fX(x) dx +

∫ ∞

u+ch+yfX(x) dx

).

Re-arranging gives

R(u, y) − R(u + ch, y)ch

= −λc

R(u + ch, y) +λc

∫ u


λc

∫u+ch+y

fX(x) dx.

Now take limits ash → 0 to obtain

−R′ (u, y) = −λc

R(u, y) +λc

∫ u


λc

∫u+y

fX(x) dx.

SinceR(u, y) ≤ ψ(u) ≤ e−Ru, both sides can be integrated with respect tou from 0 to∞.Doing this gives

R(0, y) = −λc

∫ ∞

0R(u, y) du +

λc

∫ ∞

0

∫ u

0R(u − x, y)fX(x) dx du +

λc

∫ ∞

0

∫ ∞

u+yfX(x) dx du.

Interchanging the order of integration in the double integrals shows that the first double

integral is equal to∫ ∞

0R(u, y) du, while the second double integral is equal to

∫ ∞

y

∫ x−y

0fX(x) du dx =

∫ ∞

y(x − y)fX(x) dx

=∫ ∞

yxfX(x) dx − yP[X ≥ y]

=∫ ∞

y(1− FX(x)) dx

after integration by parts. Substitution now completes the proof after usingc = (1+θ)λE[X].

This formula has two useful consequences. First, by takingy = 0, the probabilitythat the surplus ever drops below its initial level is 1/ (1 + θ). Second, an explicitformula for thesize of the drop below the initial levelis obtained as

P[L1 ≤ y|Tu < ∞] =1

E[X]

∫ y

0(1− FX(x)) dx.


This expression can be evaluated in certain cases.

Exercise 30–4.Derive this expression forP[L1 ≤ y|Tu < ∞].

Exercise 30–5.What is the conditional distribution ofL1 givenTu < ∞ if the claimsize has an exponential distribution with mean 1/δ?

Exercise 30–6.Show that the conditional moment generating function ofL1 givenTu < ∞ is (MX(t) − 1)/ (tE[X]).

This information can also be used to study the random variableL which repre-sents themaximum aggregrate lossand is defined byL = maxt≥0{

∑N(t)k=1 Xk − ct} .

Note thatP[L ≤ u] = 1−ψ(u) from which it is immediately seen that the distributionof L has a discontinuity at the origin of size 1−ψ(0) = θ/ (1 +θ), and is continuousotherwise. In fact a reasonably explicit formula for the moment generating functionof L can be obtained.

Theorem. If N(t) is a Poisson process andL = maxt≥0{∑N(t)

k=1 Xk − ct} then

ML(ν) =θE[X]ν

1 + (1 +θ)E[X]ν −MX(ν).

proof : Note from above that the size of each new deficit does not depend on the initial startingpoint of the surplus process. Thus

L =D∑

j=1

Aj

whereA1,A2, . . .are independent identically distributed random variables each having thesame distribution as the conditional distribution ofL1 given Tu < ∞, andD is a randomvariable independent of theA’s which counts the number of times a new deficit level isreached. From here it is a simple matter to compute the moment generating function ofL.

Exercise 30–7.Complete the details of the proof.

This formula for the moment generating function ofL can sometimes be usedto find an explicit formula for the distribution function ofL, and henceψ(u) =1− P[L ≤ u].

There are some other interesting consequences of the assumption that the claimnumber processN(t) is a Poisson process.

First, a bit of notation. IfA andB are random variables, writeA d= B to denotethatA andB have the same distribution.


A random variableS is said to have thecompound Poisson distributionwith Poisson parameterλ and mixing distributionF(x), denotedS

d= CP(λ , F),if S

d=∑N

j=1 Xj whereX1,X2, . . .are independent identicallydistributed random vari-ables with common distribution functionF andN is a random variable which isindependent of theX’s and has a Poisson distribution with parameterλ .

Example 30–2.For each fixedt, the aggregate claims processCP(λ t,FX).

Example 30–3. If Sd= CP(λ , F) then the moment generating function ofS is

MS(ν) = exp{ λ∫ ∞

−∞(euν − 1)dF(u)} .

This follows from the earlier general derivation of the momentgenerating functionof a random sum.

Exercise 30–8.Suppose thatSd= CP(λ , F) andT

d= CP(δ,G) and thatS andT are

independent. Show thatS + Td= CP(λ + δ, λ

λ+δF + δλ+δG).

This last property is very useful in the insurance context. Because of thisproperty the results of the analysis of different policy types can be easily combinedinto one grand analysis of the company’s prospects as a whole. A compound Poissondistribution can also be decomposed.

Example 30–4.Suppose each claim is either for $1 or $2, each event havingprobability 0.5. If the number of claims is Poisson with parameterλ then theamount of total claims,S, is compound Poisson distributed with moment generatingfunction

MS(ν) = exp{0.5λ (eν − 1) + 0.5λ (e2ν − 1)} .

HenceSd= Y1 +2Y2 whereY1 andY2 are independent Poisson random variables with

meanλ /2. Thus the number of claims of each size are independent!

Example 30–5.The collective risk model can be used as an approximation tothe individual risk model. In the individual risk model the claim amount is oftenrepresented by a productBjXj in which B is a Bernoulli random variable whichrepresents whether a claim is paid or not andX is the amount of the claim. Then

BXd=

B∑j=1

X′j ≈N∑

j=1

X′j

whereN has a Poisson distribution with parameterP[B = 1] and X′1,X′2, . . . areindependent random variables each having the same distribution asX. Thus thedistribution ofBX may be approximated by theCP(P[B = 1],FX) distribution.


Problems

Problem 30–1. If N has a Poisson distribution with parameterλ expressP[N = k]in terms ofP[N = k − 1]. This gives a recursive method of computing Poissonprobabilities.

Problem 30–2. Show that ifX takes positive integer values andSd= CP(λ , FX)

thenx P[S = x] =∑∞

k=1 λ kP[X = k]P[S = x − k] for x > 0. This is calledPanjer’srecursion formula. Hint: First show, using symmetry, thatE[Xj |S = x,N = n] =x/n for 1 ≤ j ≤ n and then write out what this means.

Problem 30–3. Suppose in the previous problem thatλ = 3 and thatX takes on thevalues 1, 2, 3, and 4 with probabilities 0.3, 0.2, 0.1, and 0.4 respectively. CalculateP[S = k] for 0 ≤ k ≤ 40.

Problem 30–4. SupposeS1 has a compound Poisson distribution withλ = 2 andthat the compounded variable takes on the values 1, 2, or 3 with probabilities 0.2,0.6, and 0.2 respectively. SupposeS2 has a compound Poisson distribution withparameterλ = 6 and the compounded variable takes on the values 3 or 4 withprobabilities 1/2 each. IfS1 and S2 are independent, what is the distribution ofS1 + S2?

Problem 30–5. The compound Poisson distribution isnot symmetric about itsmean, as the normal distribution is. One might therefore consider approximation ofthe compound Poisson distribution by some other skewed distribution. A randomvariableG is said to have the Gamma distribution with parametersα andβ if G hasdensity function

fG(x) =βα

Γ(α )xα−1e−βx1(0,∞)(x).

It is useful to recall the definition and basic properties of the Gamma function in thisconnection. One easily computes the moments of such a random variable. In factthe moment generating function isMG(ν) = (β/β − ν)α . The case in whichβ = 1/2and 2α is a positive integer corresponds to the chi–square distribution with 2αdegrees of freedom. Also the distribution of the sum ofn independent exponentialrandom variables with mean 1/β is a gamma distribution with parametersn andβ.For approximation purposes theshifted gamma distribution is used to approximatethe compound Poisson distribution. This means that anα , β, andx is found so thatx + G has approximately the same distribution as the compound Poisson variate.The quantitiesx, α , andβ are found by using the method of moments. The firstthree central moments of both random variables are equated, and the equations arethen solved. Show that when approximating the distribution of a compound Poisson


random variableS the method of moments leads to

α =4[Var(S)]3

[E[(S − E[S])3]] 2=

4λ(E[X2]

)3

(E[X3]

)2β =

2Var(S)E[(S − E[S])3]

=2E[X2]E[X3]

x = E[S] −2[Var(S)]2

E[(S − E[S])3]= λE[X] −

2λ(E[X2]

)2

E[X3].

Problem 30–6. A random variableX is amixture of exponential random variablesif the value ofX is determined in the following way. Fix a number 0< p < 1.Perform a two stage experiment. In the first stage, select a numberU at random inthe interval (0,1). For the second stage, proceed as follows. IfU < p select thevalue ofX to be the value of an exponential random variable with parameterλ1. IfU > p select the value ofX to be the value of an exponential random variable withparameterλ2. Show that the density ofX is fX(x) = pλ1e−λ1x + (1 − p)λ2e−λ2x forx ≥ 0. Show that the moment generating function ofX is E[etX ] = p

λ1−t + 1−pλ2−t .

Problem 30–7. What is the density of a random variableX with moment generatingfunctionE[etX ] = (30− 9t)/2(5− t)(3− t) for 0 < t < 3?

Problem 30–8. In the continuous time model, if the individual claimsX havedensityfX(x) = (3e−3x + 7e−7x)/2 for x > 0 andθ = 1, find the adjustment coefficientandψ(u).

Problem 30–9. In the continuous time model, if the individual claimsX are discretewith possible values 1 or 2 with probabilities 1/4 and 3/4 respectively, and if theadjustment coefficient is ln(2), find the relative security loading.

Problem 30–10.Use integration by parts to show that the adjustment coefficient in

the continuous time model is the solution of the equation∫ ∞

0erx(1−FX(x)) dx = c/λ .

Problem 30–11. In the continuous time model, use integration by parts to findML1(t). Find expressions forE[L1], E[L2

1] and Var(L1). Here L1 is the randomvariable which is the amount by which the surplus first falls below its initial level,given that this occurs.

Problem 30–12.Find the moment generating function of the maximum aggregateloss random variable in the case in which all claims are of size 5. What isE[L]?Hint: Use the Maclaurin expansion ofMX(t) to find the Maclaurin expansion ofML(t).


Problem 30–13. If ψ(u) = 0.3e−2u + 0.2e−4u + 0.1e−7u, what is the relative securityloading?

Problem 30–14. If L is the maximum aggregate loss random variable, find expres-sions forE[L], E[L2], and Var(L) in terms of moments ofX.

Problem 30–15. In the compound Poisson continous time model suppose thatλ = 3, c = 1, andX has densityfX(x) = (e−3x + 16e−6x)/3 for x > 0. Find the relativesecurity loading, the adjustment coefficient, and an explicit formula for the ruinprobability.

Problem 30–16. In the compound Poisson continous time model suppose that

λ = 3, c = 1, andX has densityfX(x) =9x25

e−3x/5 for x > 0. Find the relative security

loading, the adjustment coefficient, and an explicit formula for the ruin probability.What happens ifc = 20?

Problem 30–17.The claim number random variableis sometimes assumed to havethe negative binomial distribution. A random variableN is said to have the n

¯egative

binomial distribution with parametersp and r if N counts the number of failuresbefore therth success in a sequence of independent Bernoulli trials, each havingsuccess probabilityp. Find the density and moment generating function of a randomvariableN with the negative binomial distribution. Define thecompound negativebinomial distribution and find the moment generating function, mean, and varianceof a random variable with the compoundnegative binomial distribution.

Problem 30–18. In the case of fire insurance the amount of damage may be quitelarge. Three common assumptions are made about the nature of the loss variables in

this case. One is thatX has alognormal distribution. This means thatXd= eZ where

Zd= N(µ, σ2). A second possible assumption is thatX has aPareto distribution.

This means thatX has a density of the formαx0/xα+11[x0,∞)(x) for someα > 0. Notethat a Pareto distribution has very heavy tails, and the mean and/or variance maynot exist. A final assumption which is sometimes made is that the density ofX is amixture of exponentials, that is,

fX(t) = (0.7)λ1e−λ1t + (0.3)λ2e−λ2t

for example. After an assumption is made about the nature of the underlyingdistribution one may use actual data to estimate the unknown parameters. For eachof the three models find the maximum likelihood estimators and the method ofmoments estimators of the unknown parameters.

Problem 30–19.For automobile physical damage a gamma distribution is oftenpostulated. Find the maximum likelihood and method of moments estimators of theunknown parameters in this case.


Problem 30–20.One may also examine the benefits, in terms of risk reduction,of using reinsurance. Begin by noting thepossible types of reinsurance available.First there isproportional reinsurance. Here the reinsurer agrees to pay a fractionα , 0 ≤ α ≤ 1, of each individual claim amount. Secondly, there isstop–lossreinsurance, in which the reinsurer pays the amount of the individual claim inexcess of the deductible amount. Finally, there isexcess of loss reinsuranceinwhich the reinsurer pays the amount by which the claims of aportfolio of policiesexceeds the deductible amount. As an example, the effect of stop–loss reinsurancewith deductibled on an insurer’s risk will be analyzed. The amount of insurer’srisk will be measured by the ruin probability. In fact, since the ruin probability isso difficult to compute, the effect of reinsurance on the adjustment coefficient willbe measured. Recall that the larger the adjustment coefficient, the smaller the ruinprobability. Initially (before the purchase of reinsurance) the insurer’s surplus attime t is

U(t) = u + ct −N(t)∑j=1

Xj

where c = (1 + θ)λE[X] and N(t) is a Poisson process with intensityλ . Theadjustment coefficient before the purchase of reinsurance is the positive solution of

λ + cr = λMX(r).

After the purchase of stop loss reinsurance with deductibled the insurer’s surplus is

U′ (t) = u + c′ t −N(t)∑j=1

(Xj ∧ d)

wherec′ = c − reinsurance premium. Note that this process has the same structureas the original one. The new adjustment coefficient is therefore the solution of

λ + c′ r = λMX∧ d(r).

By examining the reinsurance procedure from thereinsurer’s standpoint it is clearthat the reinsurer’s premium is given by

(1 +θ′ )λE[(X − d)1[d,∞)(X)]

whereθ′ is the reinsurer’s relative security loading. With this information the newadjustment coefficient can be computed. Carry out these computations whenλ = 2,θ = 0.50,θ′ = 0.25,d = 750, andX has a exponential distribution with mean 500.

Problem 30–21.Repeat the previous problem for the case of proportional reinsur-ance.

Problem 30–22.The case of excess of loss reinsurance leads to a discrete timemodel, since the reinsurance is applied to a portfolio of policies and the reinsurance


is paid annually (say). The details here a similar to those in the discussion of thediscrete time gamblers ruin problem. Analyze the situation described in the previousproblem if the deductible for an excess of loss policy is 1500 and the rest of theassumptions are the same. Whichtype of reinsurance is better?

Problem 30–23. In the discrete time model, suppose theX’s have theN(10,4)distribution and the relative security loading is 25%. A reinsurer will reinsure afractionf of the total portfolio on a proportional basis for a premium which is 140%of the expected claim amount. Find the insurers adjustment coefficient as a functionof f . What value off maximizes the security of the ceding company?


Solutions to ProblemsProblem 30–2. By symmetry,E[X1|S = x,N = n] = x/n while direct com-putation givesE[X1|S = x,N = n] =

∑∞k=1 k P[X1 = k |S = x,N = n]. Now

P[X1 = k,S = x,N = n] = P[X1 = k,∑n

j=1 Xj = x,N = n] = P[X1 =k] P[∑n

j=2 Xj = x − k] P[N = n] = P[X1 = k] P[∑n

j=2 Xj = x − k] P[N = n −1]λ /n = P[X1 = k] P[S = x − k,N = n − 1]λ /n. Making this substitution givesxP[S = x,N = n] =

∑∞k=1 λ kP[X1 = k] P[S = x − k,N = n − 1]. Summing both

sides onn from 0 to∞ gives the result.

Problem 30–4. The sum has a compound Poisson distribution withλ = 8.

Problem 30–6. Compute the distribution function ofX by conditioning onU to obtainFX(x) = P[X1 ≤ x] p + P[X2 ≤ x] (1 − p) for x ≥ 0 whereX1 andX2 are exponentially distributed random variables with parametersλ1 andλ2

respectively.

Problem 30–7. Use partial fractions and the previous problem to see thatXis a mixture of two exponentially distributed random variables with parametersλ1 = 3 andλ2 = 5 andp = 1/4.

Problem 30–8. HereMX(t) = (5t − 21)/ (t − 3)(t − 7) andE[X] = 5/21. Thisleads toR = 1.69. AlsoML(t) = 1/2− 0.769/ (t − 1.69)− 0.280/ (t − 6.20) usingpartial fractions. Hence the density ofL is fL(t) = 0.769e−1.69t + 0.280e−6.20t

together with a jump of size 1/2 at t = 0. (Recall thatL has both a discrete andabsolutely continuous part.) Thusψ(u) = 0.454e−1.69u + 0.045e−6.20u.

Problem 30–9. Hereθ = 10/7 ln(2)− 1 = 1.0609.

Problem 30–11. The density ofL1 is fL1(t) = (1 − FX(t))/E[X] for t ≥ 0.Integration by parts then givesML1(t) = (MX(t) −1)/ tE[X]. Using the Maclaurin

expansion ofMX(t) = 1+tE[X]+ t2E[X2]/2+. . . then givesML1(t) = 1+E[X2]2E[X]

t+

E[X3]6E[X]

t2 + . . ., from which the first two moments ofL1 can be read off.

Problem 30–12. ML(t) = 5θt/ (1 + 5(1 +θ)t − e5t) = 1 + t E[X2]2θE[X] + . . ..

Problem 30–13. Hereθ = 2/3 sinceψ(0) = 1/ (1 +θ).

Problem 30–14. Substitute the Maclaurin expansion ofMX(t) into the expres-sion for moment generating function ofL in order to get the Macluarin expansionof ML(t).

Problem 30–15. Hereθ = 4/5 andR = 2. AlsoML(t) = 4/9+(8/9) 12−t +(4/9) 1

4−tso thatψ(u) = (4/9)e−2u + (1/9)e−4u.

Problem 30–16. HereMX(t) = 925(3/5−t)−2 so thatE[X] = 10/3 andθ = −9/10

when c = 1. Sinceθ < 0, there is no adjustment coefficient and the ruin

probability is 1. Whenc = 20,θ = 1 andR = 0.215. AlsoML(t) =12−0.119/ (t−


0.215) + 0.044/ (t − 0.834) by partial fractions. The density of the absolutelycontinuous part ofL is fL(t) = 0.119e−.0.215t − 0.044e−0.834t, and the distributionof L has a jump of size 1/2 at the origin. Soψ(u) = 0.553e−0.215u −0.053e−0.834u.

Problem 30–17. The compound negative binomial distribution is the distribu-

tion of the random sumN∑

i=1

Xi whereN and theX’s are independent,N has the

negative binomial distribution, and theX’s all have the same distribution. NowP[N = k] =

(k+r−1r−1

)pr(1− p)k for k ≥ 0 andMN(t) = pr(1− (1− p)et)−r. Now use

the general result about the moment generating function of a random sum.

Problem 30–20. HereMX(t) = (1 − 500t)−1 for t < 1/500. The adjustmentcoefficient before reinsurance is thenR = 1/1500. The reinsurance premium is(1 + 0.25)2E[(X − 750)1(750,∞)(X)] = 278.91 and the insurer’s new adjustmentcoefficient is the solution of 2 + (1500− 278.91)R = MX∧ 750(R) which givesR = 0.00143.

Problem 30–21. As in the preceding problem,R = 1/1500 before reinsurance.Suppose the insurer retains 100(1−α )% of the liability. The reinsurance premiumis then (1 +θ′ )λαE[X] = 1250α . The adjustment coefficient after reinsuranceis then the solution of 2 + (1500−1250α )R = 2M(1−α )X(R) = 2MX((1−α )R). SoR = (2−α )/500(5α 2−11α +6) which is always at least 1/1500. Notice that sinceθ′ < θ here, the insurer should pass off all of the risk to the reinsurer. By usingα = 1 the insurer collects the difference between the original and reinsurancepremiums, and has no risk of paying a claim.

Problem 30–22. The computational details here are quite complicated. Ina time interval of unit length the total claims areC =

∑Nj=1 Xj whereN is a

Poisson random variable with parameterλ . Now recall that in the discrete timesetting the adjustment coefficient is the solution of the equationE[eR(c−C)] = 1.As beforec = 1500. AlsoMC(t) = eλ (MX (t)−1). So the adjustment coefficientbefore reinsurance is 1/1500. The reinsurance premium with deductible 1500is λ (1 + θ′ )E[(C − 1500)1(1500,∞)(C)] = 568.12. This is obtained numericallyby conditioning on the value ofN and using the fact that conditional onN = k,C has a gamma distribution with parametersα = k andβ = 1/500. The newadjustment coefficient solvesE[eR(1500−568.12−C∧ 1500)] = 1.

Problem 30–23. HereMX(t) = e10t+2t2, the premium income is 12.5 for eachtime period, and the adjustment coefficient is the solution ofe−12.5tMX(t) = 1which givesR = 1.25. The reinsurance premium is 14f so that after reinsurancethe adjustment coefficient satisfiese−(12.5−14f )tMX((1 − f )t) = 1, which givesR = (5− 8f )/4(1− 2f + f 2). The valuef = 1/4 produces the maximum value ofR, namely 4/3.


Solutions to Exercises

Exercise 30–1. SinceR > 0 andu + cT −∑N(T)

k=1 Xk ≤ 0 whenT < ∞ thedenominator expectation is at least 1.

Exercise 30–2. MU(t)−u(−R) = 1 holds if and only if−ctR − λ t(MX(R) − 1) = 0,which translates into the given condition.

Exercise 30–3. If c ≤ λE[X] premium income is less than or equal to theaverage rate of the claim process. So eventually the company will be ruinedby a run of above average size claims. By Maclaurin expansion,MX(R) =1 + E[X]R + E[X2]R2/2 + . . . and all of the coefficients are positive sinceX is apositive random variable. Soλ +cR−λMX(R) = (c−λE[X])R−E[X2]R2/2−. . .is a function which is positive forR near 0 and negative for large values ofR.Thus there is some positive value ofR for which this function is zero.

Exercise 30–4. From the definition of conditional probability,P[L1 ≤ y|Tu <∞] = P[L1 ≤ y,Tu < ∞]/P[Tu < ∞] and the result follows from the previousformula and the fact thatP[Tu < ∞] = 1/ (1 +θ).

Exercise 30–5. Since in this caseFX(t) = 1− e−δt for t > 0, direct substitutiongivesP[L1 ≤ y|Tu < ∞] = 1 − e−δy for y > 0.

Exercise 30–6. GivenTu < ∞ the density ofL1 is (1−FX(y))/E[X] for y > 0. Us-ing integration by parts then gives the conditional moment generating function ofL1 as

∫ ∞0 ety(1−FX(y))/E[X] dy = ety(1− FX(y))/ tE[X]

∣∣∞0 +∫ ∞

0 etyfX(y)/ tE[X] dy =(MX(t)−1)/ tE[X]. Notice that the unconditional distribution ofL1 has a jump ofsizeθ/ (1 +θ) at the origin. The unconditional moment generating function ofL1 is θ/ (1 +θ) + (MX(t) − 1)/ (1 +θ)tE[X].

Exercise 30–7. SinceP[D = k] = (θ/ (1 + θ))(1/ (1 + θ))k for k = 0,1,2, . . .,conditioning givesML(t) = E[e

t∑D

j=1Aj ] = E[MA(t)D] =

∑∞k=0 MA(t)k(θ/ (1 +

θ))(1/ (1 + θ))k = (θ/ (1 + θ))/ (1 − MA(t)/ (1 + θ)) = θ/ (1 + θ − MA(t)) and thissimplifies to the desired result using the formula of the previous exercise.

Exercise 30–8. Using the independence,MS+T (ν) = MS(ν)MT (ν) and the resultfollows by substituion and algebraic rearrangement.

§31. Related Probability Models

In the next few sections some probability models are discussed which can beused as models for transactions other than life insurance.

Discrete and continuous time Markov chains are often used as models fora sequence of random variables which are dependent. One application of suchstochastic processes is as a model for the length of stay of a patient in a nursinghome.

The Brownian motion process is often used as a building block for a model ofstock prices. The definition and simple properties of Brownian motion are developedhere.


§32. Discrete Time Markov Chains

In many situations the random variables which serve naturally as a model arenot independent. The simplest kind of dependence allows future behavior to dependon the present situation.

Example 32–1.Patients in a nursing home fall into 3 categories, and each categoryof patient has a differing expense level. Patients who can care for themselves withminimal assistance are in the lowest expense category. Other patients require someskilled nursing assistance on a regular basis and are in the next higher expensecategory. Finally, some patients requirecontinuous skilled nursing assistance andare in the highest expense category.One way of modeling the level of care aparticular patient requires on a given day is as follows. Denote byXi the level ofcare this patient requires on dayi. Here the value ofXi would be either 1, 2, or3 depending on which of the 3 expense categories is appropriate for dayi. It isintuitively clear that the random variables{Xi} are not independent.

Possibly the simplest type of dependence structure for a sequence of randomvariables is that in which the future probabilistic behavior of the sequence dependsonly on the present value of the sequence and not on the entire history of thesequence. A sequence of random variables{Xn : n = 0,1, . . .} is said to be aMarkov chain if

(1) P[Xn ∈ { 0,1,2,3, . . .} ] = 1 for all n and

(2) for any real numbersa < b and any finite sequence ofnon-negative integerst1 < t2 < ⋅ ⋅ ⋅ < tn < tn+1,

P[a < Xtn+1 ≤ b|Xt1, . . . ,Xtn] = P[a < Xtn+1 ≤ b|Xtn ].

The second requirement is referred to as theMarkov property .

The possible values of the chain are calledstates.

Exercise 32–1.Show that any sequence of independent discrete random variablesis a Markov chain.

Because of the simple dependence structure a vital role is played by thetran-sition probabilities P[Xn+1 = j|Xn = i]. In principle, this probability depends notonly on the two statesi and j, but also onn. A Markov chain is said to havestationary transition probabilities if the transition probabilitiesP[Xn+1 = j|Xn = i]do not depend onn. In the discussion here, the transition probabilities will alwaysbe assumed to be stationary, and the notationPi,j = P[Xn+1 = j|Xn = i] will be used.

The transition probabilities together with the distribution ofX0 determine com-pletely the probabilistic behavior of the Markov chain. This is seen in the following


§32: Discrete Time Markov Chains 129

computation.

P[Xn = in, . . . ,X0 = i0]

= P[Xn = in |Xn−1 = in−1, . . . ,X0 = i0] × P[Xn−1 = in−1, . . . ,X0 = i0]

= P[Xn = in |Xn−1 = in−1] × P[Xn−1 = in−1, . . . ,X0 = i0]

= Pin−1,in P[Xn−1 = in−1, . . . ,X0 = i0]

= . . .= Pin−1,inPin−2,in−1 . . .Pi0,i1P[X0 = i0].

Exercise 32–2.Justify each of the steps here completely. Where was the Markovproperty used?

Exercise 32–3.Show that for a Markov chain with stationary transition probabilitiesP[X4 = 3,X3 ≠ 3|X2 = 3,X1 ≠ 3,X0 = 3] = P[X2 = 3,X1 ≠ 3|X0 = 3]. Generalize.

For a Markov chain with stationary transition probabilities it is useful to collectthe transition probabilities into thetransition matrix P = [Pi,j] of the chain. Thismatrix may be infinite in extent.

Exercise 32–4.Show that ifP is a transition matrix then∑

j Pi,j = 1 for eachi.

Example 32–2. In the previous nursing home example, suppose the transition

matrix isP =

0.9 0.05 0.050.1 0.8 0.10 0.05 0.95

. Then using conditioning it is easy to compute

P[X3 = 2|X0 = 1].

Example 32–3.The gambler’s ruin problem illustrates many of the features of aMarkov chain. A gambler enters a casino with $z available for wagering and sitsdown at her favorite game. On each play of the game, the gambler wins $1 withprobability p and loses $1 with probabilityq = 1 − p. She will happily leave thecasino if her fortune reaches $a > 0, and will definitely leave, rather unhappily,if her fortune reaches $0. Denote byXn the gambler’s fortune after thenth play.Clearly{Xn} is a Markov chain withP[X0 = z] = 1. The natural state space here is{0,1, . . . ,a} .

Exercise 32–5.Find the (a + 1)× (a + 1) transition matrix.

Even with the simplifying assumption of stationary transition probabilities theformula for the joint distribution of the values of the chain is unwieldy, especiallysince in most cases it is the long term behavior of the chain that is of interest.Fortunately, it is possible to find relatively simple answers to the following centralquestions.


(1) If {Xn} is a Markov chain with stationary transition probabilities, what is thelimiting distribution ofXn?

(2) If s is a state of a Markov chain with stationary transition probabilities howoften is the process in states?

As a warm up exercise for studying these questions then step transitionprobabilities defined byPn

i,j = P[Xn+m = j|Xm = i] and the correspondingn steptransition probability matrixP(n) will now be computed.

Exercise 32–6.Show thatP[Xn+m = j|Xm = i] does not depend onm.

Theorem. Then step transition probability matrix is given byP(n) = Pn whereP isthe transition probability matrix.

proof : The casen = 1 being clear, the induction step is supplied.

Pni,j = P[Xn+m = j|Xm = i]

= P[[Xn+m = j] ∩

(∞⋃

k=0

[Xn+m−1 = k]

)] |Xm = i]

=∞∑

k=0

P[[Xn+m = j,Xn+m−1 = k] |Xm = i]

=∞∑

k=0

P[Xn+m = j|Xn+m−1 = k,Xm = i] P[Xn+m−1 = k |Xm = i]

=∞∑

k=0

P[Xn+m = j|Xn+m−1 = k] P[Xn+m−1 = k |Xm = i]

=∞∑

k=0

Pk,jP(n−1)i,k .

The induction hypothesis together with the formula for the multiplication of matricesconclude the proof.

Using this lemma gives the following formula for the density ofXn in terms ofthe density ofX0.

( P[Xn = 0] P[Xn = 1] . . .) = ( P[X0 = 0] P[X0 = 1] . . .)Pn.

Exercise 32–7.Verify that this formula is correct.

Consequently,if Xn converges in distribution toY asn → ∞ then

( P[Y = 0] P[Y = 1] . . .) = limn→∞

( P[Xn = 0] P[Xn = 1] . . .)

= limn→∞

( P[X0 = 0] P[X0 = 1] . . .)Pn

= limn→∞

( P[X0 = 0] P[X0 = 1] . . .)Pn+1

= ( P[Y = 0] P[Y = 1] . . .)P


which gives a necessary condition forY to be a distributional limit for the chain,namely, the density ofY must be a left eigenvector ofP corresponding to theeigenvalue 1.

Example 32–4.For the nursing home chain given earlier there is a unique lefteigenvector ofP corresponding to the eigenvalue 1, after normalizing so that thesum of the coordinates is 1. That eigenvector is (0.1202,0.1202,0.7595). Thus apatient will, in the long run, spend about12% of the time in each of categories 1and 2 and about 76% of the time in category 3.

Exercise 32–8.Find the left eigenvectors corresponding to the eigenvalue 1 of thetransition matrix for the gambler’s ruin chain.

Example 32–5.Consider the Markov chain with transition matrixP =(

0 11 0

).

This chain will be called theoscillating chain. The left eigenvector ofP corre-sponding to the eigenvalue 1 is ( 1/2 1/2 ). If the chain starts in one of the states,there is clearly no limiting distribution.

Exercise 32–9.Show that this last chain does not have a limiting distribution.

The oscillating chain example shows that a Markov chain need not have alimiting distribution. Even so, this chain does spend half the time in each state, sothe entries in the left eigenvector do have an intuitive interpretation as properties ofthe chain. To explore this possiblilty further, some terminology is introduced. LetP be the transition probability matrix of a Markov chainX with stationary transitionprobabilities. A vectorπ = (π0, π1, . . .) is said to be astationary distribution forthe chainX if

(1)πi ≥ 0 for all i,

(2)∑∞

i=0πi = 1, and

(3)πP = π.

If a limiting distribution for the chainX exists then that limiting distribution willbe a stationary distribution. In the example above, ( 1/2 1/2 ) is a stationarydistribution even in though the chain has no limiting distribution.

From a relative frequency viewpoint a stationary distribution should arise as thelimit of the occupation times:πi = limL→∞

1L+1

∑Ln=0 1{ i} (Xn). Indeed, this is the case

even in the rather poorly behaved exampleabove. It will be easier to differentiatethe possible cases that can arise by studying the states of the chain more closely.

For each statei define the random variableNi =∑∞

n=0 1{ i} (Xn). ClearlyNi countsthe total number of visits of the Markov chainX to the statei. It is possible thatNi = ∞. A statei for which P[Ni = ∞|X0 = i] = 1 is a state which is sure to be


revisited infinitely many times. Such a state is said to berecurrent . A non-recurrentstate, that is, a statei for whichP[Ni = ∞|X0 = i] < 1 is said to betransient. It is arather amazing fact that for a transient statei, P[Ni = ∞|X0 = i] = 0. Thus for eachstatei the random variableNi is either always infinite or never infinite.

Exercise 32–10.Show that ifi is a transient state thenNi is a geometric randomvariable.

Checking each state to see whether that state is transient or recurrent is clearlya difficult task with only the tools available now. Some other useful notions cangreatly simplify the job.

One key notion is that of accessibility. The statej is said to beaccessiblefromthe statei if there is a positive probability that the chain can start in statei and reachstatej. Technically, the requirement forj to be accessible fromi is thatPn

i,j > 0 forsomen ≥ 0. Two statesi and j are said tocommunicate, denotedi↔j, if each isaccessible from the other.

Example 32–6.Consider the Markov chainX in whichXn denotes the outcome ofthenth toss of a fair coin in which 1 corresponds to a head and 0 to a tail. Clearly0↔1.

Example 32–7. In the gambler’s ruin problem it is intuitively clear that the states0 anda are accessible from any other state but do not communicate with any stateexcept themselves. Such states areabsorbing. The other states all communicatewith each other.

Exercise 32–11.Prove that the intuition of the preceding example is correct.

Example 32–8. In the nursing home example, all states communicate with eachother.

It is simple to check that the communication relation between states is an equiv-alence relation, and therefore partitions the state space of the chain into equivalenceclasses. If the chain has but a single equivalence class the chain is said to beirreducible .

It is an important fact that ifi↔j theni is recurrent if and only ifj is recurrent.(Said briefly, recurrence is a class property.)

Exercise 32–12.For the coin tossing chain, is the state 1 recurrent?

Exercise 32–13.What are the recurrent states for the gambler’s ruin chain?

Because of the possible existence of transient states, the discussion of thelimiting behavior of the chain is a bit complicated. Denote byfi,j the probability that


the chain ever enters statej given that the chain is currently in statei. Denote byµi,i

the expected number of time steps between visits to statei. (For a transient state,µi,i = ∞ and it is possible forµi,i = ∞ even for a recurrent state.) With this notation,the central result in the theory of Markov chains can be stated. For any two statesiandj, given that the chain begins at time zero in statei,

limL→∞

1L + 1

L∑n=0

1{ j} (Xn) = fi,j/µj,j.

This is the result that was anticipated based on previous examples.

This result can be used to provide an interpretation of the entries in a stationarydistribution. To do this, take expectations of both sides of the above limit to obtain

limL→∞

1L + 1

L∑n=0

Pni,j = fi,j/µj,j. This can be expressed in matrix terms by definingS to be

the matrix with entriesfi,j/µj,j. The foregoing statement is then limL→∞

1L + 1

L∑n=0

Pn = S.

If π is a stationary distribution,πP = π, and by multiplication of the limitingstatement byπ, πS = π too. So every stationary distribution is in the row space ofthe matrixS.

Now consider the case in all statesi andj are recurrent and communicate. Thenfi,j = 1 for all i andj, and all of the rows ofS are the same. Thus there is only onestationary distribution and the valueπj = 1/µj,j, which is the expected fraction of thetime the chain spends in statej. Since there is only one stationary distribution in thiscase, there is also at most one limiting distribution in this case. The oscillating chainexample shows that the stationary distribution need not be a limiting distribution.

If some of the states are transient, the rows ofS may not all be the same. Thisis in accord with the behavior of the gambler’s ruin chain which shows that thelimiting behavior may depend on the initial state.

Exercise 32–14.Show that if a Markov chain is irreducible and every state istransient then there is no stationary distribution for the chain.

Even for an irreducible and recurrent chain a limiting distribution need not existfor another reason. Ifµi,i = ∞ there is no stationary distribution. Such a chain isnull recurrent . If µi,i < ∞, then as above there is one stationary distribution. Inthis case the chain isstrongly ergodic (or positive recurrent).

To identify the cases in which the chain has a limiting distribution one additionalidea is needed. Theperiod of the statei, denoted byd(i), is the greatest commondivisor of the set{n ≥ 1 : Pn

i,i > 0} . If this set is empty, the period is defined to be0. If d(i) = 1 for all states of the chain then the chain is said to beaperiodic. If a


Markov chain is aperiodic, a limiting distribution exists (given that the chain startsin a particular state) and lim

n→∞P[Xn = j|X0 = i] = fi,j/µj,j. If all states communicate

and are recurrent, this limiting distribution will not depend on the initial state.

Exercise 32–15.What is the period of each state in the nursing home chain? Thegambler’s ruin chain? The oscillating chain?

Example 32–9.For the gambler’s ruin chain (c,0, . . . ,0,1− c ) is a stationarydistribution for any 0≤ c ≤ 1. One easily sees that only the two recurrent classescontribute non-zero probabilities here. A formula forfi,0 was found earlier.


Problems

Problem 32–1. Show thatPni,i = 0 if d(i) does not dividen. Show that ifi↔j then

d(i) = d(j). (Said briefly, period is a class property.)

Problem 32–2. Suppose the chain has only finitely many states all of which com-municate with each other. Are any of the states transient?

Problem 32–3. SupposeNi,j is the total number of visits of the chain to statej given

that the chain begins in statei. Show that fori ≠ j, E[Ni,j] =∞∑

k=0

E[Nk,j] Pi,k. What

happens ifi = j?

Problem 32–4. Suppose the chain has both transient and recurrent states. Relabelthe states so that the transient states are listed first. Partition the transition matrix in

to blocksP =(

PT Q0 PR

). Explain why the lower left block is a zero matrix. Show

that theT ×T matrix of expectationsE[Ni,j] asi andj range over the transient statesis (I − PT)−1.

Problem 32–5. Show that ifi ≠ j and both are transient,fi,j = E[Ni,i]/E[Nj,j]. Whathappens ifi = j?

Problem 32–6. In addition to the 3 categories of expenses in the nursing home ex-ample, consider also the possibilities of withdrawal from the home and death. Sup-

pose the corresponding transition matrix isP =

0.8 0.05 0.01 0.09 0.050.5 0.45 0.04 0.0 0.010.05 0.15 0.70 0.0 0.100.0 0.0 0.0 1.0 0.00.0 0.0 0.0 0.0 1.0

where the states are the 3 expense categories in order followed by withdrawal anddeath. Find the stationary distribution(s) of the chain. Which states communicate,which states are transient, and what are the absorption probabilities given the initialstate?

Problem 32–7. An auto insurance company classifies insureds in 2 classes: (1)preferred, and (2) standard. Preferred customers have an expected loss of $400 inany one year, while standard customers have an expected loss of $900 in any oneyear. A driver who is classified as preferred this year has an 85% chance of beingclassified as preferred next year; a driver classified as standard this year has a 40%chance of being classified as standard next year. Losses are paid at the end of eachyear andi = 5%. What is the net single premium for a 3 year term policy for anentering standard driver?


Solutions to ProblemsProblem 32–1. Since i↔j there are integersa and b so thatPa

i,j > 0 andPb

j,i > 0. As shown earlier,Pa+n+bj,j ≥ Pb

j,iPni,iP

ai,j. If Pn

i,i > 0 this inequality shows

thatPa+b+nj,j > 0 too and therefored(j) dividesa + b + n. But sinceP2n

i,i ≥(Pn

i,i)2

asimilar inequality shows thatd(j) dividesa + b + 2n as well. Henced(j) dividesa + b + 2n − (a + b + n) = n, and sod(j) ≤ d(i). Interchanging the roles ofi andjshows thatd(i) ≤ d(j).

Problem 32–2. No. Since all states communicate, either all are transient orall are recurrent. Since there are only finitely many states they can not all betransient. Hence all states are recurrent.

Problem 32–3. The formula follows by conditioning on the first step leaving

statei. When i = j the formula isE[Ni,i] = 1 +∞∑

k=0

E[Nk,i] Pi,k, by the same

argument.

Problem 32–4. It is not possible to go from a recurrent state to a transient state.Express the equations of the previous problem in matrix form and solve.

Problem 32–5. fi,i = (E[Ni,i] − 1)/E[Ni,i].

Problem 32–6. The 3 expense category states communicate with each other andare transient. The other 2 states are recurrent and absorbing. The probabilitiesfi,j satisfyf0,4 = 0.8f0,4 + 0.05f1,4 + 0.05f2,4 + 0.09 and 2 other similar equations,from whichf0,4 = 0.382,f1,4 = 0.409, andf2,4 = 0.601.

Problem 32–7. From the given information the transition matrix isP =(.85 .15.6 .4

). The two year transition probabilities for an entering standard

driver are found from the second row ofP2 to be (.75 .25 ). The premium is900v + (.6× 400 +.4× 900)v2 + (.75× 400 +.25× 900)v3 = 1854.90.


Solutions to ExercisesExercise 32–1. Because of the independence both of the conditional probabil-ities in the definition are equal to the unconditional probabilityP[a < Xtn+1 < b].

Exercise 32–2. This is just the definition of conditional probability togetherwith the use of the Markov property to simplify each of the conditional proba-bilities.

Exercise 32–3. Write P[X4 = 3,X3 ≠ 3|X2 = 3,X1 ≠ 3,X0 = 3] =∑

j≠3 P[X4 =3,X3 = j|X2 = 3,X1 ≠ 3,X0 = 3] =

∑j≠3 P[X4 = 3,X3 = j,X2 = 3,X1 ≠ 3,X0 =

3]/P[X2 = 3,X1 ≠ 3,X0 = 3] =∑

j≠3

∑k≠3 P[X4 = 3,X3 = j,X2 = 3,X1 = k,X0 =

3]/P[X2 = 3,X1 ≠ 3,X0 = 3] =∑

j≠3

∑k≠3 P[X4 = 3,X3 = j|X2 = 3,X1 = k,X0 =

3]P[X2 = 3,X1 = k,X0 = 3]/P[X2 = 3,X1 ≠ 3,X0 = 3] =∑

j≠3

∑k≠3 P[X4 =

3,X3 = j|X2 = 3]P[X2 = 3,X1 = k,X0 = 3]/P[X2 = 3,X1 ≠ 3,X0 = 3] =∑j≠3 P[X4 = 3,X3 = j|X2 = 3] =

∑j≠3 P[X2 = 3,X1 = j|X0 = 3] = P[X2 =

3,X1 ≠ 3|X0 = 3], where the Markov property and stationarity have been used.

Exercise 32–4.∑

j Pi,j =∑

j P[X1 = j|X0 = i] = P[X1 ∈ R |X0 = i] = 1.

Exercise 32–5.

1 0 0 0 0 . . . 0q 0 p 0 0 . . . 00 q 0 p 0 . . . 00 0 q 0 p . . . 0...

......

......

......

0 0 0 0 0 . . . 1

.

Exercise 32–6. Use induction onn. The casen = 1 is true from the definitionof stationarity. For the induction step assume the result holds whenn = k. ThenP[Xk+1+m = j|Xm = i] =

∑b P[Xk+1+m = j,Xk+m = b |Xm = i] =

∑b P[Xk+1+m =

j|Xk+m = b]P[Xk+m = b |Xm = i] =∑

b P[Xk+1 = j|Xk = b]P[Xk = b |X0 = i] =P[Xk+1 = j|X0 = i], as desired.

Exercise 32–7. P[Xn = k] =∑

i P[Xn = k |X0 = i]P[X0 = i] =∑

i Pni,kP[X0 = i]

which agrees with the matrix multiplication.

Exercise 32–8. Matrix multiplication shows that the left eigenvector conditionimplies that the left eigenvectorx = (x0, . . . ,xa) has coordinates that satisfyx0 + qx1 = x0, qx2 = x1, pxk−1 + qxk+1 = xk for 2 ≤ k ≤ a − 2, pxa−2 = xa−1 andpxa−1 + xa = xa. From these equations it follows that onlyx0 and xa can benon-zero, and that these two values can be arbitrary. Hence all left eigenvectorscorresponding to the eigenvalue 1 are of the form (c,0,0, . . . ,0,1− c) for some0 ≤ c ≤ 1.

Exercise 32–9. P[Xn = 1] is 0 or 1 depending on whethern is odd or even, sothis probability has no limit.

Exercise 32–10. Let p = P[Ni < ∞|X0 = i]. ThenP[Ni = k |X0 = i] = pk(1− p)for k = 0,1, . . ..

Exercise 32–11. If the current fortune isi, andi is not 0 ora, then it is possible


to obtain fortunej is |j − i| plays of the game by having|j − i| wins (or losses)in a row.

Exercise 32–12. Yes, 1 is recurrent since state 1 is sure to be visited infinitelyoften.

Exercise 32–13. The only recurrent states are 0 anda.

Exercise 32–14. Under these assumptions,fi,j = 0 so the limit above is alwayszero. Hence there is no stationary distribution.

Exercise 32–15. For the nursing home chain and the gambler’s ruin chaineach state has period 1. For the oscillating chain each state has period 2. Thisexplains why the oscillating chain does not have a limiting distribution.

§33. Continuous Time Markov Chains

The next step in the study of Markov chains retains a discrete state space butallows time to vary continuously.

A continuous time Markov chain is a stochastic process{Xt : t ≥ 0} with statespace{0,1,2, . . .} which satisfies theMarkov property : whenevern > 0 and 0≤t1 < t2 < . . . < tn < t thenP[Xt = xt |Xtn = xtn , . . . ,Xt1 = xt1] = P[Xt = xt |Xtn = xtn ].As in the discrete time case the discussion hereassumesthat the Markov chain hasstationary transition probabilities, that is,P[Xt+s = i|Xt = j] does not depend ont.

A discrete time chain always spent one time unit in each state before the nexttransition. For a continuous time chain thetime spent in each state before a transitionwill be random. Intuitively, this is the only difference between discrete time andcontinuous time Markov chains. The bulk of the work consists of verifying that thisintuition is indeed correct and in identifying the probabilistic properties of the timesbetween transitions for the chain.

As for discrete time chains the transition probabilitiesPi,j(t) = P[Xt = j|X0 = i]and the associatedtransition probability matrix P(t) = [Pi,j(t)] play an importantrole. By convention,P(0) = I, the identity matrix.

The first result parallels a result for discrete time chains.

Transition Semi-group Property. The transition matrices{P(t) : t ≥ 0} form asemi-group under matrix multiplication, that is,P(s + t) = P(s)P(t) for all s, t ≥ 0.

Exercise 33–1.Prove the theorem.

For a discrete time chain, the one step transition probability matrix determinedall of the interesting properties of thechain. For continuous time chains, a similarrole is played by the matrixP′ (0), which is called theinfinitesimal generator ofthe Markov chain (or of the transition semi-group{P(t) : t ≥ 0} ). Since typicallythe infinitesimal generator is specified during model building the central question ishow the infinitesimal generator determines properties of the chain.

Example 33–1.The Poisson process provides a typical example of the way in whichthe transition probabilities are specified for continuous time Markov chains. As aspecific example letXt denote the number of calls that have arrived at a telephoneswitchboard by timet. In a very short time interval essentially either 0 or 1 call canarrive. Denote byλ > 0 the average rate at which calls arrive. A model for thebehavior of the chainXt can be expressed by saying that for smallh > 0

(1) P[Xt+h − Xt = 1|Xt = i] ≈ λ h,

(2) P[Xt+h − Xt = 0|Xt = i] ≈ 1− λ h, and


§33: Continuous Time Markov Chains 140

(3) P[Xt+h − Xt ≥ 2|Xt = i] ≈ 0

or equivalently in terms of the transition probabilities

(1) Pi,i(h) ≈ 1− λ h,

(2) Pi,i+1(h) ≈ λ h, and

(3) Pi,j(h) ≈ 0 if j ≠ i, i + 1.

SinceP(0) = I these assumptions show that the derivative ofP(t) exists att = 0 andis given by

P′ (0) =

−λ λ 0 0 . . .0 −λ λ 0 . . .0 0 −λ λ . . ....

......

......

.

In the case of the Poisson process, the row sums of the generator are zero. Aninfinitesimal generator for which the row sums are all zero isconservative. Also,the off-diagonal entries of the generatorare non-negative, while the diagonal entriesof the generator are non-positive.

Exercise 33–2.Show that if the chain has only finitely many states then the in-finitesimal generator must be conservative.

Exercise 33–3.How could a generator fail to be conservative?

Exercise 33–4.Are the off-diagonal entries of an infinitesimal generator alwaysnon-negative? Are the diagonal entries ofan infinitesimal generator always non-positive?

SupposeP(t) is a Markov semi-group which is continuous at 0. Then usingresults from real analysis, the infinitesimal generatorA = P′ (0) exists. Using thesemi-group property gives

P′ (t) = limh→0

P(t + h) −P(t)h

= limh→0

P(h)P(t) − P(t)h

=(

limh→0

P(h) − Ih

)P(t)

= A P(t).

This equation is known asKolmogorov’s backward equation. The same argumentleads also toKomogorov’s forward equation P′ (t) = P(t) A.

Exercise 33–5.Why aren’t these derivations rigorous?


These two equations, especially the forward equation, are very useful in appli-cations. Unfortunately there are no simple general conditions which guarantee thatthe forward equation holds.

In the case of chains with only a finite number of states the theory is very simple:the transition semi-group is uniquely determined by the infinitesimal generator(which must be conservative) and both the forward and backward equations hold.Also, the semi-group can be obtainedfrom the generator by the formulaP(t) = etP′ (0).In this formula the exponential of a matrixis computed from the Maclaurin seriesfor the exponential function.

The case in which the chain has infinitely many states is more complex. Themain result in this case is just stated here.

Theorem. Suppose the infinitesimal generator is conservative. IfP(t) is the uniquesolution ofeither the forward or the backward equation and if

∑∞j=0 Pi,j(t) = 1 for all

i thenP(t) is the unique solution toboth the forward and the backward equation andis the unique transition semi-group with the given generator.

Example 33–2.Assume for the moment that the forward equation holds for thePoisson process. (Later this will be shown to be true.) Recall that the infinitesimalgenerator in this case is

A = P′ (0) =

−λ λ 0 0 . . .0 −λ λ 0 . . .0 0 −λ λ . . ....

......

......

.

Translating the matrix form of the forward equationP′ (t) = P(t)A into statementsabout the individual transition probabilities gives

P′i,j(t) = −λPi,j(t) + λPi,j−1(t).

To computeM(t) = E[Xt |X0 = i] use the definition of expectation and this equationto obtain

M′ (t) =∞∑

j=0

jP′i,j(t)

=∞∑

j=0

j(−λPi,j(t) + λPi,j−1(t)

)

= −λ∞∑

j=0

jPi,j(t) + λ∞∑

j=0

(j − 1)Pi,j−1(t) + λ

= λ .So M(t) = λ t + i. Thus the forward equation can be used to find the conditionalexpectation without first finding the transition matrices.


Exercise 33–6.Try the same computation using the backward equations.

A method will now be developed to directly interpret the entries of the generatorin terms of the probabilistic behavior of the chain.

DefineT0 = 0 and inductively setTn = inf{ t ≥ Tn−1 : Xt ≠ XTn−1} . TheT ’s arethe times of changes of state for the chain.

To study the behavior of the chain lett > 0 andi ≠ j be states. Then

P[T1 > t,XT1 = j|X0 = i]

= limn→∞

∞∑l=0

P[Xk/2n = i,1 ≤ k ≤ [2nt] + l,X([2nt]+l+1)/2n = j|X0 = i]

= limn→∞

∞∑l=0

(Pi,i(1/2n)

)[2nt+l] Pi,j(1/2n)

= limn→∞

(Pi,j(1/2n)1/2n

) ((Pi,i(1/2n)

)2n)[2nt]/2n ∞∑l=0

12n

Pi,i(1/2n)l

= ai,jeai,i t lim

n→∞

1/2n

1− Pi,i(1/2n)

=−ai,j

ai,ieai,i t.

(Recall thatai,i ≤ 0.) HenceP[XT1 = j|X0 = i] = −ai,jai,i

andP[T1 > t |X0 = i] = eai,it fort > 0.

Exercise 33–7.Fill in the details for all of these calculations.

This computation means that the time spent in statei until a transition occurs hasan exponential distribution with mean−1/ai,i and that the probability upon leavingstatei of entering statej is −ai,j/ai,i.

Exercise 33–8.Extend the argument above to show that

P[Tl > tl,XTl = il,0 ≤ l ≤ n|X0 = i0] =n−1∏l=0

(−ail−1,il

ail−1,il−1

)eail−1,il−1 tl .

The exercise justifies the following view of continuous time Markov chains.Begin at time 0 in statei0. Stay in statei0 for a random timeT1, whereT1 has theexponential distribution with mean−1/ai0,i0. Then move to statei1 with probability−ai0,i1/ai0,i0. Remain in statei1 an exponentially distributed time, etc.

If interest is only in the states that are visited by the chain and not in the timespent in each state one may as well study theembedded discrete time chainwith


transition matrixP = [−ai,j/ai,i(1 − δi,j)]. Note that if ai,i = 0 the correspondingtransition probability in the embedded chain isPi,i = 1 since statei is obviously anabsorbing state for the continuous time chain.

Example 33–3.For the Poisson process the embedded chain has transition matrix

P =

0 1 0 0 0 . . .0 0 1 0 0 . . .0 0 0 1 0 . . ....

......

......

...

.

Hence all states are transient and there is no stationary distribution.

The relationship between the infinitesimal generator and the chain itself has beenmade rather precise. The question of the behavior of the chain is now considered.

SupposeI = ( P[X0 = 0] P[X0 = 1] . . .) is the initial distribution of the chain.If a limiting distributionπ for the chain exists then

π = limt→∞

I P(t) = limt→∞

I P(t + s) = limt→∞

I P(t)P(s) = πP(s)

for all s ≥ 0.

Exercise 33–9.Verify that this computation is correct.

Once again the notion of a stationary distribution will play an important role.In the continuous time contextπ = (π0 π1 . . .) is said to be astationarydistribution if

(1)πi ≥ 0 for all i,

(2)∑∞

i=0πi = 1, and

(3)πP(s) = π for all s ≥ 0.

Because of the close relationship between the infinitesimal generator and thetransition semi-group it should be possible to find the stationary distribution of thechain using only the generator.

Theorem. Suppose the infinitesimal generatorA of the chain is conservative andthatπi ≥ 0 for all i and

∑∞i=0πi = 1. Thenπ is a stationary distribution if and only

if πA = 0.

There is no notion of period in the continuous time setting. Thus the distinctionbetween a stationary distribution and a limiting distribution depends only on whetherthere are transient states, and the mean recurrence time for the recurrent states. Thismakes the continuous time case somewhat simpler than the discrete time case.


Some additional examples of continuoustime chains will now be given. One ofthe main features is the way in which the behavior of the chain is determined fromthe infinitesimal generator.

Example 33–4.Look once again at the Poisson process. Here the forward equationis P′i,j(t) = −λPi,j(t) + λPi,j−1(t). This can be rewritten using an integrating factor

as ddt

(eλ tPi,j(t)

)= λ eλ tPi,j−1(t). Thus P0,0(t) = e−λ t and by inductionP0,n(t) =

(λ t)ne−λ t/n!. Similarly, Pi,0(t) = 0 if i > 0 and by inductionPi,j(t) = (λ t)j−ie−λ t/ (j −i)!1[0,∞)(j − i). This solution is unique and

∑∞j=0 Pi,j(t) = 1 for all i. From the general

theoryPi,j(t) is therefore the unique solution to the backward equation as well andis also the unique transition semi-group with this infinitesimal generator.

Exercise 33–10.Fill in the details of the induction arguments above.

Example 33–5.The next example is apure birth process. This is a variant of thePoisson process in which the probability of additional calls depends on the numberof calls received. Specifically assume that

(1) P[Xt+h − Xt = 1|Xt = k] ≈ λkh,

(2) P[Xt+h − Xt = 0|Xt = k] ≈ 1− λkh, and

(3) P[Xt+h − Xt ≥ 2|Xt = k] ≈ 0.

As before this leads to the generator

A =

−λ0 λ0 0 0 . . .0 −λ1 λ1 0 . . .0 0 −λ2 λ2 . . ....

......

......

.

Using the forward equation and the integrating factoreλj t yields

Pi,j(t) = δi,je−λj t + λj−1e−λj t

∫ t

0eλj tPi,j−1(s) ds.

This shows inductively thatPi,j(t) ≥ 0 and that the solution is unique. The remainingquestion is whether or not

∑∞j=0 Pi,j(t) = 1 for all i. Fix i and defineSn(t) =

∑nj=0 Pi,j(t).

Using the forward equation givesS′n(t) = −λnPi,n(t) and so 1− Sn(t) =∫ t

0 −S′n(s) ds =λn∫ t

0 Pi,n(s) ds. Now from the definition ofSn(t), 1−∑∞

j=0 Pi,j(t) ≤ 1− Sn(t) ≤ 1 so

1λn

1−

∞∑j=0

Pi,j(t)

≤

∫ t

0Pi,n(s) ds ≤

1λn.

Summing onn gives

∞∑n=0

1λn

1−

∞∑j=0

Pi,j(t)

≤

∫ t

0

∞∑n=0

Pi,n(s) ds ≤∞∑

n=0

1λn.


The right hand part of this inequality shows that if∑∞

n=01λn< ∞ then

∑∞j=0 Pi,j(t) can

not be 1 for allt, while the left hand part of this inequality shows that if∑∞

n=01λn

= ∞then

∑∞j=0 Pi,j(t) = 1 for all t ≥ 0. Thus the pure birth process exists if and only if∑∞

n=01λn

= ∞. Note that this condition is obviously satisfied for the Poisson process.

Exercise 33–11.Explain intuitively what goes wrong with the pure birth processwhen

∑∞n=0

1λn< ∞.

Example 33–6.TheYule processis a pure birth process for whichλn = nβ, thatis, the birth rate is proportional to the number present. This process clearly meetsthe criteria for existence which was established in the previous example. IfMi(t) =E[Xt |X0 = i] then the forward equation can be used to show thatM′i (t) = βMi(t).HenceMi(t) = ieβt.

Exercise 33–12.Fill in the details of how the forward equation is used in thiscomputation. What is the conditional variance ofXt?

Example 33–7.The final example is thebirth and death process. SupposeXt

is the size of a population at timet. Then in a short time period there will be(essentially) a single birth or a single death or neither. Formally the model is

(1) P[Xt+h − Xt = 1|Xt = k] ≈ λkh,

(2) P[Xt+h − Xt = 0|Xt = k] ≈ 1− (λk + µk)h,

(3) P[Xt+h − Xt = −1|Xt = k] ≈ µkh, and

(4) P[ |Xt+h − Xt | ≥ 2|Xt = k] ≈ 0.

Hereλk is birth rate andµk is the death rate when the population size isk.

As before this leads to the generator

A =

−λ0 λ0 0 0 . . .µ1 −µ1 − λ1 λ1 0 . . .0 µ2 −µ2 − λ2 λ2 . . ....

......

......

.

As in the case of the pure birth process the condition∑∞

n=01λn

= ∞ will guaranteethat the process is well defined.

Example 33–8.For the birth and death process the embedded chain has transitionmatrix

P =

0 1 0 0 . . .µ1

µ1+λ10 λ1

µ1+λ10 . . .

......

......

...

if λ0 > 0 so this chain behaves in a manner similar to the gambler’s ruin chain,except that when the gambler reaches fortune 0, she begins again with 1 at the


next play. Ifλ0 = 0 and 0< λi for i > 0 then finding the absorption probabilityfor the birth and death chain is exactly the same problem as finding the absorptionprobability at 0 for the gambler’s ruin chain.

Exercise 33–13.How does the first row ofP differ from that shown whenλ0 = 0?

For a continuous time Markov chain with conservative generatorA the stationarydistributionsπ are the solutions ofπA = 0. Also −1/ai,i is the mean of theexponentially distributed time that the chain spends in statei prior to a transitionand−ai,j/ai,i is the probability that when a transition occurs from statei the chainwill move to statej≠ i. The probability of a transition from statei to itself is 0 unlessai,i = 0 in which casei is an absorbing state.

One consequence of the Markov property is that the time spent in each state isexponential. In many models this is not realistic because the exponential distributionis memoryless.


Problems

Problem 33–1. Consider a continuous time version of the nursing home chain inwhich the states are (1) minimal care, (2) some skilled assistance required, (3)continuous skilled assistance required,and (4) dead. Suppose the infinitesimalgeneratorA of the chain has entriesa1,2 = 0.12,a1,3 = 0.05,a1,4 = 0.08,a2,1 = 0.05,a2,3 = 0.07, a2,4 = 0.12, a3,4 = 0.20 and all other non-diagonal entries are zero.What is the matrixA? What are the stationary distributions of the chain? What isthe expected time until absorption as a function of the initial state?

Problem 33–2. For the birth-death process show that a stationary distribution must

satisfyπj+1 =λj

µj+1πj for j ≥ 0 and thatπ0 =

1

1 +∑∞

n=1λ0⋅⋅⋅λn−1µ1⋅⋅⋅µn

. Thus the chain has a

limiting distribution as long as∞∑

n=1

λ0 ⋅ ⋅ ⋅ λn−1

µ1 ⋅ ⋅ ⋅ µn< ∞. What happens ifλ0 = 0?

Problem 33–3. Find the conditional moment generating functionE[evXt |X0 = i] fora Poisson process.

Problem 33–4. A continuing care retirement community is modeled by a contin-uous time Markov chain with five classifications for the status of a patient: (1)Individual Living Unit, (2) Skilled Nursing Facility (Temporary), (3) Skilled Nurs-ing Facility (Permanent), (4) Dead, and (5) Withdrawn from Care. The non-zerooff-diagonal entries of the generator are

0.3 0.1 0.1 0.20.5 0.3 0.3

0.4

.

If a patient is currently in the Individual Living Unit, what is the probability of morethan one visit to the Skilled Nursing Facility? What is the expected duration of avisit to the Skilled Nursing Facility?

Problem 33–5. An HMO has 3 classes of patients: (1) healthy, (2) moderatelyimparied, and (3) severely impaired.The status of a patient is modeled by acontinuous time Markov chain with generator whose non-zero off-diagonal entries

are

1/2 1/21/3 1/31/4 1/4

. What is the stationary distribution of this chain? If a

healthy person arrives today, about how long will it be until this person becomesseverely impaired?



Problem 33–1. A =

−0.25 0.12 0.05 0.080.05 −0.24 0.07 0.12

0 0 −0.20 0.200 0 0 0

, from which the sta-

tionary distribution is easily found to be ( 0 0 0 1 ). Ifei is the expectedtime until absorption starting from statei thene1 = 4 + (.12/ .25)e2 + (.05/ .25)e3

and two other similar equations also hold. Solving givese1 = 8.55, e2 = 7.40,ande3 = 5.

Problem 33–2. The relationπA = 0 givesλ0π0 = µ1π1 andλi−1πi−1 − (µi +λi)πi + µi+1πi+1 = 0 for i ≥ 1. Sum this last equality oni from 1 to j and usetelescoping and the first equation to simplify. The rest follows by normalization.

Problem 33–3. If g(t) = E[evXt |X0 = i] the forward equation shows thatg′ (t) = −λ g(t) + λ evg(t). Now solve this differential equation and use the initialconditiong(0) = evi.

Problem 33–4. The probability of no visits to the Skilled Nursing Facility is3/7. The probability of exactly one visit is 3/7(6/11+(5/11)(3/7))+1/7. Similarreasoning applies when computing the expected duration, keeping in mind thatthe probabilities are conditional on visiting the Skilled Nursing Facility.

Problem 33–5. The stationary distribution isπ1 = 2/9,π2 = 3/9, andπ3 = 4/9.If e1 ande2 are the expected waiting times until the first entry into state 3 giventhe starting state, thene1 = 1/2 + 1/2(1 +e2) ande2 = 1/2(3/2) + 1/2(3/2 + e1).Hencee1 = 7/3 (ande2 = 8/3).


Solutions to ExercisesExercise 33–1. Hint: Look at the proof of the parallel result for discrete timechains.

Exercise 33–2. Since there are only finitely many states, the derivative ofthe sum is the sum of the derivatives, so the equation

∑j Pi,j(t) = 1 can be

differentiated to show that the generator is conservative.

Exercise 33–3. If there are infinitely many states, the derivative and the sumin the previous exercise might not be interchangeable, and this could make thegenerator not be conservative.

Exercise 33–4. Yes in both cases. Look at the sign of the difference quotientsused to compute the entries of the generator.

Exercise 33–5. If there are infinitely many states there could be problems sincethese ‘derivations’ involve passing a limit through the infinite sum occurring inthe matrix multiplication.

Exercise 33–6. The backward equation givesP′i,j(t) = λPi−1,j(t)−λPi,j(t) whichdoes not produce a differential equation forM(t).

Exercise 33–10. The inductive step isddt (eλ tPi,j(t)) = λ jtj−1/ (j−1)! from which

integration giveseλ tPi,j(t) = (λ t)j/ j!, as desired.

Exercise 33–11. If the process is currently in statek the expected waiting timeuntil a birth is 1/λk. If the sum is finite then there is a finite expected waitingtime for infinitely many births to occur, that is, the population size will becomeinfinite in a finite amount of time.

Exercise 33–12. The forward equation givesP′i,j(t) = βjPi,j(t)+β(j−1)Pi,j−1(t).Now multiply both sides of this equation byj and sum onj = j − 1 + 1 to obtainthe equation.

Exercise 33–13. Whenλ0 = 0 the first row is ( 1 0 0 0 . . .).

§34. Introduction to Brownian Motion

The Brownian motion model originated as a description of the movement ofmicroscopic particles in liquid. Applications to modeling stock prices and othersituations quickly followed. Brownian motion is stochastic process for which bothtime and state space are continuous. To make this transition from earlier studiesa simpler discrete time and state space model will be constructed which mimickssome of the important features of Brownian motion. A suitable passage to the limitwill enable this simpler model to become a Brownian motion.

The discrete time and state space model will be constructed to model the motionof a particle in one dimension. LetSn denote the position of the particle at timen andassumeS0 = 0. The change in position of the particle at timei will be denotedDi.The random variablesD1,D2, . . .will be assumed to be independent and identicallydistributed random variableseach taking the values 1 and−1 with equal probability.HenceSn =

∑ni=0 Di and{Sn : n ≥ 0} is nothing more than a random walk model.

A convenient visual and conceptual aid is the path generated byD1, . . . ,Dn

which consists of the vertices{ (j,Sj) : 0 ≤ j ≤ n} together with the line segmentsjoining the vertices. Many interesting probabilistic problems can be turned into pathcounting problems by using this device.

Proposition. The number of paths joining the origin to a point(n, x) is(

nn+x2

)if

(n + x)/2 is a non-negative integer.

proof : Denote byPn the number of displacementsD1, . . . ,Dn which are positive and letNn = n−Pn

denote the number of negative displacements. TheD’s generate a path from the origin to(n, x) if and only if n = Pn + Nn andx = Pn −Nn. This shows that there are

( nPn

)such paths.

From the two equationsPn = (n + x)/2 and the result follows.

The reflection principle is another path inspired device. The notation corre-sponds to that of the picture below,which also provides the proof.

Reflection Principle. The number of paths fromA to B which touch or cross thelevel T is equal to the number of all paths fromA′ to B.

....................

............................................................................................

.....................

.............................................................................................................................

.......................

......................

...

..........................

.............

.............

..........................

.............

.............

••••••••••

••••••

Sn

T

n

AB

A′


§34: Introduction to Brownian Motion 151

As an application consider an election in which candidateA wins with a votescompared to theb < a votes for candidateB. If the ballots are counted one at a time,what is the probability that candidateA is always ahead in the count?

Ballot Theorem. Supposen andx are positive integers. The number of paths fromthe origin to(n, x) which do not touch then axis is x

n

(n

n+x2

).

proof : The number of paths meeting the requirement is the same as the number of paths from(1,1) to (n, x) which do not touch or cross then axis. Now the total number of paths from(1,1) to (n, x) is the same as the total number of paths from (0,0) to (n − 1, x − 1), which is( n−1

n−1+x−12

)by the proposition. The number of paths from (1,1) to (n, x) which touch or cross

then axis can be computed using the reflection principle. By the reflection principle thisis the same as the number of paths from (1, −1) to (n, x) which in turn is the same as thenumber of paths from (0,0) to (n −1, x + 1). Again using the proposition gives this numberas(n−1

n+x2

). Subtraction gives the desired result.

Exercise 34–1.Use the Ballot Theorem to show that the probability that candidateA is always ahead in the tally is (a − b)/ (a + b). Thus if A wins 70 votes to 30 theprobability thatA was always ahead is 0.40.

Some of the basic properties of Brownian motion will now be developed bypassing to the limit, in an appropriate way, in the discrete time and state spacemodel.

One of the first applications of the Brownian motion model was to describe thedisplacement of a small particle suspended in fluid. As a first approximation to thebehavior of the particle assume that the particle moves only because of collisionswith other particles. Assume also that collisions are as likely to come from theleft as from the right. To simplify even further suppose that each collision causesa displacement of magnitude∆ and that the time between collisions isτ . If theparticle begins at the origin at time 0 the positionXt of the particle at timet isthen ∆S[t/τ ] in the notation above. The ideais to see what happens as both∆and τ go to 0. This passage to the limit makes the approximate model becomemore realistic. Note thatE[Xt] = 0 and Var(Xt) ≈ ∆2t/ τ . To avoid disaster somerelationship between∆2 andτ must be maintained in the passage to the limit. Theassumption made here is that∆2/ τ = σ2 for someσ > 0. After passing to the limit

under this assumption the Central Limit Theorem shows thatXtd= N(0, σ2t). The

processXt obtained in this way is also a process withindependent increments.This means that ift1 < t2 < . . . < tn then the random variablesXtn −Xtn−1, . . . ,Xt2 −Xt1

are independent. The increments are alsostationary which means that fort > s

Xt − Xsd= Xt−s. Intuition further suggests that the sample paths of the processXt

should be continuous functions. This means that for each pointω in the underlyingprobability space the function oft defined byt → Xt(ω) should be continuous. Thisintuitive motivation leads to the following more formal definition. A stochasticprocess{Xt : t ≥ 0} is said to be aBrownian motion process(or Wiener process)


with parameterσ2 if

(1) Xt is a process with stationary independent increments,

(2) Xtd= N(0, σ2t) for eacht > 0, and

(3) X0 = 0.

The termstandard Brownian motion refers to the case in whichσ2 = 1.

Exercise 34–2.Use the independent increments property to show Cov(Xt,Xs) =σ2(t ∧ s).

The important but difficult to prove fact about Brownian motion is that thesample paths of the process can be assumed to be continous. This result can beintuitively deduced from the approximation argument given earlier. The samplepaths can also be shown to be nowhere differentiable. Again this fact is intuitivelyapparent from the earlier construction since the sample paths of the approximatingsums have a lot of corners.

For some purposes the notion of aBrownian motion starting from x will beuseful. Such a process is a Brownian motion as above except for the fact thatX0 = xis assumed.

Some of the basic properties of Brownian motion are derived here in an intuitiveway. The continuous time analog of the reflection principle plays an important role.

Example 34–1.Supposea > 0 and denote byTa the first time at which the standardBrownian motionXt takes the valuea. Then Ta = inf{ s > 0 : Xs = a} . Therandom variableTa is well defined because of the continuity of the sample pathsof the processX. Let t > 0 be fixed. Now by the Theorem of Total ProbabilityP[Xt ≥ a] = P[Xt ≥ a|Ta ≤ t] P[Ta ≤ t] + P[Xt ≥ a|Ta > t] P[Ta > t]. ClearlyP[Xt ≥ a|Ta > t] = 0. If Ta ≤ t the independent increment property ofX suggeststhat after timeTa the Brownian motion restarts as though from scratch except thatthe intial value isa rather than 0. HenceP[Xt ≥ a|Ta ≤ t] = P[Xt ≤ a|Ta ≤ t] = 1/2.So P[Ta ≤ t] = 2 P[Xt ≥ a] = 2

√2π

∫ ∞a/ √t e−x2/2 dx. From this formulaP[Ta < ∞] = 1

but E[Ta] = ∞.

Exercise 34–3.Show thatP[Ta < ∞] = 1 andE[Ta] = ∞.

Exercise 34–4.What happens ifa < 0?

Example 34–2.Again leta > 0 and fixt > 0. ThenP[max0≤x≤t Xs ≥ a] = P[Ta ≤ t].

Exercise 34–5.Provide the details of this argument.


Example 34–3.Suppose 0< t1 < t2. What is the probability thatXt is 0 at leastonce in the interval (t1, t2)? Let Z denote the event thatXt is zero for at least onevalue oft in (t1, t2).

P[Z] = E[P[Z |Xt1]]

=∫ ∞

−∞P[T|x| ≤ t2 − t1] dFXt1

(x)

=∫ ∞

−∞P[T|x| ≤ t2 − t1]

1

√2πt1e−x2/2t1 dx.

Exercise 34–6.Simplify this expression to obtain the Arcsine Law.

Martingales can also be exploited to obtain some results about Brownian motion.The verification that{Xt : t ≥ 0} , {X2

t − t : t ≥ 0} and{exp{ λXt − λ 2t/2} : t ≥ 0}are martingales is quite straightforward.

Exercise 34–7.Verify that these are martingales.

Example 34–4.Supposea < 0 < b and letT = inf{ s : Xs = a or Xs = b} . Applyingthe Optional Stopping Theorem gives 0 =E[X0] = E[XT ] = a P[XT = a] + b P[XT =b]. Solving givesP[XT = a] = b/ (b − a) andP[XT = b] = −a/ (b − a).

Exercise 34–8.Verify this computation. Why does the Optional Stopping Theoremapply?

Exercise 34–9.Use the martingale{X2t − t : t ≥ 0} to computeE[T].

Example 34–5.A Brownian motion with drift is a stochastic processBt of theform Bt = Xt + µt whereX is a standard Brownian motion andµ is a nonzeroconstant. Again a simple computation shows that{exp{ λBt −µλ t −λ 2t/2} : t ≥ 0}is a martingale for each fixed value ofλ . Choose for simplicityλ = −2µ so thate−2µBt is a martingale. LetT = inf{ s : Bs = a or Bs = b} wherea < 0 < b are fixed.The Optional Stopping Theorem givesP[BT = b] = (1 − e−2µa)/ (e−2µb − e−2µa). Thisis comparable to the result for the gambler’s ruin process when the game is biased.

Exercise 34–10.Fill in the details of the computation.

Example 34–6.Brownian motion in several dimensions is defined in way whichimitates the definition in dimension 1. Ad dimensional Brownian motion process{Xt : t ≥ 0} is a process with stationary independent increments for whichX0 = 0and for eacht > 0 the random vectorXt has thed variate normal distribution withmean vector 0 and covariance matrixtΣ for some positive definite matrixΣ. Thestandardd dimensional Brownian motion hasΣ as the identity matrix. This impliesthat the coordinates of a standardd dimensional Brownian motion are independent.


Example 34–7. In the multidimensional setting the distance of a standard Brownianmotion processXt from the origin is often of interest. The processYt = ||Xt || iscalled aradial Brownian motion or aBessel process.

Exercise 34–11.Find the expected time required for a standard multidimensionalBrownian motion to leave the ball of radiusa > 0 centered at the origin. Hint:Show that||Xt ||2 − dt is a martingale.


Solutions to ExercisesExercise 34–1. The number of paths from the origin to (a + b,a − b) whichdon’t touch the axis is the number of counting paths in whichA is always ahead.The Ballot Theorem gives the number of these paths. The total number of pathsfrom the origin to (a + b,a − b) is the total number of ways the ballots could becounted. This number is given by the Proposition.

Exercise 34–2. If t > s writeXt = Xt−Xs+Xs and use the independent incrementsproperty to compute as follows: Cov(Xt,Xs) = E[XtXs] = E[(Xt−Xs)Xs]+E[X2

s ] =0 + Var(Xs) = σ2s. Whens > t a similar argument shows that the covariance isσ2t.

Exercise 34–3. Let t → ∞ to obtainP[Ta < ∞] = (2/ √2π)∫ ∞

0 e−x2/2 dx = 1.

Differentiation gives the density ofTa ase−a2/2t/ √2πt3 for t > 0, from which itis easy to see thatE[Ta] = ∞.

Exercise 34–4. The Reflection Principle givesTad= T|a|.

Exercise 34–5. If the maximum exceedsa then the time required to reach levela must have been smaller thant.

Exercise 34–6. Substitute from the earlier formula forP[T|x| ≤ t2− t1] and thenintegrate by parts.

Exercise 34–7. Just writeXt = Xt −Xs +Xs and use the independent incrementsproperty.

Exercise 34–9. 0 = E[X2T − T] = a2P[XT = a] + b2P[XT = b] − E[T] so that

E[T] = −ab after substitution for the probabilities from the example above.

Exercise 34–11. Here ||Xt ||2 − dt is a martingale so ifTa is the time at whichthe Brownian motion hits the surface of the ball of radiusa, E[Ta] = a2/d by theOptional Stopping Theorem.

§35. Laboratory 11

One theory about the behavior of stock pricePt over time is thatPt shouldbehave like a stochastic process with independent ratios, that is, ift1 < . . . < tn

thenPtn/Ptn−1, . . . ,Pt2/Pt1 should be independent. A simple model for this sort ofbehaviour is given by ageometric Brownian motion processPt = eBt whereBt isa Brownian motion with drift. One use of this model of stock prices is to constructa theoretical pricing model for stock options. Acall option is a contract whichgives the owner of the contract the rightto buy one share of the underlying stock ata particular priceK called thestrike price of the option. The call option contractexpires after a fixed timeT. Thus a call that is not used by timeT becomes worthless.

1. If the drift and variance parameters of the underlying Brownian motion withdrift are 0.10 and 0.005 respectively, what is the probability that a stock with a priceof $20 today has a price exceeding $25 one year from now? What is the probabilitythat the stock price exceeds $25 atsome time during the next year?

2. The economist Robert Merton has presented an economic argument that calloptions will not be exercised before the expiration timeT. This implies that thevalue of a call option today depends only on the price of the underlying stock attimeT. Since the value of the call at timeT is max{PT −K,0} , the value of the calltoday (at time 0) is just the actuarial present value of this amount. Thus the valueof a call today isE[vT max{PT − K,0} ]. This is the Black–Scholes option pricingformula. Simplify this expression to obtain a form suitable for computation.

3. Explain how stock price data could be used in conjunction with the formulaof the previous question in order to value acall option on a particular stock. Choosea stock for which call options are traded on the exchange and estimate the parametersgoverning the stock price process. Compare the option price with that given by theBlack–Scholes formula. Most traders believe that the interest rate on 3 month U.S.Treasury bills is the interest rate to use when computing the present value.


§36. Utility Functions

In this concluding section a fundamental question will be examined. Why wouldan insurer assume a risk which the insured is unwilling to assume? To answer thisquestion, a basic understanding of how value is attached to quantities of money willbe required.

The concept of utility was invented toprovide a mathematical methodologyfor at least formally measuring the value of money (that is theutility of money).Naively, one may think that a dollar is a dollar and that’s it. A moments reflectionshows that your desire (or your measure of worth) of a payment of, say, $10,000might differ considerably from that of the person next to you. Consider also theimportance of the same$10,000 to Bill Gates.

These ideas will now be formalized. In this discussion quantities of moneywill always be measured in dollars. Denote byu(w) the utility of w dollars. Itis convenient to make a strictly mathematical assumption. Practically speaking, autility function can only be defined on a certain discrete set (the multiples of $.01).However for mathematical tractibility it is assumed that utility functions are definedon the entire real line or an interval.

As already noted above, different people may well have different utility func-tions. What features should utility functions have in common? One typical as-sumption is that for any individual the utility function should be non–decreasing.This property is the mathematical expression of the fact that having more moneyis ‘better.’ Furthermore one would also expect that as one’s wealth increases theutility of an additional dollar should decrease. This expresses the notion that tosomeone having only $10 the prospect of gaining an additional dollar is greater thanthe prospect of gaining an additional dollar if one already has $1,000,000.

Exercise 36–1.Argue that if the utility function is sufficiently smooth thenu′ (w) ≥0 andu′′ (w) ≤ 0. Thus a utility function will be concave (down).

A real valued functionu(w) is said to be arisk averseutility function if u(w) isnon–decreasing and concave down.

The reason for the terminology risk averse will be explained shortly.

Example 36–1. It is straightforward to check that each of the functions−e−αw,logw, andαw+β is a risk averse utility function as long asα ≥ 0. A utility functionof the form−e−αw is called anexponential utility function .

Example 36–2.Suppose a person with a risk averse utility function has currentwealth w and is given a choice between two investment schemes. In the first


§36: Utility Functions 158

scheme the person will receive an amountb outright. In the second scheme theperson will receive a random amountW whereE[W] = b. How does the persondecide between these two alternatives?

To answer this question, decision makers will be assumed to act according to theExpected Utility Principle : a decision maker always chooses the available optionwith the largest expected utility.

Example 36–3. In the previous example the expected utility of the first investmentscheme isE[u(w+b)] = u(w+b), while the expected utility of the second investmentscheme isE[u(w + W)].

Generally it is not easy to perform theactual computations required by theexpected utility principal. The following theorem can provide some helpful infor-mation.

Jensen’s Inequality. If X is a random variable andf is a function which is concavethen

E[f (X)] ≤ f (E[X]).

proof : This proof is valid under the additional assumption that the functionf is twice continuouslydifferentiable. For any fixedx and a, the Fundamental Theorem of Calculus and anintegration by parts gives

f (x) = f (a) +∫ x

af ′ (t) dt

= f (a) + f ′ (a) (x − a) +∫ x

a(x − t) f ′′ (t) dt

≤ f (a) + f ′ (a) (x − a)

sincef is concave. SubstitutingX for x andE[X] for a yields

f (X) ≤ f (E[X]) + f ′ (E[X]) (X − E[X]).

Taking expectations of both sides of this last inequality proves the theorem.

Exercise 36–2.Under what conditions does equality hold in Jensen’s inequality?

Continuing the example, from Jensen’s inequality and the fact that a risk averseutility function is concave

E[u(w + W)] ≤ u(E[w + W]) = u(w + b)

so that such a person will always select the sure payoffb! Would such a person buya lottery ticket?

Having discussed the methodology by which a measure of value is assigned toa given quantity of money, the question of why insurance exists can be addressed.


Consider a situation in which a person with utility functionu and current fortunew faces the prospect of a potential financial loss of amountA. For concretenesssuppose that the person is insuring a house against the possibility of a loss due tofire. It is natural to view the amount of lossA as a random variable since the amountof loss depends on the severity of the fire, the speed with which the fire departmentresponds, and other such factors which are unpredictable in nature. Suppose thatfor a fixed non–random premium amountP an insurer will indemnify the insuredagainst such a loss. What premium would the person be willing to pay? If theoffer of insurance is accepted, the expected utility isu(w − P). If the person doesnot buy insurance thenE[u(w − A)] is the persons expected utility. Because of theexpected utility principle, the person would be willing to buy insurance wheneverthe premium satisfies

u(w − P) ≥ E[u(w − A)].

This analysis applies to the person seeking insurance. Now analyze the positionof the insurer. Assume that the insurer has utility functionui and current wealthwi. Reasoning as before shows that the insurance company will offer insurance at apremiumP if

ui(wi) ≤ E[ui(wi + P − A)].

If there is a set of valuesP which satisfy simultaneously both of these inequalitiesthen there is the possibility of insurance being issued.

Exercise 36–3.What is the situation if both the person and the insurer have linearutility functions?

Exercise 36–4.What happens if the insurer and the person have the same expo-nential utility function−e−w/1000, the persons wealth is $5,000, the insurers wealth is$5,000,000 and the loss variableA has an exponential distribution with mean $500?

Exercise 36–5.What happens in the previous exercise if the utility function islogw?

Jensen’s inequality can provide some interesting information about the condi-tions under which a person will purchase insurance. Consider the largest premiumthat an individual would pay for insurance. This premiumP must satisfy

u(w − P) = E[u(w − A)].

Using Jensen’s inequality on the right member gives

u(w − P) ≤ u(w − E[A])

and since the utility function is non–decreasing,P ≥ E[A]. Thus a risk aversedecision maker would be willing to pay a premium greater than thepure premium


(expected loss) in order to obtain insurance. Similarly the premium charged by theinsurance company must also exceed the pure premium. These two facts reinforceintuition and lend a certain credibility to the analysis and assumptions.

As a final note, consider a type of insurance policy that is typically issued.A standard type of insurance policy is thestop losspolicy in which the insurerpays only the amount of the claim which exceeds a certain pre–arranged deductibleamountd. In such a policy the insurers payment when the loss isx is given by(x − d) 1[d,∞)(x). There are many other conceivable types of payment policies, andthe following discussion shows why suchpolicies do not generally occur. Supposethere is another type of payout procedure which for a claim of amountx would payR(x). Naturally, 0≤ R(x) ≤ x, but even more is true.

Theorem. Suppose a person with risk averse utility functionu is to be insuredagainst a loss of amountA. If there is a stop loss policy with expected payoutE[R(A)] which is offered for the same premium as the policy with payoutR(A) thenthe stop loss policy has greater expected utility than the policy with payoutR(A).

proof : Let d be the deductible for the stop loss policy which exists by hypothesis, let the currentwealth of the person bew, and let the premium beP. Arguing as in the proof of Jensen’sinequality gives

u(w+R(A) − A − P) − u(w + (A − d) 1[d,∞)(A) − A − P)

≤ u′ (w + (A − d) 1[d,∞)(A) − A − P)(R(A) − (A − d) 1[d,∞)(A)

)≤ u′ (w − d − P)

(R(A) − (A − d) 1[d,∞)(A)

)where the fact thatu′ is decreasing has been used. Taking expectations gives the result,since the expectation of the last term is zero (why?).


Problems

Problem 36–1. Consider a game of chance in which a fair coin is tossed until a headapprears. LetN denote the toss on which this occurs. Find the density, expectation,and variance ofN. Suppose a prize ofX = 2N is paid when a head first appears.What isE[X]? If your utility function isu(w) = log(w) what is the expected utilityof the prize?

Problem 36–2. In a typical state lotto game a person who buys a $1 ticket wins$1,000,000 with probability 10−8. Find a possible utility function for a person whoplays the lotto. Can this person have a risk averse utility function?

Problem 36–3. Suppose a person has a utility function which is increasing andconcaveup. Give an example to explain why such a person might be called a risklover.

Problem 36–4. Show that if the amount of random lossX has a uniform distributionon the interval (0,1) and if the insured has utility functionu(w) = log(w) then themaximum amount the insured will pay as a premium is

w −ww

e(w − 1)w−1.

Problem 36–5. Repeat the previous problem if the utility function isu(w) = −e−αw

and the loss random variableX has a chi square distribution withn degrees offreedom.

Problem 36–6. Use the Taylor expansion ofu(w − x) aboutx = µ to show that themaximum premium an insured will pay is approximately

µ −12

u′′ (w − µ)u′ (w − µ)

σ2.

Hereµ andσ2 are the mean and variance of the loss random variable.

Problem 36–7. A group medical insurance policy pays $D each time a member ofthe group is hospitalized. The group consists ofg distinct subgroups, which differ inthe rate of hospitalization. Suppose that the annual number of hospital admissionsfor subgroupi has a Poisson distribution with parameterλi. Find an expressionfor the expected claims payments in one year, and also find the distribution of thenumber of admissions in one year.

Problem 36–8. Suppose the loss random variableX has an exponential distributionwith mean 10. Suppose a premium of $5 will be paid. Show that thepropor-tional insurance policy with benefitX/2 and the stop loss policy with benefit


(X −10 log 2)1[10 log 2,∞)(X) are both feasible insurance policies. Which policy wouldthe insured choose, and why?

Problem 36–9. True or False: A person withan exponential utility function,−e−αw,considers wealth irrelevent when deciding the maximum premium to pay for com-plete protection against a random loss.

Problem 36–10.An insurer with wealthw insures a lossX which has the followingprobability distribution:

P[X = 0] = P[X = 16] =12.

The insurer’s utility function isu(x) = logx. The insurer is willing to pay a maximumof 6 to a reinsurer who accepts 50 percent of the loss. Findw.

Problem 36–11.An insurer and an individual have exponential utility functionswith parametersα and 2α respectively. The individual faces a random loss that isnormally distributed with mean 100. The insurer assumes that the variance isσ2.The individual assumes that the varianceis 25. Determine the largest value ofσ2

for which the insurer can charge a premium acceptable to the individual.

Problem 36–12.A decision maker has utility functionu(x) = −e−3x and initialwealthw. The decision maker faces two random losses. The lossX has a normaldistribution with meanα and variance 4. The lossY has a normal distribution withmean 10 and variance 8. Determine the maximum value ofα for which the decisionmaker prefersX to Y.

Problem 36–13.Three insurers have identical utility functionsu(w) = logw, w > 0,and a wealth of 36, 25, and 16 respectively. All three companies insure the samerisk. In the event of a loss each insurer will pay 11. The probability of loss is0.5. Another company offers to reinsure each company’s complete risk at the samepremiumπ. Each company is willing to accept the reinsurance at the premiumπif its expected utility is maximized. Determineπ such that the reinsurer maximizesits total expected profit.


Solutions to ProblemsProblem 36–1. HereP[N = n] = 2−n for n = 1,2, . . .. HenceE[X] = +∞, and

E[log(X)] =∞∑

n=1

n log(2)2−n ≈ 1.386.

Problem 36–3. See the previous problem.

Problem 36–7. Denote byNi the number of hospitalizations for groupi. Then

the amount of claims payments in one year isDg∑

i=1

Ni. Alsog∑

i=1

Ni has a Poisson

distribution with parameterg∑

i=1

λi.

Problem 36–9. True. The variablew cancels from both sides of the equation.


Solutions to ExercisesExercise 36–1. Since more wealth is better,u(w) is increasing. Thusu′ (w) ≥ 0.Since the utility of an additional dollar decreases as wealth increases,u(w + 1)−u(w) ≈ u′ (w) is decreasing. Thusu′′ (w) = (u′ (w))′ ≤ 0.

Exercise 36–2. Equality holds iff ′′ (x) = 0 for all x, which meansf (x) is alinear function ofx.

Exercise 36–3. The person requiresP ≤ E[A] while the insurer requiresP ≥ E[A] so that the only possible premium isP = E[A].

Exercise 36–4. In this case the wealth of each party is irrelevant and the onlycommon premium isP = 1000 lnE[eA/1000] = 2000.

Exercise 36–5. The person requires the premium to satisfy ln(5000− P) ≥E[ln(5000−A)] while the insurer requires ln(5000000)≤ E[ln(5000000+P−A)].To compute the expectations the utility functionu(w) is taken to have the value0 whenw ≤ 0. Then the requirement for the person isP ≤ 536.46 while forthe company,P ≥ 500.03 by numerical computation using a computer algebrapackage.

§37. Life Table at 5% Interest

Age lx dx ax 1000Ax 10002Ax 1000Axx

0 100000 1260 19.922 51.329 17.924 82.8361 98740 92 20.122 41.823 7.253 63.5282 98648 64 20.097 43.022 7.071 64.9623 98584 49 20.064 44.553 7.152 67.0004 98535 40 20.028 46.307 7.392 69.4265 98495 36 19.987 48.236 7.747 72.1446 98459 33 19.944 50.301 8.178 75.0757 98426 30 19.898 52.498 8.684 78.2118 98396 26 19.848 54.835 9.272 81.5629 98370 23 19.796 57.327 9.961 85.15610 98347 19 19.741 59.974 10.751 88.98811 98328 19 19.681 62.792 11.662 93.08712 98309 24 19.619 65.751 12.666 97.39313 98285 37 19.555 68.811 13.724 101.82414 98248 52 19.490 71.902 14.759 106.24215 98196 67 19.425 75.008 15.751 110.61316 98129 82 19.359 78.129 16.695 114.93617 98047 94 19.293 81.268 17.585 119.21218 97953 102 19.226 84.453 18.447 123.49319 97851 110 19.158 87.726 19.316 127.85220 97741 118 19.087 91.091 20.195 132.29521 97623 124 19.014 94.552 21.083 136.82622 97499 129 18.939 98.134 22.002 141.48823 97370 130 18.861 101.852 22.964 146.30524 97240 130 18.779 105.751 24.015 151.35625 97110 128 18.693 109.849 25.173 156.67026 96982 126 18.602 114.174 26.470 162.29727 96856 126 18.507 118.737 27.921 168.25228 96730 126 18.406 123.534 29.520 174.51829 96604 127 18.300 128.576 31.284 181.11230 96477 127 18.189 133.866 33.220 188.03431 96350 130 18.072 139.426 35.355 195.31932 96220 132 17.950 145.244 37.680 202.93533 96088 137 17.822 151.342 40.226 210.91834 95951 143 17.688 157.708 42.984 219.24035 95808 153 17.549 164.348 45.968 227.90236 95655 163 17.404 171.242 49.162 236.86137 95492 175 17.254 178.404 52.586 246.13838 95317 188 17.098 185.832 56.247 255.71939 95129 203 16.936 193.533 60.159 265.61140 94926 220 16.768 201.506 64.328 275.804


§37: Life Table at 5% Interest 166


41 94706 241 16.595 209.750 68.764 286.29042 94465 264 16.417 218.248 73.454 297.03143 94201 288 16.233 227.000 78.408 308.02044 93913 314 16.044 236.014 83.643 319.26545 93599 343 15.849 245.292 89.171 330.76146 93256 374 15.649 254.826 94.994 342.48947 92882 410 15.443 264.618 101.127 354.44548 92472 451 15.232 274.647 107.553 366.58849 92021 495 15.017 284.891 114.257 378.87450 91526 540 14.798 295.346 121.241 391.28651 90986 584 14.574 306.018 128.527 403.83652 90402 631 14.344 316.935 136.156 416.56353 89771 684 14.110 328.092 144.138 429.45454 89087 739 13.871 339.464 152.455 442.46355 88348 797 13.628 351.054 161.123 455.59156 87551 856 13.380 362.858 170.152 468.83057 86695 919 13.127 374.890 179.571 482.19658 85776 987 12.870 387.137 189.384 495.67059 84789 1063 12.609 399.586 199.586 509.22560 83726 1145 12.344 412.195 210.141 522.79561 82581 1233 12.076 424.941 221.027 536.34062 81348 1324 11.806 437.794 232.219 549.81463 80024 1415 11.534 450.744 243.712 563.20264 78609 1502 11.260 463.800 255.529 576.51965 77107 1587 10.983 476.997 267.729 589.82066 75520 1674 10.702 490.357 280.360 603.14367 73846 1764 10.419 503.878 293.435 616.48668 72082 1864 10.132 517.547 306.957 629.83769 70218 1970 9.843 531.305 320.858 643.10970 68248 2083 9.553 545.108 335.091 656.24771 66165 2193 9.263 558.900 349.587 669.17372 63972 2299 8.974 572.682 364.352 681.89573 61673 2394 8.684 586.454 379.395 694.42074 59279 2480 8.395 600.260 394.790 706.82275 56799 2560 8.103 614.130 410.598 719.15676 54239 2640 7.810 628.073 426.851 731.45277 51599 2721 7.517 642.054 443.518 743.68078 48878 2807 7.224 656.017 460.530 755.78679 46071 2891 6.933 669.858 477.742 767.65580 43180 2972 6.647 683.490 495.023 779.196

§37: Life Table at 5% Interest 167


81 40208 3036 6.367 696.796 512.188 790.28182 37172 3077 6.096 709.717 529.133 800.85683 34095 3083 5.834 722.208 545.770 810.88684 31012 3052 5.580 734.292 562.116 820.42385 27960 2999 5.334 746.010 578.225 829.54586 24961 2923 5.097 757.276 593.938 838.16787 22038 2803 4.873 767.969 609.034 846.15488 19235 2637 4.659 778.150 623.584 853.58589 16598 2444 4.452 787.993 637.853 860.68490 14154 2246 4.251 797.588 651.990 867.59891 11908 2045 4.057 806.812 665.785 874.22992 9863 1831 3.875 815.461 678.882 880.38493 8032 1608 3.707 823.460 691.127 886.00994 6424 1381 3.554 830.749 702.385 891.05495 5043 1159 3.416 837.313 712.595 895.50996 3884 945 3.294 843.126 721.669 899.32997 2939 754 3.184 848.396 729.931 902.71498 2185 587 3.084 853.138 737.372 905.64899 1598 448 2.992 857.517 744.245 908.260100 1150 335 2.907 861.589 750.614 910.548101 815 245 2.825 865.483 756.669 912.543102 570 177 2.740 869.539 762.975 914.484103 393 126 2.649 873.840 769.651 916.294104 267 88 2.549 878.615 777.064 917.908105 179 60 2.426 884.468 786.271 919.465106 119 41 2.253 892.737 799.737 921.800107 78 27 2.006 904.455 819.532 925.261108 51 18 1.616 923.036 852.465 933.393109 33 33 1.000 952.381 907.029 952.381

§38. Service Table

x l(τ )x d(d)

x d(w)x d(i)

x d(r)x

30 100000 100 1999031 79910 80 1437632 65454 72 985833 55524 61 570234 49761 60 397135 45730 64 2693 4636 42927 64 1927 4337 40893 65 1431 4538 39352 71 1181 4739 38053 72 989 4940 36943 78 813 5241 36000 83 720 5442 35143 91 633 5643 34363 96 550 5844 33659 104 505 6145 32989 112 462 6646 32349 123 421 7147 31734 133 413 7948 31109 143 373 8749 30506 156 336 9550 29919 168 299 10251 29350 182 293 11252 28763 198 259 12153 28185 209 251 13254 27593 226 218 14355 27006 240 213 15756 26396 259 182 16957 25786 276 178 18358 25149 297 148 19959 24505 316 120 21360 23856 313 355261 19991 298 158762 18106 284 269263 15130 271 135064 13509 257 200665 11246 204 444866 6594 147 130267 5145 119 152268 3504 83 138169 2040 49 100470 987 17 970


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