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Chapter I-2

An Overview

This is the most important chapter in Part I of this book. Here we getintroduced to the fundamental concepts of Life Contingencies such as theactuarial present value, premiums and reserves. The material in ChaptersI-6 through I-9 is mostly an elaboration of the ideas of this chapter usingactuarial symbols and some simple relations.

2.1. Two questions: The Life Contingencies part of this course addresses

two questions. They are illustrated by the following example, which happensto employ a simple, important, fundamental and extremely useful swindle.Please continue reading even after you have spotted it!

Example 2.1.1: Imagine a group of 10 people celebrating their 90-th birth-day. An insurance agent is attending the party and talks them into buyinglife insurance. He promises to pay a sum of 1 (this could be 1 thousanddollars or 1 million cents, the units really do not matter) upon the death ofeach insured. The death benefit will be paid at the end of the year of death.For this service, each person will pay a premium ofPon every birthday thatthe individual is alive, starting on the 90th birthday.

The insurer will deposit the premiums in a fund that will pay interest at anannual rate of 6%, and withdraw (and overdraw if necessary) from the fundthe requisite amounts to pay for death benefits.

It turns out that 2 people die during the first year, 3 during the second year,3 during the third year and the remaining 2 during the fourth year.

1. How much should the premium be?

2. What is the amountper survivorin the fund at the end of two years?

These two questions constitute the core of Life Contingencies part of ExamM.

Solution: You can picture what is going on with the aid of the figure andthe table below.

19

2009 by G.V. Ramanathan

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Chapter I-2 - An Overview 20

Figure 1:

Time Number Amount insurer has in the fundalive

Beginning of 1-st year 10 10 PEnd of 1-st year 8 10 P(1.06) 2Beginning of 2-nd year 8 10 P(1.06) 2 + 8PEnd of 2-nd year 5 10 P(1.06)2 2(1.06) + 8P(1.06) 3Beginning of 3-rd year 5 10 P(1.06)2 2(1.06) + 8P(1.06) 3 + 5PEnd of 3-rd year 2 10 P(1.06)3 2(1.06)2 + 8P(1.06)2 3(1.06)

+ 5P(1.06)3

Beginning of 4-th year 2 10 P(1.06)3

2(1.06)2

+ 8P(1.06)2

3(1.06)+ 5P(1.06)3 + 2PEnd of 4-th year 0 10 P(1.06)4 2(1.06)3 + 8P(1.06)3 3(1.06)2

+ 5P(1.06)2 3(1.06) + 2P(1.06)2

At the time of issue the fund has 10P. At the end of the first year thisamount will grow to 10P(1.06) and there will be benefit payments to twopersons. So the amount at the end of the first year is 10P(1.06) 2. Atthe beginning of the second year, premium payments by 8 survivors will beadded to the fund. That will grow by 6% at the end of the second year andthree benefit payments will be made and so on.

Question 1: This question, namely, what the premium should be, cannotbe answered as it is posed. The insurer may charge whatever premium hewants! In order to determine the premium uniquely we have to impose otherrequirements. One possibility is to require that the insurer must neither gain


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nor lose anything from this contract. Although this requirement is unrealistic,

it does give a rough idea of the premium. The insurer may modify it usingother considerations. This requirement is called the Equivalence Principleand the premium so calculated is called the Benefit Premium.

The equivalence principle would then require that (by the end of the contract)

10 P(1.06)4 2(1.06)3 + 8P(1.06)3 3(1.06)2

+ 5P(1.06)2 3(1.06) + 2P(1.06)2 = 0. (1)

Rearranging the terms and dividing by 10(1.06)4,we get

P1 + 0.8(1.06)1 + 0.5(1.06)2 + 0.2(1.06)3

= 0.2(1.06)1 + 0.3(1.06)2 + 0.3(1.06)3 + 0.2(1.06)4. (2)

This gives the benefit premium,P = 0.365758.Since the amount of premiumstays fixed every year, it is called a Level Premium.

Question 2: What is the amount per survivor in the fund at the end of twoyears? We will get two equivalent expressions as answers for this question.Some of the manipulations of expressions here are essential to bring out theirmeaning.

Since there are 5 survivors at the end of two years, the amount the insurer

has available per survivorat the end of the second year is (see the table)(1/5)

10 P(1.06)2 2(1.06) + 8P(1.06) 3

=

10

5(1.06)2

P

1 + 0.8(1.06)1

0.2(1.06)1 + 0.3(1.06)2

(3)

WithP = 0.365758 this expression equals 0.41826.

By noting from Eq.(2) that

P

1 + 0.8(1.06)1

0.2(1.06)1 + 0.3(1.06)2

= 0.3(1.06)3 + 0.2(1.06)4 P0.5(1.06)2 + 0.2(1.06)3we can rewrite the expression in Eq.(3) as

(1/5)10(1.06)2

0.3(1.06)3 + 0.2(1.06)4 P

0.5(1.06)2 + 0.2(1.06)3

=

(3/5)(1.06)1 + (2/5)(1.06)2 P

1 + (2/5)(1.06)1

(4)


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With P = 0.365758, the right-hand side of Eq.(4) also gives 0.41826. The

expressions in Eqs.(3) and (4) are called benefit reserves. They are equal.The reason they are equal is that we calculated the premium by the equiv-alence principle. We derived the expression in Eq.(4) from that in Eq.(3)by using Eq.(2), which is the equivalence principle. The equality of the twoexpressions says that the that the funds at the end of two years should gotowards covering the net future liability. We will say more about these twoformulas later.

But, before we go any further, I am sure you have recognized the swindlehere. What is the point of calculating the premium after they are all dead?Or do we know ahead of time how many will die each year?

The answer is at the very foundation of not only the theory of insurance butalso of a lot of areas that use Statistics. We say, yes, we know how manypeople will die.. sort of. We simply assert that the data we have in Example2.1.1 about the proportion of deaths doesnt pertain just to this group of90-year-olds, but to all 90-year-olds that we will be concerned with. An au-thoritative table that lists the number of individuals from a closed populationthat survive to various ages is called a life table. This group of population iscalled a survivorship group. How this table is constructed does not concernus here. You will be given such a table, called the Illustrative Life Table(ILT), in the exam. You can download it from the SOA web site.

Imagine for a moment that the number of survivors in Example 2.1.1 doesindeed come from such an authoritative table, and an extract is given below.Herelx stands for population aged x.

x 90 91 92 93 94lx 10 8 5 2 0

In a universal table such as this, the initial population, 10, is not very rel-evant. The table could just as well have the populations listed as 100, 80,50, 20 and 0. The reason is that as you can see from Eqs.(2)-(4) (since thepremium and the reserve are amounts per person) only the proportion of

deaths and proportion of survivors are needed. For example, in Eq.(2), 50%of the original population survives two years and each pays a premium ofP. Similarly, in Eq.(4), which represents the reserve at the end of two yearswhen the population is 5, the factor (3/5) corresponds to the proportion ofthose 5 who die within the next year and each death gets a benefit of 1.


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We identify the proportion of survivors with probability of survival and the

proportion of deaths with probability of death. For instance, if the table saysthat out of a population of 10, 2 die over a period, then we interpret thatas the probability that a randomly chosen person will die in that period is0.2. Conversely, if we are given that the probability of death over a period is0.2, it is very convenient to visualize it as, say, 20 people out of a populationof 100 dying over that period, or even 0.2 people out of a population of 1,blissfully ignoring the fact that there cannot be 0.2 people. This is the reasonwhy the swindle is very useful. The table model is called a deterministicmodel and the model resulting from the interpretation of the proportions asprobabilities is called a probabilistic model.

Let us now return to Example 2.1.1. Consider Eq.(2). First look at theright-hand side. 0.2 is the proportion of deaths (probability of death) duringthe first year. For each death a benefit of 1 is paid at the end of the firstyear. The factor (1.06)1 is the present value of unit benefit paid at the endof one year. Similarly, 0.3 is the proportion of deaths (probability of death)during the second year. The present value of the benefit of 1 paid at the endof two years is (1.06)2. And so on. Thus then-th term on the right-handside is the present value of the benefit paid at the end ofn-years multipliedby the probability of death in the n-th year. To put it differently, it is theExpected Valueof the present value of the benefit. The expected value ofthe present value of a payment is known as the Actuarial Present Value,

APVfor short, of that payment.

Now consider the expression in the left-hand side of Eq.(2). The first termcorresponds to the proportion of population that is alive at time of issuetimes the present value ofPpaid at time of issue. The second term is theproportion of population alive one year later times the present value of anamount ofPpaid at the beginning of next year and so on. The k-th term isthe present value of premium paid times the proportion of population thatsurvives k-years. If we interpret the proportion of population that is alive attimek as the probability of survival of an individual to timek, then what wehave on the left-hand side of Eq.(2) is the expected value of the present valueof the premiums paid, or the APV of the premiums paid by an individual.

Thus the equivalence principle can be stated as

APV of Premiums = APV of benefit(s).


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At this point it is worth noting that the death benefit is a one time payment,

whereas the premium payment is an annuity.2.2. General definitions: Henceforth we will denote an individual agedx by (x). Suppose an insurance policy is written for (x). Suppose that abenefit ofbk+1 will be paid at the end of year k, if death occurs during yeark, k = 0, 1, . This means, ifT is the time of death (measured from thetime of issue) and k T < k + 1, then a benefit of bk+1 will be paid attime k + 1. The present value of the benefit paid at the end of year k isbk+1v

k+1, where v = (1 +i)1 is the discount factor1 corresponding to theannual interest rate i. Then

APV of death benefit =

k=0 bk+1v

k+1

P r(k T < k+ 1). (5)

In order to calculate the probability of death in year k we can use the lifetable. Iflx denotes the population aged (x), thenP r(k T < k+ 1) is justthe proportion of the population agedx now that will die between ages x + kand x + k+ 1. Hence

P r(k T < k + 1) = (lx+k lx+k+1) /lx. (6)

Now let us look at the premium payment. Quite generally, suppose that,starting now, when a person is x years old, a payment will be made at thebeginning of every year that (x) is alive. The amount may vary from yearto year. Let k be the amount to be paid at time k if (x) is alive. Thesepayments may be made by (x) to an insurer as premiums, or they may bepaid to (x), in which case they are annuity payments. The k-th paymentis k with probability P r(T > k) and 0 with probability P r(T k). Theexpected value of the present value of the k-th payment is kv

kP r(T > k).Since the expected value is additive, the expected value of the present valueof all the payments is

APV of Premiums (or annuity) =k=0

kvk P r(T > k). (7)

SinceP r(T > k) is the proportion of people who have survived k years,

P r(T > k) =lx+k/lx. (8)

1Appendix B at the end of Chapter I-1 has a summary of definitions and symbols for

some interest theory entities.


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The equivalence principle asserts that the expression in Eq.(5) for the APV

of benefit and that in Eq.(7) for the APV of premiums should be equal.Example 2.2.1: An insurance on (80) will pay 1000 at the end of the yearof death if death occurs within 3 years (and nothing otherwise). Mortalityfollows the Illustrative Life Table. i= 0.06.Calculate the annual level benefitpremium.

Solution: The benefit isbk+1= 1000 ifk = 0, 1 or 2 and 0 otherwise2. The

present value of the benefit is 1000(1.06)(k+1) ifk = 0, 1 or 2 and 0 otherwise.From the ILT, l80 = 3, 914, 365, l81 = 3, 600, 038, l82 = 3, 284, 542 and l83 =2, 970, 496. Number of deaths in year k = 0 is 3, 914, 365 3, 600, 038 =

314, 327, in year k = 1 it is 3, 600, 0383, 284, 542 = 315, 496, and in yeark = 2 it is 3, 284, 542 2, 970, 496 = 314, 046. Using Eqs.(5) and (6), theAPV of benefits is

1000(1/3, 914, 365)

(314, 327)(1.06)1 + (315, 496)(1.06)2 + (314, 046)(1.06)3

= 214.851.

IfPis the annual premium, from Eqs.(7) and (8), the APV of premiums is

P(1/3, 914, 365)

3, 914, 365 + 3, 600, 038(1.06)1 + 3, 284, 542(1.06)2

= 2.6144 P.

Equating the two APVs we get 214.851 = 2.6144 P orP = 82.18.

2.3. Benefit Reserves: The amount in the insurers fund per survivor isthe reserve. That plus the expected future premiums must cover the expectedfuture liabilities. There are two ways to look at reserves.

The retrospective method: In Example 2.1.1, the end-of-the-year amountpresented in the last column of the table on Page 20 divided by the numberof survivors is the end-of-the-year reserve. We calculated it for the secondyear in Eq.(3). Now let us interpret it.

P{1 + 0.8(1.06)1} is the APV (at issue) of the premiums paid in the first

two years. {0.2(1.06)1

+ 0.3(1.06)2

} is the APV of the benefits paid inthe first two years. The difference is the expected present value at issue ofthe net amount in the fund. At the end of two years this will accumulate

2Note that k = 0 means that death occurs within 1 year, k 1 means death occurswithin 2 years and so on. See the beginning paragraph of Sec. 2.2.


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by a factor of 1.062. There are only half as many survivors (5 out of 10) as

the number at issue. So the reserve is obtained by dividing the accumulatedamount by 1/2 (since the reserve is the amount in the fund per survivor).Quite generally, if we denote the reserve at time k by kV, then, counting timefrom issue,

kV = 1

vk P r[T > k ]{AP Vof Premium for k years - AP Vof benefits fork years} .

(9)As before, if we use the ILT, we set

P r[T > k] = lx+k

lx.

The prospective method: Now let us look at the expressions in Eq.(4).At the end of two years there are 5 persons that are 92 years old. Three diein the next year and two the year after that. Clearly then the right-hand sidein Eq.(4) is the APV of the future benefits for each 92-year-old minus theAPV of the future premiums (determined at time of issue) received from eachperson who is 92 years old now. Thus, in general, the Prospective Formulafor the reserve at the end ofk years is

kV= APV of future benefits for (x+ k) APV of future premiums from (x+ k)(10)

Note that the premium in this equation is the premium determined at issuefor (x). We showed in Eqs.(3) and (4) that the two formulas give the sameanswer.

Example 2.3.1: A 3-year theft insurance on a new car will pay 20,000 ifthe car is stolen during the first year, 15,000 if it is stolen in the second yearand 10,000 if it is stolen in the third year. The payment will be made at theend of the year of loss.

For this insurance, the first years premium is twice the premium for thesecond and third years. Premiums will be paid at the beginning of each year

for three years so long as the car is not stolen. Each year, assuming that thecar has not been stolen by the beginning of the year, the probability that itwill be stolen during that year is 0.1. The annual interest rate is 0.06.

The premiums are based on the equivalence principle.


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Calculate the first years premium.

Solution: Suppose that K is the year that the theft occurs. K = 0 willcorrespond to the theft occurring between times 0 and 1 year. Then P r(K=0) = 0.1. P r(K= 1) is the probability that the car is not stolen in the firstyear and it is stolen in the second year, which is (0.9)(0.1) = 0.09.SimilarlyP r(K= 2) = (0.9)(0.9)(0.1) = 0.081.If the car is stolen during the first year,an amount of 20,000 will be paid at the end of the first year. The presentvalue of this is 20, 000(1.06)1.Similarly if the car is stolen in the second yearthe PV will be 15, 000(1.06)2 and if it is stolen in the third year, the PVwill be 10, 000(1.06)3. Therefore the APV of benefits is the expected valueof the present value, which is

20, 000(1.06)1(0.1)+15, 000(1.06)2(0.09)+10, 000(1.06)3(0.081) = 3768.3793.

LetPbe the second and third year premiums. The premium for the first yearis 2P. Since the probabilities that the car will not be stolen at the beginningofK = 0, K= 1 and K= 2 are 1, 0.9 and 0.92 respectively, the APV ofpremiums is (see Eq.(7))

2P+ P(1.06)1(0.9) + P(1.06)2(0.81) = 3.5699537P.

Equating the two APVs we get P = 1055.58. The first years premium is2P= 2111.16

Example 2.3.2: In Example 2.3.1 calculate the benefit reserve at the endof two years.

Solution: At time 2, there is one premium to be made and one benefit tothe amount of 10,000 if the car is stolen. Hence by the prospective formula(Eq.(10)), the reserve is

(10, 000)(1.06)1(0.1)1055.6 =112.

By the retrospective formula, Eq.(9),

2V = 1.062

P r(T >2)[2P+ P(1.06)1P r(K >1)

20, 000(1.06)1P r(K= 0)15, 000(1.06)2P r(K= 1)]

= 1.062

(0.9)(0.9)[2P+ P(1.06)1(0.9)20, 000(1.06)1(0.1)

15, 000(1.06)2(0.9)(0.1)] =112.


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Solution: Suppose thatT is the time of death (measured from the time

of issue). Then an amount of 1 will be paid at T. The present value of thepayment is et.

AP V = E(eT) =

0etf(t) dt=

0etet dt

=

+ (14)

This is an important result.

Example 2.4.2: In Example 2.4.1, suppose that the insured pays a benefitpremium at a continuousconstantrate ofP. Calculate P.

Solution: Since T is exponentially distributed, 1 F(t) = et

. FromEq.(13), the APV of premiums is

P

0etet dt=

P

+ . (15)

Since this is a benefit premium its APV should equal the APV of benefitswe calculated in Eq.(14). Hence P =.

Example 2.4.3: The future lifetime of John is exponentially distributedwith mean 50. John takes out life insurance that will pay an amount of100,000 at the moment of death. If death is due to an accident, the policy

will pay an additional amount of 400,000. The probability that John will dieof an accident is 1/10-th of the probability that he will die of non-accidentalcauses. Benefit premium will paid at a constant rate as long as John is alive.Assuming a force of interest of 0.05, calculate the annual benefit premium.

Solution: First calculate the APV of the benefit.

Step 1: LetTbe the time of death. Count the benefit in hundred thousands.The benefit is a random variable now. It is 1 with probability 10/11 (nonacci-dental) and 5 with probability 1/11 (accidental). Therefore the expected ben-efit, given that death has occurred at time T, is 1(10/11)+5(1/11) = 15/11.Its present value is (15/11)e0.05T.

Step 2: The pdf ofT is 0.02e0.02t (exponential with mean 50). Hence theexpected value of the present value is

AP V of benefit = 0.02

0e0.02t(15/11)e0.05t dt= (15/11)(0.02/0.07).


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This is in hundred thousands.

Now calculate the APV of the premium. If the annual rate is P , then byEq.(15)

AP V of premiums =P(1/0.07)

Equating this to the APV of benefit we get

P= (15/11)(0.02) = 3/110.

The annual premium rate is 300, 000/110.

Example 2.4.4: The future lifetime of (x) is exponentially distributed withmean 50 and the future lifetime of (y) is exponentially distributed with mean

40. An insurance on the two lives will pay an amount of 100,000 at themoment of the second death. Benefit premium will be paid at a constant rateuntil the moment of the first death. The force of interest is 0.03. Calculatethe premium rate.

Solution: Let the time of the second death be U. Then the PV of benefitis 100, 000e0.03U. To find the APV we need the distribution ofU, which isthe larger of the two lifetimes. For that use the results of Example 1.5.2 ofChapter I-1 with x = 1/50 = 0.02 andy = 1/40 = 0.025.

fU(u) = 0.02e0.02u + 0.025e0.025u (0.045)e0.045u.

The APV is

100, 000

0e0.03ufU(u) du

= 100, 000

0e0.03u

0.02e0.02u + 0.025e0.025(u) (0.045)e0.045u

du

= (100, 000)

0.02

0.05+

0.025

0.055

0.045

0.075

= 25454.5.

IfWis the time of the first death, then again from Example 1.5.2, P r(W >w) =e(0.02+0.025)w and the APV of the premiums is (see Eq.(12))

P 0 e

0.03w

e(0.02+0.025)w

dw=

P

0.02 + 0.025 + 0.03=P/0.075.

Equating the two APVs we get

P = (0.075)(25454.5) = 1909.


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Chapter Summary

Actuarial Present value: Express the present value in terms of thetime of payment random variable and take expectation. Examples:

The APV of an insurance that pays a benefit ofB(T) at timeT(x)the moment of death of (x), is

0B(t)etfT(x)(t) dt.

The APV of an insurance that pays a benefit ofbK+1 at the endof the year, K, K= 0, 1, , of death of (x) is

k=0

bk+1vk+1P r (k T(x)< k+ 1) =

k=0

bk+1vk+1 lx+k lx+k+1

lx.

The APV of a premium paid at a continuous rate ofP(t) as longas (x) is alive is

0P(t)et

1 FT(x)(t)

dt.

The APV of all the premiums k at the beginning of each yearthat (x) is alive is

0

kvkP r (T(x)> k) =

0

kvk lx+k

lx.

Benefit premium means the APV of premiums equals the APV of ben-efits.

The benefit reserve is the amount that is in the fund at a certain time. Itequals the APV of future benefits minus the APV of future premiums.


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Problems

1. An insurance will pay an amount of 1,000 at the end of this year if(80) dies within one year. No payment will be made if (80) survivesone year. Using the ILT with i = 0.06,, calculate the APV of thisinsurance. (75.75)

2. Peter is 10 years old now. He will receive an amount of 100,000 on his25-th birthday if he survives. Using the ILT with i = 0.06, calculatethe APV of the payment. (This payment is called pure endowment.)(41,122)

3. Pam has a computer. Its future lifetime is exponentially distributedwith mean 5. The moment the computer breaks down or at the endof 5 years from now, whichever is earlier, Pam will spend 3,000 dollarstowards a new system. Pam will save money at a continuous, constantrate ofPper year towards the cost of the system for the next five years.The force of interest is 0.1.

(a) Using the equivalence principle, calculate P. (858.5)

(b) Calculate the benefit reserve at the end of three years. (1,257.7)

4. The future life time of a new machine has the PDF f(t) = 0.05e0.05t, t >0.A warranty will pay an amount of 1 at the moment of breakdown ofthe machine if it breaks down within 5 years. The force of interest is0.05. Calculate the APV of the payment. (0.197)

5. Consider the following table of lives:

Age 90 91 92 93 94Number 100 80 50 10 0

Each of the hundred is issued a 3-year term insurance of 1 to be paidat the end of the year of death. This means that an amount of 1 will

be paid at the end of the year of death if death occurs within 3 years ofissue. Nothing will be paid if death occurs after 3 years. With i = 0.06

(a) Calculate the benefit premium. (0.35983)

(b) Calculate the benefit reserve at the end of two years. (0.395)


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6. A 3-year term insurance on a person aged 60 will pay 1000 if death

occurs during the first year, 2000 if death occurs during the secondyear and 3000 if death occurs during the third year. Payment will bemade at the end of the year of death. Premiums will be paid at thebeginning of each year that the insured is alive, for three years. Usingthe Illustrative Life Table and i = 0.06,

(a) Calculate the benefit premium. (28.41)

(b) Calculate the benefit reserve at the end of two years. (17.95)

(c) What is the benefit reserve at the end of three years? (0)

7. In addition to the death benefit in Problem 6, suppose that the insur-

ance will pay an amount of 3,000 at the end of three years if the insuredsurvives 3 years. Answer the three questions of Problem 6. (This insur-ance is called a three-yearendowment insurance.) (889.52; 1,940.67;3,000)

8. An insurance will pay an amount of 5 at the moment of Johns death.Johns future lifetime has the pdff(t) = 0.02e0.02t.The force of inter-est is 0.08.

(a) Calculate the APV of this insurance. (1)

(b) Calculate the benefit premium. (0.1)

(c) Calculate the benefit reserve at time s from issue. (0)

9. *4 A special insurance program is designed to pay a benefit in the eventa product fails. You are given:

Benefits are paid at the moment of failure

bt=

300, 0 t

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Calculate the actuarial present value of this special insurance. (168.12)

10. A continuous payment will be made at an annual rate ofP(t) as longas a life (x) is alive but for not more than n years. If the cdf of thefuture lifetime of (x) isF(t) and the force of interest is show that theactuarial present value of the payments is n

0P(s)es[1 F(s)] ds.

(This payment is called a n-year temporary annuity.)

11. Startingn- years from now a continuous payment will be made at anannual rate of P(t) as long as the life (x) is alive. No payment will

be made for the first n-years. If the cdf of the future lifetime of (x) isF(t) and the force of interest is ,derive an expression for the actuar-ial present value of the payments. (This payment is called a n-yeardeferred annuity.)

12. John is 55 now. If he survives 10 years he will then start receiving alife annuity that will pay at a continuous rate of 50,000 as long as heis alive. Johns future lifetime Thas the pdf 0.02e0.02t. Using a forceof interest of 0.06 calculate the present value of this annuity. (280,831)

13. Ann-year deferred annuity will pay at a continuous annual rate of 1 as

long as a life (x) is alive. The annuitant (x) will a pay a level benefitpremium at a continuous annual rate ofPover then-years (calledthedeferral period ) as long as he is alive. The future lifetime of (X)has an exponential distribution with mean 1/.The force of interest is.Express P in terms of and .

14. A 10-year deferred annuity will pay an amount of 1 at the beginningof every year that John is alive. In turn, John will pay a premiumof P at the beginning of every year of the deferral period (as longas he is alive). The probability that John will survive for k years is(0.98)k, k= 0, 1, You are given that v = 0.9. Using the equivalence

principle

(a) CalculateP. (0.3984)

(b) Calculate the benefit reserve at the end of 8 years. (5.843)


sample m

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