04-ko-asm-pmanual-section4

ASM Study Manual for Course P/1 Actuarial Examination. © Copyright 2004-2010 by Krzysztof Ostaszewski - 92 -

SECTION 4: RISK MANAGEMENT AND INSURANCE When a person or a business is exposed to a risk of a financial loss, the loss is generally modeled using a random variable or some combination of random variables. The loss is often related to a particular time interval. For example, in an automobile insurance, losses are generally covered for half a year, a standard duration of an automobile insurance policy (although there are some policies issued for a year). A person or a business at risk may choose to purchase insurance protection to reduce the impact of the loss, of course assuming that there is a party willing to assume the risk in return for a premium. An insurance policy is a contract between the party that is at risk (the policyholder) and an insurer, a party willing to assume all or a part of risk in return for the payment of premium. This contract generally calls for the policyholder to pay the insurer some specified amount, the insurance premium, and in return, the insurer will reimburse the policyholder for specified amounts of losses. A claim is all or part of the loss that is submitted to the insurance company for payment. There are various ways of modeling a loss and/or claim for a particular insurance policy, generally depending on the nature of the loss. The amount paid to the policyholder upon submission of a claim may be the amount of the loss that occurs, or may be less than the amount of the loss (however, in life insurance the amount of the actual economic loss is not really possible to calculate, so a fixed policy benefit amount is established at the beginning of the policy’s existence). If the loss suffered by the insured is X, and the claim submitted to the insurance firm is Y, a situation when Y < X may occur when the insured decides to not bother reporting a claim that is too small (e.g., is clearly below the deductible) to process. The actual full loss is termed the ground up loss X. If Y is the random variable representing the claim to be paid, the expected value of it E Y( ), is called the net premium, or pure premium or expected claim. The random variable Y can have a value of 0, as it is possible that no loss, or no claim, occurs. We are not only interested in the expected value E Y( ), but also in measures of risk, or variability of Y. One such measure is the variance of Y, another natural measure is the standard deviation of it. We may also use the

coefficient of variation: Var Y( )E Y( ) =

σY

µY

, also termed sometimes the unitized risk.

The random variable Y describing the claimed amount of the loss is studied carefully by actuaries, and can be described in a variety of ways. A theoretical model of it can be given, with density (or probability function) fY or cumulative distribution function FY . Alternatively, Y can be given as a function of some other random variable. Consider this simple case: let K be a certain constant, and let Y = K with probability q while Y = 0 with probability p. This is basically a description of a benefit paid in a one-year term life insurance with benefit amount of K, where q is the probability of death in that year, and p = 1 – q. A somewhat more complicated case is when the probability distribution of a loss, when it occurs, is given, and the probability q of a non-negative loss is stated. Suppose that if a loss occurs, it is described by a random variable B. Then Y, the random variable describing the amount of the

RISK AND INSURANCE


possible claim, is a mixed distribution of a non-random zero value with probability weight of p = 1 – q and the random variable B with probability weight q. If the mean and variance of B exist then the mean and variance of Y can be calculated as follows: E Y( ) = p ⋅0 + q ⋅E B( ) = q ⋅E B( ), E Y 2( ) = p ⋅02 + q ⋅E B2( ) = q ⋅E B2( ), and

Var Y( ) = E Y 2( )− E Y( )( )2 = q ⋅E B2( )− q ⋅E B( )( )2 = q ⋅E B2( )− q ⋅ 1− p( ) ⋅ E B( )( )2 =

= q ⋅ E B2( )− E B( )( )2( ) + pq ⋅ E B( )( )2 = q ⋅Var B( ) + pq ⋅ E B( )( )2 .

A more general model, called the individual risk model, considers n (a positive integer) of insurance policies, with the claim for a policy numbered i, where i = 1,2,…,n, being a random variable Xi , with the Xi ’s being independent, identically distributed with finite mean and variance. Then the random variable

S = X1 + X2 +…+ Xn = Xi

i=1

n

∑

is called the aggregate claim random variable. We have:

E S( ) = E Xi( )i=1

n

∑

and

Var S( ) = Var Xi( ).i=1

n

∑

If we denote E Xi( ) = µ and Var Xi( ) = σ 2 then E S( ) = nµ and Var S( ) = nσ 2 so that the

coefficient of variation of S is nσnµ

, and that coefficient of variation converges to 0 as n→∞.

It is also possible to consider a more general model in which the number of claims is random as well, denoted by N, resulting in

S = X1 + X2 +…+ XN = Xi

i=1

N

∑ .

This is termed the collective risk model and is studied in detail in later actuarial examinations. Since the aggregate claim random variable in the individual risk model is a sum of independent, identically distributed, random variables with finite mean and variance, its distribution can be approximated with the normal distribution with the same mean and variance (possibly using the continuity correction). If S is the total loss to be paid for an individual or a group covered by an insurance contract, the expected value of S is the pure premium or net premium for a plan of insurance. However, the actual contract premium will exceed the expected value, because the insurance company has to protect itself from random fluctuations from the mean, and must cover its expenses and earn a profit (which, economically, is also an expense, as it is the cost of capital invested in the business). If we ignore the expenses and profit component, and write the premium charged

SECTION 4


before expenses and profits as Q, and Q = 1+θ( )E S( ),

then the parameter θ is called the relative security loading. An excess-of-loss insurance specifies a deductible amount d, such that all losses are reduced by the amount d for the purpose of payment, and if the loss is less than d, then it is not paid at all. The amount paid by the insurer can be written as a new random variable:

Y =0, X ≤ d,X − d, X > d,

⎧⎨⎩

⎫⎬⎭= max 0,X − d( ) = X − d( )+ .

The expected value of the payment is

E Y( ) = E max 0,X − d( )( ) = E X −min X,d( )( ) = sX0

+∞

∫ x( )dx − sX0

d

∫ x( )dx = sXd

+∞

∫ x( )dx.

There are some variations of a concept of a deductible: • The franchise deductible d exists when the insurer pays 0 if the loss is below d but pays the full amount of loss if the loss is above d, i.e.,

Y =0, X ≤ d,X, X > d.

⎧⎨⎩

• The disappearing deductible with lower limit dL and upper limit dU , where dL < dU , refers to the situation in which the insurer pays 0 if the loss is below dL , the insurer pays the full loss if the loss amount is above dU , and the deductible amount reduces linearly from dL to 0 as the loss increases from dL to dU . The amount paid by the insurer is

Y =

0, X ≤ dL ,

dU ⋅X − dLdU − dL

, dL < X ≤ dU ,

X, X > dU .

⎧

⎨⎪⎪

⎩⎪⎪

We say that an insurance policy has a limit of u if the insurer will only pay for claims up to a maximum amount of u. The amount paid is:

Y =X, X ≤ u,u, X > u.

⎧⎨⎩

The expected value of the payment is

E Y( ) = sY y( )0

+∞

∫ dy = sX0

u

∫ x( )dx.

A policy may also combine an excess-of-loss insurance with a policy limit, with

Y =0, X ≤ d, X − d, d < X ≤ u,u − d, X > u.

⎧⎨⎪

⎩⎪

The expected payment is then

RISK AND INSURANCE


E Y( ) = sY y( )dy0

+∞

∫ = sX x( )dxd

u

∫ .

Let us produce a quick proof of this last formula, as an exercise:

sY y( ) = Pr Y > y( ) = Pr X > y + d( ), y + d < u,0, y + d ≥ u,

⎧⎨⎩

⎫⎬⎭=

sX y + d( ), y < u − d,0, y ≥ u − d.

⎧⎨⎩

Therefore,

E Y( ) = sY y( )dy0

+∞

∫ = sX y + d( )dy0

u−d

∫ =x = y + d y = 0 ⇒ x = d dx = dy y = u − d⇒ x = u

INTEGRATION BY SUBSTITUTION

= sXd

u

∫ x( )dx.

Another possible contract provision is an insurance cap. An insurance cap specifies a maximum claim amount m that would be paid if a loss occurs on the policy, so that the insurer pays the claim up to a maximum payment of m. If there is no deductible, this is the same as a policy limit, but if there is a deductible of d, then the maximum amount paid by the insurer is m = u − d. In this case, the policy limit of amount u is the same as an insurance cap of amount u − d. Proportional insurance is a contract in which a fraction α, where α ∈ 0,1( ), of a loss X is paid. For example, it is quite common for traditional medical insurance to have a design such that the insured pays 20% of a claim cost, and the insurer pays the remaining 80%. When the amount paid is not the entire loss, we may want to distinguish between random variables:

Y * : the amount paid, conditional on the event that a positive payment is made, sometimes called payment per payment, and, Y : the amount paid per loss.

For example, in a policy with a deductible d, we have: Y* = X − d( ) X > d( ),

Y =0, X ≤ d,X − d, X > d.

⎧⎨⎩

The expected value of the payment per payment in this case is called the mean excess loss

E Y *( ) = E X − d X > d( ) = x − d( ) fX x( )sX d( ) dxd

+∞

∫ =1

sX d( )E X −min X,d( )( ) =sX

d

+∞

∫ x( )dx

sX d( ) .

For a policy with a deductible of d and a limit of u Y* = min X − d,u − d( ) X > d,

Y =0, X ≤ d, X − d, d < X ≤ u,u − d, X > u.

⎧⎨⎪

⎩⎪

We can also determine the densities of the variables Y and Y * for that second case:

SECTION 4


fY * y( ) =

fX y + d( )sX d( ) ,

0 ≤ y < u − d,

sX u( )sX d( ) ,

y = u − d (point mass),

0, y > u − d,

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

fY y( ) =

FX d( ), y = 0 (point mass), fX y + d( ), 0 < y < u − d, sX u( ), y = u − d (point mass),0, y > u − d.

⎧

⎨⎪⎪

⎩⎪⎪

From the perspective of the insurance firm, the purpose of the deductibles and policy limits is to reduce the amount paid by the firm (and possibly to reduce the frequency of claims reported). The expected amount of the loss eliminated from amount paid by the insurance company divided by the expected total loss amount is called the loss elimination ratio. Consider, for example, a loss X distributed uniformly on the interval [0, 100]. Then the expected value of X is 50. Now consider a policy with deductible of 10 that pays for that loss. The expected value of the amount eliminated from payment by the insurance company (and paid by the insured) is

x0

10

∫ ⋅1100

dx + 10 ⋅ 1100

dx =10

100

∫1200

x2x=0

x=10

+ 9 = 0.5 + 9 = 9.5.

The loss elimination ratio is therefore, 9.550

= 19%. This ratio is also equal to the difference

between 1 and the ratio of the expected payment made by the insurer after modification of the contract (by a deductible or a policy limit) to the expected payment made without such a modification. Reinsurance is an insurance contract purchased by an insurance company (ceding company) from another insurance company (reinsurer). Its purpose is generally related to taxes or regulation, or to limit the exposure to catastrophic claims. The basic forms of reinsurance are very similar to the partial insurances on regular policies described above, but they apply to the insurer’s claims (either individually or in aggregate, but most often in aggregate). The claims paid by the ceding insurer are referred to as retained claims. A stop-loss reinsurance specifies a deductible amount d. If the aggregate claim S is less than d then the reinsurer pays nothing, but if the aggregate claim is greater than d then the reinsurer pays the aggregate claim in excess of d. A reinsurance contract may also have a cap, specifying a maximum amount paid by the reinsurer, u. A cap typically would be set up in conjunction with a stop-loss with deductible d. It is interesting to note that a stop-loss reinsurance with a deductible of d and a cap of u is algebraically equal to a stop loss policy with deductible d minus a stop-loss policy with a deductible of u + d. A proportional reinsurance specifies a fixed fraction of the aggregate claims that the reinsurer pays.

RISK AND INSURANCE


Exercise 4.1. May 2003 Course 1 Examination, Problem No. 15, also P Sample Exam Questions, Problem No. 99, and Dr. Ostaszewski’s online exercise posted October 31, 2009 An insurance policy pays a total medical benefit consisting of two parts for each claim. Let X represent the part of the benefit that is paid to the surgeon, and let Y represent the part that is paid to the hospital. The variance of X is 5000, the variance of Y is 10,000, and the variance of the total benefit, X + Y is 17,000. Due to increasing medical costs, the company that issues the policy decides to increase X by a flat amount of 100 per claim and to increase Y by 10% per claim. Calculate the variance of the total benefit after these revisions have been made. A. 18,200 B. 18,800 C. 19,300 D. 19,520 E. 20,670 Solution. Note that

17,000 = Var X +Y( ) = Var X( ) +Var Y( ) + 2Cov X,Y( ) = 5000 +10,000 + 2Cov X,Y( ), and hence Cov X,Y( ) = 1000. We can now calculate the answer:

Var X +100( ) +1.1Y( ) = Var X +1.1Y( ) +100( ) = Var X +1.1Y( ) = = Var X( ) + Var 1.1Y( ) + 2Cov X,1.1Y( ) = Var X( ) +1.12 ⋅Var Y( ) + 2 ⋅1.1 ⋅Cov X,Y( ) = = 5000 +12,100 + 2200 = 19,300.

Answer C. Exercise 4.2. May 2003 Course 1 Examination, Problem No. 20, also P Sample Exam Questions, Problem No. 93, and Dr. Ostaszewski’s online exercise posted September 26, 2009 A family buys two policies from the same insurance company. Losses under the two policies are independent and have continuous uniform distributions on the interval from 0 to 10. One policy has a deductible of 1 and the other has a deductible of 2. The family experiences exactly one loss under each policy. Calculate the probability that the total benefit paid to the family does not exceed 5. A. 0.13 B. 0.25 C. 0.30 D. 0.32 E. 0.42 Solution. Define X and Y to be loss amounts covered by the policies having deductibles of 1 and 2, respectively. The total benefit paid to the family is

0, if X ≤1 and Y ≤ 2,X −1, if X >1 and Y ≤ 2,Y − 2, if X ≤1 and Y > 2,X −1( ) + Y − 2( ) = X +Y − 3, if X >1 and Y > 2.

The benefit paid is less than 5 if

SECTION 4


0 < 5, if X ≤ 1 and Y ≤ 2,X −1 < 5, if X > 1 and Y ≤ 2,Y − 2 < 5, if X ≤ 1 and Y > 2,X +Y − 3 < 5, if X > 1 and Y > 2.

The first condition, 0 < 5, is true always, so the total benefit paid is less than 5 always in the region where X ≤ 1 and Y ≤ 2. The second condition gives benefit less than 5 when X < 6, X > 1, and Y ≤ 2. The third condition gives benefit less than 5 when Y < 7, X ≤ 1 and Y > 2. Finally, the last, fourth condition gives benefit less than 5 when Y < −X + 8, X > 1 and Y > 2. By combining all of these conditions, we obtained the shaded portion of the graph below showing the region over which the total benefit paid to the family does not exceed 5.

Since the joint distribution is uniform on the square 0,10[ ]× 0,10[ ], we can calculate the probability we are looking for as the ratio of the area of the shaded region and the area of the square 0,10[ ]× 0,10[ ], and that equals

Pr (Total benefit paid does not exceed 5) =1 ⋅ 7 + 6 −1( ) ⋅2 + 1

2⋅5 ⋅5

100=29.5100

= 0.295.

Answer C. Exercise 4.3. May 2003 Course 1 Examination, Problem No. 25, also P Sample Exam Questions, Problem No. 40, and Dr. Ostaszewski’s online exercise posted July 5, 2008 An insurance policy pays for a random loss X subject to a deductible of C, where 0 < C < 1. The loss amount is modeled as a continuous random variable with density function

f x( ) = 2x, for 0 < x < 1,0, otherwise.

⎧⎨⎩

x

1 6 10

2

7

y

10

RISK AND INSURANCE


Given a random loss X, the probability that the insurance payment is less than 0.5 is equal to 0.64. Calculate C. A. 0.1 B. 0.3 C. 0.4 D. 0.6 E. 0.8 Solution. Denote the insurance payment by the random variable Y. Then

Y =0, ifX − C, if

⎧⎨⎩

0 ≤ X ≤ C,C < X ≤ 1.

This relationship is illustrated in the graph below: Based on the information given:

0.64 = Pr Y < 0.5( ) = Pr 0 ≤ X < 0.5 + C( ) = 2xdx =0

0.5+C

∫ x2x=0

x=0.5+C= 0.5 + C( )2 .

Therefore, solving for C, we find C = ± 0.8 − 0.5. Since 0 < C < 1, we conclude that C = 0.3. Answer B.

Exercise 4.4. May 2003 Course 1 Examination, Problem No. 36, also P Sample Exam Questions, Problem No. 49, and Dr. Ostaszewski’s online exercise posted August 23, 2008 An insurance policy pays an individual 100 per day for up to 3 days of hospitalization and 25 per day for each day of hospitalization thereafter. The number of days of hospitalization, X, is a discrete random variable with probability function

Pr X = k( ) =6 − k15

, for k = 1,2,3, 4,5,

0, otherwise.

⎧⎨⎪

⎩⎪

Calculate the expected payment for hospitalization under this policy. A. 85 B. 163 C. 168 D. 213 E. 255 Solution.

Y

X

Y = X − C

C

0.5

0.5 + C

SECTION 4


Define H X( ) to be the total of all hospitalization payments made by the insurance policy. Then

H X( ) =

100, when X = 1, i.e., with probability 515

,


,


,


,


,

⎧

⎨

⎪⎪⎪⎪⎪⎪

⎩

⎪⎪⎪⎪⎪⎪

and

E H X( )( ) = 100 ⋅ 515

+ 200 ⋅ 415

+ 300 ⋅ 315

+ 325 ⋅ 215

+ 350 ⋅ 115

=

= 115

500 + 800 + 900 + 650 + 350( ) ≈ 213.33.

Answer D. Exercise 4.5. November 2001 Course 1 Examination, Problem No. 5, also P Sample Exam Questions, Problem No. 52, and Dr. Ostaszewski’s online exercise posted September 13, 2008 An insurance company sells a one-year automobile policy with a deductible of 2. The probability that the insured will incur a loss is 0.05. If there is a loss, the probability of a loss

amount N is KN, for N = 1,...,5 and K a constant. These are the only possible loss amounts and

no more than one loss can occur. Determine the net premium for this policy. A. 0.031 B. 0.066 C. 0.072 D. 0.110 E. 0.150 Solution. The probability distribution of the amount of the loss N, given that there is a loss is

N = 1 with probability K1, N = 2 with probability K

2,

N = 3 with probability K3, N = 4 with probability K

4,

N = 5 with probability K5.

Therefore

1 = K1+K2+K3+K4+K5= K ⋅

60 + 30 + 20 +15 +1260

⎛⎝⎜

⎞⎠⎟= K ⋅

13760.

RISK AND INSURANCE


This gives K =60137

. Since the probability of having a loss is 0.05, for the unconditional

distribution of amount of loss we obtain N = 0 with probability 0.95 and for positive losses

N = 1 with probability 0.05 ⋅ K1=3137

,


,


,


,


.

But the policy has a deductible of 2, so for N = 1 and for N = 2, the actual payment is 0. Also, this policy pays 0 when there is no loss, which happens with probability 0.95. Let Y be the amount paid after the application of the deductible. Then

Y = 0 with probability

3137

Pr N =1( )

+3

274Pr N =2( )

+ 0.95Pr no loss( ) ,

Y = 1 with probability Pr N = 3( ) = 1137

,


,


.

Therefore the net premium for this policy is

P = E Y( ) = 1 ⋅ 1137

+ 2 ⋅ 3548

+ 3 ⋅ 3685

≈ 0.0314.

Answer A. Exercise 4.6. November 2001 Course 1 Examination, Problem No. 16, also P Sample Exam Questions, Problem No. 123, and Dr. Ostaszewski’s online exercise posted May 15, 2010 You are given the following information about N, the annual number of claims for a randomly

selected insured: Pr N = 0( ) = 12, Pr N = 1( ) = 1

3, Pr N > 1( ) = 1

6. Let S denote the total annual

claim amount for an insured. When N = 1, S is exponentially distributed with mean 5. When N > 1, S is exponentially distributed with mean 8. Determine Pr 4 < S < 8( ). A. 0.04 B. 0.08 C. 0.12 D. 0.24 E. 0.25 Solution.

SECTION 4


Recall that if X has an exponential distribution with mean µ then

Pr a < X < b( ) = FX b( ) − FX a( ) = sX a( ) − sX b( ) = e−aµ − e

−bµ .

We use the law of total probability and the information on the distribution of S when N = 1 and N > 1 (when N = 0, S = 0 and this case can be disregarded), to obtain

Pr 4 < S < 8( ) = Pr 4 < S < 8 N = 1( ) ⋅Pr N = 1( ) + Pr 4 < S < 8 N > 1( ) ⋅Pr N > 1( ) =

= e−45 − e

−85

⎛⎝⎜

⎞⎠⎟⋅13+ e

−48 − e

−88

⎛⎝⎜

⎞⎠⎟⋅16≈ 0.12.

Answer C. Exercise 4.7. November 2001 Course 1 Examination, Problem No. 29, also P Sample Exam Questions, Problem No. 65, and Dr. Ostaszewski’s online exercise posted December 27, 2008 The owner of an automobile insures it against damage by purchasing an insurance policy with a deductible of 250. In the event that the automobile is damaged, repair costs can be modeled by a uniform random variable on the interval [0, 1500]. Determine the standard deviation of the insurance payment in the event that the automobile is damaged. A. 361 B. 403 C. 433 D. 464 E. 521 Solution. Let X and Y denote repair cost and insurance payment, respectively, in the event the auto is damaged. Then

Y =0, if 0 ≤ X ≤ 250, X − 250, if 250 < X ≤ 1500.

⎧⎨⎩

Furthermore

E Y( ) = x − 250( ) ⋅ 11500250

1500

∫ ⋅dx = 11500

⋅12x − 250( )2

x=250

x=1500

=12502

3000≈ 520.8333,

E Y 2( ) = x − 250( )2 ⋅ 11500

dx250

1500

∫ =1

1500⋅13x − 250( )3

250

1500

=12503

4500≈ 434,027.78,

Var Y( ) = E Y 2( ) − E Y( )( )2 ≈ 434,027.78 − 520.832 ≈ 162,763.89. and finally, the standard deviation of Y is

Var Y( ) ≈ 162,763.89 ≈ 403. Answer B. Exercise 4.8. November 2001 Course 1 Examination, Problem No. 34, also P Sample Exam Questions, Problem No. 119, and Dr. Ostaszewski’s online exercise posted January 26, 2008 An auto insurance policy will pay for damage to both the policyholder’s car and the other driver’s car in the event that the policyholder is responsible for an accident. The size of the payment for damage to the policyholder’s car, X, has a marginal density function of 1 for

RISK AND INSURANCE


0 < x < 1. Given X = x, the size of the payment for damage to the other driver’s car, Y, has conditional density of 1 for x < y < x + 1. If the policyholder is responsible for an accident, what is the probability that the payment for damage to the other driver’s car will be greater than 0.50?

A. 38

B. 12

C. 34

D. 78

E. 1516

Solution. The joint density of X and Y is:

fX ,Y x, y( ) = fX x( ) ⋅ fY y X = x( ) = 1, if 0 < x < 1, x < y < x +1,0, otherwise.

⎧⎨⎩

Since the joint density is constant where it is nonzero, the joint distribution is uniform, on the shaded region (parallelogram) illustrated in the figure below. The area of this parallelogram is one (and the joint density is one) so that all probabilities are just given by areas. The probability that the payment exceeds 0.5 is one minus the probability that Y < 0.5, and Pr Y < 0.5( ) is the area of the small triangle at the bottom of the parallelogram, i.e., 12⋅12⋅12=18. The probability sought is therefore 1− 1

8=78.

Answer D. Exercise 4.9. November 2001 Course 1 Examination, Problem No. 35, also Dr. Ostaszewski’s online exercise posted February 2, 2008 Auto claim amounts, in thousands, are modeled by a random variable with density function f x( ) = xe− x for x ≥ 0. The company expects to pay 100 claims if there is no deductible. How

many claims does the company expect to pay if the company decides to introduce a deductible of 1000? A. 26 B. 37 C. 50 D. 63 E. 74 Solution. If X is the random variable representing claim amounts (in thousands), the probability that X

x

y

1 12

2

1

y = x

y = x +1

y < 12

y > 12

SECTION 4


exceeds the deductible is Pr X > 1( ) = xe− xdx1

+∞

∫ . Recall this Darth Vader Rule version

E max X,a( )( ) = a + sX x( )dxa

+∞

∫ .

By applying this rule to an exponential random variable with hazard rate 1, denoted by Y, we get

Pr X > 1( ) = xe− x dx1

+∞

∫ = 1 ⋅ e− y dy0

1

∫ + y ⋅ e− y dy1

+∞

∫⎛

⎝⎜⎞

⎠⎟− 1 ⋅ e− y dy

0

1

∫ =

= E max Y ,1( )( ) − e− y dy0

1

∫ = E max Y ,1( )( ) − FY 1( ) =

= 1+ sY y( )dy1

+∞

∫⎛

⎝⎜⎞

⎠⎟− 1− e−1( ) = e− y dy

1

+∞

∫ + e−1 = sY 1( ) + e−1 = 2e−1 ≈ 0.7358.

The company expected 100 claims when the policy had no deductible. We now see that if a deductible of 1000 is imposed, among the losses that occur there is only 0.7358 probability that such losses exceed the deductible. It follows that the company now, with the deductible imposed, expects 100 ⋅0.7358 ≈ 74 claims. Answer E. Exercise 4.10. November 2001 Course 1 Examination, Problem No. 38, also P Sample Exam Questions, Problem No. 103, and Dr. Ostaszewski’s online exercise posted February 9, 2008 In a small metropolitan area, annual losses due to storm, fire, and theft are assumed to be independent, exponentially distributed random variables with respective means 1.0, 1.5, and 2.4. Determine the probability that the maximum of these losses exceeds 3. A. 0.002 B. 0.050 C. 0.159 D. 0.287 E. 0.414 Solution. Let X1, X2 , and X3 denote annual loss due to storm, fire, and theft, respectively. Let Y = max X1,X2 ,X3( ). Based on the information about the distributions of X1, X2 , and X3 we have:

Pr Y > 3( ) = 1− Pr Y ≤ 3( ) = 1− Pr max X1,X2 ,X3( ) ≤ 3( ) = = 1− Pr X1 ≤ 3{ }∩ X2 ≤ 3{ }∩ X3 ≤ 3{ }( ) = 1− Pr X1 ≤ 3( ) ⋅Pr X2 ≤ 3( ) ⋅Pr X3 ≤ 3( ) =

= 1− 1− e−

31

⎛⎝⎜

⎞⎠⎟⋅ 1− e

−3

1.5⎛⎝⎜

⎞⎠⎟⋅ 1− e

−3

2.4⎛⎝⎜

⎞⎠⎟≈ 0.414.

Answer E.

04-ko-asm-pmanual-section4

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