Exam P Practice Problem 10

Problem 10a
An individual is facing an outcome of an annual financial loss X (in tens of thousands of dollars) whose probability density function is given by

\displaystyle f(x)=0.003 x^2, \ \ \ \ 0<x<10

The probability of a loss in the next year is 0.08. If there is a loss, there is only one loss in any given year. An insurance policy is available to protect against the financial loss by paying in full when a loss occurs.

  1. What is the probability that the insurer’s payment to the insured will exceed $50,000?
  2. What is the mean payment made by the insurer to the insured?
  3. What is the variance of the amount of payment made by the insurer?

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Problem 10b
Suppose that instead of buying a policy that pays the loss in full, the individual buys a policy that has a 80/20 coinsurance provision, i.e., the insurance company pays 80% of the loss and the insured retains the remaining 20% of a loss. Answer the same three questions.

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Solution is found below.

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Solution to Problem 10a
\displaystyle 10a.1 \ \ \ \ \ 0.07

\displaystyle 10a.2 \ \ \ \ \ 0.6

\displaystyle 10a.3 \ \ \ \ \ 4.44

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Let X be the loss variable as described in the problem. Then the following is the probability P(X>t).

\displaystyle \begin{aligned}P(X>t)&=\int_t^{10} 0.003 x^2 \ dx \\&=1-0.001 t^3   \end{aligned}

One important thing to keep in mind is that the occurrence of a financial loss is not certain. So the answer to question #1 is not P(X>5). Let Y be the insurance payment to the insured. Note that Y is conditional on the occurrence of a loss. If the loss does not happen, Y=0. If the loss does happen, Y=X. Likewise, P(Y>t)=0 in case of no loss and P(Y>t)=P(X>t) in case of a loss. So we can use the law of total probability to obtain P(Y>5).

\displaystyle \begin{aligned}P(Y>5)&=0 \times 0.92+P(X>5) \times 0.08  \\&=(1-0.001 5^3) \times 0.08 \\&=0.07   \end{aligned}

The answers to the other two questions can also be obtained by using the law of total probability.

\displaystyle \begin{aligned}E(Y)&=0 \times 0.92+\int_0^{10} x 0.003 x^2 \ dx \times 0.08  \\&=\int_0^{10} 0.003 x^3 \ dx \times 0.08 \\&=7.5 \times 0.08 \\&=0.6 \\&=\$6000   \end{aligned}

\displaystyle \begin{aligned}E(Y^2)&=0 \times 0.92+\int_0^{10} x^2 0.003 x^2 \ dx \times 0.08  \\&=\int_0^{10} 0.003 x^4 \ dx \times 0.08 \\&=60 \times 0.08 \\&=4.8   \end{aligned}

\displaystyle Var(Y)=4.8-0.6^2=4.44

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Answers to Problem 10b
\displaystyle 10b.1 \ \ \ \ \ 0.06046875

\displaystyle 10b.2 \ \ \ \ \ 0.48

\displaystyle 10b.3 \ \ \ \ \ 2.8416


2 responses

  1. The answer for 10b #1 is wrong. The probability will be less than the .07, which was the answer for part a. I got .06046.

    1. Katy Jean, the mistake has been corrected. Thanks for pointing it out.

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