# 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

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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$\displaystyle 10b.1 \ \ \ \ \ 0.06046875$
$\displaystyle 10b.2 \ \ \ \ \ 0.48$
$\displaystyle 10b.3 \ \ \ \ \ 2.8416$