Exam P Practice Problem 9

Problem 9a
Suppose that when a policyholder incurs a loss, the size of the loss (in thousands of dollars) is Y=X^{-1} where the probability density function of X is f(x)=2.5 x^{1.5} where 0<x<1. Find the variance of the size of loss.

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Problem 9b
Suppose that when a policyholder incurs a loss, the size of the loss (in thousands of dollars) is Y=X^{-1} where the probability density function of X is f(x)=3 x^{2} where 0<x<1. Suppose that the insurance company just receives a notification that the policyholder had incurred a loss over $2000, what is the probability that the loss exceeds $3000?

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Solution is found below.

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Solution to Problem 9a

More Direct Solution

    \displaystyle E(Y)=E(X^{-1})=\int_0^1 x^{-1} \ 2.5 x^{1.5} \ dx =\frac{5}{3}

    \displaystyle E(Y^2)=E(X^{-2})=\int_0^1 x^{-2} \ 2.5 x^{1.5} \ dx =5

    \displaystyle Var(Y)=5- \biggl(\frac{5}{3}\biggr)^2=\frac{20}{9}

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The original solution

Note that the support of Y is y>1 (the values with positive probabilities). When 0<y^{-1}<1, we have y>1. First derive the cdf of X and then derive the cdf of Y=X^{-1}. Upon differentiation, the density function of Y is obtained, which is then used to compute the variance. The following shows the derivations.

\displaystyle F_X(x)=\left\{\begin{matrix}0& \ \ \ \ \ \ x \le 0 \\{\text{ }}& \\{x^{2.5}}&\ \ \ \ \ \ 0 < x <1 \\{\text{ }}& \\{1}&\ \ \ \ \ \ x \ge 1  \end{matrix}\right.

\displaystyle \begin{aligned}F_Y(y)&=P(Y \le y) \\&=P(X^{-1} \le y) \\&=P(X>y^{-1}) \\&=1-F_X(y^{-1}) \\&=1-y^{-2.5},y>1   \end{aligned}

\displaystyle f_Y(y)=2.5 y^{-3.5}.

\displaystyle E(Y)=\int \limits_{1} ^\infty 2.5 y^{-2.5} \ dy=\frac{5}{3}

\displaystyle E(Y^2)=\int \limits_{1} ^\infty 2.5 y^{-1.5} \ dy=5

\displaystyle Var(Y)=5-\frac{25}{9}=\frac{20}{9}

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Answer to Problem 9b
\displaystyle \frac{8}{27}=0.2963

Hint. It is not necessary to derive the distribution of Y before computing the probabilities concerning Y. For example, P(Y > 2)=P(X^{-1} > 2)=P(X < 2^{-1}).

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4 responses

  1. Attn Readers: It might be a helpful hint to know that the inverse is in the form of a familiar distribution. This is not specified in the solutions, but simplifies the problem immensely!

    1. Katy Jean. Thanks for all your comments. There is a more direct way to work this problem. Please the direct solution and the hint I added.

      1. I was computing the inverse, seeing it as the Gamma, and then using the CDF and variance formulas for Gamma (which I have memorized), which seemed like a shortcut. Your direct solution might be even easier. Personally, I avoid actually evaluating integrals (even easy ones) under the pressure of a test– it’s like doing long division with my 4th grade teacher watching over my shoulder– too many chances for silly mistakes.

      2. I sympathize with your desire to avoid evaluating integrals in a test. The good thing is that Exam P problems likely do not have monster integrals. If an integral in exam P turns out to be sucking up a lot of exam time, you probably do not look at the problem in a right way. So one job in exam P preparation is to learn various thought processes that can take you to the heart of problems. The way I see it, some integrals are necessary in Exam P problems. For example, integrals involving powers of x (such as \int_a^b x^t \ dx). Other examples are integrals involving the density functions of exponential distributions. Another example is the kind of integrals where you can algebraically manipulate to look like the density of a Gamma distribution.

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