Answer to Question #236628 in Statistics and Probability for john k

Question

 A manufacturer’s annual losses follow a distribution with density function

f(x) = ( 2.5(0.6)^2.5 /x^3.5 , : x > 0.6

( 0, : otherwise.

To lower its losses, the manufacturer purchases an insurance policy with an annual deductible of 2. What is the mean of the manufacturer’s annual losses not paid by the insurance policy?

Answer 1

Let us represent the manufacturers annual losses by a random variable X. Then it’s pdf f(x) is given as:

$f(x)=\left\{\begin{matrix} \frac{2.5(0.6)^{2.5}}{x^{3.5}}\;for \;x>0.6 & \\ 0\; otherwise & \end{matrix}\right.$

Let us represent losses not paid by a new random variable Y, then we can write Y in terms of X as:

$Y=\left\{\begin{matrix} x \; for \;0.6 < x< 2 & \\ 2 \; for\; x \geqslant 2 & \end{matrix}\right.$

Therefore, the mean of the manufacturer’s annual losses not paid will be equal to expected (or mean) value of the random variable Y.

Now let us recall following proposition.

If X is a continuous random variable with probability density function f(x) on interval [a,b] and h(x) be a function of x, then the expected value of h(x) is

$E[h(x)] = \int^b_a h(x) \cdot f(x) dx$

In above proposition, the random variable Y is equivalent of h(x), hence the expected value of Y can be given as:

$E(Y) = \int^2_{0.6}x \cdot f(x) dx + \int^{\infty}_2 2 f(x)dx \\ = \int^2_{0.6} x \times \frac{2.5(0.6)^2.5}{x^{3.5}}dx + \int^{\infty }_2 \frac{2 \times 2.5(0.6)^{2.5}}{x^{3.5}}dx \\ = \int^2_{0.6} \frac{0.6971}{x^{2.5}}dx + \int^{\infty }_2 \frac{1.3943}{x^{3.5}}dx \\ = 0.6971 \int^2_{0.6} x^{-2.5} dx + 1.3943 \int^{\infty }_2 x^{-.35} dx \\ = 0.6971 [\frac{x^{-1.5}}{-1.5}]^2_{0.6} + 1.3943 [\frac{x^{-2.5}}{-2.5}]^{\infty }_2 \\ = -0.4647 [x^{-1.5}]^2_{0.6} + (-0.5577)[x^{-2.5}]^{\infty }_2 \\ = -0.4647 [2^{-1.5} -0.6^{-1.5}] -0.5577 [0-2^{-2.5}] \\ = -0.4647 \times (-1.7981) -0.5577 \times (-0.1768) \\ = 0.93$