Losses, [math]X[/math], under an insurance policy are exponentially distributed with mean 10. For each loss, the claim payment [math]Y[/math] is equal to the amount of the loss in excess of a deductible [math]d \gt 0 [/math].

Calculate [math]\operatorname{Var}(Y)[/math].

• [math]100-d[/math]
• [math](10 − d)^2[/math]
• [math]100e^{-d/10}[/math]
• [math]100(2e^{-d/100}-e^{-d/5})[/math]
• [math](10-d)^2(2e^{-d/100}-e^{-d/5})[/math]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: D

We have [math]Y = 0 [/math] when [math]X \lt d [/math] and [math]Y = X-d [/math] otherwise. Then, noting that the second moment of an exponential random variable is twice the square of the mean,

[[math]] \begin{align*} \operatorname{E}(Y) &= \int_0^d 0(0.01e^{-0.1x}) dx + \int_d^{\infty} (x-d)(0.1e^{-0.1x}) dx \\ &= 0 + \int_0^{\infty}x(0.1e^{-0.1(x+d)})dx = e^{-0.1d}(10) \end{align*} [[/math]]

[[math]] \begin{align*} \operatorname{E}(Y^2) &= \int_0^d 0^2(0.01e^{-0.1x}) dx + \int_d^{\infty} (x-d)^2(0.1e^{-0.1x}) dx \\ &= 0 + \int_0^{\infty}x^2(0.1e^{-0.1(x+d)})dx = e^{-0.1d}(200) \end{align*} [[/math]]

[[math]] \operatorname{Var}(Y) = e^{-0.1d}(200) - [e^{-0.1d}(10)]^2 = 100[2e^{-0.1d}-e^{-0.2d}]. [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.