The distribution of values of the retirement package offered by a company to new employees is modeled by the probability density function

[[math]] f(x) = \begin{cases} \frac{1}{5} e^{-\frac{(x-5)}{5}}, \, x \gt5 \\ 0, \, \textrm{otherwise} \end{cases} [[/math]]

Calculate the variance of the retirement package value for a new employee, given that the value is at least 10.

• 15
• 20
• 25
• 30
• 35

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

Solution: C

The conditional variance is

[[math]] \begin{align*} \operatorname{Var}( X | X \geq 10) &= \operatorname{E} ( X^2 | X ≥ 10) − \operatorname{E} ( X | X ≥ 10)^2 \\ &= \frac{\int_{10}^{\infty} x^2(0.2)e^{-0.2(x-5)} dx}{\int_{10}^{\infty} 0.2 e^{-0.2(x-5)} dx} - \left [\frac{\int_{10}^{\infty} x(0.2)e^{-0.2(x-5)} dx}{\int_{10}^{\infty} 0.2 e^{-0.2(x-5)} dx} \right ]^2 \end{align*} [[/math]]

Performing integration (using integration by parts) produces the answer of 25. An alternative solution is to first determine the density function for the conditional distribution. It is

[[math]] f(y) = \frac{0.2e^{-0.2(y-5)}}{\int_{10}^{\infty} 0.2 e^{-0.2(x-5)} dx} = \frac{0.2e^{-0.2(y-5)}}{-e^{-0.2(x-5)} \Big |_{10}^{\infty}} = \frac{0.2e^{-0.2(y-5)}}{e^{-0.2(5)}} = 0.2 e^{-0.2(y-10)}, y \gt 10. [[/math]]

Then note that [math]Y – 10 [/math] has an exponential distraction with mean 5. Subtracting a constant does not change the variance, so the variance of [math]Y[/math] is also 25.

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