A device contains two circuits. The second circuit is a backup for the first, so the second is used only when the first has failed. The device fails when and only when the second circuit fails. Let [math]X[/math] and [math]Y[/math] be the times at which the first and second circuits fail, respectively. [math]X[/math] and [math]Y[/math] have joint probability density function

[[math]] f(x,y) = \begin{cases} 6e^{-x}e^{-2y}, \,\, 0 \lt x \lt y \lt \infty \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Calculate the expected time at which the device fails.

• 0.33
• 0.50
• 0.67
• 0.83
• 1.50

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

Solution: D

The marginal distribution of [math]Y[/math] is given by

[[math]] \begin{align*} f_2(y) = \int_0^y 6e^{-x}e^{-2y} dx &= 6 e^{-2y} \int_0^y e^{-x} dx \\ &= -6e^{-2y}e^{-y} + 6e^{-2y}e^{-y} + 6e^{-2y} = 6e^{-2y}-6e^{-3y}, \, 0 \lt y \lt \infty. \end{align*} [[/math]]

Therefore,

[[math]] \begin{align*} \operatorname{E}(Y) = \int_0^{\infty}y f_2(y) dy = \int_0^{\infty}(6ye^{-2y} - 6ye^{-3y}) dy &= 6 \int_0^{\infty}ye^{-2y} dy - 6 \int_0^{\infty}ye^{-3y} dy \\ &= \frac{6}{2} \int_0^{\infty} 2y e^{-2y} dy - \frac{6}{3} \int_0^{\infty}3y e^{-3y} dy. \end{align*} [[/math]]

But [math]\int_0^{\infty}2ye^{-2y} dy[/math] and [math]\int_0^{\infty}3ye^{-3y} dy[/math] are equivalent to the means of exponential random variables with parameters 1/2 and 1/3, respectively. In other words, [math]\int_0^{\infty}2ye^{-2y} dy = 1/2[/math] and [math]\int_0^{\infty}3ye^{-3y} dy = 1/3[/math]. We conclude that [math]\operatorname{E}(Y)[/math] equals

(6/2) (1/2) – (6/3) (1/3) = 3/2 – 2/3 =9/6 − 4/6 = 5/6 = 0.83.

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