A device containing two key components fails when, and only when, both components fail. The lifetimes, [math]T_1 [/math] and [math]T_2[/math] of these components are independent with common density function

[[math]] f(t) = \begin{cases} e^{-t}, \, 0\lt t \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

The cost, [math]X[/math], of operating the device until failure is [math]2T_1 + T_2[/math] . Let [math]g[/math] be the density function for [math]X[/math].

Determine [math]g(x)[/math], for [math]x \gt 0 [/math].

• [math]e^{−x/2} − e^{−x}[/math]
• [math]2(e^{-x/2} - e^{-x} )[/math]
• [math]\frac{x^2e^{-x}}{2}[/math]
• [math]\frac{e^{-x/2}}{2}[/math]
• [math]\frac{e^{-x/3}}{3}[/math]

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

Solution: A

The joint density of [math]T_1[/math] and [math]T_2[/math] is given by

[[math]] f(t_1,t_2) = e^{-t_1}e^{-t_2}, \, t_1 \gt 0, \, t_2 \gt 0 [[/math]]

Therefore

[[math]] \begin{align*} \operatorname{P}[X \leq x ] &= \operatorname{P}[2T_1 + T_2 \leq x] \\ &= \int_0^x \int_0^{\frac{1}{2}(x-t_2)} e^{-t_1}e^{-t_2} dt_1 dt_2 \\ &= \int_0^x e^{-t_2} \left [ -e^{t_1} \Big |_0^{\frac{1}{2}(x-t_2)} \right ] dt_2 \\ &= \int_0^x e^{-t_2} \left [ 1-e^{-\frac{1}{2}x + \frac{1}{2}t_2} \right ] dt_2 \\ &= \int_0^{x} \left (e^{-t_2} - e^{-\frac{1}{2}x} e^{-\frac{1}{2}t_2} \right) dt_2 \\ &= \left [ -e^{-t_2} + 2e^{-\frac{1}{2}x}e^{-\frac{1}{2}t_2}\right ] \Big |_0^x \\ &= -e^{-x} + 2e^{-\frac{1}{2}x} + 1 - 2e^{-\frac{1}{2}x}, \, x \gt 0 \end{align*} [[/math]]

It follows that the density of [math]X[/math] is given by

[[math]] g(x) = \frac{d}{dx} [ 1-2e^{-\frac{1}{2}x} + e^{-x} ] = e^{-\frac{1}{2}x} -e^{-x}, x \gt 0. [[/math]]

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