A device that continuously measures and records seismic activity is placed in a remote region. The time, [math]T[/math], to failure of this device is exponentially distributed with mean 3 years. Since the device will not be monitored during its first two years of service, the time to discovery of its failure is [math]X = \max(T, 2)[/math].

Calculate [math]\operatorname{E}(X)[/math].

• [math]2 + \frac{e^{-6}}{3}[/math]
• [math]2-2e^{-2/3} + 5e^{-4/3}[/math]
• 3
• [math]2+3e^{-2/3}[/math]
• 5

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

Solution: D

The density function of [math]T[/math] is

[[math]] \begin{align*} \operatorname{E}[X] &= \operatorname{E}[\max{T,2}] = \int_0^2 \frac{2}{3}e^{-t/3} dt + \int_{2}^{\infty}\frac{t}{3} e^{-t/3} dt \\ &= -2e^{-t/3} \Big |_0^2 -te^{-t/3} \Big |_2^{\infty} + \int_2^{\infty} e^{-t/3} dt \\ &= 2e^{-2/3} + 2 2e^{-2/3} -3e^{-t/3} \Big |_2^{\infty} \\ &= 2 + 3e^{-2/3}. \end{align*} [[/math]]

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