The lifetime of a machine part is exponentially distributed with a mean of five years.

Calculate the mean lifetime of the part, given that it survives less than ten years.

• 0.865
• 1.157
• 2.568
• 2.970
• 3.435

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

Solution: E

The given mean of 5 years corresponds to the pdf [math]f(t) = 0.2 e^{-0.2t}[/math] and the cumulative distribution function [math]F(t) = 1-e^{-0.2t} [/math]. The conditional pdf is

[[math]] g(t) = \frac{f(t)}{F(10)} = \frac{0.2e^{-0.2t}}{1-e^{-2}}. 0 \lt t \lt 10. [[/math]]

The conditional mean is (using integration by parts)

[[math]] \begin{align*} E(T | T \lt 10) &= \int_0^{10}tg(t) dt = \int_0^{10} t \frac{0.2 e^{-0.2t}}{1-e^{-2}} dt = 0.2313 \int_0^{10} te^{-0.2t} dt \\ &= 0.2313 \left [ t(-5e^{-0.2t}) \big |_0^{10} - \int_0^{10} -5e^{-0.2t} dt \right ] \\ &= 0.2313 \left [-6.7668 + 0 - 25e^{-0.2t} \Big |_0^{10} \right ] \\ &= 0.2313[-6.7668 - 3.3834 + 25] \\ &= 3.435. \end{align*} [[/math]]

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