Let [math]X[/math] be a random variable with density function

[[math]] f(x) = \begin{cases} 2e^{-2x}, \, x \gt0 \\ 0, \, \textrm{otherwise} \end{cases} [[/math]]

Calculate [math]\operatorname{P}[ X \leq 0.5 | X \leq 1.0].[/math]

• 0.433
• 0.547
• 0.632
• 0.731
• 0.865

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*} \operatorname{E}(T | T \lt 10) = \int_0^{10}tg(t) dt &= \int_0^{10}t \frac{0.2e^{-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 ] \\ &= 0.2323 \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.