You are given that [math]T[/math], the time to first failure of an industrial robot, has a density [math]f(t)[/math] given by

[math]f(t)= \begin{cases}0.1, & 0 \leq t\lt2 \\ 0.4 t^{-2}, & 2 \leq t\lt10\end{cases}[/math]

with [math]f(t)[/math] undetermined on [math][10, \infty)[/math].

Consider a supplemental warranty on this robot that pays 100,000 at the time [math]T[/math] of its first failure if [math]2 \leq T \leq 10[/math], with no benefits payable otherwise.

You are also given that [math]\delta=5 \%[/math].

Calculate the [math]90^{\text {th }}[/math] percentile of the present value of the future benefits under this warranty.

• 82,000
• 84,000
• 87,000
• 91,000
• 95,000

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Answer: A

The present value random variable [math]\mathrm{PV}=1,000,000 e^{-0.05 T}, 2 \leq T \leq 10[/math] is a decreasing function of [math]T[/math] so that its [math]90^{\text {th }}[/math] percentile is

[[math]] \begin{aligned} & 1,000,000 e^{-0.05 p} \text { where } p \text { is the solution to } \int_{2}^{p} 0.4 t^{-2} d t=0.10 . \\ & \int_{2}^{p} 0.4 t^{-2} d t=-\left.0.4\left(\frac{t^{-1}}{-1}\right)\right|_{2} ^{p}=0.4\left(\frac{1}{2}-\frac{1}{p}\right)=0.10 \\ & p=4 \\ & 1,000,000 e^{-0.05 \times 4}=81,873.08 \end{aligned} [[/math]]

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.