An actuary models the lifetime of a device using the random variable [math]Y = 10X^{0.8}[/math], where [math]X[/math] is an exponential random variable with mean 1. Let [math]f(y)[/math] be the density function for [math]Y[/math].

Determine [math]f(y)[/math] for [math]y\gt0[/math].

• [math]10 y^{0.8} \exp(−8 y^{−0.2} )[/math]
• [math]8 y^{−0.2} \exp(−10 y^{0.8} )[/math]
• [math]8 y^{−0.2} \exp[−(0.1y)^{1.25} ][/math]
• [math](0.1y )^{1.25} \exp[−0.125(0.1y)^{0.25} ][/math]
• [math]0.125(0.1y )^{0.25} \exp[−(0.1y )^{1.25} ][/math]

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

Solution: E

[[math]] F ( y ) = \operatorname{P}[Y ≤ y ] = \operatorname{P}[10 X 0.8 ≤ y] = \operatorname{P}[ X \leq (Y/10)^{10/8}] = 1-\exp(-(Y/10)^{10/8}). [[/math]]

Therefore,

[[math]] f(y) = F^{'}(y) = \frac{1}{8} \left ( \frac{Y}{10}\right )^{1/4} \exp(-(Y/10)^{5/4}). [[/math]]

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