Let [math]T_1[/math] and [math]T_2[/math] represent the lifetimes in hours of two linked components in an electronic device. The joint density function for [math]T_1[/math] and [math]T_2[/math] is uniform over the region defined by

[[math]] 0 \leq t_1 \leq t_2 \leq L [[/math]]

where [math]L[/math] is a positive constant. Determine the expected value of the sum of the squares of [math]T_1[/math] and [math]T_2[/math].

• [math]\frac{L^2}{3}[/math]
• [math]\frac{L^2}{2}[/math]
• [math]\frac{2L^2}{3}[/math]
• [math]\frac{3L^2}{4}[/math]
• [math]L^2[/math]

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

Solution: C

We are given

[[math]] f(t_1,t_2) = 2/L^2, \, 0 \leq t_1 \leq t_2 \leq L. [[/math]]

Therefore,

[[math]] \begin{align*} \operatorname{E}[T_1^2 + T_2^2] &= \int_0^L \int_0^{t_2} (t_1^2 + t_2^2) \frac{2}{L^2} dt_t dt_2 \\ &= \frac{2}{L^2} \left \{ \int_0^{L} \left [ \frac{t_1^3}{3} + t_2^2t_1\right ]_0^{t_2} dt_1\right \} \\ &= \frac{2}{L^2} \left \{ \int_0^{L} \left (\frac{t_2^3}{3} + t_2^3 \right ) \, dt_2 \right \} \\ &= \frac{2}{L^2} \int_0^L \frac{4}{3} t_2^3 \, dt_2 \\ &= \frac{2}{L^2} \left [ \frac{t_2^4}{3}\right ]_0^L \\ &= \frac{2}{3}L^2. \end{align*} [[/math]]

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