excans:Ac79905a29

Solution: C

Let [math]f(t_1 , t_2)[/math] denote the joint density function of [math]T_1[/math] and [math]T_2[/math]. The domain of [math]f[/math] is pictured below:

Now the area of this domain is given by

[[math]] A = 6^2 - \frac{1}{2}(6-4)^2 = 36-2 = 34. [[/math]]

Consequently,

[[math]] f(t_1,t_2) = \begin{cases} \frac{1}{34}, \quad 0 \lt t_1 \lt 6, 0 \lt t_2 \lt 6, t_1 + t_2 \lt 10 \\ 0, \, \textrm{elsewhere} \end{cases} [[/math]]

and

[[math]] \begin{align*} \operatorname{E}[T_1 + T_2 ] &= \operatorname{E}[T_1 ] + \operatorname{E}[T_2 ] = 2 \operatorname{E}[T_1 ] \quad \textrm{(due to symmetry)} \\ &= 2 \left \{ \int_0^4 t_1 \int_0^6 \frac{1}{34} dt_2 dt_1 + \int_4^6 t_1 \int_0^{10-t_1} \frac{1}{34} dt_2 dt_1\right \} \\ &= 2 \left \{ \int_0^4 t_1 \left [ \frac{t_2}{34} \Big |_0^6 \right ] dt_1 + \int_4^6 t_1 \left [ \frac{t_2}{34} \Big |_0^{10-t_1}\right ] dt_1 \right \} \\ &= 2 \left \{ \int_0^4 \frac{3t_1}{17} dt_1 + \int_4^6 \frac{1}{34} (10t_1 -t_1^2) dt_1 \right \}\\ &= 2 \left \{ \frac{3t_1^2}{34} \Big |_0^4 + \frac{1}{34}(5t_1^2 - \frac{1}{3}t_1^3) \Big |_4^6 \right \}\\ &= 2 \left \{ \frac{24}{17} + \frac{1}{34} \left [ 180-72-80 + \frac{64}{3}\right ]\right \} \\ &= 5.7 \end{align*} [[/math]]

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