For a special increasing whole life insurance on (40), payable at the moment of death, you are given:

(i) The death benefit at time [math]t[/math] is [math]b_{t}=1+0.2 t, \quad t \geq 0[/math]

(ii) The interest discount factor at time [math]t[/math] is [math]v(t)=(1+0.2 t)^{-2}, \quad t \geq 0[/math]

(iii) [math]\quad{ }_{t} p_{40} \mu_{40+t}= \begin{cases}0.025, & 0 \leq t\lt40 \\ 0, & \text { otherwise }\end{cases}[/math]

(iv) [math]Z[/math] is the present value random variable for this insurance

Calculate [math]\operatorname{Var}(Z)[/math].

• 0.036
• 0.038
• 0.040
• 0.042
• 0.044

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

Answer: A

[[math]] \begin{aligned} \operatorname{Var}(Z) & =E\left(Z^{2}\right)-E(Z)^{2} \\ E(Z) & =E\left[(1+0.2 T)(1+0.2 T)^{-2}\right]=E\left[(1+0.2 T)^{-1}\right] \\ & =\int_{0}^{40} \frac{1}{(1+0.2 t)} f_{T}(t) d t=\frac{1}{40} \int_{0}^{40} \frac{1}{1+0.2 t} d t \\ & =\left.\frac{1}{40} \frac{1}{0.2} \ln (1+0.2 t)\right|_{0} ^{40}=\frac{1}{8} \ln (9)=0.27465 \\ E\left(Z^{2}\right) & =E\left\{(1+0.2 T)^{2}\left[(1+0.2 T)^{-2}\right]^{2}\right\}=E\left[(1+0.2 T)^{-2}\right] \\ & =\int_{0}^{40} \frac{1}{(1+0.2 t)^{2}} f_{T}(t)=\frac{1}{40} \frac{1}{0.2}\left[\frac{-1}{(1+0.2 t)}\right]_{0}^{40} \\ & =\frac{1}{8}\left(1-\frac{1}{9}\right)=\frac{1}{9}=0.11111 \end{aligned} [[/math]]

[math]\operatorname{Var}(Z)=0.11111-(0.27465)^{2}=0.03568[/math]

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