For a special fully continuous whole life insurance on [math](x)[/math], you are given:

i) [math] \mu_{x+t}=0.03, t \geq 0[/math]

ii) [math]\delta=0.06[/math]

iii) The death benefit at time [math]t[/math] is [math]b_{t}=e^{0.05 t}, t \geq 0[/math]

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

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

• 0.300
• 0.325
• 0.350
• 0.375
• 0.400

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

Answer: D

[[math]] \begin{aligned} E(Z) & =\int_{0}^{\infty}{ }_{t} p_{x} \times \mu \times e^{0.05 \times t} e^{-\delta \times t} d t \\ & =\int_{0}^{\infty} e^{-0.03 \times t} \times 0.03 \times e^{0.05 \times t} e^{-0.06 \times t} d t \\ & =\frac{0.03}{0.04} \times\left. e^{-0.04 t}\right|_{0} ^{\infty}=0.75 \\ E\left(Z^{2}\right) & =\int_{0}^{\infty}{ }_{t} p_{x} \times \mu \times e^{0.05 \times 2 \times t} e^{-\delta \times 2 \times t} d t \\ & =\int_{0}^{\infty} e^{-0.03 \times t} \times 0.03 \times e^{0.05 \times 2 \times t} e^{-0.06 \times 2 \times t} d t \\ & =\frac{0.03}{0.05} \times\left. e^{-0.05 t}\right|_{0} ^{\infty}=0.6 \end{aligned} [[/math]]

[math]\operatorname{Var}(Z)=E\left(Z^{2}\right)-[E(Z)]^{2}=0.6-0.75^{2}=0.375[/math]

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