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

(i) Death benefits are payable at the moment of death

(ii) The death benefit at time [math]t[/math] is [math]b_{t}=e^{0.02 t}[/math], for [math]t \geq 0[/math]

(iii) [math]\quad \mu_{x+t}=0.04[/math], for [math]t \geq 0[/math]

(iv) [math]\delta=0.06[/math]

(v) [math]\quad Z[/math] is the present value at issue random variable for this insurance

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

• 0.020
• 0.036
• 0.052
• 0.068
• 0.083

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

Answer: E

[math]E[Z]=\int_{0}^{\infty} b_{t} \cdot v^{t} \cdot{ }_{t} p_{x} \cdot \mu_{x+t} d t=\int_{0}^{\infty} e^{0.02 t} \cdot e^{-0.06 t} \cdot e^{-0.04 t} \cdot 0.04 d t[/math]

[math]=0.04 \int_{0}^{\infty} e^{-0.08 t} d t=\frac{0.04}{0.08}=\frac{1}{2}[/math]

[math]E\left[Z^{2}\right]=\int_{0}^{\infty}\left(b_{t} \cdot v^{t}\right)^{2}{ }_{t} p_{x} \cdot \mu_{x+t} d t=\int_{0}^{\infty}\left(e^{0.04 t}\right)\left(e^{-0.12 t}\right)\left(0.04 e^{-0.04}\right) d t=\frac{0.04}{0.12}=\frac{1}{3}[/math]

[math]\operatorname{Var}[Z]=\frac{1}{3}-\left(\frac{1}{2}\right)^{2}=\frac{1}{12}=0.0833[/math]

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