For a two-year term insurance of 1 on (x) payable at the moment of death, you are given:

i) [math]q_{x}=0.04[/math]

ii) [math]q_{x+1}=0.06[/math]

iii) Deaths are uniformly distributed over each year of age

iv) [math]i=0.04[/math] v) [math]Z[/math] is the present value random variable for this insurance

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

• 0.065
• 0.069
• 0.073
• 0.077
• 0.081

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

Answer: E

[math]E(Z)=\frac{i}{\delta}\left[v q_{x}+v^{2} p_{x} q_{x+1}\right]=\frac{0.04}{\ln (1.04)}\left[\frac{1}{1.04}(0.04)+\frac{1}{1.04^{2}}(0.96)(0.06)\right]=0.09353831[/math]

[math]E\left(Z^{2}\right)=\frac{(1+i)^{2}-1}{2 \delta}\left[v^{2} q_{x}+v^{4} p_{x} q_{x+1}\right]=\frac{\left(1.04^{2}-1\right)}{2 \ln (1.04)}\left[\frac{1}{1.04^{2}}(0.04)+\frac{1}{1.04^{4}}(0.96)(0.06)\right]=0.08969072[/math]

[math]\operatorname{Var}(Z)=E\left(Z^{2}\right)-[E(Z)]^{2}=0.08969072-(0.09353831)^{2}=0.0809413[/math]

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