Exercise
Answer
Answer: A
On a unit basis, [math]\operatorname{Var}\left(L_{0}\right)=\left(1+\frac{P}{d}\right)^{2}\left[{ }^{2} A_{45}-\left(A_{45}\right)^{2}\right]=\left(1+\frac{A_{45}}{d \ddot{a}_{45}}\right)^{2}\left[{ }^{2} A_{45}-\left(A_{45}\right)^{2}\right][/math]
[math]=\left(\frac{d \ddot{a}_{45}+1-d \ddot{a}_{45}}{d \ddot{a}_{45}}\right)^{2}\left[{ }^{2} A_{45}-\left(A_{45}\right)^{2}\right]=\frac{{ }^{2} A_{45}-\left(A_{45}\right)^{2}}{(d \ddot{a})^{2}}[/math]
[math]=\frac{0.03463-0.15161^{2}}{\left(\frac{0.05}{1.05} \times 17.8162\right)^{2}}=0.016178038[/math]
The standard deviation of [math]L_{0}=0.127193[/math]
[math](200,000)\left(\right.[/math] The standard deviation of [math]\left.L_{0}\right)=25,439[/math]
Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.