For a fully discrete whole life insurance of 1 on [math](x)[/math], you are given:

(i) [math]\quad q_{x+10}=0.02067[/math]

(ii) [math]v^{2}=0.90703[/math]

(iii) [math]A_{x+11}=0.52536[/math]

(iv) [math]{ }^{2} A_{x+11}=0.30783[/math]

(v) [math]{ }_{k} L[/math] is the prospective loss random variable at time [math]k[/math]

Calculate [math]\frac{\operatorname{Var}\left({ }_{10} L\right)}{\operatorname{Var}\left({ }_{11} L\right)}[/math].

• 1.006
• 1.010
• 1.014
• 1.018
• 1.022

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

Answer: D

[math]\frac{V\left[{ }_{10} L\right]}{V\left[{ }_{11} L\right]}=\frac{\left(1+\frac{p}{d}\right)^{2}\left({ }^{2} A_{x+10}-A_{x+10}^{2}\right)}{\left(1+\frac{p}{d}\right)^{2}\left({ }^{2} A_{x+11}-A_{x+11}^{2}\right)}[/math]

[math]A_{x+10}=v q_{x+10}+v p_{x+10} A_{x+11}[/math]

[math]=(0.90703)^{1 / 2}(0.02067)+(0.90703)^{1 / 2}(1-0.02067)(0.52536)=0.50969[/math]

[math]{ }^{2} A_{x+10}=v^{2} q_{x+10}+v^{2} p_{x+10}{ }^{2} A_{x+11}[/math] [math]=(0.90703)(0.02067)+(0.90703)(1-0.02067)(0.30783)=0.29219[/math]

[math]\Rightarrow \frac{\operatorname{Var}\left({ }_{k} L\right)}{\operatorname{Var}\left({ }_{k+1} L\right)}=\frac{(0.29219)-(0.50969)^{2}}{(0.30783)-(0.52536)^{2}}=\frac{0.03241}{0.03183}=1.018[/math]

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