Revision as of 20:45, 19 January 2024 by Admin (Created page with "'''Answer: E''' 1,020 in the solution is the 1,000 death benefit plus the 20 death benefit claim expense. <math display="block"> A_{x}=1-d \ddot{a}_{x}=1-d(12.0)=0.320755 </math> <math>G \ddot{a}_{x}=1,020 A_{x}+0.65 G+0.10 G \ddot{a}_{x}+8+2 \ddot{a}_{x}</math> <math>G=\frac{1,020 A_{x}+8+2 \ddot{a}_{x}}{\ddot{a}_{x}-0.65-0.10 \ddot{a}_{x}}=\frac{1,020(0.320755)+8+2(12.0)}{12.0-0.65-0.10(12.0)}=35.38622</math> Let <math>Z=v^{K_{x}+1}</math> denote the present va...")
Exercise
ABy Admin
Jan 19'24
Answer
Answer: E
1,020 in the solution is the 1,000 death benefit plus the 20 death benefit claim expense.
[[math]]
A_{x}=1-d \ddot{a}_{x}=1-d(12.0)=0.320755
[[/math]]
[math]G \ddot{a}_{x}=1,020 A_{x}+0.65 G+0.10 G \ddot{a}_{x}+8+2 \ddot{a}_{x}[/math]
[math]G=\frac{1,020 A_{x}+8+2 \ddot{a}_{x}}{\ddot{a}_{x}-0.65-0.10 \ddot{a}_{x}}=\frac{1,020(0.320755)+8+2(12.0)}{12.0-0.65-0.10(12.0)}=35.38622[/math]
Let [math]Z=v^{K_{x}+1}[/math] denote the present value random variable for a whole life insurance of 1 on [math](x)[/math].
Let [math]Y=\ddot{a}_{\overline{K_{x}+1}}[/math] denote the present value random variable for a life annuity-due of 1 on [math](x)[/math].
[[math]]
\begin{aligned}
L & =1,020 Z+0.65 G+0.10 G Y+8+2 Y-G Y \\
& =1,020 Z+(2-0.9 G) Y+0.65 G+8 \\
& =1,020 v^{K_{x}+1}+(2-0.9 G) \frac{1-v^{K_{x}+1}}{d}+0.65 G+8 \\
& =\left(1,020+\frac{0.9 G-2}{d}\right) v^{K_{x}+1}+\frac{2-0.9 G}{d}+0.65 G+8
\end{aligned}
[[/math]]
[math]\operatorname{Var}(L)=\left[{ }^{2} A_{x}-\left(A_{x}\right)^{2}\right]\left(1,020+\frac{0.9 G-2}{d}\right)^{2}[/math]
[[math]]
\begin{aligned}
& =\left(0.14-0.320755^{2}\right)\left(1,020+\frac{0.9(35.38622)-2}{d}\right)^{2} \\
& =0.037116(2,394,161) \\
& =88,861
\end{aligned}
[[/math]]