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...")
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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]]

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

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