Exercise


Jan 18'24

Answer

Answer: A

[[math]] \begin{aligned} & \operatorname{Var}\left(Z_{2}\right)=(1000)^{2}\left[{ }^{2} A_{x: n}-\left(A_{x: n}\right)^{2}\right]=15,000 \\ & =(1000)^{2}\left({ }^{2} A_{x: n}^{1}+{ }^{2} A_{x: n}^{1}\right)-(1000)^{2}\left[A_{x: n}^{1}+A_{x n \bar{n}}^{1}\right]^{2} \\ & =(1000)^{2}{ }^{2} A_{x: n}^{1}+(1000)^{2}{ }^{2} A_{x: n}^{1}-(1000)^{2}\left(A_{x: n}^{1}\right)^{2}-(1000)^{2}\left(A_{x n \eta}^{1}\right)^{2} \\ & -2(1000)^{2}\left(A_{x: n}^{1}\right)\left(A_{x n n}^{1}\right) \\ & =(1000)^{2}\left[{ }^{2} A_{x: n}^{1}-\left(A_{x: n}^{1}\right)^{2}\right]+\left(1000^{22} A_{x: n}^{1}\right)-\left(1000 A_{x: n}^{1}\right)^{2} \\ & -\left(1000 A_{x: n}^{1}\right)^{2}-(2)\left(1000 A_{x: n}^{1}\right)\left(1000 A_{x n n}^{1}\right) \\ & =V\left(Z_{1}\right)+(1000)\left(1000{ }^{2} A_{x: n}^{1}\right)-\left(1000 A_{x \frac{1}{n}}^{1}\right)^{2}-\left(1000 A_{x: n}^{1}\right)^{2} \\ & 15,000=\operatorname{Var}\left(Z_{1}\right)+(1000)(136)-(209)^{2}-2(528)(209) \end{aligned} [[/math]]


Therefore, [math]\operatorname{Var}\left(Z_{1}\right)=15,000-136,000+43,681+220,704=143,385[/math].

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

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