exercise:A6395b007c

Jan 18'24

Exercise

You are given:

(i) [math]\quad Z_{1}[/math] is the present value random variable for an [math]n[/math]-year term insurance of 1000 issued to [math](x)[/math]

(ii) [math]\quad Z_{2}[/math] is the present value random variable for an [math]n[/math]-year endowment insurance of 1000 issued to [math](x)[/math]

(iii) For both [math]Z_{1}[/math] and [math]Z_{2}[/math] the death benefit is payable at the end of the year of death

(iv) [math]\quad E\left[Z_{1}\right]=528[/math]

(v) [math]\operatorname{Var}\left(Z_{2}\right)=15,000[/math]

(vi) [math]\quad A_{x: n} \frac{1}{}=0.209[/math]

(vii) [math]{ }^{2} A_{x: n} \frac{1}{n}=0.136[/math]

Calculate [math]\operatorname{Var}\left(Z_{1}\right)[/math].

143,400
177,500
211,200
245,300
279,300

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AAdmin

Jan 18'24

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].