For three fully discrete insurance products on the same [math](x)[/math], you are given:

(i) [math]\quad Z_{1}[/math] is the present value random variable for a 20 -year term insurance of 50

(ii) [math]\quad Z_{2}[/math] is the present value random variable for a 20 -year deferred whole life insurance of 100

(iii) [math]\quad Z_{3}[/math] is the present value random variable for a whole life insurance of 100 .

(iv) [math]E\left[Z_{1}\right]=1.65[/math] and [math]E\left[Z_{2}\right]=10.75[/math]

(v) [math]\operatorname{Var}\left(Z_{1}\right)=46.75[/math] and [math]\operatorname{Var}\left(Z_{2}\right)=50.78[/math]

Calculate [math]\operatorname{Var}\left(Z_{3}\right)[/math].

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

Answer: C

[math]Z_{3}=2 Z_{1}+Z_{2}[/math] so that [math]\operatorname{Var}\left(Z_{3}\right)=4 \operatorname{Var}\left(Z_{1}\right)+\operatorname{Var}\left(Z_{2}\right)+4 \operatorname{Cov}\left(Z_{1}, Z_{2}\right)[/math]

where [math]\operatorname{Cov}\left(Z_{1}, Z_{2}\right)=\underbrace{E\left[Z_{1} Z_{2}\right]}_{=0}-E\left[Z_{1}\right] E\left[Z_{2}\right]=-(1.65)(10.75)[/math]

[[math]] \begin{aligned} \operatorname{Var}\left(Z_{3}\right) & =4(46.75)+50.78-4(1.65)(10.75) \\ & =166.83 \end{aligned} [[/math]]

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