For a group of 100 lives age [math]x[/math] with independent future lifetimes, you are given:

(i) Each life is to be paid 1 at the beginning of each year, if alive

(ii) [math]\quad A_{x}=0.45[/math]

(iii) [math]{ }^{2} A_{x}=0.22[/math]

(iv) [math]\quad i=0.05[/math]

(v) [math]\quad Y[/math] is the present value random variable of the aggregate payments.

Using the normal approximation to [math]Y[/math], calculate the initial size of the fund needed in order to be [math]95 \%[/math] certain of being able to make the payments for these life annuities.

• 1170
• 1180
• 1190
• 1200
• 1210

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

Answer: D

Let [math]Y_{i}[/math] be the present value random variable of the payment to life [math]i[/math].

[math]E\left[Y_{i}\right]=\ddot{a}_{x}=\frac{1-A_{x}}{d}=11.55 \quad \operatorname{Var}\left[Y_{i}\right]=\frac{{ }^{2} A_{x}-\left(A_{x}\right)^{2}}{d^{2}}=\frac{0.22-0.45^{2}}{(0.05 / 1.05)^{2}}=7.7175[/math]

Then [math]Y=\sum_{i=1}^{100} Y_{i}[/math] is the present value of the aggregate payments.

[math]E[Y]=100 E\left[Y_{i}\right]=1155[/math] and [math]\operatorname{Var}[Y]=100 \operatorname{Var}\left[Y_{i}\right]=771.75[/math]

[math]\operatorname{Pr}[Y \leq F]=\operatorname{Pr}\left[Z \leq \frac{F-1155}{\sqrt{771.75}}\right]=0.95 \Rightarrow \frac{F-1155}{\sqrt{771.75}}=1.645[/math]

[math]\Rightarrow F=1155+1.645 \sqrt{771.75}=1200.699[/math]

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