For a 3-year temporary life annuity due, you are given:

i) The life annuity pays 10 at the beginning of each year

ii) [math]v=0.93[/math]

iii) [math]\quad p_{x}=0.95, p_{x+1}=0.9, p_{x+2}=0.8[/math]

Calculate the standard deviation of the present value random variable for this annuity.

• 4.4
• 4.5
• 4.6
• 4.7
• 4.8

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

Answer: B

Discounted Payment	Probability
10	[math](1-0.95)=0.05[/math]
[math]10+10 v=19.3[/math]	[math](0.95)(1-0.9)=0.095[/math]
[math]10+10 v+10 v^{2}=27.949[/math]	[math](0.95)(0.9)=0.855[/math]

[math]\operatorname{Var}(X)=E\left(X^{2}\right)-[E(X)]^{2}[/math]

[math]E(X)=10(1-0.95)+[10+10 v](0.95)(1-0.9)+\left[10+10 v+10 v^{2}\right](0.95)(0.9)=26.23[/math]

[math]E\left(X^{2}\right)=10^{2}(1-0.95)+[10+10 v]^{2}(0.95)(1-0.9)+\left[10+10 v+10 v^{2}\right]^{2}(0.95)(0.9)=708.27[/math]

[math]\operatorname{Var}(X)=708.27-(26.23)^{2}=20.26[/math]

Standard Deviation [math]=(20.26)^{1 / 2}=4.5[/math]

It's not significantly less work, but since everyone receives the first 10, you could use the formula [math]\operatorname{Var}(X[/math]-constant [math])=\operatorname{Var}(X)[/math], and work with the three payment amounts as [math]0,10 v[/math], [math]10 v+10 v^{2}[/math], and you would get the same answer.

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