For a special whole life insurance policy issued on (40), you are given:

(i) Death benefits are payable at the end of the year of death

(ii) The amount of benefit is 2 if death occurs within the first 20 years and is 1 thereafter

(iii) [math]Z[/math] is the present value random variable for the payments under this insurance (iv) [math]\quad i=0.03[/math]

(v)

[math]x[/math]	[math]A_{x}[/math]	[math]{ }_{20} E_{x}[/math]
40	0.36987	0.51276
60	0.62567	0.17878

(vi) [math]\quad E\left[Z^{2}\right]=0.24954[/math]

Calculate the standard deviation of [math]Z[/math].

• 0.27
• 0.32
• 0.37
• 0.42
• 0.47

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

Answer: A

[math]E[Z]=2 \cdot A_{40}-{ }_{20} E_{40} A_{60}=(2)(0.36987)-(0.51276)(0.62567)=0.41892[/math]

[math]E\left[Z^{2}\right]=0.24954[/math] which is given in the problem.

[math]\operatorname{Var}(Z)=E\left[Z^{2}\right]-(E[Z])^{2}=0.24954-0.41892^{2}=0.07405[/math]

[math]S D(Z)=\sqrt{0.07405}=0.27212[/math]

An alternative way to obtain the mean is [math]E[Z]=2 A_{40: 20 \mid}^{1}+{ }_{20 \mid} A_{40}[/math]. Had the problem asked for the evaluation of the second moment, a formula is

[math]E\left[Z^{2}\right]=\left(2^{2}\right)\left({ }^{2} A_{40: 20}^{1}\right)+\left(v^{2}\right)^{20}\left({ }_{20} p_{40}\right)\left({ }^{2} A_{60}\right)[/math]

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