For a fully discrete whole life insurance of 1 on [math](x)[/math], you are given:

(i) The net premium policy value at the end of the first year is 0.012

(ii) [math]\quad q_{x}=0.009[/math]

(iii) [math]\quad i=0.04[/math]

Calculate [math]\ddot{a}_{x}[/math].

• 17.1
• 17.6
• 18.1
• 18.6
• 19.1

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

Answer: A

[math]{ }_{1} V_{x}=A_{x+1}-P_{x} \ddot{a}_{x+1}=1-d \ddot{a}_{x+1}-P_{x} \ddot{a}_{x+1}[/math]

[[math]] =1-\underbrace{\left(P_{x}+d\right)} \ddot{a}_{x+1}=1-\ddot{a}_{x+1} / \ddot{a}_{x} [[/math]]

[math]\Rightarrow \ddot{a}_{x}\left(1-V_{x}\right)=\ddot{a}_{x+1}[/math]

Since [math]\ddot{a}_{x}=1+v p_{x} \ddot{a}_{x+1}[/math] substituting we get

[math]\ddot{a}_{x}\left(1-{ }_{1} V_{x}\right)=\frac{\ddot{a}_{x}-1}{v p_{x}} \Rightarrow \ddot{a}_{x}\left(1-{ }_{1} V_{x}\right) v p_{x}=\ddot{a}_{x}-1[/math]

Solving for [math]\ddot{a}_{x}[/math], we get [math]\ddot{a}_{x}=\frac{1}{1-\left(1-{ }_{1} V_{x}\right) v p_{x}}=\frac{1}{1-(1-0.012)\left(\frac{1}{1.04}\right)(1-0.009)}[/math]

[[math]] =17.07942 [[/math]]

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