Exercise


Jan 19'24

Answer

Answer: C

[math]\ddot{a}_{[x]: n]}=1+v p_{[x]} \ddot{a}_{x+1: n-1]}=1+(1+k)\left(v p_{x} \ddot{a}_{x+1: n-1}\right)=1+(1+k)\left(\ddot{a}_{x: n]}-1\right)[/math]

Therefore, we have

[math]k=\frac{\ddot{a}_{[x]: n]}-1}{\ddot{a}_{x: n]}-1}-1=\frac{21.167}{20.854}-1=0.015[/math]

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

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