Exercise
Answer
Answer: E
[math]L_{0}=100,000 v^{T}-560 \bar{a}_{\bar{T} \mid}=\left(100,000+\frac{560}{\delta}\right) e^{-\delta T}-\frac{560}{\delta}[/math]
Since [math]L_{0}[/math] is a decreasing function of [math]T[/math], the [math]75^{\text {th }}[/math] percentile of [math]L_{0}[/math] is [math]L_{0}(t)[/math] where [math]t[/math] is such that [math]\operatorname{Pr}\left[T_{35}\gtt\right]=0.75[/math].
[math]\frac{l_{35+t}}{l_{35}}=0.75[/math]
[math]l_{35+t}=0.75 l_{35}=0.75 \times 99,556.7=74,667.5[/math]
[math]l_{81} \lt 74,667.5 \lt l_{80}[/math]
[math]t=(80-35)+s[/math]
[math]l_{80+s}=s l_{81}+(1-s) l_{80}[/math]
[math]74,667.5=73,186.3 s+75,657.2(1-s)[/math]
[math]s=0.40054[/math]
[math]t=45.40054[/math]
[math]L_{0}(45.40054)=\left(100,000+\frac{560}{\ln (1.05)}\right) e^{-45.40054 \ln (1.05)}-\frac{560}{\ln (1.05)}=689.25[/math]
Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.