excans:F93c8f19ea

Answer: B

Since [math]S_{0}(t)=1-F_{0}(t)=\left(1-\frac{t}{\omega}\right)^{\frac{1}{4}}[/math], we have [math]\ln \left[S_{0}(t)\right]=\frac{1}{4} \ln \left[\frac{\omega-t}{\omega}\right][/math].

Then [math]\mu_{t}=-\frac{d}{d t} \log S_{0}(t)=\frac{1}{4} \frac{1}{\omega-t}[/math], and [math]\mu_{65}=\frac{1}{180}=\frac{1}{4} \frac{1}{\omega-65} \Rightarrow \omega=110[/math].

[math]e_{106}=\sum_{t=1}^{3}{ }_{t} p_{106}[/math], since [math]{ }_{4} p_{106}=0[/math]

[math]{ }_{t} p_{106}=\frac{S_{0}(106+t)}{S_{0}(106)}=\frac{\left(1-\frac{106+t}{110}\right)^{1 / 4}}{\left(1-\frac{106}{110}\right)^{1 / 4}}=\left(\frac{4-t}{4}\right)^{1 / 4}[/math]

[math]e_{106}=\sum_{i=1}^{i=4}{ }_{t} p_{106}=\frac{1}{4^{0.25}}\left(1^{0.25}+2^{0.25}+3^{0.25}\right)=2.4786[/math]

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

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

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