excans:F93c8f19ea: Difference between revisions
(Created page with "'''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</mat...") |
mNo edit summary |
||
| Line 10: | Line 10: | ||
<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> | <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> | ||
{{soacopyright | 2024 }} | |||
Revision as of 00:22, 18 January 2024
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.