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(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...") |
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<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> | ||
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Latest revision as of 02:34, 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.
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.