exercise:2c7c349abb

ABy Admin

Jan 15'24

Exercise

You are given:

(i) [math]\quad S_{0}(t)=\left(1-\frac{t}{\omega}\right)^{\frac{1}{4}}[/math], for [math]0 \leq t \leq \omega[/math]

(ii) [math]\quad \mu_{65}=\frac{1}{180}[/math]

Calculate [math]e_{106}[/math], the curtate expectation of life at age 106 .

2.2
2.5
2.7
3.0
3.2

Add answer Add answer

ABy Admin

Jan 15'24

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]