Jan 16'24
Exercise
[math]\mathrm{X}[/math] and [math]\mathrm{Y}[/math] are both age [math]61 . \mathrm{X}[/math] has just purchased a whole life insurance policy. [math]\mathrm{Y}[/math] purchased a whole life insurance policy one year ago.
Both [math]\mathrm{X}[/math] and [math]\mathrm{Y}[/math] are subject to the following 3-year select and ultimate table:
[math]x[/math] | [math]\ell_{[x]}[/math] | [math]\ell_{[x]+1}[/math] | [math]\ell_{[x]+2}[/math] | [math]\ell_{x+3}[/math] | [math]x+3[/math] |
---|---|---|---|---|---|
60 | 10,000 | 9,600 | 8,640 | 7,771 | 63 |
61 | 8,654 | 8,135 | 6,996 | 5,737 | 64 |
62 | 7,119 | 6,549 | 5,501 | 4,016 | 65 |
63 | 5,760 | 4,954 | 3,765 | 2,410 | 66 |
The force of mortality is constant over each year of age.
Calculate the difference in the probability of survival to age 64.5 between [math]\mathrm{X}[/math] and [math]\mathrm{Y}[/math].
- 0.035
- 0.045
- 0.055
- 0.065
- 0.075
Jan 16'24
Answer: C
[[math]]
\begin{aligned}
& { }_{3.5} p_{[61]}-{ }_{3.5} p_{[60]+1}={ }_{0.5} p_{64}\left({ }_{3} p_{[61]}-{ }_{3} p_{[60]+1}\right) \\
& =\left(\frac{l_{65}}{l_{64}}\right)^{0.5}\left(\frac{l_{64}}{l_{[61]}}-\frac{l_{64}}{l_{[60]+1}}\right) \\
& =\left(\frac{4016}{5737}\right)^{0.5}\left(\frac{5737}{8654}-\frac{5737}{9600}\right) \\
& =0.05466
\end{aligned}
[[/math]]