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

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

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]]

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.

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