Revision as of 02:14, 19 January 2024 by Admin (Created page with "For a fully discrete whole life insurance of 200,000 on (45), you are given: (i) Mortality follows the Standard Ultimate Life Table. (ii) <math>\quad i=0.05</math> (iii) The annual premium is determined using the equivalence principle. Calculate the standard deviation of <math>L_{0}</math>, the present value random variable for the loss at issue. <ul class="mw-excansopts"><li> 25,440<li> 30,440</li><li> 35,440</li><li> 40,440<li> 45,440</ul> {{soacopyright|2024}}")
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ABy Admin
Jan 19'24

Exercise

For a fully discrete whole life insurance of 200,000 on (45), you are given:

(i) Mortality follows the Standard Ultimate Life Table.

(ii) [math]\quad i=0.05[/math]

(iii) The annual premium is determined using the equivalence principle.

Calculate the standard deviation of [math]L_{0}[/math], the present value random variable for the loss at issue.

  • 25,440
  • 30,440
  • 35,440
  • 40,440
  • 45,440

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

ABy Admin
Jan 19'24

Answer: A

On a unit basis, [math]\operatorname{Var}\left(L_{0}\right)=\left(1+\frac{P}{d}\right)^{2}\left[{ }^{2} A_{45}-\left(A_{45}\right)^{2}\right]=\left(1+\frac{A_{45}}{d \ddot{a}_{45}}\right)^{2}\left[{ }^{2} A_{45}-\left(A_{45}\right)^{2}\right][/math]

[math]=\left(\frac{d \ddot{a}_{45}+1-d \ddot{a}_{45}}{d \ddot{a}_{45}}\right)^{2}\left[{ }^{2} A_{45}-\left(A_{45}\right)^{2}\right]=\frac{{ }^{2} A_{45}-\left(A_{45}\right)^{2}}{(d \ddot{a})^{2}}[/math]

[math]=\frac{0.03463-0.15161^{2}}{\left(\frac{0.05}{1.05} \times 17.8162\right)^{2}}=0.016178038[/math]

The standard deviation of [math]L_{0}=0.127193[/math]

[math](200,000)\left(\right.[/math] The standard deviation of [math]\left.L_{0}\right)=25,439[/math]

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

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