Revision as of 01:53, 20 January 2024 by Admin (Created page with "For two fully continuous whole life insurance policies on <math>(x)</math>, you are given: (i) {| class="table" ! !! Death Benefit !! Annual Premium Rate !!Variance of the Present Value of Future Loss at <math>t</math> || |- | Policy A || 1 || 0.10 || 0.455 |- | Policy B || 2 || 0.16 || - |} (ii) <math>\delta=0.06</math> Calculate the variance of the present value of future loss at <math>t</math> for Policy B. <ul class="mw-excansopts"><li> 0.9</li><li> 1.4</li><...")
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ABy Admin
Jan 20'24

Exercise

For two fully continuous whole life insurance policies on [math](x)[/math], you are given:

(i)

Death Benefit Annual Premium Rate Variance of the Present Value of Future Loss at [math]t[/math]
Policy A 1 0.10 0.455
Policy B 2 0.16 -

(ii) [math]\delta=0.06[/math]

Calculate the variance of the present value of future loss at [math]t[/math] for Policy B.

  • 0.9
  • 1.4
  • 2.0
  • 2.9
  • 3.4

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

ABy Admin
Jan 20'24

Answer: B

[math]L_{A}=v^{T}-0.10 \bar{a}_{\bar{T}}=\left(1+\frac{10}{6}\right) v^{T}-\frac{10}{6}[/math]

[math]\operatorname{Var}\left(L_{A}\right)=\left(1+\frac{10}{6}\right)^{2} \operatorname{Var}\left(v^{T}\right)=0.455 \Rightarrow \operatorname{Var}\left(v^{T}\right)=0.06398[/math]

[math]L_{B}=2 v^{T}-0.16 \bar{a}_{T}=\left(2+\frac{16}{6}\right) v^{T}-\frac{16}{6}[/math]

[math]\operatorname{Var}\left(L_{B}\right)=\left(2+\frac{16}{6}\right)^{2} \operatorname{Var}\left(v^{T}\right)=\left(2+\frac{16}{6}\right)^{2}(0.06398)=1.39[/math]

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

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