Revision as of 21:17, 19 January 2024 by Admin (Created page with "'''Answer: E''' In general, the loss at issue random variable can be expressed as: <math display="block"> L=\bar{Z}_{x}-P \times \bar{Y}_{x}=\bar{Z}_{x}-P \times\left(\frac{1-\bar{Z}_{x}}{\delta}\right)=\bar{Z}_{x} \times\left(1+\frac{P}{\delta}\right)-\frac{P}{\delta} </math> Using actuarial equivalence to determine the premium rate: <math display="block"> P=\frac{\bar{A}_{x}}{\bar{a}_{x}}=\frac{0.3}{(1-0.3) / 0.07}=0.03 </math> <math>\operatorname{Var}(L)=\l...")
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Exercise


ABy Admin
Jan 19'24

Answer

Answer: E

In general, the loss at issue random variable can be expressed as:

[[math]] L=\bar{Z}_{x}-P \times \bar{Y}_{x}=\bar{Z}_{x}-P \times\left(\frac{1-\bar{Z}_{x}}{\delta}\right)=\bar{Z}_{x} \times\left(1+\frac{P}{\delta}\right)-\frac{P}{\delta} [[/math]]


Using actuarial equivalence to determine the premium rate:

[[math]] P=\frac{\bar{A}_{x}}{\bar{a}_{x}}=\frac{0.3}{(1-0.3) / 0.07}=0.03 [[/math]]


[math]\operatorname{Var}(L)=\left(1+\frac{P}{\delta}\right)^{2} \times \operatorname{Var}\left(\bar{Z}_{x}\right)=\left(1+\frac{0.03}{0.07}\right)^{2} \times \operatorname{Var}\left(\bar{Z}_{x}\right)=0.18[/math]

[math]\operatorname{Var}\left(\bar{Z}_{x}\right)=\frac{0.18}{\left(1+\frac{0.03}{0.07}\right)^{2}}=0.088[/math]

[math]\operatorname{Var}\left(L^{*}\right)=\left(1+\frac{P^{*}}{\delta}\right)^{2} \times \operatorname{Var}\left(\bar{Z}_{x}\right)=\left(1+\frac{0.06}{0.07}\right)^{2}(0.088)=0.304[/math]

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

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