Revision as of 23:29, 13 May 2023 by Admin (Created page with "'''Key: E''' <math display = "block"> \begin{aligned} \operatorname{E}[ X \wedge 2) &= 1 f (1) + 2[1 − F (1)] = 1 f (1) + 2[1 − f (0) − f (1)] \\ &= 1(3e^{−3} ) + 2(1...")
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Exercise


May 14'23

Answer

Key: E

[[math]] \begin{aligned} \operatorname{E}[ X \wedge 2) &= 1 f (1) + 2[1 − F (1)] = 1 f (1) + 2[1 − f (0) − f (1)] \\ &= 1(3e^{−3} ) + 2(1 − e^{−3} − 3e^{−3} ) = 2 − 5e^{−3} = 1.75 \end{aligned} [[/math]]

Cost per loss with deductible is

[[math]] \operatorname{E}( X ) − \operatorname{E}( X \wedge 2) = 3 −1.75 = 1.25 [[/math]]

Cost per loss with coinsurance is [math] \alpha \operatorname{E}( X ) = 3\alpha [/math]

Equating cost: [math]3\alpha = 1.25 \Rightarrow \alpha = 0.42 [/math]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

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