May 14'23
Exercise
A risk has a loss amount that has a Poisson distribution with mean 3. An insurance policy covers the risk with an ordinary deductible of 2. An alternative insurance policy replaces the deductible with coinsurance [math]\alpha[/math], which is the proportion of the loss paid by the policy, so that the expected cost remains the same.
Calculate [math]\alpha [/math].
- 0.22
- 0.27
- 0.32
- 0.37
- 0.42
May 14'23
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]