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An auto insurance policy has a deductible of 1 and a maximum claim payment of 5. Auto loss amounts follow an exponential distribution with mean 2. Calculate the expected claim payment made for an auto loss.

Attempt

Let $L$ be the loss amount. Let $Y$ be the payment made by insurance. We know

$$ Y = \begin{cases} 0, \; \; \; 0 \leq L \leq 1 \\ \max(5, L-1), \; \; \; 1 \leq L \end{cases} $$

We are given that $L$ had density $f_L(x) = \frac{1}{2} e^{-\frac{1}{2}x} $. Thus, we want

$$ E(Y) = \int_1^{\infty} \max(5,x-1) \frac{1}{2} e^{-1/2x}dx $$

IS this the correct approach?

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  • $\begingroup$ Yea what you wrote is pretty good. Except that the integral starts from 0 I think. Then you can break the integral in two at x=6. $\endgroup$ – Max Apr 23 '18 at 19:37
  • $\begingroup$ @MaxFt Lower bound of 1 is fine. If damages are less than 1, nothing is paid. $\endgroup$ – Doug M Apr 23 '18 at 20:07
  • $\begingroup$ As the maximum payment is 5, but you have a deductible of 1, why you do not use 6 instead of 5 in the integral limit? Thanks! $\endgroup$ – Beginner Nov 21 '18 at 20:54
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It appears your statement is mostly correct. The only thing wrong in your statement is the expectation of $Y$ is the minimum of 5 and $x-1$ as once the claim is greater than 5 you will only be paying 5.

$$E(Y)=\int_{1}^{5} (x-1)f(x) dx +\int_{5}^{\infty} 5f(x) dx$$

Hope this helps

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  • $\begingroup$ As the maximum payment is 5, but you have a deductible of 1, why you do not use 6 instead of 5 in the integral limit? Thanks! $\endgroup$ – Beginner Nov 21 '18 at 20:54

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