# Find the expected claim payment made for a car loss

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?

• 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. – Max Apr 23 '18 at 19:37
• @MaxFt Lower bound of 1 is fine. If damages are less than 1, nothing is paid. – Doug M Apr 23 '18 at 20:07
• 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! – Beginner Nov 21 '18 at 20:54

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