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?
Best Answer
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 6 you will only be paying 5.
$$E(Y)=\int_{1}^{6} (x-1)f(x) dx +\int_{6}^{\infty} 5f(x) dx$$
Hope this helps