# Calculating expected value of unknown random variable

The question:

Micro Insurance Company issued insurance policies to $32$ independent risks. For each policy, the probability of a claim is $1/6$. The benefit amount given that there is a claim has probability density function $$f(y) = \begin{cases} 2(1-y) & 0<y<1, \\0 & \text{otherwise}. \end{cases}$$ Calculate the expected value of total benefits paid.

My attempt:

I'm not sure on how to define my random variable. Its expected value should sum from 1 to 32, each with probability $\frac{1}{6} \int f(y) dy$, I think.

Let $Y_{n}$ be the amount paid to the $n$-th policyholder assuming that the claim is made. Let $\mathbb{I}_{n}$ be $0$ when the $n$-th claim is not made and $1$ otherwise. Then, the total benefits paid is $$X=\sum_{n=1}^{32}\mathbb{I}_{n}Y_{n}.$$ We need to calculate $$\mathbb{E}\left[X\right]=\mathbb{E}\left[\sum_{n=1}^{32}\mathbb{I}_{n}Y_{n}\right]=\sum_{n=1}^{32}\frac{1}{6}\int_{0}^{1}yf\left(y\right)dy.$$ I think you can do the rest yourself.
• Can you explain why $X$ is defined like so? May 28 '13 at 3:37
• $X$ is the total amount paid by the insurer. There are 32 policyholders (hence the summation going from $n=1$ to $n=32$). $\mathbb{I}_n$ is the random variable associated with the $n$-th policyholder making the claim (this has a probability of $1/6$). $Y_n$ is the amount paid to the $n$-th policyholder assuming the claim has been made. May 28 '13 at 15:55