Under a group insurance policy, an insurer agrees to pay all of the medical bills incurred during the year by employees of a small company, up to a maximum total of one million dollars. The total amount of bills incurred, X, has probability density function where f(x)= $\frac{x(4-x)}{9}$ for 0 $\leq$ x $\leq$ 3, and 0 otherwise, where x is measured in millions. Let Y be the total amount of money paid. Compute $\mathop{{}\mathbb{E}}$[Y] and var[Y].
So for this question, I was thinking of computing the expectation of X, and then if it was greater than 1, the expectation for Y would be 1,000,000, otherwise it would be $\mathop{{}\mathbb{E}}$[X]*1,000,000. Is this the correct strategy? If so, would I go about computing the variance the same way? If not, what is the best way to go about this?
Edit: I actually realized that to get the expectation of Y, I need to take the integral of xf(x) from 0 to 1, and then add the integral of f(x) from 1 to infinity, in order to account for if x is less than 1,000,000 and greater than 1,000,000. I ended up getting around 0.935, but I'm still stumped on how to get the variance of Y.
No, that's not correct. For one thing, if $E[X]>1$ this only implies that sometimes $X>1$ and then sometimes (those times) $Y=1,000,000$ ; but sometimes it will still happen that $X<1$ , and in those cases $Y=X * 1,000,000$, so, the mean will be below: $E[Y]<1,000,000$.
You should first write the formula for $Y$. To avoid messing with the one millon factor, I'l denote by $Z$ the amount of money paid in millions (hence $Y= 1,000,000 Z$).
Then $$Z=\begin{cases} X & \text{if } X \le 1\\ 1 & \text{if } X >1 \end{cases} $$
Now , using $E[g(X)] = \int g(x) f_X(x)dx:$
$$E[Z] = \int_0^1 x \frac{x(4-x)}{9} dx+ \int_1^3 1 \frac{x(4-x)}{9} dx$$
and
$$E[Z^2] = \int_0^1 x^2 \frac{x(4-x)}{9} dx+ \int_1^3 1 \frac{x(4-x)}{9} dx$$
from which we can get the variance of $Z$.
Can you go on from here?