TITLE: Expected value of joint probability density functions
QUESTION [5 upvotes]: Quick question here. I'm trying to make sense of this problem...
$$f(x_1,x_2) =\begin{cases} 8x_1x_2 & \text{for } 0 < x_1 < x_2 < 1\\
0&\text{otherwise}
\end{cases}$$
If $Y = X_1(X_2)^3$ , what is the expected value of $Y$?
My thoughts:
I'm not entirely sure where to start, but here is where I would start:
\begin{align}
E(Y) &= E\left(X_1(X_2)^3\right)\\
    & = E(X_1)  E(X_2^2)
\end{align}
Then after this step I'm not sure of what to do. There must be a way to use the pdf to solve for the expected value but I'm not sure. I'm not looking for answers but guidance would be greatly appreciated!!

REPLY [7 votes]: The proposed start will not work: $X_1$ and $X_2^3$ are not independent.
I would suggest first making a name change, $X$ for $X_1$, $Y$ for $X_2$, and $W$ for $XY^3$.  You need to calculate the expectation $E(W)$ of the random variable $W$. 
Call the joint density $8xy$ over the region with $0\lt x\lt y\lt 1$.
Now draw a picture (this was the whole purpose of the name changes). The region where the density function is $8xy$ is the part of the square with corners $(0,0)$, $(0,1)$, $(1,1)$, and $(0,1)$ which is above the line $y=x$. (The density is $0$ everywhere else.) 
The region where the density is $8xy$ is a triangle. Call it $T$. Then 
$$E(W)=E(XY^3)=\iint_{T} (xy^3)(8xy)\,dx\,dy.$$
It remains to calculate the integral. This should not be hard. Express as an iterated integral. Things will be a little simpler if you first integrate with respect to $x$.