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!!
Best Answer
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 $x$.