Conditional Expectation of joint pdf I have a question about this joint distribution
$f(x,y)=4\exp(-2y^2)$ $0<x<y$ and $0<y<\infty$
$Z=1$ if $|X-Y|>2$ 
$Z=0$ otherwise
I need to find $E[Z|Y]$
Here's what I've done so far
$E[Z|Y]$
=$P[|X-Y|>2|Y]$
=$P[|X-Y|>2|Y=y]$
=$P[2>X-Y>-2|Y=y]$
Now I'm stuck because I am not able to find the pdf of X. 
 A: I think it's a good idea, as you suggest, to find $f_{X|Y=y}(x|y)$. It should be $f_{X|Y=y}(x|y) = 1/y$ for $0<x<y$ (uniform distribution on $(0,y)$), which is because for a given $y$, each possible value $x$ of $X$ has equal weight in the joint pdf. You can also get this answer by computing the marginal density of $Y$, $f_Y(y)$, which is $4ye^{-2y^2}$, and then computing $f(x,y)/f_Y(y)$.
You then have to interpret $E[Z|Y]$ correctly. Say that you are able to calculate $E[Z|Y=y]$ for each value $y$ that $Y$ can take (which in this case is any positive number). The result will depend on the $y$ you choose, so you get a function $g(y) = E[Z|Y=y]$ of $y$. $E[Z|Y]$ is then just $g(Y)$, i.e. you replace $y$ in the function $g(y)$ with the random variable $Y$. So if, for example, $E[Z|Y=y] = y^2$ for each $y$, then $E[Z|Y] = Y^2$.
So to find $P(|X-Y|>2|Y)$, it's a good strategy to compute $P(|X-y|>2|Y=y)$ for each $y \in (0,\infty)$, and then replace $y$ by $Y$ in the result. Try to figure out what $|X-y|>2$ means (if the distance between $X$ and $y$ is more than 2, then where must $X$ lie?), and remember that $X$ is always less than $y$.
