$X,Y$ are discrete r. v.'s and I know $p_{X,Y}(x,y),p_X(x),p_Y(y)$ for each $x\in \{0,1\}$ and $y \in \{0,1,2\}$.
Suppose that we also have a r.v. $Z=E[X|Y]$. Obviously, $Z$ can take values $E[X|Y=0]$, $E[X|Y=1]$,$E[X|Y=2]$. It turned out that $Z\in\{\frac{3}{7},\frac{1}{2},\frac{2}{3}\}$
I need to find the PMF of $Z$. For that I need to know the probabilities of $E[X|Y=0]$, $E[X|Y=1]$,$E[X|Y=2]$.
I've been told that in this case $p_Z(E[X|Y=0])=p_Y(0)$, $p_Z(E[X|Y=1])=p_Y(1)$, $p_Z(E[X|Y=2])=p_Y(2)$
Question: is that correct and if yes, why so? How can we know anything about the probability of $Z$ in this case?