Why is $E(Z | Y) = 0$? Let $Z$ be a random variable distributed $\mathcal{N}(0, 1)$. Let $Y = Z^2$.
Apparently, $E(Z \mid Y) = E(Z \mid Z^2) = 0$ due to "symmetry." Why is that? 

$-Y$ is also $N(0,1)$ distributed and the $\sigma$ algebras generated by $-Y$ and $Y$ are the same, so $E[-Y\mid (-Y)^2]=E[Y\mid Y^2]$, which gives the wanted result.
It can be extended to integrable random variables $X$ such that $X$ has the same law as $-X$. 
