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? 

For completion, note that for all mesurable $f$ such that $E(|f(Y)|) < \infty$, $$E(f(Y)\mid Z) = \frac{f(Z) + f(-Z)}{2}.$$
Here $f\colon x\mapsto x$ is odd, hence $E(Y\mid Z) = \frac{Z-Z}{2}=0$.
Another example of interest : if you take $f(x)=e^{i\theta x}$, you get $E(e^{i\theta Y} \mid Z) = \cos(\theta Z)$ 
