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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?

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If you are given the value of $Y = Z^2$, then you know that $Z$ can only have value $\sqrt{Y}$ or $-\sqrt{Y}$. Can you think of any reason why these two values might be equally likely? How might the answer differ if $Z \sim N(\mu, 1)$ where $\mu \neq 0$? – Dilip Sarwate Dec 4 '12 at 21:13
Ah! Is that because the normal curve is symmetric about the mean, so both $\sqrt{Y}$ and $-\sqrt{Y}$ are equally likely? I presume if the mean shifted, then we can't make conclusions since $\sqrt{Y}$ and $-\sqrt{Y}$ are no longer equally likely? – David Faux Dec 4 '12 at 21:24
up vote 3 down vote accepted

$-Y$ 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 $-X$.

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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 $E(e^{i\theta Y} \mid Z) = \cos(\theta Z)$

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