3
$\begingroup$

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?

  • 0
    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?2012-12-04

2 Answers 2

3

$-Y$ 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 $-X$.

5

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