$X,Y$ are two independent standard normal variables, calculate the value of $E[X|XY]$ I was told the result is 0 according to the symmetry, but I cannot figure out why.
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1I am pretty sure you miss out some information. – Siong Thye Goh Sep 23 '19 at 03:05
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To add to the above, $E(X \rvert XY)$ is a random variable. If this is always zero then $E(E(X \rvert XY)) = E(X) = 0$ but $E(X)$ might not be zero. – Riley Sep 23 '19 at 05:18
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yes, rally sorry that I forgot to mention, they are standard normal – ASaltedFish Sep 23 '19 at 18:39
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Since $(X, Y)$ and $(-X, -Y)$ are identically distributed, $$\mathbb E(X \mid X Y = z) = \mathbb E(-X \mid (-X) (-Y) = z).$$ – Maxim Sep 24 '19 at 10:42
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Suppose that $(X,Y)$ is a random vector such that $(-X,-Y)$ has the same law as $(X,Y)$. Each set in the $\sigma$-algebra generated by $XY$ are of the form $\{XY\in B\}$, where $B$ is a Borel subset of $\mathbb R$. Since $X\mathbf 1_{\{XY\in B\}}$ has the same law as $-X\mathbf 1_{\{(-X)(-Y)\in B\}}$ it follows that $$ \mathbb E\left[X\mathbf 1_{\{XY\in B\}} \right]=0. $$ The independent between $X$ and $Y$ and he fact that $X$ and $Y$ are both symmetric guarantees here that $(-X,-Y)$ has the same law as $(X,Y)$.
Davide Giraudo
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