Find $ E(X \mid X^2 + Y^2) $ for $X,Y$ independent standard normal

Question

Let $X,Y$ independent random variables with $ X,Y\sim \mathcal{N}(0,1) $.

I want to find: $$ E(X \mid X^2 + Y^2) $$

How can it be found? I would appreciate any tips or hints.

Answer 1

By symmetry, you'll have $$ \mathbb{E}[X\mid X^2+Y^2] = 0\,.$$ This is because $\mathbb{E}[X\mid X^2+Y^2] = \mathbb{E}[(-X)\mid (-X)^2+Y^2]=-\mathbb{E}[X\mid X^2+Y^2]$ (the first equality as $X$ and $-X$ have same distribution, and are both independent of $Y$; the second is linearity).

Answer 2

The answer is $0$. To show this what you have to show is $\int_{\{X^{2}+Y^{2} \leq t\}}XdP =0$ for all $t \in \mathbb R$. But this is immediate from the fact that $(X,Y)$ has same distribution as $(-X,-Y)$.