Why $E(X\mid X^2+Y)=0$ for $(X,Y)$ standard normal?

Question

Please help me explain why $E(X\mid X^2+Y)=0$ for $(X,Y)$ standard normal, i.e $(X, Y) \equiv (0, I_2)$.

Answer 1

By the definition of conditional expectation, it suffices to show that $\mathbb{E}(Xf(X^2+Y))=0$ for every bounded, Borel measurable function $f$. But this is $$\int\int x f(x^2+y)\,\phi(y)\,\phi(x)\, dy\, dx= \int x \left[\int f(x^2+y)\,\phi(y)\, dy\right] \phi(x)\, dx= \int x\, \psi(x)\, \phi(x)\, dx,$$ which is zero since $\psi(x):=\int f(x^2+y)\,\phi(y)\, dy$ is an even function of $x$. Here $\phi$ is the density of a standard normal distribution, also an even function.

Answer 2

The simple explanation for "why" is that knowing $X^2+Y$ tells you nothing about the sign of X ($X \mapsto X^2$ is even) and so after conditioning on $X^2+Y$, $X$ is still symmetrically distributed