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 $(X, Y) \equiv (0, I_2)$.

What does $X^2+Y$ equal? That's usually required, as the condition needs to be an event, not a random variable. — , Jul 16 '14 at 23:44
@Lord Soth: Thanks! That makes sense, I guess, but I've not run into that convention. Admittedly, it's been a while for me doing such problems. — , Jul 16 '14 at 23:58

score 5 · Accepted Answer · answered Jul 17 '14 at 00:00

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.

Why $E(Xf(X^2+Y))=0$ implies $E(X|X^2+Y)=0$? I did not see it. — bankrip, Jul 21 '14 at 15:37

score 1 · Answer 2 · answered Jul 17 '14 at 00:23

1

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

answered Jul 17 '14 at 00:23

John Fernley

1,218

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

2 Answers2