Is it true that $E[X|Y] = \rho \frac{\sigma_X}{\sigma_Y} Y$ if $X$ and $Y$ are not jointly gaussian?

Question

Let $X$ and $Y$ be centered normal (Gaussian) random variables. Let $\rho:= \frac{E[XY]}{\sigma_X \sigma_Y}$ where $\sigma_X^2 = E[X^2]$ and $\sigma_Y^2 = E[Y^2]$. It is known that if $(X,Y)$ is a Gaussian vector then $$ E[X|Y] = \frac{\rho \sigma_X}{\sigma_Y} Y. $$

However, I came across an exercise saying that this is true even without the assumption that the vector $(X,Y)$ is Gaussian ($X$, $Y$ are still centered normal r.v.s). I couldn't prove or disprove it so I wonder if this is true or not? For example, if $X = R Y$ where $R$ is a Rademacher r.v. independent of $Y$, then $E[X|Y] = 0$, so it is still true. Thank you!

Thank you!

Answer 1

Here is a counterexample when $X$ and $Y$ are centred Gaussian random variables:

Let $f$ be the inverse of the cdf of the standard Gaussian distribution and $U$ be uniformly distributed on $(0,1)$, so that $Y:=f(U)$ is standard Gaussian. Let then $X:=f(\textrm{mod}(U+\frac12,1))$, where $\textrm{mod}(x,1)$ is $x$ plus or minus a certain integer so that it stays in $(0,1)$.

Note that $\textrm{mod}(U+\frac12,1)$ is also uniformly distributed on $(0,1)$, and is a function of $U$. This tells us two things:

• $X$ has the same distribution as $Y$, hence it is standard Gaussian;
• $X$ is a map of $Y$.

So $\mathbb E[X\vert Y]=X$. But if we had $X=\lambda Y$ for some $\lambda\in\mathbb R$, as $f$ is one-to-one, we would have $U=\lambda\textrm{mod}(U+\frac12,1)$. But clearly $U$ and $\textrm{mod}(U+\frac12,1)$ do not in a linear relationship.

Answer 2

Let $X,Z$ be iid standard Gaussian. Let $Y= |Z| \, sgn(X)$. This forces $X$ and $Y$ to have same sign; in terms of the joint density function, it amounts to erasing two quadrants and multiply by two the others.

Then both $X$ and $Y$ are standard Gaussian.

But

$$E(Y|X)= sgn(X) \sqrt{\frac{2}{\pi}}$$

i.e., the conditional expectation is not linear, but constant inside each sign (because, conditioned on $X$, $Y$ follows a folded normal standard distribution).

BTW: $$\rho = E[X Y ] = E (X sgn(X) |Z|)=E (|X| |Z|)=(E|X|)^2 = \frac{2}{\pi}$$