Assume that we have two r.v. $X,Y$ that are marginally Gaussian distributed (we don't know about the joint distribution), and assume that $\mathbb{E}[X|Y] = X$ (i.e. $X$ is $Y$-measurable).
Does this mean that necessarily mean that $X=f(Y)$ for some deterministic $f$?
What do we know about the conditional density $p(x|y)$? Can it even be non-degenerate? It seems like it can not be a Gaussian, because then one would have $X|Y \sim \mathcal{N}(x;x,\sigma)$ which does not make sense.