TITLE: Conditional mean on uncorrelated stochastic variable
QUESTION [6 upvotes]: I know that $E[X|Y]=E[X]$ if $X$ is independent of $Y$. I recently was made aware that it is true if only $\text{Cov}(X,Y)=0$. Would someone kindly either give a hint if it's easy, show me a reference or even a full proof if it's short? Either will work I think :)  
Thanks.
Edit: Thanks for the great answers! I accepted Alecos simply for being first. I've made a followup question here. (When i someday reach 15 reputation I will upvote)

REPLY [7 votes]: Your assertion is false.
For two random variables $X$ $Y$ one can consider three measures of un-related-ness:
$1$: Independence:  $p(X,Y) = p(X)p(Y)$ Equivalently, $P(X|Y)= P(X)$. This is the most important and strongest one.
$2$: Unpredictability (this term is not very standard):  conditioning on one variable does not change the expectation of the other. Actually, this property is not symmetric, so there are two possible cases:
$2a$: $E(X|Y) = E(X)$
$2b$: $E(Y|X) = E(Y)$
(These properties can also be stated as: the regression line is horizontal/vertical)
$3$: Uncorrelatedness (orthogonality): $Cov(X,Y) = 0$ . Equivalently: $E(X Y) = E(X) E(Y)$
It can be shown that $1 \implies 2a$, $1 \implies 2b$ , $2a \implies 3$ , $2b \implies 3$ (hence, $1 \implies 3$). All other implicancies don't hold.

An example: consider $(X,Y)$ having uniform distribution over the triangle with vertices at $(-1,0)$, $(1,0)$, $(0,1)$. Check (you can deduce it from symmetry considerations) that $E(X|Y)=E(X)$ and $Cov(X,Y)=0$, but $E(Y|X)$ varies with $X$.