When $E(X|Y)E(Y) = E(XY)$? $X$ and $Y$ are random variables. In what cases is the following statement true

$E(X|Y)E(Y) = E(XY)$ a.s.,

other than the cases when $X$ and $Y$are independent, and when $E(X|Y)=EX$ a.s. ?
If $E(X|Y)E(Y) = E(XY)$ a.s.,, and it is not true that $E(X|Y)=EX$ a.s., then must 


*

*$E(Y)=0$,

*$ E(XY)=0$?
Is each of  the following true: 


*

*if $EY\neq 0$, then $E(X|Y)E(Y)=E(XY)$ a.s. and $E(X|Y)=EX$ a.s. are equivalent. 

*$E(X|Y)E(Y)=E(XY)$ a.s., if and only if either $E(X|Y)=EX$ a.s., or $EY = E(XY)= 0$.
Is there an example where $E(X|Y)\neq EX$ a.s., and $EY = E(XY)= 0$?
Thanks!
 A: As Did rightly points out, $X$ and $Y$ need not be independent. However, in the equality $E(X|Y) E(Y) = E(XY)$ the right side is constant, and the left side is a random variable, so the only way to have equality is if either $E(X|Y) = \mathrm{const.}$ or $E(Y) = 0$. 
If $E(Y) = 0$ then you need precisely $E(XY) = 0$. This does not tell you much about $X$; $X$ can be anything up to 'one degree of freedom'. More precisely, if you pick your favourite variable $X_0$ and you have any other variable $Z$ with $E(ZY) \neq 0$ (e.g. $Z=Y$ will work unless $Y$ is identically $0$, in which case any $X$ works anyway) then there will be a constant $c$ such that $X = X_0 + cZ$ satisfies $E(XY) = 0$. For instance, with the choice $X_0 = Y^2,\ Z = Y$ you can easily ensure that $E(Y) = 0$ and $E(XY) = 0$ but $E(X|Y) = X = Y^2 - cY$ is non-constant.
Note that $$E(XY) = E(E(XY|Y) = E(YE(X|Y)).$$ Thus, your original equation is equivalent to $E(YE(X|Y)) = E(Y)E(X|Y)$, which is equivalent to:
$$ (*) \qquad E((Y-E(Y))E(X|Y)) = 0. $$
If $E(X|Y) = \mathrm{const.}$ then $(*)$ clearly holds, because then $$E((Y-E(Y))E(X|Y)) = E(X|Y)E((Y-E(Y))) = 0. $$
The converse does not hold, and we will construct lots of counterexamples. To begin with, assume that $E(Y) = 0$ and that $X = f(Y)$ so that $(*) $ becomes $$E(Y f(Y)) = 0.$$ There a lot of ways to accomplish this, for example by ensuring that $Y$ has a symmetric distribution and that $f(y) = f(-y)$. As long as $f$ is not constant on the set of values taken by $Y$, $E(X|Y)$ isn't constant. (To be very concrete, take $Y$ uniform on $[-1,1]$ and $X = Y^2$).
