Does "not mean independent" imply "not independent"?

Question

If X and Y are two random variables such that $E(X | Y = y) \neq E (X)$ for some $y \in D_y$, then X is not mean independent from Y. Does it imply that X and Y are also not stochastically independent?

Answer 1

The answer is yes, because it is the contrapositive version of the Theorem stating that independence $\Rightarrow$ mean independence: Let $X, Y_1, Y_2$ be random variables, if $(X, Y_1)$ is independent of $Y_2$, then $$ E[X|Y_1, Y_2] = E[X| Y_1] \quad a.s., $$ (while the opposite direction is not true, see counterexample of @Peter Strouvelle).

Contraposition: Not-mean-independence $\Rightarrow$ not-independence.

Answer 2

The standard counterexample is the following. Let $U\sim \mathcal U_{(0,2\pi]}$, and define the vector $(X,Y):=(\cos(U),\sin(U))$. Then $X^2+Y^2=1$, and thus $X=\pm\sqrt{1-Y^2}$. Therefore, for $Y(\omega)=y$, $X$ is uniform distributed on $\{-\sqrt{1-y^2},\sqrt{1-y^2}\}$. Thus, $\mathbb E(X|Y=y)=0$ for all $y$. You can easily check, that $X$ and $Y$ can not be independent. Thus mean independence does not imply independence.