Krdytkyu

Is there any counterexample to show that $X,Y$ are two random variables and $E(Xmid Y)=E(X)$, but $X$ and $Y$ are not independent.

I already know that if $X,Y$ are independent, then $E(Xmid Y)=E(X)$.

begin{align}
X & = begin{cases} -1 \ phantom{-}0 & text{each with probability } 1/3 \ +1 end{cases} \[10pt]
Y & = X^2
end{align}
Then $operatorname E(Xmid Y=0) = 0$ and $operatorname E(Xmid Y=1) = 0,$ so $operatorname E(Xmid Y) = 0$ with probability $1.$ But $X$ and $Y$ are very far from independent.

Consider a two-component mixture Gaussian model with
begin{align}
& X mid Y = 0 sim N(0, 1), \
& X mid Y = 1 sim N(0, 4), \
& P(Y = 0) = P(Y = 1) = 1/2.
end{align}

Clearly $E(X mid Y = 0) = E(X mid Y = 1) = E(X) = 0$. But $X$ and $Y$ are not independent. Because on one hand,
$$P(-1 leq X leq 1, Y = 0) = P(-1 leq X leq 1 mid Y = 0) P(Y = 0) = Phi(1) -
0.5.$$
On the other hand,
begin{align}
P(-1 leq X leq 1) & = P(-1 leq X leq 1 mid Y = 0)P(Y = 0) + P(-1 leq X leq 1 mid Y = 1)P(Y = 1) \
&= Phi(1) - 0.5 + Phi(0.5) - 0.5 = Phi(1) + Phi(0.5) - 1.
end{align}
Therefore
$$P(-1 leq X leq 1)P(Y = 0) = 0.5Phi(1) + 0.5Phi(0.5) - 0.5 neq P(-1 leq X leq 1, Y = 0),$$
i.e., $X$ and $Y$ are not independent.

Per Clement's request, let me add more details to show that the random variable $E(X mid Y)$ is actually the constant $0$. In fact, as a random variable,
$Z(omega) = E[X|Y](omega)$ are constants on the the sets ${omega: Y(omega) = 1}$ and ${omega: Y(omega) = 0}$ (this is a general result holds for any conditional expectation of form $E[X | Y]$, see for example, Theorem 9.1.2. of A Course in Probability Theory). As the union of these two sets is the whole sample space $Omega$, it follows that $E[X mid Y] equiv 0$.

To avoid above technicalities which needs advanced probability theory, you may construct a discrete counterexample using the same idea. How about a $2$-by-$4$ table? The columns are possible values of $X$ while the rows are possible values of $Y$.

begin{array}{c|c c c c}
& -1 & 1 & -2 & 2 \
hline
0 & 1/4 & 1/4 & 0 & 0 \
1 & 0 & 0 & 1/4 & 1/4
end{array}

Let $X,Y$ have an uniform density over the diamond $(0,1),(1,0),(0,-1),(-1,0)$ - or over the unit circle.

Then $X,Y$ are not independent [*], but $E(X mid Y) = E(X )=0$

For another example, you could take the triangle $(-1,0),(0,1),(1,0)$. In this case $E(X mid Y) = E(X)$ but $E(Y mid X) ne E(Y)$

Notice BTW that $E(X mid Y) = E(X) implies E(X Y) = E(X)E(Y)$ (uncorrelated), but not the reverse.

[*] If the marginals have support over finite intervals, then the support of the joint density must be a rectangle for the variables to be independent. Put in other way, if for some point $f_{X,Y}(x_0,y_0)=0$ but $f_X(x_0)>0$ and $f_Y(y_0)>0$ then the variables cannot be independent.