If the expectation and variance of $X$ are both not affected by $Y$, and vice versa, then must $X$ and $Y$ be independent?

Question

I know that if $\mathbb{E}[X]=\mathbb{E}[X|Y] , \mathbb{E}[Y]=\mathbb{E}[Y|X]$, $X$ and $Y$ can be dependent, for example a ‘uniform’ distribution in a unit circle. Now we add the variance, if $$\mathbb{E}[X]=\mathbb{E}[X|Y], \mathbb{E}[Y]=\mathbb{E}[Y|X], $$$$Var(X)=Var(X|Y), Var(Y)=Var(Y|X).$$ Say the expectation and variance of $X$ are both not affected by $Y$, and vice versa, then must $X$ and $Y$ be independent? In this case I can not find a counterexample just like the uniform circle.

If they are independent, how to prove it? If not, is there a counterexample?

Thanks!

Answer 1

Consider $X\sim N(0,1)$ and $Y\sim N(0,2)$ if $X>0$ and $Y\sim t_4$ otherwise. Then $X, Y$ are dependent, but your conditions hold (the $t_4$ distribution has mean 0 and variance 2).

Answer 2

Here's a discrete example: $$100P(x,y)=\begin{array}{c|ccccc}x\backslash y&-2&-1&0&1&2\\ \hline -2&1&0&6&0&1\\-1&0&9&0&9&0\\0&6&0&36&0&6\\1&0&9&0&9&0\\2&1&0&6&0&1\end{array}$$ Condition on any particular $x$, and $y$ has mean zero and variance $1$. Condition on any particular $y$, and $x$ has mean zero and variance $1$. But, of course, they're not independent, as the distribution's support isn't a Cartesian product.