I note that when $(X,Y)$ are bivariate normal, $E(Y|X=E(X))=E(Y)$. So I wonder: is there a general relationship that exists between $E(Y|X=E(X))$ and $E(Y)$? What are some sufficient conditions to guarantee $E(Y|X=E(X))=E(Y)$?
To start, I assume $X,Y,(X,Y)$ have densities $f_X$, $f_Y$ and $f_{(X,Y)}$. Then \begin{align} E(Y|X=E(X))=\int y\frac{f_{(X,Y)}(E(X),y)}{f_X(E(X))}dy. \end{align} In the trivial case that $X$ and $Y$ are independent, $f_{(X,Y)}(E(X),y)$ factors into $f_X(E(X))f_Y(y)$ and we do have $E(Y|X=E(X))=E(Y)$. But I can't simplify further for the general case.
One situation leading to that conclusion: $E(Y|X=x)=a+bx$ for all $x$, for certain constants $a$ and $b$. (I am assuming that both $X$ and $Y$ have finite means.) In this case we must have $E(Y) = a+bE(X)$, and so $E(Y|X=x)=E(Y)+b(x-E(X))$ for all $x$.