TITLE: If $X∼F_1$, $Y∼F_2$, under what conditions on $F_1$, $F_2$ can we construct $Y=E(X\mid\mathscr{G})$ for some $\mathscr{G}$?
QUESTION [8 upvotes]: Suppose that we have distributions $F_1 $ and $F_2$. Under what conditions on $F_1,F_2$ is it possible to construct random variables $X\sim F_1,Y\sim F_2$ such that $Y=E(X|\mathscr{G})$, that is, $Y$ is the conditional expectation of $X$ with respect to some $\sigma$-field $\mathscr{G}$?

REPLY [7 votes]: The condition proposed by Nate Eldredge is not only necessary, but also sufficient. This is classical Strassen's theorem:

Theorem For integrable random variables $X$ and $Y$ the following is equivalent:

There exist $\hat X \overset{d}{=} X$ and $\hat Y \overset{d}{=} Y$ such that $E[\hat X | \hat Y] = \hat Y$.

$Y\prec X$ in convex stochastic order, i.e. $E[f(Y)]\le E[f(X)]$ for any convex function $f$.



Note that the formulation in Strassen's paper differs from this one (and Strassen attributed it himself to Hardy-Littlewood-Polya-Blackwell-Stein-Sherman-Cartier-Fell-Meyer). In the present form it is given e.g. in Ruschendorf (Theorem 6 for $i=2$).