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 $\mathscr{G}$?

7$\begingroup$ Obvious necessary conditions: $\int x\,F_1(dx) = \int x\,F_2(dx)$, and $\int \varphi(x)\,F_1(dx) \le \int \varphi(x)\,F_2(dx)$ for all $\varphi$ convex and bounded below (by conditional Jensen). $\endgroup$ – Nate Eldredge Oct 16 '15 at 2:14

$\begingroup$ There is an obvious, semitrivial answer to this for discrete distributions: you can split each probability $P(X=x_k)$ at will, say $P(X=x_k)=\sum_j w_{jk}$, and then do your partial averages to obtain $Y$, which will take the values $y_j=\sum_k w_{jk}x_k$ with the obvious probabilities. Essentially, this should tell the whole story. You can do similar things with disintegrations of measures in general (though there are measurability issues, as always). $\endgroup$ – Christian Remling Oct 18 '15 at 5:38

$\begingroup$ I of course forgot to divide by $\sum_k w_{jk}$ above. $\endgroup$ – Christian Remling Oct 18 '15 at 6:00
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 HardyLittlewoodPolyaBlackwellSteinShermanCartierFellMeyer). In the present form it is given e.g. in Ruschendorf (Theorem 6 for $i=2$).