# 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 $\mathscr{G}$?

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 $\mathscr{G}$?

• 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). – Nate Eldredge Oct 16 '15 at 2:14
• There is an obvious, semi-trivial 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). – Christian Remling Oct 18 '15 at 5:38
• I of course forgot to divide by $\sum_k w_{jk}$ above. – Christian Remling Oct 18 '15 at 6:00

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 $$i=2$$).