Сoincidence of discrete random variables

Question

Let $\xi, \eta$ be a discrete random values and $\mathbb E| ξ |$, $\mathbb E | η | < +\infty$, and any value of these values ​​are accepted with a non-zero probability. How to prove that from $\mathbb E (ξ \mid η) ≥ η$, $\mathbb E (η \mid ξ) ≥ ξ$ follows $ξ = η$?

Answer 1

Let us prove the desired conclusion generally, without assuming that the random variables $\xi$ and $\eta$ are discrete. Let $g\colon\mathbb R\to\mathbb R$ be any strictly increasing strictly convex differentiable function such that $|g(x)|\le1+|x|$ for all real $x$, so that $|g'|\le1$ and $Eg(\xi),Eg(\eta)$ exist in $\mathbb R$. (For instance, one may take $g(x):=E(x+Z)_+=\int_{-x}^\infty(x+z)\varphi(z)\,dz$, so that $0<g(x)\le|x|+EZ_+\le|x|+1$ and $g'(x)=\int_{-x}^\infty\varphi(z)\,dz=\int_{-\infty}^x \varphi(z)\,dz$, where $Z\sim N(0,1)$ and $\varphi$ is the pdf of $Z$.)

By the convexity of $g$, \begin{equation} g(\xi)\ge g(E_\eta\xi)+g'(E_\eta\xi)(\xi-E_\eta\xi), \tag{1} \end{equation} where $E_\eta\xi:=E(\xi|\eta)$. Moreover, by the strict convexity of $g$, inequality (1) is strict on the event $\{\xi\ne E_\eta\xi\}$. Taking the expectations of both sides of (1), we have \begin{equation} Eg(\xi)\ge Eg(E_\eta\xi)+EE_\eta g'(E_\eta\xi)(\xi-E_\eta\xi) =Eg(E_\eta\xi)+Eg'(E_\eta\xi)E_\eta (\xi-E_\eta\xi) =Eg(E_\eta\xi), \end{equation} so that we have the conditional Jensen inequality \begin{equation} Eg(\xi)\ge Eg(E_\eta\xi), \tag{2} \end{equation} and this inequality is strict unless $\xi=E_\eta\xi$ almost surely (a.s.). Similarly, \begin{equation} Eg(\eta)\ge Eg(E_\xi\eta), \tag{3} \end{equation} and this inequality is strict unless $\eta=E_\xi\eta$ a.s.

Using now (2), the conditions $E_\eta\xi\ge\eta$ and $E_\xi\eta\ge\xi$ given in the OP, the condition that $g$ is increasing, and (3), we have \begin{equation} Eg(\xi)\ge Eg(E_\eta\xi)\ge Eg(\eta)\ge Eg(E_\xi\eta)\ge Eg(\xi), \tag{4} \end{equation} and (since $g$ is strictly increasing) the 2nd and 4th inequalities here are strict unless, respectively, $E_\eta\xi=\eta$ a.s. and $E_\xi\eta=\xi$ a.s. However, all the inequalities in (4) must be the equalities. It follows that $\xi=E_\eta\xi=\eta$ a.s., whence $\xi=\eta$ a.s., as desired.