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 $\mathbb E (ξ | η) ≥ η$, $\mathbb E (η | ξ) ≥ ξ$ follows $ξ = η$?

Answer 1

Let $x^+=\max(x,0)$. Note that this function is monotone increasing, and convex. Therefore, $$ E[\eta^+]\le E[E[\xi|\eta]^+]\le E[E[\xi^+|\eta]]=E[\xi^+] $$ The first inequality used monotonicity, the second used the conditional Jensen's inequality.

By the same logic, $E[\xi^+]\le E[\eta^+]$, so we have equality. Furthermore, we have $$ E[\eta{\bf 1}(\eta>0)]=E[\eta^+]=E[\xi^+]=E[\xi{\bf 1}(\xi>0)]\le E[E[\eta|\xi]{\bf 1}(\xi>0)]=E[\eta{\bf1}(\xi>0)] $$ Subtracting the last from the first, you get $$ E[\eta({\bf 1}(\eta>0)-{\bf 1}(\xi>0))]\le 0. $$ A little thought shows that the random variable inside the expectation is always nonnegative, so the fact its expectation is nonpositive implies the argument is zero identically, that is, $$\eta{\bf 1}(\eta>0)=\eta{\bf 1}(\xi>0).$$ Assuming $P(\eta=0)=0$, this further implies $\eta$ and $\xi$ always have the same sign, because you can divide by $\eta$ to conclude $\eta>0$ iff $\xi>0$.

Given $a\in \mathbb R$, and applying the same logic to $\eta-a$ and $\xi-a$, you can show that $\eta$ and $\xi$ always lie on the same side of $a$, for all $a$ for which $P(\eta=a)=0$. This implies $\eta $ and $\xi$ are equal almost surely.