Conditional Expectation to show two random variables are equal almost surely

Question

I am trying to show the following:

Let $\xi, \eta$ by random variables with finite expectation such that $E(\xi \mid \eta) \ge \eta$ and $E(\eta \mid \xi) \ge \xi$. Prove that $\eta = \xi$ almost surely.

From definition I know that $E(\xi \mid \eta) = \int \xi \frac{f(\xi,\eta)}{g(\xi)}\,d\xi$, where $g$ is marginal distribution for $\xi$ and $E(\xi \mid \eta) = \int \eta \frac{f(\xi,\eta)}{g(\eta)} \, d\eta$, where $g$ is marginal distribution for $\eta.$ I am just not sure how to combine these expressions with the inequalities to show almost sure equality between the random variables.

Answer 1

A hand-made exploration here would look as follows: We can always write (I will use $X$ and $Y$)

$$X = E(X\mid Y) + e_{X|Y} \implies E(X\mid Y) = X-e_{X|Y}$$ $$Y \equiv E(Y\mid X) + e_{Y|X} \implies E(Y\mid X) = Y-e_{Y|X}$$

where $e_{X|Y},\;\;e_{Y|X}$ are the conditional expectation function errors, which by construction have expected value zero, $E[e_{X|Y}]=E[e_{Y|X}]=0$.

By the assumed inequalities we then have

$$X-e_{X|Y} \geq Y \implies X \geq Y + e_{X|Y},\;\;\; Y-e_{Y|X}\geq X$$

Combining

$$Y + e_{X|Y} \leq X \leq Y-e_{Y|X} \implies e_{X|Y}+e_{Y|X} \leq 0 $$

So under the assumed inequalities, the random variable $Z \equiv e_{X|Y}+e_{Y|X}$ is non-positive. But also $E(Z) = E(e_{X|Y})+E(e_{Y|X}) = 0$. For a non-positive random variable to have expected value zero, all probability mass must be concentrated at the value zero, making $Z$ a constant random variable equal to zero. But then

$$Z = e_{X|Y} + e_{Y|X}=0 \implies e_{X|Y} = - e_{Y|X}$$

But then we obtain

$$Y + e_{X|Y} \leq X \leq Y+e_{X|Y} \implies X = Y+e_{X|Y}$$

while also $X = E(X\mid Y) + e_{X|Y}$ which leads to $Y = E(X\mid Y)$.

By analogous manipulations, we obtain $X = E(Y\mid X)$.

Then this post contains the proof for the case of equality.