Let $(X, Y)$ be an ordered pair of real-valued random variables. Say that $(X, Y)$ is fair if $E(Y \mid X) = X$ a.s. It is shown, for example, that if $X$ has a finite mean and the pair $(X, Y)$ is fair, then $X$ and $Y$ cannot be stochastically ordered unless $X = Y$ a.s. The conclusion is in general false, if $X$ does not have a mean. On the other hand, if $X$ is independent of the increment $Y - X$, the preceding statement remains in force without any moment restrictions on $X$. The last assertion, combined with a gambling idea of Dubins and Savage, yields a simple proof of a theorem of Chung and Fuchs on the upper limit of a random walk with mean zero.
"Some Conditions Under Which Two Random Variables are Equal Almost Surely and a Simple Proof of a Theorem of Chung and Fuchs." Ann. Math. Statist. 42 (5) 1647 - 1655, October, 1971. https://doi.org/10.1214/aoms/1177693163