Doob's upcrossing inequality

Remarks

The proof uses a betting strategy: buy one unit when the process is at most $a$, and sell when it reaches at least $b$. With $Y_k = (X_k - a)^+$, which is again a submartingale, each completed upcrossing earns at least $b - a$, so the gain is at least $(b - a) U_n[a, b]$. The opposite strategy holds exactly when the first one does not. It is predictable and bounded, so its expected gain on the submartingale $Y$ is non-negative. The two gains add up to $Y_n - Y_0$. As a consequence, if $\sup_n E[X_n^+] < \infty$, then the total number of upcrossings of every interval $[a, b]$ is finite almost surely. This is the key step of Doob's martingale convergence theorems: a sequence that crosses no rational interval infinitely often has a limit in $[-\infty, \infty]$.

Going deeper

For a student

The inequality turns a question about convergence into a counting question. A real sequence fails to converge in $[-\infty, \infty]$ exactly when it crosses some interval with rational endpoints infinitely often. The upcrossing bound and monotone convergence show this happens with probability zero for $L^1$-bounded submartingales. For a supermartingale the analogous bound is $(b - a)\, E[U_n[a, b]] \le E[(X_n - a)^-]$, where $y^- = \max(-y, 0)$.