Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 201 / 2

For a finite horizon $N$, let $U_N[a,b]$ be the number of completed upcrossings by time $N$. Use the predictable strategy that holds one unit of the process after a visit below $a$ until the next visit above $b$. For a supermartingale, the expected gain of this nonnegative predictable martingale transform is nonpositive. Pathwise, the completed trades earn at least $(b-a)U_N[a,b]$, while an unfinished final trade can lose at most $(X_N-a{)}^{-}$. HenceTaking expectations and using $X_N\ge 0$ gives the Doob upcrossing inequalityAs $N\to \infty$, monotone convergence yields

For every rational $0\le a<b$, part a implies $U[a,b]<\infty$ almost surely. The intersection of these probability-one events over the countable collection of rational pairs still has probability one. On this event, ifsome rational interval $[a,b]$ lies strictly between them, forcing infinitely many upcrossings, a contradiction. Thus $X_n$ has an extended limit almost surely.

The limit cannot be $+\infty$ on a set of positive probability: Fatou lemma and the supermartingale property giveNonnegativity excludes $-\infty$. Therefore $X_n$ converges almost surely to a finite random variable, proving the almost sure supermartingale convergence theorem in this case.