Optional stopping theorem

Remarks

The stopped process $(X_{n \wedge \tau})_{n \ge 0}$ is again a martingale, so $E[X_{n \wedge \tau}] = E[X_0]$ for every $n$. Each hypothesis lets one pass to the limit $n \to \infty$. In case 1 take $n = N$. In cases 2 and 3 use dominated convergence, with the bound $\abs{X_0} + \sum_{k < \tau} \abs{X_{k+1} - X_k}$ in case 2 and the constant $c$ in case 3. Example: let $S_n$ be the simple symmetric random walk from $0$, let $a, b$ be positive integers, and let $\tau$ be the first time $S_n \in \{-a, b\}$. Case 3 applies, so $0 = E[S_\tau] = b\,P(S_\tau = b) - a\,(1 - P(S_\tau = b))$, which gives $P(S_\tau = b) = a/(a + b)$. Some hypothesis is needed: if $\tau$ is the first time the same walk hits $1$, then $\tau < \infty$ almost surely but $E[S_\tau] = 1 \ne 0 = E[S_0]$; here $E[\tau] = \infty$ and the stopped walk is unbounded below.

Going deeper

For a student

The theorem is the main tool for computing hitting probabilities and expected hitting times of random walks and Brownian motion. For example, applying it to the martingale $S_n^2 - n$ in the gambler's ruin gives $E[\tau] = ab$. It gives a short proof of Doob's martingale inequality. In continuous time the same result holds for right-continuous martingales. For uniformly integrable martingales it holds for all stopping times, with $X_\infty$ the limit from Doob's martingale convergence theorems.