Optional Stopping Theorem

The optional stopping theorem (OST) says: if $X_n$ is a martingale and $T$ is a stopping time satisfying certain regularity conditions, then: $$E[X_T] = E[X_0]$$ Intuition: A martingale has no drift — it's a fair game. The optional stopping theorem says you can't "beat" a fair game by choosing a clever time to stop. No matter how sophisticated your stopping strategy, your expected payoff equals your starting value. The catch — regularity conditions matter! The theorem requires one of: 1. $T$ is bounded ($T \leq N$ for some fixed $N$), or 2. $E[T] < \infty$ and the increments are bounded, or 3. $X_{\min(n,T)}$ is dominated by an integrable random variable Without these, the theorem fails! (The martingale $S_n = \sum X_i$ with the "double until you win" strategy illustrates the failure.) Concrete example — Gambler's ruin: Start with \$a, bet \$1 on fair coin flips. Stop when wealth hits \$0 or \$N. Wealth is a martingale. By OST: $$E[X_T] = a = 0 \cdot P(\text{ruin}) + N \cdot P(\text{reach } N)$$ So $P(\text{reach } N) = a/N$ and $P(\text{ruin}) = 1 - a/N$. Another application: For a symmetric random walk, let $T$ be the first time $S_T = 1$. The process $S_n^2 - n$ is a martingale. By OST: $E[S_T^2 - T] = 0$, so $E[T] = E[S_T^2] = 1$. The expected time to first reach $+1$ is 1 step. When to use: Computing ruin probabilities, expected hitting times, and absorption probabilities in random walks. Proving that certain gambling strategies can't work. Deriving the gambler's ruin formula. Central to pricing American options (optimal exercise is a stopping time).