Optional Stopping Theorem

The optional stopping theorem states that, under suitable conditions, the expected value of a martingale at a stopping time equals its initial value. This fundamental result shows that "you cannot beat a fair game by choosing when to quit," formalized mathematically.

Statement

Theorem3.1Optional stopping theorem

Let $(M_n)$ be a martingale and $\tau$ a stopping time. If any of the following conditions hold:

Bounded stopping time: $\tau \leq N$ almost surely for some constant $N$ .
Bounded increments: $\mathbb{E}[\tau] < \infty$ and $|M_{n+1} - M_n| \leq C$ a.s. for some constant $C$ .
Dominated: $\mathbb{E}[\tau] < \infty$ and there exists an integrable $Y$ with $|M_{n \wedge \tau}| \leq Y$ a.s. for all $n$ .

Then $\mathbb{E}[M_\tau] = \mathbb{E}[M_0]$ .

Proof strategy: Show that $\mathbb{E}[M_{n \wedge \tau}] = \mathbb{E}[M_0]$ for all $n$ (the stopped process is a martingale), then take $n \to \infty$ and use dominated or bounded convergence to pass the limit through the expectation.

Applications

ExampleSymmetric gambler's ruin

Let $X_n$ be a simple symmetric random walk starting at $k \in \{1, \ldots, N-1\}$ . Define $\tau = \inf\{n : X_n \in \{0, N\}\}$ . Then $\tau < \infty$ a.s. by recurrence, and $X_n$ has bounded increments. By optional stopping:

$\mathbb{E}[X_\tau] = k.$

But $X_\tau \in \{0, N\}$ , so if $p_k = \mathbb{P}(X_\tau = N \mid X_0 = k)$ :

$k = 0 \cdot (1-p_k) + N \cdot p_k \Rightarrow p_k = k/N.$

This gives the ruin probability for symmetric random walk.

ExampleAsymmetric walk

For an asymmetric walk with drift $\mu = p - q \neq 0$ , consider the martingale $M_n = X_n - n\mu$ . Optional stopping gives $\mathbb{E}[X_\tau - \tau \mu] = k$ , leading to formulas for both $p_k$ and $\mathbb{E}[\tau \mid X_0 = k]$ .

RemarkFailure without conditions

If $\tau$ is unbounded or has unbounded increments, the theorem may fail. Example: $M_n = X_n$ (symmetric random walk), $\tau = \inf\{n : X_n = 1\}$ . Then $\mathbb{P}(\tau < \infty) = 1$ but $\mathbb{E}[\tau] = \infty$ . Naively applying optional stopping would give $1 = \mathbb{E}[M_\tau] \overset{?}{=} \mathbb{E}[M_0] = 0$ , a contradiction.

Summary

Optional stopping connects martingales and stopping times: $\mathbb{E}[M_\tau] = \mathbb{E}[M_0]$ under appropriate conditions. This result is fundamental to sequential analysis, optimal stopping, and gambling theory.