Martingales & Optional Stopping
Martingales & Optional Stopping
A martingale is a process that models a fair game: given all current information, the best prediction of the future value is the present value. Martingales are central to modern probability — they provide the right framework for pricing derivatives (risk-neutral measure), proving convergence theorems, and analyzing stopping rules. The Optional Stopping Theorem (OST) gives precise conditions under which the martingale property is preserved at a random time.
Key examples: $W_t$ (Brownian motion), $W_t^2-t$, $e^{\theta W_t - \theta^2 t/2}$ (exponential martingale), and $S_n=\sum_{k=1}^n X_k$ for zero-mean i.i.d. $X_k$ are all martingales with respect to their natural filtrations.
Martingale
An adapted integrable process $\{M_t,\mathcal{F}_t\}$ is a martingale if $$E[M_t\mid\mathcal{F}_s]=M_s \quad \forall s\le t \text{ a.s.}$$ It is a supermartingale if $\le$ holds and a submartingale if $\ge$ holds. Any convex function of a martingale is a submartingale (Jensen).
Optional Stopping Theorem
Let $\{M_n\}$ be a martingale and $\tau$ a stopping time. Then $E[M_\tau]=E[M_0]$ provided any one of: (i) $\tau$ is bounded; (ii) $|M_n|\le C$ a.s. and $E[\tau]<\infty$; (iii) $E[\tau]<\infty$ and $E[|M_{n+1}-M_n|\mid\mathcal{F}_n]\le C$. Without such conditions $E[M_\tau]=E[M_0]$ may fail (doubling strategy).
Example 1 — Wald's Identity
Let $S_n=X_1+\cdots+X_n$ with $E[X_i]=\mu$. Then $M_n=S_n-n\mu$ is a martingale. By OST, $E[S_\tau]=\mu E[\tau]$ whenever $E[\tau]<\infty$. For $p=1/2$: $\mu=0$, so $E[S_\tau]=0$ — absorption is symmetric.
Example 2 — Gambler's Ruin
Simple random walk on $\{0,\ldots,N\}$, absorbing at $0$ and $N$. Both $M_n=S_n$ and $M_n^2-n$ are martingales. OST gives the ruin probability $P_k(\text{hit }N)=k/N$ and expected absorption time $E_k[\tau]=k(N-k)$, starting from position $k$.
Practice
- Verify $e^{\theta W_t-\theta^2 t/2}$ is a martingale using the MGF of $\mathcal{N}(0,t-s)$.
- State Doob's martingale convergence theorem and its integrability condition.
- Apply OST to find $E[\tau]$ for a simple symmetric RW hitting $\{-a,b\}$.
Show Answer Key
1. Let $M_t=e^{\theta W_t-\theta^2 t/2}$. Then $E[M_t|\mathcal{F}_s]=M_s\cdot E[e^{\theta(W_t-W_s)-\theta^2(t-s)/2}]=M_s\cdot e^{\theta^2(t-s)/2}\cdot e^{-\theta^2(t-s)/2}=M_s$ using the MGF of $\mathcal{N}(0,t-s)$.
2. If $(M_n)$ is a supermartingale bounded in $L^1$, then $M_n\to M_\infty$ a.s. For submartingales, the condition is $\sup E[M_n^+]<\infty$.
3. For symmetric RW hitting $\{-a,b\}$: the martingale $S_n$ has $E[S_\tau]=0$ by OST (bounded stopping). So $(-a)P(\text{hit }\!-a)+b\cdot P(\text{hit }b)=0$, giving $E[\tau]=ab$ (using $S_n^2-n$ martingale).