Optional Sampling Theorem¶

Martingales are fair games at deterministic times: $\mathbb{E}[M_t \mid \mathcal{F}_s] = M_s$. What happens when we replace $s, t$ by random times $\sigma \le \tau$? The Optional Sampling Theorem answers precisely this question: when does $\mathbb{E}[M_\tau \mid \mathcal{F}_\sigma] = M_\sigma$?

The Bounded Case¶

Theorem (Optional Sampling, bounded version). Let $M$ be a right-continuous martingale and let $\sigma \le \tau \le T$ be stopping times with $T < \infty$. Then

\[ \mathbb{E}[M_\tau \mid \mathcal{F}_\sigma] = M_\sigma \quad \text{a.s.} \]

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

Proof (sketch). Approximate $\tau$ from above by discrete stopping times $\tau^{(n)} = \lceil 2^n \tau \rceil / 2^n$. On each atom $\{\tau^{(n)} = k/2^n\}$ the martingale property is a direct discrete-time calculation, so $M^{\tau^{(n)}}$ is a martingale. Right-continuity and the uniform bound $|M_{t \wedge \tau^{(n)}}| \le \sup_{u \le T} |M_u| \in L^1$ let us pass $n \to \infty$ by dominated convergence, giving the result for $\sigma \le \tau \le T$. $\square$

Why Boundedness Matters¶

Counterexample: symmetric random walk

Let $S_n$ be a symmetric simple random walk ($S_0 = 0$) and $\tau = \inf\{n : S_n = 1\}$. Then $S_n$ is a martingale and $\mathbb{P}(\tau < \infty) = 1$, yet

\[ \mathbb{E}[S_\tau] = 1 \ne 0 = \mathbb{E}[S_0]. \]

The culprit is $\mathbb{E}[\tau] = \infty$. With no bound on $\tau$ and no integrability control, the "fair game" identity fails.

Sufficient Conditions Beyond Bounded τ¶

The following each suffice for $\mathbb{E}[M_\tau \mid \mathcal{F}_\sigma] = M_\sigma$ on (possibly unbounded) stopping times $\sigma \le \tau$:

Condition	Setting
$\tau \le T$ bounded	always
$M$ uniformly integrable	general
$	M_{t \wedge \tau}
bounded increments and $\mathbb{E}[\tau] < \infty$	discrete time

Each reduces to the bounded case via a truncation $\tau \wedge T$ followed by $T \to \infty$.

The Application Template¶

Optional sampling is a workhorse. The recipe:

Pick a martingale encoding the quantity of interest. Standard choices for Brownian motion (see § Brownian Motion Martingales):
$W_t$ itself,
$W_t^2 - t$,
the exponential martingale $\exp(\theta W_t - \theta^2 t / 2)$.
Pick a stopping time $\tau$ (typically a hitting or exit time).
Apply optional sampling to $\tau \wedge T$: $\mathbb{E}[M_{\tau \wedge T}] = \mathbb{E}[M_0]$.
Pass $T \to \infty$ (dominated/monotone/Fatou), and read off the answer.

Three Canonical Applications¶

1. Hitting probability. For $W_0 = x \in (a, b)$ and $\tau = \inf\{t : W_t \notin (a, b)\}$, the bounded martingale $W_{t \wedge \tau}$ gives $\mathbb{E}[W_\tau] = x$. Writing $p = \mathbb{P}(W_\tau = b)$, we get $bp + a(1-p) = x$, hence

\[ \mathbb{P}(\text{hit } b \text{ before } a \mid W_0 = x) = \frac{x - a}{b - a}. \]

2. Expected exit time. For $\tau = \inf\{t : W_t \notin (-a, a)\}$, apply optional sampling to $W_t^2 - t$ at $\tau \wedge T$ and let $T \to \infty$. Since $|W_\tau| = a$,

\[ \mathbb{E}[\tau] = \mathbb{E}[W_\tau^2] = a^2. \]

3. Laplace transform of a hitting time. For $\tau_a = \inf\{t : W_t = a\}$ with $a > 0$, apply optional sampling to $Z_t = \exp(\sqrt{2\lambda}\,W_t - \lambda t)$ at $\tau_a \wedge T$. On $\{\tau_a < \infty\}$, $W_{\tau_a} = a$, and as $T \to \infty$ the contribution from $\{\tau_a > T\}$ vanishes. The result is

\[ \mathbb{E}[e^{-\lambda \tau_a}] = e^{-a\sqrt{2\lambda}}. \]

This is the Laplace transform of the inverse Gaussian distribution.

Sub/Supermartingale Version¶

If $X$ is a supermartingale and $\sigma \le \tau \le T$ are bounded,

\[ \mathbb{E}[X_\tau \mid \mathcal{F}_\sigma] \le X_\sigma, \]

with the inequality reversed for submartingales. In particular, $|M|$ (submartingale) satisfies $\mathbb{E}[|M_\tau|] \ge \mathbb{E}[|M_\sigma|]$.

Exercises¶

Exercise 1. Let $\tau = \inf\{t : W_t \notin (-1, 1)\}$.

(a) Find $\mathbb{E}[W_\tau]$ using optional sampling. (b) Find $\mathbb{E}[\tau]$. (c) Find $\mathbb{P}(W_\tau = 1)$.

Solution to Exercise 1

(a) $W_{t \wedge \tau}$ is bounded by $1$, so optional sampling gives $\mathbb{E}[W_\tau] = W_0 = 0$.

(b) Apply optional sampling to $M_t = W_t^2 - t$ at $\tau \wedge T$:

\[ \mathbb{E}[W_{\tau \wedge T}^2] = \mathbb{E}[\tau \wedge T]. \]

As $T \to \infty$, monotone convergence gives $\mathbb{E}[\tau]$ on the right, and bounded convergence gives $\mathbb{E}[W_\tau^2] = 1$ on the left. Hence $\mathbb{E}[\tau] = 1$.

(c) By symmetry (or from (a)): $2p - 1 = 0$, so $p = 1/2$.

Exercise 2 (Gambler's ruin). A gambler starts with $$k$ and bets $$1$ per fair coin flip, stopping at $0$ or $N$.

(a) Find the probability of reaching $N$. (b) Find the expected number of bets.

Solution to Exercise 2

Let $S_n$ be wealth and $\tau$ the exit time from $\{0, \ldots, N\}$.

(a) $S_n$ is a bounded martingale on $[0, \tau]$. Optional sampling: $\mathbb{E}[S_\tau] = k$. With $p_k = \mathbb{P}(S_\tau = N)$,

\[ N p_k = k \implies p_k = k/N. \]

(b) $S_n^2 - n$ is a martingale. Optional sampling: $\mathbb{E}[S_\tau^2] = k^2 + \mathbb{E}[\tau]$. Also $\mathbb{E}[S_\tau^2] = N^2 \cdot (k/N) = kN$. Therefore $\mathbb{E}[\tau] = k(N - k)$.

Exercise 3 (Wald's identities). Let $S_n = \sum_{i=1}^n X_i$ with $X_i$ i.i.d., $\mathbb{E}[X_1] = 0$, $\mathrm{Var}(X_1) = \sigma^2$, and let $\tau$ be a stopping time with $\mathbb{E}[\tau] < \infty$.

(a) Prove $\mathbb{E}[S_\tau] = 0$. (b) Prove $\mathbb{E}[S_\tau^2] = \sigma^2 \mathbb{E}[\tau]$.

Solution to Exercise 3

(a) $S_n$ is a martingale with integrable increments; $\mathbb{E}[\tau] < \infty$ gives the discrete bounded-increments optional sampling criterion, so $\mathbb{E}[S_\tau] = \mathbb{E}[S_0] = 0$.

(b) $M_n = S_n^2 - n\sigma^2$ is a martingale:

\[ \mathbb{E}[S_{n+1}^2 \mid \mathcal{F}_n] = S_n^2 + \sigma^2. \]

Optional sampling gives $\mathbb{E}[S_\tau^2] = \sigma^2 \mathbb{E}[\tau]$.

Exercise 4 (Hitting time Laplace transform). Let $\tau_a = \inf\{t : W_t = a\}$, $a > 0$.

(a) Use optional sampling on $Z_t = \exp(\sqrt{2\lambda}\,W_t - \lambda t)$ to show $\mathbb{E}[e^{-\lambda \tau_a}] = e^{-a\sqrt{2\lambda}}$. (b) Deduce $\mathbb{E}[\tau_a] = \infty$ even though $\mathbb{P}(\tau_a < \infty) = 1$.

Solution to Exercise 4

(a) On $\tau_a \wedge T$, $Z$ is bounded by $e^{a\sqrt{2\lambda}}$, so $\mathbb{E}[Z_{\tau_a \wedge T}] = 1$. On $\{\tau_a > T\}$, $Z_T \to 0$ a.s. as $T \to \infty$ (since $W_T = o(T)$), and $\mathbb{P}(\tau_a < \infty) = 1$. Passing to the limit,

\[ \mathbb{E}[e^{\sqrt{2\lambda} \cdot a - \lambda \tau_a}] = 1 \implies \mathbb{E}[e^{-\lambda \tau_a}] = e^{-a\sqrt{2\lambda}}. \]

(b) $\mathbb{E}[\tau_a] = -\partial_\lambda \mathbb{E}[e^{-\lambda \tau_a}]\big|_{\lambda \to 0^+} = a/\sqrt{2\lambda}\big|_{\lambda \to 0^+} = \infty$.

Exercise 5 (Two-sided exit). Let $\tau = \inf\{t : W_t \notin (-a, b)\}$ with $a, b > 0$ and $W_0 = 0$.

(a) Find $\mathbb{P}(W_\tau = b)$. (b) Find $\mathbb{E}[\tau]$.

Solution to Exercise 5

(a) By optional sampling on $W$: $b p - a(1-p) = 0$, so $p = a/(a + b)$.

(b) By optional sampling on $W_t^2 - t$: $\mathbb{E}[\tau] = \mathbb{E}[W_\tau^2] = b^2 \cdot \frac{a}{a+b} + a^2 \cdot \frac{b}{a+b} = ab$.

Exercise 6 (Why the counterexample fails). Let $S_n$ be a symmetric simple random walk and $\tau = \inf\{n : S_n = 1\}$. Explain which hypothesis of optional sampling fails for this $\tau$, and why optional sampling does hold for $\tau_N = \tau \wedge N$ for every finite $N$.

Solution to Exercise 6

For the unbounded $\tau$: $\mathbb{P}(\tau < \infty) = 1$ but $\mathbb{E}[\tau] = \infty$ (and $S$ is not UI on $[0, \tau]$). None of the sufficient conditions apply, so $\mathbb{E}[S_\tau] \ne \mathbb{E}[S_0]$ is possible — and indeed $\mathbb{E}[S_\tau] = 1$.

For the truncation $\tau_N = \tau \wedge N$: this is bounded by $N$, so the bounded-case theorem applies and $\mathbb{E}[S_{\tau_N}] = 0$ for every finite $N$. The limit $N \to \infty$ is where things break: $S_{\tau_N} \to S_\tau = 1$ a.s., but $\{S_{\tau_N}\}_N$ is not uniformly integrable, so expectations do not pass to the limit.

Local martingale

A local martingale is an adapted process $M$ for which there exists a sequence of stopping times $\tau_n \uparrow \infty$ (a localizing sequence) such that each stopped process $M^{\tau_n} = (M_{t \wedge \tau_n})_{t \ge 0}$ is a (true) martingale.

Every martingale is a local martingale (take $\tau_n \equiv n$), but not every local martingale is a true martingale: the localization sequence may fail to extend to the whole time line in an integrable way. The canonical home for the theory is § Local martingales and stochastic exponential; see Exercise 8 below for a concrete failure of optional sampling beyond the bounded case.

Exercise 7. Let $M_t$ be a martingale and $\tau$ a bounded stopping time. Show that the stopped process $M_{t \wedge \tau}$ is also a martingale. Then explain in words why stopping prevents "bad behavior" in local martingales.

Solution to Exercise 7

Let $N_t := M_{t \wedge \tau}$. Since $M$ is adapted and $\tau$ is a stopping time, $N_t$ is adapted. Since $M$ is a martingale and $\tau$ is a bounded stopping time, the optional stopping theorem applies to the bounded stopping times $s \wedge \tau \leq t \wedge \tau$, giving

\[ \mathbb{E}[M_{t \wedge \tau} \mid \mathcal{F}_s] = M_{s \wedge \tau} \]

Hence $\mathbb{E}[N_t \mid \mathcal{F}_s] = N_s$. Therefore $N_t = M_{t \wedge \tau}$ is a martingale.

Why stopping helps. A process may behave badly only when it wanders into extreme regions where integrability fails or huge tail events appear. Stopping cuts the path off before it reaches those regions. So the stopped process remains "well-behaved," and martingale arguments become valid again. Localization is exactly the idea of choosing stopping times that hide the bad behavior until farther and farther out.

Exercise 8. Let $M_t$ be a local martingale and $\tau$ an unbounded stopping time. Give an example showing that $\mathbb{E}[M_\tau] = \mathbb{E}[M_0]$ may fail.

Solution to Exercise 8

A classical example uses Brownian motion $W_t$, which is a true martingale. Let

\[ \tau = \inf\{t \geq 0 : W_t = 1\}. \]

This stopping time is almost surely finite, but unbounded. Since $W_\tau = 1$ a.s. and $W_0 = 0$:

\[ \mathbb{E}[W_\tau] = 1 \neq 0 = \mathbb{E}[W_0]. \]

Why it fails. Localize by $\tau_n = \tau \wedge n$. Each $\tau_n$ is bounded, so optional stopping applies and $\mathbb{E}[W_{\tau_n}] = 0$ for every $n$. But $W_{\tau_n} \to W_\tau = 1$ a.s., so

\[ \lim_{n \to \infty} \mathbb{E}[W_{\tau_n}] = 0 \quad \text{yet} \quad \mathbb{E}\!\left[\lim_{n \to \infty} W_{\tau_n}\right] = 1. \]

The exchange of limit and expectation fails because $\{W_{\tau_n}\}_n$ is not uniformly integrable. To hit level $+1$, Brownian motion often wanders far negative first. These rare but large negative excursions keep $\mathbb{E}[W_{\tau_n}] = 0$ for each $n$. But as $n \to \infty$, the negative excursions disappear pathwise (every path eventually hits $+1$), yet their compensating contribution to expectation is lost.

When optional stopping does work. The identity $\mathbb{E}[M_\tau] = \mathbb{E}[M_0]$ holds if any of the following is satisfied: (i) $\tau$ is bounded a.s.; (ii) the family $\{M_{t \wedge \tau}\}_{t \geq 0}$ is uniformly integrable; (iii) $M$ is a uniformly integrable martingale. Without such conditions, expectation need not be preserved.

Condition	Setting
\(\tau \le T\) bounded	always
\(M\) uniformly integrable	general
$	M_{t \wedge \tau}
bounded increments and \(\mathbb{E}[\tau] < \infty\)	discrete time