A Checklist Before Applying Optional Stopping
Optional stopping is not a rule that every fair process remains fair when observed at a random time. Its conclusion depends on the relation between the stopping time and the martingale’s size or integrability. A reliable solution therefore begins by identifying the candidate time and filtration, then checking the hypotheses of a specific optional stopping result before computing an expectation.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space with filtration \((\mathcal{F}_n)_{n\ge0}\), and let \((M_n)_{n\ge0}\) be an integrable martingale adapted to that filtration. A random time \(\tau\) taking values in \(\{0,1,\ldots\}\cup\{\infty\}\) is a stopping time if \(\{\tau\le n\}\in\mathcal{F}_n\) for every \(n\). This condition says that by time \(n\), the information available then determines whether stopping has already occurred. It is not enough that \(\tau\) be a random time, or that its value be determined by the entire future.
For continuous time, the corresponding setup is a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) and an adapted martingale \((M_t)_{t\ge0}\). A random time \(\tau\) is a stopping time when \(\{\tau\le t\}\in\mathcal{F}_t\) for every \(t\ge0\). The bounded optional stopping theorem in continuous time requires the usual regularity assumptions on the filtration and martingale, such as right-continuity of the paths and filtration; these are assumed whenever that theorem is used below.
The bounded optional stopping theorem says that if \(\tau\le N\) almost surely for some deterministic finite \(N\), then \(\mathbb{E}M_\tau=\mathbb{E}M_0\) for an integrable discrete-time martingale. Its continuous-time counterpart applies to bounded stopping times under the stated regularity assumptions. These bounded-time results are standard results established earlier in the optional stopping material. For unbounded times, a useful sufficient condition is domination of all the stopped variables by one integrable random variable. The next result gives the argument that makes this condition work.
Proof. For each \(n\), \(\tau\wedge n\) is a bounded stopping time, so bounded optional stopping gives \(\mathbb{E}M_{n\wedge\tau}=\mathbb{E}M_0\). Since \(\tau<\infty\) almost surely, for almost every outcome there is an \(n\) after which \(n\wedge\tau=\tau\); hence \(M_{n\wedge\tau}\to M_\tau\) almost surely. The domination implies \(|M_\tau|\le Y\) almost surely by taking the limit, so \(M_\tau\) is integrable. Also, the family \((M_{n\wedge\tau})\) is uniformly integrable: for every \(K>0\), the event \(\{|M_{n\wedge\tau}|>K\}\) is contained in \(\{Y>K\}\), and \[ \mathbb{E}\big[|M_{n\wedge\tau}|\mathbf{1}_{\{|M_{n\wedge\tau}|>K\}}\big] \le \mathbb{E}\big[Y\mathbf{1}_{\{Y>K\}}\big]\longrightarrow 0 \] as \(K\to\infty\), uniformly in \(n\), by integrability of \(Y\). Uniform integrability and almost-sure convergence imply convergence in \(L^1\), so \(\mathbb{E}M_{n\wedge\tau}\to\mathbb{E}M_\tau\). Taking limits in the bounded-time equalities proves the claim. \(\square\)
Bounded Increments and an Integrable Horizon
In discrete time, a simple way to obtain the domination in the proposition is to control the size of each increment. This gives a practical sufficient condition for an unbounded stopping time. The condition is sufficient, not necessary: other forms of uniform integrability can also justify optional stopping.
Proof. The finite expectation of the nonnegative integer-valued \(\tau\) implies \(\tau<\infty\) almost surely. For every \(n\), the stopped value is the initial value plus at most \(\tau\) increments, so \[ |M_{n\wedge\tau}| \le |M_0|+\sum_{k=1}^{n\wedge\tau}|M_k-M_{k-1}| \le |M_0|+C\tau. \] The random variable \(Y=|M_0|+C\tau\) is integrable because \(\mathbb{E}|M_0|<\infty\), \(C<\infty\), and \(\mathbb{E}\tau<\infty\). The proposition now gives \(\mathbb{E}M_\tau=\mathbb{E}M_0\). \(\square\)
Worked Example: A Bounded Discrete-Time Stopping Rule
Let \((\Omega,\mathcal{F},\mathbb{P})\) support independent random variables \(\xi_1,\xi_2,\ldots\), each equal to \(1\) or \(-1\) with probability \(\tfrac12\). Set \(S_0=0\), \(S_n=\sum_{k=1}^n\xi_k\), and \(\mathcal{F}_n=\sigma(\xi_1,\ldots,\xi_n)\). Thus \(S_n\) is a martingale: its increments are integrable and \(\mathbb{E}[\xi_{n+1}\mid\mathcal{F}_n]=0\). Define \(\tau=\min(\inf\{n\ge1:S_n=1\},3)\), with the convention that the minimum is \(3\) if the level is not reached by time \(3\).
This is a stopping time because whether level \(1\) has been reached by time \(n\) is determined by \(S_1,\ldots,S_n\), and the cap is deterministic. Since \(\tau\le3\), bounded optional stopping applies and gives \(\mathbb{E}S_\tau=\mathbb{E}S_0=0\).
Direct check: The eight length-three increment paths have probability \(\tfrac18\) each. The four paths beginning with \(+\) stop at time \(1\), each with stopped value \(1\). Of the four paths beginning with \(-\), path \(-++\) first reaches \(1\) at time \(3\), so its stopped value is \(1\). Path \(-+-\) ends at \(-1\), path \(--+\) ends at \(-1\), and path \(---\) ends at \(-3\). Thus the stopped values have probabilities \(\tfrac58\) at \(1\), \(\tfrac28\) at \(-1\), and \(\tfrac18\) at \(-3\). Their expectation is \[ \tfrac58(1)+\tfrac28(-1)+\tfrac18(-3)=\tfrac58-\tfrac28-\tfrac38=0, \] in agreement with optional stopping.
Worked Example: An Unbounded Time with an Integrable Horizon
Let \(U_1,U_2,\ldots\) be independent fair coin outcomes, with \(U_k=1\) for heads and \(U_k=0\) for tails, and let \(\mathcal{F}_n=\sigma(U_1,\ldots,U_n)\). Define \(\xi_k=U_k-\tfrac12\) and \(M_0=0\), \(M_n=\sum_{k=1}^n\xi_k\). The increments are independent of \(\mathcal{F}_{n-1}\), have mean \(0\), and satisfy \(|\xi_k|=\tfrac12\), so \((M_n)\) is a martingale with bounded increments. Let \(\tau=\inf\{n\ge1:U_n=1\}\), the first head.
The event \(\{\tau\le n\}\) means at least one head occurred in the first \(n\) flips, so \(\tau\) is a stopping time. For \(k\ge1\), \(\mathbb{P}(\tau\ge k)=\mathbb{P}(U_1=\cdots=U_{k-1}=0)=(\tfrac12)^{k-1}\). Therefore \[ \mathbb{E}\tau=\sum_{k\ge1}\mathbb{P}(\tau\ge k) =\sum_{k\ge1}(\tfrac12)^{k-1}=2. \] The bounded-increments corollary applies, so \(\mathbb{E}M_\tau=0\).
Direct check: On \(\{\tau=k\}\), there are \(k-1\) tails followed by one head, and hence \(M_\tau=-(k-1)\tfrac12+\tfrac12=1-\tfrac{k}{2}\). Since \(\mathbb{P}(\tau=k)=(\tfrac12)^k\), \[ \mathbb{E}M_\tau =\sum_{k\ge1}(\tfrac12)^k(1-\tfrac{k}{2}) =1-\tfrac12\mathbb{E}\tau =1-\tfrac12(2)=0. \] Here the expression for \(M_\tau\) is integrable because \(\mathbb{E}|M_\tau|\le 1+\tfrac12\mathbb{E}\tau=2\). This check confirms the theorem’s conclusion without assuming that an unbounded stopping time is automatically safe.
A Continuous-Time Check—and a Failure
The same diagnosis applies in continuous time: check the filtration and stopping-time property, then match the time to a result. Boundedness is enough for the bounded optional stopping theorem under the regularity assumptions stated above. Almost-sure finiteness alone is not.
Worked Example: Brownian Motion Stopped at a Bounded Exit Time
Let \((B_t)_{t\ge0}\) be standard Brownian motion, started at \(B_0=0\), on a probability space with its usual augmented filtration \((\mathcal{F}_t)_{t\ge0}\). Its state space is \(\mathbb{R}\), and it is an integrable martingale. Define \(\sigma=\inf\{t\ge0:|B_t|=1\}\) and \(\tau=\sigma\wedge1\). Continuity and adaptation of Brownian paths imply that \(\sigma\), and hence \(\tau\), is a stopping time. Also \(\tau\le1\), so the bounded continuous-time optional stopping theorem gives \(\mathbb{E}B_\tau=\mathbb{E}B_0=0\).
Check: If \(\sigma\le1\), continuity gives \(B_\tau\in\{-1,1\}\); if \(\sigma>1\), then \(B_\tau=B_1\in(-1,1)\). Thus \(|B_\tau|\le1\), so the stopped value is integrable. In addition, Brownian motion and its negative have the same law, and the rule \(\tau\) is unchanged when the path is negated. Therefore \(B_\tau\) has a symmetric distribution. Its integrability then also gives \(\mathbb{E}B_\tau=0\), an independent check of the result.
Worked Example: Almost-Sure Finiteness Is Not Enough
On the same Brownian probability space, let \(\tau=\inf\{t\ge0:B_t=1\}\). This is a stopping time, and it is finite almost surely. A standard first-passage result for Brownian motion, obtained from the reflection principle, gives its density \[ f_\tau(t)=\frac{1}{\sqrt{2\pi t^3}}\exp\!\left(-\frac{1}{2t}\right),\qquad t>0. \] This density formula is a standard Brownian first-passage result; its derivation is not needed for the diagnosis here.
Compute the mean using the density. For \(t\ge1\), \(\exp(-1/(2t))\ge\exp(-1/2)\), and so \[ \mathbb{E}\tau =\int_0^\infty t f_\tau(t)\,dt \ge \frac{\exp(-1/2)}{\sqrt{2\pi}}\int_1^\infty t^{-1/2}\,dt =\infty. \] The last integral diverges because its integral up to \(R\) is \(2(\sqrt{R}-1)\), which grows without bound. At the stopping time, continuity gives \(B_\tau=1\) almost surely, so \(\mathbb{E}B_\tau=1\), whereas \(\mathbb{E}B_0=0\). This does not contradict optional stopping: the time is not bounded, and the integrable-horizon condition above is absent. In fact, the stopped process cannot satisfy the integrable domination condition of the proposition, since that condition would force equality of these two expectations.
Common Mistakes and a Practical Decision
A common error is to substitute a random time into the martingale identity without checking any stopping condition. For the Brownian first-passage example, the incorrect calculation is “\(B_t\) is a martingale, so \(\mathbb{E}B_\tau=\mathbb{E}B_0=0\).” The actual stopped value is \(B_\tau=1\) almost surely, so \(\mathbb{E}B_\tau=1\). The correction is to identify a theorem that applies: boundedness fails, \(\mathbb{E}\tau\) is infinite, and the integrable domination condition is unavailable. The martingale property by itself does not justify the substitution.
Another mistake is confusing almost-sure finiteness with integrability. The Brownian hitting time above is finite almost surely but has infinite expectation. Conversely, an unbounded stopping time can be safe: in the fair-coin example, \(\tau\) is unbounded but has finite expectation, and bounded increments supply an integrable bound for the stopped values.
In practice, write down the proposed conclusion and its theorem before doing any algebra. For discrete time, a bounded stopping time invokes bounded optional stopping; an unbounded time may be handled by uniform integrability or by bounded increments and an integrable horizon. For continuous time, use a continuous-time theorem with its filtration and path regularity assumptions stated. If no theorem’s hypotheses have been verified, the identity \(\mathbb{E}M_\tau=\mathbb{E}M_0\) has not yet been established.
Check Your Understanding
For each question, identify the relevant hypothesis before deciding whether an optional stopping conclusion follows.
- For a discrete-time martingale and a stopping time \(\tau\le N\), which optional stopping result applies? What changes if only \(\tau<\infty\) almost surely is known?
- Suppose the increments of a discrete-time martingale are bounded in absolute value by \(C\), and \(\mathbb{E}\tau<\infty\). Give an integrable random variable that dominates every \(|M_{n\wedge\tau}|\).
- In the first-head example, calculate \(\mathbb{P}(\tau=3)\) and the stopped value \(M_\tau\) on that event.
- Why is the Brownian exit time capped at \(1\) a stopping time, and what condition permits bounded optional stopping?
- For Brownian motion stopped at its first hitting time of \(1\), state the values of \(\mathbb{E}B_0\) and \(\mathbb{E}B_\tau\). Which missing hypothesis explains why optional stopping does not force them to agree?