What Optional Stopping Does—and Does Not—Say
Optional stopping is not a rule that allows every random time to be substituted into a martingale. The hypotheses control whether information can be gathered for an arbitrarily long time and whether the stopped values retain their expectations in the limit. A particularly instructive failure occurs when a martingale is stopped at an almost surely finite but unbounded time: the terminal value may be integrable and still have an expectation different from the initial value.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). The examples below can be constructed with independent random variables on this space; no stochastic integrals are used. In particular, there is no integrand to specify from the class \(\mathcal{L}^2\) of progressively measurable \(H\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\).
For a right-continuous martingale \(M_t\), bounded optional sampling gives \(\mathbb{E}[M_\tau]=\mathbb{E}[M_0]\) for bounded \(\tau\), provided the stopped variables are integrable. For an almost surely finite, possibly unbounded \(\tau\), the optional-stopping result under uniform integrability applies when the family \(\{M_{t\wedge\tau}:t\ge0\}\) is uniformly integrable. These are sufficient conditions, not decorations: the first example shows that merely having \(\tau<\infty\) almost surely and \(M_\tau\) integrable does not suffice.
A Finite Exit Time with Safe Optional Sampling
We first calculate a family of exit times for a symmetric random walk. This gives a controlled comparison with the counterexample. Let \(\xi_1,\xi_2,\ldots\) be independent random variables with \(\mathbb{P}(\xi_k=1)=\mathbb{P}(\xi_k=-1)=1/2\), and put \(S_0=0\), \(S_n=\sum_{k=1}^n\xi_k\). Reveal \(\xi_k\) at time \(k\), independently of the Brownian motion, and use the usual augmentation of the resulting filtration. Set \(M_t=S_{\lfloor t\rfloor}\). Its paths are right-continuous and it is a martingale: for \(s\le t\), the additional signs between \(\lfloor s\rfloor+1\) and \(\lfloor t\rfloor\) are independent of \(\mathcal{F}_s\) and have mean zero, so \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\). Each \(M_t\) is integrable because it is a finite sum of integrable signs.
Worked Example: Exit from a Finite Interval
Fix an integer \(m\ge1\), and let \(\tau_m=\inf\{n\ge0:S_n\in\{-m,1\}\}\), viewed as a stopping time in continuous time. We compute its finiteness, the probability of exiting at \(1\), and its expected value.
The event \(\{\tau_m\le t\}\) is determined by the signs revealed by time \(\lfloor t\rfloor\), so \(\tau_m\) is a stopping time. For each integer \(n\), bounded optional sampling applied to \(S_{k\wedge\tau_m}\) at \(n\) gives \(\mathbb{E}[S_{n\wedge\tau_m}]=0\). Also \(S_{k\wedge\tau_m}^2-(k\wedge\tau_m)\) is a martingale: before exit, the next sign has conditional mean zero and square one, so the conditional expected increment of \(S_k^2\) is one. Bounded optional sampling therefore gives \(\mathbb{E}[n\wedge\tau_m]=\mathbb{E}[S_{n\wedge\tau_m}^2]\le m^2\), since the stopped walk stays in \([-m,1]\). Monotone convergence implies \(\mathbb{E}[\tau_m]\le m^2<\infty\), hence \(\tau_m<\infty\) almost surely.
Because \(S_{\tau_m}\) is bounded between \(-m\) and \(1\), dominated convergence applied to \(S_{n\wedge\tau_m}\) yields \(\mathbb{E}[S_{\tau_m}]=0\). If \(p_m=\mathbb{P}(S_{\tau_m}=1)\), then \(\mathbb{E}[S_{\tau_m}]=p_m-m(1-p_m)=0\). Solving gives \(p_m=m/(m+1)\). Applying dominated convergence to the squares in the identity for \(\mathbb{E}[n\wedge\tau_m]\) gives \(\mathbb{E}[\tau_m]=\mathbb{E}[S_{\tau_m}^2]=p_m+m^2(1-p_m)=m/(m+1)+m^2/(m+1)=m\).
Check: for \(m=2\), the probability of exiting at \(1\) is \(2/3\), and the expected exit time is \(2\). The terminal mean is \((2/3)\cdot1+(1/3)\cdot(-2)=0\), as required. Although \(\tau_m\) is not bounded, the stopped walk always lies in the fixed interval \([-m,1]\), so its stopped family is uniformly integrable.
Counterexample: Hitting One Without a Uniform-Integrability Bound
Now let \(\tau=\inf\{n\ge0:S_n=1\}\), again interpreted as a continuous-time stopping time. This time there is no lower boundary. The terminal value, if reached, is always \(1\), but the time needed to reach it has no finite expectation.
Proof. For every \(m\ge1\), the finite-interval exit time \(\tau_m\) from the preceding example is finite almost surely. On the event that the walk exits at \(1\), it has hit \(1\), so \(\tau<\infty\). Thus \(\mathbb{P}(\tau<\infty)\ge\mathbb{P}(S_{\tau_m}=1)=m/(m+1)\) for every \(m\). Letting \(m\to\infty\) gives \(\mathbb{P}(\tau<\infty)=1\). By definition of the first hitting time and the unit steps of the walk, \(S_\tau=1\) almost surely, so \(\mathbb{E}[M_\tau]=1\), whereas \(M_0=0\).
To see why the stopping-time hypothesis alone cannot justify equality, suppose for contradiction that \(\mathbb{E}[\tau]<\infty\). Write \(S_\tau=\sum_{k\ge1}\xi_k\mathbf{1}_{\{\tau\ge k\}}\). The event \(\{\tau\ge k\}\) is determined by \(\xi_1,\ldots,\xi_{k-1}\), so it is independent of \(\xi_k\). Moreover, \(\sum_{k\ge1}\mathbb{E}[|\xi_k|\mathbf{1}_{\{\tau\ge k\}}]=\sum_{k\ge1}\mathbb{P}(\tau\ge k)=\mathbb{E}[\tau]<\infty\). We may therefore take expectations term by term, obtaining \(\mathbb{E}[S_\tau]=\sum_{k\ge1}\mathbb{E}[\xi_k\mathbf{1}_{\{\tau\ge k\}}]=0\), contradicting \(S_\tau=1\) almost surely. Hence \(\mathbb{E}[\tau]=\infty\).
For every \(t\), bounded optional sampling at \(t\wedge\tau\) gives \(\mathbb{E}[M_{t\wedge\tau}]=0\). Along integer times \(n\), \(M_{n\wedge\tau}\to1\) almost surely. If the family of these stopped values were uniformly integrable, almost-sure convergence would imply convergence in \(L^1\), and their expectations would converge to \(1\), contradicting that each expectation is \(0\). Thus the stopped family is not uniformly integrable. The failed expectation identity is exactly consistent with the hypotheses: \(\tau\) is finite but unbounded, and uniform integrability is absent. \(\square\)
The same failure can be seen in the tails. Since \(S_{n\wedge\tau}\le1\) and its expectation is zero, its positive part has expectation at most \(1\), while its negative part has the same expectation as its positive part. But \((S_{n\wedge\tau})^-\to0\) almost surely as \(n\to\infty\), whereas \(\mathbb{E}[(S_{n\wedge\tau})^-]=\mathbb{E}[(S_{n\wedge\tau})^+]\to1\) by bounded convergence of the positive parts. The negative tails therefore cannot be uniformly negligible.
An Unbounded Time That Does Satisfy the Hypotheses
Unboundedness alone does not cause failure. The relevant question is whether the stopped family is uniformly integrable. Here is an example in which the stopping time is unbounded but the process has already become constant before it occurs.
Worked Example: Uniform Integrability Despite an Unbounded Time
Let \(N=\inf\{k\ge1:\xi_k=1\}\), and define \(\tau=1+N\). Since \(\mathbb{P}(N>n)=2^{-n}\), the time \(\tau\) is finite almost surely and unbounded. It is a stopping time: for each \(t\), the event \(\{\tau\le t\}\) depends only on signs revealed by time \(\lfloor t-1\rfloor\), when that index is nonnegative. Define \(M_t=B_{t\wedge1}\).
This is an integrable martingale in the filtration that also reveals the independent signs. If \(s\le t\le1\), then \(\mathbb{E}[B_t\mid\mathcal{F}_s]=B_s\); if \(s<1<t\), then \(\mathbb{E}[B_1\mid\mathcal{F}_s]=B_s\); and if \(1\le s\le t\), both stopped values equal \(B_1\). Independence of the signs preserves these conditional expectations. Since \(\tau\ge2\), \(M_{t\wedge\tau}=B_{t\wedge1}\) and \(M_\tau=B_1\).
For all \(t\), \(\mathbb{E}[|B_{t\wedge1}|^2]=\mathbb{E}[B_{t\wedge1}^2]=t\wedge1\le1\). For \(K>0\), \(\mathbb{E}[|B_{t\wedge1}|\mathbf{1}_{\{|B_{t\wedge1}|>K\}}]\le\mathbb{E}[B_{t\wedge1}^2]/K\le1/K\). Taking the supremum over \(t\) and then letting \(K\to\infty\) proves uniform integrability. Finally, \(\mathbb{E}[M_\tau]=\mathbb{E}[B_1]=0=M_0\).
Check: the bound is \(1/K\) for every \(t\), not a separate bound depending on \(t\). The stopped process is frozen at time \(1\), so its values do not acquire larger tails as the unbounded stopping time grows.
Hypothesis Checks and Common Mistakes
A reliable application starts by naming the result being used and checking its actual assumptions. For bounded optional sampling, verify that the stopping time has a deterministic upper bound. For the unbounded-time result, verify almost-sure finiteness, integrability where needed, and uniform integrability of the stopped family. A finite terminal expectation by itself is not a replacement for uniform integrability.
Common mistake: “The time is finite almost surely, so expectations pass to the limit.” In the hitting example, \(M_{n\wedge\tau}\to1\) almost surely, but \(\mathbb{E}[M_{n\wedge\tau}]=0\) for every \(n\), while \(\mathbb{E}[M_\tau]=1\). The invalid step is passing expectations through almost-sure convergence without an additional condition such as uniform integrability.
Common mistake: “Every unbounded stopping time breaks optional stopping.” The Brownian example has an unbounded \(\tau\), but the stopped family has a uniform second-moment bound and is uniformly integrable, so the expectation identity holds. The time bound and the integrability control are different ways to secure the conclusion; unboundedness is not, on its own, a counterexample.
| Situation | What is verified | Conclusion |
|---|---|---|
| Finite-interval walk exit \(\tau_m\) | Stopped values lie in \([-m,1]\) | Uniform integrability; expectation is preserved |
| First hit of \(1\) by the walk | \(\tau<\infty\) a.s.; stopped family is not uniformly integrable | Terminal mean \(1\) differs from initial mean \(0\) |
| Brownian motion stopped at time \(1\), sampled at \(\tau\ge2\) | Uniform second-moment bound for the stopped family | Uniform integrability; expectation is preserved |
Check Your Understanding
For each question, identify the precise hypothesis that supports the conclusion or the precise step that fails.
- Why does \(\tau_m<\infty\) almost surely follow from \(\mathbb{E}[\tau_m]\le m^2\)?
- For the unrestricted hitting time, why does \(\mathbb{P}(\tau<\infty)\ge m/(m+1)\) for every \(m\) imply \(\mathbb{P}(\tau<\infty)=1\)?
- Which implication fails if one tries to infer \(\mathbb{E}[M_\tau]=\lim_n\mathbb{E}[M_{n\wedge\tau}]\) from almost-sure convergence alone?
- Verify the exit probability and expected time in the finite-interval example when \(m=3\).
- Why is a uniform bound \(\sup_t\mathbb{E}[|M_{t\wedge\tau}|^2]\le1\) sufficient for uniform integrability?