Why Stopping Requires Conditions
Brownian motion and its basic martingales provide useful ways to study hitting problems, but a martingale identity cannot be applied at an arbitrary random time without checking hypotheses. A bounded stopping time is one reliable setting. An unbounded stopping time can also be admissible when the stopped martingale values form a uniformly integrable family. The distinction matters: stopped paths may converge almost surely while their expectations fail to converge.
Throughout, let \((\Omega,\mathcal{F},\mathbb{P})\) carry a standard Brownian motion \((B_t)_{t\ge0}\) with state space \(S=\mathbb{R}\), and let \((\mathcal{F}_t)_{t\ge0}\) be its usual filtration. A stopping time \(\tau\) satisfies \(\{\tau\le t\}\in\mathcal{F}_t\) for every \(t\ge0\). We use the Brownian martingales \(B_t\) and \(B_t^2-t\). A family \(\mathcal{X}\) of integrable random variables is uniformly integrable if \(\sup_{X\in\mathcal{X}}\mathbb{E}[|X|\mathbf{1}_{\{|X|>K\}}]\to0\) as \(K\to\infty\).
Proof. For each positive integer \(m\), round \(\tau\) upward to the next point of the grid \(\{kT/2^m:0\le k\le2^m\}\), and call the resulting time \(\tau_m\). This is a grid-valued stopping time: whether \(\tau_m\le kT/2^m\) is determined by whether \(\tau\le kT/2^m\). For a martingale sampled at a finite grid, its expectation at a grid-valued stopping time equals its initial expectation. To see this directly, write \(M_{\tau_m}=M_0+\sum_{k=1}^{2^m}(M_{kT/2^m}-M_{(k-1)T/2^m})\mathbf{1}_{\{\tau_m\ge kT/2^m\}}\). The indicator in each summand is measurable at time \((k-1)T/2^m\), so the martingale property makes that summand's expectation zero. Thus \(\mathbb{E}[M_{\tau_m}]=\mathbb{E}[M_0]\). As \(m\to\infty\), \(\tau_m\to\tau\), and continuity gives \(M_{\tau_m}\to M_\tau\) almost surely. The variables \(M_{\tau_m}\) belong to the uniformly integrable family \(\{M_t:0\le t\le T\}\), so the standard uniformly integrable convergence theorem gives convergence in \(L^1\). Taking expectations proves the claim. For \(B_t\), \(\mathbb{E}[B_t^2]=t\le T\); for \(B_t^2-t\), \(\mathbb{E}[(B_t^2-t)^2]=2t^2\le2T^2\). Each family is bounded in \(L^2\), hence uniformly integrable: for any \(X\) in either family, \(\mathbb{E}[|X|\mathbf{1}_{\{|X|>K\}}]\le \mathbb{E}[X^2]/K\). \(\square\)
Proof. For each positive integer \(n\), \(\tau\wedge n\) is a bounded stopping time. The process \(M_{t\wedge\tau}\) is a continuous martingale, so the bounded optional stopping theorem gives \(\mathbb{E}[M_{\tau\wedge n}]=\mathbb{E}[M_0]\). Since \(\tau<\infty\) almost surely, continuity implies \(M_{\tau\wedge n}\to M_\tau\) almost surely. These random variables belong to the assumed uniformly integrable family, so they converge in \(L^1\). Taking expectations in the limit gives \(\mathbb{E}[M_\tau]=\mathbb{E}[M_0]\). \(\square\)
Bounded Stopping in Practice
Worked Example: A Stopping Rule Based on Information at Time One-Half
Define \(\tau=\tfrac12\) when \(B_{1/2}<0\), and \(\tau=1\) when \(B_{1/2}\ge0\). This is a stopping time: before time \(1/2\), the event \(\{\tau\le t\}\) is empty; from \(1/2\) until time \(1\), it is \(\{B_{1/2}<0\}\), known at time \(1/2\); and from time \(1\) onward it is the whole sample space. Compute \(\mathbb{E}[B_\tau]\) and \(\mathbb{E}[B_\tau^2-\tau]\).
Here \(S=\mathbb{R}\), time is continuous, and \(\tau\le1\). The bounded theorem applies to both Brownian martingales. To check the answers explicitly, set \(Y=B_{1/2}\) and \(Z=B_1-B_{1/2}\). The increments are independent, \(Y\) and \(Z\) have mean zero, \(\mathbb{E}[Y^2]=\mathbb{E}[Z^2]=1/2\), and symmetry gives \(\mathbb{P}(Y<0)=\mathbb{P}(Y\ge0)=1/2\).
On \(\{Y<0\}\), \(B_\tau=Y\); on \(\{Y\ge0\}\), \(B_\tau=Y+Z\). Therefore \(\mathbb{E}[B_\tau]=\mathbb{E}[Y]+\mathbb{E}[Z\mathbf{1}_{\{Y\ge0\}}]=0+0=0\), where independence of \(Z\) and \(Y\) gives the second equality. For the square, expand on the two events:
Indeed, symmetry gives \(\mathbb{E}[Y^2\mathbf{1}_{\{Y<0\}}]=\mathbb{E}[Y^2\mathbf{1}_{\{Y\ge0\}}]=\tfrac14\); the cross term has expectation zero by independence and \(\mathbb{E}[Z]=0\); and \(\mathbb{E}[Z^2\mathbf{1}_{\{Y\ge0\}}]=\tfrac12\cdot\tfrac12=\tfrac14\). Also \(\mathbb{E}[\tau]=\tfrac12\cdot\tfrac12+1\cdot\tfrac12=\tfrac34\). Thus \(\mathbb{E}[B_\tau^2-\tau]=\tfrac34-\tfrac34=0\). Both checks agree with optional stopping and \(\mathbb{E}[B_0]=\mathbb{E}[B_0^2]=0\).
Uniform Integrability at a Two-Sided Hitting Time
Worked Example: Which Side Does Brownian Motion Hit First?
Start Brownian motion at \(B_0=0\), and let \(\tau=\inf\{t\ge0:B_t\notin(-2,1)\}\). Find the probability that it exits at \(1\). This example illustrates the unbounded-time theorem through a stopped martingale whose values are uniformly bounded.
The process has state space \(\mathbb{R}\) and continuous time with the Brownian filtration. The exit time is a stopping time by the Brownian hitting-time property. It is finite almost surely: if the process has not exited by an integer time \(k\), then \(B_k\in(-2,1)\). An independent increment \(B_{k+1}-B_k>3\) forces \(B_{k+1}>1\), so exit occurs by time \(k+1\). This increment event has a fixed probability \(c=\mathbb{P}(B_1>3)>0\), independent of the past. Consequently \(\mathbb{P}(\tau>n)\le(1-c)^n\), which tends to zero.
Continuity gives \(B_\tau\in\{-2,1\}\). For every \(t\), the stopped value \(B_{t\wedge\tau}\) lies in \([-2,1]\); hence the family of stopped values is uniformly integrable. The uniformly integrable optional stopping theorem gives \(\mathbb{E}[B_\tau]=\mathbb{E}[B_0]=0\). If \(p=\mathbb{P}(B_\tau=1)\), then \(\mathbb{P}(B_\tau=-2)=1-p\), so
Check: the complementary probability is \(1-\tfrac23=\tfrac13\), and the endpoint mean is \(\tfrac23\cdot1+\tfrac13\cdot(-2)=0\), as required. The essential condition was not that \(\tau\) be bounded; it was that the stopped values stay in a fixed bounded interval.
Why a One-Sided Hitting Time Is Different
Worked Example: A Limit That Cannot Be Passed Through Expectation
For \(a>0\), let \(\tau_a=\inf\{t\ge0:B_t=a\}\), with \(B_0=0\). The reflection principle gives \(\mathbb{P}(\tau_a\le t)=2\mathbb{P}(B_t\ge a)\), and the right side tends to \(1\) as \(t\to\infty\). Thus \(\tau_a<\infty\) almost surely.
For each fixed \(t\), the time \(\tau_a\wedge t\) is bounded, so \(\mathbb{E}[B_{\tau_a\wedge t}]=0\). As \(t\to\infty\), continuity and finiteness of \(\tau_a\) imply \(B_{\tau_a\wedge t}\to B_{\tau_a}=a\) almost surely. If these stopped values were uniformly integrable, the convergence would also hold in \(L^1\), forcing their expectations to converge to \(a\). But every expectation is zero. Therefore the family \(\{B_{\tau_a\wedge t}:t\ge0\}\) is not uniformly integrable, and the unbounded-time theorem does not apply.
The incorrect calculation is to write \(\mathbb{E}[B_{\tau_a}]=\lim_{t\to\infty}\mathbb{E}[B_{\tau_a\wedge t}]=0\), or to apply optional stopping directly and conclude \(\mathbb{E}[B_{\tau_a}]=\mathbb{E}[B_0]=0\). Almost-sure convergence alone does not justify convergence of expectations. In fact \(B_{\tau_a}=a\) almost surely, so its expectation is \(a\), not zero. The missing hypothesis is uniform integrability.
A Reliable Checklist
Before using optional stopping for a Brownian hitting problem, identify the martingale, the stopping time, and the version of the theorem that could apply. For bounded stopping, check that the time is bounded by a deterministic constant. For an unbounded time, check that it is finite almost surely and that the stopped values are uniformly integrable; boundedness of those values is a useful sufficient condition. A one-sided upper bound alone does not establish uniform integrability, as the last example demonstrates.
Common mistake. A tempting but invalid argument is: “\(\tau\wedge t\) is bounded for every \(t\), so optional stopping gives \(\mathbb{E}[M_{\tau\wedge t}]=\mathbb{E}[M_0]\); now let \(t\to\infty\).” The bounded theorem justifies each finite-\(t\) equality, not the limit of the expectations. That limit requires an additional convergence condition, such as uniform integrability. In the one-sided hitting example, almost-sure convergence holds but uniform integrability fails, so the limiting expectation changes.
Check Your Understanding
For each question, identify which optional stopping condition is relevant before calculating.
- Why does the bounded optional stopping theorem apply to \(B_t^2-t\) at a stopping time \(\tau\le T\)? State the moment bound that gives uniform integrability.
- In the two-sided example, what fixed increment event ensures exit within the next unit interval whenever the process is still inside \((-2,1)\)?
- Suppose a stopped Brownian motion remains in a fixed bounded interval until an almost-surely finite stopping time. Which version of optional stopping applies, and why?
- For the one-sided hitting time \(\tau_a\), what is the almost-sure limit of \(B_{\tau_a\wedge t}\), and why does this not imply convergence of its expectations?
- Give a concrete condition that fails in the one-sided example but holds for the stopped process in the two-sided example.