Why Hitting-Time Arguments Need Care
Brownian hitting problems invite a tempting calculation: stop a martingale when it reaches a boundary, take expectations, and solve for the desired probability or mean. The calculation can be valid, but an unbounded hitting time is not automatically covered by the bounded optional stopping theorem. Even when one first stops at a deterministic time and then lets that time increase, almost-sure convergence alone does not justify passing the limit through an expectation. The key issue is whether the stopped martingale values are uniformly integrable.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) carrying a standard Brownian motion \((B_t)_{t\ge0}\), with state space \(S=\mathbb{R}\), and use its usual augmented natural filtration \((\mathcal{F}_t)_{t\ge0}\). Thus the filtration is complete and right-continuous. A random time \(\tau\) is a stopping time if \(\{\tau\le t\}\in\mathcal{F}_t\) for every \(t\ge0\). The processes \((B_t)\) and \((B_t^2-t)\) are continuous martingales for this filtration. We use the standard bounded optional stopping theorem: if \(M_t\) is a right-continuous martingale and \(\sigma\) is a bounded stopping time, then \(\mathbb{E}M_\sigma=\mathbb{E}M_0\), provided the martingale is integrable at each bounded deterministic time. Right-continuity is part of this theorem; it holds for both Brownian martingales used here.
The unbounded-time issue can be stated precisely. The result below is the useful bridge from bounded optional stopping to certain unbounded Brownian hitting times.
Proof. For each positive integer \(n\), \(\tau\wedge n\) is a bounded stopping time, so bounded optional stopping gives \(\mathbb{E}M_{\tau\wedge n}=\mathbb{E}M_0\). Since \(\tau\) is finite almost surely and \(M\) has continuous paths, \(M_{\tau\wedge n}\to M_\tau\) almost surely. Uniform integrability and almost-sure convergence imply convergence in \(L^1\): to see the relevant implication, truncate each variable to the interval \([-K,K]\); the truncated variables converge in \(L^1\) by dominated convergence, and the expected truncation errors tend uniformly to zero by uniform integrability. Fatou's lemma gives the corresponding vanishing tail for the limit. Hence \(\mathbb{E}|M_{\tau\wedge n}-M_\tau|\to0\), so their expectations converge to \(\mathbb{E}M_\tau\). Taking the limit in \(\mathbb{E}M_{\tau\wedge n}=\mathbb{E}M_0\) proves the claim. \(\square\)
A common sufficient condition is that the stopped values \(M_{t\wedge\tau}\) are bounded by one deterministic constant, uniformly over \(t\). Then they are uniformly integrable by the definition. This sufficient condition is not necessary, but it is particularly useful for Brownian motion stopped on leaving a bounded interval.
The One-Sided Hitting-Time Trap
Worked Example: Hitting a Positive Level
Start Brownian motion at \(0\), fix \(a>0\), and let \(\tau_a=\inf\{t\ge0:B_t=a\}\). This is a stopping time. The standard reflection principle gives, for \(t>0\), \(\mathbb{P}(\tau_a>t)=2\Phi(a/\sqrt{t})-1\), where \(\Phi\) is the standard normal distribution function. As \(t\to\infty\), \(a/\sqrt{t}\to0\), so \(\mathbb{P}(\tau_a>t)\to0\); therefore \(\tau_a<\infty\) almost surely and continuity gives \(B_{\tau_a}=a\).
Now apply bounded optional stopping at \(\tau_a\wedge t\). Since this is bounded, \(\mathbb{E}B_{\tau_a\wedge t}=\mathbb{E}B_0=0\). The tempting but invalid step is to let \(t\to\infty\) and conclude \(\mathbb{E}B_{\tau_a}=0\). In fact, \(B_{\tau_a\wedge t}\to a\) almost surely, and \(\mathbb{E}B_{\tau_a}=a\). Thus the expectations do not converge to the expectation of the almost-sure limit.
Check that the missing condition really fails. If the stopped family were uniformly integrable, the theorem just proved would imply \(\mathbb{E}B_{\tau_a}=\mathbb{E}B_0=0\), contradicting \(B_{\tau_a}=a>0\). Therefore \(\{B_{t\wedge\tau_a}:t\ge0\}\) is not uniformly integrable.
This hitting time also has infinite mean. For \(t\ge a^2\), the argument \(u\) in the normal integral from \(0\) to \(a/\sqrt{t}\) lies in \([0,1]\). Since the standard normal density \(\varphi(u)\) is at least \(\varphi(1)>0\) on that interval,
For a nonnegative random variable, \(\mathbb{E}\tau_a=\int_0^\infty\mathbb{P}(\tau_a>t)\,dt\), allowing the value \(+\infty\). The lower bound has divergent integral on \([a^2,\infty)\), since \(\int_{a^2}^R t^{-1/2}\,dt=2(\sqrt{R}-a)\to\infty\). Hence \(\mathbb{E}\tau_a=\infty\). This is a warning that almost-sure finiteness of a stopping time does not imply finite expectation.
A Valid Repair: Exit from a Bounded Interval
Worked Example: Exit from the Interval \((-1,1)\)
Let \(\tau=\inf\{t\ge0:|B_t|=1\}\), with \(B_0=0\). We first verify that \(\tau\) is finite almost surely, rather than assuming it. If the process has not exited by integer time \(k\), then \(B_k\in(-1,1)\). If also \(B_{k+1}-B_k>2\), continuity forces an exit before time \(k+1\). The increments over disjoint unit intervals are independent, and \(q=\mathbb{P}(B_1>2)>0\). Thus \(\mathbb{P}(\tau>m)\le(1-q)^m\) for every positive integer \(m\), which tends to zero. Therefore \(\tau<\infty\) almost surely.
For every \(t\), continuity implies \(|B_{t\wedge\tau}|\le1\). The stopped family is uniformly bounded, hence uniformly integrable. The theorem therefore gives \(\mathbb{E}B_\tau=\mathbb{E}B_0=0\). At exit, \(B_\tau\) is either \(1\) or \(-1\), so, writing \(p=\mathbb{P}(B_\tau=1)\), we have \(0=\mathbb{E}B_\tau=p-(1-p)=2p-1\). Solving gives \(p=\tfrac12\), and the probability of exiting at \(-1\) is also \(\tfrac12\). Substitution checks the expectation: \(\tfrac12(1)+\tfrac12(-1)=0\).
We can also compute the mean exit time without claiming that \(\tau\) is bounded. Apply bounded optional stopping to the continuous martingale \(B_t^2-t\) at \(\tau\wedge t\). Since \(B_0^2=0\), this gives \(\mathbb{E}[B_{\tau\wedge t}^2]=\mathbb{E}[\tau\wedge t]\). As \(t\to\infty\), \(B_{\tau\wedge t}^2\to B_\tau^2=1\) almost surely, and \(0\le B_{\tau\wedge t}^2\le1\); dominated convergence makes the left side tend to \(1\). Also, \(\tau\wedge t\) increases to \(\tau\), so monotone convergence makes the right side tend to \(\mathbb{E}\tau\). Consequently \(\mathbb{E}\tau=1\). Substitution into the stopped identity gives \(1=1\), verifying the result.
Notice the distinction: the time \(\tau\) is unbounded as a random variable, but the stopped Brownian values are bounded. That boundedness justifies the expectation limit for \(B_{\tau\wedge t}\); it does not say that \(\tau\) itself is bounded.
What the Truncated Calculation Is Hiding
Worked Example: The Residual Term at a Deterministic Horizon
Return to the one-sided time \(\tau_a\) and choose any fixed \(n>0\). The bounded stopped value splits into two cases:
Taking expectations and using bounded optional stopping gives
Thus the residual term is exactly \(\mathbb{E}[B_n\mathbf{1}_{\{\tau_a>n\}}]=-a\,\mathbb{P}(\tau_a\le n)\). As \(n\to\infty\), almost-sure finiteness gives \(\mathbb{P}(\tau_a\le n)\to1\), so this residual expectation tends to \(-a\), even though \(\mathbb{P}(\tau_a>n)\to0\). The rare paths that have not yet hit the level carry a compensating negative expectation. Their contribution cannot be discarded merely because their probability tends to zero.
Check. Adding the two terms gives \(a\,\mathbb{P}(\tau_a\le n)-a\,\mathbb{P}(\tau_a\le n)=0\), exactly the expectation required by bounded optional stopping. If one drops the residual term before taking the limit, the calculation incorrectly changes this value from \(0\) to \(a\).
A Checklist for Brownian Optional Stopping
Before using optional stopping in a hitting problem, identify which theorem applies and verify its hypotheses. For a bounded stopping time, use bounded optional stopping for a right-continuous martingale with the usual filtration assumptions. For an unbounded stopping time, do not infer expectation preservation from almost-sure finiteness alone. One valid route is to prove uniform integrability of the stopped martingale family; a uniform deterministic bound on its values is a convenient sufficient condition. Another valid route in a particular calculation may use a separate integrability or convergence argument, but that argument must be stated and checked.
Common mistake. Writing \(\mathbb{E}B_{\tau_a}=\lim_{t\to\infty}\mathbb{E}B_{\tau_a\wedge t}=0\) because \(B_{\tau_a\wedge t}\to B_{\tau_a}\) almost surely is wrong: almost-sure convergence does not in general preserve expectations. Here it would give \(0=a\). The correction is to establish uniform integrability (which fails in this example), or to retain the residual term at each finite horizon and avoid the unjustified interchange.
| Situation | What is justified | What must still be checked |
|---|---|---|
| Bounded stopping time | Bounded optional stopping for a right-continuous martingale | Martingale and filtration hypotheses |
| Almost-surely finite, unbounded time | Almost-sure convergence of \(M_{t\wedge\tau}\) to \(M_\tau\), for continuous \(M\) | Uniform integrability or another valid expectation-limit argument |
| Stopped values uniformly bounded | Uniform integrability and expectation preservation under the theorem's hypotheses | Almost-sure finiteness of \(\tau\) |
Check Your Understanding
For each question, identify the hypothesis that justifies the conclusion, or the precise step that fails.
- Why does \(\mathbb{E}B_{\tau_a\wedge t}=0\) hold for each fixed \(t\), even though \(\tau_a\) need not have finite mean?
- For \(\tau=\inf\{t\ge0:|B_t|=1\}\), which property of \(B_{t\wedge\tau}\) gives uniform integrability?
- In the one-sided hitting example, what is the limit of \(\mathbb{E}[B_n\mathbf{1}_{\{\tau_a>n\}}]\), and why does the event probability tending to zero not force that expectation to tend to zero?
- Does almost-sure finiteness of a stopping time imply that its expectation is finite? Use \(\tau_a\) to justify your answer.
- State the additional condition used in the theorem to conclude \(\mathbb{E}M_\tau=\mathbb{E}M_0\) for an almost-surely finite, possibly unbounded \(\tau\).