Why First-Passage Stopping Requires Care
Stopping a martingale at a bounded time is often safe; stopping it at an unbounded first-passage time is not automatically safe. The difficulty is not that the stopping time is random. It is that the stopped random variables may lose uniform integrability as the time bound is removed. A calculation at \(\tau\wedge n\) is justified at each finite \(n\), but passing from \(\tau\wedge n\) to \(\tau\) requires a separate argument.
Throughout, work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). An Ito integral \(\int_0^t H_s\,dB_s\) below has an adapted, progressively measurable integrand \(H_s\) in \(\mathcal{L}^2\) on every stated finite horizon: \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). A martingale has integrable values and satisfies \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\) for \(s\le t\). A local martingale has this property after stopping along a sequence of stopping times increasing to infinity; without further conditions it need not be a martingale.
What Optional Stopping Does and Does Not Say
Proof. For \(\tau\le T\), the stopped integral has the representation \(M_\tau=M_0+\int_0^T\mathbf{1}_{\{s\le\tau\}}H_s\,dB_s\); changing the indicator at the single time \(s=\tau\) does not change the integral. The stopped integrand is progressively measurable and satisfies \(\mathbb{E}\int_0^T\mathbf{1}_{\{s\le\tau\}}H_s^2\,ds\le\mathbb{E}\int_0^T H_s^2\,ds<\infty\). By the Ito integral’s mean-zero property, \(\mathbb{E}[M_\tau]=\mathbb{E}[M_0]\). For the unbounded case, uniform integrability together with almost-sure convergence implies convergence in \(L^1\). Thus \(\mathbb{E}[M_{t\wedge\tau}]\to\mathbb{E}[M_\tau]\). Each \(t\wedge\tau\) is bounded, so the bounded-time identity gives \(\mathbb{E}[M_{t\wedge\tau}]=\mathbb{E}[M_0]\); taking the limit proves the claim. \(\square\)
The bounded-time statement is a special case of the standard Optional Stopping Theorem. The theorem does not say that every martingale can be stopped at every unbounded stopping time. Localization also does not fill this gap: it provides identities at localized, bounded stages, but the limit still needs justification, such as uniform integrability or a valid domination argument.
Worked Example: A One-Sided Brownian Hitting Time
Let \(a>0\) and \(\tau_a=\inf\{t\ge0:B_t=a\}\). Brownian first-passage results give \(\tau_a<\infty\) almost surely, while \(\mathbb{E}[\tau_a]=\infty\). Explain why the calculation \(\mathbb{E}[B_{\tau_a}]=\mathbb{E}[B_0]\) is invalid.
For each fixed \(t\), \(B_{t\wedge\tau_a}=\int_0^t\mathbf{1}_{\{s\le\tau_a\}}\,dB_s\). The integrand is progressively measurable and has \(\mathbb{E}\int_0^t\mathbf{1}_{\{s\le\tau_a\}}^2\,ds\le t\). The integral is therefore square-integrable, and its mean is zero: \(\mathbb{E}[B_{t\wedge\tau_a}]=0\). But, since \(\tau_a<\infty\) almost surely, continuity gives \(B_{t\wedge\tau_a}\to B_{\tau_a}=a\) almost surely. Consequently, the expectations at finite \(t\) do not converge to the expectation of the limit.
Indeed, if \(\{B_{t\wedge\tau_a}:t\ge0\}\) were uniformly integrable, almost-sure convergence would imply convergence in \(L^1\), and hence \(0=\lim_{t\to\infty}\mathbb{E}[B_{t\wedge\tau_a}]=\mathbb{E}[B_{\tau_a}]=a\), a contradiction. The stopped process is a martingale at each finite time, but its unbounded-time stopped family is not uniformly integrable. The tempting step \(\mathbb{E}[B_{\tau_a}]=\mathbb{E}[B_0]\) has no valid optional-stopping justification.
A Bounded Spatial Exit Repairs the Calculation
One way to control a first-passage problem is to stop on leaving a bounded interval. Write \(\tau=\inf\{t\ge0:B_t\notin(-1,2)\}\). Until \(\tau\), the stopped Brownian motion stays between \(-1\) and \(2\). This spatial bound lets us establish integrability and take limits after first stopping at \(\tau\wedge n\).
Worked Example: Exit Probability and Mean Exit Time
Find the probability of exiting \((-1,2)\) through \(2\), and compute \(\mathbb{E}[\tau]\).
First apply Ito’s formula to \(f(x)=x^2\). Its derivatives are \(f'(x)=2x\) and \(f''(x)=2\), so the correction term is \(\tfrac12 f''(B_t)\,dt=dt\). Thus \(d(B_t^2-t)=2B_t\,dB_t\). Stopping at \(\tau\wedge n\), the integrand \(2B_s\mathbf{1}_{\{s\le\tau\wedge n\}}\) is progressively measurable and has square integral bounded in expectation by \(16n\), since \(|B_s|\le2\) before exit. The Ito integral is square-integrable and has mean zero. Therefore \[ \mathbb{E}[B_{\tau\wedge n}^2]=\mathbb{E}[\tau\wedge n]. \] As \(|B_{\tau\wedge n}|\le2\), this gives \(\mathbb{E}[\tau\wedge n]\le4\). Monotone convergence yields \(\mathbb{E}[\tau]\le4\), so \(\tau<\infty\) almost surely.
Now \(B_{\tau\wedge n}\) remains in \([-1,2]\), and \(B_{\tau\wedge n}\to B_\tau\) almost surely. Bounded convergence and the bounded-time stopped-integral argument give \(\mathbb{E}[B_\tau]=0\). Let \(p=\mathbb{P}(B_\tau=2)\); continuity means the exit value is either \(2\) or \(-1\). Hence \(0=2p-1(1-p)=3p-1\), so \(p=\tfrac13\).
For the mean exit time, bounded convergence gives \(\mathbb{E}[B_{\tau\wedge n}^2]\to\mathbb{E}[B_\tau^2]\), while monotone convergence gives \(\mathbb{E}[\tau\wedge n]\to\mathbb{E}[\tau]\). Taking limits in the stopped identity, \[ \mathbb{E}[\tau]=\mathbb{E}[B_\tau^2]=4\left(\tfrac13\right)+1\left(\tfrac23\right)=2. \] Check: the probability is between \(0\) and \(1\), and the expected exit time is positive and no larger than the bound \(4\) obtained before taking the limit.
The important repair was not simply to replace \(\tau\) by \(\tau\wedge n\). It was to prove an identity for each bounded stop and then justify the limit. Here boundedness of the stopped position controls the spatial terms, and monotone or bounded convergence handles the relevant limits.
An Unbounded Stop Can Still Be Valid
Unbounded stopping is not automatically wrong. A useful test is whether a uniform bound or another uniform-integrability argument controls the stopped process. The next example uses a Brownian exponential martingale and obtains a first-passage transform by taking a bounded-stop limit with an explicit dominating constant.
Worked Example: A First-Passage Laplace Transform
For \(a>0\) and \(q>0\), derive \(\mathbb{E}[e^{-q\tau_a}]\) using bounded stopping.
Set \(\lambda=\sqrt{2q}\) and \(M_t=\exp(\lambda B_t-\lambda^2t/2)\). For \(s\le t\), independent Brownian increments and the normal moment-generating function give \[ \mathbb{E}[M_t\mid\mathcal{F}_s] =M_s\,\mathbb{E}\!\left[\exp\!\left(\lambda(B_t-B_s)-\tfrac12\lambda^2(t-s)\right)\right] =M_s. \] Also \(\mathbb{E}[M_t]=1\), so \(M\) is a true martingale, not merely a local martingale. Ito’s formula confirms its stochastic differential: for \(f(t,x)=e^{\lambda x-\lambda^2t/2}\), \(f_t=-\tfrac12\lambda^2f\), \(f_x=\lambda f\), and \(f_{xx}=\lambda^2f\). The correction term is \(\tfrac12 f_{xx}\,dt=\tfrac12\lambda^2f\,dt\), which cancels \(f_t\,dt\), leaving \(dM_t=\lambda M_t\,dB_t\). On every finite horizon, \(\lambda M_s\) is progressively measurable and \(\mathbb{E}\int_0^T\lambda^2M_s^2\,ds=\int_0^T\lambda^2e^{\lambda^2s}\,ds<\infty\).
For each integer \(n\ge1\), \(\tau_a\wedge n\) is bounded, so optional stopping gives \(\mathbb{E}[M_{\tau_a\wedge n}]=1\). Before \(\tau_a\), \(B_t<a\); at the hitting time, \(B_{\tau_a}=a\). Thus \(0\le M_{\tau_a\wedge n}\le e^{\lambda a}\) for every \(n\). Since \(\tau_a<\infty\) almost surely, \(M_{\tau_a\wedge n}\to e^{\lambda a-\lambda^2\tau_a/2}\) almost surely. The uniform bound permits bounded convergence, giving \[ 1=e^{\lambda a}\mathbb{E}[e^{-\lambda^2\tau_a/2}]. \] Substitute \(\lambda^2/2=q\) to obtain \(\mathbb{E}[e^{-q\tau_a}]=e^{-a\sqrt{2q}}\). Check: as \(q\downarrow0\), the expression tends to \(1\), consistent with \(\tau_a<\infty\) almost surely and bounded convergence applied to \(e^{-q\tau_a}\).
This example differs from the invalid calculation for \(B_{t\wedge\tau_a}\). The stopped exponential has a common deterministic upper bound, so the limiting expectation is justified. The existence of a martingale before stopping is not enough; the behavior of the stopped family at the proposed limit matters.
Localization and Common Mistakes
A local martingale is a process that becomes a martingale after stopping along a sequence of stopping times increasing to infinity. For an unbounded first-passage problem, one may introduce such a sequence to make the integrals manageable and apply bounded-time arguments at each stage. But an identity for every localized process does not by itself imply the corresponding identity after removing the localization. One must still establish a limit, for example by uniform integrability, bounded convergence, or an integrable dominating variable. A local martingale can fail to be a true martingale precisely because these expectation limits fail.
Common mistake: “For every \(n\), \(\mathbb{E}[B_{n\wedge\tau_a}]=0\), and \(B_{n\wedge\tau_a}\to a\), so \(\mathbb{E}[B_{\tau_a}]=0\).” The last implication is false without a condition that allows expectations to pass through the limit. In this case the limit is the constant \(a>0\), while every finite-\(n\) expectation is \(0\); the stopped family is not uniformly integrable. The correction is to retain the bounded stopping time, find a valid uniform-integrability or domination argument, and only then pass to the limit. If no such argument is available, the calculation has not established the desired expectation.
For diffusion problems, the drift and diffusion coefficient also matter when identifying the martingale to stop. For \(dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t\), Ito’s formula applied to a twice differentiable function includes the drift contribution \(\mu f'\) and the correction \(\tfrac12\sigma^2 f''\). A function that solves the associated generator equation can remove the drift locally, but the resulting process may still be only a local martingale. Its integrability and the validity of the stopping limit must be checked separately.
Check Your Understanding
For each question, specify the stopping or limiting condition that makes the proposed expectation argument valid.
- Why does \(\mathbb{E}[B_{t\wedge\tau_a}]=0\) for every finite \(t\) not imply \(\mathbb{E}[B_{\tau_a}]=0\)?
- In the interval \((-1,2)\), which bound first gives \(\mathbb{E}[\tau\wedge n]\le4\), and how does it imply \(\tau<\infty\) almost surely?
- In the exponential-martingale example, identify the common bound that justifies passing from \(\tau_a\wedge n\) to \(\tau_a\).
- What additional property is needed to pass expectations through an almost-sure limit of stopped martingales?
- Why does localizing a local martingale not, by itself, justify an identity at an unbounded first-passage time?