Stopping a Diffusion at the Boundary
A diffusion is often studied only until it leaves a region where its coefficients or the question of interest are well controlled. The first exit time turns that restriction into a precise stopping operation: before exit, the process follows its dynamics inside the domain; at and after exit, the stopped process remains at its exit point. Generator test functions then turn the dynamics before exit into identities for quantities such as exit probabilities and expected exit times.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). For a progressively measurable integrand \(H\) with \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\), the Ito integral \(\int_0^t H_s\,dB_s\) is a square-integrable martingale on \([0,T]\). A martingale \(M_t\) is an adapted, integrable process satisfying \(\mathbb{E}[M_t\mid\mathcal{F}_s]=M_s\) for \(s\le t\). A local martingale has this property after suitable stopping, but need not itself be a martingale.
Consider the one-dimensional SDE \(dX_t=\mu(X_t)\,dt+\sigma(X_t)\,dB_t\), with drift \(\mu\), diffusion coefficient \(\sigma\), and initial state \(X_0=x\). Its generator on twice continuously differentiable functions is \(\mathcal{L}f(y)=\mu(y)f'(y)+\tfrac12\sigma(y)^2f''(y)\). For an open domain \(D\subseteq\mathbb{R}\), define \(\tau_D=\inf\{t\ge0:X_t\notin D\}\), with the infimum of the empty set taken as infinity, and \(X_{t\wedge\tau_D}\) as the process stopped at exit. We assume \(x\in D\), so the exit question is nontrivial; if \(x\notin D\), then \(\tau_D=0\).
The integral in this definition is an ordinary time integral, not a stochastic integral. The generator is evaluated only before exit; including or excluding the single time \(s=\tau_D\) does not change the time integral. The test functions are defined on the whole state space, so their values at the exit point are meaningful. In applications with locally bounded coefficients, the local martingale formulation is natural. To assert a true martingale, one must also check integrability; bounded coefficients and bounded derivatives on the relevant region provide a convenient sufficient condition.
Why Ito's Formula Gives the Stopped Identity
Proof. Ito's formula uses \(f\), \(f'\), and \(f''\), and includes the quadratic-variation correction \(\tfrac12 f''(X_s)\sigma(X_s)^2\,ds\), since \(dB_s\,dB_s=ds\), \(dB_s\,ds=0\), and \(ds\,ds=0\). Applied up to \(t\wedge\tau_D\), it gives
Here the derivatives are those of the stated test function; in particular the drift terms combine to \(\int_0^{t\wedge\tau_D}\mathcal{L}f(X_s)\,ds\). The remaining term is the Ito integral with integrand \(H_s=\mathbf{1}_{\{s\le\tau_D\}}f'(X_s)\sigma(X_s)\). It is progressively measurable, and boundedness of \(f'\) and \(\sigma\) gives \(\mathbb{E}\int_0^T H_s^2\,ds\le T\|f'\|_\infty^2\|\sigma\|_\infty^2<\infty\). The Ito isometry makes this integral square-integrable, and the Ito integral's conditional-expectation property makes it a martingale. This proves the assertion. If boundedness is removed, Ito's formula still gives a local martingale after localization; it does not by itself prove the unstopped expression is a true martingale. \(\square\)
Brownian Motion on an Interval
Take \(X_t=x+B_t\) with \(a<x<b\), and let \(\tau=\inf\{t\ge0:X_t\notin(a,b)\}\). The generator is \(\mathcal{L}f=\tfrac12 f''\). Continuity of Brownian paths implies that, when \(\tau<\infty\), \(X_\tau\) is either \(a\) or \(b\). The interval is bounded, so the stopped process stays in \([a,b]\); this boundedness will ensure the martingales used below are true martingales on each finite time interval.
Worked Example: Probability of Exiting Through the Upper Endpoint
Find the probability that Brownian motion started at \(x\in(a,b)\) exits \((a,b)\) through \(b\). The event is unambiguous because the two endpoints are distinct, and we will first establish that \(\tau\) is finite almost surely.
For \(f(y)=y\), the derivatives are \(f'(y)=1\) and \(f''(y)=0\), so \(\mathcal{L}f=0\). The stopped generator identity says \(X_{t\wedge\tau}\) is a martingale. Since \(X_{t\wedge\tau}\in[a,b]\), it is bounded and integrable, and therefore \(\mathbb{E}[X_{t\wedge\tau}]=x\). Once finiteness of \(\tau\) is proved, bounded convergence gives \(\mathbb{E}[X_\tau]=x\). Writing \(p=\mathbb{P}(X_\tau=b)\), with \(0\le p\le1\), yields \(x=pb+(1-p)a\). Solving gives
Check: both probabilities are strictly between zero and one because \(a<x<b\), and their sum is \(\frac{x-a+b-x}{b-a}=1\).
Worked Example: Expected Exit Time from an Interval
Derive the expected exit time, including a justification that \(\tau\) is finite. Use the test function \(q(y)=(y-a)(b-y)\). Its derivatives are \(q'(y)=a+b-2y\) and \(q''(y)=-2\), so Ito's formula has correction \(\tfrac12q''(X_s)\,ds=-ds\), and \(\mathcal{L}q=-1\).
The stopped identity is \(q(X_{t\wedge\tau})-q(x)+(t\wedge\tau)=\int_0^{t\wedge\tau}q'(X_s)\,dB_s\). The integrand, written on \([0,t]\) as \(\mathbf{1}_{\{s\le\tau\}}q'(X_s)\), is progressively measurable and bounded, so it belongs to \(\mathcal{L}^2\) on \([0,t]\); its Ito integral is a true martingale with expectation zero. Consequently,
because \(q\ge0\) on \([a,b]\). As \(t\) increases, \(t\wedge\tau\) increases to \(\tau\), so monotone convergence gives \(\mathbb{E}[\tau]\le q(x)<\infty\). In particular \(\tau<\infty\) almost surely. Now \(q(X_{t\wedge\tau})\) is bounded and converges to \(q(X_\tau)=0\). Bounded convergence in the identity gives \(\mathbb{E}[\tau]=q(x)\), or
Check: the expression is positive for \(a<x<b\) and tends to zero as \(x\) tends to either endpoint, as an exit time should.
Worked Example: Test the Identities on the Interval \((-1,2)\)
Let \(X_0=0\), \(a=-1\), and \(b=2\). The upper-exit probability is \(\frac{0-(-1)}{2-(-1)}=\tfrac13\); the lower-exit probability is \(\frac{2-0}{2-(-1)}=\tfrac23\). The expected exit time is \((0-(-1))(2-0)=2\).
Independently check the expected time using \(f(y)=y^2\). Its derivatives are \(f'(y)=2y\), \(f''(y)=2\), and \(\mathcal{L}f=1\), with Ito correction \(\tfrac12 f''(X_s)\,ds=ds\). Thus \(X_{t\wedge\tau}^2-(t\wedge\tau)=\int_0^{t\wedge\tau}2X_s\,dB_s\). The integrand is bounded before exit, so this is a true martingale on finite horizons. Passing to the limit is justified by boundedness of \(X_{t\wedge\tau}^2\) and \(\mathbb{E}[\tau]<\infty\), giving \(\mathbb{E}[\tau]=\mathbb{E}[X_\tau^2]\). The endpoint calculation gives \(\mathbb{E}[X_\tau^2]=\tfrac13(2^2)+\tfrac23((-1)^2)=\tfrac43+\tfrac23=2\), agreeing with the generator calculation.
Check: the exit probabilities add to \(\tfrac13+\tfrac23=1\), and the independent second-moment calculation reproduces the expected time \(2\).
General Use and a Common Pitfall
For a general diffusion, a useful exit-time identity often comes from solving \(\mathcal{L}u=-1\) inside the domain with boundary values \(u=0\). Ito's formula then suggests that \(u(X_{t\wedge\tau_D})+(t\wedge\tau_D)\) is a local martingale. To turn that observation into an expectation identity, one must justify integrability, control the stopped process, and establish that the exit time is finite or otherwise handle the event \(\{\tau_D=\infty\}\). The interval calculation did this explicitly rather than assuming it.
Common mistake: applying optional stopping directly at an unbounded exit time. A tempting but incomplete argument says that \(X_{t\wedge\tau}\) is a martingale, “so” \(\mathbb{E}[X_\tau]=x\). The martingale property at each finite time does not alone justify passage to an unbounded stopping time. In the interval example, the correction is to note \(X_{t\wedge\tau}\in[a,b]\), first prove \(\tau<\infty\) when needed, and then use bounded convergence to pass from \(X_{t\wedge\tau}\) to \(X_\tau\). For the expected-time identity, the additional bound \(\mathbb{E}[t\wedge\tau]\le q(x)\) establishes finite mean before taking the limit.
The distinction between the stopped and unstopped problems matters: after \(\tau_D\), the stopped process has no further diffusion or drift, and the generator integral also stops. The stopped martingale problem therefore encodes only the dynamics up to first exit. It is useful when analyzing local behavior in a domain without making claims about what the process does after leaving it.
Check Your Understanding
Assume throughout that Brownian motion starts strictly inside a bounded interval, so that the two exit endpoints are distinct.
- For an interval \((a,b)\), explain why the test function \(f(y)=y\) has generator zero and how this identifies the exit probabilities.
- For \(q(y)=(y-a)(b-y)\), compute \(q'\), \(q''\), and \(\mathcal{L}q\) for standard Brownian motion.
- Why does boundedness of the stopped Brownian path help justify the limit in the exit-probability calculation?
- For Brownian motion started at \(1\) in \((0,4)\), calculate the probability of exiting through \(4\) and the expected exit time.
- State the extra issue that must be resolved before replacing a finite stopped time \(t\wedge\tau\) by an unbounded exit time \(\tau\).