Two-Sided Exit Probabilities for Brownian Motion With Drift

Exit Through One Endpoint of a Bounded Interval

For Brownian motion with drift, the probability of reaching an upper boundary before a lower one depends both on the starting point and on the direction and size of the drift. A generator boundary-value problem turns this probability into a second-order ordinary differential equation. The bounded interval matters: unlike a one-sided hitting problem, the process here exits through one of two finite endpoints, and we will verify that this exit occurs almost surely.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Fix real numbers \(a<b\), a starting point \(x\in[a,b]\), and a constant drift \(\mu\in\mathbb{R}\). Define \(X_t=x+\mu t+B_t\), and let \(\tau=\inf\{t\ge0:X_t\notin(a,b)\}\). For \(x\in(a,b)\), continuity of \(X\) means that \(X_\tau\), when \(\tau\) is finite, equals \(a\) or \(b\). We seek \(p(x)=\mathbb{P}(X_\tau=b)\), the probability of exit at the upper endpoint.

The drift is the coefficient of \(dt\), while the diffusion coefficient of \(dB_t\) here is \(1\). For a twice continuously differentiable function \(f\), Ito's formula uses \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\). Thus its generator is \(\mathcal{L}f=\mu f'+\tfrac12 f''\). This is an Ito calculation; no Stratonovich integral is being used. The Ito formula and the \(L^2\) martingale property of Ito integrals are standard results of Ito calculus.

Solving the Generator Equation

The equation \(\mathcal{L}u=0\) is \(\mu u'(y)+\tfrac12u''(y)=0\) for \(a<y<b\). If \(\mu\ne0\), set \(v=u'\). Then \(v'+2\mu v=0\), so \(v(y)=C e^{-2\mu(y-a)}\) for a constant \(C\). Integrating and imposing the two boundary conditions yields the candidate below. If \(\mu=0\), the equation is \(u''=0\), so the boundary conditions give a linear function.

Proof. First suppose \(\mu\ne0\), and write \(D=1-e^{-2\mu(b-a)}\). Since \(b-a>0\) and \(\mu\ne0\), \(D\ne0\). For \(u(y)=(1-e^{-2\mu(y-a)})/D\), direct differentiation gives \(u'(y)=2\mu e^{-2\mu(y-a)}/D\) and \(u''(y)=-4\mu^2e^{-2\mu(y-a)}/D\). Consequently \(\mu u'(y)+\tfrac12u''(y)=2\mu^2e^{-2\mu(y-a)}/D-2\mu^2e^{-2\mu(y-a)}/D=0\). Substitution of \(y=a\) gives \(u(a)=0\); substitution of \(y=b\) gives \(u(b)=D/D=1\). For \(\mu=0\), \(u(y)=(y-a)/(b-a)\) has \(u''=0\) and the same boundary values.

To see uniqueness when \(\mu\ne0\), any solution has derivative \(v=u'\) satisfying \(v'+2\mu v=0\). Multiplying by \(e^{2\mu y}\) shows that \(\frac{d}{dy}(e^{2\mu y}v(y))=0\), so \(v(y)=C e^{-2\mu(y-a)}\). Integrating and using \(u(a)=0\) gives \(u(y)=C(1-e^{-2\mu(y-a)})/(2\mu)\). The condition \(u(b)=1\) fixes \(C=2\mu/D\), giving exactly the displayed expression. When \(\mu=0\), \(u''=0\) implies \(u(y)=A+By\); the two boundary conditions uniquely determine \(A\) and \(B\), giving \((y-a)/(b-a)\).

It remains to prove that the boundary-value solution is the probability, rather than only a formal solution of the differential equation. First, \(\tau\) is finite almost surely. Let \(q=\mathbb{P}(B_1>b-a-\mu)\), which is strictly positive. On the event \(\{\tau>n\}\), \(X_n\in(a,b)\). If \(B_{n+1}-B_n>b-a-\mu\), then \(X_{n+1}-X_n>b-a\), so \(X_{n+1}>b\). By continuity, exit must have occurred by time \(n+1\). Brownian increments are independent of \(\mathcal{F}_n\) and have the \(N(0,1)\) distribution, hence \(\mathbb{P}(\tau>n+1)\le(1-q)\mathbb{P}(\tau>n)\). Iteration gives \(\mathbb{P}(\tau>n)\le(1-q)^n\), which tends to zero.

Now take \(x\in(a,b)\) and use the function \(u\) just found. On the compact interval \([a,b]\), \(u'\) is bounded. For a finite deterministic \(t\), Ito's formula applied to \(u(X_{t\wedge\tau})\) gives $$ u(X_{t\wedge\tau})=u(x)+\int_0^t\mathbf{1}_{\{s<\tau\}}\mathcal{L}u(X_s)\,ds+\int_0^t\mathbf{1}_{\{s<\tau\}}u'(X_s)\,dB_s. $$ In the Ito formula, the time derivative is zero, the first derivative term is \(\mu u'(X_s)\,ds\), and the second-derivative correction is \(\tfrac12u''(X_s)\,ds\); their sum is \(\mathcal{L}u(X_s)\,ds\). The stochastic integral has the progressively measurable integrand \(H_s=\mathbf{1}_{\{s<\tau\}}u'(X_s)\), with \(\mathbb{E}\int_0^t H_s^2\,ds\le t\sup_{y\in[a,b]}|u'(y)|^2<\infty\). Thus \(H\in\mathcal{L}^2\) on \([0,t]\). The Ito integral is a true martingale: it is square integrable by the Ito isometry, and its conditional expectation satisfies \(\mathbb{E}[\int_0^t H_s\,dB_s\mid\mathcal{F}_r]=\int_0^r H_s\,dB_s\) for \(r\le t\).

Since \(\mathcal{L}u=0\), taking expectations gives \(\mathbb{E}[u(X_{t\wedge\tau})]=u(x)\). The stopped values lie in \([0,1]\); as \(t\to\infty\), \(\tau<\infty\) almost surely and \(u(X_{t\wedge\tau})\to u(X_\tau)\). Bounded convergence therefore gives \(u(x)=\mathbb{E}[u(X_\tau)]\). The exit value is \(b\) or \(a\), so \(u(X_\tau)=\mathbf{1}_{\{X_\tau=b\}}\), and \(\mathbb{E}[u(X_\tau)]=\mathbb{P}(X_\tau=b)\). Starting at either endpoint gives the same formula by immediate exit. Finally, for \(\mu\to0\), the expansion \(1-e^{-2\mu c}=2\mu c+O(\mu^2)\) gives the ratio \((x-a)/(b-a)\) for \(a\le x\le b\). This proves the probability formula and its zero-drift limit. \(\square\)

Worked Calculations

A Common Error: Losing the Factor of Two

The generator contains the correction term \(\tfrac12u''\), so its equation is \(\mu u'+\tfrac12u''=0\), or equivalently \(u''+2\mu u'=0\). Omitting the factor of two changes the exponential rate and therefore changes the answer. For example, with \(a=0\), \(b=2\), \(x=1\), and \(\mu=\ln 2\), using \(e^{-\mu y}\) in place of \(e^{-2\mu y}\) would give the incorrect candidate \(\widetilde u(y)=(1-e^{-\mu y})/(1-e^{-2\mu})\). Its value at \(1\) is \((1-\frac12)/(1-\frac14)=\frac23\), not \(\frac45\).

This candidate also fails the generator equation. Its derivatives are \(\widetilde u'(y)=\mu e^{-\mu y}/(1-e^{-2\mu})\) and \(\widetilde u''(y)=-\mu^2e^{-\mu y}/(1-e^{-2\mu})\). Hence $$ \mathcal{L}\widetilde u(y)=\mu\widetilde u'(y)+\tfrac12\widetilde u''(y)=\frac{\mu^2e^{-\mu y}}{2(1-e^{-2\mu})}\ne0. $$ The error comes from treating the Brownian contribution as though the second-order term had coefficient \(1\) instead of \(\tfrac12\).

The generator method is useful because it reduces an exit event to a boundary-value problem that can be solved and then verified probabilistically. The verification matters: solving the differential equation alone does not establish that the solution equals an exit probability. Here almost-sure exit, boundedness of the stopped function, and the square-integrable Ito integral justify that identification. We used a true martingale on each finite time interval and bounded convergence as time tends to infinity, rather than applying optional stopping directly at an unbounded exit time.