Laplace Transforms of Brownian Interval Exit Times

The Exit-Time Transform and Its Boundary Problem

Let \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) be a filtered probability space carrying a standard Brownian motion \(B_t\). Fix real numbers \(a<b\), a drift \(\mu\in\mathbb{R}\), and a starting point \(x\in(a,b)\). Consider \(X_t=x+\mu t+B_t\), whose drift coefficient is \(\mu\) and whose diffusion coefficient is \(1\). Define the first exit time \(\tau=\inf\{t\ge0:X_t\notin(a,b)\}\). For a discount parameter \(\lambda\ge0\), the Laplace transform of this exit time is \(u_\lambda(x)=\mathbb{E}[e^{-\lambda\tau}]\).

The value \(e^{-\lambda\tau}\) discounts long exits when \(\lambda>0\); when \(\lambda=0\), it equals \(1\) if \(\tau\) is finite. Since paths are continuous, the exit position is \(a\) or \(b\), and a process started at either endpoint has exit time \(0\). These boundary facts suggest the conditions \(u_\lambda(a)=u_\lambda(b)=1\). The generator of \(X\), acting on a twice continuously differentiable function \(f\), is \(\mathcal{L}f=\mu f'+\tfrac12 f''\). The corresponding boundary-value problem is

\frac12u_\lambda''(x)+\mu u_\lambda'(x)=\lambda u_\lambda(x),\qquad a<x<b,\qquad u_\lambda(a)=u_\lambda(b)=1.

Here is the precise sense in which this ODE characterizes the transform: its solution, if twice continuously differentiable on the interval and continuous on the closed interval, agrees with the expectation. We establish that claim by stopped Itô calculus, rather than assuming a representation theorem.

Definition: The interval exit-time Laplace transform is \(u_\lambda(x)=\mathbb{E}[e^{-\lambda\tau}]\) for \(x\in(a,b)\) and \(\lambda\ge0\), where \(\tau=\inf\{t\ge0:X_t\notin(a,b)\}\). The boundary-value problem uses the generator of \(X_t=x+\mu t+B_t\) and the boundary values \(1\), since exit occurs immediately when the starting point is an endpoint.

Why the Exit Time Is Finite

We first verify a fact needed for the transform and its boundary-value representation: \(\tau<\infty\) almost surely. For any integer \(n\ge0\), if \(X_n\in(a,b)\) and \(\mu+(B_{n+1}-B_n)>b-a\), then \(X_{n+1}>b\). Continuity therefore forces an exit during \([n,n+1]\). The increment \(B_{n+1}-B_n\) is independent of \(\mathcal{F}_n\) and has distribution \(N(0,1)\), so \(p=\mathbb{P}(\mu+B_1>b-a)>0\). On the event \(\{\tau>n\}\), the process is still in the interval at time \(n\). Independence of the next increment gives \(\mathbb{P}(\tau>n+1)\le(1-p)\mathbb{P}(\tau>n)\). Induction yields \(\mathbb{P}(\tau>n)\le(1-p)^n\), which tends to zero. Thus \(\mathbb{P}(\tau=\infty)=0\).

Solving the Boundary-Value Problem

Write \(L=b-a\), \(y=x-a\), and \(k=\sqrt{\mu^2+2\lambda}\). In the coordinate \(y\in[0,L]\), the ODE is \(\tfrac12u''+\mu u'=\lambda u\). When \(k>0\), its characteristic roots are \(-\mu+k\) and \(-\mu-k\). An expression satisfying both boundary values is

u_\lambda(x)=e^{-\mu y} \frac{\sinh(k(L-y))+e^{\mu L}\sinh(ky)}{\sinh(kL)}, \qquad y=x-a,\quad k=\sqrt{\mu^2+2\lambda}>0.

The denominator is nonzero because \(kL>0\). At \(y=0\), the numerator is \(\sinh(kL)\), so the formula gives \(1\). At \(y=L\), the numerator is \(e^{\mu L}\sinh(kL)\), which cancels the factor \(e^{-\mu L}\) and again gives \(1\). The formula solves the ODE because each exponential term has one of the two characteristic roots.

The only case not covered by \(k>0\) is \(\mu=0\) and \(\lambda=0\). Then the ODE is \(u''=0\), and its two boundary values force \(u\equiv1\). This also agrees with \(e^{0}=1\) and the almost-sure finiteness of \(\tau\). For zero drift and \(\lambda>0\), the expression can be written more simply as

u_\lambda(x)= \frac{\cosh\big(k(y-L/2)\big)}{\cosh(kL/2)}, \qquad k=\sqrt{2\lambda},\quad y=x-a.

Worked Example: Zero Drift on a Symmetric Interval

Take \(X_t=B_t\), the interval \((-1,1)\), starting point \(x=0\), and discount \(\lambda=2\). Here \(L=2\), \(y=1\), and \(k=\sqrt{2\lambda}=2\). Substituting these values in the zero-drift formula gives

u_2(0)=\frac{\cosh(2(1-1))}{\cosh(2\cdot1)} =\frac{\cosh(0)}{\cosh(2)} =\frac{1}{\cosh(2)}.

As a check, the unsimplified boundary formula gives \(\big(\sinh(2)+\sinh(2)\big)/\sinh(4)=2\sinh(2)/\sinh(4)\). Since \(\sinh(4)=2\sinh(2)\cosh(2)\), this is \(1/\cosh(2)\), agreeing with the result. It lies between \(0\) and \(1\), as required for \(e^{-2\tau}\).

Worked Example: Positive Drift and a Nonzero Discount

Take \(X_t=x+t+B_t\), the interval \((0,2)\), starting point \(x=1\), and \(\lambda=\tfrac32\). Then \(L=2\), \(y=1\), \(\mu=1\), and \(k=\sqrt{1+2\cdot\tfrac32}=2\). The explicit formula gives

u_{3/2}(1)=e^{-1} \frac{\sinh(2)+e^2\sinh(2)}{\sinh(4)} =e^{-1}(1+e^2)\frac{\sinh(2)}{2\sinh(2)\cosh(2)} =\frac{\cosh(1)}{\cosh(2)}.

For an independent ODE check, the characteristic roots are \(1\) and \(-3\), so a solution has form \(A e^x+C e^{-3x}\). The left boundary gives \(A+C=1\). The right boundary gives \(A e^2+C e^{-6}=1\). These are the same boundary equations used by the explicit formula; the ODE has distinct roots, and the boundary-value problem has a unique solution as shown below. Thus its value at \(x=1\) agrees with \(\cosh(1)/\cosh(2)\). In particular, the expression is not obtained by dropping the drift: \(\mu=1\) changes both characteristic roots.

Worked Example: Zero Discount

Take drift \(\mu=1\) on \((0,1)\), with any starting point \(x\in(0,1)\), and set \(\lambda=0\). Here \(L=1\), \(y=x\), and \(k=1\). The formula becomes

u_0(x)=e^{-x}\frac{\sinh(1-x)+e\,\sinh(x)}{\sinh(1)}.

Use \(\sinh(z)=\tfrac12(e^z-e^{-z})\). The numerator inside the fraction is \(\tfrac12(e^{1-x}-e^{-1+x}+e^{1+x}-e^{1-x})=\tfrac12(e^{1+x}-e^{-1+x})=e^x\sinh(1)\). Therefore \(u_0(x)=e^{-x}e^x=1\). This matches the definition: \(\lambda=0\) gives \(e^{-\lambda\tau}=1\), and the exit time is finite almost surely. It also checks the drift-dependent formula in the no-discount case.

Verification and Uniqueness

Theorem: For \(\lambda\ge0\), the exit-time transform \(u_\lambda(x)=\mathbb{E}[e^{-\lambda\tau}]\) is the unique function continuous on \([a,b]\) and twice continuously differentiable on \((a,b)\) that satisfies \(\tfrac12u''+\mu u'=\lambda u\) on \((a,b)\) and \(u(a)=u(b)=1\). In particular, the explicit formula above (with the stated \(k=0\) case) equals the transform.

Proof. Let \(v\) be any function with the stated regularity and boundary values satisfying the ODE. It is bounded on \([a,b]\), and \(v'\) is bounded on every closed subinterval of \((a,b)\). To avoid any issue with derivatives near the endpoints, stop first at \(\tau_\varepsilon=\inf\{t\ge0:X_t\notin(a+\varepsilon,b-\varepsilon)\}\), where \(0<\varepsilon<\min(x-a,b-x)\). Fix \(T>0\) and apply Itô's formula to \(f(t,z)=e^{-\lambda t}v(z)\) up to \(T\wedge\tau_\varepsilon\). Its derivatives are \(f_t=-\lambda e^{-\lambda t}v(z)\), \(f_z=e^{-\lambda t}v'(z)\), and \(f_{zz}=e^{-\lambda t}v''(z)\); the Itô correction is \(\tfrac12 e^{-\lambda t}v''(z)\,dt\). The \(dt\) coefficient is \(e^{-\lambda t}(\tfrac12v''+\mu v'-\lambda v)=0\). Hence

e^{-\lambda(T\wedge\tau_\varepsilon)} v(X_{T\wedge\tau_\varepsilon}) = v(x)+\int_0^{T\wedge\tau_\varepsilon} e^{-\lambda s}v'(X_s)\,dB_s.

The integrand, extended by zero after \(\tau_\varepsilon\), is progressively measurable and bounded on the stopped region for \(s\le T\). Thus it belongs to \(\mathcal{L}^2\): its expected squared integral over \([0,T]\) is finite. The Itô integral is a true martingale: its increments have conditional expectation zero, as follows from the martingale property of the Itô integral for \(\mathcal{L}^2\) integrands, and it is integrable by the Itô isometry. Taking expectations gives \(\mathbb{E}[e^{-\lambda(T\wedge\tau_\varepsilon)}v(X_{T\wedge\tau_\varepsilon})]=v(x)\).

As \(\varepsilon\downarrow0\), continuity of \(X\) implies \(\tau_\varepsilon\uparrow\tau\) almost surely. Since \(v\) is bounded and continuous on \([a,b]\), bounded convergence gives \(\mathbb{E}[e^{-\lambda(T\wedge\tau)}v(X_{T\wedge\tau})]=v(x)\). Now let \(T\to\infty\). On \(\{\tau\le T\}\), the stopped value is \(e^{-\lambda\tau}v(X_\tau)=e^{-\lambda\tau}\), since \(X_\tau\in\{a,b\}\). On \(\{\tau>T\}\), the absolute value is at most \(\sup_{z\in[a,b]}|v(z)|\), and this event has probability tending to zero because \(\tau<\infty\) almost surely. Bounded convergence therefore gives \(v(x)=\mathbb{E}[e^{-\lambda\tau}]=u_\lambda(x)\). This proves uniqueness. The explicit expression satisfies the ODE and boundary values by its characteristic roots and endpoint evaluations; the \(k=0\) solution does also. Applying the uniqueness just proved identifies each with the transform. \(\square\)

This verification also explains the role of discounting: the time derivative contributes \(-\lambda v\), which must balance the generator \(\mathcal{L}v\). The equation is therefore \(\mathcal{L}v=\lambda v\), not \(\mathcal{L}v=0\) unless \(\lambda=0\). The stochastic integral in the argument is a true martingale on each finite time interval because its integrand is square-integrable; no assertion that an arbitrary local martingale is a true martingale is needed.

Common Pitfall: The Discount Term and Boundary Values

A common incorrect equation is \(\tfrac12u''+\mu u'=0\) for every \(\lambda\), followed by the boundary values \(u(a)=u(b)=1\). That equation has the constant solution \(u=1\), which cannot be the positive-discount transform in general: when \(\lambda>0\) and \(\tau>0\), \(e^{-\lambda\tau}<1\). The correction is to apply Itô's formula to the discounted quantity \(e^{-\lambda t}u(X_t)\). Its time derivative is \(-\lambda e^{-\lambda t}u(X_t)\), so cancellation requires \(\mathcal{L}u-\lambda u=0\). The boundary values remain \(1\), because discounting at time \(0\) has no effect on an immediate exit.

The bounded interval matters as well. The proof that \(\tau\) is finite used a fixed positive chance of exiting during each unit time interval, uniformly over the current position in \((a,b)\). On an unbounded domain that argument does not apply, and a solution of a differential equation with boundary data need not represent a finite exit-time transform without additional conditions.

Check Your Understanding

Use the definition of the transform and the verified boundary-value problem to answer the following.

For drift \(\mu\), why are the characteristic roots of the ODE \(-\mu\pm\sqrt{\mu^2+2\lambda}\)?
For \(\mu=0\), \(a=-L/2\), \(b=L/2\), and \(x=0\), simplify the transform for \(\lambda>0\).
Explain why the stochastic integral in the verification proof is a true martingale on a fixed finite time interval.
What does the boundary condition \(u_\lambda(a)=1\) mean in terms of the exit time when the process starts at \(a\)?
Why does the proof of almost-surely finite exit use an increment event with probability strictly greater than zero?