Laplace Transforms for Two-Sided Exit Times

The Transform of a Two-Sided Exit Time

For a one-sided Brownian hitting time, the Laplace transform measures how the time to reach a level is weighted toward shorter paths. In a bounded interval there are two possible exit boundaries, but if we ask only for the time of exit, both boundaries contribute the same terminal value. This makes the boundary-value problem especially direct: its boundary values are \(1\) at both ends.

Let \((\Omega,\mathcal{F},\mathbb{P})\) carry a standard Brownian motion \((B_t)_{t\ge0}\), with its natural completed filtration \((\mathcal{F}_t)_{t\ge0}\). For a fixed starting point \(x\), set \(X_t=x+B_t\), so the state space is \(\mathbb{R}\). Fix real numbers \(a<b\), and define the exit time from the open interval \((a,b)\) by \(\tau=\inf\{t\ge0:X_t\notin(a,b)\}\). For \(x\in(a,b)\) and \(s\ge0\), the Laplace transform of this exit time is \(u_s(x)=\mathbb{E}_x[e^{-s\tau}]\), where \(\mathbb{E}_x\) means that \(X_0=x\). If \(x=a\) or \(x=b\), then \(\tau=0\), so \(u_s(x)=1\). The parameter restriction \(s\ge0\) ensures that the transform is bounded by \(1\) whenever \(\tau\) is finite.

The differential equation is the Brownian boundary-value equation for the transform. The boundary values follow from the quantity being transformed: at either boundary the exit time is already zero, and \(e^{-s\cdot0}=1\). The proof below establishes both that exit occurs in finite time and that the solution of the equation gives the desired expectation.

Why the Boundary-Value Solution Is the Transform

First, exit is almost surely finite. Put \(L=b-a>0\) and \(p=\mathbb{P}(B_1>L)\), which is strictly positive. If the process is in \((a,b)\) at an integer time \(n\), then an increment \(B_{n+1}-B_n>L\) forces \(X_{n+1}>b\), since \(X_n>a\). Thus, conditional on not having exited by time \(n\), the probability of exiting by time \(n+1\) is at least \(p\). Independent Brownian increments then give \(\mathbb{P}_x(\tau>n)\le(1-p)^n\) for every integer \(n\ge0\). The right side tends to zero, so \(\mathbb{P}_x(\tau=\infty)=0\). If \(x\) is an endpoint, the exit time is already zero.

Proof. Let \(s>0\), and let \(v\) be the solution of the stated boundary-value problem. We first solve the equation. With \(k=\sqrt{2s}>0\), it becomes \(v''=k^2v\). Writing \(m=(a+b)/2\), every solution has the form \(v(x)=C\cosh(k(x-m))+D\sinh(k(x-m))\). The endpoints are \(m-h\) and \(m+h\), where \(h=(b-a)/2>0\). The two boundary conditions give \(C\cosh(kh)-D\sinh(kh)=1\) and \(C\cosh(kh)+D\sinh(kh)=1\). Subtracting yields \(2D\sinh(kh)=0\). Since \(kh>0\), \(\sinh(kh)>0\), so \(D=0\); adding the equations gives \(C=1/\cosh(kh)\). This proves the displayed formula and its boundary values.

To identify this solution probabilistically, use Itô's formula on the stopped process, for each fixed \(t\ge0\): \[ e^{-s(t\wedge\tau)}v(X_{t\wedge\tau}) =v(x)+\int_0^{t\wedge\tau}e^{-sr}v'(X_r)\,dB_r +\int_0^{t\wedge\tau}e^{-sr}\left(\tfrac12v''(X_r)-sv(X_r)\right)\,dr. \] The last integral is zero by the differential equation. On the closed bounded interval, \(v'\) is bounded, so for fixed \(t\) the stochastic integral has a bounded square-integrable integrand and has expectation zero. Therefore \(\mathbb{E}_x[e^{-s(t\wedge\tau)}v(X_{t\wedge\tau})]=v(x)\). Almost surely \(\tau<\infty\), and continuity of Brownian paths gives \(X_\tau\in\{a,b\}\). As \(t\) tends to infinity, the random variable inside the expectation converges to \(e^{-s\tau}v(X_\tau)=e^{-s\tau}\). It is bounded in absolute value by \(\max_{y\in[a,b]}|v(y)|\), so dominated convergence gives \(v(x)=\mathbb{E}_x[e^{-s\tau}]\).

For \(s=0\), the equation is \(v''=0\), so \(v\) is affine. An affine function with value \(1\) at both distinct endpoints is the constant \(1\). Since \(\tau<\infty\) almost surely, \(u_0(x)=\mathbb{E}_x[1]=1\), as required. Finally, uniqueness holds for every \(s\ge0\). The difference \(w\) of two solutions has zero boundary values. If \(s>0\) and \(w\) had a positive value, it would attain a positive interior maximum, where \(w''\le0\); but \(w''=2sw>0\) there, a contradiction. A negative interior minimum gives the opposite contradiction. Thus \(w=0\). If \(s=0\), \(w\) is affine with zero values at both endpoints and is identically zero. \(\square\)

The formula can also be written without the midpoint. For \(s>0\), the hyperbolic addition identity gives \[ u_s(x)=\frac{\sinh(k(b-x))+\sinh(k(x-a))}{\sinh(k(b-a))}. \] Indeed, the numerator is \(2\sinh(k(b-a)/2)\cosh(k(x-m))\), and the denominator is \(2\sinh(k(b-a)/2)\cosh(k(b-a)/2)\). This form makes the two endpoint values particularly visible.

Boundary Values and a Common Pitfall

The two boundary conditions encode the quantity being transformed, not the probability of exiting at either particular endpoint. If the process starts at a boundary, its exit time from the open interval is zero, so its transform is \(1\). For a start in the interior, continuity ensures that at the exit time the process is at one of the endpoints, and the terminal factor \(e^{-s\tau}\) is multiplied by the boundary value \(1\).

Common mistake. A frequent incorrect setup is to impose \(u_s(a)=u_s(b)=0\), as if the boundary value were an absorbing payoff of zero. For \(s>0\), the differential equation with these two boundary values has only the zero solution: the uniqueness argument above applies, and the zero function is a solution. But the desired expectation is strictly positive. Indeed, \(\tau<\infty\) almost surely, so \(e^{-s\tau}>0\) almost surely and therefore \(\mathbb{E}_x[e^{-s\tau}]>0\). The correction is to use \(u_s(a)=u_s(b)=1\). Zero boundary values would describe a different terminal payoff, not the transform of the exit time alone.

The check at \(s=0\) is useful for detecting errors. The transform at zero is the expectation of \(1\), which equals \(1\) because exit is almost surely finite. The formula also gives \(1\): as \(s\downarrow0\), \(k=\sqrt{2s}\downarrow0\), and both hyperbolic cosines in the centered formula tend to \(1\). This check relies on almost-sure finiteness; without it, \(e^{-0\cdot\tau}\) on the event \(\tau=\infty\) would require a convention and would not represent the same finite-exit calculation.