Joint Exit Side and Exit-Time Transforms

Keeping Track of Both the Exit Time and the Exit Side

For Brownian motion in a bounded interval, the exit time and the endpoint of exit are related: a long path that exits at the upper endpoint contributes differently from one that exits at the lower endpoint. A single transform can record both pieces of information. It weights an exit at time \(\tau\) by \(e^{-s\tau}\), but counts that weight only when the specified endpoint is reached.

Let \((B_t)_{t\ge0}\) be standard Brownian motion on a probability space \((\Omega,\mathcal F,\mathbb P)\), with its Brownian filtration \((\mathcal F_t)_{t\ge0}\). For a starting point \(x\in(a,b)\), where \(a<b\), set \(X_t=x+B_t\) and \(\mathbb E_x=\mathbb E\) under this construction. Define the exit time \(\tau=\inf\{t\ge0:X_t\notin(a,b)\}\). This is a stopping time. By continuity, when \(\tau\) is finite, \(X_\tau\) is either \(a\) or \(b\).

The key is to encode the exit side in the boundary values. The upper-exit transform has value \(0\) at \(a\) and \(1\) at \(b\); the lower-exit transform reverses those values. The discount in time supplies the term \(s u\) in the differential equation.

The Boundary-Value Problem and Its Solution

Proof. First, \(\tau\) is finite almost surely. Write \(L=b-a>0\), and let \(c=\mathbb P(B_1\ge L)>0\). On \(\{\tau>n\}\), the position \(X_n\) lies in \((a,b)\). If \(B_{n+1}-B_n\ge L\), then \(X_{n+1}>b\), so continuity forces an exit by time \(n+1\). The increment is independent of \(\mathcal F_n\) and has the same distribution as \(B_1\). Therefore \(\mathbb P_x(\tau>n+1)\le(1-c)\mathbb P_x(\tau>n)\). Iteration gives \(\mathbb P_x(\tau>n)\le(1-c)^n\), which tends to zero.

We solve the differential equation and verify its probabilistic meaning. For \(s>0\), put \(\kappa=\sqrt{2s}\). The equation becomes \(u''=\kappa^2u\), whose solutions are linear combinations of \(\sinh(\kappa(x-a))\) and \(\cosh(\kappa(x-a))\). The boundary condition at \(a\) removes the cosine term; the condition at \(b\) fixes the remaining coefficient as \(1/\sinh(\kappa(b-a))\). This gives the displayed formula. For \(s=0\), \(u''=0\), so the solution is affine, and the two boundary values give \((x-a)/(b-a)\). These calculations also prove uniqueness: the boundary conditions determine both coefficients in each case.

Let \(u\) be the solution just found. Itô's formula, applied up to \(t\wedge\tau\), gives

As \(t\to\infty\), \(\tau<\infty\) almost surely, so the random variable inside the expectation converges to \(e^{-s\tau}u(X_\tau)\). It is bounded in absolute value by \(\max_{y\in[a,b]}|u(y)|\), so dominated convergence applies. Since \(u(a)=0\) and \(u(b)=1\), the limit is \(e^{-s\tau}\mathbf 1_{\{X_\tau=b\}}\). Thus \(u(x)=u_s(x)\). Applying the same argument to the solution with boundary values \(1\) at \(a\) and \(0\) at \(b\) proves the formula for \(\ell_s\). \(\square\)

The proof gives more than a way to guess a formula: it verifies that the boundary-value solution equals the desired expectation. It also shows why both endpoints matter. Changing which endpoint has boundary value \(1\) changes which exit event is counted.

Worked Transforms and Conditional Interpretation

Proof. Since \(x\in(a,b)\), the exit probability \(u_0(x)=(x-a)/(b-a)\) is strictly positive, so conditioning on \(\{X_\tau=b\}\) is meaningful. By the definition of conditional expectation on an event of positive probability, \(\mathbb E_x[e^{-s\tau}\mid X_\tau=b]=\mathbb E_x[e^{-s\tau}\mathbf 1_{\{X_\tau=b\}}]/\mathbb P_x(X_\tau=b)=u_s(x)/u_0(x)\). Substitution of the two formulas from the theorem gives the stated expression. At \(s=0\), the ratio \(u_0(x)/u_0(x)\) equals \(1\). The lower-exit identity follows by the same conditional-expectation formula, with \(\ell_0(x)=(b-x)/(b-a)>0\). \(\square\)

A Common Mistake and a Useful Check

A frequent error is to use the transform for exiting at either endpoint when the question asks for the upper endpoint only. In the symmetric example on \((0,2)\), starting at \(1\) with \(s=\tfrac12\), the total transform is \(2\sinh1/\sinh2\). Reporting this as the upper-exit transform is wrong: it counts lower exits as well. The correction is to impose boundary values \(0\) at the lower endpoint and \(1\) at the upper endpoint, giving \(\sinh1/\sinh2\). Adding the lower-side transform recovers the total, which is a direct check that the endpoint events have been separated correctly.

Boundary values provide another quick diagnostic. For upper exit, the transform must vanish at \(a\) and equal \(1\) at \(b\), because starting at either endpoint gives exit time zero and fixes the exit side. The formula for \(s>0\) satisfies these values: its numerator is zero at \(a\), and at \(b\) its numerator equals its denominator. At \(s=0\), the linear formula has the same endpoint values. Finally, the lower- and upper-side transforms must add to \(\mathbb E_x[e^{-s\tau}]\), since the exit is finite almost surely and occurs at exactly one endpoint.