Boundary-Value Equations for Hitting-Time Transforms

From a Markov Property to a Boundary-Value Equation

For Brownian motion, a hitting-time transform can be characterized by a differential equation in the starting point. The Markov property explains the equation locally: over a short time interval the process moves by a small Gaussian increment, and the transform of the remaining hitting time is again the same function evaluated at the new position. The boundary condition records what happens when the target has already been reached.

Let \(B_t\) be standard Brownian motion on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with its usual filtration \((\mathcal{F}_t)_{t\ge0}\). For a fixed starting point \(x<a\), write \(X_t=x+B_t\), and let \(\mathbb{P}_x\) and \(\mathbb{E}_x\) denote probability and expectation for this process. Its state space is \(\mathbb{R}\), and time is continuous. Define the first hitting time of \(a\) by \(\tau_a=\inf\{t\ge0:X_t=a\}\). It is a stopping time. For \(s>0\), define \(u_s(x)=\mathbb{E}_x[e^{-s\tau_a}\mathbf{1}_{\{\tau_a<\infty\}}]\); thus a path that never hits \(a\) contributes zero.

The boundary condition has a direct probabilistic meaning: if the process starts at \(a\), then \(\tau_a=0\), so \(e^{-s\tau_a}=1\). Boundedness at the other end is required because the transform is an expectation of a random variable between zero and one. It also selects the appropriate solution of the differential equation. The calculation below concerns a single target on a half-line; it does not impose conditions at two finite endpoints.

Deriving the Local Equation

Fix \(x<a\) and let \(h>0\) be small. On the event \(\{\tau_a>h\}\), the Markov property at deterministic time \(h\) says that the conditional transform of the remaining time is \(u_s(X_h)\). Splitting according to whether the target is hit by time \(h\) gives the exact identity

To use a Gaussian increment without restricting it to paths that have not hit the boundary, extend \(u_s\) to a bounded function \(\widetilde u_s\) on the real line, setting \(\widetilde u_s(y)=1\) for \(y\ge a\). Since \(0\le u_s\le1\), replacing the two terms in the identity by \(e^{-sh}\mathbb{E}_x[\widetilde u_s(X_h)]\) changes the value by at most \(2\mathbb{P}_x(\tau_a\le h)\).

The reflection principle gives \(\mathbb{P}_x(\tau_a\le h)=2\mathbb{P}(B_h\ge a-x)\). For \(d=a-x>0\), the Gaussian exponential bound gives \(\mathbb{P}(B_h\ge d)\le e^{-d^2/(2h)}\): apply Markov's inequality to \(e^{\theta B_h}\), use \(\mathbb{E}e^{\theta B_h}=e^{\theta^2h/2}\), and choose \(\theta=d/h\). Therefore the replacement error is at most \(4e^{-d^2/(2h)}=o(h)\).

At a point where \(u_s\) is twice continuously differentiable, the short-time Gaussian expansion is \(\mathbb{E}_x[\widetilde u_s(X_h)]=u_s(x)+\tfrac12 h u_s''(x)+o(h)\). Indeed, \(X_h-x\) has mean zero and second moment \(h\); Taylor expansion near \(x\) gives the displayed terms, while the probability of leaving any fixed neighborhood of \(x\) is exponentially small in \(1/h\). Also \(e^{-sh}=1-sh+o(h)\). Substitution into the Markov identity yields \(u_s(x)=u_s(x)+h(\tfrac12u_s''(x)-su_s(x))+o(h)\). Subtract \(u_s(x)\), divide by \(h\), and let \(h\downarrow0\). This gives the stated differential equation wherever the transform is twice differentiable. The verification below establishes that the probabilistic transform is indeed the smooth solution, rather than assuming smoothness without justification.

Why This Solution Is the Hitting-Time Transform

Proof. Let \(\lambda=\sqrt{2s}\), and define \(g(y)=e^{\lambda(y-a)}\). The basic Brownian exponential martingale gives that \(e^{-st}g(X_t)=e^{\lambda(x-a)}e^{\lambda B_t-\lambda^2t/2}\) is a martingale, since \(\lambda^2/2=s\). Apply optional stopping at the bounded stopping time \(\tau_a\wedge t\). Up to this time \(X\le a\), so \(0\le e^{-s(\tau_a\wedge t)}g(X_{\tau_a\wedge t})\le1\). Consequently,

On \(\{\tau_a>t\}\), \(X_t<a\), so the second expectation is between zero and \(e^{-st}\), which tends to zero. As \(t\to\infty\), the first expectation increases to \(\mathbb{E}_x[e^{-s\tau_a}\mathbf{1}_{\{\tau_a<\infty\}}]\), by monotone convergence. This proves that the transform equals \(e^{\lambda(x-a)}\), the formula obtained from the boundary-value equation.

For uniqueness, every solution has the form \(C e^{\lambda x}+D e^{-\lambda x}\). Boundedness as \(x\to-\infty\) forces \(D=0\), and \(u(a)=1\) forces \(C=e^{-\lambda a}\). Hence no other bounded solution satisfies the boundary condition. \(\square\)

At \(s=0\), the transform is understood as the hitting probability \(\mathbb{P}_x(\tau_a<\infty)\), rather than as \(e^{-0\tau_a}\) on paths where \(\tau_a=\infty\). Taking \(s\downarrow0\) in the formula gives \(1\); monotone convergence therefore shows that the upper level is hit with probability one. The corresponding equation is \(\tfrac12u''=0\); its bounded solutions on \((-\infty,a]\) are constant, and the boundary value makes the solution equal to one.

A Drift and Volatility Extension

The same reasoning illustrates how the differential operator changes when the process has drift or a different variance rate. Let \(X_t=x+\nu t+\sigma B_t\), where \(\nu\in\mathbb{R}\), \(\sigma>0\), and the filtration is the Brownian filtration. For an upper target \(a>x\), the short-time increment has mean \(\nu h\) and second moment \(\sigma^2h+o(h)\). Taylor expansion in the Markov identity therefore gives \(\nu u'(x)+\tfrac12\sigma^2u''(x)=su(x)\). The positive root of the characteristic equation is \(r_+=\frac{-\nu+\sqrt{\nu^2+2\sigma^2s}}{\sigma^2}\); it is positive because \(s>0\). Boundedness as \(x\to-\infty\) selects \(u_s(x)=e^{r_+(x-a)}\).

Common Pitfall: Choosing the Wrong Exponential

A frequent error is to keep the exponential solution that grows at the unbounded end. For an upper target and \(s>0\), writing \(u(x)=C e^{\lambda x}+D e^{-\lambda x}\) and choosing \(D\ne0\) violates boundedness as \(x\to-\infty\), because \(e^{-\lambda x}\to\infty\). The boundary value at \(a\) alone is not enough to determine the correct solution: it gives only \(C e^{\lambda a}+D e^{-\lambda a}=1\). Boundedness forces \(D=0\), after which \(C=e^{-\lambda a}\). The correct answer is \(e^{\lambda(x-a)}\), not the growing term.

The second check is probabilistic: a Laplace transform of a nonnegative hitting time, with the convention used here, must lie between zero and one. The upper-target expression \(e^{-\lambda(a-x)}\) has this property when \(x\le a\). If a proposed sign makes the expression exceed one far from the target, that is evidence that the wrong boundary condition or exponential branch was used.