Brownian Hitting Probabilities and Exit-Time Expectations

Two-Sided Exit from an Interval

Let Brownian motion start at \(x\) inside a finite interval \((a,b)\). A basic question is whether it reaches the upper endpoint or the lower endpoint first, and how long it takes to reach either one. The answers are determined by two martingales: the position itself and its square minus elapsed time. The key proof issue is that the exit time is not known in advance to be bounded. We therefore stop first at a bounded time, establish finiteness of the exit time, and only then pass to the limit.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) with respect to this filtration. Fix real numbers \(a<b\) and a starting point \(x\in(a,b)\), and set \(X_t=x+B_t\). Define the endpoint hitting times and the exit time by

Here the infimum of an empty set is \(+\infty\). Continuity of \(X\) implies that, on \(\{\tau<\infty\}\), the process exits at exactly one endpoint: \(X_\tau\in\{a,b\}\). The times are stopping times because \(X\) is adapted and continuous. The event of an upper exit is \(\{\tau_b<\tau_a\}\), and the lower-exit event is \(\{\tau_a<\tau_b\}\).

The Martingales Behind the Calculation

We first verify the two martingale identities used in the proof. The process \(X_t=x+B_t\) is a martingale: for \(s\le t\), the Brownian increment \(B_t-B_s\) is independent of \(\mathcal{F}_s\) and has mean zero, so \(\mathbb{E}[X_t\mid\mathcal{F}_s]=X_s\). Its variables are integrable because \(X_t\) is normally distributed with finite first moment.

For the second identity, apply Itô's formula to \(f(z)=z^2\). Its derivatives are \(f'(z)=2z\) and \(f''(z)=2\), so the correction term is \(\tfrac12 f''(X_s)\,ds=ds\). Thus

Both stochastic integrands here belong to the required class on every finite time interval. They are progressively measurable, and \(1\) satisfies \(\mathbb{E}\int_0^T 1\,ds=T<\infty\). Also, since \(\mathbb{E}[X_s^2]=x^2+s\),

Consequently these are Itô integrals with progressively measurable integrands in \(\mathcal{L}^2\). The Itô integral of an \(\mathcal{L}^2\) integrand is a true martingale: in particular, its increments have conditional expectation zero, as follows from the construction of the integral and its martingale property. It follows that \(X_t^2-t\) is a true martingale, not merely a local martingale.

Proof of the Exit Probabilities and Expected Time

Proof. For each positive integer \(n\), let \(\tau_n=\tau\wedge n\). This is a bounded stopping time. Before exit, \(X_t\in(a,b)\), and at exit it equals an endpoint. Hence \(X_{\tau_n}\in[a,b]\) and is bounded, whether or not \(\tau\) is finite. Optional sampling for the martingale \(X_t\) gives \(\mathbb{E}[X_{\tau_n}]=x\). Optional sampling for the martingale \(X_t^2-t\) gives

Set \(M=\max(|a|,|b|)\). Since \(X_{\tau_n}\in[a,b]\), we have \(X_{\tau_n}^2\le M^2\). Therefore \(\mathbb{E}[\tau_n]\le M^2-x^2\le M^2+x^2\), a finite bound independent of \(n\). As \(\tau_n\) increases to \(\tau\), the monotone convergence theorem gives \(\mathbb{E}[\tau]=\lim_{n\to\infty}\mathbb{E}[\tau_n]<\infty\). In particular, \(\tau<\infty\) almost surely: if \(\mathbb{P}(\tau=\infty)>0\), then \(\mathbb{E}[\tau]\) would be infinite.

By continuity, \(X_{\tau_n}\to X_\tau\) almost surely. The variables \(X_{\tau_n}\) lie in the bounded interval \([a,b]\), so bounded convergence yields \(\mathbb{E}[X_\tau]=\lim_n\mathbb{E}[X_{\tau_n}]=x\) and \(\mathbb{E}[X_{\tau_n}^2]\to\mathbb{E}[X_\tau^2]\). Taking limits in the second stopped identity, using monotone convergence for \(\tau_n\), gives \(\mathbb{E}[\tau]=\mathbb{E}[X_\tau^2]-x^2\).

Let \(p=\mathbb{P}(\tau_b<\tau_a)\). Since \(\tau\) is finite almost surely and the two endpoints are distinct, \(X_\tau=b\) on the upper-exit event and \(X_\tau=a\) on the lower-exit event. Thus \(x=\mathbb{E}[X_\tau]=pb+(1-p)a\), which gives \(p=(x-a)/(b-a)\). The lower-exit probability is \(1-p=(b-x)/(b-a)\). Finally,

This proves all three claims, including the finiteness needed for the limiting argument. \(\square\)

Worked Exit Problems

Why the Truncation Matters

A common mistake is to write \(\mathbb{E}[X_\tau^2-\tau]=x^2\) by applying optional stopping directly at \(\tau\), and then use this equality to conclude that \(\mathbb{E}[\tau]\) is finite. This argument is circular: optional sampling at an unbounded stopping time does not follow just because \(X_t^2-t\) is a martingale at each fixed time. Without further conditions, an unbounded stopping time can invalidate that expectation identity.

The correction is to use \(\tau_n=\tau\wedge n\), where optional sampling is valid, and to exploit the fact that \(X_{\tau_n}\) stays in the fixed bounded interval \([a,b]\). The resulting uniform bound on \(\mathbb{E}[\tau_n]\) proves that \(\mathbb{E}[\tau]\) is finite. Only after that step do bounded convergence and monotone convergence justify the limiting identities. This order of reasoning is useful well beyond Brownian exit problems: when a stopping argument has an unbounded time, first identify what controls the stopped process and then justify the limit.

The formulas also give a useful scale check. If the interval has width \(L=b-a\) and the process starts at its midpoint, each endpoint has probability one half, while the expected exit time is \(L^2/4\). Thus doubling the interval width multiplies the expected exit time by four, consistent with Brownian time scaling.