Exit Probabilities from Bounded Intervals

The Exit Event and Its Boundary Values

For Brownian motion started inside a bounded interval, the exit location can only be one of the two endpoints: continuous paths cannot jump across either boundary. This turns a path question into a calculation involving just two possible values. The key is to stop the position process at the first exit, use its martingale property, and check carefully that optional stopping is justified.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\) carrying a standard Brownian motion \((B_t)_{t\ge0}\), with filtration \((\mathcal{F}_t)_{t\ge0}\) its usual augmented natural filtration. For a fixed starting point \(x\in\mathbb{R}\), write \(X_t=x+B_t\). Then \(X_0=x\), \(X_t\) has continuous paths, and \((X_t)\) is a martingale with respect to this filtration. Fix real numbers \(a<b\), and suppose initially that \(x\in(a,b)\). Define the exit time and the two exit events by

Here the infimum of the empty set is \(\infty\). Because \(X\) is adapted with continuous paths, its first hitting time of the closed set \(\{a,b\}\) is a stopping time. Continuity also means that, if \(\tau\) is finite, \(X_\tau\) is exactly \(a\) or \(b\): the path cannot skip over an endpoint. We will first establish that \(\tau\) is finite almost surely, so \(U\) and \(L\) are disjoint events whose union has probability one.

Proof. Let \(d=b-a>0\), and let \(c=\mathbb{P}(B_1\ge d)\). A standard normal random variable has positive probability of exceeding any finite real number, so \(c>0\). For each integer \(n\ge0\), on \(\{\tau>n\}\) we have \(X_n\in(a,b)\). If also \(B_{n+1}-B_n\ge d\), then \(X_{n+1}=X_n+(B_{n+1}-B_n)>a+d=b\). By path continuity the process must have hit \(b\) by time \(n+1\), which contradicts \(\tau>n+1\). Thus \(\{\tau>n+1\}\) is contained in the event that \(\tau>n\) and \(B_{n+1}-B_n<d\). The increment \(B_{n+1}-B_n\) is independent of \(\mathcal{F}_n\), and \(\{\tau>n\}\in\mathcal{F}_n\). Consequently, \(\mathbb{P}(\tau>n+1)\le(1-c)\mathbb{P}(\tau>n)\). Iterating gives \(\mathbb{P}(\tau>n)\le(1-c)^n\), which tends to zero because \(0<c\le1\). The event \(\{\tau=\infty\}\) is contained in \(\{\tau>n\}\) for every \(n\), so \(\mathbb{P}(\tau=\infty)=0\). \(\square\)

This argument proves almost-sure finiteness; it does not require a formula for the expected exit time. To compute the exit probability, we use the bounded-time optional stopping theorem: if \((M_t)\) is an integrable martingale and \(\sigma\) is a bounded stopping time, then \(\mathbb{E}[M_\sigma]=\mathbb{E}[M_0]\). This is the standard optional stopping result for bounded stopping times. We apply it first at \(\tau\wedge t\), which is a stopping time bounded by the fixed number \(t\), and then pass to the limit.

Proof. First suppose \(x\in(a,b)\). For each fixed \(t\ge0\), \(\tau\wedge t\) is a bounded stopping time. The position process \(X_t=x+B_t\) is an integrable martingale, so bounded-time optional stopping gives \(\mathbb{E}_x[X_{\tau\wedge t}]=\mathbb{E}_x[X_0]=x\). If \(\tau\le t\), then \(X_{\tau\wedge t}\) is one of the endpoints. If \(\tau>t\), it lies in \((a,b)\). Therefore \(a\le X_{\tau\wedge t}\le b\) in all cases. By the proposition, \(\tau<\infty\) almost surely, and hence \(X_{\tau\wedge t}\to X_\tau\) almost surely as \(t\to\infty\). The variables are bounded by the fixed constants \(a\) and \(b\), so bounded convergence yields \(\mathbb{E}_x[X_\tau]=x\). Write \(p=\mathbb{P}_x(X_\tau=b)\). Since \(X_\tau=a\) on the complementary exit event, \(\mathbb{E}_x[X_\tau]=bp+a(1-p)\). Thus \(x=a+(b-a)p\), and solving gives \(p=(x-a)/(b-a)\). At \(x=a\), the process is already at the lower endpoint and the upper-exit probability is \(0=(a-a)/(b-a)\). At \(x=b\), it is already at the upper endpoint and the probability is \(1=(b-a)/(b-a)\). This proves the formula for every \(x\in[a,b]\). \(\square\)

Using the Formula

The answer is an affine function of the starting point. It equals zero at the lower endpoint, one at the upper endpoint, and changes linearly between them. The lower-exit probability follows by subtraction: since the two exit events are disjoint and exhaustive almost surely, it is \(1-(x-a)/(b-a)=(b-x)/(b-a)\). These expressions are probabilities, so for an interior starting point each lies strictly between zero and one.

Why the Stopping Argument Needs Care

The central identity is \(\mathbb{E}_x[X_\tau]=x\), but it should not be used without justification. The stopping time \(\tau\) is generally not bounded, so the bounded-time optional stopping theorem cannot be applied to \(\tau\) directly. Instead, stop at \(\tau\wedge t\), where the theorem does apply. The exit-time proposition gives \(\tau<\infty\) almost surely, and the stopped positions all remain in the bounded interval \([a,b]\); those facts justify passing to the limit by bounded convergence. In particular, this proof does not assume that \(\mathbb{E}_x[\tau]\) is finite.

Continuity is also essential to the two-endpoint calculation. At the first time the path reaches the boundary, its value must be one of the endpoints; there is no overshoot. For a process with jumps, a first crossing can land beyond a boundary, and the stopped position would not necessarily have only the values \(a\) and \(b\). The endpoint expectation equation would then need a different analysis.

Common mistake. A frequent error is to use the distance to the upper endpoint in the numerator when calculating the upper-exit probability. For \(a=0\), \(b=4\), and \(x=1\), this wrong calculation gives \(\mathbb{P}_1(X_\tau=4)=\frac{b-x}{b-a}=\frac{4-1}{4-0}=\frac34\). If that were the upper-exit probability, the expected endpoint would be \(4\cdot\frac34+0\cdot\frac14=3\), not the starting value \(1\). The corrected upper-exit probability uses the distance from the lower endpoint: \(\frac{x-a}{b-a}=\frac{1-0}{4-0}=\frac14\). Then the endpoint expectation is \(4\cdot\frac14+0\cdot\frac34=1\), as required.

The formula is useful because it reduces a continuous-time path problem to the two boundary values of a martingale. It also provides a quick consistency test: a proposed upper-exit probability must be \(0\) at \(a\), \(1\) at \(b\), and must make the weighted average of the exit endpoints equal to the starting position. For an interval and starting point as specified here, the linear formula satisfies all three checks.