Mean Time to Leave a Bounded Interval
One-sided Brownian first-passage times can have infinite mean, but requiring a path to leave a bounded interval changes the situation: either boundary is eventually reached, and the mean exit time is finite. The drift affects how quickly the process tends to one boundary or the other. An explicit formula makes this effect visible and gives useful checks through symmetry and limiting cases.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) relative to the filtration. Fix real numbers \(a<b\), a drift \(\mu\in\mathbb{R}\), and an initial point \(x\in[a,b]\). Set \(X_t=x+\mu t+B_t\). Its drift coefficient is \(\mu\), and its diffusion coefficient is \(1\). Define the first exit time by \(\tau=\inf\{t\ge0:X_t\notin(a,b)\}\). For \(x=a\) or \(x=b\), this definition gives \(\tau=0\).
The generator of \(X_t\), acting on a twice differentiable function \(f\), is \(\mathcal{L}f=\mu f'+\tfrac12f''\). The mean-exit-time equation is \(\mathcal{L}u=-1\) in the interval, with \(u(a)=u(b)=0\). Here the right-hand side \(-1\) represents the unit rate at which elapsed time accumulates. We will solve this equation and verify that its solution is indeed the expectation, including the finiteness needed for that identification.
The Explicit Formula and Its Justification
Proof. First establish finiteness of \(\mathbb{E}_x[\tau]\). Fix \(h>0\) and let \(p=\mathbb{P}(B_h+\mu h\ge L)\). Since \(B_h+\mu h\) has a normal distribution with positive variance, \(p>0\). From any point in \([a,b]\), an increment of at least \(L\) over a time interval of length \(h\) forces the process to reach or cross \(b\) by continuity. Independent increments therefore give \(\mathbb{P}_x(\tau>(n+1)h)\le(1-p)\mathbb{P}_x(\tau>nh)\), and induction yields \(\mathbb{P}_x(\tau>nh)\le(1-p)^n\). In particular, \(\tau<\infty\) almost surely. Also, splitting the integral of the tail probability into intervals of length \(h\) gives \(\mathbb{E}_x[\tau]\le h\sum_{n=0}^{\infty}\mathbb{P}_x(\tau>nh)\le h/p\).
Now solve the boundary-value equation. Write \(y=x-a\), so \(0\le y\le L\), and regard \(u\) as a function of \(y\). If \(\mu=0\), the equation is \(\tfrac12u''=-1\). Integrating twice gives \(u(y)=-y^2+C_1y+C_2\). The condition \(u(0)=0\) gives \(C_2=0\), and \(u(L)=0\) gives \(C_1=L\). Thus \(u(y)=y(L-y)\).
If \(\mu\ne0\), solve \(\tfrac12u''+\mu u'=-1\). Setting \(v=u'\) gives \(v'+2\mu v=-2\). Multiplying by \(e^{2\mu y}\) and integrating yields \(v(y)=C e^{-2\mu y}-1/\mu\), for a constant \(C\). A further integration gives \(u(y)=A(1-e^{-2\mu y})-y/\mu+D\), where \(A=C/(2\mu)\) and \(D\) is constant. The condition \(u(0)=0\) gives \(D=0\), while \(u(L)=0\) gives \(A=L/(\mu(1-e^{-2\mu L}))\). The denominator is nonzero because \(\mu L\ne0\). This is the stated formula.
It remains to show that the solution of the equation equals the expected exit time. The displayed function is twice continuously differentiable on \([a,b]\), satisfies the boundary values, and obeys \(\mu u'(z)+\tfrac12u''(z)=-1\) for \(a<z<b\). Apply Ito's formula to \(u(X_{t\wedge\tau})\). Explicitly, the function of the state is \(u\), its first derivative is \(u'\), its second derivative is \(u''\), and the quadratic-variation correction is \(\tfrac12u''\,dt\). The stochastic-integral integrand is \(u'(X_s)\mathbf{1}_{\{s<\tau\}}\), which is progressively measurable and belongs to \(\mathcal{L}^2\) on every finite time interval because \(u'\) is bounded on \([a,b]\). Thus $$ u(X_{t\wedge\tau})=u(x)+\int_0^t\mathbf{1}_{\{s<\tau\}}\big(\mu u'(X_s)+\tfrac12u''(X_s)\big)\,ds +\int_0^t\mathbf{1}_{\{s<\tau\}}u'(X_s)\,dB_s =u(x)-(t\wedge\tau)+\int_0^t\mathbf{1}_{\{s<\tau\}}u'(X_s)\,dB_s. $$ The last integral is a true martingale: its integrand is in \(\mathcal{L}^2\), so the Ito isometry gives finite second moment, and its conditional expectation at an earlier time equals its value at that time. Taking expectations therefore gives \(\mathbb{E}_x[u(X_{t\wedge\tau})]=u(x)-\mathbb{E}_x[t\wedge\tau]\). As \(t\to\infty\), \(t\wedge\tau\to\tau\), and monotone convergence applies to the expectations. Also, \(\tau<\infty\) almost surely, \(X_\tau\in\{a,b\}\), and \(u\) is bounded with \(u(X_\tau)=0\); dominated convergence gives \(\mathbb{E}_x[u(X_{t\wedge\tau})]\to0\). Hence \(u(x)=\mathbb{E}_x[\tau]\). At either endpoint both sides are zero by definition. \(\square\)
Worked Calculations
Worked Example: Zero Drift and Symmetric Starting Points
Find the mean exit time for standard Brownian motion started at \(x=1\) in the interval \((0,4)\), and compare it with the answer from \(x=3\).
Here \(a=0\), \(b=4\), \(L=4\), and \(\mu=0\). At \(x=1\), the formula gives $$ u_0(1)=(1-0)(4-1)=1\cdot3=3. $$ At \(x=3\), it gives $$ u_0(3)=(3-0)(4-3)=3\cdot1=3. $$ The two starting points are reflections of one another about the midpoint \(2\), and their mean exit times agree. As an endpoint check, the formula gives \(u_0(0)=0\cdot4=0\) and \(u_0(4)=4\cdot0=0\), as required.
Worked Example: Positive Drift from the Midpoint
Take \(X_t=1+\tfrac12t+B_t\) and the interval \((0,2)\). Compute the mean exit time.
Now \(L=2\), \(y=x-a=1\), and \(2\mu=1\). Substitution into the drifted formula gives $$ u_{1/2}(1)=2\left[\frac{2(1-e^{-1})}{1-e^{-2}}-1\right]. $$ Factor \(1-e^{-2}=(1-e^{-1})(1+e^{-1})\). The fraction becomes \(2/(1+e^{-1})=2e/(e+1)\). Consequently $$ u_{1/2}(1)=2\left[\frac{2e}{e+1}-1\right] =2\frac{e-1}{e+1}. $$ For a numerical check, \(e\approx2.7183\), so \(2(e-1)/(e+1)\approx2(1.7183)/(3.7183)\approx0.9242\). This is less than the zero-drift midpoint value \(1\), consistent with a positive drift helping the process reach the upper endpoint. The exact expression, rather than the rounded decimal, is the mean.
Worked Example: A Noncentral Starting Point and a Drift-Sign Check
Let \(X_t=\tfrac12t+B_t\) start at \(x=\tfrac12\) in \((0,2)\). Find its mean exit time, then compare it with the corresponding reflected problem with drift \(-\tfrac12\).
Here \(L=2\), \(y=\tfrac12\), and \(q=e^{-1/2}\). Since \(2\mu=1\), $$ u_{1/2}\left(\tfrac12\right) =2\left[\frac{2(1-q)}{1-q^4}-\frac12\right]. $$ Using \(1-q^4=(1-q)(1+q)(1+q^2)\), cancel the nonzero factor \(1-q\) to obtain $$ u_{1/2}\left(\tfrac12\right)=\frac{4}{(1+q)(1+q^2)}-1. $$ For a numerical check, \(q\approx0.6065\) and \(q^2=e^{-1}\approx0.3679\). Thus \((1+q)(1+q^2)\approx1.6065\cdot1.3679\approx2.1975\), giving \(4/2.1975-1\approx0.8202\).
Reflect the starting point across the midpoint: \(2-\tfrac12=\tfrac32\), and reverse the drift to \(-\tfrac12\). The symmetry proved below gives \(u_{-1/2}(\tfrac32)=u_{1/2}(\tfrac12)\). This is a check of the drift sign and the location: a positive drift from the left side and an equally strong negative drift from the reflected right side give identical mean exit times.
Symmetry, Endpoints, and the Zero-Drift Limit
Proof. If \(\mu=0\), reflection gives \(y(L-y)=(L-y)y\), proving the symmetry. For \(\mu\ne0\), write \(k=2\mu\). Substituting drift \(-\mu\) and coordinate \(L-y\) into the formula gives $$ u_{-\mu}(b-y)=-\frac1\mu\left[\frac{L(1-e^{k(L-y)})}{1-e^{kL}}-(L-y)\right]. $$ Multiplying the numerator and denominator of the fraction by \(e^{-kL}\) gives the correct identity $$ \frac{1-e^{k(L-y)}}{1-e^{kL}} =\frac{e^{-ky}-e^{-kL}}{1-e^{-kL}}. $$ Therefore $$ u_{-\mu}(b-y) =-\frac1\mu\left[\frac{L(e^{-ky}-e^{-kL})}{1-e^{-kL}}-(L-y)\right] =\frac1\mu\left[\frac{L(1-e^{-ky})}{1-e^{-kL}}-y\right] =u_\mu(a+y). $$ The displayed algebra proves the claimed symmetry without changing the ratio to a different, generally unequal expression.
For the limit, expand the exponentials at \(\mu=0\): \(1-e^{-2\mu y}=2\mu y-2\mu^2y^2+O(\mu^3)\) and \(1-e^{-2\mu L}=2\mu L-2\mu^2L^2+O(\mu^3)\). Dividing and multiplying by \(L\) gives $$ \frac{L(1-e^{-2\mu y})}{1-e^{-2\mu L}} =y\big(1+\mu(L-y)+O(\mu^2)\big). $$ Subtracting \(y\), then dividing by \(\mu\), gives \(y(L-y)+O(\mu)\), which tends to the zero-drift formula. At \(y=0\) and \(y=L\), the original formulas give zero directly, so the endpoint values agree as well. \(\square\)
The formula's endpoint behavior is also direct: when the starting point reaches either boundary, the exit time is zero. For interior points the mean is positive, as follows either from its probabilistic interpretation and the fact that continuous paths started inside cannot exit at time zero, or from the solved boundary-value problem and its representation above.
A Common Algebra Pitfall
When checking drift-reversal symmetry, a tempting but false simplification is to replace $$ \frac{L(1-e^{k(L-y)})}{1-e^{kL}} $$ by \(L(1-e^{-ky})/(1-e^{-kL})\). For example, if \(k=1\), \(L=2\), and \(y=1\), the first ratio is \(2/(1+e)\), whereas the second is \(2/(1+e^{-1})\); these are not equal. The valid transformed ratio is \(L(e^{-ky}-e^{-kL})/(1-e^{-kL})\). The correction is to keep that expression and simplify the entire bracket, including the term \(-(L-y)\), as in the symmetry proof. This avoids an incorrect intermediate identity that can produce plausible-looking but wrong symmetry calculations.
The explicit formula is useful because it separates the effects of interval width, starting position, and drift. It also connects a probabilistic quantity to a boundary-value problem without treating the differential equation alone as proof: finiteness, stopping, integrability of the Ito integral, and the boundary limit are all needed to identify the solution with the expectation. In later exit problems, the same boundary-value approach can be applied to quantities other than the mean time.
Check Your Understanding
Use the definition of the exit time and the formulas above to answer these questions.
- For driftless Brownian motion on \((1,5)\), compute the mean exit time from \(x=2\) and from its reflected point \(x=4\).
- For drift \(\mu=-1\) on \((0,1)\), write the mean exit time from \(x=\tfrac14\) using the explicit formula. Check that its endpoint values are zero.
- Explain why the event \(B_h+\mu h\ge L\) has positive probability and how it gives a uniform bound on the tail of the exit time.
- In the reflection calculation, show by multiplying numerator and denominator by \(e^{-kL}\) that the ratio becomes \((e^{-ky}-e^{-kL})/(1-e^{-kL})\).
- Use the small-\(\mu\) expansions in the proof to recover the zero-drift mean exit time for an interior point.