Exit Probabilities and Mean Exit Times on Bounded Intervals

Exit from a Bounded Interval

Consider a one-dimensional diffusion started inside a finite interval. Two basic questions are: which endpoint will it reach first, and how long will it take to reach either endpoint? Scale measures the relative tendency of the diffusion to move toward the endpoints; the speed density weights positions according to how long the diffusion spends there. Together they give explicit formulas for both quantities.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). On an open interval containing \([a,b]\), where \(a<b\), suppose \(X_t\) is a continuous solution of \(dX_t=\mu(X_t)\,dt+\sigma(X_t)\,dB_t\), with \(X_0=x\in(a,b)\). Assume that \(\mu\) and \(\sigma\) are continuous, that \(\sigma\) is continuously differentiable, and that \(\sigma(y)\ne0\) for \(y\in[a,b]\). Assume also that the solution exists up to its first exit from \([a,b]\). These conditions ensure that the coefficients are bounded on the interval and that the scale and speed densities below are well-defined. Define the exit time \(\tau=\inf\{t\ge0:X_t\notin(a,b)\}\).

The generator acting on twice continuously differentiable functions is \(\mathcal{L}f(y)=\mu(y)f'(y)+\tfrac12\sigma^2(y)f''(y)\). The scale density and a corresponding scale function are \(s'(y)=\exp\!\left(-\int_{y_0}^{y}\frac{2\mu(z)}{\sigma^2(z)}\,dz\right)\) and \(s(y)=\int_{y_0}^{y}s'(z)\,dz\), where \(y_0\) is any fixed reference point. Changing the reference point or multiplying the scale function by a positive constant does not alter the formulas. The speed density in this convention is \(m'(y)=2/(\sigma^2(y)s'(y))\).

Exit Probabilities from Scale

Proof. The finiteness of \(\tau\) will follow from the mean-exit-time argument below, which proves \(\mathbb{E}_x[\tau]<\infty\). Define \(v(y)=(s(y)-s(a))/(s(b)-s(a))\). Differentiating the scale density gives \(s''(y)=-(2\mu(y)/\sigma^2(y))s'(y)\); hence \(\mathcal{L}s(y)=\mu(y)s'(y)+\tfrac12\sigma^2(y)s''(y)=0\), and therefore \(\mathcal{L}v=0\). The derivatives of \(v\) are bounded on \([a,b]\). For each finite \(t\), Itô’s formula applied to \(v(X_{t\wedge\tau})\) gives \(v(X_{t\wedge\tau})=v(x)+\int_0^{t\wedge\tau}v'(X_r)\sigma(X_r)\,dB_r\). This is an Ito integral with progressively measurable integrand \(H_r=\mathbf{1}_{\{r\le\tau\}}v'(X_r)\sigma(X_r)\). It belongs to \(\mathcal{L}^2\) on \([0,t]\), since \(v'\) and \(\sigma\) are bounded there and \(\mathbb{E}\int_0^t H_r^2\,dr\le t\sup_{[a,b]}|v'\sigma|^2<\infty\). The Ito integral is consequently a martingale with mean zero, so \(\mathbb{E}_x[v(X_{t\wedge\tau})]=v(x)\). Since \(\tau<\infty\) almost surely, continuity gives \(X_{t\wedge\tau}\to X_\tau\) almost surely. The function \(v\) is bounded, so bounded convergence yields \(\mathbb{E}_x[v(X_\tau)]=v(x)\). At exit the continuous path is at \(a\) or \(b\), and \(v(a)=0,\ v(b)=1\). Thus \(v(X_\tau)=\mathbf{1}_{\{X_\tau=b\}}\), proving the upper-endpoint formula. The lower-endpoint formula follows because the two endpoint events are disjoint and exhaust the possibilities. \(\square\)

Mean Exit Time and the Green Kernel

Proof. Fix \(x\in(a,b)\) and define \(u(x)=\int_a^b G(x,y)m'(y)\,dy\). The integrand is nonnegative, and the coefficients, scale density and speed density are bounded on \([a,b]\); hence this integral is finite. We verify its differential equation. For fixed \(y\), as a function of its first argument \(z\), \(G(z,y)\) is a constant multiple of \(s(z)\) on each of \(z<y\) and \(z>y\). Thus \(\mathcal{L}_zG(z,y)=0\) away from \(z=y\). Its derivative with respect to \(z\) has jump \[ \partial_zG(y+,y)-\partial_zG(y-,y) =-\frac{s'(y)(s(y)-s(a))+(s(b)-s(y))s'(y)}{s(b)-s(a)} =-s'(y). \] Consequently, differentiating the integral on its two branches gives the second-derivative contribution \(-s'(x)m'(x)=-2/\sigma^2(x)\) at the join. The remaining terms are annihilated by \(\mathcal{L}\), because each branch is proportional to \(s\). Therefore \(\mathcal{L}u(x)=\tfrac12\sigma^2(x)(-2/\sigma^2(x))=-1\). Also \(u(a)=u(b)=0\), since the kernel vanishes at either endpoint, and \(u\ge0\), since \(G\ge0\) and \(m'\gt0\).

Apply Ito’s formula to \(u(X_{t\wedge\tau})\). The resulting Ito integral has integrand \(H_r=\mathbf{1}_{\{r\le\tau\}}u'(X_r)\sigma(X_r)\), which is progressively measurable and satisfies \(\mathbb{E}\int_0^t H_r^2\,dr<\infty\) because \(u'\) and \(\sigma\) are bounded on the compact interval. Its expectation is zero. Thus \(\mathbb{E}_x[u(X_{t\wedge\tau})]=u(x)-\mathbb{E}_x[t\wedge\tau]\). Since \(u\ge0\), \(\mathbb{E}_x[t\wedge\tau]\le u(x)\) for every \(t\). Monotone convergence gives \(\mathbb{E}_x[\tau]\le u(x)<\infty\), so \(\tau\) is finite almost surely. Now \(u(X_{t\wedge\tau})\to u(X_\tau)=0\) almost surely; bounded convergence applies because \(u\) is bounded on \([a,b]\). Taking \(t\to\infty\) in the identity gives \(\mathbb{E}_x[\tau]=u(x)\), as claimed. \(\square\)

This proof also justifies the finiteness needed for the exit-probability theorem. The Green kernel is piecewise linear in scale, not generally linear in the spatial coordinate. Its denominator \(s(b)-s(a)\) is essential: it normalizes the two branches so that their slopes have the derivative jump required by \(\mathcal{L}u=-1\).

Worked Calculations for Brownian Motion with Drift

Common Pitfall: Dropping the Scale Denominator

On the interval in the second example, for \(y\le x=\ln2\), using only \(s(y)-s(0)=1-e^{-y}\) as the kernel is wrong. The branch must also include \((s(\ln4)-s(\ln2))/(s(\ln4)-s(0))=(1/4)/(3/4)=1/3\). The correct branch is therefore \(\tfrac13(1-e^{-y})\), not \(1-e^{-y}\). After multiplying by \(m'(y)=2e^y\), its contribution to the mean is \(\tfrac23(1-\ln2)\). The other branch contributes \(\tfrac43(\ln2-\tfrac12)\); their sum is \(\tfrac23\ln2\). Keeping the kernel normalization and recomputing each integral prevents both a branch error and a later arithmetic error.

The formulas rely on the diffusion staying within a bounded interval until it hits an endpoint, with a nonzero diffusion coefficient throughout that interval. The bounded-interval argument establishes finite mean exit time under the stated assumptions; questions about reaching or escaping an unbounded boundary require additional analysis.