Deriving Mean Exit Times From the Generator

From an Exit-Time Expectation to a Boundary Problem

Let \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carry a standard Brownian motion \(B_t\), and fix constants \(\mu\in\mathbb{R}\), \(\sigma>0\), and \(a<b\). Starting from \(x\in(a,b)\), consider the drifted Brownian motion \(X_t=x+\mu t+\sigma B_t\) and its first exit time \(\tau=\inf\{t\ge0:X_t\notin(a,b)\}\). This explicit construction is a strong solution on the given space. The mean exit-time function is \(m(x)=\mathbb{E}_x[\tau]\), where \(\mathbb{E}_x\) denotes expectation for this process started at \(x\). The goal here is to derive the equation and boundary data that characterize \(m\), not to solve that equation explicitly.

For a twice continuously differentiable function \(f\), Ito’s formula gives the infinitesimal drift of \(f(X_t)\). Its drift coefficient is the generator \(\mathcal{L}f\); the first derivative comes from the drift of \(X\), and the second derivative comes from Brownian quadratic variation. In particular, the coefficient \(\tfrac12\sigma^2\) is essential: it is the product of the Taylor coefficient \(\tfrac12\) and \(dX_t\,dX_t=\sigma^2dt\).

Why Truncation Is Needed

To use Ito’s formula at an exit time that might be unbounded, first stop at a deterministic horizon \(t\): write \(t\wedge\tau=\min(t,\tau)\). If \(u\in C^2([a,b])\), Ito’s formula up to this bounded stopping time reads

Here the stochastic integral is an Ito integral on the stated filtered space. Its integrand, \(\mathbf{1}_{\{s\le\tau\}}\sigma u'(X_s)\), is progressively measurable. Since \(u'\) is bounded on \([a,b]\), for every finite \(t\) it satisfies \(\mathbb{E}\int_0^t\mathbf{1}_{\{s\le\tau\}}\sigma^2(u'(X_s))^2\,ds\le t\sigma^2\sup_{y\in[a,b]}|u'(y)|^2<\infty\); hence it belongs to \(\mathcal{L}^2\). By the Ito isometry, the integral has finite second moment, and by the martingale property of square-integrable Ito integrals its expectation is zero. Taking expectations therefore gives the stopped Dynkin identity \(\mathbb{E}_x[u(X_{t\wedge\tau})]=u(x)+\mathbb{E}_x\int_0^{t\wedge\tau}\mathcal{L}u(X_s)\,ds\).

The truncation is not cosmetic: it makes the integral’s square-integrability check immediate, without assuming in advance that \(\tau\) has a finite mean. For this particular process in a bounded interval, that finiteness can also be established directly.

Proof. Choose any \(h>0\) and define \(c=(b-a+|\mu|h)/\sigma\). If the process is still in \((a,b)\) at time \(nh\) and \(B_{(n+1)h}-B_{nh}>c\), then its displacement over the next interval is greater than \(\mu h+b-a+|\mu|h\ge b-a\). Since \(X_{nh}>a\), this forces \(X_{(n+1)h}>b\), so exit has occurred by \((n+1)h\). The increment is independent of \(\mathcal{F}_{nh}\) and has distribution \(N(0,h)\), so the event has a fixed probability \(p=\mathbb{P}(N(0,h)>c)>0\). Consequently \(\mathbb{P}_x(\tau>(n+1)h)\le(1-p)\mathbb{P}_x(\tau>nh)\), and induction gives \(\mathbb{P}_x(\tau>nh)\le(1-p)^n\). Integrating this geometric tail, or bounding the expectation by \(h\sum_{n\ge0}\mathbb{P}_x(\tau>nh)\), gives \(\mathbb{E}_x[\tau]\le h/p<\infty\). This holds for every \(x\in(a,b)\). \(\square\)

Deriving the Equation and Boundary Values

The derivation below states its regularity assumption openly. Suppose the mean exit-time function \(m\) is in \(C^2([a,b])\). This is the classical setting in which the generator equation can be derived pointwise by Ito’s formula; the derivation does not itself assert that regularity without proof. At either endpoint, if the process starts there, its exit time from the open interval is \(0\). Thus the boundary data are \(m(a)=m(b)=0\).

For an interior starting point, the strong Markov property gives a short-time decomposition: time spent before \(t\wedge\tau\) is followed, on the event \(\{\tau>t\}\), by a fresh expected remaining exit time \(m(X_t)\). Consequently \(m(x)=\mathbb{E}_x[t\wedge\tau]+\mathbb{E}_x[\mathbf{1}_{\{\tau>t\}}m(X_t)]\). Since \(m\) vanishes at both endpoints, the second term equals \(\mathbb{E}_x[m(X_{t\wedge\tau})]\). Applying the stopped Dynkin identity to \(m\) then identifies its generator.

Proof. For the first direction, fix \(x\in(a,b)\). The strong Markov property at the deterministic time \(t\), applied on \(\{\tau>t\}\), gives the decomposition above. The stopped Dynkin identity and \(m(X_{t\wedge\tau})=0\) on \(\{\tau\le t\}\) give \[ 0=\mathbb{E}_x[t\wedge\tau]+\mathbb{E}_x\int_0^{t\wedge\tau}\mathcal{L}m(X_s)\,ds =\mathbb{E}_x\int_0^{t\wedge\tau}\bigl(1+\mathcal{L}m(X_s)\bigr)\,ds. \] The integrand is bounded and continuous on \([a,b]\). Divide by \(t\) and let \(t\downarrow0\). By path continuity and \(x\in(a,b)\), \(\tau>0\) almost surely, and \(X_s\to x\); bounded convergence therefore makes the last expression divided by \(t\) tend to \(1+\mathcal{L}m(x)\). It must be zero, proving \(\mathcal{L}m(x)=-1\). The endpoint values follow from the definition of exit when started at an endpoint.

For the converse, use the stopped Dynkin identity with \(u\). Its assumptions hold because \(u'\) is bounded on the compact interval, so the Ito integrand is in \(\mathcal{L}^2\) on every finite horizon. Since \(u(a)=u(b)=0\), \(u(X_{t\wedge\tau})=0\) when \(\tau\le t\); otherwise it is \(u(X_t)\). The equation \(\mathcal{L}u=-1\) yields \(u(x)=\mathbb{E}_x[t\wedge\tau]+\mathbb{E}_x[u(X_t)\mathbf{1}_{\{\tau>t\}}]\). The function \(u\) is bounded, while \(\mathbb{P}_x(\tau>t)\to0\) because \(\tau<\infty\) almost surely by the finite-mean argument. Thus the last expectation tends to zero. Also \(t\wedge\tau\to\tau\) and \(0\le t\wedge\tau\le\tau\), with \(\mathbb{E}_x[\tau]<\infty\), so dominated convergence gives \(\mathbb{E}_x[t\wedge\tau]\to\mathbb{E}_x[\tau]\). Hence \(u(x)=\mathbb{E}_x[\tau]\), as required. \(\square\)

Worked Examples: Reading the Generator Equation

Common Pitfall: Dropping the Stopping or the Ito Correction

A tempting but incorrect calculation is to apply the ordinary chain rule to \(f(X_t)\), writing only \(f'(X_t)\,dX_t\). For \(f(z)=z^2\), that gives \(d(X_t^2)=2X_t\,dX_t\) and omits the term \(\sigma^2dt\). The correct Ito expansion is \(d(X_t^2)=2X_t(\mu\,dt+\sigma\,dB_t)+\sigma^2dt\), since \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\). The missing term changes the generator and therefore changes the mean-exit-time equation.

A second error is to take expectations of an unstopped stochastic integral up to \(\tau\) without checking its integrability. The bounded-horizon proof avoids that shortcut: it uses \(t\wedge\tau\), verifies the Ito integrand lies in \(\mathcal{L}^2\), and only then passes to the limit using finite mean and boundedness. This is why the boundary-value equation is derived from a justified identity rather than from formal manipulation alone.