Exit-Time Boundary Problems via Feynman-Kac

From an Exit Payoff to a Boundary Equation

Let a diffusion move inside a bounded interval until it first reaches an endpoint. Its payoff is then determined by the endpoint it reaches, possibly discounted according to the time spent inside. The corresponding boundary-value problem translates that expectation into an ordinary differential equation: the drift and diffusion coefficients enter through the generator, while the endpoint payoffs become boundary conditions.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Consider a strong solution of \(dX_t=\mu(X_t)\,dt+\sigma(X_t)\,dB_t\), starting at \(X_0=x\in(a,b)\), and define the exit time \(\tau=\inf\{t\ge0:X_t\notin(a,b)\}\). A strong solution is constructed on the given probability space using its Brownian motion; a weak solution may instead use a probability space chosen as part of the construction. Assume here that the strong solution has continuous paths and that \(\tau<\infty\) almost surely. Take \(\mu,\sigma\) bounded and Lipschitz on \([a,b]\), with \(\sigma\) bounded there. These hypotheses ensure the SDE is well posed up to exit; the separate assumption on \(\tau\) makes the exit payoff well defined.

For a twice continuously differentiable function \(f\), the generator on \((a,b)\) is \(\mathcal{L}f(x)=\mu(x)f'(x)+\tfrac12\sigma(x)^2f''(x)\). The drift \(\mu\) determines the first-derivative term, and the diffusion coefficient \(\sigma\) determines the second-derivative term. The factor \(\tfrac12\) is the Ito correction arising from \(dB_t\,dB_t=dt\), while \(dB_t\,dt=0\) and \(dt\,dt=0\).

The boundary equation associated with this value is \(\mathcal{L}u-qu=0\) on \((a,b)\), with \(u(a)=g(a)\) and \(u(b)=g(b)\). The equation says that the generator’s local change in value balances the discount. The boundary data specify the payoff once the diffusion exits.

A Verification Result

Proof. Set \(A_t=\int_0^t q(X_s)\,ds\) and \(Y_t=e^{-A_t}v(X_t)\). Ito’s formula applied to \(v\) gives \(v'(x)\) as the first derivative and \(v''(x)\) as the second derivative, including the correction \(\tfrac12\sigma(x)^2v''(x)\,dt\): \(dv(X_t)=v'(X_t)\sigma(X_t)\,dB_t+\mathcal{L}v(X_t)\,dt\). Since \(d(e^{-A_t})=-e^{-A_t}q(X_t)\,dt\), the product rule has no quadratic-covariation term with this finite-variation factor. Therefore, before exit, \(dY_t=e^{-A_t}\sigma(X_t)v'(X_t)\,dB_t+e^{-A_t}(\mathcal{L}v-qv)(X_t)\,dt\). The assumed equation makes the drift zero. For each deterministic \(t<\infty\), stopping at \(\tau\) yields \(Y_{t\wedge\tau}=v(x)+\int_0^t\mathbf{1}_{\{s<\tau\}}e^{-A_s}\sigma(X_s)v'(X_s)\,dB_s\).

The integrand is progressively measurable. Because \(q\ge0\), \(e^{-A_s}\le1\); because \(\sigma\) and \(v'\) are bounded on the interval, its squared integral has finite expectation on \([0,t]\). Thus it belongs to \(\mathcal{L}^2\) on this time horizon. The Ito integral is a true martingale: it is square-integrable by the Ito isometry, and its conditional increments have mean zero, so its conditional expectation at an earlier time equals its present value. In particular its expectation is zero, giving \(v(x)=\mathbb{E}_x[e^{-A_{t\wedge\tau}}v(X_{t\wedge\tau})]\).

As \(t\to\infty\), \(\tau<\infty\) almost surely, so \(t\wedge\tau\to\tau\) and \(X_{t\wedge\tau}\to X_\tau\) almost surely by path continuity. The integrand inside the expectation is bounded in absolute value by \(\sup_{[a,b]}|v|\), since \(q\ge0\). Dominated convergence therefore gives \(v(x)=\mathbb{E}_x[e^{-A_\tau}v(X_\tau)]\). At exit, \(X_\tau\) is an endpoint, and the boundary conditions give \(v(X_\tau)=g(X_\tau)\). This proves the representation. If \(v_1,v_2\) are two solutions, their difference \(w=v_1-v_2\) satisfies the same equation and zero boundary values. The representation just proved gives \(w(x)=\mathbb{E}_x[e^{-A_\tau}0]=0\) for every \(x\in(a,b)\); the boundary values are also zero. Hence the solution is unique. \(\square\)

This is a verification result: it proves that a sufficiently regular solution of the boundary equation has the desired probabilistic meaning. It does not assert, without further hypotheses, that every exit-payoff expectation is a classical solution. In the examples below, the exit time is almost surely finite; for bounded intervals and the uniformly nondegenerate diffusions used here, this is the standard one-dimensional exit-time property.

Worked Boundary Problems

Why the Generator Check Matters

The generator converts the diffusion’s local behavior into a deterministic equation. Once a candidate has been found, its boundary values alone are not enough: it must also satisfy the interior equation with the correct drift, diffusion coefficient, and discount term. The verification proof explains why these checks suffice. The discounted candidate evaluated along the stopped diffusion has zero drift; its remaining stochastic integral is a true martingale on each finite time interval; and bounded convergence passes from finite stopping times to the exit time.

Common pitfall: omitting Ito’s factor of one half. In the first example, replacing the correct equation \(\tfrac12u''-2u=0\) by \(u''-2u=0\) produces the boundary-matching candidate \(\widetilde u(x)=\sinh(\sqrt{2}x)/\sinh(\sqrt{2})\). But \(\widetilde u''=2\widetilde u\), so the actual generator gives \(\tfrac12\widetilde u''-2\widetilde u=\widetilde u-2\widetilde u=-\widetilde u\), which is not zero in the interval. The endpoint values are correct, but the function does not solve the required boundary problem and therefore cannot be verified by the theorem.