Feynman-Kac formula

The Feynman-Kac formula represents solutions of certain parabolic partial differential equations as conditional expectations of functionals of a diffusion. It follows by applying Itô formula to the solution along the diffusion and taking expectations.

For $\mathcal{L}=\frac{1}{2}\Delta -\lambda x\cdot \nabla$ and potential $V={\lambda }^2|x{|}^2/2$, the Feynman-Kac formula with initial value one gives $u={\mathbb{E}}_x^{OU}{e}^{{\int }_0^{t}V(X_s)ds}$. The Ornstein-Uhlenbeck likelihood relative to Wiener measure cancels the quadratic time integral, leaving a Gaussian endpoint integral. Completing the square gives the displayed answer for $1+\lambda t>0$. It is finite for all nonnegative times when $\lambda \ge 0$; when $\lambda <0$ the path expectation diverges at and beyond $t=-1/\lambda$.

For a bounded domain, a bounded classical solution with boundary data $g$, nonnegative bounded $V$, bounded $f$ and integrable diffusion exit time $\tau$ satisfiesApply the Itô product rule to the discounted stopped solution and add its source integral. The stopped martingale is dominated by $\Vert u{\Vert }_{\infty }+\Vert f{\Vert }_{\infty }\tau$, so dominated convergence permits passage to the exit. For $V=f=0$, this is a boundary-value averaging principle for functions harmonic under $L$.

For a nonexplosive diffusion process, bounded source $f$, terminal data $g$ and nonnegative bounded potential $V$, a bounded classical solution with $u(T,x)=g(x)$ isThe Itô product rule shows that the discounted solution plus its accumulated discounted source is a local martingale. Its deterministic finite-horizon bound upgrades it to a true martingale, giving the formula and uniqueness without a global derivative bound.

For a continuous Itô diffusion with diffusion generator $\mathcal{L}$, let $u$ solve $\mathcal{L}u=\lambda u$ with $\lambda >0$, be bounded on an open domain and its boundary, and equal $g$ on the boundary. Stop at the first boundary hit $\tau$. The Itô formula makes $e^{-\lambda (t\wedge \tau )}u(X_{t\wedge \tau })$ a bounded martingale. Its limit is zero on $\{\tau =\infty \}$ and equals $e^{-\lambda \tau }g(X_{\tau })$ on $\{\tau <\infty \}$. ThusBoundedness justifies passage to the terminal expectation even if hitting is not almost surely finite.

For bounded $V$ and bounded initial data $f$, a sufficiently regular classical solution bounded on every finite time slab satisfiesTo prove this, the Itô formula and Itô product rule make $u(T-t,B_t)e^{{\int }_0^{t}V(B_s)ds}$ a local martingale. Its deterministic bound on $[0,T]$ makes it a true martingale; taking expectations at the two endpoints gives the formula. This also gives uniqueness in the finite-horizon bounded class. The exponent has the same sign as the potential in the equation.

Let $\varphi$ be continuous, nonnegative, zero on $[0,\infty )$, and strictly positive on $(-\infty ,0)$. By continuity of Brownian paths, its occupation integral vanishes exactly on nonnegative paths. Dominated convergence gives the displayed limit. The Brownian reflection principle makes it $1-2{\mathbb{P}}_x(B_t\le 0)$ for $x\ge 0$ and zero for $x<0$. In particular the limit at $x=0$ is zero for every positive time. The choice $\varphi (x)=\min (1,\max (0,-x))$ is a bounded soft killing potential.

For globally Lipschitz drift $b$, couple $d{X}^{\epsilon }=b(X^{\epsilon })dt+\epsilon dW$ and $\dot{x}=b(x)$ from the same point. The Gronwall inequality gives ${\sup }_{s\le t}|X_s^{\epsilon }-x_s|\le \epsilon e^{Lt}{\sup }_{s\le t}|W_s|$. Thus convergence is uniform on each finite horizon. The Feynman-Kac formula for generator ${\epsilon }^2\Delta /2+b\cdot \nabla +c$ has weight $\exp \int c$, with a positive sign. Bounded continuous $c$ is uniformly continuous on a compact neighbourhood of the deterministic trajectory, and bounded continuous $f$ supplies a deterministic integrable bound. The dominated convergence theorem proves the displayed limit without requiring global uniform continuity of $c$.

Suppose bounded $V\ge 0$ is bounded below by a positive constant on some nonempty open interval, and $f\ge C>0$. The Feynman-Kac formula with a bounded potential gives $u(t,x)\ge C{\mathbb{E}}_x{e}^{{\int }_0^{t}V(B_s)ds}$. Infinite occupation time of one-dimensional Brownian motion makes that integral tend to infinity almost surely. The monotone convergence theorem then gives $u(t,x)\to \infty$ for every fixed $x$. Thus no globally bounded classical solution exists under these hypotheses, although boundedness on every finite time slab is compatible with the conclusion.