Proof of the Classical Feynman-Kac Theorem

From a Backward Equation to an Expectation

The Feynman–Kac formula links a parabolic partial differential equation to the distribution of a diffusion. The direction considered here starts with a classical solution of a backward equation and proves that its value equals an expectation of the terminal payoff. The central step is to apply Itô’s formula to the solution along the diffusion path. The PDE then cancels the drift, leaving a stochastic integral; localization and boundedness justify taking expectations and passing to the terminal time.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Fix $T>0$, a starting time $s\in[0,T]$, and a starting state $x\in\mathbb{R}$. Consider a continuous adapted solution of

X_t=x+\int_s^t\mu(r,X_r)\,dr+\int_s^t\sigma(r,X_r)\,dB_r,\qquad s\le t\le T.

Here $\mu$ is the drift coefficient and $\sigma$ is the diffusion coefficient. Assume both are bounded and globally Lipschitz in the state variable, uniformly in time, and continuous in time. These hypotheses give a nonexplosive strong solution; the argument below also applies to any continuous nonexplosive weak solution. A strong solution is constructed on the given Brownian probability space and is adapted to its filtration. A weak solution allows the probability space and Brownian motion to be part of the solution. The representation proved below holds for either solution whenever it has the stated SDE and assumptions.

Theorem: Let $g:\mathbb{R}\to\mathbb{R}$ be bounded and continuous. Suppose $u:[0,T]\times\mathbb{R}\to\mathbb{R}$ is bounded, has continuous derivatives $u_t,u_x,u_{xx}$, and satisfies $$ u_t(t,y)+\mu(t,y)u_x(t,y)+\tfrac12\sigma(t,y)^2u_{xx}(t,y)=0,\qquad 0\le t<T,\ y\in\mathbb{R}, $$ with terminal condition $u(T,y)=g(y)$. Then, for every $s\in[0,T]$ and $x\in\mathbb{R}$, $$ u(s,x)=\mathbb{E}\big[g(X_T)\mid X_s=x\big]. $$ Equivalently, for a solution started at the deterministic state $x$ at time $s$, $u(s,x)=\mathbb{E}[g(X_T)]$. The conditional notation refers to that initial-state interpretation; it does not assert conditioning on a generally zero-probability event.

The boundedness of $u$ is a convenient sufficient growth condition for the limit in the proof. Its derivatives need only be locally bounded: continuity makes them bounded on every compact subset of $[0,T]\times\mathbb{R}$. No uniform bound on $u_x$ or $u_{xx}$ over the whole real line is required.

Why Stopping Produces a True Martingale

For $R>|x|$, define the exit time $\tau_R=\inf\{t\ge s:|X_t|\ge R\}$, taking the infimum of the empty set to be $+\infty$. Before this time, the state remains in a compact interval, where the derivatives of $u$ are bounded. This makes the stochastic integral in Itô’s formula square-integrable, rather than merely a local martingale.

Lemma: Under the theorem’s assumptions, for $s\le t\le T$, $$ u(t\wedge\tau_R,X_{t\wedge\tau_R}) =u(s,x)+\int_s^{t\wedge\tau_R}\sigma(r,X_r)u_x(r,X_r)\,dB_r $$ is a martingale.

Proof. Apply Itô’s formula to $u(r,X_r)$ up to $t\wedge\tau_R$. The formula has time derivative $u_t$, first state derivative $u_x$, and second state derivative $u_{xx}$; its correction term is $\tfrac12u_{xx}\,d[X]_r=\tfrac12u_{xx}\sigma(r,X_r)^2\,dr$. Thus its drift is \[ \big(u_t+\mu u_x+\tfrac12\sigma^2u_{xx}\big)(r,X_r)\,dr, \] which is zero by the PDE. The remaining integrand is $H_r=\mathbf{1}_{\{r\le\tau_R\}}\sigma(r,X_r)u_x(r,X_r)$. On $[s,T]\times[-R,R]$, $u_x$ is bounded, and $\sigma$ is bounded, so $\mathbb{E}\int_s^T H_r^2\,dr<\infty$. The progressively measurable integrand $H$ is therefore in $\mathcal{L}^2$. Its Itô integral is a square-integrable martingale: in particular, its increments have conditional expectation zero given the earlier filtration. The displayed process is consequently a martingale. $\square$

This argument distinguishes a true martingale from a local martingale. Without stopping, the derivatives of $u$ could grow so fast that the stochastic integral is only locally square-integrable; a local martingale need not have the integrability needed to take expectations. Stopping controls the integrand, and the Itô isometry gives finite second moment.

Proof of the Representation

Proof. Fix $R>|x|$. By the lemma, the stopped process has constant expectation, so at $t=T$, \[ u(s,x)=\mathbb{E}\big[u(T\wedge\tau_R,X_{T\wedge\tau_R})\big]. \] As $R\to\infty$, the exit times exceed $T$ eventually on almost every path. Indeed, a continuous nonexplosive path is bounded on the compact time interval $[s,T]$; once $R$ exceeds its maximum absolute value, $\tau_R>T$. Hence $u(T\wedge\tau_R,X_{T\wedge\tau_R})\to u(T,X_T)=g(X_T)$ almost surely. The terms are bounded in absolute value by $\sup_{t,y}|u(t,y)|$. The dominated convergence theorem therefore gives \[ u(s,x)=\mathbb{E}[g(X_T)]. \] This proves the representation for the process started at $(s,x)$, and hence the stated conditional, initial-state interpretation. The terminal condition is used only after the stopped expectations have been justified; the passage to the terminal time is not an optional-stopping assertion for an unbounded local martingale. $\square$

Worked Representations

Worked Example: Brownian Motion with a Cosine Payoff

Let $X_t=x+B_t-B_s$ for $s\le t\le T$, and let $g(y)=\cos y$. Here $\mu=0$, $\sigma=1$, and $\tau=T-t$. The proposed function is \[ u(t,y)=e^{-(T-t)/2}\cos y. \] Its derivatives are $u_t=\tfrac12e^{-(T-t)/2}\cos y$, $u_y=-e^{-(T-t)/2}\sin y$, and $u_{yy}=-e^{-(T-t)/2}\cos y$. Substitution into the PDE gives \[ u_t+\tfrac12u_{yy} =\tfrac12e^{-(T-t)/2}\cos y-\tfrac12e^{-(T-t)/2}\cos y=0. \] Also $u(T,y)=\cos y=g(y)$; the function and its derivatives are bounded. Thus the theorem gives \[ \mathbb{E}[\cos(X_T)]=e^{-(T-s)/2}\cos x. \]

Independent check: $X_T=x+Z$, where $Z$ is normal with mean $0$ and variance $T-s$. Using the normal characteristic function, $\mathbb{E}[e^{iZ}]=e^{-(T-s)/2}$, so taking the real part of $e^{ix}\mathbb{E}[e^{iZ}]$ gives the same expectation. For $s=0,T=2,x=0$, this equals $e^{-1}$, as the direct normal formula also gives.

Worked Example: Drifted Brownian Motion with a Sine Payoff

Let $X_t=x+\tfrac12(t-s)+(B_t-B_s)$, with $g(y)=\sin(2y)$. Set $\tau=T-t$ and define \[ u(t,y)=e^{-2(T-t)}\sin\big(2y+(T-t)\big). \] Writing $\theta=2y+\tau$, its derivatives are $u_y=2e^{-2\tau}\cos\theta$, $u_{yy}=-4e^{-2\tau}\sin\theta$, and $u_t=e^{-2\tau}(2\sin\theta-\cos\theta)$. Since $\mu=\tfrac12$ and $\sigma=1$, \[ u_t+\tfrac12u_y+\tfrac12u_{yy} =e^{-2\tau}(2\sin\theta-\cos\theta)+e^{-2\tau}\cos\theta-2e^{-2\tau}\sin\theta=0. \] At $t=T$, $\tau=0$, so $u(T,y)=\sin(2y)$. The theorem gives $u(s,x)=\mathbb{E}[\sin(2X_T)]$.

Independent check: $X_T=x+\tfrac12(T-s)+Z$, with $Z$ normal of variance $T-s$. The imaginary part of $e^{2ix}e^{i(T-s)}\mathbb{E}[e^{2iZ}]$ is $e^{-2(T-s)}\sin(2x+T-s)$, because $\mathbb{E}[e^{2iZ}]=e^{-2(T-s)}$. This is exactly $u(s,x)$. For $T-s=1,x=0$, both calculations give $e^{-2}\sin(1)$.

Worked Example: A Degenerate Diffusion

Take $X_t=x+(t-s)$, so $\mu=1$ and $\sigma=0$, and let $g(y)=\cos(2y)$. Set $u(t,y)=\cos(2(y+T-t))$. Its derivatives are $u_t=2\sin(2(y+T-t))$, $u_y=-2\sin(2(y+T-t))$, and $u_{yy}=-4\cos(2(y+T-t))$. The PDE reduces to $u_t+u_y=0$, verified by these derivatives; the second-derivative term is multiplied by $\sigma^2=0$. The terminal value is $u(T,y)=\cos(2y)$. Thus $u(s,x)=\cos(2(x+T-s))$.

Check: the solution path is deterministic, $X_T=x+T-s$, so $g(X_T)=\cos(2(x+T-s))$, precisely the proposed value. This also shows that the theorem does not require a strictly positive diffusion coefficient.

Why the Itô Correction Cannot Be Dropped

For a process $X$ with drift $\mu$ and diffusion coefficient $\sigma$, its quadratic variation satisfies $d[X]_t=\sigma(t,X_t)^2dt$. Itô’s formula therefore contains the correction $\tfrac12\sigma^2u_{xx}\,dt$. In differential notation, the multiplication rules are $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$. These are Itô rules, not the ordinary chain rule and not Stratonovich integration rules. The integrals in the proof are Itô integrals.

Common mistake: For Brownian motion and $u(t,y)=e^{-(T-t)/2}\cos y$, applying the ordinary chain rule would retain only $u_t\,dt+u_y\,dB_t$. Since $u_t=\tfrac12e^{-(T-t)/2}\cos y$, that calculation incorrectly leaves a nonzero drift and might suggest the proposed $u$ fails. It omits the Itô correction, which is \[ \tfrac12u_{yy}\,dt=-\tfrac12e^{-(T-t)/2}\cos y\,dt. \] Adding the two drift terms gives zero. The missing term is exactly what allows the PDE and the expectation representation to agree.

The theorem’s hypotheses serve separate purposes: regularity permits Itô’s formula; the backward equation cancels its drift; localization controls the stochastic integral; and boundedness permits dominated convergence. With unbounded solutions, a growth condition and suitable moment or uniform-integrability estimates are needed instead. The bounded argument above does not by itself establish those extensions.

Check Your Understanding

For each question, identify which assumption or calculation is doing the work.

Why does stopping at $\tau_R$ make the Itô integral in the proof square-integrable?
For $\mu=0,\sigma=1$, verify the PDE for $u(t,y)=e^{-(T-t)/2}\sin y$ and state its terminal payoff.
Which step in the proof uses boundedness of $u$, rather than boundedness of its derivatives?
For $X_t=x+a(t-s)$ with $\sigma=0$, what is the representation of a bounded continuous terminal payoff $g$?
Explain why saying that the unstopped Itô integral is a local martingale would not alone justify taking its expectation.