The Feynman-Kac theorem, Kolmogorov's Forward Equation, and the Fokker-Plank Equation

Let $T>0$ , $B(t)$ be a Brownian motion, $\mathcal{F}(t)$ the filtration generated by $B(t)$ , $X(t)$ a stochastic process adapted to $\mathcal{F}(t)$ , and $h: \mathbb{R} \to \mathbb{R}$ a measurable function. Suppose $X(t)$ satisfies:

\begin{equation} dX(t) = \alpha(t,X(t)) dt + \beta(t,X(t)) d B(t), \end{equation}

where $\alpha(t,X(t))$ , $\beta(t,X(t))$ are adapted processes. For $x \in \mathbb{R}$ and $0 < t < T$ , let $E^{x,t}$ be the expected value of $h(X(T))$ starting from $X(t) = x$ , and denote this function by $g(x,t)$ . If $\alpha(t,X(t)), \beta(t,X(t))$ are sufficiently nice, then $X(t)$ is a Markov process and

E[h(X(T)) \mid \mathcal{F}(t)] = g(t,X(t)).

Theorem 1 (Feynman-Kac) The function $g(t,x)$ satisfies

\begin{align} &g_t(t,x) + \alpha(t,X(t)) g_x(t,x) + \frac{1}{2} \beta^2(t,x) g_{xx}(t,x) = 0 \\ & g(T,x) = h(x). \end{align}

Proof. We first note that the process $g(t,X(t))$ is a martingale. Indeed, for $0< s < t$ ,

E[g(t,X(t)) \mid \mathcal{F}(s)] = E[ E[h(X(T)) \mid \mathcal{F}(t)] \mid \mathcal{F}(s)] = E[h(X(T)) \mid \mathcal{F}(s)] = g(s,X(s)).

We now use Itô's formula to compute the differential of $g$ :

d g(t,X(t)) = g_t dt + g_x dX(t) + \frac{1}{2} g_{xx} dX(t)dX(t) = \left(g_t + \alpha(t,X(t)) g_x + \frac{1}{2} \beta^2(t,x) g_{xx}\right)dt + g_x \beta(t,X(t)) dB(t).

Since $g(t,X(t))$ is a martingale, the $dt$ term must vanish, so

g_t(t,x) + \alpha(t,X(t)) g_x(t,x) + \frac{1}{2} \beta^2(t,x) g_{xx}(t,x) = 0.

Finally, since $h(X(T))$ is $\mathcal{F}(T)$ -measurable,

g(T,X(T)) = E[h(X(T)) \mid \mathcal{F}(T)] = h(X(T)),

which must hold for any value of $X(T)$ .

□

Since the process $X(t)$ is a Markov process, it has a probability transition density $p(t,T,x,y)$ , so if $X(t)=x$ , then

E[h(X(T)) \mid \mathcal{F}(t)] = \int h(y) p(t,T, x, y) dy.

From the Feynman-Kac theorem, we obtain the Kolmogorov Backward Equation.

Theorem: Let $p(t,T,x,y)$ be the transition density for the solution to the SDE in equation (1). Then,

p_t + \alpha p_x + \frac{1}{2} \beta^2 p_{xx} = 0.

Proof. From the Feynman-Kac theorem we know

g_t(t,x) + \alpha(t,X(t)) g_x(t,x) + \frac{1}{2} \beta^2(t,x) g_{xx}(t,x) = 0.

Since

g(t,x) = \int h(y) p(t,T,x,y) dy,

differentiating under the integral sign yields

\int h(y) (p_t + \alpha p_x + \frac{1}{2} \beta^2(t,x) p_{xx}) dy = 0.

This holds for all measurable functions $h$ , so

p_t + \alpha p_x + \frac{1}{2} \beta^2 p_{xx} = 0.

□

What this theorem tells us is that given $X(T) = y$ , we can solve for the probability that $X(t) = x$ as a function of $x$ and $t$ . Similarly, we can establish a forward equation for the transition function. Given the current position $X(t) = x$ , we wish to find the probability that $X(T) = y$ as a function of $T$ and $y$ for $T > t$ .

Theorem. (Kolmogorov Forward Equation/Fokker-Planck Equation). Let $p(t,T,x,y)$ be the transition density for the solution to the SDE in equation (1). Then $p(t,T,x,y)$ satisfies

\frac{\partial }{\partial T} p(t,T,x,y) + \frac{\partial}{\partial y} (\beta(t,X(t),y) p(t,T,x,y)) - \frac{1}{2} \frac{\partial^2}{\partial y^2} (\beta^2(T,y) p(t,T,x,y)) = 0.

Proof. Let $a < b$ , let $h$ be any smooth function compactly supported on $[a,b]$ , and suppose $X(t) = x \in (a,b)$ . Take for example

h(x) = \begin{cases} e^{\frac{1}{(x-a)(x-b)}} & x \in (a,b)\\ 0 & \text{o.w.} \end{cases}.

For $t< u < T$ ,

d(h(X(u))) = \left(\alpha(u,X(u)) h'(X(u)) + \frac{1}{2} \beta^2(u,X(u)) h''(X(u))\right) du + \beta(u,X(u)) h'(X(u)) dB(u).

Equivalently,

\begin{align*} &h(X(T)) = h(x) + \int_t^T \left(\alpha(u,X(u)) h'(X(u)) + \frac{1}{2} \beta^2(u,X(u)) h''(X(u))\right) du \\ & + \int_t^T \beta(u,X(u)) h'(X(u)) dB(u). \end{align*}

Taking expectations of both sides and noting that $h, h', h''$ are compactly supported on $(a,b)$ ,

\begin{align*} &\int_a^b h(y) p(t,T,x,y) dy = E [h(X(T))]\\ &= h(x) + \int_t^T E \left(\alpha(u,X(u)) h'(X(u)) + \frac{1}{2} \beta^2(u,X(u)) h''(X(u))\right) du \\ &=h(x) + \int_t^T \int_a^b \alpha(u,y) h'(y) p(t,u,x,y) dy du + \int_t^T \int_a^b \frac{1}{2} \beta^2(u,y) h''(u) p(t,u,x,y) dy du. \end{align*}

We now apply integration by parts once to the second term and twice to the third term and use the fact that $h(a) = h'(a) = h'(b) = h(b) = 0$ :

\begin{align*} &\int_a^b h(y) p(t,T,x,y) dy \\ &= h(x) - \int_t^T \int_a^b \frac{\partial}{\partial y} (\alpha(u,y) p(t,u,x,y)) h(y) dy du + \frac{1}{2} \int_t^T \int_a^b \frac{\partial^2}{\partial y^2}\left(\beta^2(u,y) p(t,u,x,y) \right) h(y) dy du. \end{align*}

Finally, differentiating both sides with respect to $T$ results in

\int_a^b h(y) p_T(t,T,x,y) dy = - \int_a^b \frac{\partial}{\partial y} (\alpha(T,y) p(t,T,x,y)) h(y) dy + \frac{1}{2} \int_a^b \frac{\partial^2}{\partial y^2}\left(\beta^2(T,y) p(t,T,x,y) \right) h(y) dy.

Hence,

\int_a^b h(y) \left(p_T(t,T,x,y) + \frac{\partial}{\partial y} (\alpha(T,y) p(t,T,x,y)) - \frac{1}{2}\frac{\partial^2}{\partial y^2}(\beta^2(T,y) p(t,T,x,y))\right) dy = 0.

This holds for all smooth functions compactly supported on $(a,b)$ . Since the set of smooth functions compactly supported on $(a,b)$ is dense in the set of continuous compactly supported functions on $(a,b)$ , it follows that

p_T(t,T,x,y) + \frac{\partial}{\partial y} (\alpha(T,y) p(t,T,x,y)) - \frac{1}{2}\frac{\partial^2}{\partial y^2}(\beta^2(T,y) p(t,T,x,y)) = 0 \text{ for all } y \in (a,b).

Since $a < b$ are arbitrary, we thus have

\frac{\partial }{\partial T} p(t,T,x,y) + \frac{\partial}{\partial y} (\beta(t,X(t),y) p(t,T,x,y)) - \frac{1}{2} \frac{\partial^2}{\partial y^2} (\beta^2(T,y) p(t,T,x,y)) = 0.

□