# Feynman–Kac formula Wikipedia open wikipedia design.

The Feynman–Kac formula named after Richard Feynman and Mark Kac, establishes a link between parabolic partial differential equations (PDEs) and stochastic processes. When Mark Kac and Richard Feynman were both on Cornell faculty, Kac attended a lecture of Feynman's and remarked that the two of them were working on the same thing from different directions. The Feynman-Kac formula resulted, which proves rigorously the real case of Feynman's path integrals. The complex case, which occurs when a particle's spin is included, is still unproven.

It offers a method of solving certain partial differential equations by simulating random paths of a stochastic process. Conversely, an important class of expectations of random processes can be computed by deterministic methods.

## Theorem

Consider the partial differential equation

${\frac {\partial u}{\partial t}}(x,t)+\mu (x,t){\frac {\partial u}{\partial x}}(x,t)+{\tfrac {1}{2}}\sigma ^{2}(x,t){\frac {\partial ^{2}u}{\partial x^{2}}}(x,t)-V(x,t)u(x,t)+f(x,t)=0,$ defined for all $x\in \mathbb {R}$ and $t\in [0,T]$ , subject to the terminal condition

$u(x,T)=\psi (x),$ where μ, σ, ψ, V, f are known functions, T is a parameter and $u:\mathbb {R} \times [0,T]\to \mathbb {R}$ is the unknown. Then the Feynman–Kac formula tells us that the solution can be written as a conditional expectation

$u(x,t)=E^{Q}\left[\int _{t}^{T}e^{-\int _{t}^{r}V(X_{\tau },\tau )\,d\tau }f(X_{r},r)dr+e^{-\int _{t}^{T}V(X_{\tau },\tau )\,d\tau }\psi (X_{T}){\Bigg |}X_{t}=x\right]$ under the probability measure Q such that X is an Itô process driven by the equation

$dX=\mu (X,t)\,dt+\sigma (X,t)\,dW^{Q},$ with WQ(t) is a Wiener process (also called Brownian motion) under Q, and the initial condition for X(t) is X(t) = x.

## Proof

A proof that the above formula is a solution of the differential equation is long, difficult and not presented here. It is however reasonably straightforward to show that, if a solution exists, it must have the above form. The proof of that lesser result is as follows.

Let u(x, t) be the solution to the above partial differential equation. Applying the product rule for Itô processes to the process

$Y(s)=e^{-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau }u(X_{s},s)+\int _{t}^{s}e^{-\int _{t}^{r}V(X_{\tau },\tau )\,d\tau }f(X_{r},r)\,dr$ one gets

{\begin{aligned}dY={}&d\left(e^{-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau }\right)u(X_{s},s)+e^{-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau }\,du(X_{s},s)\\[6pt]&{}+d\left(e^{-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau }\right)du(X_{s},s)+d\left(\int _{t}^{s}e^{-\int _{t}^{r}V(X_{\tau },\tau )\,d\tau }f(X_{r},r)\,dr\right)\end{aligned}} Since

$d\left(e^{-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau }\right)=-V(X_{s},s)e^{-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau }\,ds,$ the third term is $O(dt\,du)$ and can be dropped. We also have that

$d\left(\int _{t}^{s}e^{-\int _{t}^{r}V(X_{\tau },\tau )\,d\tau }f(X_{r},r)dr\right)=e^{-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau }f(X_{s},s)ds.$ Applying Itô's lemma to $du(X_{s},s)$ , it follows that

{\begin{aligned}dY={}&e^{-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau }\,\left(-V(X_{s},s)u(X_{s},s)+f(X_{s},s)+\mu (X_{s},s){\frac {\partial u}{\partial X}}+{\frac {\partial u}{\partial s}}+{\tfrac {1}{2}}\sigma ^{2}(X_{s},s){\frac {\partial ^{2}u}{\partial X^{2}}}\right)\,ds\\[6pt]&{}+e^{-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau }\sigma (X,s){\frac {\partial u}{\partial X}}\,dW.\end{aligned}} The first term contains, in parentheses, the above partial differential equation and is therefore zero. What remains is

$dY=e^{-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau }\sigma (X,s){\frac {\partial u}{\partial X}}\,dW.$ Integrating this equation from t to T, one concludes that

$Y(T)-Y(t)=\int _{t}^{T}e^{-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau }\sigma (X,s){\frac {\partial u}{\partial X}}\,dW.$ Upon taking expectations, conditioned on Xt = x, and observing that the right side is an Itô integral, which has expectation zero[citation needed], it follows that

$E[Y(T)\mid X_{t}=x]=E[Y(t)\mid X_{t}=x]=u(x,t).$ The desired result is obtained by observing that

$E[Y(T)\mid X_{t}=x]=E\left[e^{-\int _{t}^{T}V(X_{\tau },\tau )\,d\tau }u(X_{T},T)+\int _{t}^{T}e^{-\int _{t}^{r}V(X_{\tau },\tau )\,d\tau }f(X_{r},r)\,dr\,{\Bigg |}\,X_{t}=x\right]$ and finally

$u(x,t)=E\left[e^{-\int _{t}^{T}V(X_{\tau },\tau )\,d\tau }\psi (X_{T})+\int _{t}^{T}e^{-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau }f(X_{s},s)\,ds\,{\Bigg |}\,X_{t}=x\right]$ ## Remarks

• The proof above that a solution must have the given form is essentially that of  with modifications to account for $f(x,t)$ .
• The expectation formula above is also valid for N-dimensional Itô diffusions. The corresponding partial differential equation for $u:\mathbb {R} ^{N}\times [0,T]\to \mathbb {R}$ becomes:
${\frac {\partial u}{\partial t}}+\sum _{i=1}^{N}\mu _{i}(x,t){\frac {\partial u}{\partial x_{i}}}+{\frac {1}{2}}\sum _{i=1}^{N}\sum _{j=1}^{N}\gamma _{ij}(x,t){\frac {\partial ^{2}u}{\partial x_{i}\partial x_{j}}}-r(x,t)\,u=f(x,t),$ where,
$\gamma _{ij}(x,t)=\sum _{k=1}^{N}\sigma _{ik}(x,t)\sigma _{jk}(x,t),$ i.e. $\gamma =\sigma \sigma ^{\mathrm {T} }$ , where $\sigma ^{\mathrm {T} }$ denotes the transpose of $\sigma$ .
• This expectation can then be approximated using Monte Carlo or quasi-Monte Carlo methods.
• When originally published by Kac in 1949, the Feynman–Kac formula was presented as a formula for determining the distribution of certain Wiener functionals. Suppose we wish to find the expected value of the function
$e^{-\int _{0}^{t}V(x(\tau ))\,d\tau }$ in the case where x(τ) is some realization of a diffusion process starting at x(0) = 0. The Feynman–Kac formula says that this expectation is equivalent to the integral of a solution to a diffusion equation. Specifically, under the conditions that $uV(x)\geq 0$ ,
$E\left[e^{-u\int _{0}^{t}V(x(\tau ))\,d\tau }\right]=\int _{-\infty }^{\infty }w(x,t)\,dx$ where w(x, 0) = δ(x) and
${\frac {\partial w}{\partial t}}={\frac {1}{2}}{\frac {\partial ^{2}w}{\partial x^{2}}}-uV(x)w.$ The Feynman–Kac formula can also be interpreted as a method for evaluating functional integrals of a certain form. If
$I=\int f(x(0))e^{-u\int _{0}^{t}V(x(t))\,dt}g(x(t))\,Dx$ where the integral is taken over all random walks, then
$I=\int w(x,t)g(x)\,dx$ where w(x, t) is a solution to the parabolic partial differential equation
${\frac {\partial w}{\partial t}}={\frac {1}{2}}{\frac {\partial ^{2}w}{\partial x^{2}}}-uV(x)w$ with initial condition w(x, 0) = f(x).

## Applications

In quantitative finance, the Feynman–Kac formula is used to efficiently calculate solutions to the Black–Scholes equation to price options on stocks.