# Feynman–Kac formula

°The Feynman–Kac formula named after Richard Feynman and Mark Kac, establishes a link between parabolic partial differential equations (PDEs) and stochastic processes. It offers a method of solving certain PDEs by simulating random paths of a stochastic process. Conversely, an important class of expectations of random processes can be computed by deterministic methods. Consider the PDE

${\displaystyle {\frac {\partial u}{\partial t}}(x,t)+\mu (x,t){\frac {\partial u}{\partial x}}(x,t)+{\tfrac {1}{2}}\sigma ^{2}(x,t){\frac {\partial ^{2}u}{\partial x^{2}}}(x,t)-V(x,t)u(x,t)+f(x,t)=0,}$

defined for all x in R and t in [0, T], subject to the terminal condition

${\displaystyle u(x,T)=\psi (x),}$

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

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

under the probability measure Q such that X is an Itō process driven by the equation

${\displaystyle dX=\mu (X,t)\,dt+\sigma (X,t)\,dW^{Q},}$

with WQ(t) is a Wiener process (also called Brownian motion) under Q, and the initial condition for X(t) is X(0) = x.

## Proof

Let u(x, t) be the solution to above PDE. Applying Itō's lemma to the process

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

one gets

${\displaystyle dY=d\left(e^{-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau }\right)u(X_{s},s)+e^{-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau }\,du(X_{s},s)+d\left(e^{-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau }\right)du(X_{s},s)+d\left(\int _{t}^{s}e^{-\int _{t}^{r}V(X_{\tau },\tau )\,d\tau }f(X_{r},r)dr\right)}$

Since

${\displaystyle d\left(e^{-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau }\right)=-V(X_{s},s)e^{-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau }\,ds,}$

the third term is ${\displaystyle O(dtdu)}$ and can be dropped. We also have that

${\displaystyle d\left(\int _{t}^{s}e^{-\int _{t}^{r}V(X_{\tau },\tau )\,d\tau }f(X_{r},r)dr\right)=e^{-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau }f(X_{s},s)ds.}$

Applying Itō's lemma once again to ${\displaystyle du(X_{s},s)}$, it follows that

${\displaystyle dY=e^{-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau }\,\left(-V(X_{s},s)u(X_{s},s)+f(X_{s},s)+\mu (X_{s},s){\frac {\partial u}{\partial X}}+{\frac {\partial u}{\partial s}}+{\tfrac {1}{2}}\sigma ^{2}(X_{s},s){\frac {\partial ^{2}u}{\partial X^{2}}}\right)\,ds+e^{-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau }\sigma (X,s){\frac {\partial u}{\partial X}}\,dW.}$

The first term contains, in parentheses, the above PDE and is therefore zero. What remains is

${\displaystyle dY=e^{-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau }\sigma (X,s){\frac {\partial u}{\partial X}}\,dW.}$

Integrating this equation from t to T, one concludes that

${\displaystyle Y(T)-Y(t)=\int _{t}^{T}e^{-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau }\sigma (X,s){\frac {\partial u}{\partial X}}\,dW.}$

Upon taking expectations, conditioned on Xt = x, and observing that the right side is an Itō integral, which has expectation zero, it follows that

${\displaystyle E[Y(T)|X_{t}=x]=E[Y(t)|X_{t}=x]=u(x,t).}$

The desired result is obtained by observing that

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

and finally

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

## Remarks

${\displaystyle {\frac {\partial u}{\partial t}}+\sum _{i=1}^{N}\mu _{i}(x,t){\frac {\partial u}{\partial x_{i}}}+{\tfrac {1}{2}}\sum _{i=1}^{N}\sum _{j=1}^{N}\gamma _{ij}(x,t){\frac {\partial ^{2}u}{\partial x_{i}x_{j}}}-r(x,t)u=f(x,t),}$
where,
${\displaystyle \gamma _{ij}(x,t)=\sum _{k=1}^{N}\sigma _{ik}(x,t)\sigma _{jk}(x,t),}$
i.e. γ = σσ′, where σ′ denotes the transpose matrix of σ).
• When originally published by Kac in 1949,[2] 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
${\displaystyle e^{-\int _{0}^{t}V(x(\tau ))\,d\tau }}$
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 ${\displaystyle uV(x)\geq 0}$,
${\displaystyle E\left[e^{-u\int _{0}^{t}V(x(\tau ))\,d\tau }\right]=\int _{-\infty }^{\infty }w(x,t)\,dx}$
where w(x, 0) = δ(x) and
${\displaystyle {\frac {\partial w}{\partial t}}={\tfrac {1}{2}}{\frac {\partial ^{2}w}{\partial x^{2}}}-uV(x)w.}$
The Feynman–Kac formula can also be interpreted as a method for evaluating functional integrals of a certain form. If
${\displaystyle I=\int f(x(0))e^{-u\int _{0}^{t}V(x(t))\,dt}g(x(t))\,Dx}$
where the integral is taken over all random walks, then
${\displaystyle I=\int w(x,t)g(x)\,dx}$
where w(x, t) is a solution to the parabolic partial differential equation
${\displaystyle {\frac {\partial w}{\partial t}}={\tfrac {1}{2}}{\frac {\partial ^{2}w}{\partial x^{2}}}-uV(x)w}$
with initial condition w(x, 0) = f(x).

## References

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1. http://www.math.nyu.edu/faculty/kohn/pde_finance.html
2. {{#invoke:Citation/CS1|citation |CitationClass=journal }}This paper is reprinted in Mark Kac: Probability, Number Theory, and Statistical Physics, Selected Papers, edited by K. Baclawski and M.D. Donsker, The MIT Press, Cambridge, Massachusetts, 1979, pp.268-280