Feynman–Kac formula

Consider the partial differential equation

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

defined for all ${\displaystyle x\in \mathbb {R} }$ and ${\displaystyle t\in [0,T]}$, subject to the terminal condition

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

where ${\displaystyle \mu ,\sigma ,\psi ,V,f}$ are known functions, ${\displaystyle T}$ 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

under the probability measure ${\displaystyle Q}$ such that ${\displaystyle X}$ is an Itô process driven by the equation

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

with ${\displaystyle W^{Q}(t)}$ is a Wiener process (also called Brownian motion) under ${\displaystyle Q}$, and the initial condition for ${\displaystyle X(t)}$ is ${\displaystyle X(t)=x}$.

Intuitive interpretation

Suppose we have a particle moving according to the diffusion process

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

Let the particle incur "cost" at a rate of ${\displaystyle f(X_{s},s)}$ at location ${\displaystyle X_{s}}$ at time ${\displaystyle s}$. Let it incur a final cost at ${\displaystyle \psi (X_{T})}$.

Also, allow the particle to decay. If the particle is at location ${\displaystyle X_{s}}$ at time ${\displaystyle s}$, then it decays with rate ${\displaystyle V(X_{s},s)}$. After the particle has decayed, all future cost is zero.

Then, ${\displaystyle u(x,t)}$ is the expected cost-to-go, if the particle starts at ${\displaystyle (t,X_{t}=x)}$.

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 ${\displaystyle u(x,t)}$ be the solution to the above partial differential equation. Applying the product rule for Itô processes to the process

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

one gets: ${\displaystyle {\begin{aligned}dY_{s}={}&d\left(\exp \left(-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau \right)\right)u(X_{s},s)+\exp \left(-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau \right)\,du(X_{s},s)\\[6pt]&{}+d\left(\exp \left(-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau \right)\right)du(X_{s},s)+d\left(\int _{t}^{s}\exp \left(-\int _{t}^{r}V(X_{\tau },\tau )\,d\tau \right)f(X_{r},r)\,dr\right)\end{aligned}}}$

Since

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

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

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

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

${\displaystyle {\begin{aligned}dY_{s}={}&\exp \left(-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau \right)\,\left(-V(X_{s},s)u(X_{s},s)+f(X_{s},s)+\mu (X_{s},s){\frac {\partial u}{\partial X}}+{\frac {\partial u}{\partial s}}+{\tfrac {1}{2}}\sigma ^{2}(X_{s},s){\frac {\partial ^{2}u}{\partial X^{2}}}\right)\,ds\\[6pt]&{}+\exp \left(-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau \right)\sigma (X,s){\frac {\partial u}{\partial X}}\,dW.\end{aligned}}}$

The first term contains, in parentheses, the above partial differential equation and is therefore zero. What remains is: ${\displaystyle dY_{s}=\exp \left(-\int _{t}^{s}V(X_{\tau },\tau )\,d\tau \right)\sigma (X,s){\frac {\partial u}{\partial X}}\,dW.}$

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

Upon taking expectations, conditioned on ${\displaystyle X_{t}=x}$, and observing that the right side is an Itô integral, which has expectation zero,[3] it follows that

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

The desired result is obtained by observing that

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

and finally

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

• The proof above that a solution must have the given form is essentially that of [4] with modifications to account for ${\displaystyle f(x,t)}$.
• The expectation formula above is also valid for N-dimensional Itô diffusions. The corresponding partial differential equation for ${\displaystyle u:\mathbb {R} ^{N}\times [0,T]\to \mathbb {R} }$ becomes:[5]
  $${\displaystyle {\frac {\partial u}{\partial t}}+\sum _{i=1}^{N}\mu _{i}(x,t){\frac {\partial u}{\partial x_{i}}}+{\frac {1}{2}}\sum _{i=1}^{N}\sum _{j=1}^{N}\gamma _{ij}(x,t){\frac {\partial ^{2}u}{\partial x_{i}\partial x_{j}}}-r(x,t)\,u=f(x,t),}$$

  
  where,
  $${\displaystyle \gamma _{ij}(x,t)=\sum _{k=1}^{N}\sigma _{ik}(x,t)\sigma _{jk}(x,t),}$$

  
  i.e. ${\displaystyle \gamma =\sigma \sigma ^{\mathrm {T} }}$, where ${\displaystyle \sigma ^{\mathrm {T} }}$ denotes the transpose of ${\displaystyle \sigma }$.
• More succinctly, letting ${\displaystyle A}$ be the infinitesimal generator of the diffusion process,
  $${\displaystyle {\frac {\partial u}{\partial t}}+Au-r(x,t)\,u=f(x,t),}$$

  
• This expectation can then be approximated using Monte Carlo or quasi-Monte Carlo methods.
• When originally published by Kac in 1949,[6] 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
  $${\displaystyle \exp(-\int _{0}^{t}V(x(\tau ))\,d\tau )}$$

  
  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 ${\displaystyle uV(x)\geq 0}$,
  $${\displaystyle E\left[\exp(-u\int _{0}^{t}V(x(\tau ))\,d\tau )\right]=\int _{-\infty }^{\infty }w(x,t)\,dx}$$

  
  where w(x, 0) = δ(x) and
  $${\displaystyle {\frac {\partial w}{\partial t}}={\frac {1}{2}}{\frac {\partial ^{2}w}{\partial x^{2}}}-uV(x)w.}$$

  
  The Feynman–Kac formula can also be interpreted as a method for evaluating functional integrals of a certain form. If
  $${\displaystyle I=\int f(x(0))\exp(-u\int _{0}^{t}V(x(t))\,dt)g(x(t))\,Dx}$$

  
  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
  $${\displaystyle {\frac {\partial w}{\partial t}}={\frac {1}{2}}{\frac {\partial ^{2}w}{\partial x^{2}}}-uV(x)w}$$

  
  with initial condition w(x, 0) = f(x).

• Itô's lemma
• Kunita–Watanabe inequality
• Girsanov theorem
• Kolmogorov forward equation (also known as Fokker–Planck equation)

• Simon, Barry (1979). Functional Integration and Quantum Physics. Academic Press.
• Hall, B. C. (2013). Quantum Theory for Mathematicians. Springer.