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

\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

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 Itō 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

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

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 Itō's lemma once again to du(X_s,s), it follows that


\begin{align}
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{align}

The first term contains, in parentheses, the above PDE 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 Itō integral, which has expectation zero, 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

\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 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. γ = σσ′, where σ′ denotes the transpose matrix of σ).
 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 u V(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} - u V(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} - u V(x) w
with initial condition w(x, 0) = f(x).

See also

References

  1. http://www.math.nyu.edu/faculty/kohn/pde_finance.html
  2. Kac, Mark (1949). "On Distributions of Certain Wiener Functionals". Transactions of the American Mathematical Society 65 (1): 1–13. doi:10.2307/1990512. JSTOR 1990512.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
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