Feynman Kac formula

The Feynman Kac formula is generally stated for terminal condition problems (see e.g. Wikipedia), where

\[\partial_t u(t,x) + \mu(x) \nabla_x u(t,x) + \frac{1}{2} \sigma^2(x) \Delta_x u(t,x) + f(x, u(t,x)) = 0 \tag{1}\]

with terminal condition $u(T, x) = g(x)$, and $u \colon \R^d \to \R$.

In this case the FK formula states that for all $t \in (0,T)$ it holds that

\[u(t, x) = \int_t^T \mathbb{E} \left[ f(X^x_{s-t}, u(s, X^x_{s-t}))ds \right] + \mathbb{E} \left[ u(0, X^x_{T-t}) \right] \tag{2}\]

where

\[X_t^x = \int_0^t \mu(X_s^x)ds + \int_0^t\sigma(X_s^x)dB_s + x,\]

and $B_t$ is a Brownian motion.

Intuitively, this formula is motivated by the fact that the density of Brownian particles (motion) satisfes the diffusion equation.

The equivalence between the average trajectory of particles and PDEs given by the Feynman-Kac formula allows to overcome the curse of dimensionality that standard numerical methods suffer from, because the expectations can be approximated Monte Carlo integrations, which approximation error decreases as $1/\sqrt{N}$ and is therefore not dependent on the dimensions. On the other hand, the computational complexity of traditional deterministic techniques grows exponentially in the number of dimensions.

Forward non-linear Feynman-Kac

How to transform previous equation to an initial value problem?

Define $v(\tau, x) = u(T-\tau, x)$. Observe that $v(0,x) = u(T,x)$. Further observe that by the chain rule

\[\begin{aligned} \partial_\tau v(\tau, x) &= \partial_\tau u(T-\tau,x)\\ &= (\partial_\tau (T-\tau)) \partial_t u(T-\tau,x)\\ &= -\partial_t u(T-\tau, x). \end{aligned}\]

From Eq. (1) we get that

\[- \partial_t u(T - \tau,x) = \mu(x) \nabla_x u(T - \tau,x) + \frac{1}{2} \sigma^2(x) \Delta_x u(T - \tau,x) + f(x, u(T - \tau,x)).\]

Replacing $u(T-\tau, x)$ by $v(\tau, x)$ we get that $v$ satisfies

\[\partial_\tau v(\tau, x) = \mu(x) \nabla_x v(\tau,x) + \frac{1}{2} \sigma^2(x) \Delta_x v(\tau,x) + f(x, v(\tau,x)) \]

and from Eq. (2) we obtain

\[v(\tau, x) = \int_{T-\tau}^T \mathbb{E} \left[ f(X^x_{s- T + \tau}, v(s, X^x_{s-T + \tau}))ds \right] + \mathbb{E} \left[ v(0, X^x_{\tau}) \right].\]

By using the substitution rule with $\tau \to \tau -T$ (shifting by T) and $\tau \to - \tau$ (inversing), and finally inversing the integral bound we get that

\[\begin{aligned} v(\tau, x) &= \int_{-\tau}^0 \mathbb{E} \left[ f(X^x_{s + \tau}, v(s + T, X^x_{s + \tau}))ds \right] + \mathbb{E} \left[ v(0, X^x_{\tau}) \right]\\ &= - \int_{\tau}^0 \mathbb{E} \left[ f(X^x_{\tau - s}, v(T-s, X^x_{\tau - s}))ds \right] + \mathbb{E} \left[ v(0, X^x_{\tau}) \right]\\ &= \int_{0}^\tau \mathbb{E} \left[ f(X^x_{\tau - s}, v(T-s, X^x_{\tau - s}))ds \right] + \mathbb{E} \left[ v(0, X^x_{\tau}) \right]. \end{aligned}\]

This leads to the