# Forwards Feynman–Kac formula

This might be a simple question, but I'm having trouble with it.

Consider the Cauchy problem with final condition. $$\begin{equation} \begin{cases} \frac{\partial u}{\partial t}(t,x) + \mathcal{L}u(t,x) + k(t,x)u(t,x) = g(t,x) &\textit{in}\quad\left[0,T\right]\times\mathbb{R}\\ u(T,x)=\phi(x)&\textit{in}\quad\mathbb{R} \end{cases} \label{CauchyProb} \end{equation}$$ where $$\begin{equation} \mathcal{L} = \frac{1}{2}\sigma^2(t,x)\frac{\partial^2}{\partial x^2} + \mu(t,x)\frac{\partial }{\partial x}. \end{equation}$$ I am well aware that a solution to this problem can be given in terms of the following Feynman–Kac formula: $$\begin{equation} u(t,x)=\mathbb{E}\left[\phi(X_T^{t,x})\exp\left\lbrace\int_t^Tk(s,X_s^{t,x})ds\right\rbrace -\int_t^Tg(s,X_s^{t,x})\exp\left\lbrace\int_t^s k(u,X_u^{t,x})du\right\rbrace ds\right] \end{equation}$$ where $$X_t$$ is an Itô process that is described by: $$\begin{equation} dX_t = \mu(t,X_t)dt + \sigma(t,X_t)dW_t\,, \end{equation}$$ with $$X_0=x$$.

The problem arises when I try to make the change of variables $$v(t') = u(T-t)$$. Now, the previous Cauchy problem with final condition, becomes a Cauchy problem with initial condition: $$\begin{equation} \begin{cases} -\frac{\partial v}{\partial t'}(t',x) + \mathcal{L}v(t',x) + k(T-t',x)v(t',x) = g(T-t',x) &\textit{in}\quad\left[0,T\right]\times\mathbb{R}\\ v(0,x)=\phi(x)&\textit{in}\quad\mathbb{R} \end{cases} \label{CauchyProb2} \end{equation}$$ with $$\begin{equation} \mathcal{L} = \frac{1}{2}\sigma^2(T-t',x)\frac{\partial^2}{\partial x^2} + \mu(T-t',x)\frac{\partial }{\partial x}. \end{equation}$$

What form does the Feynman–Kac formula take when we perform this change of variable?

• In the equation for $v$, the coefficients should be time reversed as well (i.e. $k(t',x)$ should be $k(T - t',x)$). – Peter Morfe Mar 1 at 13:14
• Thanks, @PeterMorfe, I just edited the question. – Paulo Rocha Mar 1 at 13:49
• $\mathcal{L}$ also needs to be transformed. – Peter Morfe Mar 1 at 14:36
• Thanks for helping improve the question. I made the necessary edits. – Paulo Rocha Mar 1 at 16:21

## 1 Answer

Given $$t \in (0,T)$$, define $$\tilde{X}^{(t),x}$$ to be the solution of the SDE $$\begin{equation*} d\tilde{X}^{(t),x}_{s} = \mu(t + s,\tilde{X}^{(t),x}_{s}) \, ds + \sigma(t + s, \tilde{X}^{(t),x}_{s}) \, d B_{s}, \quad \tilde{X}^{(t),x}_{0} = x. \end{equation*}$$ Notice that $$\tilde{X}^{(t),x}_{\cdot} = X^{t,x}_{\cdot + t}$$. Hence we can write \begin{align*} u(t,x) &= \mathbb{E} \left(\phi(\tilde{X}^{(t),x}_{T - t}) \exp \left \{ \int_{0}^{T -t} k(s + t, \tilde{X}^{(t),x}_{s} \, ds \right\}\right)\\ &\quad - \mathbb{E} \left( \int_{0}^{T - t} g(s + t, \tilde{X}^{(t),x}_{s}) \exp \left\{ \int_{0}^{t} k(u + t, \tilde{X}^{(t),x}_{u}) \,du \right\} \, ds \right) \end{align*}

Now if we define $$v(t',x) = u(T - t,x)$$, we find \begin{align*} v(t',x) &= \mathbb{E} \left(\phi(\tilde{X}^{(T - t'),x}_{t'}) \exp \left \{ \int_{0}^{t'} k(s + T - t', \tilde{X}^{(T - t'),x}_{s} \, ds \right\}\right)\\ &\quad - \mathbb{E} \left( \int_{0}^{t'} g(s + T - t', \tilde{X}^{(T - t'),x}_{s}) \exp \left\{ \int_{0}^{T - t'} k(u + T - t', \tilde{X}^{(T - t'),x}_{u}) \,du \right\} \, ds \right) \end{align*} It's worth noticing that this formula makes a certain amount of sense when $$\mu$$ and $$\sigma$$ are temporally homogeneous, in which case $$\tilde{X}^{(t)}$$ is independent of $$t$$.