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?
 A: 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 $t$.
