I have a question, which should be kind of trivial, but somehow im confused about it. Long story short, I have the following PDE which I achieved from the Feynman-Kac formula:
\begin{align*} \partial_t F+rx\partial_x F + (\alpha-\lambda v)\partial_vF + \frac{1}{2}vx^2\partial_{xx}F + \frac{1}{2}\sigma_v^2 v\partial_{vv}F + \sigma_v v\rho x\partial_{xv}F-rF = 0 \end{align*}
Then, I know that $F(t,x,v)$ solves this equation, for some $F$, and I also know that it can be represented in the form of a conditional expectation due to Feynman-Kac. The next part is where it gets confusing for me.
Define $G(\tau,x,v) = F(T-\tau,x,v)$, and show that G solves the PDE
\begin{align*} -\partial_\tau G+rx\partial_x G + (\alpha-\lambda v)\partial_vG + \frac{1}{2}vx^2\partial_{xx}G + \frac{1}{2}\sigma_v^2 v\partial_{vv}G + \sigma_v v\rho x\partial_{xv}G-rG = 0 \end{align*}
So the only thing that changed was the first term. This kinda make sense, since $\partial_x F = \partial_x G$ and $\partial_v F = \partial_v G$ because we did not change the variable. This led me to think that the only thing remaining to show is that
\begin{align*} \partial_\tau G(\tau,x,v) = -\partial_t F(t,x,v) \end{align*}
This is where it gets confusing for me. I figured it is something with the chain rule, but i don't know where to begin. How do i compute
\begin{align*} \partial_\tau G(\tau,x,v) = \partial_\tau F(T-\tau,x,v) \end{align*}
Thanks in advance
$\partial_\tau F(T-\tau,x,v)=\partial_{T-\tau}F(T-\tau,x,v)\partial_\tau(T-\tau)=-\partial_tF(t,x,v)$ where $t=T-\tau$.