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Let $F(t,x)$ be the solution to the PDE $$ F_t(t,x)=aF_x(t,x)+\frac{1}{2}F_{xx}(t,x),t>0 $$ $$F(0,x)=g(x)$$ for some function $g$. Let $X_t$ be a process defined by $$dx_t=aX(t)dt+dW(t)$$ Now consider the process $F(t-s,X_s)$.

How can I use Ito to express $dF(t-s,X(s))$?

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Welcome!

First of all, I suppose the proposed PDE should be $$ F_t=axF_x+\frac{1}{2}F_{xx}. $$ Perhaps you have missed an $x$ in front of the $F_x$ term. But please do let me know if this is not the case.

In view of $F(t-s,X_s)$, $s$ is the variable to which ${\rm d}$ is with respect, while $t$ is a parameter. Hence to be less ambiguous, let us use $F(T-t,X_t)$ instead, where $t\in\left[0,T\right]$.

Note that $F(T-t,x)$ is a function of $t$ and $x$. To make it clear, let us define $G(t,x)=F(T-t,x)$. By Ito's formula, \begin{align} &{\rm d}F(T-t,X_t)\\ &={\rm d}G(t,X_t)\\ &=\frac{\partial G}{\partial t}(t,X_t)\,{\rm d}t+\frac{\partial G}{\partial x}(t,X_t)\,{\rm d}X_t+\frac{1}{2}\frac{\partial^2G}{\partial x^2}(t,X_t)\,{\rm d}\left<X\right>_t. \end{align}

Provided that $$ {\rm d}X_t=aX_t\,{\rm d}t+{\rm d}W_t, $$ we have ${\rm d}\left<X\right>_t={\rm d}t$. Substitute these two results into the above equation, and we obtain \begin{align} &{\rm d}F(T-t,X_t)\\ &=\left(\frac{\partial G}{\partial t}(t,X_t)+aX_t\frac{\partial G}{\partial x}(t,X_t)+\frac{1}{2}\frac{\partial^2G}{\partial x^2}(t,X_t)\right){\rm d}t+\frac{\partial G}{\partial x}(t,X_t)\,{\rm d}W_t\\ &=\left(\frac{\partial G}{\partial t}+ax\frac{\partial G}{\partial x}+\frac{1}{2}\frac{\partial^2G}{\partial x^2}\right)(t,X_t)\,{\rm d}t+\frac{\partial G}{\partial x}(t,X_t)\,{\rm d}W_t. \end{align}

Recall that $G(t,x)=F(T-t,x)$, and it is obvious that \begin{align} \frac{\partial G}{\partial t}(t,x)&=-\frac{\partial F}{\partial t}(T-t,x),\\ \frac{\partial G}{\partial x}(t,x)&=\frac{\partial F}{\partial x}(T-t,x),\\ \frac{\partial^2G}{\partial x^2}(t,x)&=\frac{\partial^2F}{\partial x^2}(T-t,x). \end{align} Consequently, we obtain \begin{align} &{\rm d}F(T-t,X_t)\\ &=\left(-\frac{\partial F}{\partial t}+ax\frac{\partial F}{\partial x}+\frac{1}{2}\frac{\partial^2F}{\partial x^2}\right)(T-t,X_t)\,{\rm d}t+\frac{\partial F}{\partial x}(T-t,X_t)\,{\rm d}W_t. \end{align}

Finally, note that the PDE gives $$ -\frac{\partial F}{\partial t}+ax\frac{\partial F}{\partial x}+\frac{1}{2}\frac{\partial^2F}{\partial x^2}=0, $$ and we eventually obtain $$ {\rm d}F(T-t,X_t)=\frac{\partial F}{\partial x}(T-t,X_t)\,{\rm d}W_t. $$

That's it! Hope this could be somewhat helpful for you.

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  • $\begingroup$ Thank you very much. So helpfull. One question. why is: $\frac{\partial G}{\partial t}(t,x)=-\frac{\partial F}{\partial t}(T-t,x)$? I don't understand why we need a minus? $\endgroup$ – econmajorr Jan 11 at 10:41
  • $\begingroup$ Use the chain rule: with $\eta(t) = T-t$ and $G(t, x) = F(\eta(t), x)$ $\endgroup$ – byouness Jan 11 at 11:17
  • $\begingroup$ @econmajorr: Yes, byouness's answer clarifies. Basically, $\frac{\partial}{\partial t}F(T-t,x)$ and $\frac{\partial F}{\partial t}(T-t,x)$ mean essentially different. The former means to take the derivative of $F(T-t,x)$ with respect to the variable $t$, in the usual sense. By contrast, the latter means to take the derivative of the bivariate function $F$ with respect to its $t$-variable (originally, $F$ is defined as $F=F(t,x)$, and its $t$-variable is exactly its first variable), after which you value the derivative function at the point $\left(T-t,x\right)$. $\endgroup$ – hypernova Jan 11 at 14:24
  • $\begingroup$ @econmajorr: (cont'd) Therefore, in case of any ambiguity, a better way is to define $F=F(t,x)$, and consider $F(T-\tau,X_{\tau})$. In this case, $t$ is the name of the variable, while $\tau$ is the true variable with respect to which we take the derivative. $\endgroup$ – hypernova Jan 11 at 14:26
  • $\begingroup$ Thank you both. I have felt need to add a new question. Sorry, but I still don't understand how you normally perform ito with that kind of variables: quant.stackexchange.com/questions/43465/… $\endgroup$ – econmajorr Jan 16 at 7:41

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