How to express a process using Itos formula

Question

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 $dF(t-s,X(s))$?

Answer 1

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.