# How to express a process using Itos formula

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

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_t. \end{align}

Provided that $${\rm d}X_t=aX_t\,{\rm d}t+{\rm d}W_t,$$ we have $${\rm d}\left_t={\rm d}t$$. 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.

• 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? – econmajorr Jan 11 '19 at 10:41
• Use the chain rule: with $\eta(t) = T-t$ and $G(t, x) = F(\eta(t), x)$ – byouness Jan 11 '19 at 11:17
• @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 $\left(T-t,x\right)$. – hypernova Jan 11 '19 at 14:24
• @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. – hypernova Jan 11 '19 at 14:26
• 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/… – econmajorr Jan 16 '19 at 7:41