stochastic processes - Application of Ito's lemma

Let $X_t$ be some stochastic process driven by wiener process ($W_t)$ so it can be expressed as: $$dX_t=(...)dt+(...)dW_t$$

Let $f(t,x)$ be some $C^2$ function. Define the process $Z_s=f(t-s,X_s)$ for $0 and fixed $t$.

How can I use Ito's lemma to express $dZ_s$?

The reason for this question and my confusion is the $(t-s)$ part. Naturally $f(t,X_t)$ and $f(t-s,X_{t-s})$ would have been easy, but how does the standard Ito change when the process looks is $(t-s,X_{t-s})$?

Maybe one can show Ito is performed in general for $f(g(t),X_t)$ where in the above case: $g(t)=T-t$

Answer

Consider OP's general formula $f(g(t),X_t)$. In case of ambiguity, let us claim that

• $f=f(t,x)$ is defined with variables $t$ and $x$,


• $g=g(s)$ is defined with the variable $s$, and


• $h=h(u,x)=f(g(u),x)$ is defined with variables $u$ and $x$.

Then Ito's formula states that $$ {\rm d}h(u,X_u)=\frac{\partial h}{\partial u}(u,X_u)\,{\rm d}u+\frac{\partial h}{\partial x}(u,X_u)\,{\rm d}X_u+\frac{1}{2}\frac{\partial^2h}{\partial x^2}(u,X_u)\,{\rm d}\left_u. $$

We just need to express $h$ by using $f$ and $g$. We have \begin{align} \frac{\partial h}{\partial u}(u,x)&=\frac{\partial}{\partial u}h(u,x)=\frac{\partial}{\partial u}f(g(u),x)=\frac{\partial f}{\partial t}(g(u),x)\,\frac{{\rm d}g}{{\rm d}s}(u),\\ \frac{\partial h}{\partial x}(u,x)&=\frac{\partial}{\partial x}h(u,x)=\frac{\partial}{\partial x}f(g(u),x)=\frac{\partial f}{\partial x}(g(u),x),\\ \frac{\partial^2h}{\partial x^2}(u,x)&=\frac{\partial^2}{\partial x^2}h(u,x)=\frac{\partial^2}{\partial x^2}f(g(u),x)=\frac{\partial^2f}{\partial x^2}(g(u),x). \end{align} Therefore, $$ {\rm d}f(g(u),X_u)={\rm d}h(u,X_u)=\frac{\partial f}{\partial t}(g(u),X_u)\frac{{\rm d}g}{{\rm d}s}(u)\,{\rm d}u+\frac{\partial f}{\partial x}(g(u),X_u)\,{\rm d}X_u+\frac{1}{2}\frac{\partial^2f}{\partial x^2}(g(u),X_u)\,{\rm d}\left_u. $$

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Back to OP's original question, let us apply the above result to $f(T-u,X_u)$ (I would like to thank @Ezy for kind advices). In this case, let us take $$ g(s)=T-s. $$ Then we have $$ \frac{{\rm d}g}{{\rm d}s}(u)=-1. $$ Substitute these two expressions into the above result, and it follows that $$ {\rm d}f(T-u,X_u)=-\frac{\partial f}{\partial t}(T-u,X_u)\,{\rm d}u+\frac{\partial f}{\partial x}(T-u,X_u)\,{\rm d}X_u+\frac{1}{2}\frac{\partial^2f}{\partial x^2}(T-u,X_u)\,{\rm d}\left_u. $$