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 $0<s<t$ 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: $g(t)=T-t$


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 $$ {\rm d}h(u,X_u)=\frac{\partial h}{\partial u}(u,X_u)\,{\rm d}u+\frac{\partial h}{\partial x}(u,X_u)\,{\rm d}X_u+\frac{1}{2}\frac{\partial^2h}{\partial x^2}(u,X_u)\,{\rm d}\left<X\right>_u. $$

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, $$ {\rm d}f(g(u),X_u)={\rm d}h(u,X_u)=\frac{\partial f}{\partial t}(g(u),X_u)\frac{{\rm d}g}{{\rm d}s}(u)\,{\rm d}u+\frac{\partial f}{\partial x}(g(u),X_u)\,{\rm d}X_u+\frac{1}{2}\frac{\partial^2f}{\partial x^2}(g(u),X_u)\,{\rm d}\left<X\right>_u. $$

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 $$ {\rm d}f(T-u,X_u)=-\frac{\partial f}{\partial t}(T-u,X_u)\,{\rm d}u+\frac{\partial f}{\partial x}(T-u,X_u)\,{\rm d}X_u+\frac{1}{2}\frac{\partial^2f}{\partial x^2}(T-u,X_u)\,{\rm d}\left<X\right>_u. $$

  • $\begingroup$ Maybe you can complete the derivation $\endgroup$ – Ezy Jan 16 '19 at 13:11
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    $\begingroup$ @Ezy: Thanks you for your reminder. I edited my answer, credited to you. $\endgroup$ – hypernova Jan 16 '19 at 15:52
  • $\begingroup$ thanks. I edited my answer to point to yours and upvoted yours ;) $\endgroup$ – Ezy Jan 16 '19 at 16:00
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    $\begingroup$ @Ezy: Wow it is so beyond the generous of you! Thank you :-) $\endgroup$ – hypernova Jan 16 '19 at 16:20

$t$ is fixed to simply apply Ito Lemma to $h(s,X_s)$ with the function $h: (s,x)\rightarrow f(t-s,x)$ and you get your answer. There's nothing special about it, I think you are a bit confused by the change of variable $s\rightarrow(t-s)$.

@hypernova has laid out the complete steps below for you.

  • $\begingroup$ $dZ_t=\frac{\partial f }{\partial t}(t-s,X_t)+\frac{\partial f }{\partial x}(t-s,X_t) dX_t +\frac{1}{2} \frac{\partial^2 f }{\partial^2 x}(t-s,X_t) (dX_t)^2.$ So this is how $dZ$ wil like like? $\endgroup$ – econmajorr Jan 13 '19 at 15:22
  • $\begingroup$ no this is incorrect $\endgroup$ – Ezy Jan 13 '19 at 15:25
  • $\begingroup$ How does it look then? $\endgroup$ – econmajorr Jan 13 '19 at 20:22
  • $\begingroup$ I told you in the answer exactly how to proceed. I believe if you know about the Ito lemma then you should be able to perform this calculation yourself correctly. Hint: it's almost correct! $\endgroup$ – Ezy Jan 13 '19 at 20:24
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    $\begingroup$ @econmajorr you know enough math to know Ito which is pretty advanced maths already. I am giving you a hint to apply Ito lemma (which you know) to a new function i defined in my answer, $h$. If you are not willing to give it a try it is your decision! But if you do you will see its purpose. $\endgroup$ – Ezy Jan 16 '19 at 13:09

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