# 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: $$g(t)=T-t$$

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, $${\rm d}f(g(u),X_u)={\rm d}h(u,X_u)=\frac{\partial f}{\partial t}(g(u),X_u)\frac{{\rm d}g}{{\rm d}s}(u)\,{\rm d}u+\frac{\partial f}{\partial x}(g(u),X_u)\,{\rm d}X_u+\frac{1}{2}\frac{\partial^2f}{\partial x^2}(g(u),X_u)\,{\rm d}\left_u.$$

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 $${\rm d}f(T-u,X_u)=-\frac{\partial f}{\partial t}(T-u,X_u)\,{\rm d}u+\frac{\partial f}{\partial x}(T-u,X_u)\,{\rm d}X_u+\frac{1}{2}\frac{\partial^2f}{\partial x^2}(T-u,X_u)\,{\rm d}\left_u.$$

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

• $dZ_t=\frac{\partial f }{\partial t}(t-s,X_t)+\frac{\partial f }{\partial x}(t-s,X_t) dX_t +\frac{1}{2} \frac{\partial^2 f }{\partial^2 x}(t-s,X_t) (dX_t)^2.$ So this is how $dZ$ wil like like? – econmajorr Jan 13 '19 at 15:22
• no this is incorrect – Ezy Jan 13 '19 at 15:25
• How does it look then? – econmajorr Jan 13 '19 at 20:22
• 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! – Ezy Jan 13 '19 at 20:24
• @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. – Ezy Jan 16 '19 at 13:09