# Applying Geometric Brownian Motion itself into Ito's Lemma

I am new to this forum as well as partial differentiation. I would like to ask the following question.

Given a geometric brownian motion:

$$dS_t=(\mu S_t)dt+(\sigma S_t)dz_t$$

I would like to apply Ito's Lemma to $dS_t$ itself such that I would expect it will still give the same $dS_t$.

To do this, We need to know $\partial S_t \over \partial S_t$, $\partial^2 S_t \over \partial S_t^2$, and $\partial S_t \over \partial t$.

I have no problem of finding the first 2 derivatives, but how to calculate $\partial S_t \over \partial t$?

Thanks.

• Hint: Write the differential equation in integral form. – Chris Janjigian Jun 7 '13 at 18:38
• What is the meaning of $dz$ in $dS_t=(\mu S_t)dt+(\sigma S_t)dz$? What do you mean by "to apply Ito's Lemma to $dS_t$ itself"? How do you define $\partial S_t \over \partial S_t$, $\partial^2 S_t \over \partial S_t^2$, and $\partial S_t \over \partial t$? – Did Jun 7 '13 at 18:58
• @Did, we refer $dz_t$ to a Wiener's Process. In my attempt, I am learning to use Ito's Lemma, which can be used to find $df_t$ in terms of $dS_t$. Here if we let $f_t=S_t$, we should expect the Ito's Lemma should give us the original $S_t$. So this is what we mean by "applying Ito's Lemma to $dS_t$ itself". – Ric Yik Jun 8 '13 at 4:59
• ?? Your use of Itô's lemma below is rather... idiosyncratic. :-) For example you still did not say what you mean by $\partial S_t/\partial t$. One might mention at this point that the function $t\mapsto S_t(\omega)$ is almost surely not differentiable. – Did Jun 8 '13 at 10:42
• @Did, the $\partial f_t \over \partial t$ is one of the partial derivatives we need in order to apply Ito's Lemma to obtain $df_t$. Here I set $f_t=S_t$ and therefore we have $\partial S_t \over \partial t$. – Ric Yik Jun 10 '13 at 15:31

Thanks @chris-janjigian. Let me try to find out $\partial S_t \over \partial t$ by using the integral form of $dS_t$. I got the integral form after searching from Google: $$S_t = S_0 + \int_0^t (\mu S_u)du+\int_0^t(\sigma S_u)dz_u$$
As $t$ is just representing "t-times" in this form, ${\partial S_t \over \partial t}=0+0+0=0$. We also know that ${\partial S_t \over \partial S_t}=1$ and ${\partial^2 S_t \over \partial S_t^2}=0$. Therefore, using Ito's Lemma: $$dS_t=[{\partial S_t \over \partial S_t}(\mu S_t)+{\partial S_t \over \partial t}+{1 \over 2}{\partial^2 S_t \over \partial S_t^2}(\sigma S_t)^2]dt+[{\partial S_t \over \partial S_t}(\sigma S_t)]dz_t$$ $$dS_t=[1(\mu S_t)+0+0]dt+[1(\sigma S_t)]dz_t$$ $$dS_t=(\mu S_t)dt+(\sigma S_t)dz_t \Box$$