Geometric brownian motion - Ito's lemma

Question

I have a question about geometric brownian motion. dS = uSdt + /sigma/SdW and then we do log(S) and we want to found dlog(S). So we use Ito's lemma en I get the dt part of the lemma but I don't see why df/dt*dw is zero.

Can somebody help me?

Answer 1

If $f(t,S) = \log(S)$ then $\frac{\partial}{\partial t}f(t,S) = 0$, since $f$ does not depend on $t$. Is this what you're asking?

Using Itô's formula, you will get

\begin{align} d\log(S_t) &= \frac{\partial}{\partial t}f(t,S_t)\,dt +\frac{\partial}{\partial S}f(t,S_t)\,dS_t + \frac{1}{2}\frac{\partial^2}{\partial S^2}f(t,S)\,(dS_t)^2\\ &= 0\,dt + \frac{1}{S_t}\,dS_t - \frac{1}{2S_t^2}\,(dS_t)^2\\ &= \Bigl(\mu - \frac{1}{2}\sigma^2\Bigr)\,dt + \sigma\,dW_t \end{align}