# How to derive equivalent martingale measure using Ito's Lemma

Can someone explain how to get equation 27.14 below? I understand the first usage of Ito's Lemma to get $d(\ln f-\ln g)$ but I do not understand how to use Ito's Lemma to go from $d(\ln \frac{f}{g})$ to $d(\frac{f}{g})$. Can someone help to elucidate this process?

## 1 Answer

You have two processes, $X_t:=\log{\frac{f}{g}}$ and $Y_t=\frac{f}{g}$. Note, I use $\log$ for the natural logarithm. Hence we have $Y_t=\exp{(X_t)}$. Therefore, applying Itô:

$$dY_t=\exp{(X_t)}dX_t + \frac{1}{2}\exp{(X_t)}d\langle X,X\rangle_t$$

Using the dynamics of $X_t$, we get

$$dY_t=\frac{f}{g}[-\frac{(\sigma_f-\sigma_g)^2}{2}dt+(\sigma_f-\sigma_g)dz]+\frac{f}{g}\frac{(\sigma_f-\sigma_g)^2}{2}d\langle z, z\rangle_t$$

I think $z$ is a Brownian Motion, such that $d\langle z,z\rangle_t=dt$, which yields the desired result. If $z$ is not a BM, please provide additional information.