Changing to martingale probability world

Question

This question is really getting me annoyed and I'm struggling to do the final proof, I have no problem obtaining the adjustment to the drift rate necessary to collapse the drift term to make it a martingale measure which I perceive to be = u/sigma, however, I am struggling to find what the fucntion of S is that makes the measure a martingale, from my text books it says that the function should be S_t = S*_t + u/sigma * t ,however when I perform itos lemma on this result I cant seem to get a drift rate of zero, ( I am also assuming that S*_t = S_0), any help willl be greatly aprreciated.

Answer 1

We assume that, under the probability measure $P$, \begin{align*} dS_t = S_t(\mu dt +\sigma dW_t), \end{align*} where $\{W_t, \, t \ge0\}$ is a standard Brownian motion. Note that, \begin{align*} dS_t=\sigma S_t d\left(\frac{\mu}{\sigma}t+ W_t \right). \end{align*} You need the probability measure $Q$ such that, under $Q$, the process $\{W_t + \frac{\mu}{\sigma}t, \, t\ge 0\}$ is a standard Brownian motion. Towards that, we define the measure $Q$ such that \begin{align*} \frac{dQ}{dP}\big|_t = \exp\left(-\frac{1}{2}\left( \frac{\mu}{\sigma}\right)^2 t-\frac{\mu}{\sigma} W_t\right). \end{align*} Then, by the Girsanov theorem, $Q$ is a probability measure, and the process $\{\widehat{W}_t, \, t\ge 0\}$, where $\widehat{W}_t=W_t + \frac{\mu}{\sigma}t$, is a standard Brownian motion under $Q$. Moreover, since \begin{align*} dS_t =\sigma S_t d\widehat{W}_t, \end{align*} $\{S_t, \, t\ge 0\}$ is a martingale under $Q$.