I need help with my understanding of changing probability measure. Im not a mathematician so I hope for answers that are not too technical.
As shown in this Wikipedia article http://en.wikipedia.org/wiki/Risk-neutral_measure you can change the drift of a GBM with the following procedure:
$$dS_t = \mu S_t dt + \sigma S_t dW_t$$
Introducing a new process:
$$d\tilde{W_t} = dW_t - \frac{\mu - r}{\sigma}dt$$
I understand that now the discounted value of the following process:
$$dS_t = r S_t dt + \sigma S_t d \tilde{W_t}$$
is a martingale if $\tilde{W_t}$ is a standard Brownian motion. OK, so we change to a new probability measure Q and now $\tilde{W_t}$ is a standard Brownian motion.
My first question is, $W_t$ can no longer be a standard Brownian motion under Q, because it now has non-zero expectation, is this true?
If the probability of an event, $dW_t=x$ under the physical measure, P, is $dP(x)$, then the probability for that same event under Q is $dQ(x)=dP(x) \Phi(x)$, where $\Phi(x)$ is what i think is called the Radon-Nikodym derivative. For $d\tilde{W_t}$ to have zero expectation, then under Q $E_Q[dW_t]=\frac{\mu-r}{\sigma}t$, am I right?
If this is true, can we then find $\Phi(x)=\frac{dQ(x)}{dP(x)}$ by dividing the density function of a Brownian motion with expectation $\frac{\mu-r}{\sigma}t$ with the density function for a for a standard Brownian motion?
$$\frac{e^{\frac{-(x-\frac{\mu-r}{\sigma}t)^2}{2t}}}{e^{\frac{-x^2}{2t}}}$$ $$e^{\frac{x^2}{2t}\frac{-(x-\frac{\mu-r}{\sigma}t)^2}{2t}}$$ $$e^{\frac{x^2-(x-\frac{\mu-r}{\sigma}t)^2}{2t}}$$ $$e^{\frac{x^2-x^2+2x\frac{\mu-r}{\sigma}t-\frac{\mu^2-2\mu r+r^2}{\sigma^2}t^2}{2t}}$$ $$e^{x\frac{\mu-r}{\sigma}-\frac{\mu^2-2\mu r+r^2}{2 \sigma^2}t}$$
Denoting $\frac{\mu-r}{\sigma}$, the market price of risk, as $\lambda$ and substituting we get
$$\Phi(x)=e^{x\lambda-\frac{1}{2}\lambda^2 t}$$
The problem is that most references I have looked at states that the Radon-Nikodym derivative as something like:
$$\Phi(t)=e^{-\int^t_0 \lambda dW(u)-\frac{1}{2}\int^t_0 \lambda^2 du}$$
I cannot seem to see the link between this expressions. Is the last expression even possible to solve?