Stochastic integrals and new probability measures Let $B$ be a standard Brownian motion on $(\Omega, \mathcal{F}, P, ({\mathcal{F}_t})_{t\ge0})$, where the filtration is the one generated by $B$. Fix a time interval $[0,T]$. Define the process $X$ as the solution to the SDE
$$
\mathrm dX_t = \sigma X_t\,\mathrm dB_t,\quad X_0 = 1.
$$
Define, for each real number $\alpha$, a measure $P_{\alpha}$, such that $X$ under $P_{\alpha}$ solves the equation
$$
\mathrm dX_t = \alpha  X_t\,\mathrm dt + \sigma X_t\,\mathrm dB^{\alpha},
$$
where $B^{\alpha}$ is a Brownian motion under $P_{\alpha}$. Give an explicit expression for the Radon-Nikodym derivative (likelihood process)
$$
L^{\alpha} = \frac{\mathrm dP_{\alpha}}{\mathrm dP_0}, 
$$
on $\mathcal{F}_t$.
So this is the question. And I guess you're supposed to use the Itô formula. But I've had a hard time grasping the question. Some guidance on how I could think and where I should begin would be more than appreciated!
(This is my first post on this site and also the first time i use TeX so might not look very good, hopefully you'll understand anyway!)
 A: You have the explicit solution
$$X_t=\exp{(\sigma B_t -\frac{1}{2}\sigma^2 t)}$$
which is a martingale (important!). I assume that it is meant that $B^0=B$, and $P^0$ is the original measure? What you are actually doing is going "backwards" in the BS-model. There you start with something like $dX_t=\alpha X_tdt+\sigma X_t dB^\alpha_t$ and you want to get rid of the drift term. 
Define
$$L^\alpha=\frac{dP^\alpha}{dP^0}:=\mathcal{E}(-\int\frac{\alpha}{\sigma}dB)$$
where $\mathcal{E}(X):=\exp{(X_t-\frac{1}{2}\langle X \rangle_t)}$ denotes the stochastic exponential. So we have to verify that this density does the job.
First note, that we have $L^\alpha >0$, $L^\alpha$ is a martingale and $E[L^\alpha_T]=1$, so we can indeed define an equivalent probability measure. What we know is: $B$ is a $P^0=P$ Brownian Motion, we have $P^\alpha$ is equivalent to $P$ and the density is of the form $\mathcal{E}(\int b_s dW_s)$, where $b_s=\frac{\alpha}{\sigma}$, which does not depend on $s$. Then Girsanov's Theorem tells us that $B=B^\alpha + \int\frac{\alpha}{\sigma}$ for a Brownian Motion $B^\alpha$ under $P^\alpha$, i.e. $dB=dB^\alpha +\frac{\alpha}{\sigma}dt$. Plug this into the original SDE, yields under $P^\alpha$:
$$dX_t=\sigma X_t d(B^\alpha_t+\frac{\alpha}{\sigma}dt)=\alpha X_tdt+\sigma X_t dB^\alpha_t$$
which is exactly the desired result. 
Note:I'm just learning about this stuff myself, so please be critical!
cheers
math
