I am curious about the following problem:
Let $B_t$ be a standard Brownian motion on $(\Omega, \mathcal F, \mathcal F_t, \mathbb P_a)$, where the filtration is generated by $B_t$. On a finite interval $[0,T]$ we define $X_t$ as the one solving the SED $$\mathrm dX_t=\mu_a\,\mathrm dt+\sigma\,\,\mathrm dB_t.$$ For some other measure $\mathbb P_b$, we define $X_t$ as the solution to $$\mathrm dX_t=\mu_b\,\mathrm dt+\sigma\,\mathrm d B_t',$$ where $B_t'$ is a Brownian motion under $\mathbb P_b$, $\mu_a\ne\mu_b$ being two different real numbers, and $\sigma>0$ being a constant. Hence, the difference between the two diffusion processes lies only in the drift.
My question is: what is the Radon-Nikodym derivate (as a function of $t$ and $X_t$) $$\frac{\mathrm d\mathbb P_a}{\mathrm d\mathbb P_b}$$ on $\{\mathcal F_t\}$? What I know so far is the answer to a special case: $\mu_a=0$, where the answer can be derived explicitly. Is it possible to generalize the special case? Many thanks!
The answer to your question is the Girsanov theorem.
With some technical conditions, the Radon Nikodym derivative is given be
$ \frac{ \mathrm{d} \mathbb{P}_b}{\mathrm{d} \mathbb{P}_a} (\omega) = \exp \left( \int_0^T h_s(\omega) \, \mathrm{d}B_s(\omega) - \frac{1}{2} \int_0^T h_s(\omega)^2 \, \mathrm{d}s \right) $,
where $h$ is a (progressively measurable) function to be found. With some technical conditions it is not difficult to find $h$.