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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!

BCLC
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user225160
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    Apply the result you say you know, twice, once to pass from drift $\mu_b$ to drift $0$ and once to pass from drift $0$ to drift $\mu_a$. – Did Mar 20 '15 at 17:12
  • And why would anyone want to apply a result twice when once is sufficient? That is very, very confusing to me. – AXH Mar 21 '15 at 00:54
  • @ArbiasHashani Sorry, what was that? – Did Mar 30 '15 at 21:56

1 Answers1

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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 $h$.

AXH
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