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Roughly speaking, Girsanov's theorem says that if we have a Brownian motion $W$ on $[0,T]$, we can construct a new process with a modified drift that has an equivalent law to $W$ (subject to adaptedness constraints on the drift, integrability, etc...).

I'm wondering if there's a 'converse' to Girsanov's theorem. If we have some process $X$ whose law $\mathbb{P}$ is equivalent to Wiener measure $\mathbb{W}$, can we always express $X$ as $X_t = \alpha_t + W_t$, where $\alpha$ is adapted to the filtration of $W$? Alternatively, can the Radon-Nikodym derivative always be expressed in the form $$\frac{d\mathbb{P}}{d\mathbb{W}} = \exp\left( \int \alpha_u dW_u - \frac{1}{2}\int \alpha^2_u du \right)$$.

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  • $\begingroup$ Perhaps the Cameron-Martin theorem is what you are looking for? $\endgroup$ Feb 7, 2013 at 17:55
  • $\begingroup$ I don't think so. As far as I understand, Cameron-Martin is a special case of Girsanov. If I specify the drift, Girsanov's theorem tells me the change of measure. What I want to know is: given an equivalent change of measure, is it always expressable in the form above? In short, are changes of measure via Girsanov's theorem "the only thing you can do" to change the measure? $\endgroup$ Feb 7, 2013 at 22:05

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http://books.google.com.sg/books?id=YoMr5Mbo6coC&pg=PA164&lpg=PA164&dq=bjork+girsanov+theorem&source=bl&ots=imju5rQt6e&sig=Tp9bgNO6hDNZEXuEy6tmBs5nwZY&hl=en&sa=X&ei=HYkUUdaELoqKrgesioHYDA&redir_esc=y#v=onepage&q=bjork%20girsanov%20theorem&f=false

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  • $\begingroup$ It would be helpful if this answer could include the title/author/page of the book in question. Google Books won't show me that page. $\endgroup$ Jan 15, 2015 at 1:04
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The answer is yes:

https://fabricebaudoin.wordpress.com/2012/10/02/lecture-25-girsanov-theore/

Here is a good link. I think it is a better link than the currently most upvoted answer.

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Although I can't see the page linked by A.Kvashchuk, I presume it mostly answers your question along the lines: construct a positive martingale with the Radon-Nikodym derivative as terminal condition, represent it first as a stochastic integral and then (since it is positive) as the solution of a linear s.d.e. to yield a Doleans exponential.

In some cases, such as $X = -W$, this will yield $\alpha = 0$ , at odds with your equation $X = \alpha + W$.

It may help to clarify the terms in your question:

  1. defining $\alpha$ as $X-W$ does not get us very far, and needlessly requires to define the two processes on the same probability space.
  2. the Doleans exponential you show would correspond to $X = W - \int \alpha$ .
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