Talk:Girsanov theorem

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As stated, the aricle is wrong : (Zt) won't be a (positive) local martingale if (Xt) itself is not a local martingale (take dXt = dt as the simplest counter-example). — Preceding unsigned comment added by (talk) 00:14, 31 January 2012 (UTC)

Article is in need of attention[edit]

The statement of this difficult theorem is hard to follow and definitely wrong. For example, the article has the equation.

The left hand side is a Radon-Nikodym derivative. The notation on the right hand side is nonstandard but it is clear from what follows that it is denoting a geometric Brownian motion process. But the Radon-Nikodym derivative of one probability measure with respect to another is a random variable, not a process. Brian Tvedt 12:05, 25 January 2006 (UTC)

Good point, but you can always turn it into a process (and a martingale in fact) by taking conditional expectation with respect to the probability measure P. --DudeOnTheStreet (talk) 18:18, 27 May 2011 (UTC)

A stochastic process is just a random variable taking values in a bigger space (e.g the space of continuous functions over the non-negative reals) (talk) 10:26, 7 April 2008 (UTC)

Agreed. Lacks historical content as well. Also, there is very little mention on the Black-Scholes formula page regarding the importance of the Girsanov theorem, risk neutral measure, etc. Dmaher 00:48, 4 April 2006 (UTC)

No reference is made to quadratic variation -- the square brackets [X]_t -- which makes the article unreadable for non-specialists. Rogier (talk) —Preceding undated comment was added at 13:42, 20 November 2008 (UTC).

I agree that quadratic variation should at least be referenced; I wasn't even sure that the square-bracket notation meant quadratic variation. However, I have a more substantial issue, it seems that the page, as-is, contains a significant error. In the section "Statement of the Theorem," the process X_t is only assumed to be measurable and adapted to the Wiener filter. With only these assumptions, I do not believe that the Doleans exponential is a local martingale, as stated. In fact, I'm about 99 positive this is wrong. I think a simple assumption on X (something like X being a local martingale) was omitted. — Preceding unsigned comment added by (talk) 17:59, 16 July 2011 (UTC)