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Hi,I'm trying to evaluate an expectation of a functional of the entire path of a Brownian Bridge (i.e. it contains an integral). By using a result I found in a paper I can evaluate this expectation when the Brownian Bridge does not have any drift, however I need to generalise it to the case with drift.Thus I'm wondering if there's any way of using the C-M-G theorem (or something else?) to transform Brownian Bridges with drift into ones without. However, it's not clear how I should do this. If you try to treat the bridge as W(t)-tW(1), then you end up with quadratic terms in t, and when I tried to use the C-M-G theorem on the SDE formulation of the Brownian Bridge, I ended up having to integrate (1-u)^(-2) over the range [0,1], which is infinite, meaning the change of measure term comes out at 0 (which rather suggests some necessary condition must have failed).Any help with this would be much appreciated.Thanks in advance,Tom

It's difficult to know what you are trying to do but the distribution of W(s) given W(t) and W(0) does not depend on the drift of W. That suggests to me that you can just use the zero drift result.

Oh well by "a Brownian Bridge with drift" I meant the path of W(s) given W(0)=0 and W(1)=x where x\ne 0.I can do the x=0 case, I'd like to be able to generalise to the x\ne 0 one.Thanks,Tom

Then essentially you have that W(s) - sx is a brownian bridge pinned at zero at both ends... There is no way a change to an equivalent measure will map this to the ordinary bridge since in one case the probability of W(1) = x is one and in the other it's zero.

You understand the standard Cameron-Martin theorem right? In the standard case you have a random walk with drift mu, and you can "change measure" into one without drift. E.g. you move to a measure with drift.Your argument if I understand it would apply just as well to this case, so cannot be correct. It is certainly not obvious without doing the maths that you can't use the C-M-G theorem to do the same thing for a Brownian bridge, and yet that appears to be the case.

Indeed I am familiar with the fact that you can change the drift of a Brownian motion whilst passing to an equivalent measure. The crucial point is that passing to an equivalent measure does not change the set of events of probability zero and of probability one. (by definition...)For the Brownian bridge pinned at x, we have P(W(1)= x ) =1, and this will be true in any equivalent measure.

Ahh right, sorry I understand you now. That does make perfect sense.Thanks.

QuoteOriginally posted by: cfpYou understand the standard Cameron-Martin theorem right?fyi, mj stands for Mark Joshi

I should really have understood his original reply (sorry!), my background's not in finance/SDEs so perhaps it was a clash of language. I guess his obvious first sentence threw me into not reading the second close enough to realise he was making an argument about absolute continuity. But anyway, thanks again for being patient with me.