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Jul 4, 2010 at 19:23 vote accept weakstar
Jun 29, 2010 at 17:35 comment added weakstar Ok, I see what you're saying. Thanks.
Jun 29, 2010 at 6:37 comment added Yuri Bakhtin There must be some regularity property assumed a priori. Say, a version of this theorem, Thm.4.15 in Ch.3 of "Brownian motion and stochastic calculus" by Karatzas & Shreve (viewable at google books) assumes RCLL (cadlag) paths and then one of the conclusions is that under this assumption the paths are in fact continuous. You cannot do this with no regularity assumptions on paths. Also notice that if you have a nice process you can find its less nice modification. Since the filtration does not change, what you are looking at is not a filtration property, it depends on a concrete modification.
Jun 28, 2010 at 14:00 comment added weakstar I'm a bit confused by your response. As I understand it, a conclusion of the Martingale Representation Theorem is that all martingales with respect to the augmented Brownian filtration have continuous paths.
Jun 28, 2010 at 4:25 comment added Yuri Bakhtin @weakstar: The continuity of the martingale is a condition of the martingale representation theorem, not a conclusion.
Jun 27, 2010 at 23:22 comment added weakstar Suppose the filtration were generated by a Brownian Motion. Then at least in this special case, one can say that the actual martingale has continuous paths, by applying the Martingale Representation Theorem. This is the only way I can think of off the top of my head to show that a martingale has continuous paths, i.e. writing it as an integral of a continuous martingale. So this question could be related to when you have such representation results, although this latter condition definitely seems more restrictive.
Jun 27, 2010 at 18:48 history answered Yuri Bakhtin CC BY-SA 2.5