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I seen several sources claim that any martingale in a Brownian filtration is continuous. However while working with processes of the form $\mathbb{E}(X\mid \mathcal{F}_{s})$ for some random variable $X$ I have a hard time finding results or realise why this would be continuous(without switching to some version(in particular the cadlag version) of it.

Karatzas and Shreve assumes that they are working with cadlag processes when proving the martingale represenation theorem and thus that doesn't really apply.

https://www.springer.com/kr/book/9781468403022

Revous and Yor on the other hand only claim existence of a version on the form of a Ito integral, which in turn is continuous.

https://www.springer.com/cn/book/9783540643258

Here is the statement from Yor.

He first states that there is no discontinuous martingale in this filtration,then he only proves that there is a continuous version. This dosnt seam consistent to me, but i might be missing something.

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    $\begingroup$ Since $M_s := E[X \mid \mathcal{F}_s]$ is only well defined up to null sets, I don't see how you can expect anything stronger than "there exists a continuous version". After all, I can always choose a discontinuous version $\tilde{M}_s$, and it will be just as valid to say $\tilde{M}_s = E[X \mid \mathcal{F}_s]$. $\endgroup$ – Nate Eldredge Apr 6 '19 at 13:43
  • $\begingroup$ @Nate Eldredge that must means his statement before the theorem is not true $\endgroup$ – user1 Apr 6 '19 at 15:01
  • $\begingroup$ Well, it's a slightly informal statement, and evidently one should interpret "discontinuous" as "having no continuous version". In view of the definition of a martingale, that's really the only sensible interpretation. For the precise formal statement, one should look to Theorem 3.4 which discusses versions as it should. $\endgroup$ – Nate Eldredge Apr 6 '19 at 16:56
  • $\begingroup$ @NateEldredge right, but I was a bit confused about the fact that he says one thing then prove something else. Anyway thanks for the feedback. $\endgroup$ – user1 Apr 6 '19 at 17:51
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It is remarked just before Theorem (2.10) in Chapter II of Revuz and Yor that "unless otherwise stated, we will consider only right-continuous submartingales." Such a submartingale (in particular, martingale) will have left limits, a.s. The discussion just prior to Corollary (3.3) is therefore correct: A (by assumption right continuous) Brownian martingale is continuous, a.s.

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  • $\begingroup$ Well yes, given that one takes that paragraf(150 pages ago) to refer to rest of the book, then his statement is consistent. Thanks for pointing that out. $\endgroup$ – user1 Apr 28 '19 at 5:27
  • $\begingroup$ How else would one take it? $\endgroup$ – John Dawkins Apr 28 '19 at 16:42
  • $\begingroup$ You are right he said it there and it makes his later statement correct and consistent. I did not read this book from cover to cover and so Ive forgotten what he wrote in the beginning of it. Again thanks for pointing that it it. $\endgroup$ – user1 Apr 28 '19 at 16:46

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