6
$\begingroup$

Take a time interval $[0,T]$, and a filtered probability space $(\Omega,P,\mathcal{F},\mathcal{F}_t)$. If $X \in L^1(\mathcal{F}_T)$, then $M_t = E [X \ | \ \mathcal{F}_t]$ is a martingale. If I want the martingale $M$ to have continuous or right continuous paths, is there a condition I can impose on the filtration to ensure this?

A standard result says that if the filtration is right-continuous, meaning that $\cap_{s>t} \mathcal{F}_s = \mathcal{F}_t$, then there exists a modification of $M$ with right continuous paths (in fact right continuous with left limits). However, I want to say something about the original process, and not a modification.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ I don't see how you can enforce continuity. Let $M$ be zero until $T/2$, then $\pm 1$ with equal probabilities thereafter. Cadlag might be another story though. $\endgroup$
    – Steve Huntsman
    Commented Jun 27, 2010 at 18:46

2 Answers 2

Reset to default
3
$\begingroup$

Generally speaking, you cannot do this at the level of conditions on filtration since conditional expectation is defined up to modifications on zero measure sets. For example, take $T=1$, and the probability space be $[0,1]$ with Borel sigma-algebra and Lebesgue measure. Let $X(\omega)=\omega$ and all sigma-algebras from the filtration coincide with Borel. We can choose $M$ to be defined by $M_t(t)=0$ and $M_t(\omega)=\omega$ if $t\ne\omega$. Clearly, every trajectory is discontinuous, although the filtration is as regular as possible.

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ 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. $\endgroup$
    – weakstar
    Commented Jun 27, 2010 at 23:22
  • $\begingroup$ @weakstar: The continuity of the martingale is a condition of the martingale representation theorem, not a conclusion. $\endgroup$
    – Yuri Bakhtin
    Commented Jun 28, 2010 at 4:25
  • $\begingroup$ 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. $\endgroup$
    – weakstar
    Commented Jun 28, 2010 at 14:00
  • 1
    $\begingroup$ 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. $\endgroup$
    – Yuri Bakhtin
    Commented Jun 29, 2010 at 6:37
  • $\begingroup$ Ok, I see what you're saying. Thanks. $\endgroup$
    – weakstar
    Commented Jun 29, 2010 at 17:35
2
$\begingroup$

Hi,

The thing is that both your original process and the càldlàg modification of it, are equivalent with respect to your probability measure (if the usual hypothsesis hold).

Probabilisticly speaking, there is nothing that can be said about your original process that cannot be said from your modification (if the usual hypothsesis hold) and this is why we work with the nice one.

Why don't you ask the real question you have in mind about your original process ? (which is not continuous even for a Brownian Motion look for example at the construction by Karatzas and Shreve !!! )

Best Regards

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .