Path continuity for (closed) martingales? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T02:26:47Z http://mathoverflow.net/feeds/question/29707 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/29707/path-continuity-for-closed-martingales Path continuity for (closed) martingales? weakstar 2010-06-27T13:17:43Z 2010-09-29T14:14:29Z <p>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?</p> <p>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.</p> http://mathoverflow.net/questions/29707/path-continuity-for-closed-martingales/29726#29726 Answer by Yuri Bakhtin for Path continuity for (closed) martingales? Yuri Bakhtin 2010-06-27T18:48:08Z 2010-06-27T18:48:08Z <p>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.</p> http://mathoverflow.net/questions/29707/path-continuity-for-closed-martingales/29905#29905 Answer by The Bridge for Path continuity for (closed) martingales? The Bridge 2010-06-29T11:51:21Z 2010-06-29T11:51:21Z <p>Hi,</p> <p>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). </p> <p>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.</p> <p>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 !!! )</p> <p>Best Regards</p>