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I cannot understand the concept of it.

So a filtration is right continuous if for every $t$ it holds that:

$\mathcal{F_t}=\bigcap\limits_{\varepsilon>0}\mathcal{F_{t+\varepsilon}}$

But if for every $t$, then it also holds for $t=0$. And if I choose a large $\epsilon$, then it means that at time zero I know every information about the process?

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    $\begingroup$ No, because you're taking the intersection over $\varepsilon>0$. (To avoid the problem of uncountable intersections we can assume WLOG that $\varepsilon$ is rational.) $\endgroup$ – Math1000 Jul 14 '16 at 14:56
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    $\begingroup$ @Math1000 +1 for the comment, but I don't see any "problem" with uncountable intersections. $\endgroup$ – user940 Jul 14 '16 at 15:55
  • $\begingroup$ @Math1000 Okay but what is the intuition behind it? why is it useful? $\endgroup$ – FelB Jul 14 '16 at 17:20
  • $\begingroup$ @ByronSchmuland Good point, as we are taking the intersection of $\sigma$-algebras as opposed to the intersections of elements of $\sigma$-algebras... $\endgroup$ – Math1000 Jul 14 '16 at 20:45
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The idea is that you gain no additional information by taking an infinitesimal step forward in time.

Remember that an $\mathit{intersection}$ means that we are taking only the elements contained in EVERY set in the intersection. So, if we think of each $F_t$ as the information contained in the system up to time $t$, the intersection $\cap_{\epsilon > 0} \mathcal{F}_{t+\epsilon}$ contains only the information in EVERY $\mathcal{F}_{t+\epsilon}$ for every possible value of $\epsilon > 0$. That is, only the information contained up until $t+\epsilon$ for every $\epsilon > 0$, in particular, any arbitrarily small $\epsilon$. So, in this intersection, we have added only the information gained by taking an infinitesimally small step forward in time.

Thus, the idea of right continuity, $\mathcal{F}_t=\cap_{\epsilon > 0} \mathcal{F}_{t+\epsilon}$ is that no information is added in this infinitesimal step. In other words, there are no instantaneous developments of the system, it evolves in a continuous fashion going forward in time.

(Much credit for this answer is due to Huyen Pham, whose book I'm currently using to review some of this material.)

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  • $\begingroup$ "In other words, there are no instantaneous developments of the system, it evolves in a continuous fashion going forward in time" , can you eleborate on this? Do you mean there is no time for which we can say here somthing changed? these are so close and small that we can point at them`? $\endgroup$ – user1 Feb 18 '19 at 7:03
  • $\begingroup$ @user1 yes and no. Essentially, with right-continuity, we can know that an instantaneous change did occur in the past, we just can't know that one is about to occur in the next instant. It's easier to think of a right-continuous process than a filtration (at least for me). In a right-continuous process, you could have jumps, but the left-hand endpoint of any segment is closed. That is, if there is a jump, you only know that it happened after the fact; it can't be predicted. A right-continuous filtration is similar. I can know that something in the system has changed, just can't predict it. $\endgroup$ – Charles Beer Sep 3 '19 at 20:54
  • $\begingroup$ @user1 also, sorry for the huge delay. I had stopped checking the email address to which this account is linked (due to graduation) and hadn't been on here in a long time. And sorry for the probably remaining lack of clarity. This is my best understanding of a very abstract and difficult concept. $\endgroup$ – Charles Beer Sep 3 '19 at 20:55

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