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I learned over the years that the natural filtration of Poisson process is right-continuous as is and that for Levy processes in general it becomes right-continuous if it is augmented (completed with all P-null set in the space $(\Omega, {\cal F}, P)). $ I am aware that there are situations where even the completion does not produce that result and perhaps the simplest example is the stochastic process $Z=\{Z_t: t \in R^+\} $ where $Z_t = tX $ and $X is $ a random variable that is not constant a.s. However, I realized that I never proved that the natural filtration of Brownian motion is not right-continuous unless augmented. Is there a simple proof of this fact? Does the right-continuity fails only at $t=0 $ or does it fail at $t>0 $ as well? I have seen a proof on StackExchange (Lack of right-continuity of the filtration adapted to Brownian motion), but I am not convinced that this approach works (and in fact it was neither voted up nor accepted as an answer). Thank you.

Maurice

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Maurice
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For a fixed $t>0$, the event $\cap_{n\in\Bbb N}\{B_{t+s}>B_t$ for some $s\in\Bbb Q\cap(0,1/n)\}$ is in $\mathcal F_{t+}$ but not in $\mathcal F_t$. ($(\mathcal F_t)_{t\ge 0}$ is the natural filtration of a Brownian motion $(B_t)$.)

John Dawkins
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  • So, John, you are saying that the two terms ${\cal F}t $ and ${\cal F}{t+} $ only differs in terms of the elements contained in the tail sigma algebra and which are basically elements of the sigma algebra generated by the null sets, i.e. the same technique using to prove right-continuity of the augmented filtration for a Levy Process. I would have hoped that there was a more remarkable example of an event in one but not the other sigma algebra. However, I suppose that if augmenting does the trick, the difference cannot really amount to more than that. – Maurice Apr 03 '17 at 23:08
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    Yes, the difference between $\mathcal F_t$ and $\mathcal F_{t+}$ is the "right germ field at $t$", and for Brownian motion or any other Lévy process) these events have probability $0$ or $1$. I'm not sure how remarkable an example you are looking for. – John Dawkins Apr 04 '17 at 16:14
  • @John Dawkins : The intuition is very clear in your example but I don't see the rigourous line of argument that allows to see that if $\omega \in \cap_{n\ge 1}{B_{t+1/n}>B_t}$ then $\omega \not\in \mathcal{F}_{t}$. Could you elaborate on the details (probably by absurd reasoning argument). Best regards – TheBridge Apr 04 '17 at 18:36
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    An event $A\in\mathcal F_t$ has the property that if $\omega\in A$ and if $\omega'\in\Omega$ is such that $\omega'(s)=\omega(s)$ for all $s\in[0,t]$ then $\omega'\in A$ as well. The event that I suggested, $C:=\cap_{n\in\Bbb N}{B_{t+s}>B_t$ for some $s\in\Bbb Q\cap(0,1/n)}$ is clearly in $\mathcal F_{t+}$. The path $\omega(s):=s$, $s\ge 0$, is an element of $C$; the path $\omega'(s):= t-|s-t|$ is not an element of $C$; but but $\omega'(s)=s=\omega(s)$ for all $s\in[0,t]$. – John Dawkins Apr 04 '17 at 19:11
  • @John Dawkins : Thanks, very clear (and direct) argument. Best regards – TheBridge Apr 04 '17 at 20:40
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    @JohnDawkins For completeness let me point out that one way to see that the property that you mention holds (i.e. if $\omega(s)=\omega'(s)$ for all $s\in[0,t]$ and $\omega\in A\in F_t$ then $\omega'\in A$ as well) is to use Dynkin $\pi-\lambda$ theorem. – No-one Nov 06 '22 at 16:26
  • Also, the reasoning in @JohnDawkins comment implicitly assumes that $\Omega=C^0([0,\infty),\mathbb{R})$ since he works with a single $\omega$, but the result is true for any BM on any state space $\Omega$, since $C\cup C'\cup E=\Omega$, where $C':=\cap_n {B_{t+s}<B_t$ for some $s\in \mathbb{Q}^+\cap(0,1/n)}$ and $E$ is a null set, and hence $\mathbb{P}(C)\geq 1/2$ and the argument can be done for all $\omega\in C$ at once. – No-one Jun 19 '24 at 18:28
  • @John Dawkins: Hi, could you kindly explain why we have the property $\omega^\prime \in A$ if $\omega(s)=\omega^\prime(s)$ for all $s\in [0,t]$ ? Moreover, why you example says that $\mathcal{F}t$ and $\mathcal{F}{t+}$ only differs in terms of the elements of $\mathcal{N}$? What probability measure are we using to define the null set $\mathcal{N}$? Is it the Wiener measure?

    These problems are kind of hard to comprehend for a beginer like myself. If the answer is too tedious, could you recommend a book where I can do some self-study on them? Many thanks

     – Mingzhou Liu Jul 08 '24 at 07:55
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    Given $A\in\mathcal F_t$, there is sequence ${s_n}\subset[0,t]$ such that $A\in\sigma(B_{s_n}: n\ge 1)$. (This is a consequence of the monotone class theorem.) In other words, there exists $B\in\mathcal B(\Bbb R)^\infty$ such that $$ A={\omega\in\Omega: (B_{s_1}(\omega), B_{s_2}(\omega),\ldots )\in B}. $$ If $\omega'$ is a second element of $\Omega$ such that $B_s(\omega')=B_s(\omega)$ for all $s\in[0,t]$, then in particular $B_{s_n}(\omega') =B_{s_n}(\omega)$ for all $n$, so $\omega'\in A$ as well. – John Dawkins Jul 08 '24 at 14:31
  • It's probably best to think of the sample space in this question as being the path space $C^0([0,\infty),\Bbb R)$, so that the governing measure is the Wiener measure. Then $\mathcal N$ is the class of sets of measure $0$ with respect to the completion of the Wiener measure. – John Dawkins Jul 08 '24 at 14:32