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My question: Is the natural filtration of a right-continuous process also right-continuous?

I would say yes, but don't know where to start proving it.

Thanks for your help/ideas!

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    $\begingroup$ Sounds like homework. $\endgroup$ Nov 22, 2010 at 15:32
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    $\begingroup$ No, as Byron Schmuland mentions, this is not true. However, for many types of processes you do get right-continuity if you append zero probability sets to $\mathcal{F}_0$. This is true for many Markov processes and, in particular, it is true for Feller processes (such as Brownian motion and Levy processes). This is standard and most books on continuous-time Markov processes should prove it. Also, with apologies for the self-advertisement, I have a proof of this fact on my blog: almostsure.wordpress.com/2010/07/19/…. $\endgroup$ Nov 22, 2010 at 17:49
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    $\begingroup$ As an example of how it can fail for non-Feller processes, even after adding zero probability sets, consider the following process. Let $X_t=Ut$ where U is a standard normal. Then, $\mathcal{F}_0$ is the trivial sigma algebra, and $\mathcal{F}_{0+}=\sigma(U)$. $\endgroup$ Nov 22, 2010 at 17:52
  • $\begingroup$ Having said that, I think this kind of question is more suited to math.stackexchange. $\endgroup$ Nov 22, 2010 at 17:53
  • $\begingroup$ @Warren: Thanks for this helpful comment. @George and Byron: Thank you. $\endgroup$
    – Johannes
    Nov 23, 2010 at 9:13

1 Answer 1

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Right continuity fails even for canonical continuous processes.

The natural filtration on $C([0,\infty))$ is not right continuous. For example, the event $\{\omega: {d^+\over dt}\ \omega_t\mbox{ exists at }t=0\}$ belongs to ${\cal F}_{0+}$ but not ${\cal F}_0$. In words, you can tell whether the function $\omega_t$ has a right derivative at $t=0$ with an infinitesimal peek beyond time 0, but you cannot tell just from the value of the function $\omega_t$ at time 0.

Right continuous filtrations are nicer to work with, and since it fails for the natural filtration,
we often use the right continuous version instead. Fortunately, many of the nice properties of right continuous processes carry over even with this enlarged filtration. For example, Brownian motion is still Markov with respect to ${\cal F}_{t+}$ which leads to interesting results like Blumenthal's zero-one law.

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  • $\begingroup$ Does $C[0,\infty)$ refer to Brownian motion on $\mathbb{R}_+$? If so, then why does the set of differentiability not belong to $\mathcal{F}_0$? I agree it's measure-zero, but it could still be $\emptyset$, in which case it would still be in $\mathcal{F}_0$. I'm not doubting you're correct, I just suppose I'm asking how you prove a neglible set like the one you mentioned is nontrivial. $\endgroup$
    – VF1
    Jan 6, 2017 at 18:16
  • $\begingroup$ The space $C([0,\infty))$ is the set of all continuous functions from $[0,\infty)$ to $\mathbb{R}$. There is no measure on this space at all in the first two paragraphs, so it has nothing to do with Brownian motion. The set $\{\omega: {d^+\over dt}\omega_t \mbox{ exists at } t=0\}$ is not empty, since there are a lot of differentiable functions. $\endgroup$
    – user6096
    Jan 8, 2017 at 21:58
  • $\begingroup$ Ah - the second part of what I said was then based off an inaccurate assumption; please disregard it. $\endgroup$
    – VF1
    Jan 8, 2017 at 22:36

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