## Right-continuity of natural filtrations

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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Sounds like homework. – Warren Schudy Nov 22 2010 at 15:32
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/…. – George Lowther Nov 22 2010 at 17:49
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 $\mathcal{F}_{0+}=\sigma(U)$. – George Lowther Nov 22 2010 at 17:52
Having said that, I think this kind of question is more suited to math.stackexchange. – George Lowther Nov 22 2010 at 17:53
@Warren: Thanks for this helpful comment. @George and Byron: Thank you. – Johannes Nov 23 2010 at 9:13

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.
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.