[Math] Proving that the natural filtration of Brownian motion (not augmented) is not right-continuous

probability theorystochastic-calculusstochastic-processes

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

Best Answer

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 $(B_t)$.)

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