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Let us consider the standard Brownian motion and the natural filtration $(\mathcal{F}_t^B)$. It is known that $(\mathcal{F}_t^B)$ is not right-continuous at $t=0$. But what about $t>0$? Is it true that $(\mathcal{F}_t^B)$ is not right-continuous at $t>0$? If so, could you please explain why it is not right-continuous?

saz
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Almost sure
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I think so, surely it just follows from Markov property. Take any time $s>0$, $W_t=B_{t+s}-B_s$ is just again a standard Brownian Motion. The filtration generated by $W_t$ is not right-continuous at 0.

$\mathcal{F}_{t+s}^B=\sigma(\mathcal{F^B_t},\mathcal{F^W_{s}}) $

In particular $\mathcal{F}_t^B=\sigma(\mathcal{F^B_t},\mathcal{F^W_0}) $ and $\mathcal{F}_{t+}^B=\bigcap\limits_{s>0}\sigma(\mathcal{F^B_t},\mathcal{F^W_s})=\sigma(\bigcap\limits_{s>0}\mathcal{F^B_t},\mathcal{F^W_s})$

Now $\sigma(\mathcal{F^B_t},\mathcal{F^W_{s}})$ and $\sigma(\bigcap\limits_{s>0}\mathcal{F^B_t},\mathcal{F^W_s})$ are clearly different, because $\mathcal{F^B_t}$ and $\mathcal{F^W_{s}}$ are indepdnent, also $\bigcap\limits_{s>0}\mathcal{F^B_t}$ and $\mathcal{F^W_s}$ are independent plus the fact $\mathcal{F^B_t}$ and $\bigcap\limits_{s>0}\mathcal{F^B_t}$ generate different $\sigma$-algebras.

Lost1
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  • I am not so sure about the "plus the fact.." thing. We know that in the end the difference is only due to sets of measure zero as the augmented filtration is, in fact, right continuous. I would be more convinced if I could see one element of a class that is not in the other. – Maurice Apr 03 '17 at 18:54
  • @Maurice the set of ${B_t$ is continuous at $t}$ is measurable with respect to $\bigcap\limits_{s>0}\mathcal{F^B_t}$ , but not the other one. so are the events $B_t$ hitting an open an interval. – Lost1 Apr 20 '17 at 08:25