Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.