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.

How do you show $B_T\in\mathcal{F}_T$ for T is a stopping time?

Note the filtration is generated by the Brownian motion (and not necessarily completed, in particular, $\mathcal{F}_T\neq\mathcal{F}_{T+}$)

and a much harder question:

Are all Brownian Motion stopping times previsible? (Please point me to a proof or reference)

share|improve this question
1  
Since you assert that the first question is much easier than the second, what did you try to solve it? –  Did Feb 26 '13 at 19:52
    
@Did someone has shown me a proof, which I vaguely remember, you take a map from the sample path to $I_{(t\leq T)}, B_t$ and something else and use some composition of maps. He thinks there is an easier method and I did not quite follow the proof. If we wish to prove it is measurable with respect to $\mathcal{F}_{T+}$, then we can take a sequence of stopping time $T_n\downarrow T$, but the assertion above is a bit harder. –  Lost1 Feb 26 '13 at 20:00

1 Answer 1

up vote 1 down vote accepted

Answer to the first question is an application Proposition 2.18 on Karatzas and Shreve, where we take $X$ to be the Brownian motion in question.

The answer to the second question is difficult. It turns out this is the case for all processes, for which there exists some sort of martingale representation theorem, but unfortunately I cannot remember the reference for this.

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.