Timeline for Stopping times for Brownian motion
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 3, 2017 at 10:25 | vote | accept | James Martin | ||
| Jul 27, 2016 at 19:06 | answer | added | John Dawkins | timeline score: 8 | |
| Jul 25, 2016 at 2:23 | comment | added | Nate Eldredge | I guess the tough part is, how do we construct a candidate stopping time $\tau'$? @MartinHairer's approach seems useful in verifying that they are a.s. equal. | |
| Jul 24, 2016 at 14:02 | comment | added | Martin Hairer | I would try to combine the 0-1 law with the strong Markov property to show that the statement holds (and this would also show why the example is not a counterexample). | |
| Jul 23, 2016 at 13:28 | history | edited | James Martin | CC BY-SA 3.0 |
added 107 characters in body
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| Jul 23, 2016 at 13:24 | comment | added | James Martin | Thankyou Stéphane. Good point! I did not ask the right question. I should instead ask whether there is a $\mathcal{G}$-stopping time which is equal to $\tau$ with probability 1. In the case you mention, every such $A$ must have probability 0 or 1. I will edit. | |
| Jul 23, 2016 at 13:17 | comment | added | Stéphane Laurent | Is it true that there exists $A \in {\cal F}_0$ such that $A \not\in {\cal G}_0$ ? In this cas $\tau = {\boldsymbol 1}_A$ is a ${\cal F}$-stopping time but not a ${\cal G}$-stopping time. | |
| Jul 23, 2016 at 11:15 | history | asked | James Martin | CC BY-SA 3.0 |