Timeline for Properties of the algebraic self-difference set of Brownian motion zeros
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Aug 1, 2014 at 21:30 | vote | accept | MassiveJack | ||
| Jul 27, 2014 at 3:10 | answer | added | Pablo Shmerkin | timeline score: 4 | |
| May 27, 2014 at 21:29 | history | edited | MassiveJack | CC BY-SA 3.0 |
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| May 25, 2014 at 22:14 | history | edited | MassiveJack | CC BY-SA 3.0 |
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| May 23, 2014 at 5:30 | comment | added | MassiveJack | @Tom Hutchcroft : You're right, you have misunderstood the statement. I claim that there is no time $t$ such that both $t$ and $t+1$ are zeros of $B$. | |
| May 23, 2014 at 0:18 | comment | added | tmh | "Not surprisingly, I found that this set is almost surely empty." This isn't true: clearly there are times when $B(t) > B(t+1)$ and times when $B(t)<B(t+1)$, and so the existence of times when $B(t)=B(t+1)$ follows from the intermediate value theorem applied to $B(t)-B(t+1)$. Sorry if I have misinterpreted your original statement. | |
| May 20, 2014 at 0:22 | comment | added | Christian Remling | This looks fine to me. Thanks for the clarification. | |
| May 19, 2014 at 23:30 | history | edited | MassiveJack | CC BY-SA 3.0 |
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| May 19, 2014 at 6:53 | comment | added | MassiveJack | Well, I'm definitely not saying that this is straightforward, but I think I have a proof that the first probability is zero. I will try to post it in a few hours. | |
| May 18, 2014 at 23:46 | comment | added | Christian Remling | Is it really clear that your first probability is zero? That definitely works for fixed $t$, but then you have uncountably many $t$ to consider. Or, viewed from another angle, $D(Z)$ is a set of differences of numbers taken from a $\dim$ $1/2$ Cantor set, so it's not clear there can be a quick argument showing this is small ($C-C$ contains an interval). | |
| May 18, 2014 at 21:16 | history | asked | MassiveJack | CC BY-SA 3.0 |