Timeline for Argmax of random walk vs of Brownian motion
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| S Oct 2, 2013 at 18:53 | history | bounty ended | CommunityBot | ||
| S Oct 2, 2013 at 18:53 | history | notice removed | CommunityBot | ||
| Sep 25, 2013 at 22:07 | comment | added | ofer zeitouni | The answer is I believe yes, if $2$ and $3$ are allowed to depend on $d$. The reason is that with the statement as you wrote it, it is enough to consider $t>0$ (due to symmetry) and then all you need to check is the tail estimate $P(argmax(B_t-d t)>T)$ as $T\to\infty$ (you need a precise asymptotic). It does decay at the same rate for the RW and BM, and your factor of $2$ takes care of the rest. | |
| Sep 25, 2013 at 21:11 | comment | added | Elena Yudovina | True enough. I edited the question to clarify the sort of statement I would like to be true. | |
| Sep 25, 2013 at 21:09 | history | edited | Elena Yudovina | CC BY-SA 3.0 |
added 216 characters in body
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| Sep 24, 2013 at 20:05 | comment | added | ofer zeitouni | To elaborate on Martin's reply: if $d$ is fixed, then with positive probability the argmax's can be far from each other (a picture should convince you of that, so it is a question about probability, not about regularity of the argmax). As $d\to 0$, the probability tends to $0$. So the only sense of the question is if you make it quantitative. | |
| S Sep 24, 2013 at 17:15 | history | bounty started | Elena Yudovina | ||
| S Sep 24, 2013 at 17:15 | history | notice added | Elena Yudovina | Draw attention | |
| Sep 23, 2013 at 16:33 | comment | added | Elena Yudovina | The end goal is constructing conservative confidence intervals for the maximizer of $RW_t$ using $B_t - d|t|$ (whose maximizer has a known distribution). So a statement of the form "one is at most twice the other" (for some value of 2), or "their difference is at most blah" (for some nice blah), or any combination of the two, would work. | |
| Sep 22, 2013 at 19:00 | comment | added | Martin Hairer | If $d$ is of order one, then both arg maxes will be of order one and at distance of order one of each other. In that sense, they will not be "close". Do you have some limiting case in mind? In the limit $d \to 0$, one would expect the statement to make sense and to be true. | |
| Sep 22, 2013 at 16:27 | history | asked | Elena Yudovina | CC BY-SA 3.0 |