Timeline for Order statistics of Brownian motions
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Aug 6, 2014 at 5:19 | comment | added | Sandro | Thanks! Turns out that basically all I needed was the Lebesgue Vitali theorem that provides the right generalization of the dominated convergence theorem, so I can get from the a.s.-convergence, which quite obviously holds, to convergence in expectation. In any case, thanks a lot for your help! | |
| Aug 5, 2014 at 17:51 | comment | added | Douglas Zare | and you only have slightly more information at $T'$. So, for example, if you choose at time $1.01$ the particle that was greatest at time $1.00$, this will rarely lead to a significantly different outcome. In case you choose a different particle, use coupling to switch the motions. | |
| Aug 5, 2014 at 17:48 | comment | added | Douglas Zare | @Sandro: Lipschitz continuity should fail when the times are close to $0$, but other than that I think it should work, though I haven't written out the details. One idea is to compare $x$ at one set of times with $x$ at another set in which all of the times are a little later (not just close). If these have to be close, then you can compare $x$ at close points by comparing each with the coordinatewise maximum. Next, to compare $x$ at a point $T$ with $x$ at a point $T'$ with all times slightly greater, note that at $T'$ you can use the strategy of optimizing wrt what you knew at $T$, | |
| Aug 5, 2014 at 9:05 | comment | added | Sandro | Would you mind outlining the proof you have in mind a bit more? What do you use the coupling for? To estimate the distance to the limit for each $n$? How does the coupling imply Lipschitz continuity? Thanks a lot! | |
| Aug 4, 2014 at 20:21 | comment | added | Douglas Zare | @Sandro: I think a similar coupling argument works, showing not just continuity but Lipschitz continuity. | |
| Aug 4, 2014 at 18:30 | comment | added | Sandro | I just noted a mistake in my proof that $x_j(t_j;t_1, …, t_{j-1})$ is jointly continuous in all arguments. I think what I'm missing might be a standard argument. I take a sequence $(s_1^n, …, s_j^n)$ that converges to $(t_1, …, t_j)$, and try to show that $x_j(s_1^n, …, s_j^n) - x_j(t_1, …, t_j)$ converges to 0. So I condition the expectation on the left on $q(t_1), … , q(t_j)$ (law of it. exp.). By continuity of the Brownian paths, I then get pointwise convergence. However, I'm lacking a dominating function, so I can't apply Lebesgue's theorem. Any ideas for a workaround? Thanks! | |
| Jul 23, 2014 at 0:20 | vote | accept | Sandro | ||
| Jul 22, 2014 at 23:35 | comment | added | Douglas Zare | @Sandro: I'm happy to have helped. Yes, this is my real name. | |
| Jul 22, 2014 at 22:53 | comment | added | Sandro | Wow, I finally got it!:) Thanks! Douglas Zare is your real name? Then I can appropriately credit you in a footnote. (My Advisor, Al Roth, thinks the paper will be published well.) | |
| Jul 20, 2014 at 9:52 | comment | added | Douglas Zare | @Sandro: Continue all Brownian motions with one (inductively, some subset) translated. If you translate one BM up by removing a lower BM instead of the highest, then all of your choices at the next removal point are translations of the possibilities if you had removed the highest. | |
| Jul 19, 2014 at 19:22 | comment | added | Sandro | Could you say which BMs you couple, and what coupling you choose? | |
| Jul 19, 2014 at 9:12 | comment | added | Douglas Zare | @Sandro: No, I don't use $i$ and $j$ interchangeably. $i$ is an index of an opponent's choice. | |
| Jul 19, 2014 at 7:59 | comment | added | Sandro | You use $i$ and $j$ interchangeably, right? I guess the stumbling block is the coupling (of I know the definition, but don't have experience with). Also, I'm not sure whether $j=n$ is w.l.o.g. (but that might be because I don't yet quite understand your proof.) If $j=n$ then any gain to the adversary is a loss to player $j$. If $j<n$, then a gain to the adversary can also be due to him picking the player who will be ranked second-to-worst rather than worst at $t_j$, which will not decrease player $j$'s payoff. | |
| Jul 19, 2014 at 6:26 | comment | added | Douglas Zare | @Sandro: I assumed $j=n$ because later players don't affect player $j$ so we might as well assume player $j$ is the last player. What don't you understand about the first two lines of my answer? | |
| Jul 19, 2014 at 6:07 | comment | added | Sandro | Thanks a lot! However, either I misunderstand your reply, or you misunderstood my question. I'm interested the payoff to player $j$ holding fixed $t_j$, but varying $t_1, …, t_{j-1}$. (I'm only interested in the case $t_1\leq…\leq t_j$.) In case I misunderstood your reply, then it's probably the first two lines. | |
| Jul 19, 2014 at 5:07 | history | answered | Douglas Zare | CC BY-SA 3.0 |