Timeline for Does mixing automatically imply this seemingly stronger "uniform modulo re-ordering" version of mixing?
Current License: CC BY-SA 4.0
39 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 3, 2023 at 0:57 | vote | accept | Julian Newman | ||
| May 3, 2023 at 0:57 | answer | added | Julian Newman | timeline score: 0 | |
| Apr 30, 2023 at 21:11 | answer | added | user65023 | timeline score: 2 | |
| S Apr 5, 2022 at 19:03 | history | bounty ended | CommunityBot | ||
| S Apr 5, 2022 at 19:03 | history | notice removed | CommunityBot | ||
| Mar 30, 2022 at 19:23 | answer | added | Ronnie Pavlov | timeline score: 2 | |
| Mar 30, 2022 at 15:05 | history | edited | YCor |
edited tags
|
|
| Mar 30, 2022 at 14:13 | comment | added | Julian Newman | @RonniePavlov I have no objection myself - hopefully it will be fairly clear that what you've written isn't the complete answer but just a very useful example to point in the right direction. | |
| Mar 30, 2022 at 13:44 | comment | added | Ronnie Pavlov | I can also move my comments to answers; I just didn't want to purport to have a complete answer. But now that I did seem to answer in the bernoulli case, it might be worth moving there to get more replies. Any objection? | |
| Mar 30, 2022 at 12:11 | comment | added | Julian Newman | @Asaf We've now proved that Bernoulli systems fulfil the property I was asking about. Although my previous versions of the post did not state the question clearly at the top, and so confusion was to be expected, I have now clearly and succinctly stated the question at the top of the post. Accordingly, for the sake of reducing the excessive length of the comments thread (which will somewhat deter potentially very useful further comment-thread discussion), my own personal opinion is that it is now best if you delete your comments and I delete my replies. At this point, they are a distraction. | |
| Mar 30, 2022 at 11:56 | history | edited | Julian Newman | CC BY-SA 4.0 |
edited mixing requirement on the base of the RDS
|
| Mar 29, 2022 at 23:50 | history | edited | Julian Newman | CC BY-SA 4.0 |
added statement about Bernoulli shifts
|
| Mar 29, 2022 at 23:26 | comment | added | Ronnie Pavlov | Yes, this was a typo, I usually am thinking about the two-sided case. I don't quite understand why MO doesn't let you edit comments after five minutes, or I would fix these... | |
| Mar 29, 2022 at 23:24 | history | edited | Julian Newman | CC BY-SA 4.0 |
updated "progress update"
|
| Mar 29, 2022 at 22:25 | comment | added | Julian Newman | @RonniePavlov Very nice! By the way, you've written $\{0,1\}^\mathbb{N}$ suggesting a one-sided shift. But presumably it's fairly trivial that for the one-sided shift, the mixing holds uniformly (in the classical, much stronger sense of uniform convergence) across all $B$. Presumably it's for the two-sided shift that classical uniform convergence clearly fails and therefore your very clever arguments are necessary to get the desired result. | |
| Mar 29, 2022 at 20:47 | comment | added | Ronnie Pavlov | Thanks! I actually do think the result holds for all $A$ in the i.i.d. case. If $A$ depends on coordinates $1$ to $k$, then we can mimic the proof above by passing to a subset of $i_1,\ldots,i_n$ where all elements have distance at least $k$, which still has size at least $n/k$. Then make $\pi_n$ send coordinates $i_j,\ldots,i_j+k-1$ to $jk,\ldots,jk+k-1$ for each $j$, and everything should still work. Then, for any $A,t$, there should be clopen $A'$ approximating $A$ and $t'$ where $|\mu(A'\cap T^{-m}B)-μ(A')μ(B)|<t'$ implies $|\mu(A\cap T^{-m}B)-μ(A)μ(B)|< t$, so $N_{A,t} \leq N_{A', t'}$. | |
| Mar 29, 2022 at 19:23 | comment | added | Julian Newman | @RonniePavlov Beautiful observation! I've generalised your observation to give a possible strategy of proof - see the "progress update" after the statement of the question at the start. | |
| Mar 29, 2022 at 19:20 | history | edited | Julian Newman | CC BY-SA 4.0 |
added progress update
|
| Mar 29, 2022 at 3:06 | comment | added | Ronnie Pavlov | (above, when I say weakly convergent, I mean identify a set $S$ with the measure $\mu_S$ defined by $\mu_S(R) = \mu(S \cap R)$, and take a weak(-star) convergent sequence of those measures. Sorry if this wasn't clear, it was hard to do with character limits (and might have a mistake or two). Just trying to illustrate that maybe this isn't as easily contradicted as other similar-looking conditions. Also, this might follow from something in Julian's enormous post, but I haven't been able to make it through the whole thing. | |
| Mar 29, 2022 at 3:03 | comment | added | Ronnie Pavlov | Say $\mu$ is i.i.d. on $\{0,1\}^{\mathbb{N}}$, that $A=[0]$, $t>0$, and Julian's property fails. Then for all $n$, there exist $B_n$ and $i_0,...,i_{n-1}$ s.t. $|\mu(A\cap T^{-i_j}B_n)-\mu(A)\mu(B_n)| \geq t$ for $j<n$. But $\mu$ is i.i.d., so exchangeable. Take a self-map $\pi_n$ of $\{0,1\}^{\mathbb{N}}$ sending coordinate $i_j$ to $j$, and consider $C_n=\pi_n(B_n)$. You can check that $\mu(A\cap T^{-j} C_n)-\mu(A)\mu(C_n))| \geq t$ for $j<n$. If you take a weakly convergent sequence $C_{k_n} \rightarrow C$, then $\mu(A\cap T^{-j}C-\mu(A)\mu(C)| \geq t$ for all $j$, violating mixing. | |
| Mar 29, 2022 at 2:28 | comment | added | Ronnie Pavlov | @Asaf For what it's worth, I also thought this followed from general "you can't guarantee any explicit rate of mixing" stuff, but the way this question is phrased I don't think it's obvious. In particular, I think that whenever $\mu$ is i.i.d., $T$ the shift, and $A$ is a cylinder set, this property actually does hold (I think it holds for general $A$, but haven't thought enough). More details in next comment. | |
| Mar 28, 2022 at 17:37 | comment | added | Julian Newman | @Asaf I think you probably haven’t read the question carefully. But I would love to be proved wrong: if you’re sure you have read my question correctly, please write out a full explanation of your answer and I’d be happy to award you the 500-point bounty :) | |
| Mar 28, 2022 at 17:16 | comment | added | Asaf | I answered your question in my pervious comment, when pertaining to very general sets, one cannot quantify mixing even in Bernoulli systems. Notice that all the quantitative mixing statement usually depend on functions being in some appropriate Sobolev space... | |
| S Mar 28, 2022 at 17:04 | history | bounty started | Julian Newman | ||
| S Mar 28, 2022 at 17:04 | history | notice added | Julian Newman | Draw attention | |
| Mar 28, 2022 at 13:25 | comment | added | Julian Newman | @Asaf Hopefully things are a bit clearer now. If so, would it be okay with you if we could please refresh the comments thread, i.e. you delete your comments so far and I'll delete mine? | |
| Mar 28, 2022 at 0:44 | history | edited | Julian Newman | CC BY-SA 4.0 |
put question clearly at start
|
| Mar 27, 2022 at 23:07 | history | edited | Julian Newman | CC BY-SA 4.0 |
added comment at the start
|
| Mar 27, 2022 at 22:59 | history | edited | Julian Newman | CC BY-SA 4.0 |
edited body
|
| Mar 27, 2022 at 22:56 | comment | added | Julian Newman | @Asaf I was in no way reinventing fundamental concepts, I was expressing the completely standard definition of mixing in a slight notational shorthand. But still, your comments (and the question's two downvotes!) have been very helpful in helping me to see how the structure and presentation of my post was confusing and difficult to follow. I have made another considerable revision, that hopefully makes things a bit easier to follow :) | |
| Mar 27, 2022 at 22:52 | history | edited | Julian Newman | CC BY-SA 4.0 |
added theorem, and drastically improved structure
|
| Mar 27, 2022 at 14:46 | review | Close votes | |||
| Mar 28, 2022 at 17:04 | |||||
| Mar 27, 2022 at 14:33 | comment | added | Asaf | First $P$ wasn't a probability measure on the space of events in your previous version. Second - even in this version - $\rho_{n}$ is defined on a (product space) of the power set, again not just the space, using the term pointwise here is at best confusing, at worse probably completely off track, as it is only an $L^2$ notion for the original system... It is better not to reinvent fundamental and well-studied concepts... Regarding your actual question, it is virtually impossible to quantify mixing uniformly over ``all possible sets'', can easily be seen already for i.e. Bernoulli shifts | |
| Mar 27, 2022 at 4:51 | history | edited | Julian Newman | CC BY-SA 4.0 |
added remark after question
|
| Mar 27, 2022 at 4:33 | history | edited | Julian Newman | CC BY-SA 4.0 |
Considerably re-wrote question
|
| Mar 27, 2022 at 1:43 | comment | added | Julian Newman | @Asaf No, not the measure of their union. $P$ is already a probability measure on the space of events identified up to $\mu$-a.s. equality. But anyway, I'm currently in the process of drastically re-writing the question, since my summability requirement has a considerably simpler equivalent characterisation. | |
| Mar 27, 2022 at 1:36 | comment | added | Asaf | $B$ is a subset (event), so $Y_{n}(\epsilon,A)$ is a collection of subsets. What do you mean by $P(Y_{n}(\epsilon,A))$? The measure of their union? | |
| Mar 26, 2022 at 15:53 | history | edited | Julian Newman | CC BY-SA 4.0 |
added further thought
|
| Mar 26, 2022 at 15:13 | history | asked | Julian Newman | CC BY-SA 4.0 |