Timeline for On a von Bahr–Esseen-type inequality for pairwise independent zero-mean random variables
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| Jan 30, 2023 at 13:29 | comment | added | Iosif Pinelis | @fedja : In your latter argument, the "constant" $c(p):=2^{2-p}$ is a function of $p$, and your factor $g(z):=2^{1-2z}$ is an analytic extension of $g(t)=2^{1-2t}=c(1/t)^{-t}$ for $t\in(1/2,1)$, if I am not mistaken. So, one may say your argument is based, in part, on the analytic extension $c(z):=2^{2-z}=g(1/z)^{-z}$ of the "constant" $c(p):=2^{2-p}$. I was wondering if one can similarly extend $\tilde C_p$ to some appropriate $\tilde C_z$, using (say) approximations by the Newton or secant method. Does this make sense? | |
| Jan 29, 2023 at 23:05 | comment | added | fedja | @IosifPinelis Hmmm. I'm not sure what you mean by an "analytic extension" of a constant in an inequality but I'll take a look later. | |
| Jan 29, 2023 at 20:46 | comment | added | Iosif Pinelis | The key step here seems to extend $\tilde C_p$ analytically to all complex $p$ with $\Re\frac1p\in(\frac12,1)$ and then continuously to $p$ with $\Re\frac1p\in[\frac12,1]$ so that the extension be $\le1$ in modulus for $\Re\frac1p\in\{\frac12,1\}$. The analyticity could be attained by taking the rather straightforward analytic extensions of the approximations by the Newton or secant method of the root $x_p\in((p-1)/5,(p-1)/2)$ of equation (1.14) in that paper. What do you think? | |
| Jan 29, 2023 at 20:45 | comment | added | Iosif Pinelis | @fedja : It seems that the inequality $E\big|\sum_{j=1}^n X_j\big|^p\le E|X_1|^p+\tilde C_p\sum_{j=2}^n E|X_j|^p$ with $\tilde C_p$ as in Proposition 1.8 in projecteuclid.org/journals/annals-of-functional-analysis/… could be proved by your interpolation method for zero-mean pairwise independent $X_j$'s. | |
| Jan 27, 2023 at 15:52 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 24, 2023 at 3:47 | comment | added | Iosif Pinelis | Thank you again. I have just fixed some typos. | |
| Jan 24, 2023 at 3:47 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jan 23, 2023 at 23:57 | comment | added | fedja | @IosifPinelis Done. I tried to follow your notation but might be a bit clumsy with it, so let me know if something I wrote does not make sense. | |
| Jan 23, 2023 at 23:55 | history | edited | fedja | CC BY-SA 4.0 |
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| Jan 23, 2023 at 23:01 | comment | added | Iosif Pinelis | @fedja : Thank you. | |
| Jan 23, 2023 at 21:44 | comment | added | fedja | @IosifPinelis OK, but a few hours later :-) | |
| Jan 23, 2023 at 21:34 | comment | added | Iosif Pinelis | @fedja : 1) Of course, you are right. I don't know how I got the $+$ in the expression for $x_0$. 2) I still don't see how to modify the proof for zero-mean $X_i$'s. In particular, I don't know what you want to multiply by $2^{z-1}$. Could you just describe the needed modifications of $Y_z$, $S_z$, and $F(z)$, preferably in the explicit style they are described in my answer? | |
| Jan 22, 2023 at 22:34 | comment | added | fedja | @IosifPinelis 1) Nope, it is correct: we are looking there at the part covered by no $E_j$, so $1$ minus the sum of measures ($m\varepsilon$) plus the overlap. 2) Almost none. Just multiply by $2^{z-1}$ before applying the maximum principle. | |
| Jan 19, 2023 at 22:52 | comment | added | Iosif Pinelis | Concerning part 1 of Edit 2, I am not sure what modifications to make (say in my presentation of your proof) to get $2^{2-p}$, even though the analyticity is clear. | |
| Jan 19, 2023 at 21:40 | comment | added | Iosif Pinelis | Previous comment continued: Instead of the induction on $m$, you can use the construction on p. 4 of Ref. [9] in the paper linked in my question on this MO page, with $n=m$, $x_2=\varepsilon^2$, $x_1=\varepsilon-(n-1)\varepsilon^2$, and $x_0:=1-nx_1-\frac{n(n-1)}2\,x_2$. I had tried to use that to show that $C_{2-}=\infty$ -- of course, in vain. | |
| Jan 19, 2023 at 21:40 | comment | added | Iosif Pinelis | Thank you for Edit 2. Concerning its part 2, I think there should be $1-m\varepsilon-\frac{m(m-1)}2\varepsilon^2$ instead of $1-m\varepsilon+\frac{m(m-1)}2\varepsilon^2$, and then $1-x-\frac{x^2}2$ instead of $1-x+\frac{x^2}2$. Then, to get $C_{1+}=2$, you can let $x=p-1\downarrow0$. | |
| Jan 16, 2023 at 2:45 | comment | added | fedja | @IosifPinelis Done. I don't know if the lower bound is actually the truth in the mean zero version, but if it is, the proof will require a completely different technique, so it will take some time to figure it out anyway... | |
| Jan 16, 2023 at 2:39 | history | edited | fedja | CC BY-SA 4.0 |
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| Jan 15, 2023 at 18:03 | comment | added | fedja | @IosifPinelis Why should I mind? If someone's mother tongue is English and I send him a note in Russian, it is only natural that he would translate it before storing it :lol: But once you bumped it to the front page anyway, I'll try to edit my post once again today, adding two more remarks (in particular, showing that $C_{1+}=2$, indeed, as you said) | |
| Jan 15, 2023 at 17:44 | comment | added | Iosif Pinelis | @fedja : I have written a presentation (mathoverflow.net/a/438604/36721) of the second part of your answer, in a way that would be more accessible to me. I hope you don't mind -- please let me know otherwise. | |
| Jan 13, 2023 at 1:25 | comment | added | fedja | @IosifPinelis Let's try to find some decent lower bound for $C_p$ now :-) | |
| Jan 13, 2023 at 1:18 | comment | added | Iosif Pinelis | I see, thank you -- just Cauchy--Schwarz. This is great! | |
| Jan 13, 2023 at 1:08 | comment | added | fedja | @IosifPinelis When $\Re z=1/2$, the r.v. $X_{i,z}$ satisfy $\sum_i E|X_{i,z}|^2=1$ and they are symmetric and pairwise independent, hence orthogonal, so $\|\sum_i X_{i,z}\|_2=1$ while $\|Y_z\|_2=1$ as well and Cauchy-Schwarz finishes the story. | |
| Jan 13, 2023 at 0:59 | comment | added | Iosif Pinelis | Right, I forgot what $C_p$ actually was there. Concerning your answer, I still don't see how you get $|F(z)|\le1$ for $\Re z=1/2$. Apparently, it is here that you use the pairwise independence -- but I don't see how. Could you please explain this? | |
| Jan 13, 2023 at 0:25 | comment | added | fedja | @IosifPinelis I tried to read your example of $C_{1+}=2$ and noticed that in the paper you referenced the meaning of $C_p$ is different: the inequality there is $E|\sum_i X_i|^p\le E|X_1|^p+C_p\sum_{i\ge 2}E|X_i|^p$ and the example gives the sharp value only for that inequality and (unless I misunderstand what is written) only for $n=2$. This doesn't seem to imply that our $C_{1+}$ (which is a factor at all $E|X_i|^p$, including $i=1$) is still $2$. Probably I'm just missing something simple, but it was a long day, so can you clarify what you meant and what example you had in mind here? | |
| Jan 12, 2023 at 22:46 | comment | added | fedja | @IosifPinelis Feel free to ask more questions any time if anything remains unclear :-) | |
| Jan 12, 2023 at 22:43 | comment | added | fedja | @IosifPinelis $F(1/p)=\|\sum_i X_i\|_p$ by the construction. The function $F(z)$ is bounded by $1$ on the boundary of the strip and is continuous up to the boundary and bounded by something large but fixed a priori, so the maximum principle applies and we get $|F(1/p)|\le 1$, which is exactly what we want. Yes to the rest. | |
| Jan 12, 2023 at 22:32 | comment | added | Iosif Pinelis | $2^{2-p}$ sounds great. Can you detail "Now we can consider $F(z)=E(Y_z\sum_i X_{i,z})$ as an analytic in the strip $\frac 12\le\Re z\le 1$ function, as usual, and use the trivial endpoint estimates ($p=1$ and $p=2$) to get the whole range with constant $1$ for the symmetric case."? In particular, I am not sure what relation, if any, you mean here between $F(z)$ and $\|\sum_i X_i\|_p$. Also, you meant $H$ to be real and $|v|=1$, and $q$ dual to $p$, right? | |
| Jan 12, 2023 at 21:42 | comment | added | fedja | @PaataIvanishvili Indeed. The same complex interpolation but a cute trick of reducing the lower bound to the upper bound for some linear operator applied to a function of two variables. That was new to me. Thanks! Thus $C_p\le 2^{2-p}$. I wonder whether that is sharp now. | |
| Jan 12, 2023 at 20:56 | comment | added | Paata Ivanishvili | @fedja Yes, the value $a_{p}=2^{p-1}$ is known. See theorem 1.2 here arxiv.org/pdf/1905.01274.pdf | |
| Jan 12, 2023 at 19:22 | comment | added | fedja | @IosifPinelis $A_p=2$ for all $p\in[1,2]$ (the same complex interpolation though it might be an overkill). I'm still not sure about $a_p$. My suspicion is that it is $2^{p-1}$, i.e., Rademacher is the worst case but the proof still escapes me. But even the trivial estimate $a_p\ge 1$ gives $C_p\le 2$ for all $p\in[1,2]$ :-). | |
| Jan 12, 2023 at 14:58 | comment | added | fedja | @IosifPinelis OK, I found something interesting that you may like. See the edit :-) | |
| Jan 12, 2023 at 14:57 | history | edited | fedja | CC BY-SA 4.0 |
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| Jan 12, 2023 at 13:15 | comment | added | fedja | @IosifPinelis I'm also not sure that $1$ is the correct answer but so far all my attempts to construct a counterexample failed. Perhaps I just don't know many examples of pairwise independent random variables :-). I guess I'll spend a few more days thinking of it and will update you if I find anything interesting. | |
| Jan 12, 2023 at 4:23 | comment | added | Iosif Pinelis | Previous comment continued: When the $X_i$'s are only pairwise independent, it seems that we only can use the additivity of the 2nd (possibly truncated) moments, which appears to create a (near-)singularity when $p$ is close to $2$. | |
| Jan 12, 2023 at 4:23 | comment | added | Iosif Pinelis | Thank you for your usual generosity. Of course, I will make a proper reference to your work if I get a chance to use it (the paper is now under review). I will try to get $16+o(1)$ or, perhaps, you can share that with me. I am not sure that $1$ is the right answer. Even when the $X_i$'s are completely independent, the best constant factor $C(p)$ is discontinuous (in that case, at $p=1$, with $C(1+)=2$; Proposition 1.8 (iii) in projecteuclid.org/journals/annals-of-functional-analysis/… ). | |
| Jan 12, 2023 at 3:23 | comment | added | fedja | @IosifPinelis You are most cordially welcome! (i) Of course, but it doesn't give anything useful for $\ell<0$ (ii) That particular factor can even be replaced by $1+o(1)$ as $p\to 2$ with some more accurate considerations but I didn't try to chase the constants since I wasn't getting $1$ anyway (iii) With that modification, one can get $C_p\le 16+o(1)$ as $p\to 2$ but I don't see much difference between $16$ and $600$ when the suspected answer is just $1$. (iv) Everything that I communicate publicly can be used in any way by anybody with or without reference. | |
| Jan 12, 2023 at 1:40 | comment | added | Iosif Pinelis | May I include your proof (perhaps slightly modified) into the linked paper, at arxiv.org/abs/2210.04391? Of course, in any case, I would refer to your answer. | |
| Jan 12, 2023 at 1:31 | comment | added | Iosif Pinelis | Thank you very much for your answer! Very clever, as usual. I did think about exponential partitioning, but was far from your technique. When asking this question, I thought you would be the one who could answer it, which now has happened. :-) Just some minor remarks: (i) What you did in Step 2 for $\ell\ge0$ seems to work for all $\ell$ (even though you need a better bound for $\ell<0$). (ii) In the first display in Step 3, it seems possible to replace $4^p$ by $3^p$. (iii) Even then, the constant factor for $p=2$ is $>600$, by my calculations. Certainly $<\infty$, but still rather large. | |
| Jan 12, 2023 at 1:19 | vote | accept | Iosif Pinelis | ||
| Jan 11, 2023 at 6:16 | history | answered | fedja | CC BY-SA 4.0 |