Timeline for Anti-concentration inequalities: lower bound on realized second moment
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 26, 2023 at 19:46 | comment | added | Iosif Pinelis | @CuriousGuy : The martingale is $(E(\|S_k\|\,|\,X_1,\ldots,X_{k-1}))$, where $S_k:=X_1+\cdots+X_k$. | |
| Jun 26, 2023 at 15:07 | comment | added | CuriousGuy | Thanks for the answer: what is the considered martingale then? | |
| Jun 26, 2023 at 15:01 | comment | added | Iosif Pinelis | @CuriousGuy : The lower bound here is not on the probability to deviate from the mean -- notice the condition $u\le E\|X\|$. | |
| Jun 26, 2023 at 14:49 | comment | added | CuriousGuy | How come the Azuma inequality becomes an anti-concentration one? The bound should be a concentration bound around the mean, not a lower bound on the probability to deviate from the mean. Am I wrong? | |
| Feb 21, 2020 at 9:23 | vote | accept | ABIM | ||
| Feb 20, 2020 at 0:30 | comment | added | Iosif Pinelis | @oferzeitouni : Thank you for your comment. This is now corrected. | |
| Feb 20, 2020 at 0:19 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 19, 2020 at 20:07 | comment | added | ofer zeitouni | Hmm. The Hoeffding-Azuma inequality you quote is not the one in the reference, and is actually wrong as the example $n=1$ and $u=E|X|$ shows. I believe there is a typo also in the reference you give (Bartlett's notes) - it should be an upper bound, not a lower bound, as the example t=0 (and the proof given there) show. | |
| Feb 19, 2020 at 19:47 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 19, 2020 at 18:11 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Feb 19, 2020 at 18:03 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |