Timeline for Upper and lower bounds for a Rademacher-type expectation
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 25 at 15:06 | answer | added | van der Wolf | timeline score: 0 | |
| Oct 25 at 10:28 | comment | added | van der Wolf | This may be relevant: arxiv.org/abs/2006.16834 | |
| Oct 17 at 17:45 | comment | added | Aryeh Kontorovich | Dear @IosifPinelis, I would be happy with any analytically tractable expression in $a$. | |
| Oct 16 at 21:41 | comment | added | Iosif Pinelis | In the previous comment, Mikael de la Salle suggested $L(a)$ and $U(a)$ with $U(a)=2^{n-1}L(a)$. In the same spirit, here are $L(a)$ and $U(a)$ with $U(a)=L(a)$: $U(a)=L(a)=2^{-n}\sum_{\delta\in\{-1,1\}^n}\exp(-\,|\sum_1^n a_i\delta_i|/2)$. So, the question is this: In what terms do you want $L(a)$ and $U(a)$ to be expressed? | |
| Oct 16 at 14:28 | comment | added | Mikael de la Salle | An easy one is $U(a) = 2^{n-1} L(a) = \exp(-\frac m 2)$, where $m = \min_{\varepsilon \in \{-1,1\}^n} |\sum a_i \varepsilon_i|$. | |
| Oct 16 at 13:33 | history | edited | Aryeh Kontorovich | CC BY-SA 4.0 |
added 107 characters in body
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| Oct 16 at 11:43 | history | asked | Aryeh Kontorovich | CC BY-SA 4.0 |