Timeline for Expected centered entropy of the binomial distribution
Current License: CC BY-SA 3.0
22 events
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| Mar 21, 2015 at 15:02 | vote | accept | TMM | ||
| Mar 18, 2015 at 11:22 | comment | added | Pietro Majer | OK, so you want an estimate on the uniform convergence of the Bernstein polynomials of $h(x)$, since $\sup_{p\in[0,1]}I_n(p )=\|h−B_n h\|_\infty$. | |
| Mar 18, 2015 at 0:39 | comment | added | TMM | @Pietro: It's perhaps best described in terms of $J_n(p) \triangleq n \sup\limits_{p \in [0,1]} I_n(p)$ as $\lim\limits_{n \to \infty} J_n(p) = 0.84\dots$ where $0.84\dots$ is some numerical constant. So take the maximum over $p$ for fixed $n$, and then consider large-$n$ asymptotics of the resulting expression. | |
| Mar 18, 2015 at 0:33 | answer | added | Brendan McKay | timeline score: 4 | |
| Mar 17, 2015 at 23:50 | comment | added | Pietro Majer | What should be the meaning of "for large $n$, $\max\limits_{p \in [0,1]} I_n(p) \to I_n(\frac{\alpha}{n})$" ? | |
| Mar 17, 2015 at 20:06 | history | edited | TMM | CC BY-SA 3.0 |
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| Mar 17, 2015 at 16:32 | comment | added | TMM | @Brendan: (I've been stuck on this for days now, trying to find an airtight proof for arbitrary $p$. If you can provide an explicit proof, please do!) | |
| Mar 17, 2015 at 16:30 | comment | added | TMM | @Brendan: If I'm not mistaken, all Taylor approximations are below $I_n(p)$, so I'm not sure how exactly to obtain a nice and tight upper bound on $I_n(p)$. Also in a remainder term, where we e.g. evaluate $h^{(m)}(x)$ at a point $x = \xi$, the point $\xi$ is only known to lie in an interval $[\frac{k}{n}, p]$ which could theoretically still be very close to $\frac{k}{n} < p$ and could cause trouble for values $k$ smaller than $np$. | |
| Mar 17, 2015 at 13:20 | comment | added | Brendan McKay | If $p$ is bounded away from 0 and 1, then $nI_n(p)\to 1/(2\ln 2)$ as $n\to\infty$. I think that's also true for $p(1-p)$ larger than about $\log n/n$. To prove that, use a remainder term of order 4 for the Taylor expansion of $h(x)-h(p)$. | |
| Mar 17, 2015 at 13:13 | comment | added | Brendan McKay | Yes, it only works if $p$ is larger; I didn't try to figure out how large. | |
| Mar 17, 2015 at 9:49 | history | edited | TMM | CC BY-SA 3.0 |
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| Mar 17, 2015 at 9:42 | comment | added | TMM | @Brendan: I'm not sure that works without further assumptions on $p$. I mean $p = (\log \log n)/n$ also satisfies $n p = \omega(1)$ and I think in that case you'll be going on forever with your expansion before you can safely discard remaining terms. (And for arbitrary $p$ you cannot discard any order terms due to the non-trivial maxima near $0$ and $1$.) | |
| Mar 17, 2015 at 2:12 | comment | added | Brendan McKay | For larger $p$, restricting the sum to $\log n$ standard deviations from the mean is enough ($h(x)$ being bounded), then within those limits a few terms of the Euler-Maclaurin expansion will get it to the accuracy you need. Unfortunately the integral doesn't seem to have a closed form. | |
| Mar 17, 2015 at 0:32 | history | edited | TMM | CC BY-SA 3.0 |
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| Mar 17, 2015 at 0:13 | history | edited | TMM | CC BY-SA 3.0 |
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| Mar 16, 2015 at 22:08 | history | edited | TMM | CC BY-SA 3.0 |
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| Mar 16, 2015 at 21:32 | history | edited | TMM | CC BY-SA 3.0 |
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| Mar 16, 2015 at 19:58 | answer | added | Kevin P. Costello | timeline score: 2 | |
| Mar 16, 2015 at 18:40 | history | edited | TMM | CC BY-SA 3.0 |
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| Mar 16, 2015 at 18:31 | comment | added | TMM | @Robert: Yes, I'm sorry. I'll update it when I'm at my PC. (And the result is independent of the base of the $\log$s.) | |
| Mar 16, 2015 at 17:39 | comment | added | Robert Israel | Do you mean $h(x) = - x \log x - (1-x) \log(1-x)$? | |
| Mar 16, 2015 at 16:43 | history | asked | TMM | CC BY-SA 3.0 |