Timeline for Are the sublevel sets of Boltzmann entropy compact in Wasserstein distance?
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13 events
| when toggle format | what | by | license | comment | |
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| Dec 29, 2023 at 9:11 | vote | accept | Akira | ||
| Dec 29, 2023 at 9:04 | history | edited | Martin Sleziak |
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| Dec 29, 2023 at 8:53 | answer | added | leo monsaingeon | timeline score: 2 | |
| Dec 28, 2023 at 23:20 | answer | added | pseudocydonia | timeline score: 1 | |
| Nov 2, 2023 at 16:01 | comment | added | plm | @Kostya_I so it seems that we should go the other way around: consider discrete entropy as fundamental and recover differential entropy using some regularized limiting process. I think most of what can be said can be found here: profdoc.um.ac.ir/articles/a/1077376.pdf and math.stackexchange.com/questions/4540066/… , but i have to think about it. PS: Sorry again that i messed things up in the beginning, i only know basics and i did not realize there were bad surprises. | |
| Nov 2, 2023 at 15:26 | comment | added | plm | @Kostya_I yes im seeing now your comment, i realized after that i was computing with the formula for discrete entropy, assuming it yielded the same result for a Dirac distribution as the formula for differential entropy. Also, the entropy can indeed be negative in the differential case, but "usually" entropy is nonnegative, so i think some factor should be added to make comparisons easier. I may do a web search, later, to see what the conventions are. PS: Sorry i've been editing my comments now, as i was realizing they did not make sense, because of confusions on my part. | |
| Nov 2, 2023 at 15:24 | comment | added | plm | FIrst of all i must correct myself: the Dirac measure has differential entropy $H(\delta)=-\infty$, i was using discrete entropy, which is 0 and is generally nonnegative. To pass from differential entropy to discrete entropy i think we have to add a scaling factor like $-\log s$ when computing on intervals of size $s$. Then also @Kostya_I, i think there's a minus sign missing, to make the entropy positive in the discrete case. It is then always nonnegative with the minus sign (or nonpositive if we omit it as Akira). | |
| Nov 2, 2023 at 15:17 | comment | added | Kostya_I | @plm, sorry, I don't follow. If $\rho(x)$ is a probability density, then so is $\rho_a(x)=a^d\rho(ax)$ for any $a>0$. We then have $\mathcal{H}(\rho_a)=\mathcal{H}(\rho)-d\log a$. For large $a$, it tends to $+\infty$, while for small $a$, it tends to $-\infty$. | |
| Nov 2, 2023 at 13:58 | comment | added | Kostya_I | @Akira, the entropy in your formula is not necessarily non-negative, e.g., if the density is smaller than 1 everywhere, then the integrand is negative. | |
| Nov 2, 2023 at 12:56 | comment | added | plm | Hi Akira, do you see many people call this Boltzmann entropy ? I'd say "differential entropy", "Gibbs entropy", or just "entropy",. I think that Boltzmann was only interested in uniform measures -or rather trivial measures, on a single microstate, as the formula engraved on his tombstone indicates. I don't think this set is compact: take 2 0-entropy (Dirac) measures, $C=0$, make them move apart on $\mathbb R$, their distance tends to infinity. You can do the same with any single probability measure. You may also make up examples with classical distributions whose entropies are known and match. | |
| Nov 2, 2023 at 12:51 | comment | added | Akira | @Kostya_I Thank you so much for your help! I follow the convention in optimal transport that the Boltzmann entropy is non-negative. In $\mathcal{P}^2_{ac}(\mathbb{R}^d)$, I identify a measure with its density. | |
| Nov 2, 2023 at 12:47 | comment | added | Kostya_I | Clearly not, since you can take $\rho_n(x)=\rho(x-x_n)$ with $x_n\to\infty$, and this has a fixed entropy but no convergent subsequence. Also, is you formula for entropy missing a - sign? And shouldn't you call $\rho$ densities rather than measures? | |
| Nov 2, 2023 at 12:27 | history | asked | Akira | CC BY-SA 4.0 |