Timeline for When is the entropy of a $\sigma$-algebra finite?
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14 events
| when toggle format | what | by | license | comment | |
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| Sep 17, 2016 at 17:49 | comment | added | Chill2Macht | @StéphaneLaurent That is true, I just thought that was only a feature of bijections, not also other types of isomorphism. Now I know better. | |
| Sep 17, 2016 at 16:12 | comment | added | Stéphane Laurent | @William An isomorphism between measure-algebras. For a usual $\sigma$-algebra, i.e. an essentially separable one, that means that the two $\sigma$-algebras have the same list of atoms. About your second question, isn't there a bijection between $\mathbb{N}$ and $\mathbb{Z}$ ? | |
| Aug 31, 2016 at 18:34 | comment | added | Chill2Macht | @StéphaneLaurent What does isomorphism mean for $\sigma$-algebras? How can there be an isomorphism between two $\sigma-$algebras, one of which is a strict subset of the other? | |
| Aug 31, 2016 at 7:42 | comment | added | Stéphane Laurent | It would be strange to find a quantity $H({\cal A})$ such that $H({\cal F}_n) < H({\cal F}_{n+1})$ for your filtration ${\cal F}$. Because the ${\cal F}_n$ are isomorphic whenever they are countably generated and have no atoms. | |
| Aug 27, 2016 at 16:29 | comment | added | Chill2Macht | entropy functions for a sigma algebra are defined in terms of functions on the atoms $$I_{\mu}(\mathscr{C}|\mathscr{A})(x) = -\log \mu_x^{\mathscr{A}} ([x]_{\mathscr{C}}) $$ I apologize for the poor job I am doing of explaining all of this -- these are the only four references for the subject which I have been able to find, and honestly speaking I don't really understand any of them well or how to apply them to the specific case I am interested in. maths.dur.ac.uk/~tpcc68/entropy/Ch5November2013.pdf, ncatlab.org/nlab/show/… | |
| Aug 27, 2016 at 16:27 | comment | added | Chill2Macht | @DuchampGérardH.E. You're right, when I say event space I am talking about the first element in the triple defining a probability space $(X, \mathscr{F},\mathbb{P})$, i.e. $X$ is the space such that the sigma algebra $\mathscr{F} \subset \mathcal{P}(X)$. With regards to the statement about the atoms of $\mathcal{F}$ being the points of $\mathbb{R}$ I am trying to refer to the 2nd page of maths.dur.ac.uk/~tpcc68/entropy/Ch5November2013.pdf, where the author says that the atoms of a countably generated sigma algebra have a certain form, and then on the fourth page the information and... | |
| Aug 27, 2016 at 15:08 | comment | added | Alexander Shamov | Disintegration (= conditioning) of probability measures is explained in any decent textbook on probability, see e.g. Kallenberg's "Foundations of modern probability". For the case of $\sigma$-finite measures I don't know of a good reference, but the basic idea is to reduce to the finite measure case: just multiply by some positive $L^1$ density $\phi$ to get an equivalent finite measure, then do the disintegration, then multiply the conditional measures by $\phi^{-1}$. | |
| Aug 27, 2016 at 6:13 | comment | added | Duchamp Gérard H. E. | I do not understand the foundations of your post : "on the event space $\mathbb{R}$ be given. I believe we also need the atoms of $\mathcal F,G$ to be the points of $\mathbb{R}$." It seems that it is ill-typed. Am I right ? could you elaborate a bit ? | |
| Aug 27, 2016 at 3:46 | comment | added | Chill2Macht | @AlexanderShamov It seems pretty clear I don't understand this as well as I thought I did. Do you have any suggested references for a text which discusses these issues (conditional measures) at this level? My understanding of conditional expectations is OK at best, but I really don't understand conditional (probability) measures well at all and was hoping I could avoid them -- it seems like this is not possible. | |
| Aug 27, 2016 at 2:55 | comment | added | Alexander Shamov | ... The expression $\mathbb{E}[f \mid \mathscr{G}]$ that you suggested would give exactly this density, if you were taking the conditional expectation w.r.t. the normalized conditional Lebesgue measure, which would only make sense if it was "normalizable", i.e. finite. | |
| Aug 27, 2016 at 2:54 | comment | added | Alexander Shamov | Another issue with your post seems to be that since the reference measure is infinite, some of the conditional measures on the fibers of the partitions $\mathscr{F}$ and $\mathscr{G}$ (are still well-defined, but) may be infinite as well. Thus even if an absolutely continuous measure $\mu$ has an $\mathscr{F}$-measurable density when restricted to $\mathscr{F}$, it doesn't necessarily have a $\mathscr{G}$-measurable density when restricted to $\mathscr{G}$ (think about the trivial $\sigma$-algebra)... | |
| Aug 27, 2016 at 2:42 | comment | added | Chill2Macht | @AlexanderShamov I imagine $\mathbb{E}[f|\mathscr{G}]$ or any other reasonable/standard/canonical projection of $f$ onto the space of $\mathscr{G}$-measurable functions would work. Do you think I should update/change the post to try to correct this error? | |
| Aug 27, 2016 at 2:15 | comment | added | Alexander Shamov | What do you mean by $\intop (- \log f) d \mu^{\mathscr{G}}$ when $f$ is not $\mathscr{G}$-measurable? | |
| Aug 27, 2016 at 1:14 | history | asked | Chill2Macht | CC BY-SA 3.0 |