Timeline for Is "conditioning to a sub-$\sigma$-algebra" a measurable operation?
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| S Mar 15, 2017 at 19:58 | history | bounty ended | CommunityBot | ||
| S Mar 15, 2017 at 19:58 | history | notice removed | CommunityBot | ||
| Mar 12, 2017 at 0:31 | comment | added | Julian Newman | @NateEldredge: Thank you for your idea about considering $\mathbb{E}_\mathcal{G}$ for $\sigma$-algebras of the form $\mathcal{G}=\sigma(\{x\}:x\in A)$ for arbitrary $A \subset [0,1]$. This has turned out to be exactly the basis of my proof. | |
| Mar 12, 2017 at 0:25 | vote | accept | Julian Newman | ||
| Mar 12, 2017 at 0:23 | answer | added | Julian Newman | timeline score: 4 | |
| Mar 8, 2017 at 17:56 | comment | added | Julian Newman | Sorry, I don't know why I can't get the hyperlink to work: the url for my PhD thesis is https-colon-//spiral.imperial.ac.uk-colon-8443/handle/10044/1/39569 | |
| Mar 8, 2017 at 17:49 | comment | added | Julian Newman | I believe this can be found in Proposition 3.6 of Crauel, Random probability measures on Polish spaces, although it cites some other book for the proof (if I recall correctly). I also have a proof in <a href="spiral.imperial.ac.uk:8443/handle/10044/1/39569">my PhD thesis</a>, Lemma 3.27. | |
| Mar 8, 2017 at 17:46 | comment | added | Julian Newman | @user95282 Yes $\nu$ is really $\mathcal{G}$-measurable. The version of the disintegration theorem which I am using is that for any probability space $(\Omega,\mathcal{F},\mathbb{P})$, for any probability measure $\mu$ on $(\Omega\times [0,1],\mathcal{F}\otimes\mathcal{B})$ with $\mu( \,\cdot\, \times [0,1])=\mathbb{P}$, there exists an $\mathcal{F}$-measurable function $\nu\colon\Omega\to [0,1]$ (unique up to $\mathbb{P}$-a.s. equality) such that $\mu(A)=\int_\Omega\int_{[0,1]}\mathbf{1}_A(\omega,x)\,\nu(\omega)(dx)\,\mathbb{P}(d\omega)$ for all $A\in\mathcal{F}\otimes\mathcal{B}$. | |
| Mar 8, 2017 at 11:50 | comment | added | user95282 | @JulianNewman Please give a reference for the disintegration theorem used in the statement of the question. Is the function $\nu$ really $\mathcal{G}$-measurable, or is it measurable with respect to a completion of $\mathcal{G}$? | |
| Mar 8, 2017 at 3:21 | history | edited | Julian Newman | CC BY-SA 3.0 |
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| Mar 8, 2017 at 3:01 | comment | added | Julian Newman | So, since the number of equivalence classes of $2^{[0,1]}$ under countable symmetric difference is larger than $\mathfrak{c}$ (I presume!), there exists $\mathcal{G}$ such that $\mathbb{E}_\mathcal{G}$ is not Borel-measurable. The question of universal measurability remains. | |
| Mar 8, 2017 at 2:58 | comment | added | Julian Newman | Well, at the least, for any $A,B\subset [0,1]$ with $|B\setminus A|\geq 2$, we have that $\mathbb{E}_{\mathcal{G}_A}\neq\mathbb{E}_{\mathcal{G}_B}$: Fix distinct $x,y\in B\setminus A$. Then $\mathbb{E}_{\mathcal{G}_B}(\frac{1}{2}(\delta_{(x,x)}+\delta_{(y,y)}))=\frac{1}{2}(\delta_{(x,x)}+\delta_{(y,y)})$ but $\mathbb{E}_{\mathcal{G}_A}(\frac{1}{2}(\delta_{(x,x)}+\delta_{(y,y)}))=\frac{1}{4}(\delta_{(x,x)}+\delta_{(x,y)}+\delta_{(y,x)}+\delta_{(y,x)})$. | |
| Mar 8, 2017 at 2:00 | comment | added | Julian Newman | Wow, good question! If so, then this proves that $\mathbb{E}_\mathcal{G}$ is not necessarily a Borel map. However, what I am really keen to know is whether $\mathbb{E}_\mathcal{G}$ is at least universally measurable (in the sense that the pre-images of Borel sets are universally measurable). I think there are $2^\mathfrak{c}$ universally measurable maps. | |
| Mar 7, 2017 at 23:44 | comment | added | Nate Eldredge | One thing that's potentially alarming is that there are $2^{\mathfrak{c}}$ sub-$\sigma$-algebras of $\mathcal{B}$, whereas there are only $\mathfrak{c}$ many Borel maps from $\mathcal{M}_2$ to itself. (For instance, given any $A \subset [0,1]$, let $\mathcal{G}_A$ be generated by all the countable subsets of $A$.) Is the map $\mathcal{G} \mapsto \mathbb{E}_{\mathcal{G}}$ 1-1? | |
| S Mar 7, 2017 at 18:49 | history | bounty started | Julian Newman | ||
| S Mar 7, 2017 at 18:49 | history | notice added | Julian Newman | Draw attention | |
| Mar 7, 2017 at 14:25 | history | edited | Julian Newman | CC BY-SA 3.0 |
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| S Mar 7, 2017 at 1:00 | history | bounty ended | CommunityBot | ||
| S Mar 7, 2017 at 1:00 | history | notice removed | CommunityBot | ||
| Mar 5, 2017 at 21:54 | history | edited | Julian Newman | CC BY-SA 3.0 |
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| Feb 27, 2017 at 2:59 | history | edited | Julian Newman | CC BY-SA 3.0 |
further thoughts
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| S Feb 26, 2017 at 23:29 | history | bounty started | Julian Newman | ||
| S Feb 26, 2017 at 23:29 | history | notice added | Julian Newman | Draw attention | |
| Feb 26, 2017 at 23:24 | history | edited | Julian Newman | CC BY-SA 3.0 |
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| Feb 25, 2017 at 22:49 | history | edited | Julian Newman | CC BY-SA 3.0 |
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| Feb 22, 2017 at 17:40 | comment | added | Julian Newman | Thank you, that is interesting to know. It does not surprise me; my intuition behind why I expect continuity to fail in the general setup (where the marginal can vary) is that the sub-$\sigma$-algebra $\mathcal{G}$ is unrelated to the topology of $[0,1]$. This seems unlikely to be problematic when we fix the marginal; but when we vary the marginal, I expect Crauel's arguments will only generalise if we equip the base space with a topology like that induced by the total-variation distance (i.e. ignoring the topology of the base space), but this will probably induce the wrong $\sigma$-algebra. | |
| Feb 22, 2017 at 9:02 | comment | added | Michael Greinecker | If one looks at the measures in $\mathcal{M}_2$ with a fixed marginal in the first coordinate, the answer is yes. This follows from arguments in Random Probability Measures on Polish Spaces by Hans Crauel, where that case is treated under the heading "Conditional Expectation of Random Measures". In that case, the function is actually continuous in the weak topology. Maybe one can adapt the arguments there to your problem, but it might take a lot of work. | |
| Feb 22, 2017 at 2:32 | comment | added | Julian Newman | Certainly, at the least, I know that for general measurable spaces $G$ and $X$ with $X$ standard, there typically does not exist a measurable function $p:P\!r(G\times X)\times G\to P\!r(X)$ such that $p(\mu,\,\cdot\,)$ is a disintegration of $\mu$ for every $\mu$. | |
| Feb 22, 2017 at 2:00 | comment | added | Julian Newman | The idea that "everything that can be constructed is measurable" I think only really applies to standard Borel spaces, and the $\sigma$-algebra $\mathcal{G} \otimes \mathcal{B}$ on which we apply the disintegration theorem is not standard. (Now this idea that constructivity implies measurability is precisely why I expect my question to have the answer "yes"; but I suspect that proving it via measurability of the disintegration procedure itself will not work.) | |
| Feb 22, 2017 at 1:58 | comment | added | Julian Newman | As for the proof of the disintegration theorem, I have a feeling there are genuine issues with measurability here, which can probably only be remedied if perhaps $\mathcal{G}$ is countably generated. | |
| Feb 22, 2017 at 1:53 | comment | added | Julian Newman | Yes, the evaluation $\sigma$-algebra is precisely the Borel $\sigma$-algebra of the topology of weak convergence. I thought about the possibility of trying to use some kind of continuity property, but it felt unlikely to me that we would have such continuity, since $\mathcal{G}$ is an arbitrary sub-$\sigma$-algebra. Some kind of continuity with respect to total-variation distance seems more likely, but this generates the wrong $\sigma$-algebra. | |
| Feb 21, 2017 at 23:46 | comment | added | Nate Eldredge | We probably just need to look at a constructive proof of the disintegration theorem, and the measurability may become obvious. By the way, isn't the evaluation $\sigma$-algebra equal to the Borel $\sigma$-algebra of the weak topology? That could be helpful; the conditioning map may turn out to be (semi)continuous or otherwise nice with respect to that topology. | |
| Feb 21, 2017 at 22:44 | comment | added | Julian Newman | For each $x \in [0,1]$, $\nu(x)$ is a measure on $[0,1]$. An alternative way to write the integral is $\ \int_{A_1} \nu(x)(A_2) \, \pi_{1\ast}\mu(dx)\,$, where $\pi_1 \colon [0,1] \times [0,1] \to [0,1]$ is the projection $\pi_1(x,y)=x$. | |
| Feb 21, 2017 at 22:37 | comment | added | Michael Greinecker | In the integral, $\nu$ seems to be a function of both coordinates. | |
| Feb 21, 2017 at 17:08 | history | asked | Julian Newman | CC BY-SA 3.0 |