Timeline for Fixed objects of the M endofunctor on category Meas
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Dec 28, 2012 at 19:23 | comment | added | Michael Greinecker | @Tom Taht would be great! I'm from Europe, but I'm sure going there one day. | |
| Dec 28, 2012 at 18:46 | comment | added | Tom LaGatta | @Michael Okay, now I see where you're coming from, and I understand why it's not as straightforward as I'd hoped. Thanks so much for your comments! Let me know if you ever pass through NYC, btw. I'd love to chat more in person. | |
| Dec 28, 2012 at 11:25 | comment | added | Michael Greinecker | ...the problem with adapting this to the case of $M(X)$ is that the measurable functions $f:X\to P(Y)$ will in general not be integrable anymore, even with respect to very simple probability measures. | |
| Dec 28, 2012 at 11:21 | comment | added | Michael Greinecker | @Tom Can you explain how you construct your canonical section? My impression is that is quite different from the type space construction in epistemic game theory. For probability measures, the original problem seems to be rather easy. If $P(X)\subseteq M(X)$ consist of the probability measures on $X$, then $P$ is idempotent in that one can integrate with respect to a probability measure over probability measures to get a probability measures. As a matter of fact, measurable functions $f:X\to P(Y)$ correspond to transition probabilities or Markov kernels $k:X\times\mathcal{Y}\to[0,1]$... | |
| Dec 28, 2012 at 4:56 | comment | added | Tom LaGatta | Returning to my question: are the counterexamples you have provided obstructions to the map $\varphi$ being one-to-one and/or onto? | |
| Dec 28, 2012 at 4:55 | comment | added | Tom LaGatta | Also, @Michael, I am not necessarily trying to construct a particular measure on a product space which agrees with an arbitrary collection of marginal distributions, as with Daniell-Kolmogorov. I would rather study the space of measures on the product, then project down and see what kinds of marginals we are left with. Now, if you argue that there are no interesting measures on the product space (there is always the zero measure), then so be it, but that does not seem to be your point. | |
| Dec 28, 2012 at 4:52 | comment | added | Tom LaGatta | (oops $2c$, but the point stands) | |
| Dec 28, 2012 at 4:52 | comment | added | Tom LaGatta | I certainly agree that probability measures are convenient, but I don't see why you insist that probability measures are necessary. I also don't know what you mean by, "in general, all sets have infinite or zero measure." There are plenty of product measures which do not satisfy that property, e.g., the product measure on $[0,1]^{\mathbb N}$ whose $i$th marginal is $c^{2^{-i}}$ times the uniform measure. Its total mass is $c$, so it is not a probability measure when $c\ne 1$. If you insist that the marginal projections are IID random variables then yes, I agree that one runs into difficulties. | |
| Dec 27, 2012 at 17:57 | comment | added | Michael Greinecker | ...due to Heifetz and Samet: tau.ac.il/~samet/papers/coherent.pdf The problem does however not occur when one simply takes products of probability spaces. Every direct product of prob spaces can be extended to a prob measure on the product $\sigma$-algebra, something that kakutani has shown a long, long time ago. This result does hold only for probability measures. | |
| Dec 27, 2012 at 17:52 | comment | added | Michael Greinecker | @Tom Yes, the infinite product of measurable spaces clearly exists and the problem lies in identifying it with a measure on the product. The problem one has when working with non-probability measures that in general, all sets have infinte or zero measure (just think about $\mathbb{R}^\infty$ with Lebesgue measure on each factor). In the type space constructions used in game theory (and in the projective limit theorems of Daniell, Kolmogorov, Bochner etc.), only probability spaces are used. There is also a version of the Andersen-Jessen counterexample that is adapted to epistemic game theory... | |
| Dec 27, 2012 at 17:04 | comment | added | Tom LaGatta | @Michael, your counterexample is sound but I am not certain it is relevant here. First, note that we are working with measurable spaces and not measure spaces. As a set, $X_{\infty}$ certainly exists, and it should be a measurable space when equipped with the minimal $\sigma$-algebra so that the projection maps are measurable. The example by Andersen & Jessen might be an obstruction to showing that $X_{\infty} \cong M(X_{\infty})$ in general; I do not readily see why this is the case. Why does the projective limit seem to make sense only when we deal with probability measures? | |
| Dec 27, 2012 at 11:25 | comment | added | Michael Greinecker | One needs additional assumption for the projective limit to exist, usually of a quasi-topological kind. The classical counterexample can be found here: sdu.dk/media/bibpdf/Bind%2020-29%5CBind%5Cmfm-25-4.pdf Also, the projective limit seems to make only sense when we deal with probability measures. | |
| Dec 27, 2012 at 4:22 | history | edited | Tom LaGatta | CC BY-SA 3.0 |
added 14 characters in body
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| Dec 27, 2012 at 4:16 | history | answered | Tom LaGatta | CC BY-SA 3.0 |