Timeline for The property of a Markov measure
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 30, 2013 at 11:08 | answer | added | Stéphane Laurent | timeline score: 0 | |
| Mar 15, 2013 at 15:15 | vote | accept | Anton | ||
| Mar 15, 2013 at 15:07 | comment | added | Anton | To Ilya: by Markov measure I mean exactly the concept defined here: en.wikipedia.org/wiki/Subshift_of_finite_type | |
| Mar 15, 2013 at 15:07 | comment | added | Anton | Yes, I assume that $(A^\mathbb{N}, \sigma, m)$ is a measure-preserving dynamical system, namely shift of finite type with Markov Measure (sometimes it is called Markov shift). | |
| Mar 15, 2013 at 15:06 | answer | added | R W | timeline score: 1 | |
| Mar 15, 2013 at 13:24 | comment | added | SBF | I would support HW in his question about the framework (even though it perhaps not related to probability) - seems to be similar to what I am doing. | |
| Mar 15, 2013 at 13:11 | comment | added | R W | Another question: are you assuming that $m$ is shift invariant? | |
| Mar 15, 2013 at 13:03 | answer | added | Anton | timeline score: 0 | |
| Mar 15, 2013 at 13:01 | comment | added | SBF | By a Markov measure you mean, that for some stochastic kernel $K$ on the measurable space $(A,2^A)$ it holds that $P$ is the induced measure on $A^\Bbb N$? | |
| Mar 15, 2013 at 12:53 | comment | added | Anton | to R W: yes, exactly | |
| Mar 15, 2013 at 12:45 | comment | added | R W | What are $a$ and $b$ in your cylinder sets? Are you talking about one-dimensional cylinders determined by first letters? | |
| Mar 15, 2013 at 11:58 | comment | added | Anton | To Ilya: I consider a shift system $(A^\mathbb{N}, \sigma, m)$ over some finite "alphabet" $A$, therefore "open" means open in product topology | |
| Mar 15, 2013 at 11:53 | comment | added | Anton | To HW: the context is not easy to describe and it does not seem to be helpful - this is a small separate problem. The global problem is to show that some special partition has good property (similar to the properties of cylinder sets partition). "Doesn't it follow immediately from the case of $P$ a cylinder set" - this would be great, but I'm not sure that the statement of interest obviously follows from this case (may be I just do not understand some simple thing). | |
| Mar 15, 2013 at 11:13 | comment | added | HJRW | Can you give some context for this question? It looks a little like homework. Also, I'm no expert, but doesn't it follow immediately from the case of $P$ a cylinder set, just because $m$ is determined by its values on cylinder sets? | |
| Mar 15, 2013 at 10:38 | comment | added | SBF | What does an open set men in your context? | |
| Mar 15, 2013 at 9:54 | history | asked | Anton | CC BY-SA 3.0 |