Timeline for Borel-Cantelli Lemma on MCs (absorbing states)
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Feb 16, 2011 at 21:56 | vote | accept | sigma_z_1980 | ||
| Feb 8, 2011 at 6:59 | history | edited | Did | CC BY-SA 2.5 |
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| Feb 7, 2011 at 22:47 | comment | added | Did | @sigma: Like Ori said. The essentials are in chapter 1 of Markov chains by James Norris statslab.cam.ac.uk/~james/Markov. More detailed is Markov chains: Gibbs fields, Monte Carlo simulation, and queues by Pierre Brémaud books.google.com/books?id=KF0LgxRCgQsC. | |
| Feb 7, 2011 at 22:28 | comment | added | Ori Gurel-Gurevich | @sigma: I don't know about a source, but the proof is pretty straightforward. If the probability of returning to some state, when starting at it is $p<1$ then the expected number of visits to this state when started at it is $1/(1-p)$, since the distribution is just geometric with probability $1-p$ of success (=not returning). In particular, if the expected number of visits is infinite, then $p=1$. | |
| Feb 7, 2011 at 21:13 | comment | added | sigma_z_1980 | @DIdier: I think I got what you mean. Can you recommend some good source that discusses this proof in details? | |
| Feb 7, 2011 at 11:27 | comment | added | Did | @sigma: 1. The recurrence of a state of a Markov chain is equivalent to the divergence of a series but the proof of this result does not rely on a naive application of B-C lemma because the successive times of visits of a given state are not independent (to recover an independent structure one usually considers successive cycles of the path). 2. The absorbing property has nothing to do with B-C lemma nor with the convergence of a series: the fact that $x$ is absorbing simply means that the transition from $x$ to $x$ has probability one. | |
| Feb 7, 2011 at 10:38 | comment | added | sigma_z_1980 | @Didier: so, if I got you right, it is not enough to show divergence of seies to prove recurrence, not to mention absorbing property? | |
| Feb 7, 2011 at 9:32 | comment | added | Did | @sigma: Indeed, the sum of probabilities must diverge for recurrence. That is, recurrence implies divergence of the series, that is, convergence of the series implies transience. This part is correct / does not require independence / uses the (so-called) easy part of Borel-Cantelli lemma. But to show recurrence would require the (so-called) difficult part. | |
| Feb 7, 2011 at 9:18 | comment | added | sigma_z_1980 | Correct me if I'm wrong, but the issue with independence is solved using BCL+Kolmogorov 0-1 Law. Therefore, 'visit state A i.o.' is an intersect of events 'visit state A k or more times', and the sum of probailities of these events A_{k} must diverge for recurrence. At least this is how Iosifescu and Shiryaev treat the matter. Nevertheless the problem with absorbing states remains. | |
| Feb 7, 2011 at 8:59 | history | answered | Did | CC BY-SA 2.5 |