Timeline for Frequency of visiting states in Markov chains
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Jul 4, 2016 at 21:21 | history | edited | R W | CC BY-SA 3.0 |
deleted 3 characters in body
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| Jul 4, 2016 at 21:21 | comment | added | R W | Sorry - got lost in the inequalities :) So, if $T=1$, then for any $a$ a.e. sample path contains a length $a$ interval which consists of $m$'s, so that $p=0$ for $b=0,1,\dots,a-1$ and $p=1$ for $b=1$. | |
| Jul 4, 2016 at 21:00 | comment | added | maomao | Many thanks for pointing out the mistakes in the formulation of the problem. I do not quite understand the statement "if T(m)=1, then p=1 if and only if b≤a". Say, the chain has only state m, and loop there; a=2>b=1. Obviously T(m)=1, and according to your claim, p=1, but obviously in this case, p=0 as the only sample path is m forever. | |
| Jul 4, 2016 at 16:15 | history | answered | R W | CC BY-SA 3.0 |