Timeline for Product of a transient and a positive recurrent Markov chain
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 18, 2011 at 18:06 | comment | added | Ori Gurel-Gurevich | I meant for any fixed sequence $t_n$. It then follows for any random sequence which is independent of $Y$. As for an obvious way to see it: this fact is true also conditioned on any finite history, so you can always choose $n$ large enough so that $\mathbb{P}(Y_{t_n}=y)\ge \pi_y /2$ conditioned on what you observed so far. Thus, you have infinitely many "trials", each having at least some fized probability, given all the previous ones. | |
| Sep 18, 2011 at 10:28 | history | edited | Did | CC BY-SA 3.0 |
added 14 characters in body
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| Sep 18, 2011 at 0:43 | vote | accept | Elena Yudovina | ||
| Sep 18, 2011 at 0:43 | comment | added | Elena Yudovina | I think you mean for $t_n$ to be independent of $Y$? In that case, yes, I think an application of Levy's extension of the Borel-Cantelli lemma (for weakly dependent random variables) should do it. Or is there a more obvious reason why this is true? | |
| Sep 17, 2011 at 19:22 | comment | added | Ori Gurel-Gurevich | If Y is aperiodic than $P(Y_n=y) \to \pi_y$, so for any increasing sequence of times $t_n$, the probability of $Y_{t_n}=y$ for some (infinitely many) $n$'s is 1. | |
| Sep 17, 2011 at 16:36 | comment | added | Elena Yudovina | Agree with 1., thanks. 2. What if I add that Y is aperiodic? (And, of course, independent of X.) | |
| Sep 17, 2011 at 15:39 | history | answered | Did | CC BY-SA 3.0 |