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Let $X$ be a transient Markov chain with countable state space $S(X)$. Let $Y$ be a positive recurrent Markov chain with countable state space $S(Y)$. (Time is discrete.)

Let $A \subseteq S(X)$ be such that, with probability 1, $X$ visits $A$ infinitely many times. In fact, if it helps, let $A$ be such that, with probability 1, there exists a sequence $T_n \to \infty$ such that $\frac{1}{T_n}$(number of visits to $A$ during $[0,T_n]$) $\to a > 0$. (Obviously such sets $A$ exist, e.g. $A = S(X)$.)

1) Is it true that, in fact, $\lim_{T \to \infty} \frac{1}{T}$ (number of visits to $A$ during $[0,T]$) $= a$?

2) Let $y \in S(y)$ be any state; recall that $Y$ was positive recurrent. Is it true that the product chain $(X,Y)$ will visit $(A,y)$ with probability 1? Is there an estimate on $\frac{1}{T}$(number of visits to $(A,x)$ during $[0,T]$), at least for some sequence of $T$?

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The answer to 1. is negative. Consider the transient Markov chain $X_n=X_0+n$ on $\mathbb Z$ and choose for $A$ any subset of $\mathbb Z_+$ with different inferior and superior densities.

As regards 2., I assume you want $X$ and $Y$ to be independent. Then the answer to the first question is negative. Same $X$ as before, $X_0$ deterministic, $Y_n=(-1)^n$, $y=(-1)^{X_0}$ and $A=2\mathbb Z+1$. Then $(X,Y)$ never visits $A\times\{y\}$.

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  • $\begingroup$ Agree with 1., thanks. 2. What if I add that Y is aperiodic? (And, of course, independent of X.) $\endgroup$
    – Elena Yudovina
    Commented Sep 17, 2011 at 16:36
  • $\begingroup$ 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. $\endgroup$
    – Ori Gurel-Gurevich
    Commented Sep 17, 2011 at 19:22
  • $\begingroup$ 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? $\endgroup$
    – Elena Yudovina
    Commented Sep 18, 2011 at 0:43
  • $\begingroup$ 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. $\endgroup$
    – Ori Gurel-Gurevich
    Commented Sep 18, 2011 at 18:06

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