Question regarding positive recurrence Suppose that $\{Y_n\}$ is an irreducible discrete time Markov chain with transition matrix $P = \{P_{i,j}\}_{i,j \in S}$. Let $\{Z_n\}$ be a discrete time Markov chain such that if it is in state $i$, then it either moves according to $P$ with probability $p_i$ or does nothing with probability $1-p_i$. If $\{Y_n\}$ is positive recurrent, is it true that $\{Z_n\}$ is positive recurrent?
I managed to show that $\{Z_n\}$ is irreducible, but I can't seem to find a counterexample/proof to this. Is there any suggestion?
 A: Not necessarily, though a sufficient condition is that there exists $p$ such that $p_i>p>0$
Pick a state and follow the chain $Z_n$ as it moves, counting up the number of transitions according to $P$ (call them $Y$-transitions) and the number of extra waiting steps. The expected number of $Y$-transitions before returning to the initial state is the same as it would be for the chain $Y$, so it is finite. Call it $N$. The expected number of extra waiting steps before a $Y$-transition in state $i$ is $(1-p_i)/p_i$. If $p_i>p>0$, the expected total number of extra waiting steps is less than $N(1-p)/p$, which is finite.
This argument fails if $p_i$ are not bounded away from zero, so we can look for a counterexample. Let $Y_n$ be the random walk with reflection at zero, where the transitions are

*

*i to i-1, with probability 3/4 (for i>1)

*i to i+1 with probability 1/4

*0 to 0 with probability 3/4

This is positive recurrent (see eg).
Now consider $Z_n$. A recurrence that gets from 0 to a maximum value of $j$ will have expected extra waiting of at least $\sum_{i=0}^{j-1} (1-p_i)/p_i+ (1-p_j)/p_j +\sum_{i=j-1}^{0} (1-p_i)/p_i $ steps, and this can be made arbitrarily large by choosing $p_i$ decreasing fast enough, so that the expectation over all $j$ will not be finite.  The zero state is null recurrent, and hence so is the chain.
