Persistent Markov Chain with infinite mean recurrence time Given a Markov Chain $(X_n)_{n \ge 0}$, state $i \in S$ is defined as persistent if $P(T_i < \infty | X_0 = i) = 1$ (where $T_i$ is the first passage time to state $i$). Moreover, the mean recurrence time $\mu_i$ of state $i$ is $E[T_i | X_0 = i]$ which equals $\sum_n n \cdot P(T_i = n|X_0 = i)$ if the state is persistent and $\infty$ if the state is transient.
From what I understand, just because a state is persistent that does not necessarily imply that $\mu_i$ is finite. I was wondering if someone could please provide an example of a persistent state with infinite mean recurrence time?
Thank you.
 A: while I think that studying the simple symmetric random walk is the standard example here, I'll explicitly construct a different one.
Consider
$P:=\begin{pmatrix}p_{1,1} & p_{2,2} & p_{3,3} & p_{4,4} & p_{5,5} & p_{6,6} & \dots \\
1 & 0 & 0 & 0 & 0 & 0 & \dots\\ 
0 & 1 & 0 & 0 & 0 & 0 & \dots\\
0 & 0 & 1 & 0 & 0 & 0 & \dots\\
0 & 0 & 0 & 1 & 0 & 0 & \dots\\
\vdots & \vdots & \vdots & \vdots & \ddots & \ddots & \ddots\\
\end{pmatrix}$
This chain describes the Residual Life of an integer valued Renewal Process and sometimes goes by names like Residual Waiting Time chain.  (The original post uses the term 'Persistent' -- the standard term I think is Recurrent, but it reminds me that Feller uses the term Persistent instead of Recurrent, so in case that OP   uses Feller as a reference, this chain shows up in Feller vol 1, 3rd edition, page 381, and elsewhere subsequently in that chapter.)
Given its simple structure, we can easily construct a case where this chain is null recurrent.
1.) For simplicity we'll require each  $p_{i,i}\gt 0$ and by inspection, there is a single communicating class -- i.e. state 1 may reach any state with positive probability and state $j\geq 2$ may reach state 1 with positive probability (in fact it reaches state 1 deterministically in exactly $j-1$ iterations).
2.) To verify recurrence, it we need to ensure that given a start in state one, we return with probability 1.
3.) To be null recurrent, we need to ensure that given a start in state one, the expected time until revisiting state one is $\infty$.
the finish
Set $p_{i,i}:=\frac{1}{i(i+1)}$
noting that
$\sum_{i=1}^{n-1}\frac{1}{i(i+1)}=1-\frac{1}{n}$
if you don't know this result -- it's a simple and worthwhile telescoping exercise.
thus
$\sum_{i=1}^{\infty}\frac{1}{i(i+1)} = 1$
so the chain is recurrent, but
$\sum_{i=1}^{\infty}i\cdot\frac{1}{i(i+1)} = \sum_{i=1}^{\infty}\frac{1}{i+1} = \infty$
because the harmonic series is divergent.  Thus this is an example of a null recurrent chain.
