Integrability of first return time of Markov chain

Question

Consider a Markov chain with a possibly infinite state space $S$. Assume that $E_i(\tau_i)$ is finite for some $i$ in $S$, where $\tau_i = \inf\{k \geq 1 \colon X_k = i\}$. Is it true that $E_i(\tau_i^2)$ is finite?

Answer 1

The result is wrong but a construction showing that it is, might be worth remembering.

Consider a Markov chain on the nonnegative integers whose transitions are from $0$ to $1$ with probability $p_0$, from $0$ to $0$ with probability $1-p_0$, and, for every positive $n$, from $n$ to $n+1$ with probability $p_n$ and from $n$ to $0$ with probability $1-p_n$. Thus, the paths of the Markov chains are the concatenation of some excursions $0\to0$ or $0\to1\to0$ or $0\to1\to2\to0$ or $0\to1\to2\to3\to0$ or... $0\to1\to2\to\cdots\to n-1\to n\to0$ for some nonnegative $n$. We assume that every $p_n$ is in $(0,1)$, to avoid degenerate cases.

It happens that one can tune the family of parameters $(p_n)$ to adjust the integrability properties of the return times.

To wit, the Markov chain described above is recurrent if and only if $\tau_0$ is finite $P_0$-almost surely if and only if $q_n\to0$, where, for every $n$, $$q_n=\prod_{k=0}^{n-1}p_k$$ Furthermore, for every $n$, $$P_0(\tau_0\geqslant n+1)=q_n$$ hence the chain is positive recurrent if and only if $E_0(\tau_0)$ is finite if and only if the series $$\sum_nq_n$$ converges. Likewise, for every $i$, $E_i(\tau_i^2)$ is finite if and only if $E_0(\tau_0^2)$ is finite if and only if the series $$\sum_nnq_n$$ converges. Thus, one obtains positive recurrent Markov chains such that $E_0(\tau_0^2)$ diverges if (every $p_n$ is in $(0,1)$ and), for example, $$p_n=1-\frac{c}n+o\left(\frac1n\right)$$ when $n\to\infty$, for some $c$ in $(1,2)$.