Ergodicity implies finite first moment in discrete time Markov chains?

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Suppose that $X(n)=(X_1(n),X_2(n))\in\mathbb{N}^2$ is a discrete time homogeneous Markov chain with uniformly bounded jumps (see below). Assume that the chain is ergodic and let $\pi$ be its invariant probability measure.

If $Y$ is random variable distributed as $\pi$, is it true that $\mathbb{E}[Y]<\infty$? In other words, I'm interested in understanding whether ergodicity and uniformly bounded jumps ensure a finite first moment.


Uniformly bounded jumps. I mean that there exists some finite $c>0$ such that the transition $x\mapsto x+(\Delta_1,\Delta_2)$ can occur only if $-c\le \Delta_1,\Delta_2<c$, for all $x$.

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Even in one dimension, the answer is no: you can cook up an ergodic Markov chain on $\mathbb N$ with an arbitrary limiting distribution, even if you require jumps to be bounded by $\pm1$. You ask a question about 2D chains, but we can extend a 1D example to 2D in a number of ways, such as:

  • Fold up a 1D chain to make it 2D, zigzagging to cover all of $\mathbb N^2$.
  • From each state of the 2D chain, choose with probability $\frac12$ whether to walk in the $x$ or the $y$ direction, and follow the 1D chain for whichever direction you choose.
  • Just embed the 1D chain as states $(x,1) \in \mathbb N^2$ so that states $(x,y)$ with $y>1$ are unreachable. (Have them go to $(x,y-1)$ with probability $1$ if you like, so that the states of the 1D chain are the only recurrent states.)

For an example of how to attain an arbitrary probability distribution, take the following Markov chain. Here, each state loops back to itself with whatever the leftover probability is, and $C$ is chosen to make sure that that leftover probability is positive ($C = \frac12$ will do).

enter image description here

This Markov chain has limiting distribution $\pi_n = \frac6{\pi^2 n^2}$ because this satisfies the detailed balance equations $\pi_n p_{n,n+1} = \pi_{n+1} p_{n+1,n}$ for all $n$. So here, the expected value is $$\sum_{n\ge 1}n \pi_n = \sum_{n \ge 1} \frac{6}{\pi^2n} = \infty.$$