I am reading the 2004 MCMC paper by Roberts and Rosenthal and was hoping someone could shed some further light on Example 3. I restate Theorem 4 first, which the example relies on.

Theorem 4. If a Markov chain on a state space with countably generated $\sigma$-algebra is $\phi$-irreducible and aperiodic, and has a stationary distribution $\pi$, then for $\pi$-a.e. $x\in \mathcal{X}$, $$ \lim_{n\to\infty}||P^n(x, \cdot)-\pi(\cdot)||=0. $$ In particular, $\lim_{n\to\infty}P^n(x, A)=\pi(A)$ for all measurable $A\subseteq \mathcal{X}$.

Now for the example (due to Charles Geyer):

Example 3. Let $\mathcal{X}=\{1, 2, \dots\}.$ Let $P(1, \{1\})=1$, and for $x\geq 2$, $P(x, \{1\})=1/x^2$ and $P(x, \{x+1\})=1-1/x^2$. Then chain has stationary distribution $\pi(\cdot)=\delta_1(\cdot)$ and it is $\pi$-irreducible and aperiodic. On the other hand, if $X_0=x\geq 2$, then $\mathbf{P}[X_n=x+n \text{ for all }n]=\prod_{j=x}^\infty(1-1/j^2)>0$ so that $||P^n(x, \cdot)-\pi(\cdot)||/{\to}0$. Here Theorem 4 holds for $x=1$ which is indeed $\pi$-a.e. $x\in\mathcal{X}$, but it does not hold for $x\geq 2$.

(In the example, I used $/\to$ to denote "does not converge to"; $||\cdot||$ denotes total variation distance.)

It is the final sentence of the example I cannot wrap my head around. Can someone explain the example differently?

After some more thought, I realized $\delta_1(\cdot)$ must be the Kronecker delta. Hence, $\pi(1)=1$ and $\pi(x)=0$ for $x\geq 2$ and the theorem holds if $x=1$ which happens with probability 1 (with respect to $\pi$). So the issue is simply that the theorem is for $\pi$-a.e., whereas elements with $\pi$-measure zero may have positive measure for finite $n$ (and in this particular case, the chain may keep on going). Is this (fairly vague) reasoning correct?

  • $\begingroup$ Do you know of a proof of Theorem 4? Thanks. $\endgroup$ Commented Dec 21, 2018 at 9:24
  • $\begingroup$ Sorry, it seems the paper has given an argument. $\endgroup$ Commented Dec 21, 2018 at 9:33

1 Answer 1


Yes, your edit indicates correct reasoning.

Like you said, Example 3 aims to show the role of $\pi$-a.e. $x \in \mathcal{X}$. Theorem 4 applied here would mean that if you start the Markov chain at values where the stationary distribution has positive mass, then the Markov chain converges to that stationary distribution. In the example, $x = 1$ is the only point of mass for $\pi$, so if we start the chain from $x = 1$, then since $P(1, \{1\}) = 1$, we deterministically stay at this point, which is also the stationary distribution.

The behavior of the chain starting at any other point is immaterial since those points are not $\pi$-a.e. $x \in \mathcal{X}$. Another thing to note is that $\phi$ is chosen to be $\pi$ in this case, which leads to $\pi$-irreducibility. For other measures that chain may not be irreducible (for e.g., counting measure), but that doesn't matter, because for Theorem 4 to hold, there needs to exists one such $\phi$.

As R+R (2014) say in their paper, to replace $\pi$-a.e $x \in \mathcal{X}$ with $x \in \mathcal{X}$, the Markov chain needs to be Harris recurrent. This would mean that for all starting values in the space $\mathcal{X}$, the chain will eventually reach all sets of positive mass with probability 1. This is not true for this example, because for starting value say $x = 2$, the chain has positive probability of not going to $\{1\}$ for all $n$. So there exists no such $n$, such that $P^n(2, \{1\}) = 1$. Hence, Example 3 descries a chain that is not Harris recurrent.


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