Assume the chain $\{X_n\}_{n\in\mathbb{N}}$ on the statespace $(S,\mathcal{F})$ (we may assume it is countable) is aperiodic, irreducible and positive recurrent. We denote with $\pi$ its (unique) stationary distribution. Is it true that for any $x\in S$ $$\underset{n\rightarrow\infty}{\text{lim}}~\underset{\text{A}\in\mathcal{F}^{\mathbb{Z}_+}}{\text{sup}}~|P_x((X_{n},X_{n+1},\dots)\in\text{A})-P_\pi(X\in A)| =0?$$

I would assume this holds true due to the similarity of the statement to the usual convergence theorem $$\underset{n\rightarrow\infty}{\text{lim}}~||P_i(X_n\in\cdot)-\pi(\cdot)||_V=0$$ and the feeling that the Markov chain is reproducing itself. But I can't figure out how to prove it. I tried to come up with coupling which seems to be the way to deal with those Variance norms. Thanks a lot for any suggestion.


You are asking about the (one half of) the total variation distance between the measure $\mathbf P_\pi$ and the shifted by $n$ measure $\mathbf P_x$. The latter measure is Markov with the initial distribution $\delta_x P^n$ (here $P$ is the transition operator of the Markov chain). Now, for Markov measures determined by two different initial distributions of the same chain their total variation distance is the same as the total variation distance between their initial distributions.

  • $\begingroup$ Thanks for your answer. Hence my intuition was right. I assume your definition of total Variation is the absolute distance summed over the single elements and hence you are speaking of half the total variation distance. Do you have a reference for your statement about the total variation distance of Markov measures? $\endgroup$ – MathProb Mar 12 '15 at 1:13
  • $\begingroup$ My definition of the total variation is really that of "total variation" - namely, one takes the total variation of the difference between two measures. The statement you are asking about follows from the fact that Markov measures are uniquely determined by their initial distributions. $\endgroup$ – R W Mar 12 '15 at 1:32
  • $\begingroup$ Sorry for bugging you again. If you got a second I would be very grateful if you could check this question (Last thing I ask you). math.stackexchange.com/questions/1186381/… $\endgroup$ – MathProb Mar 12 '15 at 5:56
  • $\begingroup$ Check the book by Norris, Theorem 1.8.5 (it is available online at statslab.cam.ac.uk/~james/Markov/s18.pdf) $\endgroup$ – R W Mar 12 '15 at 8:54

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