I have this question I have been struggling with for a while. It seems rather intuitive, however, I was not able to proof it yet:

Let $\Omega = \{1,2,\cdots,N\}$ a finite alphabet, $\Sigma \subset \Omega^{\mathbb Z}$ be an irreducible Markov shift (i.e. an irreducible 1-step subshift of finite type). Denote by $(\Sigma,\mathcal B)$ the measure space with the usual product $\sigma$-algebra. Further, let $\mu$ be an ergodic, shift-invariant probability measure with full support on $(\Sigma,\mathcal B)$.

Question: Is $\mu$ necessarily a Markov measure? More precisely: Is $\mu$ a measure that takes the value $p_{\omega_0} \prod_{t=1}^{\tau}\Pi_{\omega_{t},\omega_{t+1}}$ on the cylinder sets specified by the finite string $(\omega_0,\omega_1,\dots,\omega_\tau)$, where $\Pi_{i,j}$ are the elements of a compatible stochastic matrix and $p$ is its unique unity stochastic eigenvector?

It is quite easy to construct counter-examples if I drop certain assumptions. However, in the described set-up I could not find one. It feels like that question should have an easy answer, but somehow I don't get it.

Any help would be really appreciated!

Cheers, Bernhard

  • $\begingroup$ Is a Markov measure what I think it is, namely a kind of Fubini product measure derived from eigenvectors of the transition matrix? $\endgroup$
    – Lee Mosher
    Apr 24, 2014 at 11:26
  • $\begingroup$ I think we mean the same, so I edited the OP to be more precise. However, it is not really a product measure, or the appearance of subsequent symbols would be independent of each other. $\endgroup$
    – beralt
    Apr 24, 2014 at 16:21

2 Answers 2


Not at all. The simplest example is provided by the so-called $d$-Markov measures. These are Markov measures for the associated shift whose alphabet is the subset of $\Omega^d$ which consists of all $d$-tuples of symbols that occur in $\Sigma$. It is easy to see that for $d>1$ there are more $d$-Markov measures than plain Markov ones.

If you are not satisfied with this example, then you should look up the notion of a Gibbs measure, see the chapter on symbolic dynamics from the lecture notes of Rufus Bowen (Springer LN, vol.470).

  • $\begingroup$ Thank you so much. Now I really feel embarrassed for even asking. Sometimes the answer is so close... On a related issue: Under the assumptions stated, can you think of a condition on $\mu$ such that it is a Markov measure than specifying its values on a $\pi$-system (that is a system which is closed under finite intersections)? $\endgroup$
    – beralt
    Apr 24, 2014 at 16:29
  • $\begingroup$ Have never thought about $\pi$-systems. However one can formulate such a condition in terms of cylinder sets. Namely, an invariant measure $\mu$ is Markov if and only if the ratios of measures of cylinder sets $\mu(C_{\omega_0,\omega_1,\dots,\omega_t})/\mu(C_{\omega_1,\dots,\omega_t})$ only depend on $\omega_0$ and $\omega_1$. $\endgroup$
    – R W
    Apr 24, 2014 at 17:34
  • $\begingroup$ The cylinder sets themselves form a $\pi$-system, but a rather large one. The question is more if there would be some smaller $\pi$-system, which allows for that...in any case, thank you very much already! $\endgroup$
    – beralt
    Apr 24, 2014 at 19:31

As "R W"s answer points out, Gibbs measures (of regular enough potentials, such as Holder continuous potentials) are a strict generalization of Markov measures. Another strict generalization I'd like to mention is (stationary) hidden Markov chains, for which see "Hidden Markov processes in the context of symbolic dynamics" by Boyle & Petersen. I'll refer to the arxiv v2 of the preprint.

In the preprint, as in many related literature, "Markov measure" is defined to be any invariant probability measure on any irreducible SFT (subshift of finite type) with full support and with n-step Markov property for some n, and "sofic measure" is defined to be images of such measures under factor codes. Following this terminology, we can say that the image measure $\pi\mu_p$, for $0 < p < 1/2$, in Example 2.9 in the preprint is a sofic, non-Markov measure on the full two-shift. It is probably one of the simplest counter-examples for your question. The measure is not Gibbs either. In fact, any time you have two different ergodic measures $\mu, \mu'$ on an mixing SFT projecting to the same measure $\pi\mu = \pi\mu'$ via some finite-to-one factor code onto a mixing SFT, the image $\pi\mu$ cannot be a Gibbs state in the sense of "R W"'s answer, let alone a Markov measure.

Yet another way of producing a counter-example is to look inside the class of zero entropy processes because Markov measures always have positive entropy and if IIRC, this is also true for Gibbs measures. So you only need to construct an example of a fully supported, ergodic measure of zero entropy. Such measures actually form a residual subset of the space of all invariant probability measures on $\Sigma$, for which see "ergodic theory on compact spaces" by Denker & Grillenberger & Sigmund. That's the only proof that I know of for the existence of fully supported, ergodic, zero entropy measure on $\Sigma$. But I suspect that it might be straightforward to have a direct construction, more direct than that from unwinding the proof of residuality, by exploiting the fact that whenever you have any ergodic system of zero entropy, such as irrational rotation on the unit circle, you obtain a zero entropy process just by observing the ergodic system through any finite partition.


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