Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X=\{0,1\}^{\mathbb{N}}$. For simplicity I consider measures on $X$ only.

A measure $\mu$ is quasi-Bernoulli if there is a constant $C\ge 1$ such that for any finite sequences $i,j$, $$ C^{-1} \mu[ij] \le \mu[i]\mu[j] \le C\mu[ij]. $$

(Here as usual $ij$ is the juxtaposition of $i$ and $j$ and $[k]$ is the cylinder of all infinite sequences starting with $k$.)

Let $f:X\to \mathbb{R}$ be continuous. The measure $\mu$ is a Gibbs measure with potential $f$ if there are $C>0$ and $P\in\mathbb{R}$, such that for every infinite sequence $i_1 i_2\ldots$ and all natural $n$, $$ C^{-1} \le \frac{\mu[i_1\ldots i_n]}{\exp(-nP+f(i)+f(\sigma i)+\cdots+f(\sigma^{n-1}i))} \le C, $$ where $\sigma$ is the left shift.

Of course, there are other definitions of Gibbs measure, but they all agree if the potential $f$ is Hölder. In this case, it follows readily that a Gibbs measure is quasi-Bernoulli.

If a measure is quasi-Bernoulli, there is an equivalent measure (mutually absolutely continuous with bounded densities) which is invariant and ergodic under the shift.

Question: Are all quasi-Bernoulli measures Gibbs? (for some continuous potential, not necessarily Hölder). If not, what is a counterexample?

Motivation: Gibbs measures (with Hölder potentials) enjoy many nice statistical properties. Sometimes I have a measure that is quasi-Bernoulli or satisfies some similar but weaker property. I would like to understand if and to what extent good statistical properties continue to hold in that setting.

share|cite|improve this question
Just a thought that doesn't necessarily go anywhere. If $\mu$ is a Gibbs measure, then you can recover the corresponding potential $f$ (up to a constant) by $f = \log(\frac{d\mu}{d(\mu\circ\sigma)})$. So if we take an arbitrary quasi-Bernoulli measure and take $f$ to be its log-Jacobian, can we deduce anything about the regularity of $f$ and/or a Gibbs relationship based on the quasi-Bernoulli property? – Vaughn Climenhaga Jan 26 '11 at 22:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.