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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.

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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

Let me point out that there is a standard procedure associating to each almost additive sequence of functions a measure on the algebra up to each level (say $n$ if you consider the element $\varphi_n$ of the sequence of functions). Moreover, any weak sublimit of an almost additive sequence with bounded variation is a Gibbs measure.

This implies that your question has a positive answer at each level $n$, that is, indeed any quasi-Bernoulli measure (in your sense) is Gibbs at the algebra of level $n$. Moreover, any of its weak sublimits is also Gibbs, now on the whole $\sigma$-algebra.

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