Product Measure Only Possible Measure? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T07:26:39Z http://mathoverflow.net/feeds/question/37010 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/37010/product-measure-only-possible-measure Product Measure Only Possible Measure? J D 2010-08-28T23:46:00Z 2010-08-29T18:52:37Z <p>Let $X$ be a separable complete metric space and $Z$ be the set of all integers. Let $\nu$ be a Borel probability measure on $X^Z$ invariant under the shift function $S:X^Z \to X^Z$. Is it necessarily the case that $\nu = \mu^Z$ for some Borel probability measure $\mu$ on X?</p> <p>Thanks</p> http://mathoverflow.net/questions/37010/product-measure-only-possible-measure/37011#37011 Answer by rpotrie for Product Measure Only Possible Measure? rpotrie 2010-08-29T00:01:28Z 2010-08-29T00:09:47Z <p>The answer is no. A trivial example is to concentrate the measure in a "periodic orbit", this will give an invariant measure for the shift. </p> <p>But there are a whole lot of invariant measures (including full support measures which probably are more interesting). </p> <p>The measure which is a product measure, has though some important features. For example, if you look at its "entropy".</p> <p>(See K. Sigmund, Generic properties for Axiom A diffeomorphisms, Inventiones Math 11 (1970) for the case of the space X being finite)</p> http://mathoverflow.net/questions/37010/product-measure-only-possible-measure/37012#37012 Answer by Vaughn Climenhaga for Product Measure Only Possible Measure? Vaughn Climenhaga 2010-08-29T00:34:51Z 2010-08-29T03:37:16Z <p>Measures with the property you describe are called <em>Bernoulli measures</em>. There are many, many invariant measures that are not Bernoulli: one class of examples is given by the measures concentrated on periodic orbits (as rpotrie points out in another answer); another important class is the <em>Markov measures</em>. These are given by a measure $\mu$ on $X$ (which if $X$ is finite is simply a probability vector) together with a function $p\colon X \to \mathcal M(X)$ that represents transition probabilities, where $\mathcal{M}(X)$ is the space of Borel probability measures on $X$. Then one defines a measure $\nu$ on $X^\mathbb{Z}$ by \begin{multline} \nu(X_1 \times X_2 \times \cdots \times X_n) = \\ \int_{X_1} \int_{X_2} \cdots \int_{X_{n-1}} p(x_{n-1},X_n) dp(x_{n-2},x_{n-1}) \cdots dp(x_1,x_2) d\mu(x_1), \end{multline} where $\int dp(x,\cdot)$ represents integration with respect to $p(x)$. Note that this simplifies quite a bit if $X$ is finite, in which case $\mu$ is a probability vector, $p$ turns into a stochastic matrix, and you just need to write down the measure of an arbitrary cylinder. In any case, these give you a broad class of invariant measures that are not Bernoulli, but are very important for many applications.</p> http://mathoverflow.net/questions/37010/product-measure-only-possible-measure/37032#37032 Answer by mr.gondolier for Product Measure Only Possible Measure? mr.gondolier 2010-08-29T07:19:59Z 2010-08-29T07:19:59Z <p>Every such a $\nu$ is the law of some stationary process on $X^{\mathbb{Z}}$. Of course not every stationary process is i.i.d.</p> http://mathoverflow.net/questions/37010/product-measure-only-possible-measure/37033#37033 Answer by Pietro Majer for Product Measure Only Possible Measure? Pietro Majer 2010-08-29T07:24:47Z 2010-08-29T18:52:37Z <p>No (unless X is a one-point space). The mean of two distinct shift-invariant product probability measures is a shift-invariant probability measure, though not a product. </p>