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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?


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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.

But there are a whole lot of invariant measures (including full support measures which probably are more interesting).

The measure which is a product measure, has though some important features. For example, if you look at its "entropy".

(See K. Sigmund, Generic properties for Axiom A diffeomorphisms, Inventiones Math 11 (1970) for the case of the space X being finite)

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This also works for N in place of Z. – Carl Mummert Aug 29 '10 at 1:03

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.

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Measures with the property you describe are called Bernoulli measures. 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 Markov measures. 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.

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You need $\mu$ to be the stationary distribution of kernel $p$, right? – mr.gondolier Aug 29 '10 at 7:19
Yes, sorry, I left that out. $p$ induces a map $\mathcal M(X) \to \mathcal M(X)$, and you need $\mu$ to be a fixed point for this map. (Or in the finite case, you need the probability vector $\mu$ to be an eigenvector for the stochastic matrix $p$ with eigenvalue $1$.) – Vaughn Climenhaga Aug 29 '10 at 12:49

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.

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