Bernoulli scheme

Example

Let $A = \set{1, \dots, k}$ be a finite alphabet and let $p = (p_1, \dots, p_k)$ be a probability vector, with $p_i \ge 0$ and $p_1 + \cdots + p_k = 1$. Let $X = A^{\Z}$ be the set of two-sided sequences $x = (x_n)_{n \in \Z}$, with the σ-algebra generated by the cylinder sets $$[a_m, \dots, a_n]_m = \set{x \in X : x_i = a_i \text{ for } m \le i \le n}.$$ Let $\mu$ be the infinite product measure of copies of $p$, so $\mu([a_m, \dots, a_n]_m) = p_{a_m} p_{a_{m+1}} \cdots p_{a_n}$. The shift $\sigma \colon X \to X$, $(\sigma x)_n = x_{n+1}$, is the Bernoulli scheme or Bernoulli shift $B(p_1, \dots, p_k)$.

Check that $\sigma$ is a measure-preserving transformation: $x \in \sigma^{-1}[a_m, \dots, a_n]_m$ means $x_{i+1} = a_i$ for $m \le i \le n$. So $\sigma^{-1}[a_m, \dots, a_n]_m = [a_m, \dots, a_n]_{m+1}$, which has the same measure $p_{a_m} \cdots p_{a_n}$. Cylinder sets form a π-system generating the σ-algebra, so $\mu(\sigma^{-1} E) = \mu(E)$ for every measurable $E$.

The coordinate functions $x \mapsto x_n$ form an i.i.d. sequence with law $p$, a two-sided version of the Bernoulli process when $k = 2$. The shift is invertible, mixing, and a Kolmogorov automorphism. Its Kolmogorov–Sinai entropy is $-\sum_{i=1}^{k} p_i \log p_i$, computed with the generating partition into the sets $\set{x : x_0 = i}$. The one-sided version uses $A^{\N}$ instead of $A^{\Z}$ and gives a non-invertible measure-preserving map.