stationary and ergodic

Question

Let $p\in \mathbb N$ and $a_1,\ldots,a_p\in \mathbb R $. Denote by $x$ the sequence $$x=(x_k)_{k\in\mathbb N }=(a_{k \bmod p})_{k\in\mathbb N }$$ where $(k \bmod p)\in{1,\ldots,p}$ is the remainder when dividing $k$ by $p$. Let $\sigma$ be a random variable that is uniformly distributed on $\{1,\ldots,p\}$, defined on some probability space $(\Omega,\mathcal{F},\mathbb P)$. We consider the stochastic process $\mathbb X$ defined by $$\mathbb X(\omega):=(x_{\sigma(\omega)},x_{\sigma(\omega)+1},x_{\sigma(\omega)+2},\ldots),\quad\omega \in\Omega$$ We had that as an example whithout proof, but why exactly is $\mathbb X$ stationary and why is $\mathbb X$ ergodic with respect to the shift operator $\theta$.

Answer 1

I prefer to put $N=\sigma$ and $X_n=x_{N+n}$. You are interested in the stochastic process ${(X_n)}_{n \geq 0}$. It is not difficult to see that each $X_n$ has the uniform distribution on the set $\{a_1, \ldots, a_p\}$ and $X_{n+1} = a_{(j+1) \bmod p}$ when $X_n=a_{j \bmod p}$. Thus $(X_n)$ is a stationary Markov process (with deterministic Markov transition).

The space of trajectories $E$ of $(X_n)$ is finite (this is the set of the $p$ possible shifted sequences of the periodic sequence $(a_1, \ldots, a_p, a_1, \ldots, a_p, \ldots)$), and for any $x \in E$ one has $E = \{\theta^k(x), k=0, \ldots, p-1\}$, wherefrom it is easy to deduce that the shift operator $\theta$ is ergodic.