Let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $(E,\mathcal E)$ be a measurable space
- $(X_n)_{n\in\mathbb N_0}$ be an $(E,\mathcal E)$-valued i.i.d. process on $(\Omega,\mathcal A,\operatorname P)$
- $\mu$ denote the distribution of $X_0$ under $\operatorname P$
Assume that there is a Markov kernel $\pi$ with source $(E,\mathcal E)$ and target $\left(E^{\mathbb N_0},\mathcal E^{\otimes\mathbb N_0}\right)$ with $$\operatorname P\left[X\in B\mid X_0\right]=\pi(X_0,B)\;\;\;\text{almost surely for all }B\in\mathcal E^{\otimes\mathbb N_0}\tag1.$$ Now, let $$\tau:E^{\mathbb N_0}\to E^{\mathbb N_0}\;,\;\;\;(x_n)_{n\in\mathbb N_0}\mapsto(x_{n+1})_{n\in\mathbb N_0}$$ and $\operatorname P_\mu:=\mu\pi$ denote the composition of $\mu$ and $\pi$, i.e. $$\operatorname P_\mu[B]=\int\mu({\rm d}x_0)\pi(x_0,B)\;\;\;\text{for all }(x_0,B)\in E\times\mathcal E^{\otimes\mathbb N_0}\tag2.$$
Note that $\tau$ is $\operatorname P_\mu$-preserving, i.e. the distribution of $\tau$ under $\operatorname P$ is $\operatorname P$.
How can we show that $\operatorname P_\mu$ is $\tau$-ergodic, i.e. $$\mathcal I:=\left\{B\in\mathcal E^{\otimes I}:\tau^{-1}(B)=B\right\}$$ is $\operatorname P_\mu$-trivial (each event has $\operatorname P_\mu$-measure $0$ or $1$)?
Seems to be an easy task, but I don't see where we can make use of the i.i.d.-property of $X$.
Best Answer
Set $A:=\{X \in B\}$. Since $B=\tau^{-1}(B)$ we have
$$A = \{X \in \tau^{-1}(B)\} = \{\tau(X) \in B\}.$$
Iterating the procedure we get
$$A = \{\tau^n(X) \in B\},$$
for any $n \in \mathbb{N}$ and therefore
$$A \in \sigma(X_n,X_{n+1},\ldots), \qquad n \in \mathbb{N}.$$
Since the sequence $(X_n)_{n \in \mathbb{N}}$ is iid, it follows from Kolmogorov's 0-1 law that $P(A) \in \{0,1\}$. As
$$P(A) = P(X \in B) = \int_E P(X \in B \mid X_0 = x_0) \, \mu(dx_0) = \int_E \pi(x_0,B) \, \mu(dx_0) = P_{\mu}(B)$$
this proves the assertion.