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This question might be easy but I am really stuck on it.

Let $M$ be compact metric space and $\mathcal B(M)$ the Borel $\sigma$-algebra of M. Consider the discrete-time Markov process, $$\mathbf{X} =\left(\Omega,\{\mathcal F_n\}_{n\in\mathbb N}, \{X_n\}_{n\in\mathbb N}, \{P_n\}_{n\in\mathbb N} , \{\mathbb P_x\}_{x\in M}\right), $$ with state space $(M,\mathcal B(M)$ (I am considering that $0\in\mathbb N$) i.e.

  1. $(\Omega,\mathcal F_n),$ is a filtered measurable space,
  2. $X_n:\Omega\to M$ is $\mathcal F_n$ measurable,
  3. $\mathbb P_x [X_0 = x] =1,$ for every $x\in M,$
  4. For every $0 \leq n \leq m \in \mathbb N,$ $f:M\to\mathbb R$ bounded measurable function, and $x\in M$ $$\mathbb E_x [f(X_{n+m}) \mid \mathcal F_n] = (P_m f)(X_n), \ \mathbb P_x\ \mathrm{a.s.}, $$ where $P_n$ is a transition function on $(M,\mathcal B(M)),$ i.e. a family of probability maps $P_n : M\times \mathcal B(M) \to [0,1],$ such that
  • $P_0(x,\mathrm{d} y) = \delta_x(\mathrm{d}y),$
  • $P_n(x,\cdot)$ is a Borel probability measure for every $x\in M.$
  • For every $n,m \in \mathbb N$ and $A\in\mathcal B(M),$ $$P_{n+m}(x, A) = \int_{M} P_n(y,A) P_m(x,\mathrm{d} y). $$

Assume that $\mathbf{X}$ admits an ergodic stationary measure $\mu$ on $M,$ i.e. $$\int_{M} P_n(x,A) \mu(\mathrm d x) = \mu(A),\ \forall \ A\in\mathcal B(M), $$ and if $$P_1(x,A) = 1,\ \forall \ x \ \mu\text{-a.s.}\ \in A \Rightarrow \mu(A) = 0\ \text{or }1. $$

Question: I would like to know if under this setup we would have the following ergodic theorem. For every $f\in L^1(M,\mathcal B(M), \mu),$ we obtain $$ \lim_{n\to\infty} \frac{1}{n} \sum_{i=0}^{n-1}f (X_n(\omega)) = \int_{M} f(x)\mu(\mathrm{d} x),\ \forall\ \omega \text{-}\mathbb P\ \text{a.s.,} $$ where $\mathbb P(\mathrm{d} y) := \int_{M}\mathbb P_x (\mathrm{d} y) \mu(\mathrm{d} x).$

Comments regarding my question

Consider $\mathbf{X}$ being an ergodic Markov process (using the above notation). For every $n\in\mathbb N$ let us consider the projection map \begin{align*} \pi_n : M^{\mathbb N}&\to M\\ (x_m)_{m\in\mathbb N}&\mapsto x_n. \end{align*}

If we define (via Komolgorov extension Theorem) the Borel probability measure $P_\mu$ on $M^\mathbb N$ as the unique Borel probability, such that given $A_0,\ldots,A_n \in M,$ then $$P_{\mu}\left(\{\omega_n\}_{n\in\mathbb N} \in M^{\mathbb N}; x_i\in A_i, \ \forall \ i\in\{0,1,\ldots,n\}\right) = \int_{A_0}\int_{A_1} \ldots \int_{A_{n-1}} P_1(x_{n},A_n) P_1(x_{n-1},\mathrm{d}x_n) \ldots P_1(x_0,\mathrm{d} x_1) \mu(\mathrm{d}x_0). $$

We have that the shift \begin{align*} \theta: (M^{\mathbb N},\mathcal B(M^{\mathbb N}) , P_\pi)&\to (M^{\mathbb N},\mathcal B(M^{\mathbb N}),P_\pi) \\ (x_{n})_{n\in\mathbb N}&\to (x_{n+1})_{n\in\mathbb N}, \end{align*} is an ergodic dynamical system and we have that for every $f\in L^1(M,\mathcal B(M),\mu)$ $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1} f(\pi_i(\omega)) = \int_M f(x) \mu(\mathrm{d} x),\ \forall \ \omega\text{-}P_\pi\ a.s.. $$

How do I translate the information of the canonical process (the one above) to the original Markov process $\mathbf{X}$? For every $\omega \in \Omega$, we have that $$ \left(X_n(\omega)\right)_{n\in\mathbb N}\in M^{\mathbb N}, $$
and $$\pi_i\left(\left(X_n(\omega)\right)_{n\in\mathbb N}\right) = X_i(\omega).$$

But it is not clear that $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1} f( X_i(\omega) ) = \int_M f(x) \mu(\mathrm{d} x),\ \forall \ \omega\text{-}\mathbb P\ a.s., $$ where $\mathbb P (\mathrm{d} y) = \int_M \mathbb P_x(\mathrm{d} y) \mu(\mathrm{d}x), $ can anyone help me? Or simply provide a reference for this ergodic Markov chains result without considering the canonical space.


1 Answer 1


I believe I found an answer. Note that in a similar way that we constructed $P_\mu,$ we may define $P_x$ as the unique Borel probability on $M^{\mathbb N}$, such that given $A_0,\ldots,A_n \in M,$ then $$P_{x}\left(\{\omega_n\}_{n\in\mathbb N} \in M^{\mathbb N}; x_i\in A_i, \ \forall \ i\in\{0,1,\ldots,n\}\right) = \int_{A_0}\int_{A_1} \ldots \int_{A_{n-1}} P_1(x_{n},A_n) P_1(x_{n-1},\mathrm{d}x_n) \ldots P_1(x_0,\mathrm{d} x_1) \delta_x(\mathrm{d}x_0), $$ it is clear that $P_\mu(\mathrm d y) = \int_M P_x(\mathrm d y) \mu(\mathrm x).$

Consider the measurable inclusion \begin{align} \iota : \Omega &\to M^\mathbb N\\ \omega&\mapsto (X_n(\omega))_{n\in\mathbb N}. \end{align}

We will prove one auxiliary lemma.

Lemma 1: Let $A\in \mathcal F = \sigma\left(\bigcup_{n\in\mathbb N}\mathcal F_n\right),$ then $$ P_\mu [\iota (A) ] =0 \Rightarrow \mathbb P [A] = 0. $$ Remember that $\mathbb P = \int_M \mathbb P_x [A] \mu (\mathrm{d}x).$

Proof. Let $A$ be such that $P_\mu(\iota(A) ) = 0.$ This means that given $\varepsilon >0,$ there exists $n_0\in\mathbb N,$ such that, for every $n>n_0$. $$P_\mu[\pi_i \in \pi_i(\iota (A)),\ \forall \ i\in\{0,\ldots, n\}] <\varepsilon.$$

Note that $$X_i(A) = \pi_i(\iota(A)), \ \forall \ i\in\mathbb N. $$

Note that for every $n_0 < n\in\mathbb N,$ \begin{align*} \varepsilon \geq P_\mu[\pi_i \in \pi_i(A_i),\ i\in\{0,\ldots, n\}] &= \int_M P_x[\pi_i \in X_i(A),\ i\in\{0,\ldots, n\} ] \mu(\mathrm{d}x)\\ &=\int_{M} \mathbb P_x[ X_i \in X_i(A),\ \forall \ i\in\{0,\ldots,n\}] \mu(\mathrm{d}x)\\ &\geq \int_M \int_M \mathbb P_x[A] \mu(\mathrm d x) = \mathbb P[A]. \end{align*} Implying that $\mathbb P[A]=0.$

Let $f\in L^1(M,\mathcal B(M), \mu),$ consider the set $$B= \left\{y\in M^\mathbb N; \lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1} f \circ \pi_i (y) \neq \int_M f(x) \mu(\mathrm{d}x)\right\}.$$

We have that $P_\mu(B) = 0.$

Note that for every $n\geq 0,$ we have that $$f (X_n(\omega)) = f (\pi_n ( \iota(\omega)). $$

We have that $$ \lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1} f \circ X_i (\omega) \neq \int_M f(x) \mu(\mathrm{d}x) \Leftrightarrow \iota(\omega) \in B\Leftrightarrow \omega \in \iota^{-1}(B).$$

By Lemma 1 we have that $$ P_\mu (\iota (\iota^{-1}(B))) \leq P_\mu(B) = 0 \Rightarrow \mathbb P [\iota^{-1}(B)] = 0. $$

Therefore, $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1} f \circ X_i (\omega) =\int_M f(x) \mu(\mathrm{d}x),\ \forall \ \omega \text{-}\mathbb P \ \text{a.s.}. \ $$


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