3
$\begingroup$

This question was previously posted on MSE.

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.

$\endgroup$

1 Answer 1

0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.