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Given a $\sigma$-additive measure space $(E,\Sigma,\mu)$. A Markov operator $P : L^1(\mu) \to L^1(\mu)$ is a linear operator with

  • $ f \geq 0 \Rightarrow Pf \geq 0 $
  • $ f \geq 0 \Rightarrow \|Pf\| = \|f\| $.

The operator posses an adjoint operator $P^* : L^\infty(\mu) \to L^\infty(\mu)$ with $$\langle Pf,g\rangle_\mu = \langle f,P^* g\rangle_\mu$$ for $f \in L^1(\mu), g \in L^\infty(\mu)$ where $\langle f,g\rangle_\mu := \int_E f(x) \, g(x) \, \mu(dx)$. $P$ is called an ergodic Markov operator if $P^*f = f $ implies that $f$ is constant. Further we assume that $P$ has a invariant measure $\mu_g(A):=\int_A g(x) \, \mu(dx)$, i.e. $P g = g$.

Now the operator is associated to a Markov kernel $p$ which is loosely spoken $p(x,A)=$"Probability to move from point x to set A". If the associated Markov kernel is related to a deterministic process determined by a map $S: E \to E$ $$ p(x,A) = \begin{cases} 1 &\text{if } x \in S^{-1}(A) ,\\0 &\text{else,}\end{cases} $$ then the conditions that $P$ is ergodic and $\mu_g$ invariant imply that Birkhoffs theorem holds and we have $$ \lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{n-1} f(S^k(x)) = \int_E f(x) \, \mu_g(dx).$$

Question 1: Is there an equivalent form of writing this when $p$ is a non-deterministic Markov kernel? Something like

$$ \lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{n-1} P^n f = \int_E f(x) \, \mu_g(dx) \text{?}$$

Question 2: Does ergodicity imply that any density will convergete to $g$, i.e. for $f\in L^1(\mu)$ with $\|f\|=1$ and $f \geq 0 $ we get $$ \|P^nf - g\|_{L^1(\mu)} \to 0 $$ for $n\to \infty$?

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  • $\begingroup$ I think there are a couple of typos. That the measure $\mu_g$ is invariant means $\mu_g(Pf)=\mu_g(f)$ for $f\in L^1(\mu)$, which is equivalent to $P^*g=g$ (almost everywhere). If you speak of probabilities, then $g$ must be non-negative with integral $1$. Is that what you assume? Also, I suppose your definition for ergodicity should be that if $f\in L^1(\mu)$ satisfies $Pf=f$, then $f$ must be almost everywhere constant. $\endgroup$
    – Algernon
    Commented Jul 20, 2015 at 10:09

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