I have a couple of questions regarding ergodicity for Markov processes in continuous time. (In particular, the first question seems like it should be particularly basic, and yet I haven't managed to find a proof [or counterexample!].)
We have a family $(P_x^t)_{x \in \mathbb{R},t \geq 0}$ of Borel probability measures on $\mathbb{R}$ such that
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for all $A \in \mathcal{B}(\mathbb{R})$, the map $(x,t) \mapsto P_x^t(A)$ is Borel-measurable;
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for all $x \in \mathbb{R}$, $P_x^0=\delta_x$;
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for all $A \in \mathcal{B}(\mathbb{R})$ and $s,t \geq 0$, $P_x^{s+t}(A)=\int_\mathbb{R} P_y^t(A) \, P_x^s(dy)$.
[We can refer to $(P_x^t)_{x \in \mathbb{R},t \geq 0}$ as a "measurable stochastic semigroup". In general, "stochastic semigroups" only need to be measurable in $x$ for each $t$.]
We will say that a probability measure $\rho$ on $\mathbb{R}$ is stationary if $\rho(A)=\int_\mathbb{R} P_x^t(A) \, \rho(dx)$ for all $A \in \mathcal{B}(\mathbb{R})$ and $t \geq 0$. We will say that a probability measure on $\mathbb{R}$ is ergodic if it is an extremal point of the convex set of stationary probability measures.
Q1. Let $(\Omega,\mathcal{F},(\mathcal{F}_{t \geq 0}),\mathbb{P})$ be a filtered probability space, and let $(X_t)_{t \geq 0}$ be a progressively measurable real-valued homogeneous Markov process with transition probabilities given by $(P_x^t)_{x \in \mathbb{R},t \geq 0}$ — that is to say, $P_{X_s(\cdot)}^t(A)$ is a conditional probability of $X_{s+t}^{-1}(A)$ with respect to $\mathcal{F}_s$ (for all $s,t,A$). Suppose also that $\rho:=X_{0\ast}\mathbb{P}$ is stationary. Fix any bounded measurable $f:\mathbb{R} \to \mathbb{R}$; is it the case that
$\hspace{5mm} \lim_{T \to \infty} \frac{1}{T} \int_0^T f(X_t(\omega)) \, dt$
exists for $\mathbb{P}$-almost all $\omega \in \Omega$?
(Please note that we do not assume any kind of continuity of $(X_t)$, but only that it is progressively measurable.)
Now in terms of my motivation, what I am really after is an ergodic decomposition theorem for the setting that I'm currently working with; I think that a positive answer to Q1 will be enough for me to prove this. However, I would ideally like to know if ergodic decompositions exist more generally:
Q2. Suppose $\rho$ is a stationary probability measure. Does there exist a probability measure $Q$ on the set $\mathcal{M}$ of probability measures on $\mathbb{R}$ (equipped with the usual $\sigma$-algebra, which is known to be standard) such that
$Q$-almost every $\mu \in \mathcal{M}$ is ergodic;
for all $A \in \mathcal{B}(\mathbb{R})$, $\rho(A) = \int_\mathcal{M} \mu(A) \, Q(d\mu)$?
The following might be useful:
Equivalent definitions of ergodicity: Given a stationary probability measure $\rho$, we will say that a set $A \in \mathcal{B}(\mathbb{R})$ is $\rho$–almost stationary if for all $t \geq 0$, $\rho(x \in A: P_x^t(A)=1)=\rho(A)$.
(1) In analogy to Proposition 7.2.4 of books.google.co.uk/books?isbn=0521515971 (p378) for deterministic systems, we have that a stationary probability measure $\rho$ is ergodic if and only if every $\rho$-almost stationary set has $\rho$-trivial measure: If $\rho(A) \in (0,1)$ and $A$ is $\rho$-almost stationary, then $\rho$ conditioned on $A$ and $\rho$ conditioned on $\mathbb{R} \setminus A$ are stationary probability measures which can be linearly combined in the obvious way to give $\rho$. In the other direction, it suffices to show that if every $\rho$-almost stationary set has trivial measure and $\tilde{\rho}$ is a stationary probability measure that is absolutely continuous with respect to $\rho$, then $\rho=\tilde{\rho}$. Take a density $g$ of $\tilde{\rho}$ with respect to $\rho$. For each $t$, define the probability measure $\rho_t$ on $\mathbb{R} \times \mathbb{R}$ by $\rho_t(A \times B) = \int_A P_x^t(B) \, \rho(dx)$. The stationarity of $\tilde{\rho}$ implies that
$\hspace{5mm} \int_{A \times (X \setminus A)} g(x_1) \, \rho_t(d(x_1,x_2)) \ = \ \int_{(X \setminus A) \times A} g(x_1) \, \rho_t(d(x_1,x_2))$
for any $A \in \mathcal{B}(\mathbb{R})$ and $t \geq 0$. Setting $A:=\{x \in X : g(x) \geq 1\}$, the above equation (combined with the stationarity of $\rho$) implies that $A$ is $\rho$-almost stationary, so $A$ has trivial measure. It follows that $\tilde{\rho}=\rho$.
(2) We will say that a set $A \in \mathcal{B}(\mathbb{R})$ is invariant if for all $t \geq 0$ and all $x \in A$, $P_x^t(A)=1$. Given a set $A$ that is $\rho$-almost stationary, there exists a set $A'$ that is invariant, with $\rho(A \triangle A')=0$. Namely, set
$\hspace{5mm} A' \ := \ \{ x \in X : \textrm{Leb}(t \geq 0 : P_x^t(A)<1) = 0 \}$
where $\textrm{Leb}$ denotes the Lebesgue measure. So a stationary probability measure $\rho$ is ergodic if and only if every invariant set has $\rho$-trivial measure.
(It is perhaps worth pointing out that (1) does not rely on the stochastic semigroup $(P_x^t)$ being a "measurable" stochastic semigroup, but the construction in (2) does rely on this.)
Update: I'm pretty sure the answer to Q2 is yes, because I think I can prove it using an ergodic theorem for measurable stochastic semigroups; namely, letting $\rho$ be a stationary probability measure, I think I can first prove that for any bounded measurable $f:\mathbb{R} \to \mathbb{R}$,
$\hspace{5mm} \lim_{T \to \infty} \frac{1}{T}\int_0^T \int_\mathbb{R} \! f(y) P_x^t(dy) \; dt$
exists for $\rho$-almost all $x \in \mathbb{R}$, with the limit (as a function of $x$) being a conditional expectation of $f$ over the probability space $(\mathbb{R},\mathcal{B}(\mathbb{R}),\rho)$ with respect to the $\sigma$-algebra of $\rho$-almost stationary sets. (As mentioned in Kifer's book "Ergodic Theory of Random Transformations", the discrete-time analogue of the above statement can be obtained as a special case of the Chacon-Ornstein ergodic theorem.) Using this fact, it should be possible to prove the ergodic decomposition theorem (by a similar approach as in the proof for deterministic dynamical systems).
However, I suspect that the answer to Q1 is no (although I do not have a counterexample!!). More precisely, I suspect that the answer to Q1 is the same as the answer to my question Is it true that all stationary measurable stochastic processes are "measurably stationary"? – and I expect that the answer to that question is no (although again, I do not have a counterexample).
If the answer to Q1 is no, I wonder whether perhaps it becomes yes in the particular case that $(X_t)_{t \geq 0}$ is a strong Markov process.
Best Answer
I have the answers to my two questions. (I've actually had them for a while; apologies for the delay in posting.) I will give them in reverse order:
The answer to Q2 is yes; the structure of the proof is exactly as I outlined in the update. Details can be found in section 5 (in particular, Corollary 100) of my notes http://wwwf.imperial.ac.uk/~jmn07/Ergodic_Theory_for_Semigroups_of_Markov_Kernels.pdf.
The answer to Q1 is also yes: Since $f$ is bounded, it is sufficient just to consider the limit as $T$ tends to $\infty$ in the integers. By the positive answer to the question Is it true that all stationary measurable stochastic processes are "measurably stationary"?, the discrete-time stochastic process $\left(\int_n^{n+1} f(X_t(\cdot)) \, dt \right)_{n \geq 0}$ is stationary, and therefore Birkhoff's ergodic theorem (applied to the shift map on $\mathbb{R}^{\mathbb{N}_0}$) gives the desired convergence.