1
$\begingroup$

Informal description.

Suppose I have a dynamical system $f$ defined on the product of a compact space $X$ representing the state space of an "experimentally visible" variable and a compact space $Y$ representing the state space of an "experimentally invisible" variable. Although $Y$ is "invisible", I assume that it is equipped with an equivalence class of probability measures that - like the Riemannian volume measure on a compact smooth manifold - defines "physically accessible" sets of initial conditions in $Y$. I assume that $f$ has no skew-product structure, meaning that there is allowed to be bi-directional feedback between the visible variable and the invisible variable. A heuristic interpretation of my question below is: If in my experiments I consistently observe the same ergodic statistics for the visible variable, is it necessarily a "physical possibility" that this ergodic statistics comes from an underlying stationary dynamics for the process as a whole?


Precise formulation.

Let $X$ and $Y$ be compact metric spaces, let $\pi_X \colon X \times Y \to X$ be the first coordinate projection, let $\mu$ be a Borel probability measure on $X$ of full support, let $\lambda$ be a Borel probability measure on $Y$ of full support, and let $$ f \colon X \times Y \to X \times Y $$ be a homeomorphism. Suppose that for every continuous function $g \colon X \to \mathbb{R}$, $$ \mu \otimes \lambda \left( (x,y) \in X \times Y \, : \, \frac{1}{N} \sum_{n=0}^{N-1} g(\pi_X(f^n(x,y))) \to \int_X g \, d\mu \ \text{ as } \, N \to \infty \right) = 1. $$ Does it follow that there exists an $f$-invariant probability measure on $X \times Y$ whose $X$-marginal is $\mu$?

If so, does there necessarily exist a Borel probability measure $\nu$ on $X \times Y$ with $\lambda$-absolutely continuous $Y$-projection such that $\frac{1}{N} \sum_{n=0}^{N-1} f^n_\ast\nu$ has a subsequence converging weakly to a measure whose $X$-marginal is $\mu$?


Remark. In the case that $f$ has the skew-product structure $f(x,y)=(\theta(x),\varphi_x(y))$ [in which case the measure $\lambda$ is irrelevant for the first question], an affirmative answer to the first question is given by Corollary 6.13 of Hans Crauel, Random Probability Measures on Polish Spaces - although quoting that result is probably overkill, and I think an affirmative answer to both questions is provided (with $\nu=\mu \otimes \lambda$) by just slightly adapting the regular proof of the Krylov-Bogolyubov theorem applied to $f$. In terms of application, typically this skew-product setup concerns the scenario that $X$ is the state space of an unknown "noise" and $Y$ is the "visible" state space - which is opposite to what I'm thinking about in my question.

$\endgroup$

1 Answer 1

1
$\begingroup$

Okay, I've seen that the proof of an affirmative answer to both questions in the general case (with $\nu=\mu \otimes \lambda$) is not very hard:

Since $X \times Y$ is compact, we can let $(k_n)$ be a strictly increasing sequence of positive integers such that $\frac{1}{k_n} \sum_{i=0}^{k_n-1} f^i_\ast(\mu \otimes \lambda)$ is weakly convergent as $n \to \infty$. Let $\mathbb{P}$ denote the limit. As in the usual proof of the Krylov-Bogolyubov theorem, $\mathbb{P}$ is $f$-invariant. It remains to show that $\pi_{X\ast}\mathbb{P}=\mu$. For any continuous function $g \colon X \to \mathbb{R}$, we have \begin{align*} \int_X g \, d(\pi_{X\ast}\mathbb{P}) &= \int_{X \times Y} g \circ \pi_X \, d\mathbb{P} \\ &= \lim_{n \to \infty} \frac{1}{k_n} \sum_{i=0}^{k_n-1} \int_{X \times Y} g \circ \pi_X \, d(f^i_\ast(\mu \otimes \lambda)) \\ &= \lim_{n \to \infty} \frac{1}{k_n} \sum_{i=0}^{k_n-1} \int_{X \times Y} g(\pi_X(f^i(x,y))) \, (\mu \otimes \lambda)(d(x,y)) \\ &= \lim_{n \to \infty} \int_{X \times Y} \frac{1}{k_n} \sum_{i=0}^{k_n-1} g(\pi_X(f^i(x,y))) \, (\mu \otimes \lambda)(d(x,y)) \\ &= \int_{X \times Y} \int_X g \, d\mu \, (\mu \otimes \lambda)(d(x,y)) \\ &\hspace{20mm} \text{by Dom. Conv. Thm. and assumption on $\mu$ and $\lambda$} \\ &= \int_X g \, d\mu. \end{align*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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