Given random dynamical system $(X, \mathcal{B}, (T_{\omega})_{\omega\in \Omega}, \mu)$ where $(\Omega, \mathbb{P})$ is probability space with ergodic transformation $\sigma: \Omega \to \Omega$. Define random composition $T_\omega^n:=T_{\sigma^{n-1}\omega} \circ \dots \circ T_\omega$. Assume we have quasi-invariant absolutely continuous probability $\mu_\omega:= h_\omega d\mu$ such that $(T_\omega)_{*} \mu_\omega=\mu_{\sigma \omega}$.

There are two kinds of mixing rate I know so far, but could not tell the reason how it comes from.

the first type of mixing rate is : a.s. $\omega \in \Omega$, and any function $\phi, \psi \in L^{\infty}(\Omega \times X)$ (denote $\phi_{\omega}, \psi_{\omega}: X \to \mathbb{R}$ as function restricted on fiber $\omega$),

$$ \left|\int \varphi_{\sigma^n \omega} \circ T^n_\omega \cdot \psi_\omega \, d\mu_\omega -\int \varphi_{\sigma^n \omega} \, d\mu_{\sigma^n \omega} \int \psi_\omega \, d\mu_\omega\right| \le C_{\omega} \cdot \|\varphi\|_\infty \cdot C_\psi \cdot e^{-n}$$

the second type of mixing rate is : a.s. $\omega \in \Omega$, and any function $\phi, \psi \in L^{\infty}(\Omega \times X)$, $$ \left|\int \varphi_{\sigma^n \omega} \circ T^n_\omega \cdot \psi_\omega \, d\mu_\omega -\int \varphi_{\sigma^n \omega} \, d\mu_{\sigma^n \omega} \int \psi_\omega \, d\mu_\omega\right| \le \|\varphi\|_\infty \cdot C_\psi \cdot e^{-n}$$

The first mixing rate depends on the random environment $\omega$, while the second one doesn't.

what is the reason making $C_{\omega}$ show up in the mixing rate? what kind of system has homogenous mixing rate, independent of $\omega$, i.e. does not have $C_{\omega}$. can we find examples in the real world to explain such difference? Thanks in advanced!

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    $\begingroup$ The $C_\omega$ is completely natural. You should imagine that you are applying two dynamical systems:$T_0$ and $T_1$ (so $\omega$ is a sequence of 0’s and 1’s). Imagine $T_1$ mixes much faster than $T_0$. Now if $\omega$ has a large number of initial 0’s, the mixing will be slow, so $C_\omega$ should be large. You would expect $C$ not to depend on $\onega$ if all of the fiber maps mix at the same rate. $\endgroup$ – Anthony Quas Apr 13 at 19:34
  • $\begingroup$ @Anthony Quas Thanks for the example. But with the same setting of your example, assume decay rate of all $T_1$ is $e^{-n}$ , decay rate of all $T_0$ is $e^{-\sqrt{n}}$. Then the decay rate of any mix of $T_0$ and $T_1$ should be lying between them, and less than $e^{-\sqrt{n}}$. Can I deduce the decay rate of any mix of $T_0$ and $T_1$ is less than $e^{-\sqrt{n}}$ without $C_{\omega}$? $\endgroup$ – jason Apr 14 at 13:25
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    $\begingroup$ At the level of intuition, yes. At the level of proof, no. You could, no doubt, build counter examples to that statement. But they would be artificial, but if you had a concrete system of that type in front of you, you should expect the behaviour you’re talking about. You would still have to work hard to prove it. $\endgroup$ – Anthony Quas Apr 14 at 16:52
  • $\begingroup$ Thank you very much! $\endgroup$ – jason Apr 14 at 18:25

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