If we have quenched decay of correlation, can we transfer it to annealed decay of correlation? To be precise, let us consider following setting:

Given transformations $T_{\omega}: (S^1, dm) \to (S^1, dm)$ indexed by $\omega \in (\Omega, \sigma, \mathbb{P})$ where $\mathbb{P} $ is ergodic (or Bernoulli) probability on $\Omega$ and $dm $ is Leb measure on circle $S^1$. define random composition $T_\omega^n:=T_{\sigma^{n-1}\omega} \circ \dots \circ T_\omega$. Assume we have quasi-invariant absolutely continuous probability $\mu_\omega:= h_\omega\, dm$ such that $(T_\omega)_{*} \mu_\omega=\mu_{\sigma \omega}, h_\omega \in \operatorname{Lip}(S^1)$.

So we have invariant transformation $\tau $ defines on skew product space $(\Omega \times S^1, \mu)$:

$$ \tau (\omega, x): =(\sigma \omega, T_{\omega}x) \text{ where } \mu = d\mu_{\omega} \, d\mathbb{P} \text{ is invariant probability on } \Omega \times S^1. $$

we study two decay of correlations:

$\forall \phi, \psi \in L^\infty(\Omega \times S^1)$ and $\psi_\omega \in \operatorname{Lip}(S^1)$ for all $\omega \in \Omega$

Quenched decay of correlation:

$$ \left|\int \varphi_{\sigma^n \omega} \circ T^n_\omega \cdot \psi_\omega \, d\mu_\omega -\int \varphi_{\sigma^n \omega} \, d\mu_{\sigma^n \omega} \int \psi_\omega \, d\mu_\omega\right| \le C_{\omega} \cdot \|\varphi\|_\infty \cdot C_\psi \cdot e^{-n},$$ where $\mathbb{P}(C_\omega >m) \le \frac{1}{m^2}$

Anneal decay of correlation:

$$\left| \int \varphi \circ \tau^n \cdot \psi \, d\mu-\int \varphi \, d\mu \cdot \int \psi \, d\mu \right| \le \|\varphi\|_\infty \cdot C_\psi \cdot e^{-n} $$

I tried to prove Quenched decay of correlation $\implies $ Annealed decay of correlation, but got stuck because $\psi-\int \psi \, d\mu$ is no longer fiber-wise mean zero. I guess we should assume certain mixing on $\sigma$.

is it really true or has counter-example? Thanks in advanced.


Your Answer

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

Browse other questions tagged or ask your own question.