Given dynamic $f: S^1 \to S^1$ with Lebegue measure $dm$ on $S^1$. Assume it has unique SRB probability measure $\frac{d\mu_f}{dm} dm $. Given left shift space $([-\epsilon, \epsilon]^{\otimes \mathbb{N}}, \theta_{\epsilon}^{\otimes \mathbb{N}}, \sigma )$, where $\theta_{\epsilon}$ is probability on $[-\epsilon,\epsilon ]$.

For any $\omega \in [-\epsilon, \epsilon]^{\otimes \mathbb{N}}$, define $f^n_{\omega}:=f_{\omega_{n-1} } \circ f_{\omega_{n-2} } \circ...\circ f_{\omega_{0} } $ where $ f^1_{\omega}=f_{w_0}:=f+\omega_0$. Assume for each $\epsilon$, we have stationary probability $\mu_{\epsilon}$ on $S^1$, which means $ \mu_{\epsilon}E=\int 1_{E} \circ f^1_{\omega}(x) d\mu_{\epsilon}(x) d\theta_{\epsilon}(\omega).$

Assume we have quasi-invariant probability $\mu_{\omega}:= h_{\omega}dm$ s.t. $(f^1_{\omega})_{*} \mu_{\omega}=\mu_{\sigma(\omega)}$. Assume we have decay of correlation: $\int \phi \circ f^n_{\omega} \cdot \psi dm-\int \phi d \mu_{\sigma^n \omega} \int \psi dm \precsim C_{\epsilon, \phi, \psi}(\omega) \cdot \frac{1}{n^{\alpha}}. $

I found two ways to define stochastically stable:

  1. $C_{\epsilon, \phi, \psi}(\omega)$ does not blow up when $\epsilon \to 0$. See Page 5 second paragraph of Almost Sure Rates of Mixing for I.i.d. Unimodal Maps (1999) Baladi (http://citeseerx.ist.psu.edu/viewdoc/summary?doi=

  2. $\mu_{\epsilon} \to \mu_f$ in weak * topology as $\epsilon \to 0$. See page 2 of On stochastic stability of expanding circle maps with neutral fixed points, Sebastian van Strien (https://arxiv.org/abs/1212.5671)

So if we assume we have SRB $\mu_f$, stationary probability $\mu_{\epsilon}$, quasi-invariant probability $\mu_{\omega}$, and decay of correlation ( not loss of memory since it may depend on $\omega$ ), definition 1 and 2 are equivalent? or they are different terminologies? We all know the second definition is the official one. But why Baladi uses the same terminology? Thanks!

  • 1
    $\begingroup$ These are certainly not equivalent. I believe that rotations by a random perturbation of an irrational number satisfy 2) but not 1). That is: $f(x)=x+\alpha$. $\endgroup$ – Anthony Quas Dec 6 '18 at 17:46
  • $\begingroup$ I don’t think you get any decay of correlation: imagine $\phi$ and $\psi$ are supported on small intervals. $\endgroup$ – Anthony Quas Dec 7 '18 at 3:36

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