I have a sequence of stationary ergodic random variables $(\epsilon_t)_{t\in\mathbb{Z}}$ and a function $\phi:\mathbb{R}\times \mathbb{R} \rightarrow \mathbb{R}$. Define the sequence of random functions $(\phi_t(x) = \phi(x,\epsilon_t))_{t\in\mathbb{Z}}$.

Now suppose there exists a stationary ergodic sequence of random variables $(X_t)_{t\in\mathbb{Z}}$ such that

  1. $X_{t+1} = \phi_t(X_t)$ for all $t\in\mathbb{Z}$.
  2. For any random variable $Y$ there exists a $\rho>1$ such that almost surely $$ \rho^t\left|X_{t+1} - \phi_t \circ\phi_{t-1}\circ\ldots\phi_0(Y)\right| \stackrel{t\rightarrow\infty}{\rightarrow} 0. $$


The second property is called invertibility and can be interpreted as the process forgetting about its past. Does anyone know about the relation of this condition with mixing, explicitly whether it implies

  1. Mixing in the ergodic sense
  2. $\alpha$-mixing

Thank you in advance!

  • $\begingroup$ Why "invertibility" if it forgets its past? Do you mean "non-invertibility"? $\endgroup$
    – Algernon
    Jun 17 '17 at 19:11
  • $\begingroup$ If you choose $\phi(x,y):=y$, then $(X_t)_t$ will simply be the same as $(\epsilon_t)_t$ shifted one step to the future, and the second condition is trivially satisfied. So, you would at least need to assume that $(\epsilon_t)_t$ is mixing. I suspect you would also need $\phi$ to be continuous. $\endgroup$
    – Algernon
    Jun 17 '17 at 20:43
  • $\begingroup$ I just read about an example that an autoregressive process of order one with independent Bernoulli errors and autoregressive constant $< 1$ is not mixing, so that unfortunately destroys any potential implications. $\endgroup$
    – Marc
    Jun 19 '17 at 9:32

I don't think your condition implies anything much. Let $(\epsilon_t)$ be your favourite ergodic 0--1 valued process; and let $X_t=\sum_{j=1}^\infty 2^{-j}\epsilon_j$. Then the function $\phi$ is $\phi(x,y)=(x+y)/2$. I believe this satisfies all of your conditions for any $\rho<2$.

  • $\begingroup$ Hi Anthony, is the upper limit of the sum supposed to be $t$? What about the condition $X_{t+1}=\phi(X_t,\epsilon_t)$? $\endgroup$
    – Algernon
    Jun 17 '17 at 19:09

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