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Consider the random circle rotation $x \to x + Z \text{ mod 1}$ on $([0, 1], \text{Lebesgue})$ where at each rotation, $Z$ is uniformly distributed on $[0, 1]$ and independent of previous rotations.

Is it true that almost surely, the system is weak mixing?

Remark: Note that the rotation by a fixed number is never weak mixing due to being an isometric system.

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  • $\begingroup$ What do you mean by "almost surely" on $[0,1]^{\mathbb{N}}$? $\endgroup$
    – Alessandro Della Corte
    Commented Oct 26 at 8:48
  • $\begingroup$ I mean almost surely with respect to the probability space underlying the independent random variables $Z_i$. @AlessandroDellaCorte $\endgroup$
    – Nate River
    Commented Oct 26 at 9:06
  • $\begingroup$ Note that this is a different $[0, 1]$ than the underlying space of the dynamics. $\endgroup$
    – Nate River
    Commented Oct 26 at 9:07
  • $\begingroup$ If you take $[0, 1]^{\mathbb N}$ as a model for the probability space, then the natural measure is the countable product of the uniform measure, and almost surely means with respect to this. $\endgroup$
    – Nate River
    Commented Oct 26 at 9:10
  • $\begingroup$ I would have asked what you meant by weak-mixing. It's normally defined for a fixed transformation, when it is equivalent to the non-existence of non-constant eigenfunctions. Did you mean the condition that Will Sawin showed does not hold in his answer? (this seems the most reasonable guess). Maybe it would make more sense to define the random rotation as a skew product $T\colon [0,1)^\mathbb Z\times [0,1)\to [0,1)^\mathbb Z\times [0,1)$ that is $T(\omega,t)=(\sigma(\omega),t+\omega_0\bmod1)$. $\endgroup$
    – Anthony Quas
    Commented Oct 27 at 4:38

1 Answer 1

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No,

$$ \frac{1}{n} \sum_{k=1}^{n}| \mu (A) \cap T^{-k}(B) - \mu(A)\mu(B)|$$ is a sum of i.i.d random varables, since $T^{-k}$ are i.i.d uniform circle rotations so by the law of large numbers it converges almost surely to the expectation of one of these variables, which is $$ \mathbb E | \mu (A) \cap F^{-1}(B) - \mu(A)\mu(B)|$$ for $F$ a random rotation.

This is nonzero for $A,B$ any two intervals. In fact, it is equal to the long run average in the case that $T$ is an irrational rotation, so this reduces to the case you already know.

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    $\begingroup$ Just noting for myself - the key here is a composition of iid random circle rotations is independent of its earlier iterates, just due to the niceness of uniformly chosen rotations. Very cool. $\endgroup$
    – Nate River
    Commented Oct 26 at 12:04
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    $\begingroup$ @NateRiver Yes, although independence is not so strictly necessary since the law of large numbers is robust to many forms of dependence between the random variables and we only need a very weak form of the law of large numbers. $\endgroup$
    – Will Sawin
    Commented Oct 26 at 12:56

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