Does Bernoulli imply exponential mixing?

Question

This question comes from this paper where the authors proved that exponential mixing implies Bernoulli. They also mentioned in the introduction that Bernoulli is the strongest ergodic property and Bernoulli implies mixing of all orders. But they didn't mention whether Bernoulli also imply exponential mixing.

Answer 1

This is something that has confused me too, but let me try to answer anyway.

This question itself is a bit tricky to interpret. The reason is that "exponential mixing" really depends on the metric/topology, in the sense that no nontrivial system is exponentially mixing for ALL functions. Usually you have to assume $f,g$ are Holder or something like this, which depends on the metric. Even true i.i.d systems fail exp. mixing for some functions; here's a silly example.

Take $X = \{0,1\}^\mathbb{N}$, $\sigma$ the left shift, $\mu$ the unif. distributed i.i.d measure. Take $n(x)$ to be the minimum $n$ so that the $2^n$-th bit of $x$ is a $1$, i.e. $x(2^i) = 0$ for $i < n(x)$ and $x(2^{n(x)}) = 1$. Then define $f(x) = (n(x))^{-1}$ (with the convention that $\infty^{-1} = 0$). $f$ is integrable; $\int f = \sum_{n = 1}^{\infty} \frac{1}{n} \mu(\{x : x(2^1), x(2^2), \ldots, x(2^{n-1}) = 0, x(2^n) = 1\}) = \sum_{n = 1}^{\infty} \frac{2^{-n}}{n} = \ln 2$.

Take $g(x)$ to be the first bit of the sequence $x$.

Then for any $N$, $g(\sigma^N x)$ is just the $(N+1)$th bit of $x$, which is conditionally independent of $f$ unless $N+1$ is a power of $2$. If $N+1 = 2^k$, then $g(\sigma^N x)$ is still conditionally independent of $f$ if $f(x) > k^{-1} \Longleftrightarrow n(x) < k$. On the other hand, if $f(x) = k^{-1}$, then $g(\sigma^N x) = 1$, and if $f(X) < k^{-1}$, then $g(\sigma^N x) = 0$.

Therefore, when $N+1 = 2^k$, $\int f(x) g(\sigma^N x) d\mu = \sum_{n = 1}^{k-1} 0.5 \frac{2^{-n}}{n} + \frac{2^{-k}}{k}$. This means that $\int f(x) g(\sigma^N x) d\mu - \int f(x) g(x) d\mu = 0.5(\frac{2^{-k}}{k} - \sum_{n = k+1}^{\infty} \frac{2^{-n}}{n})$, which is greater than $\frac{2^{-k}}{4k(k+1)}$, which is on the order of $N^{-1}$. So you do not have exponential mixing for $f,g$.

For the usual metric on $\{0,1\}^{\mathbb{N}}$ (which is $2^{-i}$ for $i$ the first bit where $x,y$ disagree), this is not surprising, since $f$ is very much not Holder.

However, the term "Bernoulli" is usually understood as "measure-theoretically isomorphic to an i.i.d. system." So now you have the following silly example: change the metric $d$ on $\{0,1\}^{\mathbb{N}}$ to be $1/\log_2 i$, where $i$ is the first bit where $x,y$ disagree. It induces the same topology, so the system is still Bernoulli. But the class of Holder functions has changed, and the $f$ from above is now Holder! (To see this, note that if $d(x,y) < 1/k$, then $x,y$ must agree on $2^k$ bits, so the only way to have $f(x) \neq f(y)$ is if they are both less than $1/k$). So on this Bernoulli system, you have a pair of Holder functions which don't exhibit exponential mixing.