Autocorrelation function of the logistic map $x_{n+1}=4x_n(1-x_n)$

Question

Is there a proof that for the sequence $x_{n+1}=4x_n(1-x_n)$, the lag-$m$ autocorrelation for $m=1,2$ and so on, is zero if you start with a random seed $x_0$, or say $x_0=\frac{1}{3}$?

It is defined as follows:

$$ \rho(m) = \lim_{n\rightarrow\infty} \frac{1}{8n}\sum_{k=1}^n \Big(x_k-\frac{1}{2}\Big)\Big(x_{k+m}-\frac{1}{2}\Big). $$

Just asking as there are plenty of authors spending a lot of time doing computations to show it is very close to zero, in the context of random number generation. I have an elegant proof that it is exactly zero regardless of $m>0$, I'd like to publish it but I'm wondering if such a fundamental result was not already proved long ago. If you start with a random seed, then $\rho(m)$ is a.s. zero. If you start with $x_0=\frac{1}{3}$, it would be zero if the number $\pi^{-1}\arcsin(\sqrt{1/3})$ is normal in base 2. Proving the latter is a very hard, unsolved problem, although everyone strongly believe this conjecture (about normalcy) to be true.

Update

To avoid confusion, here is a different (simplified) version of the problem. If it has been proved (which I suspect, but could not find any reference), would love to see a reference. If not, I have a proof but rather than sharing it, I'd encourage readers to try to prove it. It's not that hard, but not trivial either. Happy to share my proof too if there is interest.

Assume $X_0$ has a $\text{Beta}[\frac{1}{2},\frac{1}{2}]$ distribution. Let $X_{k+1}=4X_k(1-X_k)$. Then $\text{E}[X_0 X_k] = \text{E}[X_0] E[X_k]$ for all $k>0$.

Answer 1

A derivation of $\mathbf{E[X_0X_k] = 1/4}$

The solutions of this special case of the logistic map can also be written parameterized by $\theta$

$$x_n(\theta) = \frac{1}{2} - \frac{1}{2} \cos(2^n\pi \theta )$$

And the case $X_0 \sim Beta(\frac{1}{2},\frac{1}{2})$ would be similar to $\theta \sim U(0,1)$ (a typical transformation between the uniform distribution and the arcsine distribution).

Below are some graphs of this transformation from $\theta$ to $x_n$ for different $n$

The expectation of $x_0x_k$ can be written as the integral

$$\begin{array}{} E[x_kx_n] &=& \int_{0}^1 x_0(\theta) x_n(\theta)f(\theta) \,\text{d}\theta\\& = &\int_{0}^1 \left( \frac{1}{2} + \frac{1}{2} \cos(2^k\pi \theta ) \right) \left( \frac{1}{2} + \frac{1}{2} \cos(2^n\pi \theta ) \right) \text{d}\theta \\ &=& \begin{cases} \frac{1}{4} & \quad \text{if $k\neq n$} \\ \frac{3}{8} & \quad \text{if $k = n$} \end{cases} \end{array}$$

Where the last equation is found by using the fact that the product of the cosine terms with different frequencies cancel because they are orthogonal.

Other distributions than $\mathbf{X_0 \sim Beta[1/2,1/2]}$

Also when we would start with a different distribution as $\theta = U(0,1)$ if it is a continuous distribution, then eventually we will approach the same result.

We can cut the domain of $\theta$ into $2^n$ evenly spaced intervals and each interval will eventually be distributed as an arcsine distribution This is because the image $x_n$ of $\theta \in (k \frac{1}{2^n}, (k+1) \frac{1}{2^n})$ such intervals is similar to the image $x_0$ of $\theta \in (0,1)$.

Then by letting $n \to \infty$ the domain of any continuous distribution will eventually be able to be split up into intervals, except for some part whose density approaches zero, that all map to the arcsine distribution.

What if $\mathbf{x_0 = 1/3}$

I don't think that the above result is strong enough to be able to say anything about specific cases.

We already have for the set $x_0 = q^2$ with $q$ a rational number between zero and one, that the $x_n$ will have a cycling behaviour.

In addition we may have that for other numbers $x_0$ we have $\rho(m) \neq 0$. The above result only tells that $E[\rho(m)] = 0$ when we average over all numbers. We could have a set of numbers, with non-zero density, for which $\rho(m) \neq 0$ as long as the negative and positive cases are cancelling out in an average over a distribution.