Consider the random variable $S=(s_0, \dots ,s_{N-1})$, a sequence of signs uniformly distributed on the hypercube $\{-1,1\}^N$. With the Fourier transform we can define $N$ random walk variables $$ \hat{s}_q=\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1} \zeta^{q k} s_k, \qquad q=0, \ldots, N-1, $$ where $\zeta=e^{2\pi i/N}$ and to keep things simple we take $N$ to be prime. As we take $N\to\infty$ all of these have normal distributions: $\hat{s}_0$ on the real line and the others in the complex plane.

I would like to be able to claim that these random variables become independent as $N\to\infty$ in a certain qualified sense. First, since $\hat{s}_q$ and $\hat{s}_{-q}$ are complex conjugates we only consider $q\in\{1,\ldots,(N-1)/2\}$. Second, I am only interested in "fixed-information" counterparts of these. We define these random variables as $f(\hat{s}_q)$, where $f$ maps the complex plane to a finite set. For example (in fact the case that generated this question), $f$ might be take the magnitude of the complex number and round it to 7 significant digits, or to $\infty$ if the magnitude exceeds a certain bound (in order to keep the cardinality of the range of $f$ fixed as $N\to\infty$).

Is it true, say for particular kinds of fixed-information maps, that $f(\hat{s}_q)$ and $f(\hat{s}_{q'})$, $q\ne q'$, are independent as $N\to\infty$?


With help, offline, from P. Diaconis:

I believe my "fixed information" map is really just a license to take the limit $N\to \infty$ before asking if the random variables are independent. Now, if I exercise this license, I notice that $\hat{s}_q$ and $\hat{s}_{q'}$ have a bivariate normal distribution (in the limit), and when I work out the covariance matrix, which I can do for finite $N$, I see that indeed the two variables are independent for $q\ne q'$ (taking the $N\to \infty$ limit of the covariance matrix).


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