TITLE: A case of nested central limits
QUESTION [7 upvotes]: Consider the random variable $S=(s_0, \dots ,s_{N-1})$, a sequence of signs uniformly distributed on the hypercube $\{-1,1\}^N$. We are interested in $N$ large and prime. The Fourier transform $\hat{S}=(\hat{s}_0, \dots ,\hat{s}_{N-1})$, where 
$$
\hat{s}_q=\frac{1}{\sqrt{N}}\sum_{k=0}^{N-1} \zeta^{q k} s_k, \qquad \zeta=e^{2\pi i/N},
$$
defines a new multivariate random variable. Finally, consider the univariate random variable
$$
h=\left(\prod_{q=1}^{N-1}\hat{s}_q\right)^\frac{1}{N-1}=\left(\prod_{q=1}^{(N-1)/2}|\hat{s}_q|^2\right)^\frac{1}{N-1}.
$$
This variable is the scaled algebraic norm of the cyclotomic integer whose coefficients are $S$.
I would like to prove that $h$ has the central limit property for $N\to \infty$, that is, its distribution becomes increasingly concentrated at one value. The plausibility argument goes like this. Consider a "baby" estimator of the variable given by the geometric mean of a fixed number $M$ of the $|\hat{s}_q|^2$, instead of all $(N-1)/2$. For this fixed number of Fourier transform components we have a multivariate central limit theorem, for $N\to \infty$. When we apply CLT and compute the covariance matrix, we find that these $|\hat{s}_q|^2$ are independent and identically distributed random variables. The baby estimators therefore have the central limit property for $M\to\infty$. Unfortunately, the $M\to\infty$ limit of the baby estimators may be a different random variable than the $N\to \infty$ limit of $h$.

REPLY [3 votes]: The central limit property of $h$ follows from the main result of:
David Freedman and David Lane, "The Empirical Distribution of Fourier Coefficients", Ann. Statist. Volume 8, Number 6 (1980), 1244-1251.
They show that all the Fourier coefficients (not trivially related as complex conjugates) are asymptotically independent.