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Re: Binomial variance for DFT
Posted:
Dec 27, 2012 10:18 AM


"Cristiano" wrote in message news:kbhdes$9f3$1@dontemail.me...
I have n/2 real numbers.
I calculate a threshold T (a real number) for which p * n/2 numbers (0 < p < 1) are expected to fall below T.
Then I count how many numbers are less than T.
That count should have a binomial distribution, right? Hence, its variance should be n/2 * p * (1p).
But the n/2 numbers are the modulus of the first n/2 complex numbers obtained from a discrete Fourier transform and the variance should be n/2 * p * (1p) / 2 as explained here (page 10): http://eprint.iacr.org/2004/018.pdf
Unfortunately, the value given in the paper is significantly different from the one obtained by extensive simulations.
Does anybody now a procedure or a formula to calculate the exact variance of the counts?
Thanks Cristiano

There seems very little "explanation" in the paper you cite.
However, you might like to start from the nonasymptotic theory of the periodogram, such as given by Priestly (MB Priestly, 1981, "Spectral Analysis and Time Series Analysis", Academic Press, Theorem 6.1.3 and surrounding text). Things to consider are (i) the nonzero excess kurtosis of the starting random variables in the test case: (ii) the correlation at adjacent/neighbouring frequencies.



