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Re: Binomial variance for DFT
Posted:
Dec 27, 2012 10:18 AM
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"Cristiano" wrote in message news:kbhdes$9f3$1@dont-email.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 * (1-p).
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 * (1-p) / 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
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There seems very little "explanation" in the paper you cite.
However, you might like to start from the non-asymptotic 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 non-zero excess kurtosis
of the starting random variables in the test case: (ii) the correlation at
adjacent/neighbouring frequencies.
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