Someone argues: "I have a record of samples of pure noise

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Scientific Computing for Physicists, Ch Spectra with error bars: the fast transform, convolution and spectral estimation

Problem

Someone argues: "I have a record of N=4096N=4096 samples of pure noise. Its periodogram ordinates are exponentially distributed with mean equal to the true noise level SS, so P(S^k>5S)=e−5=0.67%P(\hat S_k > 5S) = e^{-5} = 0.67\%; with 20472047 independent interior bins I expect about 1414 bins above this threshold, which is what I see, so nothing is unusual. To be more confident in this null result, I now record N=65 536N = 65\,536 samples, sixteen times longer. This narrows each bin (better frequency resolution) and, since each periodogram value now averages over sixteen times more data, its variance must fall by a factor of 1616. So a bin at 5×5\times the mean level in the new record corresponds to an effective ratio of 5×16=805\times16=80 at the old threshold probability, i.e. the same 0.67%0.67\% chance now guards against a bin 8080 times the mean rather than 55 times, which would mean a bin at only 5×5\times the mean is far more significant with the longer record." ((a)) Identify the error in this argument. ((b)) State the correct probability that a bin of the N=65 536N=65\,536 periodogram exceeds 5S5S. ((c)) In at most 80 words, explain why a longer record alone does not reduce this probability.

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