5
$\begingroup$

Suppose I have $N$ independent random numbers with a uniform distribution in $[+1, -1]$. Suppose I take the discrete Fourier transform of these numbers. What would the probability distribution of each frequency coefficient be?

Each frequency coefficient is the weighted sum of $N$ independent random numbers with the same distribution. If it were an unweighted sum, then the distribution would presumably tend towards a normal distribution as $N$ increases.

But since it's a weighted sum... The only fact I have which seems relevant is that the PDF of the sum of some random variables supposedly equals the convolution of the PDFs of the individual variables. (Assuming they are all independent, obviously.) I'm not sure what you get if you convolve several unequal uniform distributions thought...

$\endgroup$
2
  • $\begingroup$ It is easy to compute the characteristic function of a weighted sum of independent uniform random variables. But in general it is not the characteristic function of a standard distribution. $\endgroup$ Apr 27 '12 at 17:24
  • 1
    $\begingroup$ Even a weighted sum of i.i.d. variates with finite mean and variance tends towards Normality as long as, loosely speaking, the maximum weight (when the weights sum to 1 regardless of sample size) goes to zero as the sample size goes to infinity. $\endgroup$
    – jbowman
    Apr 27 '12 at 19:52
4
$\begingroup$

The complex Fourier coefficients of a random series form a 2-D normal distribution in the complex plane (a Gaussian rotated around zero). When taking the magnitude of the complex spectrum, at each magnitude $r$ (the distance from the origin) the probability density will be $\int 2 \pi r dr$, multiplied by the Gaussian, which gives (up to some constants) $r\times e^{-r^2}$. Incidentally, this does not depend on the shape of the original random distribution (from the central limit theorem, I guess). I tried in Matlab randn(...) (normally distributed), rand(...) (uniformly distributed) or rand(...)>.5 (zeros and ones only). Magic!

$\endgroup$
4
$\begingroup$

You may find this topic dealt with in Brillinger, D.R. Time Series Analysis and Theory, in Chapter 4, particularly Theorem 4.4.2. I think in your case the answer is that the Fourier coefficients will have asymptotically a complex normal distribution, as pointed in the response by @micork. This will be the case rather generally, the crucial assumption being that the dependence in the original time series is mild enough (for precise conditions, see op. cit. Assumption 2.6.1).

$\endgroup$
2
  • $\begingroup$ Hi, The reference was very useful for me. Thank you for your answer. I wanted a slight clarification here. You mentioned in the last line a reference to Assumption 2.6.1. Is it from the same book by Brillinger ? $\endgroup$
    – sinner
    Sep 5 '16 at 4:35
  • $\begingroup$ Yes, it is from that book. $\endgroup$
    – F. Tusell
    Sep 8 '16 at 11:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.