Suppose I have 1,000 independent random values with a uniform distribution $[+1, -1]$. Now suppose I take the discrete Fourier transform of this data. What the heck is the probability distribution of the resulting Fourier coefficients?
(I'm guessing it's either uniformly distributed again, or else some manner of exotic distribution which converges to a normal distribution if the number of samples is high enough…)
Edit:
Each coefficient produced by the Fourier transform is the weighted sum of the input samples.
It appears that if $x$ and $y$ are both random variables, then the probability distribution of $x+y$ is equal to the convolution of the probability distribution for $x$ and the probability distribution for $y$. [Obviously, this applies iff $x$ and $y$ are independent.] I think I read that right, anyway!
If we were just summing $N$ I.I.D. variables, the result would be a normal distribution. But because it's a weighted sum, each weighted variable has a different [uniform] distribution. So… that means the answer is… uh…??
Best Answer
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 $2 \pi r dr$, multiplied by the Gaussian, which gives (up to some constants) $r*\exp(-r^2)$. 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!