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...)
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...??