# PDF of discrete fourier transform of a sequence of gaussian random variables

I have a set of numbers drawn from iid gaussian random variables:

$P(d_0, ..., d_{N-1}) = (\sigma \sqrt{2 \pi})^{-N} exp\left(\frac{-1}{2\sigma^2} (d_0^2 + ... + d_{N-1}^2)\right)$

What is the pdf for the discrete fourier transform $f_0, ... f_{N-1}$ of the $d_k$?

It seems like this should be a fairly strightforward calculation but I'm getting bogged down. One thing that seems to make it tricky is the periodicity property of the $f_k$ means that you need to choose N particular real numbers to work with. I'd actually be happy with an answer that marginalizes over the phase angles, but a full answer would be great.

Is there a good text that addresses this type of question?

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## 1 Answer

Hi Dan,

maybe I misunderstand your question, but I would go about this the following way:

The discrete Fourier transform is an invertible linear transform $\mathbf f = W\mathbf d$, so you can just transform the density via $p_F(\mathbf f) = p_D(\mathbf W^{-1}f)|\det W^{-1}|$. The particular structure of $W$ depends on the type of DFT you are using (sometimes the differ by constant factors).

If you are really interested in the distribution of the phase variables for the Gaussian above, the answer is: they are uniformly distributed. One can see this as follows: When computing the complex coefficient of the Fourier transform you do something like (ignoring constants) $\sum_t d_t (\cos(\frac{2\pi }{N} k t) + i\sin(\frac{2\pi }{N} k t)) = a_k + ib_k$. Your Gaussian variables are white, meaning $\mathbf d \sim \mathcal N(0,\sigma^2 I)$. Since $\sin$ and $\cos$ are orthogonal $a_k$ and $b_k$ are white Gaussian as well. In particular, this means that $a_k,b_k$ are spherically symmetric distributed and $(a_k,b_k)/\|(a_k,b_k)\|)$ follows a uniform distribution on the unit sphere. Therefore, the angles are uniformly distributed.

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Oh, of course, I was was being dense late at night. I had already been following your approach, had realized that the determinant was just some function of N (that gets you the right normalization), but didn't see that you can just say that sum d_k^2 = 1/N^2 sum |f_k|^2 (by Parseval) = 1/N^2 sum a_k^2 + b_k^2, so you immediately get that both the real and imag components are gaussian distributed. Only half of the a_k and b_k are distinct, so you do have to be careful to get the right std dev for each particular component. – Dan Becker Aug 26 '11 at 20:08