Fourier transform of random binary vector Consider a uniformly chosen random binary vector $V$ with $n$ elements. That is we say $V_i = 0$ with probability $1/2$ and $V_i=1$ with probability $1/2$. What is the probability distribution of the Fourier transform of $V$? 
I have searched online but have not managed to find an answer.
 A: Well, discrete Fourier transform is a linear transform of a random variables. You can write $W = FV$. Because $F$ is invertible, $V=F^{-1}W$ and you get that  $f_W(w)=f_V(F^{-1}w)/\det(F)$. This assumes you know the pdf of V. Is this what you were looking for? 
A: The DC component is a one-dimensional random walk with step size 1/2 and drift 1/2, of $n$ steps, so is binomially distributed with mean $n/2$ and variance $n/4$.
The other components are two-dimensional random walks of $n$ steps.  Splitting these into real and imaginary components, we find that the mean step size is $\sqrt{2}/2$.  The real part has mean $n/4 \cdot \sqrt{2}/2$ and variance $n/8 \cdot \sqrt{2}/2$.  The imaginary part has mean $0$ (via symmetry) and variance $n/8 \cdot \sqrt{2}/2$.  (In each quadrant, say the first, half of the vectors are left of $\pi/4$ and half are right of it.  These can be paired to make approximately $\pi/4$-directed sums.  There is at most one unpaired vector, leading to the $1/n$ comment below.)
There is a small correction of order $1/n$ to the mean step size for even $n$ versus odd $n$.  Since interesting FFTs have large $n$s, thus large variances, this should be negligible (contrast $O(n)$ with $O(1/n)$).
Edited to correct pre-coffee-brain errors:


*

*The DC component is not a random walk with step size 1.

*"Half-way up the circle" is $\sqrt{2}/2$, not $1/2$.

*Justifying those estimates would be good.

