# What is the variance of DFT of Fourier coefficient of difference of a vector of white noise?

Consider $$\big\{x[0], x[1], \ldots, x[N-1]\big\}$$. Suppose, $$\begin{cases} x[n] \sim \mathcal N\left(0, \sigma^2\right)\\ \big\langle x[n], x[n-1]\big\rangle = \frac12 & \forall \ n\\ \big\langle x[n_1],x[n_2]\big\rangle = 0 & \text{whenever}\quad\lvert n_1-n_2\rvert > 1 \end{cases}$$ I need to compute $$\operatorname{Var}\{X[k]\}$$ where, $$X[k] = \sum\limits_{n=0}^{N-1} x[n] \ e^{-i 2 \pi \frac{nk}{N}}.$$ The definition of variance I am using is $$\operatorname{Var}\big\{X[k]\big\} = \operatorname{E}\big\{\big\lvert X[k]\big\rvert^2\big\} - \operatorname{E}\big\{\big\lvert X[k]\big\rvert\big\}^2$$ The second term is zero. But I am failing to compute the variance. I am not sure how to do it.

All samples in a white noise process are independent and identically distributed (IID) so it is helpful to understand how the variance grows as you sum N IID random variables: the sum of $$N$$ IID random variables with variance $$\sigma^2$$ will have a variance of $$N\sigma^2$$. Note that the exponential in the DFT equation only changes the phase of each sample without changing the amplitude. Importantly, all the samples were independent from each other and identically distributed; how does this change when we change the phase of those samples?