Distribution of Gaussian noise in the frequency domain

Question

I have $N$ samples of Gaussian noise independently drawn from a normal distribution $n(t) \sim \mathcal{N}(0, \sigma^2)$. How would the noise be distributed in the frequency domain? I.e., what is $n(f)$? I understand it should also be Gaussian, and centered at $0$. Using the unitary DFT matrix,

$$E(n(f)) = \frac{1}{\sqrt{N}}\sum_{t=0}^{N-1}E(n(t))e^{-\frac{2\pi ift}{N}} = 0$$

However, what would the variance of the distribution be?

Answer 1

Take the Fourier transform of the PDF to get

$ \mathcal{F} \left( \frac {1} {\sigma \sqrt{2\pi}} e^{ - \frac {x^2} {2\sigma^2} } \right) = \frac {1} {\sigma \sqrt{2\pi}} \cdot \mathcal{F} \left( e^{ - \frac {x^2} {2\sigma^2} } \right) = \frac {1} {\sigma \sqrt{2\pi}} \cdot \left( \sqrt{ 2\pi \sigma^2} \cdot e^{ -2\pi^2 \sigma^2 f^2 } \right) = e^{ - \frac {f^2} {2 / (2\pi\sigma)^2}}. $

Thus, the variance of the noise in the frequency domain is $(2\pi\sigma)^{-2}$.

Answer 2

Suppose we have $n$ IID Gaussian random variables $w_1, w_2, \dots, w_n \sim \mathcal{N} \left( 0, \sigma^2 \right)$. From independence,

$$ {\Bbb E} \left( w_i w_j \right) = \sigma^2 \delta_{ij} $$

where $\delta_{ij}$ is the Kronecker delta. Let ${\bf w} := \begin{bmatrix} w_1 & w_2 & \dots & w_n\end{bmatrix}^\top$. Note that ${\bf w} \sim \mathcal{N} \left( {\bf 0}_n, \sigma^2 {\bf I}_n \right)$. Hence, the noise signal $w$ is white (which is why I called it $w$).

Let ${\bf F} {\bf w}$ be the discrete Fourier transform (DFT) of $\bf w$, where $\bf F$ is the $n \times n$ (unitary) DFT matrix. Since the transformation ${\bf w} \mapsto {\bf F} {\bf w}$ is linear, fortunately, Gaussian-ness is preserved and, quite amusingly, we have ${\bf F} {\bf w} \sim \mathcal{N} \left( {\bf 0}_n, \sigma^2 {\bf I}_n \right)$ as well because ${\bf F} {\bf F}^* = {\bf I}_n$.