I would like to discuss and ask a question regarding the Fourier transform of a Gaussian process, if it makes sense.

For that purpose, let me describe the following situation.Let $z(s)$ be a Gaussian process in some space X (e.g., $\mathbb{R}^2$).

We can construct such a Gaussian process (following Higdon, 1998) when:$$z(s) = k(s)\ast\nu(s) = \int_{X}k(u-s)\nu(s)du,$$

where $\ast$ is the convolution operator, $k$ is the smoothing kernel function and $\nu(s)\sim\mathcal{N}(0,\lambda_s^{-1})$ is a white noice process.

Then, if $k$ is isotropic (dependent on distance only), one can compute its covariance simply by using the Convolution theorem:$$\Sigma = cov(z(s), z(s’)) = \mathcal{F}^{-1}[\mathcal{F}[k(s)\ast k(s’)]] = \mathcal{F}^{-1}[K^2(w)],$$

where $\mathcal{F}$ and $\mathcal{F}^{-1}$ are the Fourier and inverse Fourier transform operator respectively, and $K(w)$ is the Fourier transform of the kernel function.

I would like to actually work with the Fourier spectra of my data, meaning I would like to sample the parameters space of my Gaussian process in reciprocal space. The reason is that I know that $k$ is isotropic and I have the full form of $K$, so working in Fourier space would save me the computation cost of taking inverse Fourier Transform of my kernel function (the data itself is already in Fourier component so working in Fourier space would save me two transforms).

We know that the Fourier transform of a Gaussian $\mathcal{N}(0, \sigma)$ is itself Gaussian $\mathcal{N}(0, \frac{1}{\sigma})$.

Does that mean that if $z(s)\sim\mathcal{N}(0, \Sigma)$, its fourier transform is $Z(w)\sim\mathcal{N}(0, K^2(w))$?

Any comments, discussions, and/or reading recommendations will be more than welcome!

Thank you in advances! Looking forward to the replies!

**Answer**

Chapter 11 of the book *Probability, Random Variables, and Stochastic Processes* has this result:

The Fourier transform of a stochastic process $x(t)$ is a stochastic process $\mathbf{X}(w)$ given by

$$

\mathbf{X}(\omega)=\int_{-\infty}^{\infty} \mathbf{X}(t) e^{-i \omega t} d t

$$

I changed out the $j$ they had for the more common $i$ to denote $\sqrt{-1}$.

**Attribution***Source : Link , Question Author : LeFlan , Answer Author : Adam Elder*