First, the definitions:

Definition (Gaussian process)

A random process $X(t)$ is a Gaussian process if for all time points, $t_1,\ldots,t_n$ the random variables $X(t_1 ),\ldots,X(t_n)$ have a multivariate normal distribution. In particular, the distribution of $X(t)$ at any time point $t$ is normal.

Alternate Definition: $X(t)$ is a Gaussian process iff all linear combinations of $X(t_1),\ldots,X(t_n)$ have a normal distribution and all time-points $t_1,\ldots,t_n$.


Suppose $X(t)$ is a Gaussian process. Then $X(t)$ is wide-sense stationary if, and only if, $X(t)$ is strict-sense stationary.

But in book "Gravitational Waves - Volume 1: Theory and Experiments" (page 337, equation 7.6) it says:

$$ \text{E}\Big[\tilde{n}^*(f)\,\tilde{n}(f')\Big] = \delta(f - f')\frac{1}{2}S_n(f), $$

where $\tilde{n}(f)$ is the Fourier transform of $n(t)$, $S_n(f)$ is the power spectral density of $n(t)$ and $\delta(f-f')$ is the delta distribution.

This definition also seems correct, see the following paper: https://d-nb.info/1116766698/34 (equation 44).

I first thought that this is white noise, ie:

$$\text{E}\Big[n(t)\Big] = 0, \hspace{30pt} S_n(f) = \frac{n_0}{2} \hspace{30pt} R_n(t,t') = \frac{n_0}{2}\delta(t-t'),$$

where $R_n(t,t')$ is the autocorrelation function of $n(t)$. Namely, then holds:

$$R_n(t,t') = \text{E}\Big[n(t)\,n(t')\Big] = \frac{n_0}{2}\delta(t-t') = S_n(f)\,\delta(t-t')$$

But then this is still not identical with the term to be drawn. Moreover, the given term seems to be much more general, as I assume white noise here.

Hence my actual question:

Where does the following expression for stationary Gaussian Noise come from: $\langle \tilde{n}(f)\tilde{n}(f')\rangle = \delta(f-f')\frac{1}{2}S_n$?

Is there any derivation for this expression? Any book or paper that defines or even derives this expression?

  • $\begingroup$ I have no idea if this makes sense, but is this correct? $$S_n(f) = \mathcal{F}_\tau(R_n(\tau)), \hspace{30pt} \tau = t'-t$$ $$\text{E}\Big[n(t)\,n(t')\Big] = \text{E}\Big[n(t)\,n(t+\tau)\Big]$$ $$\mathcal{F}_\tau\left\{\text{E}\Big[n(t)\,n(t+\tau)\Big]\right\} = \text{E}\left[\mathcal{F}_\tau\Big\{n(t)\,n(t+\tau)\Big\}\right] = \text{E}\left[n(t)\,e^{i2\pi f t}\, \tilde{n}(t+\tau)\Big\}\right]$$ And the question is of course how to proceed now... $\endgroup$ Sep 5, 2022 at 15:22
  • $\begingroup$ Instead of just saying "a book" could you edit your question to properly cite the book -- name, author, edition (if not first), publisher, copyright date? And perhaps let us know what point the author is trying to make with their re-casting of the definitions? Sometimes it makes a ton of sense to recast something (fairly) ordinary into something oddball in order to illuminate some point -- knowing the point the author is trying to illuminate may help make sense of what they're saying. $\endgroup$
    – TimWescott
    Sep 5, 2022 at 16:21
  • $\begingroup$ @TimWescott I did. $\endgroup$ Sep 5, 2022 at 16:37

1 Answer 1


From Probability, Random Variables, and Stochastic Processes, A. Papoulis, McGraw-Hill 1984, p. 306:

If the process $\mathbf x(t)$ is WSS with power spectrum $S(\omega)$ then its transform $\mathbf X(\omega)$ is white noise with average intensity $2 \pi S(u) \delta (u - v)$

$$E\left \{ \mathbf X(u) \mathbf X^*(v) \right\} = 2 \pi S(u) \delta (u - v) \tag {10-153}.$$

(Don't let the $2\pi$ vs. $\frac 1 2$ scaling factors throw you -- different authors scale things differently. Each will have used consistent systems, so the scaling will work for the other stuff that author's doing.)

Probably any 4th-year introductory text in stochastic processes that covers the Fourier transforms of stochastic processes is going to cover this.

  • $\begingroup$ Thank you for your question. I had forgotten this tidbit (blush) -- if things go true to form, I'll run into an application for it and thank you again. $\endgroup$
    – TimWescott
    Sep 5, 2022 at 17:57
  • $\begingroup$ Thing is that white noise has infinite power. If your wide-sense stationary process has finite power, then it can't be white and it doesn't have any $\delta(f-f_0)$ or $\delta(\tau-\tau_0)$ in any of these inner products. $\endgroup$ Sep 6, 2022 at 2:38

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