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

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 $$t_1,\ldots,t_n$$.

Proposition

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?

• 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... Sep 5, 2022 at 15:22
• 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. Sep 5, 2022 at 16:21
• @TimWescott I did. Sep 5, 2022 at 16:37

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.)
• 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. Sep 6, 2022 at 2:38