Suppose we have a noise process $V(t)$ which is the result of passing Gaussian white noise through a filter with frequency response function $H(\omega)$. Can we represent realizations of this process as a Fourier transform $$V(t) = \int \frac{d\omega}{2\pi} M(\omega) e^{i \phi(\omega)}$$ where $M(\omega)$ and $\phi(\omega)$ are random variables? If so, what are the statistics of $M(\omega)$ and $\phi(\omega)$?


I don't think it makes sense to talk about the Fourier transform of a (realization of a) random process, because it generally doesn't exist if no further restrictions are imposed. This problem is also discussed here.

What can be done is analyze the statistics of the discrete Fourier transform (DFT) and the discrete-time Fourier transform (DTFT) of a windowed version of a discrete-time stationary random process. This has been done in this paper.

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  • $\begingroup$ Indeed, the discrete case can be understood using elementary rules of statistics. Could you elaborate on your statement that the FT doesn't exist unless further restrictions are imposed? $\endgroup$ – DanielSank Jun 14 '15 at 16:06
  • $\begingroup$ @DanielSank: What I meant is that for the realization of a random process, especially if it's white, i.e. it has infinite power, there is no reason to believe that the conditions for the existence of the Fourier transform are met. In general they aren't (e.g. the function is not $L_1$, etc.). $\endgroup$ – Matt L. Jun 14 '15 at 20:00
  • $\begingroup$ Of course. Would this question make more sense if we ask about filtered white noise? I changed the question to ask about filtered white noise. I realize this "pulls the carpet" out from under your answer but I think, given your comments, that the filtered case is the only one worth discussing. $\endgroup$ – DanielSank Jun 14 '15 at 20:02
  • $\begingroup$ @DanielSank: I think that even in that case my original answer still holds, because also realizations of a filtered white noise process will generally not be $L_1$ or $L_2$, so their Fourier transform will not exist. I added a link to my answer where this problem is discussed. $\endgroup$ – Matt L. Jun 14 '15 at 20:23
  • $\begingroup$ Awesome. I actually discovered the construction shown in that book myself when trying to understand this issue. I suppose as you say it doesn't make sense to define a Fourier transform for a noise process. This leaves a funny feeling though, because I thought I'd heard that it's possible to think of white noise as a Fourier integral where the phases $\phi(\omega)$ are all independent and uniformly distributed. $\endgroup$ – DanielSank Jun 14 '15 at 20:35

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