I have been told that for deterministic signals, it makes sense to look at their respective Fourier transforms/spectra.

For stochastic processes on the other hand, I am supposed to work with power spectral density in terms of qualitative analysis.



Because a stochastic process itself doesn't have a Fourier transform.

That's really all there is to it.

You can only transform signals (i.e. functions over a body isomorphic to $\mathbb R$, for example, functions of time). You can't transform a random variable whose individual realizations are such functions!

  • $\begingroup$ Muller You know you're stuff for sure so I believe you. But my question is then why do I have 20 or so books ( haven't read much of them yet ) on the spectral density in the time-series framework by Priestley, Granger, Brillinger, Parzen, Tukey, Percival and other giants. Clearly they are dealing with estimation of the spectral density given a stochastic process, so what's up with that ? Thanks. $\endgroup$ – mark leeds Mar 11 '18 at 2:48
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    $\begingroup$ I don't understand that question $\endgroup$ – Marcus Müller Mar 11 '18 at 9:29
  • $\begingroup$ Hi: Sorry for lack of clarity. What I meant was that I I think ( have to look again at one of the books ), when they discuss estimating the spectral densirty, they use some kind of version of the fouirier transform where the frequency ( they use $\lambda$ rather than $\pi$ ) is assumed to be a random variable with orthogonal increments. that assumption allows for the handling of the concept that the process is a stochastic process. $\endgroup$ – mark leeds Mar 11 '18 at 13:13
  • $\begingroup$ $\lambda$ instead of $\pi$? That sounds unlikely. Really, maybe hit the books again and then ask clearer, please? But yeah, spectral estimation is the estimation of parameters, amongst them frequency, from a stochastic signal. That makes the estimate they get a random variable, too. $\endgroup$ – Marcus Müller Mar 11 '18 at 13:16
  • $\begingroup$ Hi Marcus: I'll try to find something relevant and send a link. thanks. $\endgroup$ – mark leeds Mar 11 '18 at 17:59

As @Marcus Müller mentions a stochastic process does not have a Fourier transform.

From a practical point of view, assume you have $N$ realizations of a signal, either deterministic or stochastic. Then, in order to estimate the power spectral density, it is common practice to average the power spectral densities of the individual realizations. For both deterministic and stochastic signals this will result in a more accurate estimate of the power spectral density as $N$ increases.

On the other hand, if you try the same with the Fourier transforms of the individual realizations this will only work for deterministic signals. For stochastic signals the phase of the Fourier transform is random and averaging generally result in a spectrum converging to zero as $N$ increases.

  • $\begingroup$ No, your summation argument for Fourier transforms of the individual realizations doesn't hold in general. Nothing says the phase of your stochastic process is random. This is not the reason! $\endgroup$ – Marcus Müller Mar 11 '18 at 0:11
  • $\begingroup$ Also, no, for deterministic signals averaging will not increase accuracy. How should it? There's nothing inaccurate! $\endgroup$ – Marcus Müller Mar 11 '18 at 0:13
  • $\begingroup$ @MarcusMüller I cannot imagine a stochastic process where, when comparing individual realizations, this is not the case. Any examples? $\endgroup$ – user883521 Mar 11 '18 at 6:46
  • $\begingroup$ @MarcusMüller I should have formulated this more carefully. I think the above still holds when performing actual measurements; i.e. when measuring realizations of an actual deterministic signal in noise. As the measured realizations are finite and (generally) discrete it is then of course no longer correct to talk about the Fourier transform. Applying the DFT to “measured” realizations of a stochastic process typically result in the described behavior (at least I’m not familiar with any counter examples). Admittedly, this is probably not what the OP asked about. $\endgroup$ – user883521 Mar 11 '18 at 7:10
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    $\begingroup$ no, not necessarily in the way you mean "random". For example, receiving the same noisy signal (assuming perfect synchronization) and overlaying would have a perfectly coherently constructing average. $\endgroup$ – Marcus Müller Mar 11 '18 at 15:04

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