Suppose I have a random process $X(t)$ and I know the power spectral density of $X(t)$, $S_{XX}(f)$.

What can be said about the distribution of $Y(t) = \int_{t'=0}^T X(t') dt'$?

Bear in mind I have a physicists background and little formal knowledge on stochastic integration.

As a crude physicist example, if $X(t)$ is white noise with a flat power spectral density $S_{XX}(f) = \sigma^2$ then $Y(t)$ is like a Weiner process with $Y(t) \sim \mathcal{N}(0, \sigma^2t)$.

I'm curious if this observation can be generalized. If $S_{XX}(f)$ is not enough information to determine how $Y(t)$ is distributed then what additional information is needed?

  • $\begingroup$ What do exactly you mean by "distribution "? Are you talking about the amplitude statistics (distribution function or probability density function?) or the spectrum of $Y(t)$ ? $\endgroup$
    – Hilmar
    Commented Feb 15, 2022 at 17:06
  • $\begingroup$ @Hilmar by distribution of $Y(t)$ I mean probability distribution function. $\endgroup$
    – Jagerber48
    Commented Feb 15, 2022 at 17:08

1 Answer 1


Given the power spectral density (PSD) of a white noise process, you can not infer anything about the distribution of that waveform in time. For all white noise processes, the power spectral density is a constant and this is given by the autocorrelation function being an impulse, or equivalently the unit sample for discrete-time waveforms: any zero mean waveform with independent identically distributed samples (regardless of the distribution of the magnitude and phase of those samples) will have an autocorrelation function that is a unit sample scaled by the variance of the samples. The Fourier Transform of an impulse is 1 for all frequencies and the Discrete Fourier Transform of the unit sample function ($x[n] = 1$ for $n=0$, and 0 for all other n) is similarly 1 for all frequencies, and as further derived in other posts here and elsewhere the Fourier Transform of the Autocorrelation function is the PSD.

As an example, consider the samples $x[n]$ as samples of either a Gaussian Distributed Random White Noise process or a Uniformly Distributed White Noise process each having the same variance and distribution (Independent and Identically Distributed or I.I.D.): as long as each of these samples are uncorrelated, the autocorrelation function will be a unit sample scaled by the variance and therefore, given that the PSD is the Fourier Transform of the autocorrelation function, will have the same PSD regardless of the distribution. This is very similar to the fact that we are not able to infer anything about the distribution if we are only given the variance of the waveform.

Note too, interestingly how in the Discrete Fourier Transform $X(k)$ of the samples of an I.I.D. random variable in time will tend toward a Gaussian distributed in frequency regardless of the distribution in time, as each of the samples $X(k)$ are created by the sum of I.I.D. samples as given by the formula for the Discrete Fourier Transform. Regardless of the distribution of the magnitude of those samples in time, as long as they are uncorrelated and have a constant mean and standard deviation, then according to the Central Limit Theorem as a sum of the samples from an I.I.D random process, the result will tend toward Gaussian. In general for a continuous time white noise waveform extending to $t=\infty$, the Fourier Transform will equivalently be a white noise waveform given all samples on the Fourier Transform will be independent regardless of how closely spaced any two frequency samples are (which is not the case when the time domain waveform is windowed, which creates dependence on adjacent samples in the frequency domain).

This is also discussed in these other related posts:

Fourier Transform of a PSD and response of a PSD input

Effect of windowing on noise

  • $\begingroup$ What if I specify that $X(t)$ is stationary and Gaussian distributed at all times? Does that allows us to begin to say something about $Y(t)$? $\endgroup$
    – Jagerber48
    Commented Feb 16, 2022 at 0:26
  • $\begingroup$ Yes, if Y(t) is simply a time domain integration process, then the Laplace transform of time domain integration is $1/s$, or has a PSD that goes down as $1/f^2$ (-20 dB/decade) Is that what you are looking for? $\endgroup$ Commented Feb 16, 2022 at 0:33
  • $\begingroup$ Will $Y(t)$ be normally distributed in time? if so what will it's variance be over time? $\endgroup$
    – Jagerber48
    Commented Feb 16, 2022 at 0:41
  • $\begingroup$ Yes I believe so; it's variance will depend on the rate of integration: It's a low pass filter so depending on the gain of integration will set how much of that white noise gets removed (filtered). You can see this in a Bode plot for your specific case but will depend on your actual parameters. $\endgroup$ Commented Feb 16, 2022 at 0:44
  • $\begingroup$ Wow! Regarding the power spectrai density of white noise. (which you have averred elsewhere has constant value for all frequencies), you now say "....will both have a power spectrum that is Gaussian distributed, given the law of large probabilities and that each frequency is independent if t goes to infinity," so the power spectral density of a white noise process has a Gaussian shape now? not flat over the entire band? because of some laws of large numbers? or did you mean the central limit theorem? and somehow dependent on time going to infinity? $\endgroup$ Commented Feb 16, 2022 at 3:45

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