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The computation of the mean, variance, skewness and kurtosis of a time series in the time domain is straightforward from their definition formulae.

I need these statistic values in different frequency bands. Currently I apply different bandpass filters to the signal and repeat to compute the statistics. I'm curious if it is possible to compute them from the spectrum in the frequency domain?

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    $\begingroup$ Well I know off hand that the first two can be calculated, but I am not personally certain without researching regarding the last two. The mean is the DC component of the signal in the frequency domain, and the variance of the signal in the frequency domain is equal to the variance of the signal in the time domain (as given by Parseval's theorem). $\endgroup$
    – Dan Boschen
    Jun 27, 2017 at 1:34
  • $\begingroup$ The DC value at f = 0 Hz is one single value, only for the full signal. Seems it's impossible to get the mean values of different frequencies from the spectrum. $\endgroup$
    – wsdzbm
    Jun 27, 2017 at 1:37
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    $\begingroup$ But I see your new edit now. What does the mean value in a different frequency band signify to you? The average of what? The average power over that band of frequencies? Or something else? You should probably update your question to be more specific as to what units you want these statistics for in any given band. $\endgroup$
    – Dan Boschen
    Jun 27, 2017 at 1:42
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    $\begingroup$ Oh, correct for the DC. The mean value is zero for all bandpassed signals. $\endgroup$
    – wsdzbm
    Jun 27, 2017 at 1:53
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    $\begingroup$ Then yes, at least for variance (power) the power in the frequency bands of interest is equal to the power in the time domain signal if you passed that signal through a filter and then used your time domain approach to computing variance (given Parseval's theorem). $\endgroup$
    – Dan Boschen
    Jun 27, 2017 at 2:13

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If your time series $x(t)=s(t)+n(t)$ where $s(t)$ is a deterministic signal,and $n(t)$ is independent Gaussian noise, it has a pdf of the form $$ p(x(t))=\frac{1}{\sqrt{2 \pi} \sigma} \exp\left( -\frac{1}{2} \left(\frac{x(t) - s(t)}{\sigma}\right)^2\right) $$ the first moment is $$ E\left\{ x(t) \right\} = s(t) $$ which is NOT the DC component of the Fourier Transform of a sample. The first moment is not a constant and the time series is not WSS, and the first moment is not equal to the time average.

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  • $\begingroup$ well, this is a response to the comments, not an answer to the Q $\endgroup$
    – wsdzbm
    Jun 28, 2017 at 13:26
  • $\begingroup$ How it not an answer to the question? $\endgroup$
    – user28715
    Jun 28, 2017 at 14:09

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