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For some given signals, I have these parameters:

  • Mean
  • Standard Deviation
  • Skewness
  • Kurtosis
  • Length (number of samples)

Now I would like to know if I can rebuild the signal (an estimation) based on these parameters. (If the solution can be implemented in MATLAB, I will appreciate it even more, but that is not necessary)

Thanks.

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    $\begingroup$ The only thing you can do with those properties is to draw random numbers having a distribution with the said properties. In other words, you can reconstruct noise with a given distribution. Perhaps you know something more about the signal you want reconstruct. Add that information to the question. $\endgroup$
    – mmh
    Commented Jul 15, 2015 at 20:33
  • $\begingroup$ That is what I have for my signal, but I do not know how leverage on Skewness and Kurtosis. Rebuilding a set of random numbers with such distribution is just fine. $\endgroup$
    – Matin Kh
    Commented Jul 16, 2015 at 15:49

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As mmh commented, you cannot reconstruct the exact signal with these numbers. You can generate data that are distributed such that they have these moments, but the distribution is not uniquely determined by them; that would only be the case if you had all moments of arbitrarily high order (and even in this case not necessarily, see comment by Glen_b below). Moreover, these moments do not provide any information about the temporal structure (autocorrelation, spectrum) of the signal.

If this is not a problem, you could opt to generate white noise using some distribution that matches the moments. A reasonable choice might be to use a distribution from the Pearson family. The Matlab function pearsrnd from the Statistics and Machine Learning Toolbox provides a way to do that.

I would like to stress though that this provides in no way an estimate of the signal; rather it produces realizations of a stochastic process that shares some properties with your given signal.

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    $\begingroup$ +1,.. However, even if you have moments of every (positive integer) order, you don't necessarily pin down the distribution. There are distinct distributions that share all their moments. $\endgroup$
    – Glen_b
    Commented Jul 17, 2015 at 10:07
  • $\begingroup$ @Glen_b, I didn't know that, thanks. What kind of distributions are that? And doesn't this contradict the method of moments as a tool to prove convergence in distribution? $\endgroup$
    – A. Donda
    Commented Jul 17, 2015 at 14:33
  • $\begingroup$ Ah OK, I found some stuff about Carleman's condition & Co. I'd still be curious about a concrete example of a distribution that is not uniquely determined by its moments, if you happen to have one. $\endgroup$
    – A. Donda
    Commented Jul 17, 2015 at 14:43
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    $\begingroup$ It can happen with distributions for which the MGF doesn't exist; it's possible for all moments to exist but for the MGF not to converge in a neighborhood of 0. In that case, two different distributions can share a complete sequence of moments. It doesn't contradict method of moments at all. There are some examples in answers on site here. $\endgroup$
    – Glen_b
    Commented Jul 18, 2015 at 1:39
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    $\begingroup$ Just tried it, and it absolutely did help. Thank you. $\endgroup$
    – Matin Kh
    Commented Jul 23, 2015 at 18:35

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