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However, I don't understand what the subscript '4x' or the parameter (0,0) stand for. Could anyone explain ?


For a stationary process $x$, moments and cumulants, when they exist, depend on relative time lags, hence $n-1$ variables when the moment or cumulant is of order $n$ (see for instance Basic Signal Processing and Statistics, p. 446 sq.):

$ \mathcal{M} \{ x(k),x(k+\tau_1),x(k+\tau_2),x(k+\tau_3) \} = \mathrm{E} \{ x(k)x(k+\tau_1)x(k+\tau_2)x(k+\tau_3) \}\,. $

A standard notation is $m_{4x}(\tau_1,\tau_2,\tau_3)$, where $x$ denotes the process, $4$ the $n$-th order, $\tau_1,\tau_2,\tau_3$ the lags, with a triple $0$ for the $0$-lag cumulants. The first $0$-lag cumulants are classicaly named variance, skewness and kurtosis. Your version seems a normalized kurtosis, which could be written, with the above notations: $$ \kappa_{4x}(0,0,0) = \frac{m_{4x}(0,0,0)}{m_{2x}^2(0)}\,. $$ I am quite surprised thant you only have two zeroes in your parenthesis.

On the origin side, kurtosis comes for a greek word meaning arched, curved, measuring the tailedness of a distribution.

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  • $\begingroup$ I'm computing the kurtosis for a wavelet sub band matrix. I'm directly using the fourth central moment and the standard deviation for this purpose. I'm supposed to modify my sub band matrix using a threshold on kurtosis. $\endgroup$ – Manish Ramesh Dec 26 '15 at 18:01
  • $\begingroup$ Strange: the normalized kurtosis is scale-invariant, while thresholds not in general. $\endgroup$ – Laurent Duval Dec 26 '15 at 20:14
  • $\begingroup$ Documentation on the thresholding process is really scant and it's kind of part of a bigger process. As a newbie, this is turning out to be hard. $\endgroup$ – Manish Ramesh Dec 26 '15 at 20:38
  • $\begingroup$ There are quantities of papers related to wavelet thresholding, I hope you will find hints in them. Was your initial kurtosis question answered properly? $\endgroup$ – Laurent Duval Dec 26 '15 at 21:14
  • $\begingroup$ Yes, I'm trying to. The initial question's been answered. Thank you very much. $\endgroup$ – Manish Ramesh Dec 27 '15 at 19:03

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