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A book reports that ICA cannot be used if the independent components of the analyzed data are Gaussian (at most one can be Gaussian, but no other). However, in the same book, the following example is reported:

"A number of 1024 samples of a two-dimensional normal distribution was generated with mean μ and covariance matrix Σ. Similarly, 1024 samples from a second normal pdf were generated with the same covariance matrix and mean −μ. For the ICA, the method based on the second- and fourth-order cumulants, presented in this section, was used. The resulting transformation matrix W [...]"

The example continues using the ICA showing the indipendent component found. However, my question is: how was it possible to use ICA if the data comes from two normal/gaussian distributions (and not just one)?

UPDATE: The complete example is available here https://www.google.it/books/edition/Pattern_Recognition/gAGRCmp8Sp8C?hl=it&gbpv=1&dq=pattern+recognition&printsec=frontcover at pages 283,284

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If there are two different groups of data within each of which there is a Gaussian distribution, this amounts to having a mixture of two Gaussian distributions. A mixture of two Gaussian distributions is not a Gaussian (e.g., a single Gaussian has a single mode, most mixtures of two Gaussians have two modes). Therefore the non-Gaussian assumption of ICA is not violated.

Note by the way that it is not correct to say that ICA "cannot be applied" to Gaussian data. In fact it can, however it will be unstable and not be informative.

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  • $\begingroup$ Thank you for your last clarification! Indeed, I wondered why ICA is often reported as "not applicable" on Gaussian. $\endgroup$
    – volperossa
    Oct 18, 2022 at 12:42

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