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This I found on google while I was going through the Independent Component Analysis in unsupervised learning.

Let x = As where A is the Mixing Matrix.

So, Lets assume that s here is Gaussian and A is an orthogonal matrix. Then x will become an orthogonal transformation of the Gaussian s.Implying that x is also Gaussian.This is due to the isotopic behavior of the Gaussian s which has the same distribution in any orthonormal basis.

when x is Gaussian, due to the symmetric nature of the distribution , we cannot figure out the Nature of the independent components. As the projections will not give us a good degree of information.

Can anyone explain me the following statement given in the above statements?

when x is Gaussian, due to the symmetric nature of the distribution , we cannot figure out the Nature of the independent components. As the projections will not give us a good degree of information.

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  • $\begingroup$ Btw, you should tag these questions with self study $\endgroup$ – ReneBt Jun 16 '18 at 5:36
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    $\begingroup$ And this comes back to the circular joint distribution having no unique rotation solution. $\endgroup$ – ReneBt Jun 16 '18 at 5:38
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    $\begingroup$ This question is answered here stats.stackexchange.com/questions/97043/… $\endgroup$ – jpmuc Jun 16 '18 at 7:41
  • $\begingroup$ @jpmuc Thank you sir, but I have one more doubt in statement "x is Gaussian, due to the symmetric nature of the distribution , we cannot figure out the Nature of the independent components". Would you clear this with geometric figure. As I am not able to visualize it. $\endgroup$ – ironman Jun 16 '18 at 7:50
  • $\begingroup$ Gaussian distributions are completely defined by its mean and covariance matrix. If you centered the data, the covariance matrix (which is symmetric and defined by the 2nd order correlations) suffice. And that is the source of ambiguity. For ICA it means that it cannot solve X = AS, as explained in that question. Recall that both, A and S are unknown. If you only know about correlations, it is not possible to solve it. Please have a look at these slides: wwwf.imperial.ac.uk/~nsjones/TalkSlides/HyvarinenSlides.pdf. They contain some illustrations of the problem. $\endgroup$ – jpmuc Jun 16 '18 at 8:20

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