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The following statement is given in Unsupervised Learning chapter of the book Elements of Statistical Learning.

Since the multivariate Gaussian distribution is determined by its second moments alone, it is the exception, and any Gaussian independent components can be determined only up to a rotation, as before.

I just want to know,

How is multivariate Gaussian distribution is determined by its second moments alone? And

how does any Gaussian independent components can be determined only up to a rotation?

This is not making any sense to me.

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    $\begingroup$ Good question - I am curious to see the answer - see here for some slides which seem to hint at the answers $\endgroup$
    – Xavier Bourret Sicotte
    Jun 15, 2018 at 12:06
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    $\begingroup$ Do they restrict to a centered multivariate Gaussian distribution ? The multivariate Gaussian distribution is determined by its mean and its variance-covariance matrix (= the second moments), this is well-known. For the second point, if you rotate a multivariate Gaussian with independent components around its mean, you don't change the distribution. $\endgroup$
    – Stéphane Laurent
    Jun 15, 2018 at 13:05
  • $\begingroup$ I think they so. But I am not clear from the text. You can go through this section in the book. It is availble for free. $\endgroup$
    – ironman
    Jun 15, 2018 at 13:24
  • $\begingroup$ The second question was answered for your question about non Gaussianality for ICA stats.stackexchange.com/questions/351294/… two Gaussians have a circular joint distribution, so there is no single rotation solution to optimise. I think the first is only strictly true for mean centred data, although after accounting for the largest source of variance it will the be true of any other smaller sources of variance. I'd need to think about it a bit more to make sure my intuition isn't off whack. $\endgroup$
    – ReneBt
    Jun 16, 2018 at 5:35
  • $\begingroup$ In many (if not most) texts, the multivariate Gaussian distribution is defined in terms of parameters that are obviously coincident with its first two moments (usually by writing its characteristic or moment generating function explicitly). It then is an easy step to show that independent and uncorrelated are equivalent conditions. The statement about "Gaussian independent components," as I interpret it, is false: it requires you to assume all covariances are equal as well. Perhaps some such limitation is implied by the context of this quotation. $\endgroup$
    – whuber ♦
    Jun 16, 2018 at 12:01

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