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Mar 27, 2019 at 12:09 comment added Arun @Confounded Gaussian random vectors have the property that if they are uncorrelated, they are independent. This is because when the cross terms of the covariance matrix is zero, the covariance matrix is block diagonal and hence the probability distribution factors into two parts, each being the pdfs of the subvectors
May 6, 2018 at 1:25 comment added Lella @Confounded $\mathbf{X}$ is transformed to $\mathbf{Y}$ such that the sub-vectors of $\mathbf{Y}$ are independent. The sub-vectors of $\mathbf{Y}$ being linear transformations of $\mathbf{X}$ are in turn multivariate normal. The sub-vectors of $\mathbf{Y}$ are independent since their covariance matrix is $\mathbf{O}$.
May 6, 2018 at 0:16 history edited Lella CC BY-SA 4.0
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May 5, 2018 at 20:40 comment added Confounded Thank you for your reply. I am not clear, however, on how you deduce that they are independent. As far as I can see you just showed that they are uncorrelated, but for this to imply independence, they would have to be jointly normal, which is not clear that they are (as I stated in the OP).
May 5, 2018 at 13:42 history answered Lella CC BY-SA 4.0