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I have a question that probably will sound stupid, but I am approaching myself to statistic concepts. I saw that the multivariate standard normal has a distribution characterized by the parameters $0$ (that is, a vector of 0s) and $I$ as a matrix of variance-covariance. I understand the fact that each marginal has mean 0 and variance 1 (because they are standard normal) but why they are uncorrelated? I saw that 2 standard normal can have an association. Is there any way to prove it?

Fynn Heinrich
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    They are uncorrelated because here the identity matrix is the covariance matrix whose $(i,j)$th entry denotes the covariance $\mathbb{C}\text{ov}(X_i,X_j)$. In fact for multivariate normals, they are independent not just uncorrelated, a fact that is an exercise in most mathematical statistics books. – Yashaswi Mohanty Mar 07 '23 at 21:07
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    Because "multinomial" and "multivariate" have very different meanings, may we presume you are asking about a multivariate standard Normal distribution? As far as your question goes, if by "uncorrelated" you mean "have zero correlation coefficient," then observe that the correlation coefficients are proportional to the off-diagonal entries in the variance-covariance matrix (by their very definition), which in the matrix $I$ are all zeros (by the definition of $I$). Do you perhaps mean something a little different by "uncorrelated," then? – whuber Mar 07 '23 at 22:21

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It follows pretty directly from the various definitions of the terms.

The off-diagonal elements of the variance-covariance matrix are the covariances and the diagonal elements are variances. When the diagonal elements are all 1's, the off-diagonal elements are then correlations (see the definition of correlation in terms of covariance and variances.)

With an identity variance-covariance matrix, $I$, the diagonals are indeed 1, and the off-diagonals (which are all 0) will then be correlations. i.e. if the variance-covariance matrix is $I$, all the variables are uncorrelated. With a standard multivariate normal, the variance-covariance matrix is defined to be $I$. That's the whole thing, in effect, just looking at two definitions.

If any off the off diagonal elements were not-zero, then the distribution will be multivariate normal, but the variance-covariance matrix won't be $I$ and then it's not multivariate standard normal. Further, if all the off-diagonals are 0 but not all the diagonals are 1, then the variance-covariance matrix is again not $I$ (it is multivariate normal and uncorrelated but the margins are not standard normal) and it's therefore not multivariate standard normal.

[It's also possible to have a distribution where the margins are standard normal and the variables are all uncorrelated but the distribution is not multivariate normal -- that is, to have the variables related by some non-Gaussian copula that leaves the variables uncorrelated but not independent.]

Glen_b
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