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Lets say that I have a vector $X$ for which the marginal distribution of each element $x_1,x_2,...$ is normal with variance = 1. Additionally, assume that $X$ is pairwise independent.

Does this prove that the covariance matrix of $X$ is the identity matrix? What is the difference in the distribution of $X$ if instead I assume that $X$ is mutually independent? Is the distribution any different?

Thanks,

Clark

Clark
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For many people, calling a vector a normal vector is equivalent to assuming that the random variables in question are jointly normal random variables. For others, such as yourself, that the marginal density of each random variable is a normal density is sufficient reason to call the vector a normal vector.

That being said, the covariance matrix doesn't care diddly-squat as to whether the random variables are normal or not. The pairwise independence of the (unit variance) random variables suffices to ensure that the covariance matrix is the identity matrix regardless of the actual multivariate distribution, and regardless of whether the random variables are mutually independent or just pairwise independent. And no, pairwise independence of random variables does not guarantee mutual independence of the random variables even if the random variables happen to be normal. An example of three pairwise independent standard normal random variables that are not mutually independent random variables can be found in the latter half of this answer of mine.

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Dilip Sarwate
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  • Really helpful. One other comment. So marginal normality and pairwise independence is not enough to prove multivariate normality (i.e., your example). Is marginal normality and mutual independence enough to prove multivariate normality? – Clark Nov 21 '15 at 00:45
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    @Clark Under the assumption of mutual independence, you can get $f(x_1, \ldots, x_p) = f_{X_1}(x_1) \cdots f_{X_p}(x_p)$, so that multivariate normality holds. – Zhanxiong Nov 21 '15 at 06:40
  • Yes, marginal normality and mutual independence give multivariate normality (with a diagonal covariance matrix). But be aware that there are several readers (and some askers) on this forum who do not include mutually independent normal random variables under the rubric of multivariate normal random variables. Perhaps they insist on the covariance matrix of a multivariate normal distribution having nonzero off-diagonal entries... – Dilip Sarwate Nov 21 '15 at 14:45