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Can we define a three dimensional multivariate Gaussian that is pairwise-independent but not collectively independent?

U.Rush
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Read literally, the answer to the OP's question

Can we define a three dimensional multivariate Gaussian that is pairwise-independent but not collectively independent?

is No. If $X, Y, Z$ enjoy a three-dimensional multivariate Gaussian (a.k.a. multivariate normal or MVN) distribution, then it is not possible for $X, Y, Z$ to be pairwise independent random variables but not mutually independent random variables. As pointed out by whuber, if $X, Y, Z$ are pairwise independent, then $$\operatorname{cov}(X,Y)=\operatorname{cov}(X,Z)=\operatorname{cov}(Y,Z)=0$$ so that the covariance matrix is a diagonal matrix (and so is its inverse a diagonal matrix). Consequently, the three-dimensional multivariate Gaussian density of $X, Y, Z$ factors into the product of the three marginal densities, which is the condition for mutual independence of $X, Y, Z$.

A slightly different question is whether it is possible for $X, Y, Z$ individually to be Gaussian random variables, for them to be pairwise independent, that is, for the pairs $(X,Y), (X,Z), (Y,Z)$ of random variables to be pairwise independent Gaussian random variables (and thus for each pair to have a bivariate Gaussian distribution), but for $X, Y, Z$ to not be independent Gaussian random variables or to enjoy a trivariate Gaussian distribution. The answer to this question is Yes and an example of three such standard Gaussian (or standard normal) random variables is given in my answer to an earlier question on a similar topic.

Dilip Sarwate
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    +1 It might help to make your initial reasoning a little plainer: to wit, pairwise independence implies *each* covariance is zero; and when *all* covariances are zero, the marginals are independent. – whuber Dec 28 '20 at 22:52
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    @whuber Thanks for the suggestion, which I have incorporated into my answer. – Dilip Sarwate Dec 28 '20 at 23:13