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I have K normal r.v. $Z_1,\cdots,Z_K$ but they are correlated with each other. Which condition should be added for them to be joint multivariate distributed?

Richard Hardy
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Gracie
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1 Answers1

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The condition that must be added is (wait for it!)

$Z_1, Z_2, \ldots, Z_K$ enjoy a multivariate normal distribution (also called a jointly normal or jointly Gaussian distribution) with mean vector $\big[E[Z_1], E[Z_2], \ldots, E[Z_K]\big]$ and $K\times K$ covariance matrix $R$ whose $(i,j)$-th element is $\operatorname{cov}(Z_j, Z_j)$, $1 \leq i \leq K, 1 \leq j \leq K$.

Note that both $\big[E[Z_1], E[Z_2], \ldots, E[Z_K]\big]$ and $R$ are known to you; in fact, they exist regardless of whether the random variables are normal or not, as long as all the random variables have finite variance. What is being added is the insistence the random variables are jointly normal, that is, the "condition that must added for them to have a multivariate normal distribution" is just that they have a multivariate normal distribution!

That's it. The joint normality implies that the random variables are marginally normal with the specified means and variances, and the covariance matrix is the same as that which you already know -- it is implicitly specified in the phrase "they are correlated with each other".

Any other formulation of the extra condition (e.g., "every linear combination of the $Z_i$'s is a normal random variable") is essentially equivalent to the above. Note that $Z_1, Z_2, \ldots, Z_K$ don't have a multivariate normal density function when $R$ is a singular matrix, but they nonetheless do enjoy a multivariate normal distribution function as stated above.

See also this answer of mine from a few years ago.

Dilip Sarwate
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  • Thanks a lot for your clarification. According to your answer, is that true that if $R$ is defined above, then the joint distribution of $Z_1,\cdots,Z)K$ must be a multivariate normal? No exceptions? Because I know that if we have 2 normal random variables, their joint distribution may not be a bivariate normal one. – Gracie Apr 22 '22 at 14:12
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    This looks perfectly circular rather than constructive. See our thread at https://stats.stackexchange.com/questions/4364 for some interesting and helpful answers. – whuber Apr 22 '22 at 19:07
  • @Gracie No, you are missing the whole point of my answer. Random variables with finite variance have finite means and finite covariances regardless of whether the random variables are normal or not. Even if the random variables are normal snd so have a mean vector and covariance matrix, they need not be jointly normal and there is no additional condition that must be satisfied in order to have joint normality other than the obvious: they must have a jointly normal distribution – Dilip Sarwate Apr 22 '22 at 19:08
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    Dilip, about "no additional condition:" that's simply incorrect. There are plenty of interesting characterizations of multivariate normality that aren't so tautological. Perhaps the best well known is that the univariate normality of all linear combinations implies multivariate normality. – whuber Apr 22 '22 at 19:09
  • @whuber I do mention the "all linear combinations" in my answer (though not in my comment to Gracie). The thread that you mention (questions/4364) is, in my opinion, primarily concerned with characterizations of univariate normality other than the standard one via the pdf. – Dilip Sarwate Apr 22 '22 at 19:17
  • Yes, you do mention it, only to dismiss it as "essentially equivalent." I would be interested to see the extremely short obvious proof of that! One direction is trivial, but what about the other one? The simplest I can think of requires the machinery of cumulant generating functions. – whuber Apr 22 '22 at 19:22
  • If you found this answer helpful, then please consider upvoting and/or accepting it. – kjetil b halvorsen Apr 25 '22 at 13:52