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Are the marginal distributions of a multivariate distribution necessarily the corresponding univariate distributions?

For example:

  • Every marginal distribution of a multivariate normal distribution must be a univariate normal distribution.
  • So it is for the multinomial distribution (which is the multivariate generalization of binomial distribution).
  • So it is for Dirichlet distribution (which is the multivariate generalization of beta distribution)
  • Not sure for multivariate t distribution.

Those are all the multivariate distributions I have seen so far. I am not sure for other multivariate distributions I haven't seen either. So I wonder if it is true for all the multivariate distributions?

This may involve a related question:
How is a univariate distribution generalized to multivariate generally?

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  • $\begingroup$ Every marginal distribution of a multivariate uniform distribution is a univariate uniform distribution. $\endgroup$ – Dilip Sarwate Mar 23 '14 at 14:48
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    $\begingroup$ @DilipSarwate: Thanks! How do you understand that a uniform distribution within a 2-dim origin-centred circle doesn't have uniform marginal distribution? Or a multivariate distribution must be supported by a set like $[a,b]^K$? $\endgroup$ – Tim Mar 24 '14 at 0:15
  • $\begingroup$ I meant the support of the joint pdf must be $[a,b]^K$ and the joint pdf has constant value on the support. $\endgroup$ – Dilip Sarwate Mar 24 '14 at 1:45
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The notion of a multivariate-$<$*whatever*$>$ is not uniquely defined.

People assign the name of a univariate distribution to some corresponding multivariate distribution based on whatever features are most relevant/important to them. So there's not one multivariate gamma, but many (and there may well be more get the name in the future).

Usually one of the features people consider to be 'basic' to the multivariate form is retaining the distribution of the univariate as the distribution of the marginals, so it's nearly always the case that something called a multivariate-$<$*whatever*$>$ has univariate-$<$*whatever*$>$ margins. But there's no extant rule about how the multivariate distributions are named, it's just how people conventionally tend to anticipate the use of the term.

It's quite possible (and I imagine has likely happened several times) that some other features may be considered more basic on occasion, and quite likely there are a few cases where something called a multivariate-$<$*whatever*$>$ doesn't have univariate-$<$*whatever*$>$ margins - but it would be surprising enough that it would be expected that this unusual feature would be pointed out explicitly.

You can pretty safely assert that at the very least it would almost always be the case that it has univariate-$<$*whatever*$>$ margins if it's called a multivariate-$<$*whatever*$>$.

More specific answers would require a more specific question.


Edit to address the particular case of the multivariate-t

The term 'multivariate-t' is usually applied to the object obtained when a multivariate normal is divided by the square root of (an independent $\chi^2$ divided by its degrees of freedom), as described on the Wikipedia page.

Consequently, a margin will consist of the marginal normal divided by the square root of (an independent $\chi^2$ divided by its degrees of freedom). This implies that the univariate margins are t-distributed. [Indeed it implies that any $q-$dimensional subset of the variables should have a t-distributed margin.]


Edit: See also Kotz & Nadarajah Multivariate t-distributions and their applications, p15-16

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  • $\begingroup$ thanks, Glen_b! My question is both for general, and for specific (i.e. counterexamples) $\endgroup$ – Tim Mar 23 '14 at 23:37
  • $\begingroup$ Off the top of my head, I have no counterexamples, but if I think of one I'll post it here. $\endgroup$ – Glen_b Mar 23 '14 at 23:39
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    $\begingroup$ Also as mentioned in my post, I am not sure if multivariate t distribution satisfies it or not. $\endgroup$ – Tim Mar 23 '14 at 23:40
  • $\begingroup$ Precisely. I didn't see its marginal distribution in Wikipedia. $\endgroup$ – Tim Mar 23 '14 at 23:46
  • $\begingroup$ I've updated my answer - which is the argument from the comments I gave earlier. It is not invalidated by the concern I raised. $\endgroup$ – Glen_b Mar 24 '14 at 4:24

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