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Bivariate and multivariate distribution relationship.

If we have say 3 variables where any two variables follow a normal bivariate distribution, then does it necessarily follow a multivariate normal distribution?

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    $\begingroup$ Short answer: No. $\endgroup$
    – Glen_b
    Jan 7, 2014 at 9:20
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    $\begingroup$ @Andre The questions are clearly different. The question here is about a trivariate distribution where each of the bivariate marginals is bivariate normal. The suggested duplicate if about a bivariate with univariate normal margins. While both answers are negative, for much the same reasons, the questions are asking different things; this one is somewhat subtler than the other. $\endgroup$
    – Glen_b
    Sep 5, 2014 at 4:24

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Here's a counter-example:

Let $X$, $Y$, $Z$ be independent standard normal, and let $W = |Z|\cdot \text{sign}(XY)$.

Then $(W,X)$, $(W,Y)$ and $(X,Y)$ are bivariate normal, but $(W,X,Y)$ is not trivariate normal, since $WXY$ is never negative.

What's happening is that the trivariate distribution has been constructed so that probability is only in four of the eight octants, in such a way that each quadrant of the pairwise margins gets an octant with probability and an octant without.

To help visualize what's going on, see the following simulation:

 x=rnorm(1000)
 y=rnorm(1000)
 z=rnorm(1000)
 w=abs(z)*sign(x*y)

Here are the pairwise samples:
enter image description here

Here's the sample bivariate distribution of $X$ and $Y$ when $W$ is restricted to be positive:
enter image description here

(when $W$ is restricted to be negative, the $(X,Y)$ values are in the other two quadrants)

And here's a particular projection of the trivariate distribution; you should, for example, be able to make out that there's a low-density "gap" at the bottom.
enter image description here


That might seem like a somewhat artificial counterexample, but it's not an issue just with some odd edge-cases. More generally, the trivariate distribution may be quite different from trivariate normal, in any number of smooth or not-smooth ways. The same goes for more than three variables.

Copulas give us a way of constructing infinities of such counterexamples with various characteristics.

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No. In theory, exceptions like @Glen_b's answer may apply.

In practice, multivariate normality partly depends on how precisely your variables follow their normal uni/bivariate distributions. It is rare that any real distribution has exactly zero skewness and zero excess kurtosis, after all. Therefore, if you're inferring that all the bivariate distributions are normal because you fail to reject the null hypothesis of a multivariate normality test on each, the result of the same test for all three variables might just cross your rejection threshold if the bivariate results are all $p = .06$ and your $\alpha = .05$. This answer may not matter to you if you've already avoided such problems with significance tests of normality though, and have determined through better means that your variables are normally distributed (whether exactly, or close enough).

Even so, in practice, one may find that subsets of data exhibit problematic relationships like that in Glen_b's example. Acceptable deviation from bivariate normality, once mixed with even limited instances of such multivariate problems, may exceed your purpose's tolerance for non-normality, which is bound to appear to some extent in most real sample data.

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    $\begingroup$ I will undo my downvote when I can, but I still think your answer is irrelevant to what is a theoretical question. $\endgroup$
    – mark999
    Jan 7, 2014 at 10:13
  • $\begingroup$ @mark999: Thanks for clarifying! I think I see your point. I've edited to motivate my answer a bit, and distinguish it slightly from a purely theory-oriented answer. Purely theoretical questions are slightly irrelevant to real problems if they almost never apply. Exactly normal distributions rarely apply exactly to sample data, and analysts who don't specify how they determine whether a distribution is normal are sometimes (if not often) judging this through inappropriate use of significance tests. The OP does not specify how normality is determined, nor that this is a purely theoretical issue $\endgroup$ Jan 7, 2014 at 10:44
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    $\begingroup$ This is interesting commentary on the question. I am disturbed by the apparent suggestion that (a) the departure of real data from theoretical ideals implies (b) theoretical considerations are "slightly irrelevant." Exactly the opposite is true: theory is our touchstone and basis for understanding reality and so understanding its implications becomes especially relevant when reality departs from theoretical assumptions. Moreover, Glen_b's example is such a profound departure from multivariate normality that in most cases it surely would be an unacceptable deviation from multivariate normality. $\endgroup$
    – whuber
    Jan 7, 2014 at 15:21
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    $\begingroup$ Only purely theoretical questions, such as exactly normal distributions, and only slightly irrelevant if they almost never apply, in as much as some implications might not apply to less-than-perfectly-normal distributions...I certainly didn't mean to imply anything about Glen_b's answer there, and I admit I'm unconfident that I could think of an actual example. I think my intention was to imply that the question is more valuable if it is also considered in the context of real data, which may never be exactly normal, but variably close. $\endgroup$ Jan 7, 2014 at 15:55
  • $\begingroup$ @NickStauner I've undone the downvote (and sorry for not commenting initially) but I'm curious to know whether you would have taken the same approach had the question been "Is the sum of two normally distributed variables also normally distributed?". $\endgroup$
    – mark999
    Jan 7, 2014 at 19:59

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