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Possible Duplicate:
Is it possible to have a pair of Gaussian random variables for which the joint distribution is not Gaussian?

In the Wikipedia entry on the multivariate normal distribution, it says that one definition

is that a random vector is said to be k-variate normally distributed if every linear combination of its k components has a univariate normal distribution.

However, since it's also true that any linear combination of normally distributed variables is itself normal, does this mean that any vector of univariate random normals is itself multivariate normal? Is there ever any situation where a vector of random normals is not multivariate normal?

Update

What I should have said was "any linear combination of independent normally distributed variables is itself normal." All the answers below are good examples of variables that are not independent and thus not multivariate normal. So I should rephrase my question to be: is there ever any situation where a vector of independent random normals is not multivariate normal? I'm leaving the headline as is, to reflect history and the nature of the answers to this question, but I will alter it if you think I should. Sorry for any confusion I may have caused.

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marked as duplicate by mpiktas, cardinal, gung, whuber Sep 23 '12 at 21:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Yes, you can construct the example where the vector marginals are normal, but the vector itself is not. $\endgroup$ – mpiktas Sep 23 '12 at 6:24
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    $\begingroup$ See great pictures of this at stats.stackexchange.com/questions/30159/… $\endgroup$ – Douglas Zare Sep 23 '12 at 7:11
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    $\begingroup$ voting to close, as the question is exact duplicate to the question pointed out by Douglas. $\endgroup$ – mpiktas Sep 23 '12 at 12:43
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    $\begingroup$ sparc_spread, the conceptual error in your question comes at the beginning of the last parameter, i.e., it is not true that any linear combination of (marginally) normally distributed variables is itself normal. Second, the example given in the answer below corresponds to the top-right image at the link Douglas provides. $\endgroup$ – cardinal Sep 23 '12 at 15:18
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    $\begingroup$ A vector of independent normal random variables is always multivariate normal. One relatively straightforward way to see this is to look at the moment-generating function (or characteristic function). In fact, a common definition of multivariate normality is that $\mathbf X$ is multivariate normal if $\mathbf X = \mathbf A \mathbf Z + \mathbf b$ where $\mathbf Z$ is a vector of iid standard normals, $\mathbf A$ is any real-valued matrix and $\mathbf b$ is a real-valued vector. The Wikipedia article mentions this definition. $\endgroup$ – cardinal Sep 23 '12 at 20:45
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Does this mean that any vector of univariate random normals is itself multivariate normal?

No.

Is there ever any situation where a vector of random normals is not multivariate normal?

Yes.

Consider two independent standard normals, $X$ and $Y$. Now let $Z = |X| \cdot \mathrm{sign}(Y)$. Then, $Z$ is normal.

Consider either the vector $(X,Z)$ or $(Y,Z)$. The margins are normal, the vector is not multivariate normal. (e.g. consider that $Y$ and $Z$ must have the same sign and that $X$ and $Z$ must have the same absolute value)

It's basically a matter of constructing something that isn't a linear combination that's nevertheless normal, and you can get something that is not multivariate normal.

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    $\begingroup$ The answer is correct, but for the OP it would be at least get an idea why the vector $(X,Z)$ is not normal, or why $Z$ is normal. $\endgroup$ – mpiktas Sep 23 '12 at 12:41
  • $\begingroup$ Note that $(Y,Z)$ is shown in the top-right panel of the figure at the link that Douglas provided above. $\endgroup$ – cardinal Sep 23 '12 at 15:19

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