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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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sparc_spread
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    Yes, you can construct the example where the vector marginals are normal, but the vector itself is not. – mpiktas Sep 23 '12 at 06:24
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    See great pictures of this at http://stats.stackexchange.com/questions/30159/is-it-possible-to-have-a-pair-of-gaussian-random-variables-for-which-the-joint-d – Douglas Zare Sep 23 '12 at 07:11
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    voting to close, as the question is exact duplicate to the question pointed out by Douglas. – mpiktas Sep 23 '12 at 12:43
  • Apologies for not spotting that other one before - I did try to look. I do think the answers that are appearing her are different enough from the answers there to merit the survival of this question. However you're right this really is a duplicate and I totally accept however the voting turns out. – sparc_spread Sep 23 '12 at 13:34
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    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. – cardinal Sep 23 '12 at 15:18
  • I have also voted to close as an exact duplicate. (The answer here would likely be merged.) – cardinal Sep 23 '12 at 15:24
  • Somehow, *paragraph* became *parameter*! Muscle memory, I guess. Sorry about that. – cardinal Sep 23 '12 at 15:53
  • @cardinal thanks for posting re the error. This is actually a big misconception I've been operating under and I'm glad you pointed it out. I know it's true of two normals but I assumed I could generalize to `n` of them. Both your comment and the question that @Douglas linked to show that two normals is really just a special case. If you know of any additional discussions of linear transformations of a vector of more than two normals, I'd be really grateful for links. Either way, thanks again for clearing this up for me. – sparc_spread Sep 23 '12 at 16:38
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    Hi, sparc_spread. I think my comment was probably misinterpreted. Even in the case of two normals, as the answers at Douglas' link show and the answer to this question also shows, the linear combination of two or more (marginally) normal random variables is not necessarily normally distributed. If the entire vector is *multivariate* normal (bivariate being a subcase), then linear combinations of the components will be normal. – cardinal Sep 23 '12 at 17:00
  • Argh, I think my main misconception was forgetting that the variables need to be **independent** in order for their linear combinations to be always normal! You guys have been very patient with me, I appreciate it . . . – sparc_spread Sep 23 '12 at 18:41
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    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. – cardinal Sep 23 '12 at 20:45

1 Answers1

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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.

Glen_b
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    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. – mpiktas Sep 23 '12 at 12:41
  • Note that $(Y,Z)$ is shown in the top-right panel of the figure at the [link that Douglas provided](http://stats.stackexchange.com/questions/30159/is-it-possible-to-have-a-pair-of-gaussian-random-variables-for-which-the-joint-d) above. – cardinal Sep 23 '12 at 15:19