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We know that $X$, $Y$ are normal does not guarantee $(X, Y)$ is jointly normal. A typical example is: $X=Z$, and $Y=ZU$, where $Z$ standard normal, $P(U=1)=P(U=-1)=1/2$, and $Z, U$ are independent.

My question is: what would be a condition for $(X, Y)$ to be jointly normal given that both $X$, and $Y$ are normal?

Glen_b
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bankrip
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    Answers appear at http://stats.stackexchange.com/questions/4364. See, for instance, http://stats.stackexchange.com/a/4373 and http://stats.stackexchange.com/a/24591 (the Herschel-Maxwell theorem). – whuber Apr 10 '14 at 18:31

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Any Linear combination of $X$ and $Y$ is still normal would be a condition for $(X,Y)$ to be jointly normal. Actually, it is a characterization of bivariate normal and could be generalize to $\mathbb{R}^p$.

PS: The third link @whuber provided talks about the Herschel-Maxwell Theorem. And it seems to me that this can not serve as a characterization for general multivariate normal, but could serve as a characterization for standard multivariate normal.

sadapple
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  • I don't have enough reputation so it seems I can not comment directly in that link. – sadapple Apr 11 '14 at 05:59
  • Thanks a lot for the answer. I was confused about this too. Prompted by your answer, I searched a bit and found a proof: https://people.eecs.berkeley.edu/~ananth/223Spr07/jointlygaussian.pdf – syeh_106 Feb 11 '20 at 15:39
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I guess it depends what you're trying to do. In many cases, the simplest approach is just to make $X$ and $Y$ jointly normal by construction. E.g., suppose that you have independent standard Gaussians $U$, $V$, $W$, and then you set $$ X = a\,U + b\,V; \ Y = c\,U + d\,W. $$ In this case, it's easy to check that $X$ and $Y$ are jointly normal.

Stefan Wager
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