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

For our upcoming exam we had to calculate the joint density of two normally distributed random variables a few times.

Say $X \sim N(0,1)$ and $Y \sim N(2,3)$.

We just assumed that $Z=(X,Y)^\top $ have a joint bivariate normal distribution $ Z \sim N\left( \left(\begin{array}{c} 0\\ 2 \end{array}\right) , \left(\begin{array}{cc} 1 & a\\ a & 3 \end{array}\right) \right) $ and calculated the only missing parameter, the covariance $a$ of the two random variables.

Now, wikipedia says that this is not always true and brings a counter example. Here is the image of the example's resulting plot.

My question: Under which conditions are they jointly normally distributed?

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Alexander Engelhardt
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    **Related**: http://stats.stackexchange.com/q/30159/2970 – cardinal Jul 30 '12 at 18:17
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    The condition is that *every* linear combination of $X$ and $Y$ is normally distributed. The wikipedia page includes this characterization under the [Definition](http://en.wikipedia.org/wiki/Multivariate_normal_distribution#Definition) subsection. – cardinal Jul 30 '12 at 18:20
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    Given $X\sim N(0,1)$ and $Y \sim N(2,3)$ and that the random variables are jointly normal, _how_ did you calculate the covariance $a$? There _must_ have been some other information provided to you that enabled you to calculate $a$ because given just that $X\sim N(0,1)$ and $Y \sim N(2,3)$ (whether jointly normal or not), $a$ can have _any_ value in $[\sqrt{3}, -\sqrt{3}]$. – Dilip Sarwate Jul 30 '12 at 18:31
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    Also of interest: [surprising characterizations of the Gaussian distribution](http://stats.stackexchange.com/questions/4364/) – whuber Jul 30 '12 at 18:47
  • Dilip, I didn't calculate $a$, and there normally is more information. I just made up a few quick numbers :) – Alexander Engelhardt Jul 30 '12 at 18:49
  • cardinal: Thanks for your answer. How would I show that every linear combination is normally distributed? Would I calculate the density of $a + bX + cY$? – Alexander Engelhardt Jul 30 '12 at 19:23
  • Alexx, that's a little tricky to do well, because linear combinations can be degenerate. The fastest, most elegant, rigorous demonstrations look at the cumulant generating function. – whuber Jul 30 '12 at 20:45
  • In addition to what cardinal and whuber say, I would say that when two variables have normal marginals and a normal copula, then they are a multivariate normal distribution. Other copulae could potentially descrie the distribution. – John Jul 31 '12 at 00:48

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