Q:

If $X_1, X_2$ are Gaussian, is $(X_1, X_2)$ necessarily a Gaussian vector?

I recently learned about Gaussian random vectors and am not so sure about my answer to this question.

Let $X_1$ and $X_2$ be Gaussian random variables. Does this imply that
  $(X_1, X_2)$ is a Gaussian random vector?

For your reference, here's my definition of Gaussian:

A random vector $X = (X_1, X_2, \ldots X_n)$ on $(\Omega, \mathcal{F}, P)$ is called Gaussian if there is a vector $\xi = (\xi_1, \xi_2, \ldots, \xi_n)$ of independent Gaussian random variables with parameters $(0,1 )$ which may be defined on a different probability space $(\tilde{\Omega}, \tilde{\mathcal{F}}, \tilde{P})$, an $n\times n$ matrix $A$, and a vector $a = (a_1, \ldots, a_n)$ such that the vectors $X$ and $A\xi + a$ have the same distribution. 

This definition is really confusing to me. I guess it doesn't make so much sense intuitively. 
So to show that $(X_1, X_2)$ is Gaussian, I need another vector of Gaussian parameters such that the condition above holds. I think that the answer is yes, $(X_1, X_2)$ is a Gaussian random vector, but I haven't been able to construct these quantities in the general case. I would really appreciate any help.

A:

No. For example, suppose $X_1\sim\operatorname N(0,1)$ and
$$
X_2 = \begin{cases} \phantom{-}X_1 & \text{each with probability }1/2 \\ -X_1 & \text{independently of } X_1. \end{cases} 
$$
Then each of $X_1,X_2$ is normally distributed, but $(X_1,X_2)$ is not jointly normally distributed. To see that this pair is not jointly normal, consider that $\Pr(X_1+X_2=0) = 1/2.$

