# 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.

• Are $X_1$ and $X_2$ independent? – user247327 Dec 1 '19 at 19:03
• No, they do not have to be independent. @user247327 – hom Dec 1 '19 at 19:03
• I am not sure what is meant by parameter $(0,1)$. And $X_1$, $X_2$ need not be independent for $(X_1, X_2)$ to be a Gaussian random vector (or a multivariate Gaussian normal distribution). In this case we would have mean vector $\mathsf \mu$ the means of $X_1$ and $X_2$ and covariance matrix $\mathsf \Sigma$ the variances of $X_1$ and $X_2$ on the diagonal and the covariance of $(X_1,X_2)$ on the antidiagonal. – Math1000 Dec 1 '19 at 19:05
• @Math1000 I supose "parameter $(0,1)$" means that the random variables have mean zero and variance 1. – saz Dec 1 '19 at 19:07
• @hom Take a look at this question – saz Dec 1 '19 at 19:10

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 $$\Pr(X_1+X_2=0) = 1/2.$$
• I think I am confused by your definition of $X_{2}$. So $X_{2}$ equals $X_{1}$ and $-X_{1}$ each with probability $1/2$? What do you mean by $X_{2} = -X_{1}$ independently of $X_{1}$? – hom Dec 2 '19 at 2:01
• @hom : The plus or minus sign to be put in front of $X_1$ is chosen randomly, and that choice is probabilistically independent of the value of $X_1.$ In effect, you toss a coin to decide whether $X_2$ will be equal to $X_1$ or to $-X_1. \qquad$ – Michael Hardy Dec 2 '19 at 2:07
• Ok thanks. What do you mean by "Consider that $P(X_1 + X_2) = 1/2$? Did you mean $P(X_1 + X_2 = 1/2)$? – hom Dec 2 '19 at 2:11
• @hom : Sorry --- I meant $\Pr(X_1+X_2=0) = 1/2. \qquad$ – Michael Hardy Dec 2 '19 at 20:07