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.

  • $\begingroup$ Are $X_1$ and $X_2$ independent? $\endgroup$ – user247327 Dec 1 '19 at 19:03
  • $\begingroup$ No, they do not have to be independent. @user247327 $\endgroup$ – hom Dec 1 '19 at 19:03
  • $\begingroup$ 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. $\endgroup$ – Math1000 Dec 1 '19 at 19:05
  • $\begingroup$ @Math1000 I supose "parameter $(0,1)$" means that the random variables have mean zero and variance 1. $\endgroup$ – saz Dec 1 '19 at 19:07
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    $\begingroup$ @hom Take a look at this question $\endgroup$ – 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 $\Pr(X_1+X_2=0) = 1/2.$

  • $\begingroup$ 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 $X_{1}$? $\endgroup$ – hom Dec 2 '19 at 2:01
  • $\begingroup$ @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 $ $\endgroup$ – Michael Hardy Dec 2 '19 at 2:07
  • $\begingroup$ Ok thanks. What do you mean by "Consider that $P(X_1 + X_2) = 1/2$? Did you mean $P(X_1 + X_2 = 1/2)$? $\endgroup$ – hom Dec 2 '19 at 2:11
  • $\begingroup$ @hom : Sorry --- I meant $\Pr(X_1+X_2=0) = 1/2. \qquad$ $\endgroup$ – Michael Hardy Dec 2 '19 at 20:07
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    $\begingroup$ Thanks. That's what I thought but just wanted to be sure. @Michael Hardy $\endgroup$ – hom Dec 3 '19 at 16:34

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