5
$\begingroup$

I don't have any basis on multivariate gaussian distributions and amid one exercise I was solving about a non-directly related topic, the following question came to my mind:

Let's say we have some random variables $X_1,\dots,X_n$ such that each $X_i$ follows a gaussian distribution of some kind. Furthermore, assume that $X_1,\dots,X_n$ are independent. Then, is it true that $(X_1,\dots,X_n)$ also follows a gaussian distribution?

Also, an aditional question: in the case that the result is true, is the same statement also true if we don't require that $X_1,\dots,X_n$ are independent?

I am not looking for an elaborate explanation of this facts, just a yes/no so that I can proceed with solving my exercise (again, multivariate gaussian distribution is not the main topic of the exercise and I don't have any basis on it).

$\endgroup$
2
  • $\begingroup$ Related: Can "doubt" sometimes mean "question"? $\endgroup$ Commented Nov 11, 2023 at 3:47
  • $\begingroup$ Nice topic @PeterMortensen . I am from Portugal and here doubt translate to "dúvida" which relates to something we're uncertain about, or something we're suspicious about. It's nice to learn English on a maths forum too! Thanks! $\endgroup$
    – xyz
    Commented Nov 11, 2023 at 11:25

2 Answers 2

3
$\begingroup$

Yes, if they are independent, then the distribution of the random vector is a multivariate Gaussian.

If they are not independent, it does not need to be multivariate Gaussian. For example, take $X_1 \sim N(0,1)$ and $X_2 = X_1$ if $|X_1| > c$ and $X_2 = -X_1$ otherwise. Then $X_1$ and $X_2$ are Gaussian, but not independent, and the vector $(X_1, X_2)$ is not Gaussian.

$\endgroup$
3
$\begingroup$

The set of Multivariate Gaussian distributions can be defined as the set of all distributions of $\mu + AZ$, where $\mu$ is a vector, $A$ is a matrix, and $Z$ is a vector of I.I.D. $N(0, 1)$ random variables. In your case, $X_i = \mu_i + \sigma_{i}Z_i$, so $X = \mu + \text{diag}(\sigma_1, \dots, \sigma_n)Z$ is multivariate Gaussian.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .