Jointly Gaussian? Given a joint probability density function of the two random variables$\ X_1,X_2 $ how can I determine if they are Gaussian and further, how can I say if they are jointly Gaussian?
Jointly Gaussian means that under any linear combination of $\ X_1,X_2 $ they shall remain Gaussian, but how can I use the joint pdf to determine this property?
$\ f_{x,y}(x,y)="something that looks Gaussian", x*y>=0 $ and zero otherwise. So my pdf is non-zero when both x and y and positive or both are negative.
 A: $X_1$ and $X_2$ being Gaussian just means that each of their individual (marginal) pdf has the form:
$$\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$
Being jointly Gaussian (or you can say $(X_1,X_2)$ is a Gaussian vector) is much more. There are two equivalent formulations:


*

*each linear combination of $X_1,X_2$ is Gaussian

*their joint distribution is a multivariate Gaussian (Normal) distribution


The mutlivariate Normal distribution is a generalization of one dimensional Normal distribution where instead of using a 1 variable degree 2 polynomial $\frac{(x-\mu)^2} {2\sigma^2}$ in the exp, you use a quadratic form:
\begin{align}
f_{\mathbf X}(x_1,\ldots,x_k)
& = \frac{\exp\left(-\frac 1 2 ({\mathbf x}-{\boldsymbol\mu})^\mathrm{T}{\boldsymbol\Sigma}^{-1}({\mathbf x}-{\boldsymbol\mu})\right)}{\sqrt{(2\pi)^k|\boldsymbol\Sigma|}}
\end{align}
You can read more on Wikipedia's Multivariate Normal Disbribution
There is no such formula in the special "degenerate" case where $X_1$ and $X_2$ are collinear. In this case, the pair $(X_1,X_2)$ has support on a line and thus has no density. Visually, it is when the usual "ellipse" because thinner and thinner until it's just a line. Actually a "point" is an even more degenerate case: both constant variables.
The first formulation is most often preferred because it also covers the "degenerate" cases. This text gives further definitions and proofs: Gaussian Random
Vectors.
