# Degenerate distribution

If $X \, \sim \, \mathcal{N}(m,\sigma^{2})$, I know that $\displaystyle \begin{bmatrix} X \\ X \end{bmatrix}$ is not a Gaussian vector since its entries are not independent. However, what can we say about the distribution of the random vector $\displaystyle \begin{bmatrix} X \\ X \end{bmatrix}$ ? I believe it is a degenerate (in the sense that it does not have a density with respect to the Lebesgue measure on $\mathbb{R}^{2}$) Gaussian distribution (intuitively, it should follow '' a Gaussian distribution on $\mathrm{Span}\Big( \begin{bmatrix} 1 \\ 1 \end{bmatrix} \Big) \subset \mathbb{R}^{2}$ ). Is that right ?

• Well, a Gaussian with non-trivial covariance matrix will have correlated (and hence not independent) co-ordinates. In particular "is not a Gaussian vector since its entries are not independent" is a non-sequitur.. – P.Windridge Apr 17 '15 at 11:30
• It is probably more correct to say it is not Gaussian because there exist a linear combination of the two entries which does not follow a Gaussian distribution, namely the first entry minus the second entry. – jibounet Apr 17 '15 at 11:37
• $(X,X)$ is Gaussian but degenerate. – Xi'an Apr 17 '15 at 12:13

What you say is right: $(X,X)$ follows a Gaussian distribution on the linear subspace you indicated, and it does have a density with respect to Lebesgue measure on that subspace, but not with respect to Lebesgue measure in the plane.