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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 ?

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    $\begingroup$ 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.. $\endgroup$ – P.Windridge Apr 17 '15 at 11:30
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    $\begingroup$ 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. $\endgroup$ – jibounet Apr 17 '15 at 11:37
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    $\begingroup$ $(X,X)$ is Gaussian but degenerate. $\endgroup$ – Xi'an Apr 17 '15 at 12:13
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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.

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