Let $X:=\left(X_1,... ,X_n\right)\sim N_n(\mu,\Sigma)$, $\mu\in\mathbb{R}^n$, $\Sigma\in\mathbb{R}^{n\times n}$ symmetric and positive semi-definite and $\bar{X}:=\frac{1}{n}\sum_{i=1}^n X_i$ as usual. Is now $Z:=\left(X_1,... ,X_n,\bar{X}\right)$ multivariate normal distributed?

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    $\begingroup$ The question has been answered by Narut, but you should be aware that $Z$ will be degenerate even if $X$ isn't. $\endgroup$ – Andreas Blass May 16 '13 at 21:26

Yes. One possible definition of multivariate normal vector is that every linear combination of its components is (univariate) normally distributed. Since $X$ is multivariate, every linear combination of $(X_1, ..., X_n)$ is normally distributed. But every linear combination of components in $Z$ is also a linear combination of $(X_1, ..., X_n)$ because $\bar{X}$ is linearly dependent on $(X_1, ..., X_n)$. Thus, $Z$ is also multivariate normal.

(There are a few definitions of multivariate normal. They're equivalent: http://en.wikipedia.org/wiki/Multivariate_normal_distribution )


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