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It seems to me that if a random vector $X$ is to have a multivariate normal distribution, then it is necessary and sufficient that $X$ is a vector of independent Gaussians -- is this correct?

In my understanding, a random vector $X$ of length $k$ is said to have a multivariate normal distribution if for any constant vector $a \in \mathbb{R}^k$, the random variable $Y = a^TX$ has a univariate normal distribution.

Then $e_i^TX$ must be a Gaussian for the elementary basis vectors $e_i$, implying that each $X_i$ must be a Gaussian.

In the other direction, any linear combination of two (or more, by induction) Gaussians should produce another Gaussian.

hessian
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    In your argument where does independence come from? Jointly Gaussian random variables need not be independent. Fort example, if $X$ has normal distribution so does $(X,X)$ because $a^{T}(X,X)$ has Gaussian distribution for any $a$. – Kavi Rama Murthy Dec 03 '18 at 06:01
  • I see, thank you! – hessian Dec 03 '18 at 15:48

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