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I always thought a gaussian vector and a multivariate gaussian distribution were more or less the same thing but I've remembered that a gaussian vector have a more complex definition than that.

A gaussian vector is a vector such that every linear combination of its coefficients follows a gaussian distribution.

If the coefficients of a vector follow a multivariate distribution, the vector should be gaussian.

For the converse, each coefficient of a gaussian vector (as a trivial linear combination) should follow a gaussian distribution. The only difference I can see is about an assumption on correlation structure. A gaussian vector does not imply directly a linear correlation structure. But I feel that the condition "EVERY linear combination..." is strong enough so that we can't have the correlation structure we want.

Is the converse true? If so what is the difference between the two?

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The two conditions (or definitions if you prefer) are equivalent

A random vector $x = (X_1, \ldots, X_k)'$ is said to have the multivariate normal distribution if it satisfies the following equivalent conditions.

  • Every linear combination of its components $Y = a_1 X_1 + \ldots + a_k X_k$ is normally distributed. That is, for any constant vector $a \in R^k$, the random variable $Y = a′x$ has a univariate normal distribution, where a univariate normal distribution with zero variance is a point mass on its mean.

...

Wikipeida is the source.

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  • $\begingroup$ So what is the point of introducing two terms and definitions that do not always match. I feel like there may be a problem on a very particular case (like infinite dimension...) $\endgroup$
    – lcrmorin
    Sep 25, 2015 at 15:08
  • $\begingroup$ Sorry, I must have been unclear. They always DO match. You can either specify the pdf explicitly or declare that all linear combinations of the vector components are univariate normal, and you get the same class of objects. Both characterizations are useful in different situations. $\endgroup$
    – Matthew Drury
    Sep 25, 2015 at 15:12
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    $\begingroup$ No, they do not always match. A multivariate normal distribution defined via the density excludes the case of singular normal distributions which is the case when the covariance matrix is singular. Then the density do not exist, but the distribution is still multinormal according to the definition using "all linear combinations ...". Then, with probability one, the random vector lies in some affine subspace, (and a density exists, with respect to Lebegue measure in that subspace, but not in the full space). $\endgroup$
    – kjetil b halvorsen ♦
    Sep 25, 2015 at 15:50
  • $\begingroup$ Is this not properly covered by the "where a univariate normal distribution with zero variance is a point mass on its mean."? Is there no way to specify a cdf using distributions (in the measure theory sense) in place of a function? $\endgroup$
    – Matthew Drury
    Sep 25, 2015 at 15:57
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Just to add Matthew's answer and give you some intuition. The covariance matrix of a multivariate normal distribution is symmetric positive definite. That means you can take the singular-value-decomposition of this matrix, and change coordinates so that the coordinates are independent normally distributed random variables. A linear combination of the original normal coordinates can be rewritten as a linear combination of these independent normal coordinates and will therefore be normal.

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  • $\begingroup$ I think that the covariance matrix need only be positive semi-definite (that is, the quadratic form can evaluate to $0$ instead of always having strictly positive value) unless the phrase "positive definite" allows for $0$ values and one uses "positive definite" and "strictly positive definite" for what some (including myself) would call positive semi-definite and positive definite respectively. $\endgroup$
    – Dilip Sarwate
    Sep 25, 2015 at 18:21
  • $\begingroup$ I tend not to focus too much on the degenerate case of the multivariate normal, because I don't think it brings anything interesting to the table. That being said, for what I said, we only need a symmetric matrix. $\endgroup$
    – jlimahaverford
    Sep 25, 2015 at 18:31

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