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Assuming a distribution on $\mathbb{R}^n$ has a density function $$f(x) = \frac{1}{\alpha} \exp(g(x))$$

I wonder what is the condition for the distribution to be a multivariate normal distribution?

Some books says the distribution is a multivariate normal distribution, if and only if $g(x)$ is a quadratic form of $x$.

  1. But as far as I know, a quadratic form is defined as a homogeneous polynomial of degree $2$. For a multivariate normal distribution, its density function, when exists (note that a multivariate normal distribution has a density function, if and only if its covariance matrix is nonsingular), may allow $g(x) = \sum_{ij} a_{ij} x_i x_j + \sum_{i} b_ix_i +c$, which is not a homogeneous polynomial in $x_i$'s.

  2. when $g(x) = \sum_{ij} a_{ij} x_i x_j + \sum_{i} b_ix_i +c$, it can always (?) be written as $g(x) = (x-\mu)^T A (x-\mu) + d$. Since the distribution is a multivariate normal distribution and has a density function, doesn't $A$ need to be nonsingular? So the distribution is multivariate normal, if and only if $A$ is nonsingular?


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up vote 1 down vote accepted

You can easily convince yourself that $A$ has to be negative definite in order that $f(x)$ is in fact a probability density. If $A$ where to have a single vanishing or positive eigenvalue then the distribution is not normalizable.

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+1 Thanks! (1) Does "the distribution is not normalizable" mean the distribution is not multivariate normal? What does "normalizable" mean? (2) $A$ has to be negative definite, because $-A^{-1}$ must be the covariance matrix (which is positive semidefinite) and nonsingular? (3) If $A$ has a zero or positive eigenvalue, the distribution is not multivariate normal. But is it still a probability distribution, given that its density function has a similar form as the density of a multivariate distribution? If yes, what name is this kind of distributions? – Tim Nov 22 '12 at 3:21
(4) would you call $g(x) = \sum_{ij} a_{ij} x_i x_j + \sum_{i} b_ix_i +c$ a quadratic form of $x$? – Tim Nov 22 '12 at 3:25
@Tim: with the wording "not normalizable" I just wanted to express the fact that you cannot find $\alpha$ such that $\int\!dx\,f(x)=1$ (which should answer (1)-(3) -> if $A$ is not negative definite $f(x)$ is not a probability distribution). To (4) no, I would not call this a quadratic form but a quadratic polynomial. – Fabian Nov 22 '12 at 3:25
Thanks! (3) Why if A is not negative definite, $∫dx \exp(g(x))$ is unbounded? – Tim Nov 22 '12 at 3:41
@Tim: if $A$ has a positive eigenvalue then the integrand $\exp(g(x))$ is unbounded in the direction of the corresponding eigenvector. – Fabian Nov 22 '12 at 6:19

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