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I read that if you have a quadratic $f(x) = ax^2+bx+c$ and providing the leading coefficient $a < 0$, then $e^{f(x)}$ is the pdf of a normal distribution with mean $\mu = -\frac{b}{2a}$ and $\sigma^2 = -\frac{1}{2a}$.

Does this generalise to $e^{f(\mathbf{x})}$ defining a multivariate normal distribution where $f(\mathbf{x}) = \mathbf{x}'\mathbf{A}\mathbf{x} + \mathbf{x}'\mathbf{b} + \mathbf{c}$. If so what are the conditions for this to be true and what is the mean and covariance?


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Just a little clarification, $e^{f(x)}$ is proportional to the normal pdf. – user10525 Jul 5 '12 at 14:37
All these questions are answered near the top of the Wikipedia article on the multivariate normal distribution. – whuber Jul 5 '12 at 15:39
Thanks for the link – NapierL Jul 5 '12 at 16:02
up vote -1 down vote accepted

The result in 1 dimension that you gave above comes about by equating exponents namely ax$^2$ +bx +c =-(x-μ)$^2$/(2σ$^2$)= -1/(2σ$^2$) x$^2$ +μx/(σ$^2$)-μ$^2$/(2σ$^2$) which gives


b=μ/(σ$^2$) and

c=-μ$^2$/(2σ$^2$) Now solving for μ and σ$^2$ we get

μ =-b/(2a) and

σ$^2$= -1/(2a) The equations involving a and b were all that was needed.

Now similarly for the multivariate normal the quadratic form in the exponent is

-(x-μ)' ∑$^-$ $^1$ (x-μ)/2 and you equate it to

x′Ax+x′b+c and solve for ∑ and μ in terms of the given A, b and c.

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Thanks, seems obvious once it's been explained! – NapierL Jul 5 '12 at 15:40
Perhaps obvious but still incomplete, because (1) you cannot always find solutions $\Sigma$ and $\mu$ and (2) even when you do they do not necessarily determine a valid distribution. Isn't that what the question asks, though: what are the conditions for $f(\mathbf{x})$ to determine a multivariate normal? – whuber Jul 5 '12 at 17:24
@whuber No based on the OPs satisfaction with my answer. I think he was only asking how to determine the mean and covariance of an MVN if it is expressed in the form exp(f(x)) where f is given in the form x′Ax+x′b+c. In some respects the question seems to be consistent with your interpretation excpet that in the 1 D case he mentioned the restriction that a must be less than 0. So he probably recognizes that there may be restrictions requires on A b and c for Σ and μ to exist. – Michael Chernick Jul 5 '12 at 17:32

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