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Math Help - is this a gaussian?

  1. #1
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    is this a gaussian?

    Is the following density a bivariate gaussian?:
    Code:
    f(x,y)=[sqrt(1-p^2)/(2*pi)]*exp[-1/2*(x^2+y^2-2*p*x*y)]
    where p is the correlation between x and y. It does integrate to 1 and the marginal densities for x and y are indeed normal (both x and y are distributed Normal(0,1/(1-p^2)), as you can see from completing the squares). I have a book that claims that f(x,y) is 2d gaussian, but I can't figure out what the covariance matrix would be... it isn't [[1 p] [p 1]] because f doesn't have the 1/(1-p^2) in the denominator of the exponential... help?

    More general question: are all exponentials of quadratic functions gaussian? This seems to be true for the univariate case but can't confirm for multivariate.
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  2. #2
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    Sorry, stupid question, the covariance is obviously [[1/(1-p^2) p/(1-p^2)] [p/(1-p^2) 1//(1-p^2)]].

    But my other question still stands: are all distributions of the form "f(X) = A*exp(quadratic of X)" gaussian (where X = vector of variables)
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  3. #3
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    Hello,

    Well, see if your pdf looks like the one that is given here : Multivariate normal distribution - Wikipedia, the free encyclopedia
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  4. #4
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    Quote Originally Posted by tomgrayson View Post
    More general question: are all exponentials of quadratic functions gaussian? This seems to be true for the univariate case but can't confirm for multivariate.
    I would think that the fact that the quadratic must be of the form (x - \mu)' \Sigma ^{-1} (x - \mu) for some symmetric matrix \Sigma would be pretty restrictive. If my algebra is right, in the bivarate case you should not be able to get

    \frac{x^2}{2} + \frac{y^2}{2} + xy + x - y + K

    from that operation (K being more or less irrelevant). I constructed the polynomial in the exponent for general coefficients, and then picked a polynomial that would result in the set of linear equation being inconsistent when you go to solve for the undetermined coefficients, but you can feel free to check it since I did it pretty thoughtlessly.
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