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)
Is the following density a bivariate gaussian?:
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?Code:f(x,y)=[sqrt(1-p^2)/(2*pi)]*exp[-1/2*(x^2+y^2-2*p*x*y)]
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.
Hello,
Well, see if your pdf looks like the one that is given here : Multivariate normal distribution - Wikipedia, the free encyclopedia
I would think that the fact that the quadratic must be of the form for some symmetric matrix would be pretty restrictive. If my algebra is right, in the bivarate case you should not be able to get
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.