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Sep 28, 2010 at 14:26 vote accept RadonNikodym
Aug 23, 2010 at 1:13 comment added Richard Borcherds Sazonov's theorem</a> is a somewhat fancier way of saying what Peter Shor wrote in his answer; see en.wikipedia.org/wiki/Sazonov_theorem
Aug 23, 2010 at 1:03 comment added RadonNikodym Aha! This is much clearer! Yes, of course, such a "uniform" Gaussian could not exist. In fact, you would need the variances in all dimensions to have a finite sum, which is equivalent to saying that the Covariance operator of the Gaussian measure is trace class. So, given a positive symmetric trace class operator, there is a Gaussian measure that has that covariance operator, however I think what you lose is unitary invariance, e.x if I permute the basis elements the corresponding Gaussian will be different. Thank you very much!
Aug 23, 2010 at 0:46 comment added Peter Shor By "non-uniform" I meant that the variance isn't the same in all dimensions. I don't think this is standard terminology. I also think you're stuck with the generalized $\chi$-squared distribution, unfortunately.
Aug 23, 2010 at 0:40 comment added RadonNikodym As regards the generalized chi-square distribution idea. This is exactly what I chasing, however even on finite dimensional spaces the generalized chi-squared distribution has a quite complicated pdf, and things become hairy quite fast.
Aug 23, 2010 at 0:32 comment added RadonNikodym Ok, your definition of "a" Gaussian measure matches mine. Sorry if I'm asking a dumb question, but what does a "non-uniform" Gaussian mean exactly? Thanks for the pointer on quantum optics, I am very interested in learning of applications of probability on infinite dimensional spaces.
Aug 22, 2010 at 23:53 history edited Peter Shor CC BY-SA 2.5
removed editorializing; added 110 characters in body
Aug 22, 2010 at 23:22 history answered Peter Shor CC BY-SA 2.5