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Re: uniform prior
Posted:
Dec 3, 2012 1:41 PM


On 20121203, majeti dinesh <dinesh.majeti@gmail.com> wrote: > I am estimating the parameters of a gaussian mixture model in a Bayesian framework. So I need the prior probabilities of the parameters which I am going to estimate. I am using uniform priors for the parameters (mean, covariance etc) . So according to the Bayes formula, I need the value of P(mean) for computations.
> Hence I would like to know what value to take for P(mean) i.e, uniform prior probability of mean.
I suggest you learn the meaning of the terms you are using. I also see no way in which covariance occurs; do you mean variance?
There is no such thing as P(mean); one can ask for P(mean is in a set). Also, the prior probabilities are provided by the user,
Is this what you want? The observations are independent, with their marginal distributions P(X in S) = p*N(Sm_1,v_1) + (1p)N(Sm_2,v_2), with the joint prior "distribution" of the five parameters being the measure with density 1/(v_1*v_2)^2. This is considerered by some to be the "noninformative" prior. The exponent 2 in the density is correct; this is the best invariant prior for the parameters of the normal distributions.
In this formulation, there are two equivalent solutions, as the parameters can be interchanged.
 This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)4946054 FAX: (765)4940558



