On 2012-12-03, majeti dinesh <email@example.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(S|m_1,v_1) + (1-p)N(S|m_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 firstname.lastname@example.org Phone: (765)494-6054 FAX: (765)494-0558