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Re: uniform prior
Posted:
Dec 3, 2012 1:41 PM
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On 2012-12-03, 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(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
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
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