Re: to prove that Multivariate Gaussian distribution is self-conjugate via Bayesian inference
- From: hrubin@xxxxxxxxxxxxxxxxxxxx (Herman Rubin)
- Date: 17 Feb 2008 14:17:34 -0500
In article <542eee2f-1a59-4595-9c3e-62e1a2e723d8@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<semenov.pk@xxxxxxxxx> wrote:
On 15 =C6=C5=D7, 05:03, hru...@xxxxxxxxxxxxxxxxxxxx (Herman Rubin) wrote:
The proof is sufficiently obvious, and the theorem is
usually stated in textbooks with only comment on the
use.
It is possible that I don't know some positions in mathematical
analysis because I'm only the
undergraduate student.
Let \theta be normal with mean m and covariance matrix
T, and given \theta, let X be normal with mean \theta
and covariance matrix S. Then the vector (\theta X)'
is normal with mean (m m)' and covariance matrix
T T
T T+S
This can be seen in the non-singular case by the usual
integration methods; working with multivariate normal
problems requires "speaking" matrix.
The expression of the conditional distribution of
\theta given X is a little easier with mean 0.
In that case, \theta is normally distributed with
mean (T+S)^{-1}TX and covariance matrix
T-T(T+S)^{-1}T, just as in the univariate case.
This last matrix is also S(T+S)^{-1}T.
--
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@xxxxxxxxxxxxxxx Phone: (765)494-6054 FAX: (765)494-0558
.
- References:
- to prove that Multivariate Gaussian distribution is self-conjugate via Bayesian inference
- From: semenov . pk
- Re: to prove that Multivariate Gaussian distribution is self-conjugate via Bayesian inference
- From: Herman Rubin
- Re: to prove that Multivariate Gaussian distribution is self-conjugate via Bayesian inference
- From: semenov . pk
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