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This is a very simple question but I can't find the derivation anywhere on the internet or in a book. I would like to see the derivation of how one Bayesian updates a multivariate normal distribution. For example: imagine that

$$ P({\bf x}|{\bf μ},{\bf Σ}) = N({\bf \mu}, {\bf \Sigma}) \\ P({\bf \mu}) = N({\bf \mu_0}, {\bf \Sigma_0}) $$

After observing a set of ${\bf x_1 ... x_n}$, I would like to compute $P({\bf \mu | x_1 ... x_n})$. I know that the answer is $P({\bf \mu | x_1 ... x_n}) = N({\bf \mu_n}, {\bf \Sigma_n})$ where

$$ \bf \mu_n = \Sigma_0 (\Sigma_0 + \frac{1}{n}\Sigma)^{-1}(\frac{1}{n}\sum_{i=1}^{n}{\bf x_i}) + \frac{1}{n}\Sigma(\Sigma_0+\frac{1}{n}\Sigma)^{-1}\mu_0 \\ \bf \Sigma_n = \Sigma_0(\Sigma_0 + \frac{1}{n}\Sigma)^{-1}\frac{1}{n}\Sigma $$

I am looking for the derivation of this result with all the intermediate matrix algebra.

Any help is much appreciated.

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It is also solved in our book Bayesian Core, Chap. 3, Section 3.2, pages 54-57 with what we think is detailed matrix algebra! – Xi'an May 18 '12 at 20:10
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The OP said it was not a homework problem and even explained why he asked it and how he wants to use the answer. Why not post it for others? I understand why we do not want to provide a homework problem solving service but this is taking it a little too far. – Michael Chernick May 18 '12 at 23:11
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@Alex: Sorry, wrong link, I meant Bayesian Core. Note that we also posted solutions to all problems on arXiv. So posting a complete solution here would not hurt! – Xi'an May 19 '12 at 7:56
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I have deleted the portion of the comments that amount to a private exchange between individuals with an arrangement to share a private answer to the question. That sort of thing is abusing this site, which is all about public questions and public answers. – whuber May 19 '12 at 14:20
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Just as an FYI, the derivation is in Pattern Classification by Duda, Hart and Stork. However, I was having difficulty following some of their steps which only matters to me. If this was simply homework one could just write down exactly what they have. – Alex May 19 '12 at 16:46
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