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I'd like to calculate the posterior distribution given the prior distribution $w\sim N(0,\Sigma_p)$ and the likelihood $y|X,w\sim N(X^\top w,\sigma_n^2I).$

Ignoring everything that does not contain w:

$p(w|X,y)\propto p(y|X,w)p(w) \propto exp(-{1 \over 2} (\sigma_n^{-2}(y-X^\top w)^\top (y-X^\top w) -w^\top\Sigma_p^{-1}w))= exp(-{1 \over 2}(\sigma_n^{-2}(w^\top Xy + y^\top X^\top w)-w^\top(\sigma_n^{-2}XX^\top -\Sigma_p^{-1} )w))$

Now the answer should be $exp(-{1 \over 2}(w - \bar w)^\top(\sigma_n^{-2}XX^\top -\Sigma_p^{-1})(w - \bar w))$ where $\bar w =\sigma_n^{-2}(\sigma_n^{-2}XX^\top -\Sigma_p^{-1})^{-1}Xy.$

Could anyone help how to do this?

Thanks!

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Ok so I figured it out. Notice $w^\top X y = y^\top X^\top w$. So continuing the above expression: $p(w|X,y) \propto exp(-{1 \over 2}(2\sigma_n^{-2}y^\top X w -w^\top(\sigma_n^{-2}XX^\top -\Sigma_p^{-1} )w))= exp(-{1 \over 2}(w^\top A^{-1}w-2 \bar w^\top A^{-1}w))$ where $A=(\sigma_n^{-2}XX^\top -\Sigma_p^{-1})^{-1} $ and the answer follows.

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