Con las distribuciones en nuestros vectores aleatorios:
xi|μ∼N(μ,Σ)
μ∼N(μ0,Σ0)
Por la regla de Bayes, la distribución posterior se ve así:
p(μ|{xi})∝p(μ)∏Ni=1p(xi|μ)
Entonces:
lnp(μ|{xi})=−12∑Ni=1(xi−μ)′Σ−1(xi−μ)−12(μ−μ0)′Σ−10(μ−μ0)+const
=−12Nμ′Σ−1μ+∑Ni=1μ′Σ−1xi−12μ′Σ−10μ+μ′Σ−10μ0+const
=−12μ′(NΣ−1+Σ−10)μ+μ′(Σ−10μ0+Σ−1∑Ni=1xi)+const
=−12(μ−(NΣ−1+Σ−10)−1(Σ−10μ0+Σ−1∑Ni=1xi))′(NΣ−1+Σ−10)(μ−(NΣ−1+Σ−10)−1(Σ−10μ0+Σ−1∑Ni=1xi))+const
Which is the log density of a Gaussian:
μ|{xi}∼N((NΣ−1+Σ−10)−1(Σ−10μ0+Σ−1∑Ni=1xi),(NΣ−1+Σ−10)−1)
Using the Woodbury identity on our expression for the covariance matrix:
(NΣ−1+Σ−10)−1=Σ(1NΣ+Σ0)−11NΣ0
Which provides the covariance matrix in the form the OP wanted. Using this expression (and its symmetry) further in the expression for the mean we have:
Σ(1NΣ+Σ0)−11NΣ0Σ−10μ0+1NΣ0(1NΣ+Σ0)−1ΣΣ−1∑Ni=1xi
=Σ(1NΣ+Σ0)−11Nμ0+Σ0(1NΣ+Σ0)−1∑Ni=1(1Nxi)
Which is the form required by the OP for the mean.