D-dimensional Gaussian posterior distribution

Question

Let $X$ be a $D$-dimensional Gaussian random variable with distribution $N_D(x|\mu, \Sigma)$ in which the covariance $\Sigma$ is known. Given a vector of observations $x = (x_1,\ldots,x_n)$, we wish to infer the mean $\mu$.

Given a prior distribution $p(\mu) = N_D(\mu| \mu_0, \Sigma_0)$, find the corresponding posterior distribution $p(\mu|x)$.

I want to solve the problem above but I don't know how. Since I should get the $p(x|\mu)$ to find the posterior distribution and I don't have $\mu$, should I try maximum likelihood first?

Answer 1

First, apply Bayes Rule: $p(\mu|X)=\frac{p(X|\mu)p(\mu)}{p(X)}$. In the numerator, $p(X|\mu)$ and $p(\mu)$ are already given. Normally, you should calculate the denominator as $\int{p(X|\mu)p(\mu)d\mu}$. This integral might be cumbersome and sometimes can be avoided, i.e. when the format of the numerator is known, $p(X)$ can easily be guessed thanks to normalization. So, we have:

$$p(\mu|X)\propto p(X|\mu)p(\mu)\propto\exp(-\frac{1}{2}(x-\mu)^T\Sigma^{-1}(x-\mu))\exp(-\frac{1}{2}(\mu-\mu_0)^T\Sigma_0^{-1}(\mu-\mu_0))$$

In this expression, terms without $\mu$ can be treated as constants. So, $$p(\mu|X)\propto \exp((x^T\Sigma^{-1}+\mu_0^T\Sigma_0^{-1})\mu-\frac{1}{2}\mu^T(\Sigma^{-1}+\Sigma_0^{-1})\mu)\propto\exp(-\frac{1}{2}(\mu-a)^T\Sigma_\mu^{-1}(\mu-a))$$

which is in normal form. So, we actually found the posterior, just need to calculate the norm. constant wrt these parameters.