`BPE - Bayesian Parameter Estimation from EC662, Lecture 7
BPE FOR MULTIVARIATE GAUSSIAN :
Estimation of mean, given a known covariance
Consider a set of iid samples $ \{X_i\}_{i=1}^N $ where $ X_i \in\mathbb{R}^n $ is such that $ X_i \sim N(\mu,\Sigma) $. Suppose we know $ \Sigma $, but wish to estimate $ \mu $ using BPE. If we assume a prior distribution for the unknown mean to be distributed as a Gaussian random variable, we will obtain a posterior distribution for the mean which is also Gaussian, i.e. $ p(\mu|X_1,X_2,\ldots,X_N) = N(\mu_N,\Sigma_N) $, where $ \mu_N $ and $ \Sigma_N $ are calculated to utilize both our prior knowledge of $ \mu $ and the samples $ \{X_i\}_{i=1}^N $. Fukunaga p. 391 derives that the parameters $ \mu_N $ and $ \Sigma_N $ are calculated as follows:
$ \mu_N = \frac{\Sigma}{N}(\Sigma_\mu + \frac{\Sigma}{N})^{-1}\mu_0 + \Sigma_\mu(\Sigma_\mu + \frac{\Sigma}{N})^{-1}\left(\frac1N\sum_{i=1}^NX_i\right) $,
where $ \mu_0 $ is the initial "guess" for the mean $ \mu $, and $ \Sigma_\mu $ is the "confidence" in that guess. In other words, we can consider that $ N(\mu_0,\Sigma_\mu) $ is the prior distribution for $ \mu $ that we would assume without seeing any samples. For the covariance parameter, we have
$ \Sigma_N = \Sigma_0(\Sigma_0+\frac{\Sigma}{N})^{-1}\frac{\Sigma}{N} $.
We find that as the number of samples increases, that the effect of the prior knowledge ($ \mu_0 $,$ \Sigma_\mu $) decreases so that
$ \lim_{N\rightarrow\infty}\mu_N = \frac1N\sum_{i=1}^NX_i $, and $ \lim_{N\rightarrow\infty}\Sigma_N = 0 $.
Estimation of covariance, given a known mean
Again, given iid samples $ \{X_i\}_{i=1}^N $, $ X_i \in\mathbb{R}^n $, $ X_i \sim N(\mu,\Sigma) $, let us now estimate $ \Sigma $ with $ \mu $ known. As in Fukunaga p. 392, we assume that both the posterior distribution of $ \Sigma $ is normal (i.e. $ p(X|\Sigma) = N(\mu,\Sigma) $), and it can be shown that the sample covariance matrix follows a Wishart Distribution. Fukunaga p.392 shows the distribution $ p(K|\Sigma_0,N_0) $, where $ K = \Sigma^{-1} $, and parameter $ \Sigma_0 $ represents the initial "guess" for $ \Sigma $ and $ N_0 $ represents "how many samples were used to compute $ \Sigma_0 $". Note that we compute the distribution for $ K = \Sigma^{-1} $ instead of $ \Sigma $ directly, since the inverse covariance matrix is used in the definition for a normal distribution. It can be shown, then, that
$ p(K|\Sigma_0,N_0) = c(n,N_0)\left|\frac12N_0\Sigma_0\right|^{(N_0-1)/2}|K|^{(N_0-n-2)/2}\exp(-\frac12\mathrm{trace}(N_0\Sigma_0K)) $,
where $ c(n,N_0) = \left\{\pi^{n(n-1)/4}\prod_{i=1}^n\Gamma\left(\frac{N_0-i}{2}\right)\right\}^{-1} $.
Simultaneous estimation of unknown mean and covariance
Finally, given iid samples $ \{X_i\}_{i=1}^N $, $ X_i \in\mathbb{R}^n $, $ X_i \sim N(\mu,\Sigma) $, we now wish to estimate both $ \mu $ and $ \Sigma $ (or $ K = \Sigma^{-1} $). Fukunaga p. 393 gives that the joint distribution follows the Gauss-Wishart distribution as follows
$ p(\mu,K|\mu_0,\Sigma_0,\mu_{\Sigma},N_0) = (2\pi)^{-n/2}|\mu_{\Sigma} K|^{1/2}\exp\left(-\frac12\mu_{\Sigma}(\mu-\mu_0)^TK(\mu-\mu_0) \right)\times c(n,N_0)|\frac12N_0\Sigma_0|^{(N_0-1)/2}|K|^{(N_0-n-2)/2}\exp\left(-\frac12\mathrm{trace}(N_0\Sigma_0K\right) $, where $ \mu_0 $, $ \Sigma_0 $, $ N_0 $, and $ c(n,N_0) $ are as above.
