Maximisation step

Assuming the prior distribution of the parameters $p(\boldsymbol{\theta}\mid\mathcal{H})$ to be formed from Dirichlet distributions, makes the maximisation step easy. Let $\rho(j)\in[0,1]$, $j=1,\dots,k$ be random variables such that $\sum_{j=1}^k\rho(j)=1$. In our case, $\rho(j)$ are the transition probabilities $p_{cl}$ corresponding to a certain body or the selection probabilities $\mu$ corresponding to a certain type. A Dirichlet distribution is defined as

$${P}(\rho)=\prod_j\rho(j)^{z(j)-1}/C,$$ (10)

where $z(j)$, the pseudocounts, are parameters controlling the distribution and $C$ is a normalisation constant, which guarantees that $\int {P}(\rho)d\rho=1$. Note that a Dirichlet distribution with $\forall j, z(j)=1$ is uniform.

The terms of log likelihood $\log p(\boldsymbol{X}\mid \boldsymbol{\theta}, \mathcal{H})$ that depend on $\rho(j)$ can be written in the form $\sum_j s(j) \log \rho(j)$, where $s(j)$ are the expected counts of how many times $\rho(j)$ was used. If we assume the counts $s(j)$ to be constant and vary only the probabilities $\rho(j)$, we can find the maximum of the a posteriori density analytically:

$$\rho(i)=\frac{s(i)+z(i)-1}{\sum_j \left[s(j)+z(j)-1\right]}.$$ (11)

The solution of the componentwise Bayes estimate [8]

$$\rho(i)=\frac{s(i)+z(i)}{\sum_j \left[s(j)+z(j)\right]}$$ (12)

resembles closely the map solution (11). The cB estimate is actually a map-estimate with a modified Dirichlet prior (the pseudocounts $z(j)$ increased by one), i.e. the same convergence results hold.