Details in proof for convergence of Expectation Maximization Algorithm

Question

I am going through the paper provided here http://www.cs.cmu.edu/~dgovinda/pdf/recog/EM_algorithm-1.pdf

I could not make out how the following was derived

$\sum_z \mathcal P(\mathbf z|X, \theta_n) \ln \big( \frac{\mathcal P(X| \mathbf z, \theta)\mathcal P(z |\theta)}{\mathcal P(\mathbf z|X, \theta_n)}\big ) - \ln\mathcal P(X|\theta_n)$

=$\sum_z \mathcal P(\mathbf z|X, \theta_n) \ln \big( \frac{\mathcal P(X| \mathbf z, \theta)\mathcal P(z |\theta)}{\mathcal P(\mathbf z|X, \theta_n) \mathcal P(X|\theta_n)}\big ) $

Considering that the left summation in first equation has several terms how is $\ln \mathcal P(X|\theta_n)$ distributed over it?

Answer 1

Isn't that simply that $$\sum_z\mathcal P(z|X,θ_n)=1$$ and hence that \begin{align*} \sum_z &\mathcal P(\mathbf z|X, \theta_n) \ln \left( \frac{\mathcal P(X| \mathbf z, \theta)\mathcal P(z |\theta)}{\mathcal P(\mathbf z|X, \theta_n)}\right ) - \ln\mathcal P(X|\theta_n)\\ &= \sum_z \mathcal P(\mathbf z|X, \theta_n) \ln \left( \frac{\mathcal P(X| \mathbf z, \theta)\mathcal P(z |\theta)}{\mathcal P(\mathbf z|X, \theta_n)}\right ) - \sum_z \mathcal P(\mathbf z|X, \theta_n) \ln\mathcal P(X|\theta_n)\\&=\sum_z \mathcal P(\mathbf z|X, \theta_n) \ln \left( \frac{\mathcal P(X| \mathbf z, \theta)\mathcal P(z |\theta)}{\mathcal P(\mathbf z|X, \theta_n) \mathcal P(X|\theta_n)}\right) \end{align*}