Bayes: markov chain, serial connection, marginalization

Question

Goal is to check if p(a) is unconditionally independent to p(c) in the markov chain - serial connection.

$$ p(a,b,c) = p(a) p(b|a) p(c|b) $$ $$ p(a,c) = \sum_b p(a) p(b|a) p(c|b) = p(a) p(c|a) \neq p(a) p(c) $$

How can the marginalization in the second equation be evaluated to $p(a) p(c|a)$?

Answer 1

That equality is justified by the following:

\begin{eqnarray*} p(c\mid a) &=& \sum_b p(c\mid b,a)p(b\mid a) \qquad\qquad\qquad\text{(conditioning on $b$)} \\ &=& \sum_b p(c\mid b)p(b\mid a) \\ && \text{(Markov property: given $b$ we know $a,c$ are conditionally independent)} \\ && \\ \therefore\quad p(a)p(c\mid a) &=& \sum_b p(a)p(c\mid b)p(b\mid a). \end{eqnarray*}