graphical models: integration rules over conditional prob distns

Suppose we have a graphical model over three RVs $a,b,c$

[![enter image description here][1]][1]

whose conditional independence structure gives  $$p(a,b,c) = p(a)\,p(c|a)\,p(b|c), \quad p(a,b)= \sum_c p(a,b,c) = p(a) \sum_c p(c|a)\, p(b|c).$$ From this, how can I see that $\sum_c p(c|a)\, p(b|c) = p(b|a)$? I tried \begin{align} \sum_c p(c|a)\, p(b|c) = \sum_c \frac{p(a|c)\, p(c)}{p(a)} p(b|c) \end{align} and \begin{align} \sum_c p(c|a)\, p(b|c) = \sum_c p(c|a)\, \frac{p(c|b)\, p(b)}{p(c)} \end{align} but still don't see it.

Is it the case that any time we have a sum over a RV's values (i.e., when summing over $c$ for arbitrary RVs $x,y$, we need to keep conditionals of the form $p(x|c)$ and $p(c|y)$ inside the summation?

Bayes' rule is unnecessary; it is simpler than that. Hint: use conditional independence. Or in other words, use: $$p(b \mid c) = p(b \mid c, a).$$ What happens when you multiply both sides by $p(c \mid a)$?

• I feel it should be $p(c|a)p(b|c,a)=p(b,c|a)$ (so that summing over $c$ gets rid of the $c$, but im not clear on why it is equal – dunno Jun 2 '18 at 23:02
• @dunno that's it. you're done – Taylor Jun 2 '18 at 23:31
• thanks! can you provide some intuition/explanation to help me see why this is (outside of my trying to back into the solution)? – dunno Jun 3 '18 at 2:06
• @rrrrr I wouldn’t call that Bayes’ – Taylor Jun 3 '18 at 12:43
• Ok ya, it's just definition I guess. – rrrrr Jun 3 '18 at 17:57