# graphical models: integration rules over conditional prob distns

Suppose we have a graphical model over three RVs $$a,b,c$$  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 Jun 2 '18 at 23:02