# Graphic model factorizing, marginalization

This is actually a probability marginalization question that I encountered in graphic models section of PRML by Bishop (question about equation 8.26 page 391). Assume I have the following graphic model

Therefore the joint probability density of the variables will factor to $$p(a,b,c)=p(a)p(c|a)p(b|c)$$. Now assume I want to marginalize over $$c$$, the book says

$$\sum_c p(a,b,c) = \sum_c p(a)p(c|a)p(b|c) = p(a) \sum_c p(c|a)p(b|c)= p(a)p(b|a)$$

This means that $$p(b|a) = \sum_c p(c|a)p(b|c)$$, how to prove this?! If it was like this $$\sum_c p(c|a)p(b|c,a)$$ then one can reason that $$\sum_c p(c|a)p(b|c,a) = \sum_c \frac{p(a,c)}{p(a)}\frac{p(a,b,c)}{p(a,c)} = \sum_c p(b,c|a)=p(b|a)$$. But I can't conclude the same result with $$\sum_c p(c|a)p(b|c)$$. What am I getting wrong?

You need to use the fact that $$p(b|c,a)=p(b|c)$$ i.e. $$a$$ and $$b$$ conditionally independent given $$c$$. This is directly visible on the graph and also proved with:
$$p(b|c,a)=\frac{p(b,c,a)}{p(c,a)}=\frac{p(a)p(c|a)p(b|c)}{\sum_{b}p(a)p(c|a)p(b|c)}=\frac{p(b|c)}{\sum_{b}p(b|c)}=p(b|c).$$
• Thank you for your answer, I think you should replace $c$ with $a$ in your answer and it will match the notation in my question but overall nice idea for proof. – K.K.McDonald Jul 27 '20 at 14:08