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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?

  [1]: https://i.stack.imgur.com/04vDN.png

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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 $p(c \mid a)$?

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  • $\begingroup$ 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 $\endgroup$ – dunno Jun 2 '18 at 23:02
  • $\begingroup$ @dunno that's it. you're done $\endgroup$ – Taylor Jun 2 '18 at 23:31
  • $\begingroup$ thanks! can you provide some intuition/explanation to help me see why this is (outside of my trying to back into the solution)? $\endgroup$ – dunno Jun 3 '18 at 2:06
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    $\begingroup$ @rrrrr I wouldn’t call that Bayes’ $\endgroup$ – Taylor Jun 3 '18 at 12:43
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    $\begingroup$ Ok ya, it's just definition I guess. $\endgroup$ – rrrrr Jun 3 '18 at 17:57

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