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


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 $p(c \mid a)$?

  • $\begingroup$ 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
  • 1
    $\begingroup$ @rrrrr I wouldn’t call that Bayes’ $\endgroup$ – Taylor Jun 3 '18 at 12:43
  • 1
    $\begingroup$ Ok ya, it's just definition I guess. $\endgroup$ – rrrrr Jun 3 '18 at 17:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.