Given the probability: $$p(a,b,c)=p(a)p(c|a)p(b|c)$$
How to prove $$ p(a,b)=p(a)\sum_{c}p(c|a)p(b|c)=p(a)p(b|a) $$
Does anyone have idea how to prove the above sequence? I am confused how is the sum rule applied.
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1$\begingroup$ This is not true in general. If $c$ and $b$ are independent, you'll have $p(b) = p(b|a)$, which implies than $a$ and $b$ are independent, too. The formula is correct if and only if $b$ depends on $a$ through $c$, that is $p(b|a,c) = p(b|c)$. $\endgroup$ – Artem Sobolev Apr 25 '15 at 11:07
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$\begingroup$ @Barmaley.exe Sorry I haven't given all conditions. I editted once again. Would you please give me some inspiration? Thanks a lot^_^ $\endgroup$ – spacegoing Apr 25 '15 at 11:18
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1$\begingroup$ Ok, then just note that $p(a,b,c) = p(b,c|a) p(a) $ (by definition) which leads to $ p(b,c|a) = p(c|a) p(b|c)$ (from the first statement). The right-hand side of the last equation is what you have in the sum, so the sum is equal to $\sum_c p(b,c|a)$. This is marginalization over $c$. $\endgroup$ – Artem Sobolev Apr 25 '15 at 11:24
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$\begingroup$ Thats really helpful, thanks a lot! Would you please tell me how did you come up with that idea! I was looking at this like half hour notching my head :P $\endgroup$ – spacegoing Apr 25 '15 at 11:30
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With multiple applications of the conditional probabilty rule you get:
$$p(a,b) = \sum_{c} p(a,b,c) = \sum_{c} p(b,c|a) p(a) = p(a) \sum_{c} p(b,c|a) = p(a) \sum_{c} p(b|a,c) p(c|a).$$
This is as far as you can actually go without more assumptions. In the special case where $b \text{ } \bot \text{ } a | c$ (i.e., $b$ is conditionally independent of $a$ given $c$) you have $p(b|a,c) = p(b|c)$ which gives you the desired result.
