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I read two different papers on some similar problems. In one of the papers this statement is written:

$ p(a|b) = \sum_{c \in C}p(a,c|b) $

While in the other it is written as:

$ p(a|b) = \sum_{c \in C}p(a|c)p(c|b) $

Therefore, this expression should hold:

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

I tried to prove it but failed at some point. Are they the same? Or am I missing something? Or maybe each of them are correct under a specific assumption?

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    $\begingroup$ second line should read $p(a,c|b) = \sum p(a|c,b)p(c|b)$ right? $\endgroup$ – bdeonovic Apr 27 '16 at 11:14
  • $\begingroup$ @bdeonovic Yes it should be. $\endgroup$ – Moh Apr 27 '16 at 13:53
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At the second article they should actually write: $$ p(a|b) = \sum_{c\in C} p(a|c,b) p(c|b) $$ (unless $p(a|c)=p(a|c,b)$ for all $b$).

The equation $p(a,c|b)=p(a|c,b)p(c|b)$ is correct.

By definition, $$p(x|y)=N(x,y)/N(y) = p(x, y) / p(y)$$ where $N(x,y)$ is the number of cases when both $x$ and $y$ hold, while $N(y)$ is the number of cases when $y$ holds.

Therefore, $$ p(a,c|b) = N(a,c,b)/N(b) $$ $$ p(a|c,b) = N(a,c,b)/N(c,b) $$ $$ p(c|b) = N(c,b)/N(b) $$

We can easily see that the equation is correct.

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  • $\begingroup$ I read somewhere this statement: $p(a|c,b)=p(a|c)$ under the assumption that $a$ is conditionally independent of $b$ given $c$. Does it make any sense to you in this example? It might have been their assumption. $\endgroup$ – Moh Apr 27 '16 at 13:51
  • $\begingroup$ By definition, $p(a|b) = p(a,b)/p(b)$. If a and b independent, $p(a,b) = p(a)p(b)$, hence $p(a|b) =p(a)$. Condition on $c$, you get what you want... $\endgroup$ – dv_bn Apr 27 '16 at 13:52
  • $\begingroup$ "I read somewhere this statement: p(a|c,b)=p(a|c) under the assumption that a is conditionally independent of b given c." Yes, you are right, the statement is true. $\endgroup$ – user31264 Apr 27 '16 at 13:54
  • $\begingroup$ In the second article it is written that we write this marginalization based on the independence assumption of $c$ and $b$. In that case $p(c|b)$ will be $p(c)$ and makes the whole statement wrong. Am I right? $\endgroup$ – Moh Apr 27 '16 at 13:55
  • $\begingroup$ @dv_bn Could you please give me an example? $\endgroup$ – Moh Apr 27 '16 at 13:57
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Start from the identity $P(a,c) = P(a|c)P(c)$.

Now condition both sides with respect to $b$ and obtain

$P(a,c|b) = P(a|c,b)P(c|b)$

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