Marginalize over C, can I say $\sum_c P(A,B,C)=\sum_c (P(A|C)P(B|C)P(C))=\sum_c P(A|C) \sum_c P(B|C) \sum_c P(C)$

Question

When reading lectures about Probabilistic Graphic Models, I saw a simple belief nets like this: which encodes $P(A,B,C)=P(A|C)P(B|C)P(C)$. But then it says we can marginalize over C to erase C. Which means $\sum_c P(A,B,C)=\sum_c P(A|C)P(B|C)P(C)$. But what made me confused is that how can we write a mathematical justification for this. I wrote down one here: $$\sum_c P(A|C)P(B|C)P(C)=\sum_c P(A|C) \sum_c P(B|C) \sum_c P(C)$$ But this does not make sense from the algebra perspective. I am really confused. Thanks for any helpful reply.

Answer 1

In general we have $$ P(A,B,C) = P(A,B \mid C) P(C). $$ But your picture seems to imply that $A$ and $B$ are conditionally independent of $C$ and thus $$ P(A,B \mid C) = P(A \mid C)P(B \mid C). $$ Now summing both sides over $c$ we get $$ \sum_{c} P(A,B,C) = \sum_{c} P(A \mid C)P(B \mid C) P(C). $$