How can we prove this equation using marginalization and conditioning?

Question

I want to prove $$P(A|C) = \sum_{B} P(AB|C) $$ How can we prove this using marginalization and conditioning?

Answer 1

I preassume that $AB$ is an abbreviation of $A\cap B$ here, and that we are dealing with disjoint events $B_1,B_2,\dots$ that cover the whole space.

Also I just don't know what is meant by marginalization and conditioning, but here is a direct route to prove the statement.

We have the general rule: $$P(S|R)P(R)=P(SR)\tag1$$

leading to:

$$\left[\sum_{n=1}^{\infty}P\left(AB_{n}|C\right)\right]P\left(C\right)=\sum_{n=1}^{\infty}P\left(AB_{n}|C\right)P\left(C\right)=$$$$\sum_{n=1}^{\infty}P\left(AB_{n}C\right)=P\left(AC\right)=P\left(A\mid C\right)P\left(C\right)$$ The third equality is a consequence of $AC=\bigcup_{n=1}^{\infty}AB_{n}C$ where the $AB_{n}C$ are disjoint.

So if $P\left(C\right)>0$ then we can divide by it and are allowed to conclude that: $$\sum_{n=1}^{\infty}P\left(AB_{n}|C\right)=P\left(A\mid C\right)$$

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I'll try to shed some light on what I saw in the video. According to the rule (1) we have:$$P(A|BC)P(B|C)P(C)=P(A|BC)P(BC)=P(ABC)=P(AB|C)P(C)$$

Again dividing by $P(C)$ we find:$$P(A|BC)P(B|C)=P(AB|C)$$