# How to expand $p(A|B)$ when $p(C)$, $p(A|C)$ and $p(B|C)$ is known?

This is a silly and basic question however I got myself confused. Suppose $A, B, C$ are r.v, how to expand $p(A|B)$ when $p(C)$, $p(A|C)$ and $p(B|C)$ is known? Does it hold

$$p(A|B) = \int p(A|C)p(B|C)p(C)$$

Would you please show me some derivation steps? Thanks!

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Are $A,B,C$ events or random variables? Does $p$ represent a probability or a density? –  Henry Mar 15 '12 at 23:24
@Henry Edited. They're r.v. –  shuaiyuancn Mar 16 '12 at 9:32
@shuaiyuancn: Then you shouldn't accept my answer, which assumes they are events. –  Henning Makholm Mar 16 '12 at 16:15
@HenningMakholm and Henry: Both answers are helpful to me. Thanks! –  shuaiyuancn Mar 16 '12 at 16:31

Your expression does not hold, and indeed it is not toally clear what you mean by the integral. What does hold is

$$p(A=a|B=b) = \int_c p(A=a|C=c,B=b) \, p(C=c|B=b) \, dc$$

or similarly

$$p(A=a|C=c) = \int_b p(A=a|C=c,B=b) \, p(B=b|C=c) \, db.$$

The former is an application of

$$p(A=a) = \int_c p(A=a|C=c) \, p(C=c) \, dc$$

conditional on $B=b$.

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(This answer assumes that $A$, $B$ and $C$ are events; the OP later clarified that they are random variables).
You don't have enough information to derive $P(A\mid B)$ in general, since you don't know anything about what the probabilities outside $C$ are.
But it's worse than that: Even supposing that $A\subseteq C$ and $B\subseteq C$, you don't know anything about the relation between $A$ and $B$.
For example, if $P(A\mid C) = P(B\mid C)=\frac 12$ it could either be that $A=B$, so $P(A\mid B)=1$, or that $A=C\setminus B$, in which case $P(A\mid B)=0$ -- or anything in between.