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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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    @HenningMakholm and Henry: Both answers are helpful to me. Thanks!2012-03-16

2 Answers 2

1

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 $B=b$.

2

(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.