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Given P(A|B) and P(A|C), how to get or strategically approach P(A|(B & C))?

Is there a way to approach this if it is not known whether B and C are dependent? If not, how to get P(A|(B & C)) assuming B and C are independent, or, dependent?

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Try expanding P(A,B,C) using the law of total probability in different ways. –  Aditya Apr 26 '11 at 20:24
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From a purely mathematical viewpoint, I don't think there's a way to analytically derive one from the other. However, since you say "strategically", I'll link to a fairly well-known (but ugly) hack in machine learning circles called product of experts: cs.toronto.edu/~hinton/absps/nccd.pdf. It makes the assumption that $p(A|B,C) \propto P(A|B)P(A|C)$ and lives with it. It's utterly wrong, but it works for engineering purposes. –  JasonMond Apr 26 '11 at 23:15

2 Answers 2

Suppose two $0/1$ coins $X_1,X_2$ are thrown. Let $B$ be the event $X_1 = 0$, $C$ be the event $X_2 = 0$, $A^1$ be the event $X_1 + X_2 = 0$, and $A^0$ be the event $X_1 + X_2 = 1$. Then $$P(A^0|B) = P(A^1|B) = P(A^0|C) = P(A^1|C) = 1/2$$ whereas $P(A^0|B \land C) = 0$ and $P(A^1|B\land C) = 1$.

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To make Yuval's point completely explicit, this proves that there is no hope to determine P(A|B,C) from P(A|B) and P(A|C) alone without some additional hypothesis. –  Did May 1 '11 at 8:24

If A and B are independent from C, then P(C|A) = P(C|B). If they are dependent, then just follow Bayes' theorem for a three event senerio.

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The question was about $P(A\vert B\cap C)$... –  tomasz Oct 30 '12 at 18:57

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