Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
Try expanding P(A,B,C) using the law of total probability in different ways. –  Aditya Apr 26 '11 at 20:24
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 $P(A^1|B\land C) = 1$.

share|improve this answer
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.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.