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One thing I don't understand for Bayes' theorem is that I end up in a circular position sometimes when I try to solve it. $$P(B|C) = \frac{P(C\mid B) \times P(B)}{P(C)}$$ However, I don't know $P(C\mid B)$!

If I try to use Bayes' theorem to calculate it, then I end up right back where I started.

I know $P(B)$ and $P(C)$, but they are dependent on a variable $A$ whose probability I also know. How do I get out of the circle if I don't have a prior for $P(C\mid B)$?

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A more appropriate expression for Bayes' theorem is $$P(B\mid C) = \frac{P(C\mid B)P(B)}{P(C\mid B)P(B)+P(C\mid B^c)P(B^c)}$$ where the denominator is the law-of-total-probability expression for $P(C)$. If you know $P(B)$, you also know $P(B^c)$. You also know $P(C)$. If you don't know $P(C\mid B)$ but do know $P(C\mid B^c)$, then you know four of the five probabilities in the law-of-total-probability expression $$P(C)=P(C\mid B)P(B)+P(C\mid B^c)P(B^c)$$ and can solve for $P(C\mid B)$. If you don't know $P(C\mid B^c)$ either, then you don't have enough information to apply Bayes' theorem. – Dilip Sarwate Nov 21 '11 at 3:09
So you know P(B | A) and P(C | A) and you want to compute P(B | C, A)? If so, you'll need to either have access to the joint of B and C (with or without A). From the joint, you can compute the conditional since: P(A | B) = P(A, B) / P(B). – Nick Nov 21 '11 at 3:10
up vote 4 down vote accepted

To use this formula, you need to know $P(C|B)$.

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If you think of Bayes's theorem with the denominator cleared, i.e., $P(B|C)P(C)=P(C|B)P(B)$, you see that it's just two expressions for $P(B\cap C)$. "Applying" it in the customary fraction form (as in the question) amounts to replacing one of these two expressions by the other. So it's not surprising that, if you do it again, replacing the "other" expression by the "one", you just undo what you did the first time.

Furthermore, if you know only $P(B)$ and $P(C)$, you won't be able to "get out of the circle" and find $P(B|C)$ and/or $P(C|B)$, because these are not determined by the information you have. Imagine, for example, flipping a fair coin and letting both $B$ and $C$ be "heads"; so $P(B)=P(C)=\frac12$ and $P(B|C)=P(C|B)=1$. Now consider the same $B$ with a different $C$, namely "tails". You still have $P(B)=P(C)=\frac12$ but now $P(B|C)=P(C|B)=0$.

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