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This is probably almost standard textbook Bayes rule, but I wanted to double check a formula that I am applying:

$P(A | B,C) = P(A|C) \cdot \frac{P(B|A,C)}{P(B|C)}$.

Is this correct?

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It's correct.

$$P(A | B,C) = P(A|C) \cdot \frac{P(B|A,C)}{P(B|C)}$$

Expand:

$$P(A | B,C)=\frac{P(A,C)}{P(C)}\cdot\frac{\frac{P(B,A,C)}{P(A,C)}}{\frac{P(B,C)}{P(C)}}$$

And then remove like terms:

$$P(A | B,C) = \frac{P(B,A,C)}{P(B,C)}$$

Rearrange:

$$P(A | B,C) = \frac{P(A,B,C)}{P(B,C)}$$

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$$P(A | B,C) = \frac{P(A \cap B \cap C)}{P(B\cap C)} = \frac{P(A \vert B, C)P(B \vert C)P(C)}{P(B \vert C)P(C)}$$

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    $\begingroup$ When one algebraically cancels repeated factors on the right hand side, the left hand side reappears, suggesting this series of equations is trivial. Is the OP's formula correct or not? $\endgroup$ – whuber Jun 17 '16 at 13:48
  • $\begingroup$ Yes,following this logic, $P(A|B,C)=\frac{P(A|B,C)xyz}{xyz}$ and all. $\endgroup$ – FisherDisinformation Jun 17 '16 at 13:52

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