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I have three events A,B and C and I have the following relationship:

P(A|B,C) = P(A|B) P(A|C)

Does this imply that event B and C are conditionally independent given A?

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You have, from given condition, $P(BC\mid A)=\frac{P(ABC)}{P(A)}=\frac{P(AB)P(BC)P(CA)}{P(A)P(B)P(C)}\ne \frac{P(AB)}{P(A)}\cdot\frac{P(CA)}{P(A)}$, in general.

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  • $\begingroup$ How did you go from \frac{P(ABC)}{P(A)} to \frac{P(AB)P(BC)P(CA)}{P(A)P(B)P(C)} ? $\endgroup$ – Lanc Mar 11 '15 at 10:50
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Using the definition of conditional probability and some algebraic manipulation of the original identity we find. $\begin{align} \tag{1} {\sf P}(A\mid B, C) &= {\sf P}(A\mid B){\sf P}(A\mid C) \\ \tag{2} {\sf P}(A, B, C)/{\sf P}(B, C) & = {\sf P}(A, B) {\sf P}(A, C) / {\sf P}(B) {\sf P}(C) \\ \tag{3} {\sf P}(B, C\mid A){\sf P}(A)/{\sf P}(B, C) & = {\sf P}(A, B) {\sf P}(A, C) / {\sf P}(B) {\sf P}(C) \\ \tag{4} {\sf P}(B, C\mid A)& = {\sf P}(A, B) {\sf P}(A, C) {\sf P}(B, C)/ {\sf P}(A) {\sf P}(B) {\sf P}(C) \\ \tag{5} {\sf P}(B, C\mid A)& = {\sf P}(B\mid A) {\sf P}(C\mid A)\times{\sf P}(A) {\sf P}(B, C)/ {\sf P}(B) {\sf P}(C) \end{align}$

Since ${\sf P}(A) {\sf P}(B, C)/ {\sf P}(B) {\sf P}(C)$ is not generally equal to $1$, then we do not necessarily have conditional independence of $B$ and $C$ given $A$.

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