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Given two independent events A and B, what can we say about $P(X\mid A,B)$ ?

Is the following correct ?

$$P(X\mid A,B) = \frac{P(A,B\mid X)P(X)}{P(A,B)}\tag{Bayes}$$

$$P(X\mid A,B) = \frac{P(A\mid X)P(B\mid X)P(X)}{P(A)P(B)} \tag{Independence}$$

$$P(X\mid A,B) = \frac{\frac{P(X\mid A)P(A)}{P(X)}\frac{P(X\mid B)P(B)}{P(X)}P(X)}{P(A)P(B)} \tag{Bayes}$$

$$P(X\mid A,B) = \frac{P(X\mid A)P(X\mid B)}{P(X)}$$

Did
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GuillaumeThomas
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1 Answers1

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Is the following correct ?

No, the "independence" step is bogus since it requires that A and B are independent conditionally on X, a condition which is not implied by the independence of A and B.

Did
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    For an example, suppose $A$ and $B$ are the first and second outcomes of a roll of a fair six-sided die, respectively. Clearly, $A$ and $B$ are independent. However, now suppose $X$ is the event that the sum of the rolls is $7$. Then the conditional probability $\Pr[A = 1 \cap B = 1 \mid X]$ is not equal to $\Pr[A = 1 \mid X] \Pr[B = 1 \mid X]$: the former is obviously zero, but the latter is nonzero. – heropup Jan 14 '14 at 22:18
  • This answers only that it is incorrect - but is there any other way to solve it so that A & B do not appear in the same condition, e.g. P(X|A)P(X|B)...? I started a new question for that: https://math.stackexchange.com/questions/2344912/pab-c-pab-pac?noredirect=1#comment4828513_2344912 – Thomas Jul 03 '17 at 11:58
  • "This answers only that it is incorrect " Sure, since this was the question asked. Kind of the very principle of the site... – Did Jul 03 '17 at 16:48