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

Is the following correct ?

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

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

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

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


1 Answer 1


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.

  • 8
    $\begingroup$ 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. $\endgroup$
    – heropup
    Commented Jan 14, 2014 at 22:18
  • $\begingroup$ 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: math.stackexchange.com/questions/2344912/… $\endgroup$
    – Thomas
    Commented Jul 3, 2017 at 11:58
  • $\begingroup$ "This answers only that it is incorrect " Sure, since this was the question asked. Kind of the very principle of the site... $\endgroup$
    – Did
    Commented Jul 3, 2017 at 16:48

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