Is $P(X|A,B)=\frac{P(A|X)P(B|X)P(X)}{P(A)P(B)}$ when $A$ and $B$ are conditionally indepdent?

Question

I am trying to evaluate $P(X|A,B)$ given that $A$ and $B$ are conditionally independent given $X$. I know that this can be expressed as,

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

Using the conditional independence of $A$ and $B$ given $X$ this is, $$\frac{P(A|B,X)P(B,X)}{P(A)P(B)}=\frac{P(A|X)P(B,X)}{P(A)P(B)}$$ Using the chain rule this is $$\frac{P(A|X)P(B|X)P(X)}{P(A)P(B)}$$

Is this correct? Does this hold with conditional independence of $A$ and $B$ given $X$, or does it also require independence alone of $A$ and $B$?

Answer 1

You could save some steps using Bayes' formula: $$ P(X \mid A, B) = \frac{P(A, B \mid X)P(X)}{P(A, B)} $$ Using the conditional independence, this is $$ \frac{P(A \mid X) P(B \mid X) P(X)}{P(A, B)} $$ But as has been pointed out, you cannot factor the denominator without assuming independence of $A$ and $B$

Answer 2

Almost correct: $$P(X|A,B) = \frac{P(A|X)P(B|X)P(X)}{P(A,B)}$$

You know that A and B are conditionally independent given X but not that A and B are independent (that cannot be assumed).