TITLE: Conditional probability given independent random variables
QUESTION [5 upvotes]: I have a fundamental question regarding conditional probability. Lets say I have $n$ independent random variables $X_1, X_2, \ldots, X_n$. Another random variable, $W$ is conditioned on the conjunction of these random variables: $P(W | X_1, X_2, X_3, \ldots, X_n)$. Given that $X_1, \ldots, X_n$ are independent, is it possible to write
$$P(W | X_1, X_2, X_3, \ldots, X_n) =  P(W | X_1) \cdot P(W | X_2) \cdot \cdots \cdot P(W | X_n).$$
Or is the only way to rewrite is the Bayes theorem?

REPLY [2 votes]: The answer is no, it is no possible in general. Here a counterexample:
Let the sample space be $\{(0,0),(0,1),(1,0),(1,1)\}$ and the events:

$A$: The first element is one, or $\{(1,0),(1,1)\}$. 
$B$: The second element is one, or $\{(0,1),(1,1)\}$. 

Then is easy to see that:

$P(A)=\frac{1}{2}$ (First element one)
$P(B)=\frac{1}{2}$ (Second element one)
$P(A,B)=\frac{1}{4}$ (Both ones)
$P(A)P(B)=\frac{1}{4}$

Hence, $A$ and $B$ are independents. Now, let:
$W$: Both ones, or $\{(1,1)\}$ (this is the same as A,B)
Then:

$P(W|A)=\frac{1}{2}$
$P(W|B)=\frac{1}{2}$

But
$$P(W|A,B)=1$$
Since W={A,B} (Both elements one). Also, 
$$ P(W|A)P(W|B)=\frac{1}{4}$$
so
 $$P(W|A,B) \neq P(W|A)P(W|B)$$