Factorization of conditional probability I read a paper where the authors factorized a conditional probability as follows:
$P(a|b, c)\propto P(a|b)P(a|c)$.
They say that they can do that because $b$ and $c$ are causally independent (they are using graphical models), and cite the paper [1] to justify this. Under which assumptions can this be true?  Honestly, I don't see how this statement is true.
Thanks for your comments.
[1]: Zhang, N. L., & Poole, D. (1996). Exploiting causal independence in Bayesian network inference. Journal of Artificial Intelligence Research, 5, 301-328.
 A: $P(A\mid B, C) = 
\displaystyle \frac{P(ABC)}{P(BC)} = \frac{P(ABC)}{P(B)P(C)}~~$ if $B$ and $C$ are independent.
$P(A\mid B)P(A\mid C) =\displaystyle \frac{P(AB)}{P(B)}\cdot \frac{P(AC)}{P(C)}$ regardless of whether $B$ and $C$ are independent or not.
So, equality will hold if $P(ABC) = P(AB)P(AC)$ and $B$ and $C$ are independent events. But, knowing that $B$ and $C$ are independent events
does not allow us to infer that $P(ABC)$ equals $P(AB)P(AC)$. In particular, if $B$ and $C$ are independent, it does not follow that $AB$ and $AC$ are independent events too.
But if $AB$ and $AC$ are indeed independent events, then of course we would have that
$P(AB)P(AC) = P(AB \cap AC) = P(ABC)$ and it would be true that
$$P(A\mid B, C) = P(A\mid B)P(A\mid C)$$
What about $P(A\mid B, C) \propto P(A\mid B)P(A\mid C)$ instead of equality? Well, given
any two nonzero real numbers $x$ and $y$, it is always the case
that $x \propto y$: just choose the constant of proportionality to be
$\frac xy$ !!  And so, for independent $B$ and $C$, it is always the case
that $P(ABC) \propto  P(AB)P(AC)$ and hence $P(A\mid B, C) \propto P(A\mid B)P(A\mid C)$. It must be this Naive Bayesian Network Inference stuff that is going around.
