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.

  • 1
    $\begingroup$ Thanks. I provided the full citation of the article that they cited. However, I cannot provide the citation of the article I read because I'm actually revieweing it for publication. $\endgroup$ Commented Mar 2, 2017 at 21:40

1 Answer 1


$P(A\mid B, C) = \displaystyle \frac{P(ABC)}{P(BC)} = \frac{P(ABC)}{P(B)P(C)}~~$ 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.


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