Can $p(Y|a,b)$ ever be equal to $p(Y|a) \cdot p(Y|b)$? This strikes me as a simple question, but in re-visiting how the Naive Classifier works I started wondering if there is any probabilistic model that under certain independency assumptions obtains:
$p(Y |  a,b) = p(Y|a) \cdot p(Y|b)$
This could be used, for example, to build a classifier that gets the class probability as the product of the conditional class probabilities (in the case above, $Y$ would be the class, and $a$ & $b$ the features).
Intuitively something tells me that this is doable, but if I try to find under what conditions I could reach the equation above, I don't get anywhere.
For example, if I asssume that $Y$ is independent of $b$, given $a$, I obviously simply get: $p(Y|a,b)  = p(Y|a)$ so I am missing the second term. I can't think of any independency assumptions that gives me the original equation. Is that because they don't exist?
 A: The equation in the question only happens in the trivial case where both $p(Y|a)$ and $p(Y|b)$ are point masses on a single $Y$ value.  If it were true, then $\sum_Y p(Y|a) p(Y|b) = 1$.  However, consider the following bound:
$$
\sum_Y p(Y|a) p(Y|b) \leq \sum_Y p(Y|a) \max_{Y'} p(Y'|b) = \max_{Y'} p(Y'|b)
$$
For the LHS to be 1, there must be a value of Y for which $p(Y|b)=1$, which means that value of $Y$ is the only one possible.  By symmetry, this must also be true for $p(Y|a)$.
A: Sure, if $a$ and $b$ are assumed to be non-interacting events yielding independent information about $Y$ and $Y$ has a uniform prior, then this is the only reasonable way of combining information.  Geoff Hinton calls this a product of experts.  One caveat, if $Y$ doesn't have a uniform prior, then you'll double-count it when you do the pointwise multiplication.  So you should really do $$P(Y\mid a,b) \propto \frac{p(Y\mid a) \cdot p(Y\mid b)}{p(Y)}$$.
Maybe you could say that the likelihoods induced on $a$ and $b$ are independent given $Y$?
