I've been searching for almost a day now and I'm running out of steam... can someone give me a nice, easy to grasp explanation/example of what a weighted product does compared to what a weighted sum does? I'm having trouble coming up with an interpretation I can work with, especially when looking at weighted joint probabilities.
To be perfectly clear, I'm talking about $P(X) = \prod_i p(x_i)^{w_i}$ vs. $S(X) = \frac{\sum_i w_i * p(x_i)}{\sum_i w_i}$.
Note that the full formula for a weighted product is $$ P(X) = \left[\prod_{i}p(x_i)^{w_i}\right]^{1/\sum_{i}w_i} $$ of course, if $\sum_i w_i = 1$, then this reduces to your formula. Similarly, if $\sum_{i}w_i = 1$, then the formula for the weighted sum is $\sum_{i} w_i x_i$. Of course, every weighted sum can be taken using the $\sum_i w_i = 1$ if we simply redefine our weights to be $\alpha_i = \frac{w_i}{\sum_i w_i}$.
There is a direct connection between the weighted sum and weighted product. In particular, let $\exp(x) = e^x$, let $\exp(X) = \{\exp(x): x \in X\}$, and let $\log(Y) = \{\log(y) : y \in Y\}$ (log here is the natural log). Then we have $$ \log(P(X)) = S(\log(X)) $$ In other words, taking the weighted product produces a value whose logarithm is the weighted sum of the logarithms. Per the earlier discussion, it suffices to prove that this is the case when $\sum_i w_i = 1$; I don't think that you will find the proof particularly difficult.
Another piece of intuition: a weighted sum generalizes the arithmetic mean in precisely the same way that a weighted product generalizes the geometric mean.
