# Weighted convolution?

Suppose I have two arbitrary discrete probability distributions with the same domain.

I want to convolve the two together to come up with third distribution, however I want them to be weighted.

Here's what I have so far.

Let's say that the distributions' domain goes from $[1,N]$. I get the $N$ probability values of both distributions and perform a DFT. I now have two lists of complex numbers of length $N$.

I now multiply the elements of both lists together $c_i = a_i \times b_i$ and preform a reverse DFT to get the convolved distribution.

This is a normal, unweighted convolution.

Let's say I want "more of distribution $a$ than distribution $b$ inside the convovled distribution $c$". I cannot simply say $c_i = (w_1 \times a_i) \times (w_2 \times b_i)$ where $w_1 > w_2$ and $w_1 + w_2 = 1$ because the individual weights lose their meaning and are applied to both terms.

I'll tell you what I'm doing and why I want this:

I have a list of elements, and I need to select one. I have two very different ways of choosing which element is the "fittest", which absolutely cannot be compared.

I create two probability distribution functions for the list of elements for each of the different fitness metrics and convolve them together. I then choose a random element from the list based on the convolved distribution.

This is fine if I want to give equal weighting to both metrics, but, in fact, I want to arbitrarily specify the weighting of the two metrics when deciding which element to choose.

Any ideas? Can this even be done or am I approaching this from an entirely wrong direction?

If I want to give metric $a$ a weighting of $2$ and metric $b$ a weighting of $1$, would I need to do $c = (a \star(a \star b))$? In which case performing arbitrary weightings would end up being a little cumbersome?

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Have you considered a [mixture model | en.wikipedia.org/wiki/… ] for solving this problem? –  Tim Duff Nov 12 '12 at 6:00
The (unweighted) convolution yields the distribution for the sum of the two variables (modulo $N$ if you calculate it with a DFT of length $N$). If $1$ through $N$ represent $N$ different items from which to choose, what meaning does the sum of the indices of two items have? –  joriki Nov 12 '12 at 7:57
@TimDuff I implemented what you suggested. Adding together weighted probabilities does seem a lot simpler than convolving them together. –  Ozzah Nov 13 '12 at 22:25