I am facing a statistical problem that I am not sure whether it's solvable. Simply put, I am given multiple weight vectors and here are two examples:

$w_1 = [.2, .3, .4, .1]$ for items A, B, C, D, respectively.

$w_2 = [.1, .1, .05, .05, .03, .03, .02, .5, .12]$ for items A, B, C, D, E, F, G, H, I, respectively.

and I have many weight vectors like this (sum of all elements is 1), but of different sizes.

Now I want to construct an "importance vector" for all 26 items (A ~ Z), each entry represents each item's importance.

Since many weight vectors are of different sizes, their weights are of different scale. I don't think it's a good idea to simply add them up and take average because of the reason stated before. Any ideas? Or any pieces of literature that I can refer to? Or what is this kind of problems called?

  • $\begingroup$ So the importance vector would use the weight information for each item across all these vectors, right? $\endgroup$ – Lucas Farias Feb 2 at 14:18
  • $\begingroup$ @LucasFarias Yes $\endgroup$ – zcylywde Feb 2 at 15:20

You're right. Mathematically speaking, your vectors belong to different vector spaces, so they can't be compared or aggregated directly. Unless you somehow project them into the same vector space.

In this sense, given the data in these vectors are probabilities, isn't it the case that the probability of an item that does not belong to a particular vector is $0$? Namely, $w_1 = \left[0.1,0.2,0.7\right]$ for $A,B,C$ imply $\mathbb{P}(D)=0$.

If that's so, you can project them all onto $\mathbb{R}^{26}$, and this would allow you to aggregate (e.g. sum, average, etc.) them properly. By this I mean that the resulting projection of the $w_1$ of my example would be $w_1^{*} = \left[0.1,0.2,0.7,0,...,0\right]$.

  • $\begingroup$ I am afraid I cannot simply infer P(D) = 0. In my weight vector, e.g. $w_1 = [0.1, 0.2, 0.7]$ for A, B, C, one piece of information should be "C is more important than B", but I do not know whether C is more important than D or than the rest of letters that are not included in this vector. $\endgroup$ – zcylywde Feb 4 at 3:22
  • $\begingroup$ @zcylywde But if the entries are probabilities, and the importance is a function of these probabilities, unless the array entries don't sum to one, how come an item would have any importance? $\endgroup$ – Lucas Farias Feb 4 at 13:06

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