2011/7/8 Toby Pereira <[email protected]> > While discussing median-based range voting - > http://rangevoting.org/MedianVrange.html, Warren Smith says "Average-based > range voting generalizes to a *multiwinner* proportional > representation<http://rangevoting.org/PropRep.html>voting system called > reweighted > range voting <http://rangevoting.org/RRV.html>. (See papers 78 and 91 > here<http://www.math.temple.edu/~wds/homepage/works.html>.) > But there currently is no known way to generalize median-based range voting > to do that." >
I've told Warren to change that, and he hasn't given me a clear criterion for what I have to do so he will. I've created a system called AT-TV which is PR and reduces to a median-based system in the single-winner case. It's Bucklin-like, in that there is a falling approval threshold, and when a candidate gets enough approvals to be elected (a Droop quota) they are, which "uses up" those votes (except for the excess). So in a one-winner case, it's based on 50th percentile (median), but in, for instance, a 3-winner case, it would be (pseudo-)maximizing the elected candidates' 75th-percentile score, not their 50th-percentile. I think this is the appropriate thing to do in the multi-winner "median" case. JQ > > So I was thinking about how you might get a median-based PR system, using > range voting, or some other score system, such as Borda Count. I don't think > there is necessarily a "perfect" method but I did come up with something > (possibly ridiculous). You find a way to convert the scores of the > candidates so that a candidate's median score becomes their mean score. For > example, if a candidate's mean was 5 (out of 10) and their median was 7, > their scores would undergo some sort of transformation so that their mean > score became 7. Likewise if someone had a mean of 7 and a median of 5, their > scores would undergo a transformation to reduce the mean to 5. > > One way to do this is as follows: Convert the range so that it becomes 0 to > 1 (so in a 0-10 case, just divide all scores by 10). Then for each candidate > you convert their score s to s^n where n is the number for that particular > candidate that will make the original median score the mean of the > transformed scores. For n over 1 the score will be reduced and for n under > 1, the score will increase. So each candidate has their own value of n. > > Once all the scores have been converted, you can just do whatever you would > have done in your non-median-based PR system to find the winning candidates. > > Obviously, this is a bit of a fudge because although we are fixing the mean > for each candidate to what we want, the rest of the scores just end up how > they end up. There would be different conversion systems that convert median > to mean but give different values for the other scores. > > Just looking at the median and mean here could be seen as a bit arbitrary. > As well as converting median to mean, we would ideally also want to convert > other percentiles accordingly. We'd want to convert the 25th percentile > score to the 25th "permeantile"*, or whatever the term is. (Is there a > term?) But it would actually be impossible to do this properly. With > repeated scores (which would always happen where there are more voters than > possible scores), different percentile values will have the same score. For > example, if someone's median score is 5, it's likely to also be 5 at the > 51st percentile. But, as far as I understand it, the "permeantile" would not > be able to have a flat gradient at any point, unless it's flat all the way > across. So we couldn't have a "perfect" system that worked on this basis. So > for simplicity we can just use the system as described. > > Of course, with range voting, people might vote approval style, so many > candidates might simply have a median of 0 or 10. In that case the only > "reasonable" conversion would be to convert all their scores to 0 or 10 > respectively. This problem wouldn't occur to the same degree under Borda > Count, however. > > *I was thinking about how you would calculate permeantiles. In a uniform > distribution between 0 and 1, the 25th permeantile would be 0.25. If you > weight the averages of each side 3 to 1 in favour of the smaller side of the > permeantile (0 to 0.25), and average these, then you get 0.25. (3*0.125 + > 1*0.625) / 4 = 0.25. So for the 10th permeantile, you have (9*0.05 + 1*0.55) > / 10 = 0.1 and so on. I imagine this would work for non-uniform > distributions too. (Sorry for going off topic.) > > ---- > Election-Methods mailing list - see http://electorama.com/em for list info > >
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