[EM] Median-based Proportional Representation

[Jameson Quinn]

2011/7/8 Toby Pereira <tdp201b at yahoo.co.uk>

> 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
>
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.electorama.com/pipermail/election-methods-electorama.com/attachments/20110708/aea9ab1b/attachment-0004.htm>