So to go from median to mean you just keep adding one datum above
and one below. Thanks for that.
Your idea about a "gradual Black's" method is interesting
for a different reason. Black's method actually has some
mometum behind it at the moment, as it is advocated by
Dasgupta and Maskin. I couldn't find the original paper online,
but below is a link to their article in Scientific American.
Their justification is what they call the "majority dominance
theorem" which says that Black's method is a counter-example
to Arrow's theorem, and is the only such method, all this provided
that certain not-too-onerous profile restrictions are satisfied.
http://www.scientificamerican.com/article.cfm?id=ranking-candidates-more-accurate
(What they call "majority rule" is Condorcet, and what they propose is Black.
The
article is verging on non-technical.)
I wonder how "gradual Black" would fit in.
--- El mar, 11/1/11, Kristofer Munsterhjelm escribió:
De: Kristofer Munsterhjelm
Asunto: Re: [EM] a Borda-Condorcet relation
Para: "Stephen Turner"
CC: election-methods@lists.electorama.com
Fecha: martes, 11 de enero, 2011 00:47
Stephen Turner wrote:
> Kathy, yes the Borda winner assigns r points to the (n-r)th
> ranked candidate. It's a method with important problems.
>
> Robert: yes, that was the point, a sort of
> Condorcet:Borda::median:mean, if you like. There is no
> new method or algorithm here.
>
> Kristofer: yes, it is a very simple observation. Your comments
> about burial are very interesting, and show how
> we might be able to apply analysis of (one of) Condorcet
> or Borda methods to the other.
>
> Your way of passing from median to mean, I'm not sure
> I understood completely. Is it something like this?
>
> - median
> - mean of central 5%
> - mean of central 10%
> - etc
>
> where "central 5%" would mean that you discard the 47.5%
> highest and the 47.5% lowest values (with multiplicity
> of course), and when these don't give whole numbers,
> interpolate linearly. Is this right?
More or less, but since the number of voters are exact, there's little need to
interpolate.
Say that (for the sake of simplicity) the number of voters is odd, so that the
median is a single value rather than the mean of two, and that you have one
array for every candidate pair, containing the rank differences/distances as
you showed in your previous post. Then, define
f(x, y, 0) to be the median rank difference, your c(x, y).
f(x, y, 1) is the mean of the element just above the median, the median, and
the element just below median.
f(x, y, k) is the mean of the element just above the upper element used in f(x,
y, (k-1)), the elements used in f(x, y, (k-1)), and the element just below the
lower element used in f(x, y, (k-1)).
This assumes that all the elements (rank differences) have been sorted so that
finding the median is as easy as picking the central element (value) in the
sorted array.
The method would then work like this, in the absence of true ties:
find the minimal p >= 0 so that there exists a candidate X for which, for any
and all other candidates Y, f(X, Y, p) > 0.
If there's a Condorcet winner, then X is the CW and p = 0, because when p = 0,
f is equivalent to your c function. We also know that this method can't be any
less decisive than Borda, simply because at the maximal value of p, the Borda
winner will satisfy the stopping rule above.
An actual algorithm would gradually increase p until the stopping rule is met.
The method concedes just enough in the direction of Borda, widens the truncated
mean just enough, to get an unambiguous result. When dealing with weighted
votes, the problem becomes a little more complex and that's where one would use
a sweep-line algorithm.
Practically speaking, one would also use sums instead of means, so that
expanding the truncated mean is as simple as adding a few more values.
Also note that I haven't dealt with the case where there are true ties (e.g.
everybody votes A = B). In the case of a true tie, one wouldn't want the method
to go all the way to Borda, trying to break a tie that shouldn't (and can't) be
broken.
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