[EM] The voter median candidate generalized to multidimensional issue spaces. (fwd)

[Forest Simmons]

Rob LeGrand pointed out that although my example gave a different order
than Ranked Pairs, they both had the same winner B.

A better example is the following closely related one:

5 A>B>C
4 B>C>A
2 C>A>B

Then

A beats B 7 to 4,
B beats C 9 to 2, and
C beats A 6 to 5.

When the weakest win is deleted, we get the order A>B>C, which is the
winning order according to Ranked Pairs, Beatpath, Kemeny, MinMax, etc.

But the VMO (Voter Median Order) is B>A>C in agreement with Borda.

I'll leave this as an exercise for now with the hint that normalized
eigenvectors spanning the row space of the matrix A are approximately

        [-.814,.352,.462] and [.064,-.737,.673].

Does the VMO always agree with Borda?  No, in fact I showed in another
posting that

60 A>B>C
40 B>C>A

yields A>B>C as the VMO, which is pretty obvious in hindsight, since a
majority faction will always dictate the VMO.

In all examples so far, the VMO order agrees with Black.

Is there an example with three candidates in which Black does not yield
the VMO?

---------- Forwarded message ----------
Date: Thu, 9 Jan 2003 19:23:41 -0800 (PST)
From: Rob LeGrand <honky1998 at yahoo.com>
Reply-To: robl at aggies.org
To: Forest Simmons <fsimmons at pcc.edu>
Subject: Re: [EM] The voter median candidate generalized to
    multidimensional issue spaces.

> Example:
>
> 4 A>B>C
> 5 B>C>A
> 2 C>A>B
. . .
> Since 2.4 > 2.1 > 1.5 the winning order is B>A>C, in this case the same
> as the Borda order, rather than the Ranked Pairs order of A>B>C.

I think the Ranked Pairs order is B>C>A.  B is also the decisive winner
under Bucklin, Coombs, Dodgson, Nanson, Schulze and Simpson (PC, Minmax, whatever).


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