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

Forest Simmons fsimmons at pcc.edu
Fri Jan 10 17:49:33 PST 2003

One of Markus' clone examples shows that the Black order is not always the
same as the VMO (Voter Median Order):


There is no CW so Black picks the Borda winner B.

The Borda order (strongly influenced by the clone group {C,D,E}) is

                B > C=D=E > A.

If my calculations are right, the VMO order is  A=C > E > B=D.

Here's some insight into this result:

In the natural coordinate system based on the two eigenvectors
corresponding to the two positive eigenvalues, and with center at the
center of gravity, the three respective factions reside at the points

              P1=(-9,-4.1), P2=(0,8.2), and P3=(9,-4.1).

This is an isosceles triangle with the long side parallel to the first
eigenvector (the eigenvector with the greatest eigenvalue).  The two
shorter sides have a length of about 15.3, while the long side has a
length of 18, so the triangle is nearly equilateral, which is one way of
understanding the difficulty of deciding a winner.

In this coordinate system the median coordinates are 0 and -4.1,
respectively, so in the taxicab metric relative to this coordinate system,
the point X=(0,-4.1) minimizes the total distance to the voter positions.

In the taxicab metric

          d(P1,X)=9=d(P3,X), while d(P2,X)=12.3 .

So the median position strongly favors the first and third factions which
are represented by the endpoints of the long side of the isosceles

In fact, the median position for the triangle is the midpoint of the long
side, so it makes sense that the VMO should correspond to the average of
the two faction orders represented by the endpoints of that side.

If we restrict the Borda Count to those two factions, then the order is
precisely A=C > E > C=D, the VMO order.

So the result makes sense when you know the shape of the distribution of
voters in Voter Space, which presumably reflects the distribution of
voters in issue space.

It remains to be seen if there is a three candidate example in which Black
does not agree with the VMO.


On Fri, 10 Jan 2003, Forest Simmons wrote:

> 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?

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