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

Forest Simmons fsimmons at pcc.edu
Mon Jan 13 15:42:02 PST 2003

A simple example illustrates the Achilles heel of the VMO (Voter Median


The axis of symmetry is the line from ABC through the midpoint X of the
line joining BCA and CAB.

This midpoint X is closer to CBA than to ABC on the line joining these two
opposite orders, so the VMO is BCA.

MinMax, Ranked Pairs, Kemeny, and Borda (hence Black) all yield the
opposite order ABC, which makes sense because this example is obtained
from a perfectly symmetrical set of ballots by adding one ABC ballot.

Why does the VMO give the exact opposite of the "right" answer?

Because the symmetry breaker ballot ABC reveals an issue axis that was
hidden before that ballot was included.

Is that issue axis real? Or is it just the result of random fluctuation?

It is impossible to answer this question with any certainty in such a
statistically small sample.

For this reason, it appears that the VMO is not reliable for small

Before leaving this example, consider the symmetrical subset of ballots

2 CAB,

and suppose that you had a way of knowing (independent of the ballots)
that the main issue axis was parallel to the perpendicular bisector of the
segment joining BCA and CAB.

Since the BCA and CAB ballots are closer to the CBA end of issue space
than they are to the ABC end, it is clear that a 4:2 majority prefers the
CAB outcome to the ABC outcome on the main issue, in spite of the symmetry
of the ballot distribution.

In general, there is no reliable way to determine the issue space axes
in the case of symmetry or near symmetry of ballot distribution, so the
VMO is not reliable except when the ballot distribution itself reveals
statistically significant issue dimensions.

A statistical test of the hypothesis that the eigenvalue estimators are
distinct would show us the level of confidence that we could have in the

Here's a similar example where the computations are easy:


Any method based on the pairwise matrix rightly yields a precise three way
tie.  This includes Borda and all of the popular Condorcet Methods.

However, the VMO order is CBA. In fact, the unique median position X is
exactly the same as in our first example above.

The positive eigenvalues corresponding to eigenvectors

              [1 0 -1] and [-1 2 -1], respectively,

are seven and three.

The ratio 7/3 appears to be significantly greater than unity, but with
only a few of degrees of freedom to be shared between the numerator
and denominator, we cannot have much confidence in this conclusion.

Note that the top VMO candidate is also the IRV winner.  That's not too


On Fri, 10 Jan 2003, Forest Simmons wrote:

> One of Markus' clone examples shows that the Black order is not always the
> same as the VMO (Voter Median Order):
> 40 ABCDE
> 40 BDECA
> 40 ECDAB
> 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
> triangle.
> 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.
> Forest
> 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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