[EM] The voter median candidate generalized to multidimensional issue spaces.
Forest Simmons
fsimmons at pcc.edu
Fri Jan 10 12:32:44 PST 2003
The Borda order is the mean voter order in voter space. If the voters are
distributed symmetrically about that position, then that will also be the
median voter order.
But consider the non-symmetrical example
60 A>B>C
40 B>C>A
The Borda order is B>A>C, but the median voter order is A>B>C because
any time there is a majority faction, the median point in voter space (in
any direction) is automatically the point represented by that faction.
Let's go through the procedure on this example to see how it works:
The first matrix has 60 rows of [1,0,-1] and 40 rows of [-1,1,0].
The average is v0=[.2,.4,-.6].
The matrix A (after subtracting out v0 from each row) is 60 rows of
[.8,-.4,-.4] and 40 rows of [-1.2,.6,.6].
Since these rows are at 180 degrees from the others, the row space of A is
one dimensional, and the product of the transpose of A with A has only one
positive eigenvalue, for which a normalized eigenvector is
v1=[2,-1,-1]/sqrt(6).
The product A*v1 has sixty components of 2.4/sqrt(6) and forty components
of -3.6/sqrt(6).
So the median component value is c1=2.4/sqrt(6).
[Here's where it is most obvious that the procedure will always pick the
majority faction point if there is one.]
Then the Voter Median Order is given by the point X=v0+c1*v1, which is
[.2,.4,-.6]+(2.4/sqrt(6))*[2,-1,-1]/sqrt(6),
which simplifies to [1,0,-1], the numerically encoded order of the
majority faction.
Forest
On Fri, 10 Jan 2003 stephane.rouillon at sympatico.ca wrote:
> Forest --
>
> I like the mathematical approach...
> But maybe I did not follow all.
> It seems to me that you will always find the Borda order.
> Could you summarize in words why you could not?
> In addition, your starting weights makes me believe
> your method is crowding dependent...
>
> Hope I am wrong, show it to me.
> Steph.
>
> >
> > Date: 2003/01/09 jeu. PM 08:46:46 GMT-05:00
> >
> > Example:
> >
> > 4 A>B>C
> > 5 B>C>A
> > 2 C>A>B
> >
> > We start with a matrix whose rows represent the ballots. It has four rows
> > of [3,2,1], five rows of [1,3,2], and two rows of [2,1,3].
> >
> > The average of all eleven rows is the center of gravity vector
> > V0=[21,25,20]/11. We can see from this that B is the Borda winner and C
> > the Borda loser.
> >
> > Subtracting this V0 from every row we get a matrix A with four rows of
> > [12,-3,-9]/11, five rows of [-10,8,2]/11, and two rows of [1,-14,13]/11.
> >
> > The product of the transpose of A with A is a three by three matrix whose
> > rows are [98,-52,-46]/11, [-52,68,-16]/11, and [-46,-16,62]/11.
> >
> > The normalized eigenvectors corresponding to positive eigenvalues are
> > v1=[.814...,-.462...,-.3518...] and v2=[.06369...,.6731...,-.7367...], so
> > the issue space is two dimensional.
> >
> > The vector A*v1 has four entries of approximately 1.3, five of -1.14, and
> > two of .246, so the median is c1=.246 .
> >
> > The vector A*v2 has four entries of .489, five of .298, and two of -1.72,
> > so the median is c2=.298.
> >
> > Then X=V0+c1*V1+c2*V2 is approximately [2.1,2.4,1.5].
> >
> > Since 2.4 > 2.1 > 1.5 the winning order is B>A>C, in this case the same as
> > as the Borda order, rather than the Ranked Pairs order of A>B>C.
> >
> > Forest
> >
> > On Wed, 8 Jan 2003, Forest Simmons wrote:
> >
> > > Suppose that (in a certain election) candidate X is preferred over any
> > > other candidate Y on any issue Z by some majority (depending on Y and Z).
> > >
> > > Such a candidate would seem like a logical choice for winner of the
> > > election, if there were such a candidate.
> > >
> > > How could we locate such a candidate if there happened to be one?
> > >
> > > Otherwise, how might one find the candidate closest to this ideal?
> > >
> > > Here's an idea along these lines:
> > >
> > > Form the matrix whose entry in row i and column j is the rank or rating of
> > > candidate j by voter i.
> > >
> > > Let V0 be the average of all the row vectors of the resulting matrix.
> > > Subtract V0 from all of the rows of this matrix, and call this new matrix
> > > A. Let B be the transpose of A.
> > >
> > > Let V1, V2, etc. be normalized eigenvectors of the matrix product B*A, in
> > > descending order of the magnitudes of the eigenvalues (until the
> > > eigenvalues are too small to represent anything other than roundoff
> > > error).
> > >
> > > V0 is the center of gravity vector, and the other V's are the
> > > principal axes of rotation, i.e. those axes (through the center of
> > > gravity) about which the rigid body (with unit masses positioned by the
> > > row vectors) can rotate without wobble.
> > >
> > > Statistically, the eigenvectors give the uncorrelated directions in voter
> > > space (the row space of A offset by the cg vector V0). They may be thought
> > > of as the voter space shadows of the decorrelated issues, in order of
> > > importance to the voters.
> > >
> > > Sort (by numerical value) the components of the vector given by the
> > > product A*V1 to find the median value c1.
> > >
> > > Sort (by numerical value) the components of the vector given by the
> > > product A*V2 to find the median value c2.
> > >
> > > Etc.
> > >
> > >
> > > Let X be the vector V0+c1*V1+c2*V2+...
> > >
> > > Then X is the voter space position of the ideal candidate we are looking
> > > for. In other words, if voter space were identical to issue space, then a
> > > candidate positioned at X would be unbeaten on any of the (uncorrelated)
> > > issues by any of the other candidates.
> > >
> > > What do we do in the almost sure case that no candidate vector (image in
> > > voter space) points in the direction of X ?
> > >
> > > Here are two suggestions out of many possibilities:
> > >
> > > (1) Find the candidate vector (image) that has the largest dot product
> > > with X.
> > >
> > > (2) Look at the components of X, and award the win to the candidate
> > > corresponding to the largest component. In other words, let the
> > > components of X order the candidates from winner to loser.
> > >
> > > After experimenting I prefer the second approach for reasons which I will
> > > explain if this becomes a hot topic.
> > >
> > > Remark. I'm sure that this method doesn't satisfy the Local IIAC, because
> > > each candidate, no matter how marginal in support contributes something to
> > > the issue space profile. In other words, even the losing candidates help
> > > to plumb the depths of issue space.
> > >
> > > Any method that makes intentional, intelligent use of the issue space
> > > information inherent in the distribution of votes in voter space should
> > > not be concerned with satisfying any version of the IIAC, because the
> > > voter response to each candidate helps outline the shape of issue space.
> > >
> > > Sometimes I think that the best way to look at Arrow's Paradox is that the
> > > IIAC was just one of those "good" ideas that didn't pan out. It was based
> > > on a lack of imagination; nobody thought of using the profiles of the
> > > losing candidates to help determine the relative positions of stronger
> > > candidates in issue space.
> > >
> > > On the other hand, intelligent issue space methods should automatically
> > > get rid of clone problems, because clones just give redundant information
> > > about the shape of issue space. The singular value decomposition
> > > decorrelates whatever residual information there might be when the
> > > redundancy is factored out, just as it finds uncorrelated linear
> > > combinations of random variables in taxonomy problems associated with
> > > measuring and recording various dimensions and features of related plants,
> > > for example.
> > >
> > > [In that context the SVD brings out the distinguishing features for
> > > classification of related species. Similar uses are made in image
> > > processing. If nostril width is directly proportional to ear lobe
> > > thickness, the SVD will find that out automatically and count these two
> > > measurements as only one dimension.]
> > >
> > > Thanks for your patience in wading through these rather technical musings.
> > >
> > > Forest
> > >
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> > >
> > >
> >
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