[EM] Arrow's Theorem - The Return (again)

Alex Small asmall at physics.ucsb.edu
Mon Aug 4 11:36:01 PDT 2003


Eric Gorr said:
> I have found no evidence, in the original work, to support this claim.
>
> His proofs are based upon the relation 'R' takes into full account
> equal or distinct rankings.
>
> While I could write to Dr. Arrow again, I would only want to do so if
> presented with a compelling reason based on the original work.

Consider this argument:

If ALL voters have, or at least choose to report, dichotomous rankings (a
voter places the candidates within two groups, and is indifferent among
the candidates within each group but has a preference between the 2
groups), then there is ALWAYS a Condorcet winner, i.e. a candidate who
defeats all other candidates pairwise.  Moreover, Approval Voting always
finds this winner.

(For more on this claim, go look at Brams and Fishburn's book _Approval
Voting_.  I believe it's in Chapter 2, but I don't have the book in front
of me.)

Anyway, if we regard Approval Voting as a system that uses ranked ballots
(a claim that I dispute, but I'll go along with it for now to make my
point), then we find the following:

1)  Approval Voting is non-dictatorial.  (I HOPE nobody argues with this
one.)
2)  Approval Voting is Pareto efficient, since if every single voter
prefers A to B then B will get no approval votes but A will get unanimous
approval.  B will therefore lose, and A will win, unless A ties with
another unanimously approved candidate.
3)  Approval Voting is a ranked method (I know, most other people here are
skeptical of that claim, but Eric isn't, and I'm trying to make a point
here).
4)  Approval voting can handle 3 or more candidates.
5)  Approval Voting satisfies IIA.

That last claim needs proof.  Take as a given for now that when ALL voters
have dichotomous preferences Approval Voting finds the candidate who
defeats all other candidates pairwise (although he probably won't get a
majority in all pairwise contests, since a lot of voters might be
indifferent between 2 candidates).  This claim is documented and proven in
Brams and Fishburn.  If you send me to the library, I'll send you to the
library ;)

Anyway, if we add another candidate and assume that all voters still have
dichotomous preferences (they just put him in one of their two categories)
then there will still be a Condorcet candidate, and it will either be the
previous Condorcet winner, or the new guy.  It all comes down to whether
the new guy defeats the previous winner pairwise.


Uh, oh, we now have a ranked voting system that satisfies IIA, Pareto, and
non-dictatorship when we have 3 or more candidates.  What went wrong?

Nothing.  We limited ourselves to the case where ALL voters have
dichotomous or binary preferences.  We have a mapping from the set of all
voters with binary preferences to the set of all candidates, and we find
that it satisfies all of Arrow's criteria.  But Arrow specified that the
mapping should be from ANY possible set of voter preferences (including
sets in which at least one voter has trichotomous preferences) to the set
of all candidates.  If we try to "enlarge" our function, to handle a
larger domain that includes electorates where at least one person has
trichotomous preferences (or perhaps even more finely differentiated
preferences), we will find that at least one criterion will have to give
way.

We might decide to get rid of the assumption that the method allows each
voter to report his true preferences, WHATEVER THEY MIGHT HAPPEN TO BE. 
Then we have Approval Voting.

We might get rid of IIA, which is flunked by virtually any Pareto
Efficient and non-dictatorial method that we can think of.


Finally, a question for Eric:

What will it take to persuade you that Approval is not a ranked method in
the sense that voters with ANY preference order are ABLE to report their
actual preferences, WHATEVER those preferences might happen to be?  Will
anything short of Arrow's assurances do the job for you?



Alex





More information about the Election-Methods mailing list