[EM] Consensus and PR methods

Richard Lung voting at ukscientists.com
Wed Mar 7 17:11:39 PST 2018

If my old memory serves me tolerably well, isn't this paper something 
like an article entitled The Best of Both Worlds, where the authors did 
a survey of a tendency for European electoral systems, over the decades, 
to have decreased their average magnitude. I forget the details, just 
about everything actually. But it may have gone something like: the 
constiuencies shrank and the thresholds got higher.
It was an informative statistical survey.
But I think it went awry on what academics are fond of calling 
"normative" considerations. Or on the stricture of David Hume, that what 
is, is not necessarily right.
I would have put to the authors, as a critic. That was this trend, they 
so diligently exposed, but the moving to a "sweet spot" for political 
incumbents, with precious little to do with democracy and effective 
elections for the voters?

Richard Lung.

On 07/03/2018 19:19, Jack Santucci wrote:
> Consensus in academia? Maybe that cigarettes cause cancer. Maybe.
> I jest.
> This paper may be helpful: 
> http://personal.lse.ac.uk/hix/Working_Papers/Carey-Hix-AJPS2011.pdf
> On Wed, Mar 7, 2018 at 2:14 PM, Richard Lung <voting at ukscientists.com 
> <mailto:voting at ukscientists.com>> wrote:
>     So, the academic world has no consensus or standard model of
>     election method?
>     On 03/03/2018 19:57, Kristofer Munsterhjelm wrote:
>         Say we have a consensus method M that works by choosing the
>         council C that minimizes the maximum penalty p(C, v) for the
>         voter that maximizes this penalty. That is, the method finds C
>         according to
>         C = arg min max p(c, v)
>                  c   v
>         where ties are broken in a leximax fashion (i.e. considering
>         next to max, then next to next to max and so on). Furthermore
>         let the penalty "nonnegative" in the sense that any voter with
>         a real preference has at least as great a penalty as a voter
>         with no preference (the zero voter, as it were).
>         Now let the modified consensus method M' be one that has the
>         same optimization objective, but the method is permitted to
>         remove a Droop quota of votes to help minimize the penalty.
>         So M says "what council displeases the most displeased voter
>         the least?", while M' says "what council displeases the most
>         displeased voter the least, if we can discard a Droop quota of
>         voters from consideration?"
>         Then, are there any properties for p that makes M' satisfy
>         Droop proportionality? Can we in general turn consensus
>         methods of this form into PR methods by adding a "you can
>         discard a Droop quota" rule?
>         If we can, then we easily get a multiwinner version of
>         Bucklin/MJ by doing the following:
>         Let g(c, v) be the grade voter v gives to the least preferred
>         candidate in c.
>         Let the consensus method M be
>         C = arg max min g(c, v)
>                  c   v
>         Let M' permit the method to remove a Droop quota, i.e. if |V|
>         is the number of voters, and V is the set of voters itself:
>         C' = arg max c:
>             max x subset of V so that |x| = |V|/(seats+1):
>                 min v in V \ x:
>                     g(c, v)
>         For a single-winner election, M' is (up to tiebreaker) just
>         MJ, because for each potential winner c, the removal step will
>         remove the voters who grade c the worst, and the Droop quota
>         for single-winner is a majority. Then the voter grading the c
>         the worst after half of the voters have been removed is just
>         the median voter.
>         Some thoughts about two-winner remove-voter minimax Approval
>         follow. Reasoning about what voter removal actually does can
>         get kinda hairy, thus I may very well be wrong:
>         In minimax Approval, p(c, v) is the Hamming distance between c
>         and voter v's ballot, i.e. the number of candidates in c but
>         not approved by v plus the number of candidates approved by v
>         not in c.
>         Say we have an analogous Droop criterion for Approval: if more
>         than k Droop quotas approve of a set of j candidates (and
>         nobody else), then at least min(k, j) of these must be elected.
>         For two winners, there are these possibilities:
>             1. no Droop constraints
>             2. k = 2, j >= 2
>             3. k = 2, j = 1
>             4. k = 1, j >= 1
>             5. k = 1, j = 1
>         1. is no problem, because we can elect anyone we want without
>         running afoul of the Approval DPC.
>         2. Since there can only be three Droop quotas in total, when
>         we're considering A = {C_1, C_2} with C_1 and C_2 in the set
>         of j candidates (call it J), we can eliminate all but the
>         J-voters and the maximum penalty is j-2.
>         In contrast, for some B = {C_x, C_y} not a subset of j, the
>         best it can do is eliminate a Droop quota of the J-voters. In
>         the best case (for B), everybody but the J-voters approve of B
>         alone. But there still remains a Droop quota (plus one voter)
>         of the J-voters, and each of them gives penalty j. So A is
>         preferred to B.
>         If B = {C_1, C_x}, then even if everybody but the J-voters
>         approve of B alone, the J-voters give penalty j-1. So A is
>         still preferred to B.
>         3. Same as in 2, but let A = {C_1, C_x}, J = {C_1}. With A, we
>         eliminate so that only the J-voters are left, and then max
>         penalty is 1 (for C_x). Furthermore, every remaining voter
>         gives penalty 1. Let B = {C_x, C_y}. In the best case for B, a
>         Droop quota of J-voters are eliminated and we have a Droop
>         quota plus one left. These all give penalty 2, which is worse
>         than penalty 1. So A is preferred to B.
>         5. Let A = {C_1, C_x} and B = {C_x, C_y}. In the best case for
>         B here, two Droop quotas minus a voter approve only of B, and
>         the remaining Droop quota plus one voter approves of J =
>         {C_1}. Eliminating all but one of the J-voters gives a max
>         penalty of 3 from that one J-voter: one point for not having
>         C_1, and two points for having C_x and C_y. A eliminates one
>         of the two B-approving Droop quotas and gets a penalty of 1
>         from every remaining voter, which is better.
>         Note that I assume that C_x is approved by the B-voters. If
>         that were not the case, then {C_x, C_y} would already be
>         beaten by some {C_z, C_y} where C_z is. Note also that I don't
>         consider the case where the B-voters also approve of a whole
>         load of other candidates, with the idea of raising the penalty
>         under A. The problem is that because only two candidates can
>         be elected, this would also raise their penalty under B.
>         4. Let A = {C_1, C_x} and B = {C_x, C_y}. The best case for B
>         has worst penalty j+2, since after a Droop quota of J-voters
>         have been eliminated, there remains a single voter who only
>         approves of J. After eliminating some of the B-voters, A gets
>         penalty j from the J-voters (j-1 for the members of J not part
>         of {C_1, C_x} and one more for C_x which is not approved by
>         them), and one penalty point from the B-voters.
>         Here it'd seem that adding loads of candidates to the B-voters
>         would make things hard. Can it be salvaged?
>         Suppose there are J-voters and C-voters. B is a subset of C.
>         When considering outcome B, before excluding a Droop quota,
>         every J-voter gives a penalty of j+2 and every C-voter gives a
>         penalty of c-2 where c=|C|.
>         Under outcome A, before excluding, every J-voter gives j, and
>         every B-voter gives c (-1 for having C_x, +1 for having C_1).
>         If j+2 > c, then we're in the domain above, and no problem.
>         If c > j+2, then the excluded candidates under both A and B
>         are C-voters.
>         So under B we have a Droop quota of C-voters with penalty c-2,
>         and a Droop quota plus one of J-voters at j+2.
>         Under A we have a Droop quota of C-voters with penalty c, and
>         a Droop quota plus one of J-voters at j.
>         So unless I made a mistake, Hamming distance is not good
>         enough. But I might have made a mistake, because it seems
>         strange that even in ordinary minimax Approval, a coalition
>         can increase its power by approving a lot of clones. E.g.
>         suppose in ordinary minimax Approval that there are two
>         coalitions of almost equal size:
>         n+1: A B
>         n: C1 C2 C3 ... Cq
>         {A, B} gets worst penalty q+2 (there are n of these and n+1
>         zeroes)
>         {A, C1} gets worst penalty q (n voters like C1 but not A)
>         {C1, C2} gets worst penalty q-2 (n voters give this penalty,
>         and then n+1 give penalty 4).
>         ... does that mean an arbitrarily small minority can force its
>         own council to win by just approving enough clones that they
>         set the worst penalty in every outcome? That feels rather wrong.
>         ----
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>     -- 
>     Richard Lung.
>     http://www.voting.ukscientists.com
>     <http://www.voting.ukscientists.com>
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> -- 
> Jack Santucci, Ph.D.
> Independent scholar
> http://www.jacksantucci.com

Richard Lung.
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