Movatterモバイル変換


[0]ホーム

URL:


Jump to content
WikipediaThe Free Encyclopedia
Search

Consistency criterion

From Wikipedia, the free encyclopedia
Property of electoral systems
A jointPolitics andEconomics series
Social choice andelectoral systems
iconMathematics portal

Avoting system satisfiesjoin-consistency (also called thereinforcement criterion) if combining two sets of votes, both electingA overB, always results in a combined electorate that ranksA overB.[1] It is a stronger form of theparticipation criterion. Systems that fail the consistency criterion (such asinstant-runoff voting orCondorcet methods) are susceptible to themultiple-district paradox, apathological behavior where a candidate can win an election without carrying even a single precinct.[1] Conversely, it can be seen as allowing for a particularly egregious kind ofgerrymander: it is possible to draw boundaries in such a way that a candidate who wins the overall election fails to carry even a singleelectoral district.[1]

Rules susceptible to the multiple-districts paradox include allCondorcet methods[2] andinstant-runoff (or ranked-choice) voting. Rules that are not susceptible to it include allpositional voting rules (such asfirst-preference plurality and theBorda count) as well asscore voting andapproval voting.

Variants

[edit]

There are three variants of join-consistency:

  1. Winner-consistency—if two districts elect the same winnerA,A also wins in a new district formed by combining the two.
  2. Ranking-consistency—if two districts rank a set of candidates exactly the same way, then the combined district returns the same ranking of all candidates.
  3. Grading-consistency—if two districts assign a candidate the sameoverall grade, then the combined district assigns that candidate the same grade.

A voting system is winner-consistent if and only if it is a point-summing method; in other words, it must be apositional voting system orscore voting (includingapproval voting).[3][2]

As shown below for theKemeny rule andmajority judgment, these three variants do not always agree with each other (which contrasts with most other voting criteria). Kemeny is the only ranking-consistent Condorcet method, and no Condorcet method can be winner-consistent.[2]

Examples

[edit]

Copeland

[edit]
Main article:Copeland's method

This example shows that Copeland's method violates the consistency criterion. Assume five candidates A, B, C, D and E with 27 voters with the following preferences:

PreferencesVoters
A > D > B > E > C3
A > D > E > C > B2
B > A > C > D > E3
C > D > B > E > A3
E > C > B > A > D3
A > D > C > E > B3
A > D > E > B > C1
B > D > C > E > A3
C > A > B > D > E3
E > B > C > A > D3

Now, the set of all voters is divided into two groups at the bold line. The voters over the line are the first group of voters; the others are the second group of voters.

First group of voters

[edit]

In the following the Copeland winner for the first group of voters is determined.

PreferencesVoters
A > D > B > E > C3
A > D > E > C > B2
B > A > C > D > E3
C > D > B > E > A3
E > C > B > A > D3

The results would be tabulated as follows:

Pairwise preferences
ABCDE
A
9
5
6
8
3
11
6
8
B
5
9
8
6
8
6
5
9
C
8
6
6
8
5
9
8
6
D
11
3
6
8
9
5
3
11
E
8
6
9
5
6
8
11
3
Pairwise election results
(won-tied-lost)
3–0–12–0–22–0–22–0–21–0–3
  1. ^indicates voters who preferred the candidate listed in the column caption to the candidate listed in the row caption
  2. ^indicates voters who preferred the candidate listed in the row caption to the candidate listed in the column caption

Result: With the votes of the first group of voters, A can defeat three of the four opponents, whereas no other candidate wins against more than two opponents. Thus,A is elected Copeland winner by the first group of voters.

Second group of voters

[edit]

Now, the Copeland winner for the second group of voters is determined.

PreferencesVoters
A > D > C > E > B3
A > D > E > B > C1
B > D > C > E > A3
C > A > B > D > E3
E > B > C > A > D3

The results would be tabulated as follows:

Pairwise election results
X
Y
ABCDE
A
6
7
9
4
3
10
6
7
B
7
6
6
7
4
9
7
6
C
4
9
7
6
7
6
4
9
D
10
3
9
4
6
7
3
10
E
7
6
6
7
9
4
10
3
Pairwise election results
(won-tied-lost)
3–0–12–0–22–0–22–0–21–0–3

Result: Taking only the votes of the second group in account, again, A can defeat three of the four opponents, whereas no other candidate wins against more than two opponents. Thus,A is elected Copeland winner by the second group of voters.

All voters

[edit]

Finally, the Copeland winner of the complete set of voters is determined.

PreferencesVoters
A > D > B > E > C3
A > D > C > E > B3
A > D > E > B > C1
A > D > E > C > B2
B > A > C > D > E3
B > D > C > E > A3
C > A > B > D > E3
C > D > B > E > A3
E > B > C > A > D3
E > C > B > A > D3

The results would be tabulated as follows:

Pairwise election results
X
Y
ABCDE
A
15
12
15
12
6
21
12
15
B
12
15
14
13
12
15
12
15
C
12
15
13
14
12
15
12
15
D
21
6
15
12
15
12
6
21
E
15
12
15
12
15
12
21
6
Pairwise election results
(won-tied-lost)
2–0–23–0–14–0–01–0–30–0–4

Result: C is the Condorcet winner, thus Copeland choosesC as winner.

Instant-runoff voting

[edit]
Main article:Instant-runoff voting

This example shows that Instant-runoff voting violates the consistency criterion. Assume three candidates A, B and C and 23 voters with the following preferences:

PreferencesVoters
A > B > C4
B > A > C2
C > B > A4
A > B > C4
B > A > C6
C > A > B3

Now, the set of all voters is divided into two groups at the bold line. The voters over the line are the first group of voters; the others are the second group of voters.

First group of voters

[edit]

In the following the instant-runoff winner for the first group of voters is determined.

PreferencesVoters
A > B > C4
B > A > C2
C > B > A4

B has only 2 votes and is eliminated first. Its votes are transferred to A. Now, A has 6 votes and wins against C with 4 votes.

CandidateVotes in round
1st2nd
A46
B2
C44

Result:A wins against C, after B has been eliminated.

Second group of voters

[edit]

Now, the instant-runoff winner for the second group of voters is determined.

PreferencesVoters
A > B > C4
B > A > C6
C > A > B3

C has the fewest votes, a count of 3, and is eliminated. A benefits from that, gathering all the votes from C. Now, with 7 votes A wins against B with 6 votes.

CandidateVotes in round
1st2nd
A47
B66
C3

Result:A wins against B, after C has been eliminated.

All voters

[edit]

Finally, the instant runoff winner of the complete set of voters is determined.

PreferencesVoters
A > B > C8
B > A > C8
C > A > B3
C > B > A4

C has the fewest first preferences and so is eliminated first, its votes are split: 4 are transferred to B and 3 to A. Thus, B wins with 12 votes against 11 votes of A.

CandidateVotes in round
1st2nd
A811
B812
C7

Result:B wins against A, after C is eliminated.

Conclusion

[edit]

A is the instant-runoff winner within the first group of voters and also within the second group of voters. However, both groups combined elect B as the instant-runoff winner. Thus, instant-runoff voting fails the consistency criterion.

Kemeny method

[edit]
Main article:Kemeny method

This example shows that the Kemeny method violates the consistency criterion. Assume three candidates A, B and C and 38 voters with the following preferences:

GroupPreferencesVoters
1stA > B > C7
B > C > A6
C > A > B3
2ndA > C > B8
B > A > C7
C > B > A7

Now, the set of all voters is divided into two groups at the bold line. The voters over the line are the first group of voters; the others are the second group of voters.

First group of voters

[edit]

In the following the Kemeny winner for the first group of voters is determined.

PreferencesVoters
A > B > C7
B > C > A6
C > A > B3

The Kemeny method arranges the pairwise comparison counts in the following tally table:

Pairs of choicesVoters who prefer
XYX over YNeitherY over X
AB1006
AC709
BC1303

The ranking scores of all possible rankings are:

Preferences1 vs 21 vs 32 vs 3Total
A > B > C1071330
A > C > B710320
B > A > C613726
B > C > A136928
C > A > B931022
C > B > A39618

Result: The ranking A > B > C has the highest ranking score. Thus,A wins ahead of B and C.

Second group of voters

[edit]

Now, the Kemeny winner for the second group of voters is determined.

PreferencesVoters
A > C > B8
B > A > C7
C > B > A7

The Kemeny method arranges the pairwise comparison counts in the following tally table:

Pairs of choicesVoters who prefer
XYX over YNeitherY over X
AB8014
AC1507
BC7015

The ranking scores of all possible rankings are:

Preferences1 vs 21 vs 32 vs 3Total
A > B > C815730
A > C > B1581538
B > A > C1471536
B > C > A714728
C > A > B715830
C > B > A1571436

Result: The ranking A > C > B has the highest ranking score. Hence,A wins ahead of C and B.

All voters

[edit]

Finally, the Kemeny winner of the complete set of voters is determined.

PreferencesVoters
A > B > C7
A > C > B8
B > A > C7
B > C > A6
C > A > B3
C > B > A7

The Kemeny method arranges the pairwise comparison counts in the following tally table:

Pairs of choicesVoters who prefer
XYX over YNeitherY over X
AB18020
AC22016
BC20018

The ranking scores of all possible rankings are:

Preferences1 vs 21 vs 32 vs 3Total
A > B > C18222060
A > C > B22181858
B > A > C20202262
B > C > A20201656
C > A > B16181852
C > B > A18162054

Result: The ranking B > A > C has the highest ranking score. So,B wins ahead of A and C.

Conclusion

[edit]

A is the Kemeny winner within the first group of voters and also within the second group of voters. However, both groups combined elect B as the Kemeny winner. Thus, the Kemeny method fails the reinforcement criterion.

Ranking consistency

[edit]

The Kemeny method satisfies ranking consistency; that is, if the electorate is divided arbitrarily into two parts and separate elections in each part result in the same ranking being selected, an election of the entire electorate also selects that ranking. In fact, it is the onlyCondorcet method that satisfies ranking consistency.

Informal proof
[edit]

The Kemeny score of a rankingR{\displaystyle {\mathcal {R}}} is computed by summing up the number of pairwise comparisons on each ballot that match the rankingR{\displaystyle {\mathcal {R}}}. Thus, the Kemeny scoresV(R){\displaystyle s_{V}({\mathcal {R}})} for an electorateV{\displaystyle V} can be computed by separating the electorate into disjoint subsetsV=V1V2{\displaystyle V=V_{1}\cup V_{2}} (withV1V2={\displaystyle V_{1}\cap V_{2}=\emptyset }), computing the Kemeny scores for these subsets and adding it up:

(I)sV(R)=sV1(R)+sV2(R){\displaystyle {\text{(I)}}\quad s_{V}({\mathcal {R}})=s_{V_{1}}({\mathcal {R}})+s_{V_{2}}({\mathcal {R}})}.

Now, consider an election with electorateV{\displaystyle V}. The premise of reinforcement is to divide the electorate arbitrarily into two partsV=V1V2{\displaystyle V=V_{1}\cup V_{2}}, and in each part the same rankingR{\displaystyle {\mathcal {R}}} is selected. This means, that the Kemeny score for the rankingR{\displaystyle {\mathcal {R}}} in each electorate is bigger than for every other rankingR{\displaystyle {\mathcal {R}}'}:

(II)R:sV1(R)>sV1(R)(III)R:sV2(R)>sV2(R){\displaystyle {\begin{aligned}{\text{(II)}}\quad \forall {\mathcal {R}}':{}&s_{V_{1}}({\mathcal {R}})>s_{V_{1}}({\mathcal {R}}')\\{\text{(III)}}\quad \forall {\mathcal {R}}':{}&s_{V_{2}}({\mathcal {R}})>s_{V_{2}}({\mathcal {R}}')\end{aligned}}}

Now, it has to be shown, that the Kemeny score of the rankingR{\displaystyle {\mathcal {R}}} in the entire electorate is bigger than the Kemeny score of every other rankingR{\displaystyle {\mathcal {R}}'}:

sV(R) =(I) sV1(R)+sV2(R) >(II) sV1(R)+sV2(R) >(III) sV1(R)+sV2(R) =(I) sV(R)q.e.d.{\displaystyle s_{V}({\mathcal {R}})\ {\stackrel {(I)}{=}}\ s_{V_{1}}({\mathcal {R}})+s_{V_{2}}({\mathcal {R}})\ {\stackrel {(II)}{>}}\ s_{V_{1}}({\mathcal {R}}')+s_{V_{2}}({\mathcal {R}})\ {\stackrel {(III)}{>}}\ s_{V_{1}}({\mathcal {R}}')+s_{V_{2}}({\mathcal {R}}')\ {\stackrel {(I)}{=}}\ s_{V}({\mathcal {R}}')\quad q.e.d.}

Thus, the Kemeny method is consistent with respect to complete rankings.

Majority Judgment

[edit]
Main article:Majority Judgment

This example shows that majority judgment violates reinforcement. Assume two candidates A and B and 10 voters with the following ratings:

CandidateVoters
AB
ExcellentFair3
PoorFair2
FairPoor3
PoorFair2

Now, the set of all voters is divided into two groups at the bold line. The voters over the line are the first group of voters; the others are the second group of voters.

First group of voters

[edit]

In the following the majority judgment winner for the first group of voters is determined.

CandidatesVoters
AB
ExcellentFair3
PoorFair2

The sorted ratings would be as follows:

Candidate   
 Median point
A
 
B
 
  
 

  Excellent  Good  Fair  Poor

Result: With the votes of the first group of voters, A has the median rating of "Excellent" and B has the median rating of "Fair". Thus,A is elected majority judgment winner by the first group of voters.

Second group of voters

[edit]

Now, the majority judgment winner for the second group of voters is determined.

CandidatesVoters
AB
FairPoor3
PoorFair2

The sorted ratings would be as follows:

Candidate   
 Median point
A
 
B
 
  
 

  Excellent  Good  Fair  Poor

Result: Taking only the votes of the second group in account, A has the median rating of "Fair" and B the median rating of "Poor". Thus,A is elected majority judgment winner by the second group of voters.

All voters

[edit]

Finally, the majority judgment winner of the complete set of voters is determined.

CandidatesVoters
AB
ExcellentFair3
FairPoor3
PoorFair4

The sorted ratings would be as follows:

Candidate   
 Median point
A
  
B
 
  
 

  Excellent  Good  Fair  Poor

The median ratings for A and B are both "Fair". Since there is a tie, "Fair" ratings are removed from both, until their medians become different. After removing 20% "Fair" ratings from the votes of each, the sorted ratings are now:

Candidate   
 Median point
A
   
B
 

Result: Now, the median rating of A is "Poor" and the median rating of B is "Fair". Thus,B is elected majority judgment winner.

Conclusion

[edit]

A is the majority judgment winner within the first group of voters and also within the second group of voters. However, both groups combined elect B as the Majority Judgment winner. Thus, Majority Judgment fails the consistency criterion.

Ranked Pairs

[edit]

This example shows that theranked pairs method violates the consistency criterion. Assume three candidates A, B and C with 39 voters with the following preferences:

PreferencesVoters
A > B > C7
B > C > A6
C > A > B3
A > C > B9
B > A > C8
C > B > A6

Now, the set of all voters is divided into two groups at the bold line. The voters over the line are the first group of voters; the others are the second group of voters.

First group of voters

[edit]

In the following the ranked pairs winner for the first group of voters is determined.

PreferencesVoters
A > B > C7
B > C > A6
C > A > B3

The results would be tabulated as follows:

Pairwise election results
X
Y
ABC
A
6
10
9
7
B
10
6
3
13
C
7
9
13
3
Pairwise election results
(won-tied-lost)
1–0–11–0–11–0–1

The sorted list of victories would be:

PairWinner
B (13) vs C (3)B 13
A (10) vs B (6)A 10
A (7) vs C (9)C 9

Result: B > C and A > B are locked in first (and C > A can't be locked in after that), so the full ranking is A > B > C. Thus,A is elected ranked pairs winner by the first group of voters.

Second group of voters

[edit]

Now, the ranked pairs winner for the second group of voters is determined.

PreferencesVoters
A > C > B9
B > A > C8
C > B > A6

The results would be tabulated as follows:

Pairwise election results
X
Y
ABC
A
14
9
6
17
B
9
14
15
8
C
17
6
8
15
Pairwise election results
(won-tied-lost)
1–0–11–0–11–0–1

The sorted list of victories would be:

PairWinner
A (17) vs C (6)A 17
B (8) vs C (15)C 15
A (9) vs B (14)B 14

Result: Taking only the votes of the second group in account, A > C and C > B are locked in first (and B > A can't be locked in after that), so the full ranking is A > C > B. Thus,A is elected ranked pairs winner by the second group of voters.

All voters

[edit]

Finally, the ranked pairs winner of the complete set of voters is determined.

PreferencesVoters
A > B > C7
A > C > B9
B > A > C8
B > C > A6
C > A > B3
C > B > A6

The results would be tabulated as follows:

Pairwise election results
X
Y
ABC
A
20
19
15
24
B
19
20
18
21
C
24
15
21
18
Pairwise election results
(won-tied-lost)
1–0–12–0–00–0–2

The sorted list of victories would be:

PairWinner
A (25) vs C (15)A 24
B (21) vs C (18)B 21
A (19) vs B (20)B 20

Result: Now, all three pairs (A > C, B > C and B > A) can be locked in without a cycle. The full ranking is B > A > C. Thus, ranked pairs choosesB as winner, which is the Condorcet winner, due to the lack of a cycle.

Conclusion

[edit]

A is the ranked pairs winner within the first group of voters and also within the second group of voters. However, both groups combined elect B as the ranked pairs winner. Thus, the ranked pairs method fails the consistency criterion.

References

[edit]
  1. ^abFranceschini, Fiorenzo; Maisano, Domenico A. (2022-06-01)."Analysing paradoxes in design decisions: the case of "multiple-district" paradox".International Journal on Interactive Design and Manufacturing (IJIDeM).16 (2):677–689.doi:10.1007/s12008-022-00860-x.ISSN 1955-2505.
  2. ^abcYoung, H. P.; Levenglick, A. (1978)."A Consistent Extension of Condorcet's Election Principle"(PDF).SIAM Journal on Applied Mathematics.35 (2):285–300.doi:10.1137/0135023.ISSN 0036-1399.JSTOR 2100667.
  3. ^Balinski, Michel; Laraki, Rida (2011-01-28).Majority Judgment. The MIT Press.doi:10.7551/mitpress/9780262015134.001.0001.ISBN 978-0-262-01513-4.
  1. ^John H Smith, "Aggregation of preferences with variable electorate",Econometrica, Vol. 41 (1973), pp. 1027–1041.
  2. ^D. R. Woodall, "Properties of preferential election rules",Voting matters, Issue 3 (December 1994), pp. 8–15.
  3. ^H. P. Young, "Social Choice Scoring Functions",SIAM Journal on Applied Mathematics Vol. 28, No. 4 (1975), pp. 824–838.
Part of thepolitics andEconomics series
Single-winner
Proportional
Systems
Allocation
Quotas
Mixed
Semi-proportional
Criteria
Other
Comparison
Retrieved from "https://en.wikipedia.org/w/index.php?title=Consistency_criterion&oldid=1304628929"
Category:
Hidden categories:

[8]ページ先頭

©2009-2025 Movatter.jp