Independence of covered alternatives: Difference between revisions

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===Monotonicity===
 
===Monotonicity===
   
ICA is incompatible with monotonicity for any voting method where a perfect tie can be broken in favor of a candidate W by using ballots that rank W first. The proof by Forest Simmons is as follows:<ref>{{cite web|url=http://lists.electorama.com/pipermail/election-methods-electorama.com/2010-July/091997.html|title=independence form covered alternatives is incompatible with monotonicity|website=Election-methods mailing list archives|date=2010-07-16|last=Simmons|first=Forest}}</ref><ref>{{cite web|url=https://rangevoting.org/PuzzUncovMono.html|title=Puzzle #??: Independence from covered alternatives & monotonicity |website=Rangevoting.org|last=Smith|first=Warren D.}}</ref>
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ICA is incompatible with monotonicity for any voting method where a perfect tie can always be broken in favor of a candidate W by adding one or more identical ballots that rank W first. The proof by Forest Simmons is as follows:<ref>{{cite web|url=http://lists.electorama.com/pipermail/election-methods-electorama.com/2010-July/091997.html|title=independence form covered alternatives is incompatible with monotonicity|website=Election-methods mailing list archives|date=2010-07-16|last=Simmons|first=Forest}}</ref><ref>{{cite web|url=https://rangevoting.org/PuzzUncovMono.html|title=Puzzle #??: Independence from covered alternatives & monotonicity |website=Rangevoting.org|last=Smith|first=Warren D.}}</ref>
   
 
The election
 
The election

Latest revision as of 19:11, 25 March 2022

Independence of covered alternatives is a voting criterion that states that if an alternative X wins an election, and a new alternative Y is added, X will sitll win the election if Y is not in the Uncovered set. Independence of covered alternatives implies ISDA (and hence Smith and the Condorcet criterion) because the uncovered set is a subset of the Smith set.

Criterion compliances and failures[edit | edit source]

Monotonicity[edit | edit source]

ICA is incompatible with monotonicity for any voting method where a perfect tie can always be broken in favor of a candidate W by adding one or more identical ballots that rank W first. The proof by Forest Simmons is as follows:[1][2]

The election

1: B>C>A
1: C>A>B
1: A>B>C

is a perfect tie. By the tiebreaking property, the election (call it election 1)

2: B>C>A
1: C>A>B
1: A>B>C

must give B a greater chance of winning than C. Hence, in this election (election 2)

2: D>B>C
1: B>C>D
1: C>D>B

B must have a lower chance of winning than in election 1, because all we've done is relabel {A, B, C} in election 1 to {C, D, B} to get election 2.

Now consider election 3:

2: D>B>C>A
1: C>A>D>B
1: A>B>C>D

The uncovered set is {A, B, C}, and eliminating D (which is covered) gives us election 1. So by independence of covered alternatives, D can't win and the other candidates' winning probabilities must be those of election 1. Now raise B on the last ballot to get

2: D>B>C>A
1: C>A>D>B
1: B>A>C>D

Now the uncovered set is {B, C, D}, and eliminating A gives us election 2. By independence of covered alternatives, A can't win and the other winning probabilities must be the same as in election 2. But B's winning probability in election 2 is lower than in election 1. Hence raising B harmed B, which contradicts monotonicity.

References[edit | edit source]

  1. Simmons, Forest (2010-07-16). "independence form covered alternatives is incompatible with monotonicity". Election-methods mailing list archives.
  2. Smith, Warren D. "Puzzle #??: Independence from covered alternatives & monotonicity". Rangevoting.org.