Dear Forest, you wrote: > So practically speaking, elimination of Pareto dominated alternatives > is extremely unlikely to have any effect, although UncAAO technically > fails the criterion.
I believe so, too... > Note that if approval is measured in relation to a virtual "approval > cutoff candidate," or if the approval order is automatically refined > by the rank order on each ballot to enforce a "sincere approval" > requirement, then the alternative Y could tie Y' but could not beat > it. If all such ties were broken by random ballot, then Y' would beat > Y, and that would make UncAAO Independent from Pareto Dominated > Alternatives. I'm not sure I grasped that but will give it a try tomorrow .-) > But if the approval winner A is not a member of Smith, then it is > possible (however unlikely) that the first member of the sequence A, > f(A), f(f(A)), ... that resides in Smith is not the highest approval > member of Smith, and that this sequence would lead to a different > UncAAO winner if the highest approval Smith member were taken as the > initial point. So we could consider taking the most approved Smith set member as the starting point, couldn't we? That would still be monotonic, right? > Is TACC monotone? It seems to me that the winner W could improve in > approval enough to overtake and surpass some W' in approval without > defeating W' pairwise, though W' covers W. Recalling my proof from March 3, 2005, I believe it is: > Let us assume that the actual winner X is raised on some individual > ballots by moving either the approval cutoff or another candidate from > directly above X to directly below X. Then what can happen is twofold: > First, the order in step 1 either does not change or does only change > in that X gets moved up one position. Second, the defeats do not > change or do only change in that X now defeats some candidate Y which > she was defeated by before. In either case, X still must win: It was > the last candidate added to the chain. The new chain developes exactly > as before: As the order did not change left of X, the chain evolves > just as before until the original position of X. If X did not change > position, it still defeats all candidates in the chain and so gets > added. If X did change position with Y, then Y was beaten by X or some > other candidate already in the chain, since it was not added to the > chain originally. If it is added now, it must hence be beaten by X, so > that X still gets added after Y was added. In either case, when X was > considered, the resulting chain is the old one except that perhaps Y > is added. As no later candidate beat was added originally and X beats > everything it beat before, also now no further candidate can be added, > so that X is again the winner. QED. Still seems correct to me... I wonder whether the two methods are more closely related than we think since UncAAO also constructs a chain of options in which all defeats are transitive. Perhaps the winner is not only uncovered but even in the Banks set, just like the TACC winner? Yours, Jobst
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