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Social Choice Session 2

Social Choice Session 2. Carmen Pasca and John Hey. Arrow’s Impossibility Theorem.

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Social Choice Session 2

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  1. Social ChoiceSession 2 Carmen Pasca and John Hey

  2. Arrow’s Impossibility Theorem • In social choice theory, Arrow’s impossibility theorem ... states that, when voters have three or more discrete alternatives (options), no voting system can convert the ranked preferences of individuals into a community-wide ranking while also meeting a certain set of criteria. These criteria are called unrestricted domain, non-dictatorship, Pareto efficiency, and independence of irrelevant alternatives. (Wikipedia) • Is this obvious or not? • This lecture we give a proof, borrowed from John Bone at the University of York. We have also included another proof borrowed from Kevin Feasel. Use whichever you prefer.

  3. Arrow • Arrow asks the question: “can we aggregate individual preferences into social preferences?” • He invokes the following conditions: • Universal Domain: is applicable to any profile. • Consistency: produces a complete, transitive ordering of available alternatives. • Pareto: if everyone prefers x to y then so should society. • Independence of Irrelevant Alternatives: orders x and y on the basis only of individual preferences on x and y.

  4. The result • Arrow shows that if one wants these conditions then the only way to get them is through Dictatorship. • On the one hand this seems trivial (“how can you aggregate different preferences?”)... • ... but one should look at the result as telling us that we need to drop something if we want something that in some way represent the preferences of society – whatever that means. • We need to drop some of the assumptions... • ... or change our way of looking at the problem.

  5. The Story and Two of the Axioms • We consider a very simple society with two members, Jen and Ken, and three things to order: a, b and c. Think of these as ways to organise society. • Universal Domain: is applicable to any profile. • This means that whatever the preferences are we should be able to aggregate them. • Pareto: if everyone prefers x to y then so should society. • This seems unquestionable.

  6. aPb bPc aPc Universal Domain & Pareto Ken a b c a c b b a c b c a c a b c b a Jen 11 12 13 14 15 16 a b c 21 22 23 24 25 26 a c b 31 32 33 34 35 36 b a c 41 42 43 44 45 46 b c a 51 52 53 54 55 56 c a b 61 62 63 64 65 66 c b a

  7. aPb aPb bPc aPc aPc Universal Domain & Pareto Ken a b c a c b b a c b c a c a b c b a Jen 11 12 13 14 15 16 a b c 21 22 23 24 25 26 a c b 31 32 33 34 35 36 b a c 41 42 43 44 45 46 b c a 51 52 53 54 55 56 c a b 61 62 63 64 65 66 c b a

  8. aPb aPb bPc bPc aPc aPc aPc Universal Domain & Pareto Ken a b c a c b b a c b c a c a b c b a Jen 11 12 13 14 15 16 a b c 21 22 23 24 25 26 a c b 31 32 33 34 35 36 b a c 41 42 43 44 45 46 b c a 51 52 53 54 55 56 c a b 61 62 63 64 65 66 c b a

  9. aPb aPb bPc bPc aPc aPc aPc Universal Domain & Pareto Ken a b c a c b b a c b c a c a b c b a Jen aPb 11 12 13 14 15 16 a b c bPc 21 22 23 24 25 26 a c b 31 32 33 34 35 36 b a c 41 42 43 44 45 46 b c a 51 52 53 54 55 56 c a b 61 62 63 64 65 66 c b a

  10. aPb aPb bPc bPc bPc aPc aPc aPc aPb aPb aPb cPb cPb cPb aPc aPc aPc bPa bPa bPa bPc bPc bPc aPc aPc aPc bPa bPa bPa bPc bPc bPc cPa cPa cPa aPb aPb aPb cPb cPb cPb cPa cPa cPa bPa bPa bPa cPb cPb cPb cPa cPa cPa Universal Domain & Pareto Ken a b c a c b b a c b c a c a b c b a Jen aPb 11 12 13 14 15 16 a b c 21 22 23 24 25 26 a c b 31 32 33 34 35 36 b a c 41 42 43 44 45 46 b c a 51 52 53 54 55 56 c a b 61 62 63 64 65 66 c b a

  11. Now the other two axioms • Consistency: produces a complete, transitive ordering of available alternatives. • We have a complete transitive ordering for society. • Independence of Irrelevant Alternatives: orders x and y on the basis only of individual preferences on x and y. • This is particularly important – we are going to use it frequently. • In the slides that follow we first impose consistency and then repeatedly we invoke the independence of irrelevant alternatives.

  12. aPb aPb aPb bPc bPc bPc aPc aPc aPc aPb aPb aPb cPb cPb cPb aPc aPc aPc bPa bPa bPa bPc bPc bPc aPc aPc aPc bPa bPa bPa bPc bPc bPc cPa cPa cPa aPb aPb aPb cPb cPb cPb cPa cPa cPa bPa bPa bPa cPb cPb cPb cPa cPa cPa Consistency & IIA Ken requires (at least) one of these a b c a c b b a c b c a c a b c b a Jen 11 12 13 14 15 16 a b c cPb cPb cPb aPc 21 22 23 24 25 26 a c b 31 32 33 34 35 36 b a c cPb cPb cPb 41 42 43 44 45 46 b c a cPb cPb cPb 51 52 53 54 55 56 c a b 61 62 63 64 65 66 c b a

  13. Now what? • We can now use an implication of what we have found. • Notein cell 36 that bPa and cPb; and hence we can deduce that cPa. • Let us insert that in cell 36 and explore the implications. • This allows us to use the Independence of Irrelevant Alternatives Axiom again.

  14. aPb aPb aPb bPc cPb bPc bPc cPb cPb aPc aPc aPc aPb aPb aPb cPb cPb cPb aPc aPc aPc bPa bPa bPa bPc cPb bPc bPc cPb cPb aPc aPc aPc bPa bPa bPa bPc cPb bPc bPc cPb cPb cPa cPa cPa aPb aPb aPb cPb cPb cPb cPa cPa cPa bPa bPa bPa cPb cPb cPb cPa cPa cPa Consistency & IIA Ken a b c a c b b a c b c a c a b c b a Jen 11 12 13 14 15 16 a b c cPa cPa cPa 21 22 23 24 25 26 a c b cPa cPa cPa 31 32 33 34 35 36 b a c cPa cPa cPa 41 42 43 44 45 46 b c a 51 52 53 54 55 56 c a b 61 62 63 64 65 66 c b a

  15. Now what? • Once again we can now use an implication of what we have found. • Notein cell 14 that bPc and cPa; and hence we can deduce that bPa. • Let us insert that in cell 14 and explore the implications. • This allows us to use the Independence of Irrelevant Alternatives Axiom again.

  16. aPb aPb aPb bPc cPb bPc bPc cPb cPb aPc aPc aPc cPa cPa cPa aPb aPb aPb cPb cPb cPb aPc aPc aPc cPa cPa cPa bPa bPa bPa bPc cPb bPc bPc cPb cPb aPc aPc aPc cPa cPa cPa bPa bPa bPa bPc cPb bPc bPc cPb cPb cPa cPa cPa aPb aPb aPb cPb cPb cPb cPa cPa cPa bPa bPa bPa cPb cPb cPb cPa cPa cPa Consistency & IIA Ken a b c a c b b a c b c a c a b c b a Jen bPa bPa bPa 11 12 13 14 15 16 a b c bPa bPa bPa 21 22 23 24 25 26 a c b 31 32 33 34 35 36 b a c 41 42 43 44 45 46 b c a bPa bPa bPa 51 52 53 54 55 56 c a b 61 62 63 64 65 66 c b a

  17. Now what? • Once again we can now use an implication of what we have found. • Notein cell 23 that bPa and aPc; and hence we can deduce that bPc. • Let us insert that in cell 23 and explore the implications. • This allows us to use the Independence of Irrelevant Alternatives Axiom again.

  18. aPb aPb bPa bPa bPa aPb bPc cPb bPc bPc cPb cPb aPc aPc aPc cPa cPa cPa aPb aPb bPa bPa aPb bPa cPb cPb cPb aPc aPc aPc cPa cPa cPa bPa bPa bPa bPc cPb bPc bPc cPb cPb aPc aPc aPc cPa cPa cPa bPa bPa bPa bPc cPb bPc bPc cPb cPb cPa cPa cPa aPb aPb bPa bPa aPb bPa cPb cPb cPb cPa cPa cPa bPa bPa bPa cPb cPb cPb cPa cPa cPa Consistency & IIA Ken a b c a c b b a c b c a c a b c b a Jen 11 12 13 14 15 16 a b c 21 22 23 24 25 26 a c b bPc bPc bPc 31 32 33 34 35 36 b a c 41 42 43 44 45 46 b c a 51 52 53 54 55 56 c a b bPc bPc bPc 61 62 63 64 65 66 c b a bPc bPc bPc

  19. Now what? • Once again can now use an implication of what we have found. • Notein cell 51 that aPb and bPc; and hence we can deduce that aPc. • Let us insert that in cell 51 and explore the implications. • This allows us to use the Independence of Irrelevant Alternatives Axiom again.

  20. aPb aPb bPa bPa bPa aPb bPc cPb bPc bPc cPb cPb aPc aPc aPc cPa cPa cPa aPb aPb bPa bPa aPb bPa bPc cPb bPc bPc cPb cPb aPc aPc aPc cPa cPa cPa bPa bPa bPa bPc cPb bPc bPc cPb cPb aPc aPc aPc cPa cPa cPa bPa bPa bPa bPc cPb bPc bPc cPb cPb cPa cPa cPa aPb aPb bPa bPa aPb bPa bPc cPb bPc bPc cPb cPb cPa cPa cPa bPa bPa bPa bPc cPb bPc bPc cPb cPb cPa cPa cPa Consistency & IIA Ken a b c a c b b a c b c a c a b c b a Jen 11 12 13 14 15 16 a b c 21 22 23 24 25 26 a c b 31 32 33 34 35 36 b a c 41 42 43 44 45 46 b c a aPc aPc aPc 51 52 53 54 55 56 c a b aPc aPc aPc 61 62 63 64 65 66 c b a aPc aPc aPc

  21. Finally • For the final time can now use an implication of what we have found. • Notein cell 62 that cPb and aPc; and hence we can deduce that aPb. • Let us insert that in cell 62 and explore the implications. • This allows us to use the Independence of Irrelevant Alternatives Axiom again.

  22. aPb aPb bPa bPa bPa aPb bPc cPb bPc bPc cPb cPb aPc aPc aPc cPa cPa cPa aPb aPb bPa bPa aPb bPa bPc cPb bPc bPc cPb cPb aPc aPc aPc cPa cPa cPa bPa bPa bPa bPc cPb bPc bPc cPb cPb aPc aPc aPc cPa cPa cPa bPa bPa bPa bPc cPb bPc bPc cPb cPb aPc aPc aPc cPa cPa cPa aPb aPb bPa bPa aPb bPa bPc cPb bPc bPc cPb cPb aPc aPc aPc cPa cPa cPa bPa bPa bPa bPc cPb bPc bPc cPb cPb aPc aPc aPc cPa cPa cPa Consistency & IIA Ken The Dictator a b c a c b b a c b c a c a b c b a Jen 11 12 13 14 15 16 a b c 21 22 23 24 25 26 a c b aPb aPb aPb 31 32 33 34 35 36 b a c aPb aPb aPb 41 42 43 44 45 46 b c a 51 52 53 54 55 56 c a b aPb aPb aPb 61 62 63 64 65 66 c b a

  23. Two questions for the break • (1) You may recall that we started this line of logic with one of two possible ways of getting consistency. • What do you think happens if we take the second possible way? • (2) Why did we not pick up any intransitivities on the way? That is, for example, why did we not find something like • aPb bPc cPa • in one of the cells? • Magic?

  24. aPb aPb aPb bPc bPc bPc aPc aPc aPc aPb aPb aPb cPb cPb cPb aPc aPc aPc bPa bPa bPa bPc bPc bPc aPc aPc aPc bPa bPa bPa bPc bPc bPc cPa cPa cPa aPb aPb aPb cPb cPb cPb cPa cPa cPa bPa bPa bPa cPb cPb cPb cPa cPa cPa Consistency & IIA Ken a b c a c b b a c b c a c a b c b a Jen 11 12 13 14 15 16 a b c cPb aPc aPc aPc 21 22 23 24 25 26 a c b aPc aPc aPc 31 32 33 34 35 36 b a c aPc aPc aPc 41 42 43 44 45 46 b c a 51 52 53 54 55 56 c a b 61 62 63 64 65 66 c b a

  25. aPb aPb aPb bPc bPc bPc aPc aPc aPc aPc aPc aPc aPb aPb aPb cPb cPb cPb aPc aPc aPc aPc aPc aPc bPa bPa bPa bPc bPc bPc aPc aPc aPc aPc aPc aPc bPa bPa bPa bPc bPc bPc cPa cPa cPa aPb aPb aPb cPb cPb cPb cPa cPa cPa bPa bPa bPa cPb cPb cPb cPa cPa cPa Consistency & IIA Ken a b c a c b b a c b c a c a b c b a Jen aPb aPb aPb 11 12 13 14 15 16 a b c aPb aPb aPb 21 22 23 24 25 26 a c b 31 32 33 34 35 36 b a c 41 42 43 44 45 46 b c a aPb aPb aPb 51 52 53 54 55 56 c a b 61 62 63 64 65 66 c b a

  26. aPb aPb aPb aPb aPb aPb bPc bPc bPc aPc aPc aPc aPc aPc aPc aPb aPb aPb aPb aPb aPb cPb cPb cPb aPc aPc aPc aPc aPc aPc bPa bPa bPa bPc bPc bPc aPc aPc aPc aPc aPc aPc bPa bPa bPa bPc bPc bPc cPa cPa cPa aPb aPb aPb aPb aPb aPb cPb cPb cPb cPa cPa cPa bPa bPa bPa cPb cPb cPb cPa cPa cPa Consistency & IIA Ken a b c a c b b a c b c a c a b c b a Jen 11 12 13 14 15 16 a b c 21 22 23 24 25 26 a c b cPb cPb cPb 31 32 33 34 35 36 b a c 41 42 43 44 45 46 b c a 51 52 53 54 55 56 c a b cPb cPb cPb 61 62 63 64 65 66 c b a cPb cPb cPb

  27. aPb aPb aPb aPb aPb aPb bPc bPc bPc aPc aPc aPc aPc aPc aPc aPb aPb aPb aPb aPb aPb cPb cPb cPb cPb cPb cPb aPc aPc aPc aPc aPc aPc bPa bPa bPa bPc bPc bPc aPc aPc aPc aPc aPc aPc bPa bPa bPa bPc bPc bPc cPa cPa cPa aPb aPb aPb aPb aPb aPb cPb cPb cPb cPb cPb cPb cPa cPa cPa bPa bPa bPa cPb cPb cPb cPb cPb cPb cPa cPa cPa Consistency & IIA Ken a b c a c b b a c b c a c a b c b a Jen 11 12 13 14 15 16 a b c 21 22 23 24 25 26 a c b 31 32 33 34 35 36 b a c 41 42 43 44 45 46 b c a cPa cPa cPa 51 52 53 54 55 56 c a b cPa cPa cPa 61 62 63 64 65 66 c b a cPa cPa cPa

  28. aPb aPb aPb aPb aPb aPb bPc bPc bPc aPc aPc aPc aPc aPc aPc aPb aPb aPb aPb aPb aPb cPb cPb cPb cPb cPb cPb aPc aPc aPc aPc aPc aPc bPa bPa bPa bPc bPc bPc aPc aPc aPc aPc aPc aPc bPa bPa bPa bPc bPc bPc cPa cPa cPa cPa cPa cPa aPb aPb aPb aPb aPb aPb cPb cPb cPb cPb cPb cPb cPa cPa cPa cPa cPa cPa bPa bPa bPa cPb cPb cPb cPb cPb cPb cPa cPa cPa cPa cPa cPa Consistency & IIA Ken a b c a c b b a c b c a c a b c b a Jen 11 12 13 14 15 16 a b c 21 22 23 24 25 26 a c b bPa bPa bPa 31 32 33 34 35 36 b a c bPa bPa bPa 41 42 43 44 45 46 b c a 51 52 53 54 55 56 c a b bPa bPa bPa 61 62 63 64 65 66 c b a

  29. aPb aPb aPb aPb aPb aPb bPc bPc bPc aPc aPc aPc aPc aPc aPc aPb aPb aPb aPb aPb aPb cPb cPb cPb cPb cPb cPb aPc aPc aPc aPc aPc aPc bPa bPa bPa bPa bPa bPa bPc bPc bPc aPc aPc aPc aPc aPc aPc bPa bPa bPa bPa bPa bPa bPc bPc bPc cPa cPa cPa cPa cPa cPa aPb aPb aPb aPb aPb aPb cPb cPb cPb cPb cPb cPb cPa cPa cPa cPa cPa cPa bPa bPa bPa bPa bPa bPa cPb cPb cPb cPb cPb cPb cPa cPa cPa cPa cPa cPa Consistency & IIA Ken a b c a c b b a c b c a c a b c b a The Dictator Jen 11 12 13 14 15 16 a b c bPc bPc bPc 21 22 23 24 25 26 a c b 31 32 33 34 35 36 b a c bPc bPc bPc 41 42 43 44 45 46 b c a bPc bPc bPc 51 52 53 54 55 56 c a b 61 62 63 64 65 66 c b a

  30. Arrow’s Theorem The only principle that satisfies: Universal Domain Consistency Pareto Independence of Irrelevant Alternatives (IIA) and that requires only individual rankings is the Dictator Principle i.e., where the social ordering simply replicates, across all profiles, some given individual’s ranking.

  31. Given the four conditions (U, C, P, IIA) If, for some x and y, xPy at some profile where everyone in V ranks x above y and everyone else ranks y above x ... .. then, for any w and z, wPz at any profile where everyone in V ranks w above z, irrespective of how anyone else ranks w and z. We have already shown this in the 2-person, 3-alternative, strict ranking case. Generalising this to n-persons, m-alternatives is not very difficult.

  32. Arrow’s Impossibility Theorem • We have proved it in the context of a two-person society, but as John Bone says, it is not difficult to extend it to a larger society. This is what he does in the next three slides. • It is not necessary for you to know this in detail, just understand the principle. • If you would like another proof of the theorem, you might like this by Kevin Feasel which I have shamelessy downloaded from the Internet. This follows after the generalisation to n people.

  33. Given the four conditions (U, C, P, IIA) There must be some x and y, and some individual J, such that xPy at some profile where J ranks x above y and everyone else ranks y above x. Consider any set V of individuals such that aPb at some profile where everyone in V ranks a above b and everyone else ranks b above a. Suppose that V contains more than one person. (We know there must be such a V, since from Pareto one such V is the set of all individuals.) From the previous slide, we know that therefore aPb at any profile where everyone in V ranks a above b. Such as ..

  34. a c b b a c c b a everyone in V1 everyone in V2 everyone else where V1 and V2 are subsets of V (i.e. everyone in V is in exactly one of V1 and V2 , and no-one else is). aPb implies that for any c either aPc or cPb. It follows that there is a set smaller than V, and a pair {x,y}, such that xPy at some profile where everyone in that set ranks x above y and everyone else ranks y above x. Since we can make the same argument for any V containing more than one person, it follows that the smallest such V is an individual (let’s say J).

  35. x x x x P P P P z y y z z x y w y z x w x z x z w P z If there is any profile in which Jen ranks x above y and Ken ranks y above x, and at which , then from IIA: at every profile in which Jen ranks x above y and Ken ranks y above x Jen such as at: Ken at which, from Pareto and Consistency: Jen So, again from IIA, also at: Ken at which, from Pareto and Consistency:

  36. Kevin Feasel December 10, 2006 http://36chambers.wordpress.com/arrow/ Arrow's Impossibility Theorem

  37. The Rules • 2 individuals with 3 choices (x, y, z) --> 6 profiles for each. • x > y > z, x > z > y, y > x > z, y > z > x, z > x > y, z > y > x • 36 profiles in all for our two-person example (6 * 6)

  38. The 36 Profiles

  39. Explanation Of Symbols • Colored Text • Blue – First choice for both • Green – Second choice for both • Red – Third choice for both • Black – Inconsistent choices (e.g., in profile 10, Z > Y for person A but Y > Z for person B) • Numbers along the left-hand side depict the #1, #2, and #3 choices for person A. • Numbers along the bottom depict the #1, #2, and #3 choices for person B. • Each profile has a number (e.g., 1 and 10 here). • How to read this: for profile 10, person A (left-hand side) likes Z > X > Y. Person B (bottom) prefers X > Y > Z.

  40. The Assumptions • Completeness – All profiles must be solvable. In this case, we have 3 choices among 2 people, so 36 total sets of preferences could arise. We must have a solution for each one of the 36.

  41. The Assumptions • Unanimity – If all individuals agree on a single position, that position will be guaranteed. • In our case, if an option is red, both individuals rank this option as the #3 choice, so it will end up as #3 in the preference ranking setup.

  42. The Assumptions • Independence of Irrelevant Alternatives – The relationship between X and Y should be independent of Z. In general terms, the relationship between two elements will not change with the addition of another element. • Example: in profile 25, X > Y for person A (X > Y > Z) and for person B (X > Z > Y). Adding in option Z, we assume, does not alter this relationship. • Example: in profile 26, Y > X for person A (Y > Z > X) and for person B (Z > Y > X). If we removed option Z, we would still expect Y > X for both.

  43. Proving The Final Rule • Non-dictatorship. We want to understand whether we can come up with a social rule which follows the rules of completeness, unanimity, and independence of irrelevant alternatives, and which is simultaneously non-dictatorial. “Dictatorial” here means that all 36 profiles will match one person's profiles exactly. In other words, all social choices will precisely match the individual's preferences.

  44. Solving The Problem • 6 preference sets are already solved, thanks to unanimity. In profile 1, for example, all parties agree that X > Y > Z, so X > Y > Z is the social rule. These six completed profiles have the “>” highlighted in yellow.

  45. Solving The Problem

  46. Solving The Problem • In addition to this, we know that any profile with colored text is also solved—both people agree on where to place this option. So these can all be filled in as well.

  47. Solving The Problem

  48. Solving The Problem • There are two remaining sets of relationships, as shown in profile 23. • a > b: in this example, Y > Z for both individuals. Y is the #2 choice for person A and Z #3, whereas Y is the #1 choice for person B and Z #2. In a case such as this, we know that Y > Z in the social preferences because both people prefer Y to Z. • a ? b: in this example, Z > X for person B, but X > Z for person A. A's preferences are X > Y > Z and B's are Y > Z > X. In this type of situation, we do not yet know if Z or X will be socially preferred because there is no agreement between individuals.

  49. Solving The Problem • In this case, because we know that Y > Z socially, we can put that down. But we do not know if Y > Z > X, Y > X > Z, or X > Y > Z, so we will not write these down just yet, but once things start clearing up, these rules will be used. • The key here is that Z is above and to the right of Y, which means that there is agreement that Y > Z. However, Z is above and to the left of X, meaning there is conflict. Y is also above and to the left of X, so there is another conflict, due to the fact that X > Y > Z for person A, but Y > Z > X for person B.

  50. The Single Choice • If we look at profile 7, there is a single choice to be made. • Person A (left-hand column): X > Z > Y • Person B (bottom row): X > Y > Z • We know that X will be the #1 preference, but we have to decide whether Y > Z or Z > Y here. We can choose either, but let us pick that Y > Z. In other words, we support person B's preference here. • This means that the social preference will be X > Y > Z. In addition, because of the Independence of Irrelevant Alternatives axiom, any time we see a conflict between Y and Z similar to the one in profile 7, we know to choose Y > Z. • Let us fill in the chart with this new information...

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