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Mechanism Design without Money Lecture 12 1. Individual rationality and efficiency: an impossibility theorem with a (discouraging) worst-case bound For.

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Presentation on theme: "Mechanism Design without Money Lecture 12 1. Individual rationality and efficiency: an impossibility theorem with a (discouraging) worst-case bound For."— Presentation transcript:

1 Mechanism Design without Money Lecture 12 1

2 Individual rationality and efficiency: an impossibility theorem with a (discouraging) worst-case bound For every k> 3, there exists a compatibility graph such that no k-maximum allocation which is also individually rational matches more than 1/(k-1) of the number of nodes matched by a k-efficient allocation. 2

3 Proof (for k=3) 3 a3a3 a2a2 c d a1a1 e b

4 “Cost” of IR is very small - Simulations No. of Hospitals IR,k= Efficient, k=

5 But the cost of not having IR could be very high if it causes centralized matching to break down 5

6 But current mechanisms aren’t IR for hospitals Current mechanisms: Choose (~randomly) an efficient allocation. Proposition: Withholding internal exchanges can (often) be strictly better off (non negligible) for a hospital regardless of the number of hospitals that participate. O-A A-O 6 And hospitals can withhold individual overdemanded pairs

7 What if we have a prior? Infinite horizon In each timestep, a hospital samples its patients from some known distribution Then there exists a truthful mechanism with efficiency 1 – o(1) 7

8 Matching Initially the hospital has zero credits In the beginning of the round, if the hospital has zero credits, each patients enters the match with probability 1 – 1/k 1/6 For each positive credit, the hospital increases this probability by 1/k 2/3 and the credit is gone For each negative credit, the hospital decreases this probability by 1/k 2/3 and the credit is gone. The probability is always > ½ 8

9 Gaining credit For each patient over k, the hospital gets 1 credit For each patient below k, the hospital looses 1 credit These credits only affect the next rounds 9

10 Proof idea Hiding a patient can give an additive advantage, but causes a multiplicative loss Number of credit doesn’t matter – you always care about the future Can work for every distribution of patients 10

11 Voting 11

12 Terminology Voting rule – Social choice: mapping of a profile onto a winner(s) – Social welfare: mapping of a profile onto a total ordering Agent – Sometimes assume odd number of agents to reduce ties Vote – Total order over outcomes Profile – Vote for each agent Extensions include indifference, incomparability, incompleteness

13 Voting rules: plurality Otherwise known as “majority” or “first past the post” – Candidate with most votes wins With just 2 candidates, this is a very good rule to use – (See May’s theorem)

14 Voting rules: plurality Some criticisms – Ignores preferences other than favourite – Similar candidates can “split” the vote – Encourages voters to vote tactically “My candidate cannot win so I’ll vote for my second favourite”

15 Voting rules: plurality with runoff Two rounds – Eliminate all but the 2 candidates with most votes – Then hold a majority election between these 2 candidates Consider – 25 votes: A>B>C – 24 votes: B>C>A – 46 votes: C>A>B – 1st round: B knocked out – 2nd round: C>A by 70:25 – C wins

16 Voting rules: plurality with runoff Some criticisms – Requires voters to list all preferences or to vote twice – Moving a candidate up your ballot may not help them (monotonicity) – It can even pay not to vote! (see next slide)

17 Voting rules: plurality with runoff Consider again – 25 votes: A>B>C – 24 votes: B>C>A – 46 votes: C>A>B C wins easily Two voters don’t vote – 23 votes: A>B>C – 24 votes: B>C>A – 46 votes: C>A>B Different result – 1st round: A knocked out – 2nd round: B>C by 47:46 – B wins

18 Voting rules: single transferable vote STV – If one candidate has >50% vote then they are elected – Otherwise candidate with least votes is eliminated – Their votes transferred (2nd placed candidate becomes 1st, etc.) Identical to plurality with runoff for 3 candidates Example: – 39 votes: A>B>C>D – 20 votes: B>A>C>D – 20 votes: B>C>A>D – 11 votes: C>B>A>D – 10 votes: D>A>B>C – Result: B wins!

19 Voting rules: Borda Given m candidates – ith ranked candidate score m-i – Candidate with greatest sum of scores wins Example – 42 votes: A>B>C>D – 26 votes: B>C>D>A – 15 votes: C>D>B>A – 17 votes: D>C>B>A – B wins Jean Charles de Borda,

20 Voting rules: positional rules Given vector of weights, – Candidate scores si for each vote in ith position – Candidate with greatest score wins Generalizes number of rules – Borda is – Plurality is

21 Voting rules: approval Each voters approves between 1 and m-1 candidates Candidate with most votes of approval wins Some criticisms – Elects lowest common denominator? – Two similar candidates do not divide vote, but can introduce problems when we are electing multiple winners

22 Voting rules: other Cup (aka knockout) – Tree of pairwise majority elections Copeland – Candidate that wins the most pairwise competitions Bucklin – If one candidate has a majority, they win – Else 1st and 2nd choices are combined, and we repeat

23 Voting rules: other Coomb’s method – If one candidate has a majority, they win – Else candidate ranked last by most is eliminated, and we repeat Range voting – Each voter gives a score in given range to each candidate – Candidate with highest sum of scores wins – Approval is range voting where range is {0,1}

24 Voting rules: other Maximin (Simpson) – Score = Number of voters who prefer candidate in worst pairwise election – Candidate with highest score wins Veto rule – Each agent can veto up to m-1 candidates – Candidate with fewest vetoes wins Inverse plurality – Each agent casts one vetor – Candidate with fewest vetoes wins

25 Voting rules: other Dodgson – Proposed by Lewis Carroll in 1876 – Candidate who with the fewest swaps of adjacent preferences beats all other candidates in pairwise elections – NP-hard to compute winner! Random – Winner is that of a random ballot …

26 Voting rules So many voting rules to choose from.. Which is best? – Social choice theory looks at the (desirable and undesirable) properties they possess – For instance, is the rule “monotonic”? – Bottom line: with more than 2 candidates, there is no best voting rule

27 Axiomatic approach Define desired properties – E.g. monotonicity: improving votes for a candidate can only help them win Prove whether voting rule has this property – In some cases, as we shall see, we’ll be able to prove impossibility results (no voting rule has this combination of desirable properties)

28 May’s theorem Some desirable properties of voting rule – Anonymous: names of voters irrelevant – Neutral: name of candidates irrelevant

29 May’s theorem Another desirable property of a voting rule – Monotonic: if a particular candidate wins, and a voter improves their vote in favour of this candidate, then they still win Non-monotonicity for plurality with runoff – 27 votes: A>B>C – 42 votes: C>A>B – 24 votes: B>C>A Suppose 4 voters in 1st group move C up to top – 23 votes: A>B>C – 46 votes: C>A>B – 24 votes: B>C>A

30 May’s theorem Thm: With 2 candidates, a voting rule is anonymous, neutral and monotonic iff it is the plurality rule – May, Kenneth "A set of independent necessary and sufficient conditions for simple majority decisions", Econometrica, Vol. 20, pp. 680–68 – Since these properties are uncontroversial, this about decides what to do with 2 candidates!

31 May’s theorem Thm: With 2 candidates, a voting rule is anonymous, neutral and monotonic iff it is the plurality rule – Proof: Plurality rule is clearly anonymous, neutral and monotonic – Other direction is more interesting

32 May’s theorem Thm: With 2 candidates, a voting rule is anonymous, neutral and monotonic iff it is the plurality rule – Proof: Anonymous and neutral implies only number of votes matters – Two cases: N(A>B) = N(B>A)+1 and A wins. – By monotonicity, A wins whenever N(A>B) > N(B>A)

33 May’s theorem Thm: With 2 candidates, a voting rule is anonymous, neutral and monotonic iff it is the plurality rule – Proof: Anonymous and neutral implies only number of votes matters – Two cases: N(A>B) = N(B>A)+1 and A wins. – By monotonicity, A wins whenever N(A>B) > N(B>A) N(A>B) = N(B>A)+1 and B wins – Swap one vote A>B to B>A. By monotonicity, B still wins. But now N(B>A) = N(A>B)+1. By neutrality, A wins. This is a contradiction.

34 Condorcet’s paradox Collective preference may be cyclic – Even when individual preferences are not Consider 3 votes – A>B>C – B>C>A – C>A>B – Majority prefer A to B, and prefer B to C, and prefer C to A! Marie Jean Antoine Nicolas de Caritat, marquis de Condorcet (1743 – 1794)

35 Condorcet principle Turn this on its head Condorcet winner – Candidate that beats every other in pairwise elections – In general, Condorcet winner may not exist – When they exist, must be unique Condorcet consistent – Voting rule that elects Condorcet winner when they exist (e.g. Copeland rule)

36 Condorcet principle Plurality rule is not Condorcet consistent – 35 votes: A>B>C – 34 votes: C>B>A – 31 votes: B>C>A – B is easily the Condorcet winner, but plurality elects A

37 Condorcet principle Thm. No positional rule with strict ordering of weights is Condorcet consistent – Proof: Consider 3 votes: A>B>C 2 votes: B>C>A 1 vote: B>A>C 1 vote: C>A>B – A is Condorcet winner

38 Condorcet principle Thm. No positional rule with strict ordering of weights is Condorcet consistent – Proof: Consider 3 votes: A>B>C 2 votes: B>C>A 1 vote: B>A>C 1 vote: C>A>B – Scoring rule with s1 > s2 > s3 Score(B) = 3.s1+3.s2+1.s3 Score(A) = 3.s1+2.s2+2.s3 Score(C) = 1.s1+2.s2+4.s4 Hence: Score(B)>Score(A)>Score(C)

39 Arrow’s theorem We have to break Condorcet cycles – How we do this, inevitably leads to trouble A genius observation – Led to the Nobel prize in economics

40 Arrow’s theorem Free – Every result is possible Unanimous – If every votes for one candidate, they win Independent to irrelevant alternatives – Result between A and B only depends on how agents preferences between A and B Monotonic

41 Arrow’s theorem Non-dictatorial – Dictator is voter whose vote is the result – Not generally considered to be desirable!

42 Arrow’s theorem Thm: If there are at least two voters and three or more candidates, then it is impossible for any voting rule to be: – Free – Unanimous – Independent to irrelevant alternatives – Monotonic – Non-dictatorial

43 Arrow’s theorem Can give a stronger result – Weaken conditions Pareto – If everyone prefers A to B then A is preferred to B in the result – If free & monotonic & IIA then Pareto – If free & Pareto & IIA then not necessarily monotonic

44 Arrow’s theorem Thm: If there are at least two voters and three or more candidates, then it is impossible for any voting rule to be: – Pareto – Independent to irrelevant alternatives – Non-dictatorial

45 Arrow’s theorem With two candidates, majority rule is: – Pareto – Independent to irrelevant alternatives – Non-dictatorial So, one way “around” Arrow’s theorem is to restrict to two candidates

46 Proof of Arrow’s theorem If all voters put B at top or bottom then result can only have B at top or bottom – Suppose not the case and result has A>B>C – By IIA, this would not change if every voter moved C above A: B>A>C => B>C>A B>C>A => B>C>A A>C>B => C>A>B C>A>B => C>A>B Each AB and BC vote the same!

47 Proof of Arrow’s theorem If all voters put B at top or bottom then result can only have B at top or bottom – Suppose not the case and result has A>B>C – By IIA, this would not change if every voter moved C above A – By transitivity A>C in result – But by unanimity C>A B>A>C => B>C>A B>C>A => B>C>A A>C>B => C>A>B C>A>B => C>A>B

48 Proof of Arrow’s theorem If all voters put B at top or bottom then result can only have B at top or bottom – Suppose not the case and result has A>B>C – A>C and C>A in result – This is a contradiction – B can only be top or bottom in result

49 Proof of Arrow’s theorem If all voters put B at top or bottom then result can only have B at top or bottom Suppose voters in turn move B from bottom to top Exists pivotal voter from whom result changes from B at bottom to B at top

50 Proof of Arrow’s theorem If all voters put B at top or bottom then result can only have B at top or bottom Suppose voters in turn move B from bottom to top Exists pivotal voter from whom result changes from B at bottom to B at top – B all at bottom. By unanimity, B at bottom in result – B all at top. By unanimity, B at top in result – By monotonicity, B moves to top and stays there when some particular voter moves B up

51 Proof of Arrow’s theorem If all voters put B at top or bottom then result can only have B at top or bottom Suppose voters in turn move B from bottom to top Exists pivotal voter from whom result changes from B at bottom to B at top Pivotal voter is dictator (need to show)

52 Proof of Arrow’s theorem Pivotal voter is dictator between A and C – Consider profile when pivotal voter has just moved B to top (and B has moved to top of result) – For any AC, let pivotal voter have A>B>C – By IIA, A>B in result as AB votes are identical to profile just before pivotal vote moves B (and result has B at bottom) – By IIA, B>C in result as BC votes are unchanged – Hence, A>C by transitivity

53 Proof of Arrow’s theorem Each two alternatives {A,C} have a voter which dictates which one of them will be higher. Let i be the dictator for {A,C} Let j be the dictator for {A,B} Let k be the dictator for {B,C} If i  j and j  k and i  k we can create a cycle: – i prefers A to C – k prefers C to B – j prefers B to A Similar argument for i  j=k, i=j  k, j  i=k 53

54 Proof of Arrow’s theorem Pivotal voter is dictator – Consider profile when pivotal voter has just moved B to top (and B has moved to top of result) – For any AC, let pivotal voter have A>B>C – Then A>C in result – This continues to hold even if any other voters change their preferences for A and C – Hence pivotal voter is dicatator for AC – Similar argument for AB

55 Arrow’s theorem How do we get “around” this impossibility – Limit domain Only two candidates – Limit votes Single peaked votes – Limit properties Drop IIA …

56 Single peaked votes In many domains, natural order – Preferences single peaked with respect to this order Examples – Left-right in politics – Cost (not necessarily cheapest!) – Size – …

57 Single peaked votes There are never Condorcet cycles Arrow’s theorem is “escaped” – There exists a rule that is Pareto – Independent to irrelevant alternatives – Non-dictatorial – Median rule: elect “median” candidate Candidate for whom 50% of peaks are to left/right

58 What about dynamics? What is the tradeoff between waiting and number of matches? Dynamic matching in dense graphs (Unver, ReStud,2010).

59 Matching over time 59 Simulation results using 2 year data from NKR* In order to gain in current pools, we need to wait probably “too” long *On average 1 pair every 2 days arrived over the two years Matches

60 Matching over time 60 Simulation results using 2 year data from NKR* In order to gain in current pools, we need to wait probably “too” long *On average 1 pair every 2 days arrived over the two years Matches – high PRA

61 Matching over time 61 Simulation results using 2 year data from NKR* In order to gain in current pools, we need to wait probably “too” long *On average 1 pair every 2 days arrived over the two years

62 Match the pair right away?  A H-node forms an edge with each node u of U with probability ξ/n.  A L-node forms an edge with each node u of U with probability π 62 Arriving pair Lemma: the online algorithm matches almost all pairs when p is a constant and n is large enough (even with just 2- way cycles) Online: match the arrived node to a neighbor; remove cycles when formed. Online: match the arrived node to a neighbor; remove cycles when formed. Either a sparse finite horizon model or an infinite horizon model and analyze steady state

63 Dynamic matching in dense-sparse graphs n nodes. Each node is L w.p. q< 1/2 and H w.p. 1-q incoming edges to L are drawn w.p. L H 63

64 Dynamic matching in dense-sparse graphs n nodes. Each node is L w.p. q< 1/2 and H w.p. 1- q incoming edges to L are drawn w.p. L H 64 At each time step 1,2,…, n, one node arrives.

65 Heterogeneous Dynamic Model  (PRA).  PRA determines the likelihood that a patient cannot receive a kidney from a blood-type compatible donor.  PRA < 79: Low sensitivity patients (L-patients).  80 < PRA < 100: High sensitivity patients (H-patients).  Most blood-type compatible pairs that join the pool have H-patients.  Distribution of High PRA patients in the pool is different from the population PRA. 65 pc/n

66 Chunk Matching in a heterogeneous graph 66 At time steps Δ, 2Δ, …, n: Find maximum matching in H-L; remove the matched nodes. Find maximum matching in L-L; remove the matched nodes.

67 Chunk Matching in a heterogeneous graph 67 Theorem (Ashlagi, Jalliet and Manshadi): When matching only 2- way cycles: 1. If Δ = o(n), M(Δ) = M(1) + o(n) 2. Δ = αn, then M(Δ) = M(1) + f(q,p)n for strictly increasing f()>0. Chunk matching finds a maximum matching at time steps Δ, 2Δ, …, n. M(Δ) - expected number of matched pairs at time n.

68 Chunk Matching in a heterogeneous graph 68 When matching 2 and 3-way cycles: 1.If Δ = M(Δ) = M(1) + f(q,p) (n) (formally this is still a conjecture )

69 Denser Pools 69 Need to wait less time to gain… If the graph is dense (large) – no need to wait at all…

70  Special structure: Sparse H-L and dense L-L.  (PRA).  PRA determines the likelihood that a patient cannot receive a kidney from a blood-type compatible donor.  PRA < 79: Low sensitivity patients (L-patients).  80 < PRA < 100: High sensitivity patients (H-patients).  Most blood-type compatible pairs that join the pool have H-patients.  Distribution of High PRA patients in the pool is different from the population PRA.  Compare the number of H-L matchings. Proof Ideas 70 pξ/n

71 In H-L graph, Δ = o(n):  No edge in the residual graph.  Tissue-type compatibility: Percentage Reactive Antibodies (PRA).  PRA determines the likelihood that a patient cannot receive a kidney from a blood-type compatible donor.  PRA < 79: Low sensitivity patients (L-patients).  80 < PRA < 100: High sensitivity patients (H-patients).  Most blood-type compatible pairs that join the pool have H-patients.  Distribution of High PRA patients in the pool is different from the population PRA.  Decision of online and chunk matching are the same on depth- one trees. M(Δ) = M(1) + o(n). arrived chunk residual graph 71 Proof Ideas

72 In H-L graph, Δ = αn:  Find f(α)n augmenting paths to the matching obtained by online.  Given M the matching of the online scheme:  Chunk matching would choose (l1,h1) and (l2,h2). M(Δ) = M(1) + f(α)n, 72 Proof Ideas h1 l2 l1 h2

73 Chunk Matching in a heterogeneous graph 73 Theorem (Ashlagi, Jalliet and Manshadi): MC(1) = M(1) + f(q)n M(Δ) - expected number of matches using only 2-ways MC(Δ) - expected number of using 2-ways and allowing an unbounded chain.


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