# 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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Mechanism Design without Money Lecture 12 1

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

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

“Cost” of IR is very small - Simulations No. of Hospitals246810121416182022 IR,k=36.818.3735.4249.363.6881.4397.82109.01121.81144.09160.74 Efficient, k=36.8918.6735.9749.7564.3481.8398.07109.41122.1144.35161.07 4

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

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

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

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

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

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

Voting 11

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

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)

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”

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

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)

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

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!

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, 1733-1799

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

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

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

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}

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

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 …

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

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)

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

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

May’s theorem Thm: With 2 candidates, a voting rule is anonymous, neutral and monotonic iff it is the plurality rule – May, Kenneth. 1952. "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!

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

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)

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.

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)

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)

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

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

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)

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

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

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

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

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

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

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

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!

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

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

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

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

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)

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

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

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

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

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 – …

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

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

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

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

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

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

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

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.

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

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.

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.

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 )

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

 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

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

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

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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