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Mechanism Design without Money
Lecture 12
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.
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Proof (for k=3)
3
a3
a2
cd
a1
e b
“Cost” of IR is very small - Simulations
No. of Hospitals 2 4 6 8 10 12 14 16 18 20 22
IR,k=3 6.8 18.37 35.42 49.3 63.68 81.43 97.82109.01121.81144.09 160.74
Efficient, k=3 6.89 18.67 35.97 49.75 64.34 81.83 98.07109.41 122.1 144.35 161.07
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But the cost of not having IR could be very high if it causes centralized matching to break down
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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
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And hospitals can withhold individual
overdemanded pairs
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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)
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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/k1/6
• For each positive credit, the hospital increases this probability by 1/k2/3 and the credit is gone
• For each negative credit, the hospital decreases this probability by 1/k2/3 and the credit is gone. The probability is always > ½
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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
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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
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Voting
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, <s1,..,sm>– Candidate scores si for each vote in ith position– Candidate with greatest score wins
• Generalizes number of rules– Borda is <m-1,m-2,..,0>– Plurality is <1,0,..,0>
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
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
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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 ij and jk and ik we can create a cycle:
– i prefers A to C– k prefers C to B– j prefers B to A
• Similar argument for ij=k, i=j k, ji=k
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
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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
1 5 10 20 32 64 100 260 520 1041300
350
400
450
500
550
2-ways3-ways2-ways & chain3-ways & chain
Waiting period between match runs
Matches
Matching over time
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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
1 5 10 20 32 64 100 260 520 104190
110
130
150
170
190
210
230
2-ways3-ways2-ways & chain3-ways & chain
Matching over time
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1D 1W 2W 1M 3M 6M 1Y250255260265270275280285290295
Matches
Simulation results using 2 year data from NKR*
1D 1W 2W 1M 3M 6M 1Y100
120
140
160
180
200
220
240
Waiting Time
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
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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
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.
Either a sparse finite horizon modelor 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 .
• incoming edges to L are drawn w.p .
L
H
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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 .
• incoming edges to L are drawn w.p .
L
H
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At each time step 1,2,…, n, one node arrives .
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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.
pc/n
𝑝2
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Chunk Matching in a heterogeneous graph 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.
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Chunk Matching in a heterogeneous graph
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
.
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Chunk Matching in a heterogeneous graph
When matching 2 and 3-way cycles:
1. If Δ = M(Δ) = M(1) + f(q,p) (n)
(formally this is still a conjecture)
𝜔 (𝑛)
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Denser Poolsξ:
Theorem: 1 .If Δ < 1 ,/
M(Δ) = M(1) + o(n)2 .If Δ = α/
M(Δ) = M(1) + f(q)nfor strictly increasing f()>0.
Need to wait less time to gain …If the graph is dense (large) – no need to wait at all…
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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
pξ/n
𝑝2
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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
Proof Ideas
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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,
Proof Ideas
h1
l2 l1
h2
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Chunk Matching in a heterogeneous graph
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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