1. Economics and Computation Week 6: Assignment games
Prof. Pingzhong Tang
Fall 2016
9/22/2018
Pingzhong Tang

2. References Two-sided matching
A study in game theoretical modeling and analysis
Alvin Roth & Marilda Sotomayor
Chapters 7 (one seller and many buyers) and 8 (assignment games)
9/22/2018
Pingzhong Tang

3. One seller and many buyers
A market (thought of as an auction)
One seller s
Indivisible item for sale
Values at rs: will not sell the item below price rs
Interpretation: has an outside option that worth rs
A set of buyers
Each buyer b values it at rb: will not buy it above rb
Transaction happens between b and s at price p
If, no other monetary transfers are made
Then, s gains p (not p- rs), b gains rb -p
9/22/2018
Pingzhong Tang

4. Transferable utility model
Money is freely transferable
Not only possible between seller and buyers
But also possible among buyers!
Utility of a buyer can be anything
A rigorous game-theoretical analysis is in need
Cooperative game theory
Stability, core
Spoiler alert:
At stable outcomes, no money transfer for losers
Losers always get utility 0
9/22/2018
Pingzhong Tang

5. Formulation A cooperative game N={1, …, n, n+1}
1 to n are buyers, n+1 is the seller
ri is the valuation of i
A subset S of N is called a coalition
Coalition function: v(S), for each S
The maximum amount of utility S can create
v(S)=0, if n+1 not in S, otherwise
v(S)=max{ri|i in S},
v(N)=max{ri|i in N}
Knowing v(S) summarize possible ways S divides the money
Money transferable after transaction
9/22/2018
Pingzhong Tang

6. Outcome(=Payoff) of the game
Payoff vectors:
The amount of money each player ends up with
Notation: x=(x1,x2,…,xn+1)
x(S)=∑xi|i in S
Feasibility constraints
X={x=(x1,x2,…,xn+1) in Rn+1 |x(N)≤v(N)}
Question: why don’t we have x(S)≤v(S) for each S?
9/22/2018
Pingzhong Tang

7. Core Define a payoff vector x dominates y
If there exists a coalition S, such that
xi>yi, for all i in S, and x(S)≤v(S)
In other words, every member of S prefer x to y
S can afford x (can create x on their own)
Two vectors can dominate each other
Core is the set of undominated payoff vectors
Theorem:
Core={x in X|x(S)≥v(S) for all S }
If for some S, we have a <, easily construct y that dominates x
Can be empty
9/22/2018
Pingzhong Tang

8. Stability A payoff vector is stable, if
It is individually rational (nonnegative)
There does not exist a buyer i and price p such that
p > xn+1
ri-p > xi
if such i and p exists, they will both deviate and block x
i and n+1 will make transaction, both better off than at x
IOW, stability =core constraints applied to |S|=1, 2
9/22/2018
Pingzhong Tang

9. Main theorem Renaming N: {1*, 2*, …, (n+1)*} Theorem:
so that ri*≥ri+1*
Theorem:
1. The core is always non-empty 2.
If n+1 ≠ 1*, and r1*≥r2*, Core={x in X| such that xn+1=p and x1*=r1*-p for r1*≥ p ≥r2*, and xi=0 for other}
Otherwise, Core={(0, …,0, v(N))}
3. The set of stable vectors equals core
9/22/2018
Pingzhong Tang

10. Statement 1 Non-emptyness of the core
(0, …,0, v(N)) is always in the core
Easy to check: every S, x(S) no less than v(S)
9/22/2018
Pingzhong Tang

11. Statement 2
If n+1 ≠ 1*, and r1*≥r2*, Core={x in X| such that xn+1=p and x1*=r1*-p for r1*≥ p ≥r2*, and xi=0 for loser}. Otherwise, Core={(0, …,0, v(N))}
If unique highest buyer, core divides the benefit between seller and this buyer. Losers gets nothing.
Intuition of proof
If non-zero for loser, S={1*, n+1} dominated
If p less than r2*, S={2*, n+1} dominated
9/22/2018
Pingzhong Tang

12. Statement 3 Core  Stability Not core  not stable
By statement 2, when the price is above the second highest price, no other buyer wants to buy the item
Not core  not stable
Then
either not individually rational
Or {1*, n+1} dominated (transfer losers’ payoff to {1*, n+1} )
Thus, unstable
9/22/2018
Pingzhong Tang

13. Truthfulness
Define a stable matching mechanism to be a function h that selects, for every valuation input r, a payoff vector h(r) in the core
Theorem: No stable matching mechanism exists for which truthful report of valuation is a dominant strategy for every agent
9/22/2018
Pingzhong Tang

14. Proof If exists unique highest buyer
By previous theorem, core = continuum set
Let h(r) be in the core, and is not buyer optimal
buyer 1* obtains the item at p>r2*
1* could have reported p>r’>r2*
Since h selects member in the core, in h(r’), 1* obtains the item at price less or equal to r’. Better off for 1*
Let h(r) be in the core and is not seller optimal
buyer 1* obtains the item at p<r1*
Seller could have reported p<r’≤r1*
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Pingzhong Tang

15. Truthfulness
There does exist a stable matching mechanism in which truthful report of valuation is a dominant strategy for every buyer
Why?
Exercise, check whether third price auction satisfies this theorem.
9/22/2018
Pingzhong Tang

16. Assignment games Similar to the previous model
But with multiple sellers
P: set of m buyers; Q: set of n sellers
αij: how much buyer i values item of seller j
Seller has valuation 0 of her own item
Thus, the coalition function v is given by
V({i,j})= αij for i in P and j in Q.
V(S)=0, if S consists only of P agents or Q agents
V(S)= value of maximum matching in S
9/22/2018
Pingzhong Tang

17. Payoff vectors Payoff vector (u,v) in Rm x Rn
If no other monetary transfer are made or received by i or j:
ui= αij-p
vj=p
Buyer i purchased item from seller j at price p
Seller’s valuation is normalized to 0. WLOG.
Again, money is freely transferable
Question: find core/stable outcome of this game
9/22/2018
Pingzhong Tang

18. Worth of P∪Q v(P∪Q) Find via LP Maximize ∑ αijxij
Maximum total payoff achievable
At core, ∑ui+∑vi= v(P∪Q)
Find via LP
Maximize ∑ αijxij
Subject to
(a) ∑i xij≤1
(b) ∑j xij≤1
(c) xij ≥ 0
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Pingzhong Tang

19. About the LP Guarantee integer solution
xij is either 0 or 1
Denote whether there is a transaction between (i, j)
v(P∪Q) = ∑ αijxij where x is the solution of LP
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Pingzhong Tang

20. Feasible and optimal assignment
Feasible: integer x that satisfies a), b), c)
Optimal: feasible and is solution of the LP
Optimal assignment always exists, not unique
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Pingzhong Tang

21. Feasible payoff vectors
A payoff vector (u,v) is feasible, if there exist a feasible assignment x such that
∑ui+∑vi= ∑ αijxij
In this case, we say (u,v) and x are compatible and ((u,v),x) is a feasible outcome
9/22/2018
Pingzhong Tang

22. Stable outcomes
Definition: A feasible outcome ((u,v),x) is stable, if for all (i, j):
1) ui, vj ≥ 0
2) ui+vj ≥ αij
Lemma: Let ((u,v),x) be a stable outcome, then
1) ui+vj = αij for all pairs matched in x (i.e., xij=1)
2) ui, vj = 0 for unassigned i,j
Proof
∑ui+∑vi= ∑ (ui+vj) xij|xij=1 + ∑i unmachedui + ∑j unmachedvj
= ∑ αijxij
Now apply stability. QED.
Again, losers get 0 payoff in stable matching
First =, by decomposition, second = by feasibility ∑ui+∑vi= ∑ αijxij
9/22/2018
Pingzhong Tang

23. Core outcomes Dual of the maximum matching LP: Minimize ∑ui+∑vi
Subject to
a) ui, vj ≥ 0
b) ui+vj ≥ αij
By strong duality:
If x is an optimal assignment (solution of primal) and (u,v) is a solution of the dual, we have
∑ui+∑vi = ∑ αijxij = v(P∪Q)
Thus,
1) ((u,v) ,x) is feasible and optimal.
2) ((u,v) ,x) is stable.
9/22/2018
Pingzhong Tang

24. The core - continued Minimize ∑ui+∑vi
Subject to
a) ui, vj ≥ 0
b) ui+vj ≥ αij
b) says ui+vj ≥ v({i,j})  for any S: ∑ui+∑vi ≥ v(S)
This is exactly the constraints of core!
i) ∑ui+∑vi = ∑ αijxij = v(P∪Q) ensures feasibility
ii)∑ui+∑vi ≥ v(S) ensures non-dominance for any S
Or simply by the theorem…
Conversely, any payoff in the core is a solution of the dual
9/22/2018
Pingzhong Tang

25. To summarize…
Theorem (Shapley and Shubik, 1972). In any assignment game,
The set of stable outcomes and the core are the same
The core is the set of solutions of the dual
At the core/stable outcomes, primal maximized, dual minimized, sum of payoffs equals the value of maximum matching
In other words, any algorithmically optimal solution satisfies desirable economic properties!
9/22/2018
Pingzhong Tang

26. Corollaries
Corollary: If x is an optimal assignment, it is compatible with any stable/core payoff.
In stable payoff, ∑ui+∑vi = ∑ αijxij = v(P∪Q), for any optimal x
Corollary: if ((u,v),x) is a stable outcome, then x is optimal.
Again, ∑ui+∑vi = v(P∪Q)= ∑ αijxij
First =, feasible, second feasible
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Pingzhong Tang

27. More corollaries
Corollary: Let ((u,v),x) and ((u’,v’),x’) be stable outcomes of an assignment game. Then if x’ij=1, ui’>ui implies vj’<vj.
Interpretation: opposed interests between seller and buyer.
Proof: Suppose otherwise vj’≥vj. Then, αij = u’i+v’j > ui+vj = αij
Corollary: The core forms a lattice.
Exists a P optimal stable payoff.
Exists a Q optimal stable payoff.
= is from a previous Lemma. The order on the lattices is defined by relations on the two stable outcomes.
9/22/2018
Pingzhong Tang