TU Game Theory. v(s ) + v(t) v(s T), S, T N with S T =,

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1 TU Game Theory Advances in Game Theory Basic Properties A game v is said to be super additive if v(s ) + v(t) v(s T), S, T N with S T =, inessential if it is additive so that v(s ) = v({i}), S N 1

2 A super additive game v is inessential iff v(n) = i N v({i}). A game is essential if it is not inessential. monotonic if v(s ) v(t), T S N, constant-sum or zero-sum if v(s ) + v( S ) = v(n), S N, where S denotes the complement of S relative to N, symmetric if v(s ) = v(t), S, T N with S = T. convex if v(s ) + v(t) v(s T) + v(s T), S, T N, or equivalently, if for all S, T, and i such that S T N \ {i}, v(s {i}) v(s ) v(t {i}) v(t). 2

3 Theorem A game v is convex if and only if for all S, T, and i such that S T N \ {i}, v(s {i}) v(s ) v(t {i}) v(t). Proof. The direction = is straightforward, because by considering S T N \ {i}, we have v(s {i}) + v(t) v(s {i} T) + v((s {i}) T) = v(t {i}) + v(s ). To see the converse, let S, R N, and let S \ R = {i, j, k,..., l 1, l} 3

4 Then, v(s ) v(s R) = v((s R) {i}) v(s R) + v((s R) {i} { j}) v((s R) {i}) + v((s R) {i} { j} {k}) v((s R) {i} { j}) + + v(s ) v((s R) {i} { j} {k} {l 1}) v(r {i}) v(r) + v(r {i} { j}) v(r {i}) + v(r {i} { j} {k}) v(r {i} { j}) + + v(r S ) v(r {i} { j} {k} {l 1}) = v(r S ) v(r). 4

5 Two games v and v with an identical player set N are said to be strategically equivalent each other if for some a > 0 and (b 1,..., b n ), v (S ) = av(s ) + b i, S N. (Show that any constant-sum game is strategically equivalent to the game with the constant being replaced by zero. ) Given any game v, we may obtain strategically equivalent normalizations of v as follows: 0-normalization 0,1-normalization v (S ) = v(s ) v({i}), S N v (S ) = v(s ) v({i}) v(n) i N v({i}) 5

6 1.1.4 Imputations Given a coalitional game (N, v), the vector x = (x 1,..., x n ) R n will be called an imputation if it satisfies total rationality i N x i = v(n) individual rationality x i v({i}), i N If x is an imputation of v, then it should be obvious that corresponding imputations x and x for 0-normalization v and 0,1-normalization v, respectively, are given by x i = x i v({i}), i N x i = x i v({i}) v(n) i N v({i}), i N 6

7 1.2 Stable Sets Let (N, v) be a coalitional game, and let A be the set of imputations. For any x, y A, we define consecutively as follows: x dom y S N s.t. x i v(s ), and x i > y i Dom x = {y y A, x dom y} For K A, Dom K = x K Dom x i S Definition K A is called a stable set if K = A \ Dom K. Remark K is a stable set internal stability x, y K [x dom y] external stability z K x K [x dom z] 7

8 Definition The set C = {x A z A[z dom x]} is called the core of a game. Problem Show that for a nonempty core C, 1. C K for all nonempty stable sets K, 2. C is a unique stable set if C itself is a stable set Stable Sets of Zero-Sum Three Person Games v({1, 2, 3}) = v({1, 2}) = v({1, 3}) = v({2, 3}) = 1; v({i}) = 0, i {1, 2, 3} objective solution K = {(1/2, 1/2, 0), (1/2, 0, 1/2), (0, 1/2, 1/2)} discriminatory solution K i (c), for 0 c < 1/2 and i = 1, 2, 3, where K i (c) = {x A x i = c, x j + x k = 1 c}, and {i, j, k} = {1, 2, 3} 8

9 1.2.2 One Seller and Two Buyers player 1: the seller of a house with value a player 2: buyer with evaluation b player 3: buyer with evaluation c Assumption: a < b c v({1}) = a; v({2}) = v({3}) = 0; v({1, 2}) = b; v({1, 3}) = c; v({2, 3}) = 0; v({1, 2, 3}) = c. K = {(p, 0, c p) b p c} {(p, d(p), c p d(p)) a p b, d(p) = 0 if p = b}, where the graph (p, d(p), c p d(p)) goes down within 30 degrees relative to the vertical direction. 9

10 A x B + x C = v({b, C}) = 0 w w = (p, 0, c p) where b p c p = b z D Y = (y A, y B, y C ) p = b e B e F z Y w E z C = (p, d(p), c p d(p)) = (b e, d(p), c b + [e d(p)]) where a p = b e b x A + x B = v({a, B}) = b x A + x C = v({a, C}) = c The payoff d(p) is a compensation paid from player 3 with which player 2 ceases to be a competitor of player 3. The segment [(c, 0, 0), (b, 0, c b)] is the core and contained in every one of infinite number of stable sets. 10

11 1.3 The Core In the previous section, the core was defined to be the set of all undominated imputations. When the game is super additive, the core can be obtained as follows: Theorem If v is super additive, the core C is given by C = { x A x i v(s ), S N }. Problem Prove this theorem. We may, however, define the core directly to be the set C given in the above theorem. Hereafter, the core of a game v will be denoted by C(v). 11

12 Example (The Core of 3-Person Games). Let (N, v) be a 0 - normalized 3-person game. Then: C(v) 1. v({1, 2}) + v({1, 3}) + v({2, 3}) 2v({1, 2, 3}) 2. v(s ) v(n) S N, S = 2 Proof. The necessity follows simply from adding both sides of the core inequality. To prove the sufficiency, letting v({1, 2}) = a 3, v({1, 3}) = a 2, v({2, 3}) = a 1, we may take ɛ i 0 (i = 1, 2, 3) such that a 1 + a 2 + a 3 + ɛ 1 + ɛ 2 + ɛ 3 = 2v(N). Define the payoff vector x by x i = v(n) (a i + ɛ i ), i = 1, 2, 3. Then, x 1 + x 2 + x 3 = 3v(N) 2v(N) = v(n), 12

13 implying x to satisfy the total rationality. By condition (2), we may take 0 ɛ i v(n) a i for i = 1, 2, 3 to obtain x i 0 for i = 1, 2, 3. Hence, x is an imputation. That x belongs to the core follows from x i v(n) a i for i = 1, 2, 3. Theorem (the core of symmetric games). Let v be a symmetric game. Then, C(v) v(s ) S v(n), S N. n Proof. The sufficiency part is obvious. To show the necessity, let s be any integer with 1 s n and let S N be any coalition with S = s. Then, for some payoff vector x, v(s ) x(s ) := x i, S N. 13

14 Adding both sides for all S N with S = s, we have ( ) n ( ) n 1 v(s ) x(s ) = x(n). s s 1 Hence, which leads to the conclusion. S =s n! (n s)!s! v(s ) (n 1)! (n s)!(s 1)! v(n), Theorem (the core of convex games, Shapley (1972)). The core of a convex game is nonempty. 14

15 Proof. (Shapley [40, 1972]). Let x be a payoff vector defined by x i = v({1,.., i 1, i}) v({1,..., i 1}), i N. It is easy to see that x is an imputation. Let S N, and let N \ S = { j(1),..., j(n s)}, where j(1) j(n s), and s = S. For j(1), put T = {1,..., j(1) 1, j(1)}. Then, S T = S { j(1)}, S T = {1,..., j(1) 1}. Hence, v(s ) + v({1,..., j(1) 1, j(1)} v(s { j(1)}) + v({1,..., j(1) 1}), 15

16 so that x j(1) = v({1,..., j(1) 1, j(1)}) v({1,..., j(1) 1}) v(s { j(1)}) v(s ). Next, putting S = S { j(1)} and considering T = T { j(2)}, we obtain x j(2) v(s { j(1)} { j(2)}) v(s { j(1)}). Continuing this process, we finally obtain x j(n s) v(n) v(s { j(1)} { j(n s 1)}). Adding both sides of these n s inequalities, we obtain x j(k) v(n) v(s ). Hence, v(s ) x i. k=1,...,n s 16

17 1.3.1 Balanced Games Definition A family B of nonempty, proper subsets of N is balanced iff there exist positive weights (balancing weights) δ S for S B such that δ S = 1 f or all i N S B, S i Example N = {1, 2, 3}, B= {{1, 2}, {2, 3}, {1, 3}}, δ S = 1 2 Example Take a game (N, v) and any nonempty S N. Then, the family B = {R N R = S } is balanced with the balancing weights δ R = ( 1 n 1 ), R B. R 1 17

18 Definition A game (N, v) is balanced iff for every balanced family B, δ S v(s ) v(n). S B Theorem (Bondareva [8, 1963], Shapley [37, 1967]) The core of a TU coalitional game is nonempty if and only if it is balanced Proof by LP Duality Theorem Primal. min n j=1 c j x j s.t. matrix form: min cx s.t. Ax b nj=1 a i j x j b i, i = 1,..., m. Dual. max m i=1 b i y i s.t. mi=1 y i a i j = c j, j = 1,..., n, y i 0, i = 1,..., n. matrix form: max by s.t. ya = c, y 0. 18

19 Theorem (Duality Theorem). If either the Primal or the Dual has a finite optimal solution, then the other also has a finite optimal solution, and the optimal values are the same. In order to make use of this theorem, consider the following problems. Primal. min n j=1 x j s.t. j S x j v(s ) S N. Dual. max S N δ S v(s ) s.t. S N S j δ S = 1, j = 1,..., n; δ S 0 S N. 19

20 To put in the matrix form, let us define: A = (a S j ) is the (2n 2) n matrix with rows in a given order. row vector a S is the characteristic vector of S N, i.e., a S = (a S 1,..., as n ), a S j = 1 if j S and a S j = 0 if j S. 1 = (1,..., 1) R n +. v R 2N \({N} { }) with elements v(s ) arranged in the given order. δ R 2N \({N} { }) with elements δ S arranged in the given order. Then: Primal: min 1x s.t. Ax v Dual: max δv s.t. δa = 1, δ 0 20

21 The game (N, v) is thus balanced if and only if the problem Dual has an optimal solution {δ S } S N satisfying S N δ S v(s ) v(n). Proof. (Proof of Bondareva=Shapley Theorem). If v is balanced, then there exists an optimal solution {δ S }. Hence by the duality theorem the Primal also has an optimal solution x = (x1,..., x n) satisfying n x j = v(s ) v(n), j=1 S N δ S which shows that the core is nonempty. 21

22 Conversely, if the core of v is nonempty, then the Primal has an optimal solution such that n j=1 x j v(n). Thus, by the duality theorem, the Dual has an optimal solution {δ S }, which implies that the game v is balanced, because n δ S v(s ) δ S v(s ) = x j v(n) S N S N j=1 for all solutions {δ S } of Dual. Problem Prove the necessary and sufficient condition for a symmetric game to have a nonempty core by showing the balancedness of the game. 1.4 TU Market Games Assumption The preference relation for each i N over X = R m + R is represented by a continuous function u i : R m + R such that for each i N, (x i, ξ) i (x i, ξ ) iff u i (x i ) + ξ u i (x i ) + ξ 22

23 Remark The value u i (x i ) is called a transferable utility (TU) for x i, and ξ is called money. Proposition Let U : R m + R R be the utility function given by U(x, m) = u(x) + m, where u is a transferable utility. Then, U is quasiconcave u is concave. Problem Prove this proposition. Definition A TU market game is a coalitional game (N, v) defined as follows: For each S N, v(s ) = max { } u i (x i ) x = (x 1,..., x n ) R m +, x i = w i, i N where u i ( ) is a concave transferable utility function of player i. The vector w i R m + is the initial endowments of player i. The vector x such that x i R m + for all i N is called an allocation. The allocation x that 23

24 satisfies the resource constraint x i = in coalition S will be called an S allocation. Theorem (Shapley and Shubik [41]) Every TU market game is balanced. Proof. Let B be any balanced collection, and for each S N, let f S = ( fi S ) be the S -allocation that gives v(s ). Define the allocation f by f i = S B, S i w i δ S f S i, which is a convex combination of f S i, S B, S i. 24

25 The allocation f is an N allocation, since f i = δ S fi S = δ S fi S i N i N S B, S i S B = δ S w i = S B By the concavity of u i, we have δ S v(s ) = δ S S B S B u i i N u i ( f S i ) = S B, S i which shows that the game is balanced. δ S f S i i N i N S B, S i S B, S i δ S u i ( f S i ) δ S w i = = u i ( f i ) v(n), i N w i. i N Definition A game (N, v) is totally balanced iff every subgame of (N, v) is balanced. 25

26 Crollary Every market game is totally balanced. This is so because every subgame of a market game is a market game; and then, it is balanced by the above theorem The Direct Market Definition A direct market (DM, for short) is a market (N, e, u ) where e = (e i ) i N, e i R+ N is the characteristic vector of i N, and u : R+ N R is the common utility function of the players which is homogeneous of degree 1, concave and continuous. Remark The common utility function u is superadditive, i.e., Problem Prove this. u (x + y) u (x) + u (y), x, y R N + 26

27 Remark If (N, v) is the market game generated by the DM, then v(s ) = u (e S ). To see this, the superadditivity of u implies, v(s ) = max u (x i ) u ( ) x i = u (e S ), S N where max runs over all the S-allocations x (i.e., x i = e i := e S ). On the other hand, since u (e S ) = S u (e S / S ) and x i = e S / S for all i S is an S -allocation, we have v(s ) u (e S ). Hence, v(s ) = u (e S ). that We now associate with every game (N, v) a direct market (N, e, u ) such Note that u (x) = max δ S v(s ) s.t. S N δ S e S = x, and δ S 0 S N. S N S N δ S e S = e R = S R δ S e S = e R Problem Prove that u is concave. Homogeneity and continuity are obvious. 27

28 Definition Given a game (N, v) and the associated DM (N, e, u ), the game (N, v) will be called the totally balanced cover of (N, v) iff for any R N, v(r) = max δ S v(s ) S R s.t. δ S e S = e R and δ S 0 S R. S R Remark v(r) v(r) R N. Remark The totally balanced cover (N, v) is a market game. This is so by Remark and the fact that the market game (N, v M ) defined on the DM which is generated from a given game (N, v) satisfies that v M (S ) = u (e S ) = v(s ) for all S N. 28

29 Lemma A game (N, v) is balanced if and only if v(n) = v(n). Proof. By definition, the game (N, v) is balanced if and only if v(n) v(n), which is equivalent to v(n) = v(n). Crollary A game (N, v) is totally balanced if and only if v = v. Proof. For any nonempty R N, the totally balanced cover of the subgame (R, v) is (R, v). Hence by lemma the result follows. Theorem (Shapley and Shubik [41, 1969]). A game is a market game if and only if it is totally balanced. Proof. Every market game is totally balanced (Corollary 1.4.1). A totally balanced game coincides with its totally balanced cover (Corollary 1.4.2). The totally balanced cover is a market game. Crollary A convex game is totally balanced. 29

30 Proof. A convex game has a nonempty core; and hence, it is balanced. Every subgame of a convex game is convex, which implies that a convex game is totally balanced. Crollary A convex game is a market game. Remark Let BAL, T BAL and CON be the collections of balanced games, totally balanced games and convex games, respectively. Then, BAL T BAL CON The inclusions are proper. The games v 1 and v 2 given below satisfy that v 1 BAL \ T BAL and v 2 T BAL \ CON. v 1 is a symmetric game satisfying 1. v 1(S ) v 1(N) for all S N; so that v 1 is balanced. S N 2. v 1(S ) S > v 1(T) T for some S T N; so that v 1 is not totally balanced. 30

31 v 2 is a symmetric game defined by 0 if S = 1 v 2 (S ) = a S if S > 1 where a > 0. Then: 1. v 2 is totally balanced, since for all R = 1, S > 1 and S T N, 0 = v 2(R) R < v 2(S ) S = a = v 2(T) T 2. v 2 is not convex, since v({1, 2}) v({2}) = 2a > v({1, 2, 3}) v({2, 3}) = a. 31

32 1.4.2 A Public Good Game Definition Let v be defined by v(s ) = max ( u i (y) c(y)) f or all S N, y 0 where y is a quantity of a public good, u i ( ) is an increasing transferable utility function, and c : R + R is an increasing linear cost function. Then, (N, v) is said to be a TU public good game. Theorem A TU public good game (N, v) is convex. Problem Prove this theorem. 32

33 1.4.3 Competitive Outcomes and the Core of Market Games Let x i R m + be an allocation to player i N, and let u i (x i ) + ξ i be the transferable utility of i for x i, with ξ i being the net amount of money of i after the trade. Lemma Allocation (x, ξ) = ((x 1, ξ 1),..., (x n, ξ n )) i N R m+1 + and price vector (p, 1) = (p 1,..., p m, 1) R m+1 + constitute a TU competitive equilibrium if and only if for each i N, 1. u i (x i ) p (x i w i ) u i (x i ) p (x i w i ) x i R m + 2. ξ i = p w i p x i Proof. Note first that the price for money is normalized to 1. Since (x, ξ) = ((x 1, ξ 1),..., (x n, ξ n )) i N R m+1 + is a competitive equilibrium under p, we 33

34 have for all i N that p x i + ξ i p w i, and (1) p y i + η i p w i implies u i (y i ) + η i u i (x i ) + ξ i (2) Then, since p w i = p xi = p xi + ξ i p i N i N i N i N we have 2, that is p xi + ξ i = p w i i N. i N w i To see 1, letting η i = p w i p y i in (2), we have u i (y i ) u i (x i ) ξ i η i = (p w i p x i ) (p w i p y i ), or, u i (x i ) p (x i w i ) u i (y i ) p (y i w i ). 34

35 Conversely, let p and (x i, ξ i) satisfy 1 and 2 for all i N. Then, p y i +η i p w i implies that u i (y i ) + η i u i (y i ) + p w i p y i u i (xi ) p (xi w i ) = u i (xi ) + ξ i. Hence, the price vector (p, 1) and the allocation ((x1, ξ 1),..., (xn, ξ n )) constitute a competitive equilibrium. Theorem Let b = (b 1,..., b n ) be the competitive equilibrium payoff vector defined by b i = u i (xi ) p (xi w i ) i N, where p and (x1,..., x n) constitute a TU comtetitive equilibrium. Then, b belongs to the core of the market game. 35

36 Proof. For any S N, let S allocation y R m + satisfy that u i (y i ) = v(s ) = max { u i (x i ) x is an S allocation }. Then, by the definition of b i we have that b i u i (y i ) p (y i w i ). Hence, ( b i u i (y i ) p y i = u i (y i ) = v(s ). ı S ) w i 36

37 Moreover, b i = u i (xi ) max{ u i (x i ) } = v(n). i N i N i N Hence, b is in the core. Theorem Every payoff vector b = (b 1,..., b n ) in the core of a game (N, v) is competitive in the direct market (N, e, u ) of that game. Remark If the game v is not balanced, the theorem is vacuously true. If the game is balanced but not totally balanced, the theorem is stil true because it is shown in Shapley and Subik [41, 1969] that any balanced game is d-equivalent to its cover v; and hence, the core coincides with each other. Proof. By the above remark, we shall assume that the game v is a market game. Let b be in the core of v. We show that b itself can be used as a competitive price vector in the direct market of v. Recall that u (e R ) = v(r) for 37

38 all R N, which is because a market game is totally balanced and coincides with its cover v. Then, u (e N ) = v(n) = b e N Take any x R+ N and let {δ S : S N} be any nonnegative weights with S N δ S e S = x. Then, by the fact that b is in the core of v, u (x) = max δ S v(s ) max δ S (b e S ) = max b δ S e S = b x. S N S N Now, define prices by π i = b i for all i N, and take z N satisfying z i = f i e N and f i 0 for all i N with i N f i = 1. Then it is easy to see that the pair (π, z N ) constitutes a competitive equilibrium. In fact, for each i N we have, S N u (z i ) π (z i e i ) = 0 + π i u (x i ) π (x i e i ) x i R N +. 38

39 Dominance-Equivalence. Two games are called d-equivalent (domination - equivalent) if 1. they have the same sets of imputations 2. they have the same domination relations on the imputation sets It is known that d-equivalent games have the same NM-stable sets, or lack of them the same cores, or lack of them the property of being balanced is preserved under d-equivalence within the class of games satisfying v(s ) + v({i}) v(n) S N i N\S every balanced game is d-equivalent to its cover. 39

40 The last fact may be seen as follows. 1. The imputation sets are equal since v(n) = v(n), and v({i}) = v({i}). 2. For imputations x and y, if x dom y in v but not in v, we must have some nonempty R N such that a x i > y i i R, and x(r) v(r) b x(s ) > v(s ) S R 3. Recalling the definition of v that v(r) = δ S v(s ) and δ S e S = e R, S R S R it follows from 2-(b) that v(r) < δ S x(s ) = δ S x i = x(r). 4. This contradicts 2-(a). S R i R S i, S R 40

41 1.4.4 Glove Market Games Recall that a market game is defined as follows. Definition A game (N, v) is a market game if there is a market (N, R m +, A = (a i ) i N, W = (w i ) i N ) with v(s ) = max w i (x i ) x S X S, S N, where X S := { x S = (x i ) x i R m + i S and x i = a i}. 41

42 Definition A glove market game (N, v) is given by v(s ) = min { S N 1, S N 2 }, S N, where {N 1, N 2 } is a partition of N with N 1, N 2 ; m = 2, a i = (1, 0) for all i N 1, a i = (0, 1) for all i N 2 ; w i (x) = min{x 1, x 2 } for all x R 2 + and for all i N. Proposition The glove market game is a market game; and hence, totally balanced. 42

43 Proof. We have to show that a glove market game satisfies the definition of a market game. That is, we have to show v(s ) = max min{x 1 i, xi 2 } x S X S, S N. Note first that since min{x1 i, xi 2 } xi 1, xi 2, we have for any S N, min{x1 i, xi 2 } x1 i, x2 i. Hence, min{x1 i, xi 2 } min x 1 i, that is, min{, } is constant; so that we have v(s ) max min x 1 i, 43 x i 2 x i 2 = min a i 1, = min a i 1, a i 2 a i 2.,

44 On the other hand, since the allocation y S = (y i ) given by y i = a i 1, a i 2, for some i S, y j = (0, 0), for all j S \ {i} is a feasible S allocation, we have Hence, v(s ) w i (y i ) = min a i 1, v(s ) = min a i 1, a i 2 a i 2. = min{ S N 1, S N 2 }. 44

45 Remark The function w(x) = w(x 1, x 2 ) = min{x 1, x 2 } is concave in R 2 +. Proof. Let h [0, 1], and take any x, y R 2 +. Then, hw(x) = min{hx 1, hx 2 } hx 1, hx 2, (1 h)w(y) = min{(1 h)y 1, (1 h)y 2 } (1 h)y 1, (1 h)y 2 Then, hw(x) + (1 h)w(y) hx 1 + (1 h)y 1, hx 2 + (1 h)y 2. Hence, hw(x) + (1 h)w(y) min{hx 1 + (1 h)y 1, hx 2 + (1 h)y 2 } = w(hx + (1 h)y). 45

46 Remark The function w(x 1, x 2 ) = min{x 1, x 2 } is homogeneous of degree 1, that is, w(tx 1, tx 2 ) = tw(x 1, x 2 ) for all t > 0. Remark Since the function w(x 1, x 2 ) = min{x 1, x 2 } is concave and homogeneous of degree 1, it is superadditive; that is, min{x 1 + y 1, x 2 + y 2 } min{x 1, x 2 } + min{y 1, y 2 } x, y R 2 +. The proof of the above proposition will be made more concise by exploiting this property. 46

47 1.4.5 Assignment Games Let N = S B, where S, B and S B =. S is the set of sellers of indivisible goods, and B is the set of potential buyers. Each seller has exactly one item, and each buyer wants at most one item. a i 0 is the worth of the good to the seller i S, b i j 0 is buyer j s worth for the good of seller i, c({i, j}) = max{b i j a i, 0} is the joint net profit of {i, j} for all i S and j B. 47

48 Let T N. Then an assignment for T is a collection of pairs of a seller and a buyer in T; that is, a collection T 2 T such that P S = P B = 1 and P Q = for all P, Q T with P Q. Example Let N = {1, 2,..., 9}; S = {1, 2, 3, 4, 5}, B = N \ S, and let T = {2, 4, 5, 6, 9}. Then T = {{2, 6}, {4, 9}} is an assignment for T. Note that the seller 5 is excluded from the trade in the assignment T. Definition (N, v) is an assignment game if v(t) = max c(p) T is an assignment for T, for all T N. P T By convention, the empty sum is zero, so that v(t) = 0 for all T S and T B. For any T N, let τ(t) be the set of all assignments for T. 48

49 Remark The glove market game v(s ) = min{ S L, S R } for all S N is an assignment game, where every member in L has one left glove, and every member in R has one right glove. To see this, let L be the set of sellers of left gloves, and let R be the set of buyers of left gloves. Take any set T N and an assignment T ; and any seller i and buyer j paired in the assignment T. Then, letting b i j = min{1, 1} = 1, a i = 0 and 0 p 1 be the price of left gloves, the joint net profit for P = {i, j} is 1 p + p = 1; namely, c({i, j}) = max{b i j a i, 0} = 1. Hence, v(t) = max P T c(p) T τ(t) = min{ T L, T R } for all T N. 49

50 1.4.6 The LP Problem and its Dual of the Assignment Game Denoting by c i j = c({i, j}) for all i S and j B, the LP problem to obtain v(n) is given as follows. max c i j x i j s.t. x i j 1, i S ; x i j 1, j B, j B j B x i j 0, i S, j B. Remark Clearly, all the pairs {i, j} in the objective function do not constitute the assignment in N. But it is known that the optimum is attained with every x i j taking 0 or 1, yielding an assignment in N *1. Hence, the optimum attains v(n). *1 See, e.g., Danzig,G.B.:Linear Programming and Extensions, Princeton University Press, Princeton, Any optimum is generically attained at an extreme point, i.e., at a point that cannot be expressed by a proper convex combination in the convex set. 50

51 The primal problem was : max c i j x i j s.t. j B x i j 1, i S ; j B x i j 0, i S, j B. x i j 1, j B, The dual of this LP problem is then given by: min u i + v j s.t. u i + v j c i j, i S, j B j B u i 0, i S, v j 0 j B. That the dual is given as above can be seen from a simple case. Consider the case: S = {1, 2, 3}, B = {4, 5, 6}. The matrix form of the LP problem is as follows: 51

52 Letting c = (c 14, c 15, c 16, c 24,..., c 36 ) and x = (x 14, x 15, x 16, x 24,..., x 36 ), and max cx s.t. x 0 (x 14, x 15, x 16, x 24,..., x 36 ) (1, 1, 1, 1, 1, 1) 52

53 Then, denoting by F and w = (u, v) the matrix and the dual variables, respectively, the dual problem can be obtained as follows. min u 1 + u 2 + u 3 + v 4 + v 5 + v 6 s.t. w 0, and Fw T c T The constraint is thus u i + v j c i j, i = 1, 2, 3; and j = 4, 5, The Core of the Assignment Game Theorem The core of the assignment game (N, v) coincides with the set of payoff vectors w = (u, v) achieving the optimum of the corresponding dual LP problem. 53

54 Proof. Let (u, v) = ((u i ), (v j ) j B ) be a vector of optimal solution of the dual. By the duality theorem, we have u i + v j = v(n). j B Hence, (u, v) is an imputation of the assignment game. Moreover, by the constraints u i + v j c i j, i S, j B, we have that for any optimal assignment T in T N, (u P S + v P B ) c P S,P B = c(p) = v(t). P T P T P T Since this is true for all T N, the vector (u, v) is in the core of the assignment game (N, v). Conversely, by definition, any core element satisfies the constraints in the dual problem. 54

55 Theorem The core of the assignment game coincides with the set of competitive equilibrium allocations. Proof. Outline of the proof *2 : Take any payoff vector (u, v) in the core, and let p i = a i + u i i S be the price at which seller i S sells i s item. Then, the typical buyer j B will face a choice among the S possible net gains: b i j p i, i S. If these numbers are all negative, then j will stay out of the market, and obtain a profit of zero; otherwise j will seek to maximize the net profit. Then, since c i j = b i j a i, this is equivalent to maximizing c i j u i i S. *2 see, e.g. Suzuki,M and S.Muto: Theory of Cooperative Game Theory (Japanese), University of Tokyo Press

56 But, since the core coincides with the dual optimal solutions, any of these values cannot exceed v j ; and, at least one of them is equal to v j. Hence, buyer j s maximum profit is precisely v j by trading with the seller i for which c i j u i = v j. The converse can be seen from the fact that the assignment game is a market game, so that the core contains competitive allocations. 56

57 1.5 Simple Games Definition A pair (N, W) is a simple game if N is the set of players and W is a set of subsets of N satisfying 1. N W 2. W 3. If S T N, then S W = T W The collection W of coalitions is the set of winning coalitions. Definition A simple game (N, W) is proper if S W N \ S W strong if S W N \ S W weak if V = S W S 57

58 The members of V are called veto players or vetoers. The simple game is dictatorial if there exists j N such that S W j S. The player j N is called the dictator. Remark Consider the simple game with a unique vetoer that is not the dictator. Then this game is weak and proper, but is not strong. Definition Given a simple game g = (N, W), the associated coalitional game (N, v) is defined by 1 if S W v(s ) = 0 otherwise 58

59 Remark Let g = (N, W) be a simple game and let G = (N, v) be the associated coalitional game. Then: 1. G is monotonic. 2. G is constant-sum if and only if g is strong. 3. G is superadditive if and only if g is proper. proof : S, N W and the superadditivity imply that v(n \ S ) = 0 i.e., g is proper. Conversely, if v(s ) + v(t) > v(s T) for some S and T with S T =, then S, T and S T are all winning. But, then N \ S is also winning because T N \ S and g is monotone, which contradicts the assumption that g is proper. 4. Any monotonic coalitional game (N, v) satisfying v(s ) {0, 1} for all S N and v(n) = 1 is the associated coalitional game of some simple game. proof: Just take W = {S N v(s ) = 1}. 59

60 Definition A simple game (N, W) is a weighted majority game if there exist a quota q > 0 and weights w i 0 for all i N such that for all S N S W w(s ) q The ( N + 1) tuple (q; (w i ) i N ) is called a representation of g. Definition Let g = (N, W) be a weighted majority game. The representation (q; (w i ) i N) of g is a homogeneous representation of g if S W min w(s ) = q, where W min is the set of minimal winning coalitions. A weighted majority game is homogeneous if it has a homogeneous representation. 60

61 Remark A symmetric simple game g = (N, W) has the homogeneous representation (k; 1,..., 1), where k = min S W S. Such a game is also denoted by ( N, k). The simple majority game is then given by ( N, [ n 2 ] + 1), where [z] is the integer satisfying z 1 < [z] z for all real numbers z. Example The UN Security Council is a weighted majority game with the representation g := (39; 7, 7, 7, 7, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1). This game is weak and homogeneous. The vetoers are the Big Five. 61

62 Problem Prove the following assertions. 1. A weak simple game is proper. 2. A simple game is dictatorial if and only if it is both weak and strong. 3. A dictatorial simple game gives rise to an inessential game. 62

63 1.5.1 Social Choice Games The Social Choice Game (N, v) is the coalitional game defined by v(s ) = max u i (y) S W, y Y where N = {1,..., n}, n 3 Y is a finite set of social alternatives W is a nonempty set of winning coalitions with the monotonicity condition S W, S T = T W 63

64 The core C(W) is given by C(W) = { x R+ N i N x i = v(n), x i v(s ) S W } Theorem (Kaneko *3 ). Let (N, v) be the social choice game with the simple majority rule W m : W m = {S N S > n/2}. Then, x C(W m ) if and only if there exists y N Y with v(n) = i N u i (y N ) satisfying 1. x i = u i (y N ) i N 2. v(s ) = u i (y N ) S W m *3 Kaneko, M.: Necessary and sufficient conditions for the existence of nonempty core of a majority game, International Journal of Game Theory 4 (1974),

65 Proof. Sufficiency is obvious. To see (1), suppose to the contrary that there exist x C(W m ) and i N such that x i > u i (y N ). Then, for this i we have v(n) x i v(n \ {i}). Hence, v(n) u i (y N ) > v(n) x i v(n \ {i}) or j N\{i} u j (y N ) > v(n \ {i}). Since N \ {i} W m, this inequality contradicts the definition of v(n \ {i}) that v(n \ {i}) j N\{i} u j (y N ). Hence (1). Next, suppose that there existed S W m with v(s ) j S u j (y N ). Then, by definition of v and (1) v(s ) > u j (y N ) = x j, j S j S which contradicts the assumption that x C(W m ). 65

66 This theorem indicates that it is rather an exceptional case to obtain a nonempty core of a simple majority game. If it happened that there existed a social alternative y N Y in the core C(W), then the core consists of a unique payoff vector, and every member i of the society would acquire exactly the payoff equal to u i (y N ). In this sense, the majority rule would yield a fair social state involving no sidepayments at all. In real world social decision making situations, however, illegal sidepayments such as bribery or vote purchases are frequently observed. The next theorem clarifies the structural relation between bribery and power. 66

67 Theorem (Nakayama *4 ). Suppose there exist a payoff vector x and i N satisfying that x C(W) and x i > u i (y ), where y Y is one that attains v(n), i.e., v(n) = i N u i (y ). Then, the player i is a vetoer. Proof. We must have N \ {i} W, because we would otherwise have u j (y ) = u j (y ) u i (y ) > v(n) x i v(n \ {i}), j N\{i} j N contradicting the definition of v(n\{i}). It then follows from the monotonicity that S N \ {i} = S W. Hence, this i is a vetoer. *4 Nakayama, M.: Note on the core and compensation in collective choice, Mathematical Social Sciences 2 (1982),

68 This theorem states that if some player obtains a payoff in the core greater than the amount obtainable under a Pareto efficient alternative y, then the player is necessarily a veto player. While two payoff vectors (u i (y )) i N and x with x (u i (y )) i N cannot be both in the core under the majority rule, this is possible under a weak social choice game, bribing all veto players. 1.6 The Information Trading Game Technological information may have a negative externality in the sense that the profit from the information decreases as the the number of the holders increases. A typical example will be the information on a new technology that reduces the production cost of a firm in an oligopolistic industry. Below, we consider the trade of such an information. 68

69 E(k) is the profit or benefit of the information holder when the number of holders of information is k 1 E(k) is decreasing in k. Player 1 is the sole initial holder of the information, and faces the choice whether or not to sell the information. Player 1 may sell to nonholders under the agreement not to resell the acquired information. This agreement will be generally not self-enforcing, since resale is costless to the reseller. We show, however, that the information can be traded without any resale occuring (the resale-proof trading of information). 69

70 Example N = {1, 2, 3}: E(1) = 30, E(2) = 16, E(3) = 9 1. E(1) < 2E(2) 2. E(2) < 2E(3) 3. E(1) 3E(3) 4. E(1) 2E(3) + E(3) > E(2) + E(3). Thus, by 3, the initial holder cannot sell to the two remaining nonholders. But, even if the holder sells to just one player at price E(2), 2 implies that the no-resale agreement will be violated and the information will be necessarily resold to the last player. Then, 4 indicates that the profit of the initial holder will go down to E(2) + E(3), which is less than E(1). Hence, anticipating this, the initial holder will not sell the information to any nonholder. 70

71 Example 2 N = {1, 2, 3, 4}: E(1) = 30, E(2) = 16, E(3) = 9, E(4) = 5 1. E(1) < 2E(2) 2. E(2) < 2E(3) 3. E(2) 3E(4) 4. E(3) < 2E(4) 5. E(2) 2E(4) + E(4) > E(3) + E(4). Thus: By 2 and 3, when there are two holders, any one of holders may try to resell to just one player. But, 4 indicates that when the number of the holders becomes 3, one of the holders surely deviates from the agreement and secures the profit by reselling to the last nonholder, thereby allowing the information disseminated to all. 71

72 Then, the profit of the first reseller will be reduced less than E(2). That is, any one of the two holders cannot resell to anyone: the no-resale agreement is enforcing in this case. Hence, the original holder can sell the information to just one player. Definition (Dissemination-Proof Coalition M) Let M N be any coalition with 1 M. 1. If M = N, then we say M is dissemination-proof. 2. Suppose the definition is completed for all M N with n M > m 1. Then, for all M with M = m, we say M is dissemination-proof if E( M ) (1 + T )E( M T ) for all T N M such that M T is dissemination-proof. 72

73 Theorem (Nakayama et al. [31]). Let M and M be any two dissemination-proof coalitions satisfying 1 M < M. Then, M E( M ) M E( M ) where the inequality is strict if M > 1. Proof. By the symmetry we may assume that M M. Since M and M are both dissemination-proof, it follows by definition that E( M ) (1 + M M )E( M ). Then, summing over all i M, we have M E( M ) M E( M ) + M ( M M )E( M ) = M E( M ) + ( M 1)( M M )E( M ) M E( M ), with a strict inequality if M > 1. 73

74 1.7 Dissemination-Proof Information Trading Game v v(s ) = M E( M ) if 1 S and M S E(1) if 1 S and M > S 0 if 1 S where, M is the minimal-size dissemination-proof coalition. Theorem Let C(v) be the core of the game v, and let x be the imputation given by x = ( M E( M ), 0,..., 0 ). Then, C(v) = {x } M N Proof. ( =): Take any S with 1 S N. Then, x (S ) = M E( M ) v(s ); so thatx C(v) If an imputation x satisfies that x 1 < x1, some j N \ {1} has x j > 0 Hence, whenever x(m ) = M E( M ), x k = 0 for any k N \ M. Then, x ( (M \ { j}) {k} ) < v( M ). Hence, x C(v). (= ): M = N implies that v(s ) = E(1) or 0 for any S N, it follows that any imputation x satisfies x(s ) v(s ) for any S N. 74

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