In this paper we construct sets of marginal vectors of a TU game with the property that if the marginal vectors from these sets are core elements, then the game is convex.This approach leads to new upperbounds on the number of marginal vectors needed to characterize convexity.An other result is that the relative number of marginals needed to characterize convexity converges to zero.game theory;convexity;marginal vectors

Hamers, H.J.M.

Norde, H.W.

Velzen, S. van

English

Research Papers in Economics

No. 2002-53
CONVEXITY AND MARGINAL VECTORS
By Bas van Velzen, Herbert Hamers, Henk Norde
May 2002
ISSN 0924-7815Convexity and marginal vectors
Bas van Velzen, Herbert Hamers, Henk Norde
CentER and Department of Econometrics and Operations Research,
Tilburg University, P.O. Box90153, 5000 LE Tilburg, The Netherlands.
E-mail: S.vanVelzen@kub.nl,H.J.M.Hamers@kub.nl,H.Norde@kub.nl
Abstract
In this paper we construct sets of marginal vectors of a TU game with the property that if the
marginal vectors from these sets are core elements, then the game is convex. This approach
leads to new upperbounds on the number of marginal vectors needed to characterize convexity.
Another result is that the relative number of marginals needed to characterize convexity con-
verges to zero.
Keywords: convexity, marginal vectors, TU game
1 Introduction
This paper shows that if speciﬁc sets of marginal vectors of some TU game are core elements, then
the game is convex. Convexity is an appealing property in cooperative game theory. For convex
games it is established that the Shapley value is the barycentre of the core (Shapley (1971)),t h e
bargaining set and the core coincide, the kernel coincides with the nucleolus (Maschler et al. (1972))
and the τ-value can easily be calculated (Tijs (1981)).
It is well-known that a game is convex if and only if all marginal vectors are core elements
(Shapley (1971), Ichiishi (1981)). This result is strengthened in Rafels, Ybern (1995).T h e y
showed that if all even marginal vectors are core elements, then all odd marginal vectors are core
elements as well, and vice versa. Hence, if all even or all odd marginal vectors are core elements,
then the game is convex.
In this paper we establish new bounds for the number of marginal vectors needed to characterize
convexity and we present a convergence theorem which states that the relative number of marginal
vectors needed to characterize convexity converges to zero.
2 Preliminaries
In this section we recall some notions from cooperative game theory and introduce some notation.
A cooperative TU game is a pair (N,v)w h e r eN = {1,...,n} is a ﬁnite (player) set and v,t h e
characteristic function, is a map v :2 N → R with v(∅)=0 . T h em a pv assigns to each subset
S ⊂ N, called a coalition, a real number v(S), called the worth of S.T h ecore of a game (N,v)i s
the set
C(v)={x ∈ RN|
￿
j∈S
xj ≥ v(S) for every S ⊂ N,
￿
j∈N
xj = v(N)}.
1Intuitively, the core of a game can be interpreted as the set of payoﬀ vectors for which no coalition
has an incentive to leave the grand coalition N. Note that the core of a game can be empty. A
game (N,v)i sc a l l e dconvex if for all i,j ∈ N and S ⊂ N\{i,j} it holds that
v(S ∪{ i}) − v(S) ≤ v(S ∪{ i}∪{ j}) − v(S ∪{ j}). (1)
Hence, for convex games the marginal contribution of a player to any coalition is less than his
marginal contribution to a larger coalition.
Before we introduce marginal vectors, we ﬁrst introduce orders. For any T ⊂ N,a norder σ
of T is a bijection σ : {1,...,|T|} → T. This order is denoted by σ(1)···σ(|T|), where σ(i)=j
m e a n st h a tw i t hr e s p e c tt oσ,p l a y e rj is in the i-th position. The i-th neighbour of σ is the
order σi which is obtained by switching the players at the i-th and (i + 1)-th position of σ,i . e .
σi = σ(1)···σ(i−1)σ(i+1)σ(i)σ(i+2)···σ(|T|) .A no r d e ri sc a l l e deven if it can be turned into
the identity order eT by an even number of neighbourswitches, where eT is such that for every
i,j ∈{ 1,...,|T|} with i <j , it holds that eT(i) <e T(j). An order which is not even is called odd.
Note that every neighbour of an even order is odd and vice versa and that the set of all orders ST
contains as many even orders as odd orders.
Let (N,v) be a game. For any σ ∈ SN,t h emarginal vector mσ(v)i sd e ﬁ n e db y
mσ
i (v): =v([σ,i]) − v((σ,i)) for all i ∈ N
where [σ,i]={j ∈ N : σ−1(j) ≤ σ−1(i)} is the set of predecessors of i with respect to σ including
i,a n d( σ,i)={j ∈ N : σ−1(j) <σ −1(i)} is the set of predecessors of i excluding i.A m a r g i n a l
vector is called even (odd) if the corresponding order is even (odd). Furthermore, if τ is the i-th
neighbour of σ,t h e nmτ(v)i sc a l l e dt h ei-th neighbour of mσ(v).
Let {A1,...,A k} be a partition of N and let σi ∈ SAi for each i ∈{ 1,...,k}.T h e nt h ec o m b i n e d
order σ1σ2 ···σk of N is that order which begins with the players in A1 ordered according to σ1,
followed by the players in A2 ordered according to σ2, etcetera, i.e.
σ1σ2 ···σk := σ1(1)···σ1(|A1|)σ2(1)···σ2(|A2|)···σk(1)···σk(|Ak|).
The set A1|A2|...|Ak contains those orders which begin with the players in A1,f o l l o w e db yt h e
players in A2, etcetera, i.e. A1|A2|...|Ak := {σ1 ···σk : σi ∈ SAi for every i ∈{ 1,...,k}}.
Example 1 Let N = {1,2,3,4,5}, A1 = {1,5}, A2 = {3} and A3 = {2,4}.I fσ1 =5 1 ,σ2 =3a n d
σ3 =2 4 ,t h e nσ1σ2σ3 = 51324, and 15|3|24 = {15324,15342,51324,51342}.
3 Characterizing convexity using core marginals
In this section we present our main results. First we recall the theorem of Shapley (1971) and
Ichiishi (1981) which states that a game is convex if and only if all marginal vectors are core
elements. We recall a strengthening of this theorem by Rafels, Ybern (1995) and we prove this
strengthening in an alternative way. We obtain sharper bounds for the number of marginal vectors
needed to characterize convexity and we prove a convergence theorem.
Theorem 1 (Shapley (1971),Ichiishi (1981)) Let (N,v) be a game. Then v is convex if and
only if mσ(v) ∈ C(v) for all σ ∈ SN.
The strengthening posed in Rafels, Ybern (1995) states that if all even marginal vectors are core
elements, then all odd marginal vectors are core elements as well, and vice versa. Hence, Theorem
2 states that if n!
2 speciﬁc marginal vectors are core elements, then the game is convex. Therefore,
Theorem 2 provides an upperbound of n!
2 for the number of marginal vectors needed to characterize
convexity.
2Theorem 2 (Rafels,Ybern (1995)) Let (N,v) be a game. Then the following statements are
equivalent:
1. (N,v) is convex
2. mσ(v) ∈ C(v) for all even σ ∈ SN
3. mσ(v) ∈ C(v) for all odd σ ∈ SN.
Rafels, Ybern (1995) proved this theorem by showing that if all even or all odd marginal vectors
are core elements, then (1) is satisﬁed for every i,j ∈ N and every S ⊂ N\{i,j}. The following
lemma states that if for a marginal vector two consecutive neighbours are core elements, then this
marginal vector is a core element as well.
Lemma 1 Let (N,v) be a game with |N|≥3,a n dl e tσ ∈ SN.S u p p os et h e r ei sa n h ∈{ 1,...,n−2}
such that mσh
(v),m σh+1
(v) ∈ C(v).T h e nmσ(v) ∈ C(v).
Proof: Without loss of generality we assume that σ =1 2···n.T h e n
σh =1 ···h − 1 h +1hh+2···n,a n dσh+1 =1 ···hh+2h +1h +3···n. To show that
mσ(v) ∈ C(v), we need to show that
￿
i∈S mσ
i (v) ≥ v(S) for every S ⊂ N.S o l e t S ⊂ N,w e
distinguish between four cases.
Case 1: h,h +1￿∈ S.
Then it follows that
￿
j∈S mσ
j(v)=
￿
j∈S mσh
j (v) ≥ v(S), where the inequality holds because
mσh
(v) ∈ C(v).
Case 2: h ￿∈ S and h +1∈ S.
Let T = {1,...,h} =[ σ,h]. It follows that
￿
j∈T
mσh
j (v)=v([σ,h +1 ] )− v([σ,h − 1] ∪{ h +1 })+v([σ,h − 1]) ≥ v(T)=v([σ,h]), (2)
where the inequality holds because mσh
(v) ∈ C(v). Therefore
￿
j∈S
mσ
j(v)=
￿
j∈S
mσh
j (v) − mσh
h+1(v)+mσ
h+1(v)
=
￿
j∈S
mσh
j (v) − (v([σ,h − 1] ∪{ h +1 }) − v([σ,h − 1]))
+(v([σ,h +1 ] )− v([σ,h])) ≥ v(S),
where the inequality follows from mσh
(v) ∈ C(v) and (2).
Case 3: h ∈ S and h +1￿∈ S. We distinguish between two subcases.
Subcase 3a: h +2￿∈ S.
Then it follows that
￿
j∈S mσ
j(v)=
￿
j∈S mσh+1
j (v) ≥ v(S).
Subcase 3b: h +2∈ S.
By taking T = {1,...,h+1 } and using that mσh+1
(v) ∈ C(v), it follows similar to case 2 that ￿
j∈S mσ
j(v) ≥ v(S).
Case 4: h,h +1∈ S.
Then it follows that
￿
j∈S mσ
j(v)=
￿
j∈S mσh
j (v) ≥ v(S). ￿
3Example 2 Consider a game (N,v)w i t hN = {1,2,3}. Then 213 and 132 are the ﬁrst and second
n e i g h b o u ro f1 2 3 .F r o mL e m m a1i tf o l l o w st h a ti fm213(v),m 132(v) ∈ C(v), then m123(v) ∈ C(v).
By applying Lemma 1 we have, for n ≥ 3, an alternative proof of Theorem 2 by using that each
neighbour of an even marginal vector is odd and vice versa. In fact, using Lemma 1, even sharper
bounds than n!
2 c a nb eo b t a i n e d .T oo b t a i nt h e s eb o u n d sw ei n t r o d u c et h eo p e r a t o rb :2 ST → 2ST
for every T ⊂ N.F o r|T| =1 ,2, we deﬁne b(A): =A for all A ⊂ ST.F o r|T|≥3, we deﬁne
b(A): =A ∪{ σ ∈ ST :t h e r ei sa ni ∈{ 1,...,|T|−2} such that σi,σi+1 ∈ A}.
By applying Lemma 1 it follows that if mσ(v) ∈ C(v) for all σ ∈ A ⊂ SN,t h e nmτ(v) ∈ C(v)f o r
all τ ∈ b(A). Hence b can be used to generate larger sets of core marginals.
Example 3 Let A = {213,132},a n dl e tσ =1 2 3 . T h e nσ1 =2 1 3∈ A and σ2 =1 3 2∈ A.
Therefore, 123 ∈ b(A).
The following example shows that by repetetive appliance of b larger sets of core marginals can be
obtained.
Example 4 Let A = {2134,1324,1423}, σ =1 2 3 4a n dτ = 1243. It follows that σ1 =2 1 3 4∈ A,
and σ2 =1 3 2 4∈ A. Therefore it follows that σ ∈ b(A). Note that because τ ￿∈ A, τ1 =2 1 4 3￿∈ A
and τ3 =1 2 3 4￿∈ A it follows that τ ￿∈ b(A). However τ2 =1 4 2 3∈ b(A)a n dτ3 =1 2 3 4∈ b(A).
Therefore, τ ∈ b(b(A)) = b2(A).
Let T ⊂ N.D e ﬁ n et h eclosure of A ⊂ ST,d e n o t e db yb∗(A), to be the largest set of orders that can
be obtained by repetetive appliance of b, i.e. b∗(A)=bk(A)f o rk ∈ N such that bk(A)=bk+1(A).
Let C ⊂ ST.I fA ⊂ C is such that C ⊂ b∗(A), then A is called complete in C.I fA ⊂ ST is complete
in ST,t h e nA is called complete. From Lemma 1 it follows that if A ⊂ SN and mσ(v) ∈ C(v)f o r
every σ ∈ A,t h e nmτ(v) ∈ C(v) for every τ ∈ b∗(A). Hence, if A is complete, then mσ(v) ∈ C(v)
for every σ ∈ A implies that v is convex. The following lemma provides a set of orders which is
complete in T1|...|Tk.
Lemma 2 Let {T1,...,T k} be a partition of N and let Ai ⊂ STi.I fAi is complete in STi for each
i ∈{ 1,...,k},t h e nA = {τ1 ···τk : τi ∈ Ai for each i ∈{ 1,...,k}} is complete in T1|...|Tk.
Proof: Let σ1 ···σk ∈ T1|···|Tk. We use induction to show that σ1 ···σk ∈ b∗(A).
Let j ∈{ 1,...,k+1} be such that σ1 ···σj−1τj ···τk ∈ b∗(A) for all (τj,...,τ k) ∈ Aj×...×Ak.
Note that j = 1 satisﬁes this property.
Now let (τj+1,...,τ k) ∈ Aj+1 × ...× Ak and let C = {σ1 ···σj−1ττj+1···τk| τ ∈ Aj}.B y
assumption we have C ⊂ b∗(A). Because σj ∈ b∗(Aj)i tf o l l o w st h a tσ1 ···σj−1σjτj+1 ···τk ∈
b∗(C) ⊂ b∗(A). We conclude that σ1 ···σk ∈ b∗(A). ￿
Example 5 Let N = {1,...,6}, T1 = {124} and T2 = {3,5,6}.L e t A1 = {124,241,412} and
A2 = {356,563,635}.N o t e t h a t A1 and A2 are complete sets. It follows from Lemma 2 that
A = {στ : σ ∈ A1,τ ∈ A2} is complete in 124|356.
We now focus on the cardinality of complete sets. A set of orders A ⊂ ST is called minimum
complete if A is a complete set and for every other complete set B it holds that |B|≥| A|,i . e .
there exists no complete set with smaller cardinality. Because of symmetry, the cardinality of a
minimum complete set only depends on the cardinality t = |T|,a n dn o to nT. Therefore, let
the neighbour number Qt denote the cardinality of a minimum complete set in ST,i . e . Qt :=
4minA⊂ST:A is complete |A|. By deﬁnition it holds that Q1 =1a n dQ2 =2 . F r o mT h e o r e m2i t
follows that Qn ≤ n!
2 for each n ≥ 3.
In the ﬁnal part of this section we establish upperbounds and lowerbounds for Qn. Moreover, it
is shown that the relative neighbour number Fn :=
Qn
n! → 0i fn →∞ . The following theorem, which
is a direct consequence of Lemma 2, gives a strengthening of the bound obtained from Theorem 2.
Theorem 3 Let n1,...,n k,n∈ N,b es u c ht h a t
￿k
i=1ni = n.T h e nQn ≤ n!
n1!···nk!
￿k
i=1Qni.
Proof: The set SN can be partitioned into n!
n1!···nk! sets of the form T1|...|Tk with |Ti| = ni for each
i ∈{ 1,...,k}.N o wl e t{T1,...,T k} be a partition of N for which |Ti| = ni for each i ∈{ 1,...,k}.
A c c o r d i n gt oL e m m a2t h e r ei sas e tA containing
￿k
i=1 Qni elements such that A is complete in
T1|...|Tk.T h e r e f o r eQn ≤ n!
n1!···nk!
￿k
i=1 Qni. ￿
From Theorem 3 it follows that Qn+1 ≤
(n+1)!
n!1! QnQ1 =( n+1)Qn. Hence it follows that Fn+1 ≤ Fn
for each n, i.e. the relative neighbour number Fn is nonincreasing. The following lemma gives a
lowerbound on the neighbour number.
Lemma 3 If n is even then Qn ≥ n! 1
2
n−2
2
,a n di fn is odd then Qn ≥ n! 1
2
n−1
2
.
Proof: Let n be even, and let k = n+2
2 .
The set SN can be partitioned into n!
1!2!···2!1! = n!
2k−2 sets of the form T1|...|Tk with |T1| = |Tk| =1
and |Ti| =2f o re v e r yi ∈{ 2,...,k− 1}.L e t C ⊂ SN be one of those sets, i.e. there exists a
partition {T1,...,T k} of N with |T1| = |Tk| =1a n d|Ti| =2f o re v e r yi ∈{ 2,...,k−1} such that
C = T1|...|Tk.
It follows that for every σ ∈ C it holds that C = σ(1)|σ(2)σ(3)|...|σ(n). For every even
i ∈{ 1,...,n− 1} it holds that σi ∈ C.N o wl e tA ⊂ SN such that A ∩ C = ∅. Because for every
σ ∈ C there is no h ∈{ 1,...,n− 2} with σh,σh+1 ∈ A, it follows that σ ￿∈ b(A). This holds for
every σ ∈ C, hence b(A) ∩ C = ∅. By repetition it follows that b∗(A) ∩ C = ∅.
We conclude that if A ⊂ SN is complete, then |A ∩ C|≥1. This holds for every C such
that C = T1|...|Tk with |T1| = |Tk| =1 ,|Ti| =2f o re v e r yi ∈{ 2,...,k− 1}. It follows that
|A|≥n! 1
2
n−2
2
.
Now let n be odd, and let k = n+1
2 .T h e s e t SN can be partitioned into n!
1!2!···2!2! = n!
2k−1 sets of
the form T1|...|Tk with |T1| =1a n d|Ti| =2f o re v e r yi ∈{ 2,...,k}.N o wl e tC ⊂ SN be such
that C = T1|...|Tk with |T1| =1a n d|Ti| =2f o re v e r yi ∈{ 2,...,k}. Similar to the case where
n is even it follows that if A ⊂ SN is complete, then A ∩ C ￿= ∅.T h i sh o l d sf o re v e r yC such that
C = T1|...|Tk with |T1| =1 ,|Ti| =2f o re v e r yi ∈{ 2,...,k}.I tf o l l o w st h a t|A|≥n! 1
2
n−1
2
.H e n c e ,
|A|≥n! 1
2
n−1
2
. ￿
From Theorem 2 and Lemma 3 it follows that Q3 =3a n dQ4 = 12. We also obtain from Theorem 2
that Q5 ≤ 60 and from Lemma 3 we obtain that Q5 ≥ 30. Therefore Q5 ∈ [30,60]. However, using
ad hoc methods, we established that Q5 = 30. The proof of this lemma is given in the appendix.
Lemma 4 The neighbour number Q5 =3 0 .
Now, taking n1 = n2 = 3 and using Theorem 3, it follows that Q6 ≤ 6!
3!3! · Q3 · Q3 =1 8 0 . F r o m
L e m m a3w ed e r i v et h a tQ6 ≥ 6! 1
22 = 180, and hence Q6 = 180. Similar, by taking n1 =3
and n2 = 4, it follows that Q7 ≤ 1260. From Lemma 3 it follows that Q7 ≥ 630, and therefore
Q7 ∈ [630,1260]. Hence for Q7 we do not have a sharp bound. However, for Q8 we have a sharp
bound. By taking n1 =3a n dn2 =5w eo b t a i nt h a tQ8 ≤ 5040. From Lemma 3 it follows that
5n 34 5 6 7 8 9 1 0 1 1 1 2
n! 6 24 120 720 5040 40320 362880 3628800 39916800 479001600
n!
2 3 12 60 360 2520 20160 181440 1814400 19958400 239500800
Qn 3 12 30 180 [630,1260] 5040 [22680,45360] 22680 [1247400,2494800] [14968800,29937600]
Fn
1
2
1
2
1
4
1
4 [ 1
8, 1
4] 1
8 [ 1
16, 1
8] 1
16 [ 1
32, 1
16][ 1
32 , 1
16 ]
Table 1: New bounds
Q8 ≥ 5040, and hence Q8 = 5040. Some other new bounds are given in Table 1.
The following theorem states that the relative neighbour number Fn converges to 0.
Theorem 4 The relative neighbour number Fn → 0 if n →∞ .
Proof: Let k ∈ N,l e tn =3 k and let ni =3f o re v e r yi ∈{ 1,...,k}. From Theorem 3 we deduce
that
Q3k ≤
(3k)!
(3!)k 3k,
by using that Q3 ≤ 3. Therefore F3k ≤ (1
2)k for every k ∈ N.I tf o l l o w st h a tF3k → 0i fk →∞ .I t
follows from Theorem 3 that Fn+1 ≤ Fn for all n ∈ N. Hence, Fn → 0a sn →∞ . ￿
4 Appendix
P r o o fo fL e m m a4 :Let N = {1,...,5} and let
A = {13245,21345,31452,41523,52134,
14235,21453,31524,42135,53124,
14352,21534,32145,42513,53241,
15234,24351,32451,43125,54123,
15342,25341,32514,43251,54231,
15423,25413,35412,43512,54312}.
We will show that A is a complete set. Consider the sets
B = {13452,24513,31245,41235,52341,
14523,25134,35124,42351,53412}
and
C = {15324,21435,32541,43152,54213}.
Let σ ∈ B.T h e nσ1,σ2 ∈ A. Hence σ ∈ b(A). Let τ ∈ C.T h e nτ3,τ4 ∈ A. Hence τ ∈ b(A). It
follows that (A ∪ B ∪ C) ⊂ b(A).
Now let
D = {13524,24135,35241,41352,52413}.
Let σ ∈ D.T h e n σ1 ∈ A ⊂ b(A)a n dσ2 ∈ C ⊂ b(A). It follows that σ ∈ b(b(A)). Hence
(A ∪ B ∪ C ∪ D) ⊂ b2(A).
6Now let
E =1 3 |245 ∪ 14|235 ∪ 24|135 ∪ 25|134 ∪ 35|124 ∪ 1|5|234 ∪ 2|1|345 ∪ 3|2|145 ∪ 4|3|125 ∪ 5|4|123.
For each σ ∈ E, it holds that either that σ ∈ A∪B ∪C ∪D or that σ3,σ4 ∈ A∪B ∪C ∪D. Hence
σ ∈ b(A ∪ B ∪ C ∪ D) ⊂ b(b2(A)). Now let
F = SN\E =1 |2|345 ∪ 2|3|145 ∪ 3|4|125 ∪ 4|5|123 ∪ 5|1|234.
Then for all σ ∈ F it holds that σ1,σ2 ∈ E. Hence it follows that E is a complete set. From this
we conclude that A is a complete set. Because |A| =3 0i tf o l l o w st h a tQ5 ≤ 30. ￿
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Convexity and Marginal Vectors

Convexity and Marginal Vectors

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