T . Quint, M. Shubik Mathematical Social Sciences 41 2001 89 –102
91
2. On status games vs. matching games
Below, we will define status games formally. Readers versed in the theory of matching games will see a strong resemblance between status games and the models
presented by Shapley and Scarf 1974; Kaneko 1982; Quinzii 1984 and Quint 1997. Indeed, all of the above constructs use permutation matrices in order to define
feasible distributions of a set of indivisible goods. In the matching games listed above, the indivisible goods are thought to be large objects such as houses, or else service-
binding contracts. In status games, they are of course the different ‘positions’ that a society has to offer.
However, there are two important differences between status games and the matching models listed above. First, in matching games there is an initial ownership of goods,
while in status games there is not. In this sense, status games are more general than matching games, because the capabilities of coalitions are not limited by their members’
initial endowments. On the other hand, in matching games the rankings of the players over the objects is allowed to be more general than in status games. For example, in
Shapley and Scarf’s houseswapping game, the traders are each allowed to value different houses differently – while in status games, all players rank ‘first place’ first, ‘second
place’ second, etc.
In Section 5, we will define one-to-one ordinal preference games, which are a large class of games including both status games and the above matching games.
3. Two models
We have distinguished two ways of modeling games of status, based on what is assumed about the capabilities of coalitions smaller than N. In ‘exo-status games’, we
assume that such small coalitions are capable of enacting orderings over all n players. [The adjective ‘exo’ refers to the fact that a coalition S might have the power to directly
determine the fates of players outside of S.] Alternatively, in ‘endo-status games’, we assume that a small coalition S is only capable of guaranteeing certain positions for its
own members within the n-player hierarchy, but has no control over how the other n 2
uSu players are placed into the remaining positions. To see the difference between these two approaches, consider an example of an organization with a defacto leader. If
modeled as an endo-status game, the most powerful he could be would be if he were able to guarantee himself ‘first place’. On the other hand, in an exo-status game, he
would also be able to place the other players into any positions he wished in the hierarchy.
In this paper, we are concerned primarily with endo-status games. For a discussion of exo-status games, see Quint and Shubik 1999.
4. Endo-status games
4.1. The model We now define endo-status games formally. Let N 5
h1, . . . ,nj be the player set, and
92 T
. Quint, M. Shubik Mathematical Social Sciences 41 2001 89 –102
N
let 2 be the set of subsets of N. In this game, an outcome for a player is either a a
ranking or status or position, i.e., an ordinal number from the set h1, . . . ,nj, or b to
4
be ‘unranked’ . The utility to any player of being unranked is defined to be zero. For rankings, define the n 3 n matrix R, where r represents the utility to player i of ending
ij
up with ranking j. Since we assume that a player always weakly prefers a higher status to a lower status, we have j , k
⇒ r r . We also assume that any player weakly
ij ik
prefers any ranking over the ‘unranked state’, i.e. r 0 ;i, j.
ij N
Let S be an element of 2 . Then, an S-ordering P is an n 3 n 0–1 matrix in which a
S n
n
if i [ ⁄ S, then p
5 0 for j 5 1, . . . ,n, b for all j, o
p 1, and c for all i,
o
S uij
i 51 S
uij j 51
p 1. In words, an S-ordering describes an outcome that the coalition S could
S uij
potentially effect, under the interpretation p 5 1 ⇔
player i receives position j.
ij
Condition a is the characteristic assumption of endo-status games, that coalition S has
5
no power to determine the placement of players in N 2 S. Condition b states that two
6
players cannot be placed in the same position . Finally, c states that no player can be placed in more than one position. Note that there is no requirement that individual
members of S be given exactly one position, i.e., it is possible for there to exist an i [ S
n
with o
p 5 0. Such a player would be unranked, and would receive payoff zero.
j 51 S
uij
Let P be the set of all N-orderings, and, for any S let P denote the set of all
S
S-orderings that S can actually effect. We assume the n 3 n zero-matrix Z is an element of P for all S. Let p i i [ S denote the position that i obtains under S-ordering p ,
S S
S
with the convention p i 5 5 meaning that i is not given a position under p .
S S
We assume that the sets hP j
satisfy the following assumptions:
S S N
1 GRAND COALITION POWER: P 5 P. The grand coalition can institute any
N
N-ordering of the players. 2 ‘CAN DO NOTHING’: Z [ P for all S.
S
3 RANK CONSISTENCY: Suppose S and T are disjoint, and let p [ P , p [ P .
S S
T T
Then the matrix p 1 p has all of its column sums equal to zero or one.
S T
The idea of rank consistency is to forbid disjoint coalitions from having the ability to carry out mutually inconsistent outcomes. To wit, if the condition stated above did not
hold, this would imply the existence of a pair of disjoint coalitions, each able to guarantee a common ‘‘position j’’ for one of its constituents.
4 SUPERADDITIVITY: Suppose S and T are disjoint, and let p [ P and
S S
4
We need to allow for the ‘unranked’ outcome in this game, in order to ensure that the characteristic function of the resultant NTU game evaluated at any coalition is nonempty.
5
Except via the ‘filling up of positions’ with its own players. For instance, suppose that the coalition ]]]
S 51, . . . ,n 2 1 can guarantee position 1 for player 1, position 2 for player 2, . . . , position n 2 1 for player n 2 1. Then necessarily S can limit player n to position n even though n [
⁄ S because that position is the
only one left.
6
So we rule out status games with ties, in which outcomes are allowed in which players ‘tie’ for positions. For example, an outcome in a four-player status game with ties could be [3 1T4 2], which means that player 3
comes in ‘first place’, 2 comes in ‘last’, but players 1 and 4 come in tied for second place. We remark that situations in which there are a large number of ‘equivalent positions’ e.g. societies with many equal positions
in the ‘peasant class’ can be modeled with a large number of distinct positions, no ties allowed, but with all players ranking such ‘equivalent positions’ the same.
T . Quint, M. Shubik Mathematical Social Sciences 41 2001 89 –102
93
p [ P . Then given its column sums are all 0 or 1, the matrix p 1 p is an element
T T
S T
of P .
S T
5 MONOTONICITY: If S T, then P P . Note that 5 follows from 2 and
S T
4. We now define a cooperative NTU game, called an endo-status game, in which the
N n
7
characteristic function V :2 →
R is given by :
u r ,
for i [ S: p i ± 5
i ip i
S
S
VS 5 hu , . . . ,u j: p [ P satisfying
H
1 n
S S
u 0, for i [ S: p i 5 5
i S
Note that we may define an endo-status game by the triple N, hP j
, R.
S S N
4.2. The core of an endo-status game
n
The core of an endo-status game is defined as the set of vectors u [ R such that a
u [ VN and b there does not exist S, w with w [ VS and w . u ;i [ S. The strict
i i
core is defined similarly, except condition b is amended to read ‘there does not exist S, w with w [ VS and w u ;i [ S, with w . u only required for one i is S.’ These
i i
i i
are the usual definition of core and strict core for a NTU game.
Example 4.1. Suppose n 5 3, with P 5
hZj for all S with uSu 5 1 Z is the three-by-three
S
zero-matrix and P 5 P. For S with uSu 5 2, define the matrices
N
1 M 5 0
1
S D
1
1 M 5 1
S D
2
1 M 5 0
S D
3
1 1
M 5 0
S D
4
1 M 5 1
S D
5
1 M 5 0
1
S D
6
1
] ]
]
and suppose P 5
hZ, M , M j, P 5 hZ, M , M j, and P 5 hZ, M , M j. Finally, let
12 1
2 13
3 4
23 5
6
7
Note that the characteristic function below assumes the free disposal of utility.
94 T
. Quint, M. Shubik Mathematical Social Sciences 41 2001 89 –102
the matrix R 5 hr j be any 3 by 3 matrix in which r . r . r for i 5 1,2,3 i.e., each
ij i 1
i 2 i 3
player strictly prefers ‘Rank 1’ to ‘Rank 2’ to ‘Rank 3’. Then we claim the core is empty. This is because in any element of P there will be one player who gets ‘Rank 2’
N
or worse, and another who gets ‘Rank 3’ or worse. These two players will be able to form a coalition in which the former player can get ‘Rank 1’ and the latter ‘Rank 2’.
]
Example 4.2. Let us modify Example 4.1, so that P is equal to
hZ, M j instead of hZ,
23 5
M , M j. Then the outcome where player 1 gets ‘Rank 1’, player 3 gets ‘Rank 2’, and
5 6
8
player 2 gets ‘Rank 3’ is in the core. Hence here we have a nonempty core. We next turn to classical game theory to find sufficient conditions for core
nonemptiness. To some readers, the ‘balancedness’ conditions that we present below will seem complicated. To some extent, we agree; however, we also note that throughout
the theory of matching games, such balancedness conditions have been used again and again to prove that the cores of certain games are empty. See Shapley and Scarf 1974,
Kaneko 1982, Quinzii 1984 and Quint 1997.
Suppose G 5 N,V is an NTU game, with N the player set and VS the characteristic
n
function whose values are regions in R . Let T be a set of coalitions for which there exist a set of nonnegative weights
hd j with
o d 5 1 for all i [ N. Then T is a
S S [T S [T :S ]i
S
balanced family of coalitions, and hd j
its balancing weights. It is a minimal
S S [T
balanced family if furthermore it has no proper subset which is balanced. Note that if T is minimal balanced, then d . 0 for all S [ T.
S
Theorem 4.3. Scarf, 1967 Let G 5 N,V be an NTU game. If
a VS is closed for each S N; b x [ VS and x y ;i [ S implies y [ VS ;
i i
S
c [VS 2 int V
hij] R is bounded and nonempty;
i [S
d VS VN for all minimal balanced families T;
S [T
Then the core of the game is nonempty. Definition 4.4. Suppose N and
hP j are the player set and ‘feasible permutation set’
S S N
from an endo-status game. Suppose that for every minimal balanced family T and every set b
; hP j P is any element of P , there is at least one player who gets ‘position
S S [T S
S
n or worse’ in some element of b, at least two players who get ‘position n 2 1 or worse’ in some element of b, . . . , and at least n 2 1 players who get ‘position 2 or worse’ in
9
some element of b. Then we say N, hP j
is balanced.
S S N
Theorem 4.5. If G 5 N,
hP j ,R is a status game in which N,
hP j is balanced,
S S N S S N
its core is nonempty for any R. ] ] ]
Example 4.6. In Example 4.1 above, consider T 5
h12, 13, 23j with d 5 12, 12,
8
And no player disposes utility. See Footnote 7.
9
Lest this is unclear, it is permissible for various players to get their ‘position k or worse’ out of different elements of b.
T . Quint, M. Shubik Mathematical Social Sciences 41 2001 89 –102
95
1 2. Consider b 5 hM , M , M j. There is no player who gets ‘position n or worse’ in
1 3
5
any element of b, and so N, hP j
is not balanced. In fact, we have seen that the
S S N
core of this game is empty.
Example 4.7. Suppose we now consider the game of Example 4.2. Using the logic of Example 4.6, we observe that again N,
hP j is not balanced. Yet, as we have seen,
S S N
the core of the game is nonempty. Hence we conclude that the converse of Theorem 4.5 will not hold.
Finally, before embarking on the proof of Theorem 4.5 we need two more pieces of notation. Given an S-ordering P , define the n 3 n 1 1 matrix P
by:
S S
P ,
if j n;
S uij
n
P 5 1 2
O
P ,
if j 5 n 1 1 and i [ S;
S uij
S uij
j 51
5
0, if j 5 n 1 1 and i [
⁄ S.
Essentially, the matrix P is an extension of P in which an extra the n 1 1st column
S S
is added that represents the ‘unranked’ status. Let P represent the set of members of
S
P , written in this fashion.
S
In keeping with this interpretation, we define r 5 0 for i 5 1, . . . ,n.
in 11
Proof of Theorem 4.5. We use Theorem 4.3. So suppose G 5 N,
hP j ,R is an
S S N
endo-status game in which hN, P j is balanced, and consider any minimal balanced
S
family T with balancing weights hd j
. Suppose u [ VS ;S [ T. Our aim is to show
S S [T
that u [ VN as well.
u u
u
Define the n by n 1 1 0–1 matrix F in which f 5 0 if r u and f 5 1 otherwise.
ij ij
i ij
Also, define the operation ‘‘ 3 ’’ on n by n 1 1 matrices by: A 3 B 5 a b . Given this
ij ij
ij n
n 11 u
notation, one may say that u [ VS iff o
o P 3 F 5 0 for some P [ P .
i 51 j 51
S ij
S S
We have
n n 11 u
u [ VS ;S [ T ⇒
OO
P 3 F 5 0 ;S [ T
S ij
i 51 j 51 u
⇒
O
d
O
P 3 F 5 0
S S
ij
S D
S [T i, j
4.1
u
⇒
OO
d P 3 F 5 0
s d
ij S
S S [T i, j
u
⇒
O O
d P 3 F
5 0.
S S
SS D D
ij i, j
S [T
Now the matrix o
d P has all row sums equal to 1 proof similar to that in Shapley
S [T S
S 10
and Scarf, 1974. Hence it can be written as a convex combination of stochastic 0–1
k m
ˆ matrices, i.e.
o d P 5
o a P for some positive constants
ha j and n 3 n 1 1
S [T S
S k 51
k k
10
By a stochastic matrix, we mean a nonnegative matrix whose row sums are all 1.
96 T
. Quint, M. Shubik Mathematical Social Sciences 41 2001 89 –102
k
ˆ stochastic 0–1 matrices
hP j. Continuing from 4.1, we have
m u
k u
ˆ
O O
d P 3 F
5 0 ⇒
O O
a P 3 F
5 0
SS D D
S S
k
SS D D
ij ij
i, j S [T
i, j k 51
m k
u
ˆ ⇒
O O
a P 3 F 5 0
k ij
k 51 i, j k
u
ˆ ⇒
O
P 3 F 5 0 for all k
ij i, j
In addition, due to the ‘at least one player gets position n or worse . . . ’ condition, in matrix
o d P there is at least one row with a positive element in column n or n 1 1,
S [T S
S
at least two rows with a positive element in column n 2 1, n, or n 1 1, at least three
k m
ˆ rows . . . etc. Hence when we decompose
o d P
into o
a P , we can do so in
S [T S
S k 51
k k
K
ˆ ˆ
such a way so that one of the P ’s say P will have a row with a ‘1’ in column n or n 1 1, another row with a ‘1’ in column n 2 1, n, or n 1 1, another row with . . . and
another row with a ‘1’ in column 1, 2, . . . , or n 1 1. Without loss of generality, assume these are rows n, n 2 1, . . . , and 1 respectively. Then, if we define the N-ordering P by
N K
u
ˆ P i 5 i for i 5 1, . . . ,n, we have that r
r ;i. So
o P 3 F 5 0
K
ˆ N
iP i iP i
i, j ij
N
u
implies o P 3 F 5 0. But by the ‘grand coalition power’ axiom, P [ P . This
i, j N
ij N
N u
together with o P 3 F 5 0 implies that u [VN. h
i, j N
ij
Remark 1. Unfortunately, we believe that the balancedness condition is much stronger than core nonemptiness. In matching games, it is desirable at least to have the
‘balancedness condition’ be equivalent to guaranteed core nonemptiness in the case of three players. This is in fact the case with restricted houseswapping games with ordinal
preferences Quint, 1997. We have already seen in Example 4.7 that we do not have this equivalence.
Remark 2. However, let us now consider the class of exchangeable endo-status games,
2 2
i.e., games in which if P [ P and if P is another S-ordering with o P
5 o P
S S
S i
S uij
i S
uij 2
for all j, then P [ P as well. In words, an exchangeable endo-status game is one in
S S
which, if a certain S can guarantee a certain set of statuses for its members, then it is possible for those members to freely distribute those rankings among themselves in any
way they like.
Note that the game in Example 4.1 4.6 is exchangeable, while that of Example 4.2 4.7 is not. We also remark that exchangeability like balancedness is a property of
N, hP j
and not of R.
S S N
For exchangeable endo-status games, it is now true that for n 5 3, the balancedness condition is equivalent to guaranteed core nonemptiness. For higher n, we have the
following ‘almost converse’ of Theorem 4.5:
Theorem 4.8. Suppose N,
hP j is exchangeable but not balanced. Then there is a
S S N
ranking matrix R for which G 5 N, hP j
,R has an empty strict core.
S S N
In order to prove Theorem 4.8, we first state and prove a small lemma:
T . Quint, M. Shubik Mathematical Social Sciences 41 2001 89 –102
97
1
Lemma 4.9. Let T be a balanced family of coalitions for an n-person game. Let S and
2 1
2
S be coalitions not necessarily members of T satisfying a S S 5 5 and b
1 2
1 2
uS u . uS u. Then there is an element S of T for which uS S u . uS S u.
1
Proof. Suppose the Lemma was not true. Then there is a balanced family T with S and
2 1
2
S as above, but with
uS S u uS S u ;S [ T. Let hd j be the set of balancing
S S [T
weights. Trivially, we have o
d o
d for all S [ T. Summing, we have
1 2
i [S S S
i [S S S
O O
d
O O
d .
S S
1 2
S [T S [T
i [S S i [S S
1
By the definition of the set of balancing weights, the sum on the left is equal to uS u and
2
the sum on the right is uS u. Hence we have violated our initial assumption that
1 2
uS u . uS u.
Proof of Theorem 4.8. Since N,
hP j is not balanced, there exist a T, a
S S N
b 5 hP j
P [ P ;S [ T , and a k such that at least k 1 1 players get position k
S S [T S
S
or better in every element of b that contains them. Let Y be the set of such players. We
k
have uY u 5 k 1 p, where p 1.
k
Now suppose the ranking matrix is given by: 1,
if j k; r 5
H
ij
0, otherwise.
We claim the resultant status game has an empty strict core. For, suppose P is any
N 11
element of P . [We aim to show that the payoff under P is not in the strict core.]
N N
Define set of players in Y
who get
k
Y 5
H
k
a payoff of zero out of P .
N
set of players in Y who get
k 1
Y 5
H
k
a payoff of one out of P .
N
set of players not in Y who get
k 1C
Y 5
H
k
a payoff of one out of P .
N 1
1 1C
Now define s ; uY u. Then uY u 5 k 1 p 2 s. Furthermore, since uY u 1 uY u 5 k, we
k k
k k
1C
have uY u 5 s 2 p , uY u.
k k
Now, by Lemma 4.9 there is an element S [ T which contains more members of Y
k 1C
than it does of Y . Consider any i [ S Y . [Such a player exists because
k k
1C
uS Y u . uS Y u 0.] This player gets payoff 0 out of P , but gets 1 out of P .
k k
N S
11
When we refer to ‘the payoff under from out of P ’ we mean the payoff in which the positions are
N
distributed according to P , and there is no disposal of utility. [Again see Footnote 7.] Clearly, showing that
N
these particular payoff vectors are not in the strict core is sufficient for proving the emptiness of the strict core.
98 T
. Quint, M. Shubik Mathematical Social Sciences 41 2001 89 –102
12
Hence, the only possible way in which S would not ‘block’ P via P is if there were
N S
another player ji [ S for which 1 ji gets a payoff of 1 out of P ; and
N
2 ji gets a payoff of 0 out of P .
S 1C
Observe that 1 and 2 together imply that ji is an element of Y .
k 1
Now form P from P
by having i and ji switch positions. By the exchangeabili-
S S
1 1
ty property, we have P [
P . Also note that ji now gets a payoff of 1 out of P .
S S
S
Next, we observe that there must be another element of S Y , other than i , because
k 1C
1C 1
ji was an element of S Y and
uS Y u . uS Y u. Call this element i .
k k
k 1
1
Again, we see that i gets payoff 1 from P , but 0 from P . Hence, the only possible
S N
1 1
way in which S will not ‘block’ P via P is if there is a ji [ S for which
N S
1
3 ji gets a payoff of 1 out of P ; and
N 1
1
4 ji gets a payoff of 0 out of P .
S 1
Now ji is not equal to i because of 3; and it is not equal to any other element of
1 1
1C
Y because of 4. Hence ji [
⁄ Y . Indeed, by 3 we have ji [ Y . Also, since
k k
k 1
ji does not satisfy 4, it must be that ji is different from ji .
2 1
1 1
Again, form P from P
by having i and ji switch positions. Again by the
S S
2 1
2
exchangeability property, P [
P . Furthermore, ji gets a payoff of 1 out of P .
S S
S
And again we observe that there must be a third element of S Y , because we have
k 1C
2
already seen two distinct elements of S Y . Call this element i . . .
k m
Continuing in this fashion, we see that eventually we will find a i for which there
m m
will not exist a corresponding ji . In this case, S will ‘block’ P via P . h
N S
5. One-to-one ordinal preference OOP games
