A theory of fairness, competition, and cooperation

s 2 ^ai (1 2 2s) 2 ^{n 2 2} b s $ 0 2 ^ai

a_i
(1 2 s) 2 s.

n 2 1
n 2 1 ⁱ

n 2 1

n 2 1

The right-hand side of this inequality is smaller than 0. We know already that the left-hand side is greater than 0 since (A11) is violated. Therefore, responder i prefers to accept s as well. We conclude that if (A11) does not hold for at least one i, then at least one responder will accept s. Hence, (A11) is also necessary. h If b₁ , (n 2 1)/ n, an equilibrium offer must be sustained by

the threat that any smaller offer s˜ will be rejected by everybody. But we know from Lemma 2 that an offer s˜ may be rejected only if (A11) holds for all i. Thus, the highest offer s that can be sustained in equilibrium is given by (8).

QED


Proof of Proposition 4

1. 
   Suppose that 1 2 a . b_i for player i. Consider an arbitrary contribution vector ( g₁, . . . , g_i21, g_i11, . . . , g_n) of the other play- ers. Without loss of generality we relabel the players such that i 5 1 and 0 # g₂ # g₃ # ? ? ? # g_n. If player 1 chooses g₁ 5 0, his payoff is given by
   

n _b n
(A14) U₁( g₁ 5 0) 5 y 1 a _o g_j 2 _{n 2 1 o} g_j.
j52 j52
Note Žrst that if all other players choose g_j 5 0, too, then g₁ 5 0 is clearly optimal. Furthermore, player 1 will never choose g₁ . max g_j . Suppose that there is at least one player who chooses g_j .
0. If player 1 chooses g₁ . 0, g₁ [ [ g_k, g_k11], k [ 2, . . . , n , then his payoff is given by
U₁(g₁ . 0)
n _b1 n _a1 k
5 y 2 g₁ 1 ag₁ 1 a _o g_j 2 _{n 2 1 o} (g_j 2 g₁) 2 _{n 2 1 o}(g₁ 2 g_j)

j52
n
j5k11
_b1 n
j52
_b1 k

, y 2 g₁ 1 ag₁ 1 a _o g_j 2 _{n 2 1 o} (g_j 2 g₁) 1 _{n 2 1 o}(g₁ 2 g_j)

j52
n
j5k11
_b1 n
j52

5 y 2 g₁ 1 ag₁ 1 a _o g_j 2 _{n 2 1 o} g_j

j52
b₁
j52

1 _{n 2 1} (n 2 1)g₁
n


_b1 n

5 y 2 (1 2 a 2 b₁)g₁ 1 a _o g_j 2 _{n 2 1 o} g_j

n _b1
j52
n
j52

, y 1 a _o g_j 2 _{n 2 1 o} g_j 5 U₁(g₁ 5 0).
j52 j52
Hence, g_i 5 0, is indeed a dominant strategy for player i.

2. 
   It is clearly an equilibrium if all players contribute nothing because to unilaterally contribute more than zero reduces the monetary payoff and causes disadvantageous inequality. Suppose that there exists another equilibrium with positive contribution levels. Relabel players such that 0 # g₁ # g₂ # ? ? ? # g_n. By part (a) we know that all k players with 1 2 a . b_i must choose g_i 5 0. Therefore, 0 5 g₁ 5 . . . g_k. Consider player l . k who has the smallest positive contribution level; i.e., 0 5 g_l21 , g_l # g_l11 # ? ? ? # g_n. Player 1’s utility is given by
   

o

n 2 1 ^o
n _bl n

(A15) U_l( g_l) 5 y 2 g_l 1 ag_l 1 a g_j 2
j5l11 j5l11
( g_j 2 g_l)

_al l21 n _bl n
2 _{n 2 1 o} g_l 5 y 1 a _o g_j 2 _{n 2 1 o} g_j

j51


n 2 l
j5l11


l 2 1
j5l11

2 (1 2 a) g_l 1 b_{l n 2 1} g_l 2 a_{l n 2 1} g_l
n 2 l l 2 1
5 U_l(0) 2 (1 2 a) g_l 1 b_{l n 2 1} g_l 2 a_{l n 2 1} g_l,
where U_l(0) is the utility player 1 gets if he deviates and chooses
g_l 5 0. Since a_l $ b_l, l $ k 1 1, and b_l , 1, we have
n 2 l l 2 1
(A16) U_l( g_l) # U_l(0) 2 (1 2 a) g_l 1 b_{l n 2 1} g_l 2 b_{l n 2 1} g_l
n 2 2(k 1 1) 1 1

# U_l(0) 2 (1 2 a) g_l 1 b_l
n 2 1 ^gl

, U_l(0) 2 (1 2 a) g_l 1

n 2 2k 2 1
n 2 1 ^gl

Thus if

5 U_l(0) 2

(1 2 a)(n 2 1) 2 (n 2 2k 2 1)
n 2 1 ^gl.

(A17)
(1 2 a)(n 2 1) 2 (n 2 2k 2 1)
_{n 2 1} $ 0,

player l prefers to deviate from the equilibrium candidate and to

choose g_l 5 0. But this inequality is equivalent to (A18) (1 2 a)(n 2 1) $ n 2 2k 2 1

n 2 2k 2 1

Û a # 1 2


n 2 1

n 2 1 2 n 1 2k 1 1 2k
^{Û a #} n 2 1 ⁵ n 2 1
k a
^Û n 2 1 ^$ 2 ^,
which is the condition given in the proposition.

3. 
   Suppose that the conditions of the proposition are satis- Žed. We want to construct an equilibrium in which all k players with 1 2 a . b_i contribute nothing, while all other n 2 k players contribute g [ [0, y]. We only have to check that contributing g is indeed optimal for the contributing players. Consider some player j with 1 2 a , b_j. If he contributes g, his payoff is given by
   

(A19) U_j( g) 5 y 2 g 1 (n 2 k) ag 2 [a_j/(n 2 1)] kg.
It clearly does not pay to contribute more than g. So suppose that player j reduces his contribution level by D . 0. Then his payoff is
U_j( g 2 D) 5 y 2 g 1 D 1 (n 2 k) ag 2 Da
a_j b_j
2 _{n 2 1} k( g 2 D) 2 _{n 2 1} (n 2 k 2 1)D
a_j
5 y 2 g 2 (n 2 k) ag 2 _{n 2 1} kg
a_j b_j
1 D 1 2 a 1 _{n 2 1} k 2 _{n 2 1} (n 2 k 2 1)
a_j b_j
5 U_j( g) 1 D 1 2 a 1 _{n 2 1} k 2 _{n 2 1} (n 2 k 2 1) .
Thus, a deviation does not pay if and only if
a_j b_j
1 2 a 1 _{n 2 1} k 2 _{n 2 1} (n 2 k 2 1) # 0,
which is equivalent to
(A20) k/(n 2 1) # (a 1 b_j 2 1)/(a_j 1 b_j).

Thus, if this condition holds for all (n 2 k) players j with 1 2 a , b_j, then this is indeed an equilibrium. It remains to be shown that (a 1 b_j 2 1)/(a_j 1 b_j) # a/ 2. Note that a_j $ b_j implies that (a 1 b_j 2 1)/(a_j 1 b_j) # (a 1 b_j 2 1)/(2b_j). Furthermore,

a 1 b_j 2 1 2b_j