aa r X i v : . [ ec on . T H ] J un Extractive Contest Design ∗ Tomohiko Kawamori † Abstract
We consider contest success functions (CSFs) that extract contestants’ values ofthe prize. In the case in which the values are observable to the contest designer, inthe more-than-two-contestant or common-value subcase, we present a CSF extractivein any equilibrium; in the other subcase, we present a CSF extractive in some equi-librium, but there exists no CSF extractive in any equilibrium. In the case in whichthe values are not observable, there exists no CSF extractive in some equilibrium.In the case in which the values are observable and common, we present extractivea CSF extractive in any equilibrium; we present a class of CSFs extractive in someequilibrium, and this class can control the number of active contestants.
Keywords: contest success function; extraction of values; observability of valuesaggregate effort equivalence across equilibria
JEL classification codes:
C72; D72 ∗ The author is grateful to Kazuo Yamaguchi for his valuable comments. This work was supported byJSPS KAKENHI Grant Number JP19K01563. † Faculty of Economics, Meijo University, 1-501 Shiogamaguchi, Tempaku-ku, Nagoya 468-8502, Japan. [email protected] Introduction
In this paper, we define extractiveness of contest success functions (CSFs). In a con-test, which is formalized by Tullock (1980), contestants make an effort, and a winnerof a prize is determined according to a probability distribution, which depends onefforts. A function that maps from effort tuples to winning probability distributionsis called a contest success function (CSF). We consider the design of CSFs that ex-tract values of the prize through contestants’ efforts. We say that a CSF is extractive if under this CSF, there exists a Nash equilibrium such that the aggregate effort isequal to the maximum value (the maximum of contestants’ values of the prize). Wealso say that a CSF is strictly extractive if this CSF is extractive, and under thisCSF, in every Nash equilibrium, the aggregate effort is equal to the maximum value.Focusing on observability of contestants’ values of the prize, we present CSFssatisfying extractiveness. Firstly, we consider the case in which contestants’ values areobservable for the contest designer. In the subcase in which the number of players isgreater than 2, or players have a common value, we present a strictly extractive CSF.In the subcase in which the number of players is 2, and the players have heterogeneousvalues, we present an extractive CSF, and show that there does not exist a strictlyextractive CSF, even though the contest designer can fully use the information of thevalues. The aggregate effort equivalence between Nash equilibria holds in the formersubcase but does not in the latter subcase. Secondly, we consider the case in whichthe values are unobservable for the contest designer. We show that there does notexist a CSF extractive under every value tuple. We consider the subcase in whichplayers have a common value unobservable for the contest designer. We examinethe CSF such that the winning probabilities are proportional to the aa − th powers ofefforts, where a is an integer in [2 , n ] ( n is the number of contestants). We show thatunder every common value, this CSF is extractive, this CSF with a = 2 is strictlyextractive, and this CSF with a > a = 2 but does not under a > he other contestant wins otherwise (Subsection 3.1). In the 2-contestant case (the n -contestant case, resp.), Nti (2004) (Franke et al. (2018), resp.) presented a CSFsuch that a contestant with the maximum value wins if their effort is greater than orequal to their value, and a contestant with the second highest value, which may beequal to the maximum value, wins otherwise (Proposition 2 (Proposition 4.7, resp.)).These CSFs are extractive. However, they are not strictly extractive, because thereexists a Nash equilibrium such that every contestant’s effort is zero. Meanwhile, wepresent strictly extractive CSFs in the 3-or-more-contestant or common-value case.Several papers have presented CSFs that are reduced to CSFs extractive in theunobservable-common-value case. In the 2-contestant case, Nti (2004) (Epstein et al.(2013); Ewerhart (2017), resp.) presented a CSF that maximizes the aggregate effortin a class of CSFs (Section 4 (Subsection 4.2; Proposition 6, resp.)). In the n -contestant common-value case, Michaels (1988) did so (Subsection 2.1). Each CSFin Nti (2004), Epstein et al. (2013) and Ewerhart (2017) in the common-value case(the CSF in Michaels (1988), resp.) is the CSF such that the winning probabilities areproportional to the 2nd ( nn − th, resp.) powers of efforts, and it is in the unobservable-common-value case because it does not depend on the common value. The CSF usingthe 2nd power is strictly extractive in the 2-contestant case. We show that this CSFis also strictly extractive in the n -contestant case. The CSF using the nn − th poweris extractive. We show that this CSF is not strictly extractive. We also show thatin the n -contestant case, CSFs such that the winning probabilities are proportionalto the aa − th powers of efforts (2 ≤ a ≤ n ) is extractive. Under this CSF, any setof a contestants is the set of active contestants (i.e., contestants whose efforts arepositive) in some Nash equilibrium. Thus, by selecting a and suggesting a Nashequilibrium, the contest designer can choose an arbitrary set of two-or-more activecontestants. If the contest designer has a preference such that contestants’ efforts arecomplementary with each other, they may most prefer the CSF with a = n .Several papers have shown extraction of values in mixed-strategy Nash equilibria.Hillman and Riley (1989) (Baye et al. (1996), resp.) showed that in the all-pay auc- Nti (2004) suggested that if the CSF is modified as the threshold of effort is slightly lowered, the Nashequilibrium such that every contestant’s effort is zero is removed. However, under this modified CSF, in aunique Nash equilibrium, the aggregate effort is slightly smaller than the maximum value. ion, if the highest two values are equal, the expected aggregate effort is equal to themaximum value in any mixed-strategy Nash equilibrium (the second last paragraphin Section 3 (Theorem 1, resp.)). Alcalde and Dahm (2010) showed that under CSFssatisfying some conditions, there exists a mixed-strategy Nash equilibrium such thatif the highest two values are equal, the expected aggregate effort is equal to themaximum value (Theorem 3.2). We show extraction of values in pure-strategy Nashequilibria.Several papers have considered maximization of the aggregate effort in a classof CSFs. CSFs using the following devices have been examined: concave technolo-gies and power technologies in the lottery contest (Nti (2004)); power technologies inthe lottery contest (Michaels (1988)); biases multiplying efforts in the lottery contest,(Franke et al. (2013)); biases multiplying efforts with power technologies in the lot-tery contest and biases multiplying efforts the all-pay auction auction (Epstein et al.(2013)); biases multiplying efforts in the lottery contest and the all-pay auction(Franke et al. (2014a)); head starts added to efforts in the lottery contest and all-payauction (Franke et al. (2014b)); biases multiplying efforts given a power technologyin the lottery contest (Ewerhart (2017)); biases multiplying efforts and head startsadded to efforts in the lottery contest and the all-pay auction (Franke et al. (2018)).Fang (2002) compared the simple lottery contest and the simple all-pay auction. Ow-ing to restriction on forms of CSFs, the maximized aggregate effort is not equal tothe maximum value except for the above-mentioned results. In our paper, becauseno restriction is imposed on forms of CSFs, the extraction of values is achieved.Several papers have considered the aggregate effort under asymmetric informa-tion. In Kirkegaard (2012), P´erez-Castrillo and Wettstein (2016), Matros and Possajennikov(2016), Drugov and Ryvkin (2017) and Olszewski and Siegel (2020), the values orthe productivities of efforts are private information. In our paper, in the observable-value case, the contest designer and contestants know all contestants’ values; in theunobservable case, the contest designer knows none of the contestants’ values, butcontestants know all contestants’ values. A concave (power, resp.) technology is a concave (power, resp.) function that transforms efforts. Thewinning probabilities are determined according to the transformed efforts. Epstein et al. (2011) considered the same class of CSF but a different objective of the contest designer,which is the weighted sum of the aggregate effort and the welfare. s well as the existing papers on maximization of the aggregate effort, our paperprovides CSFs a positive foundation. , In rent-seeking interpretation, the contestdesigner (politician) wants to maximize contestants’ efforts, because if the effortsare political contributions, they obtain monetary benefit from the efforts, and if theefforts are political lobbying, they flaunt their power by the efforts. Thus, theyshould determine the CSF as it maximizes the efforts. Hence, they should choose anextractive CSF if it exists.The remainder of this paper is organized as follows. Section 2 describes the model.Section 3 presents our results. Section 4 discusses questions for future research. Theproofs of all propositions are provided in Appendix.
For any sets X , Y and I , any f : X → Y I and any x ∈ X and i ∈ I , let f i ( x ) be thevalue f ( x ) for i .Let N be a finite set such that | N | ≥ N is the set of contestants. Let n := | N | . Let X := R N ≥ : X is the set of tuples of contestants’ efforts. Let∆ := (cid:8) p ∈ R N ≥ | P i ∈ N p i = 1 (cid:9) : the set of tuples of contestants’ success probabili-ties (for any p ∈ ∆ and any i ∈ N , p i is the probability of contestant i ’s winning).Let F := { f | f : X → ∆ } : F is the set of contest success functions (CSFs). Let V := R N> : the set of tuples of contestants’ values of the prize. For any f ∈ F and any v ∈ V , let u fv : X → R N such that for any x ∈ X and i ∈ N , u fvi ( x ) = f i ( x ) v i − x i : u fvi ( x ) is contestant i ’s utility from effort tuple x ( f i ( x ) v i is the expected value thatthey obtain, and x i is the cost of their effort).For any f ∈ F and any v ∈ V , (cid:0) N, X, u fv (cid:1) is a strategic form game: N is the setof players, X is the set of strategy tuples, and u fv is the function that maps eachstrategy tuple to the payoff tuple by it. For any f ∈ F and any v ∈ V , let E fv bethe set of Nash equilibria in (cid:0) N, X, u fv (cid:1) .Let ˆ V := { v ∈ V | ∀ i, j ∈ N ( v i = v j ) } : the set of value tuples such that all Jia et al. (2013) referred to this as the optimally-derived foundation , which is one of four types offoundations. Some papers have provided CSFs axiomatic foundations (e.g., Skaperdas (1996) and Clark and Riis(1998)). ontestants have a common value. For any v ∈ V , let m v := max i ∈ N v i and M v := arg max i ∈ N v i : m v is the maximum of contestants’ values of the prize, and M v is the set of contestants who have the maximum value. First, we show that the maximum value of the prize is an upper bound of the equi-librium aggregate effort. Subsequently, we seek CSFs under which the equilibriumaggregate effort is equal to the maximum value in the case in which the values areobservable to the contest designer and the case in which they are unobservable, re-spectively. Note that the values are observable to contestants.
For any CSF and any value tuple, in any Nash equilibrium, the aggregate effort isless than or equal to the maximum value.
Proposition 1.
Let f ∈ F and v ∈ V . Let x ∗ ∈ E fv . Then, P i ∈ N x ∗ i ≤ m v . We say that a CSF is extractive if in some Nash equilibrium, the aggregate effortis equal to the maximum value. We say that a CSF is strictly extractive if it isextractive, and in any Nash equilibrium, the aggregate effort is equal to the maximumvalue.
Definition 1.
Let f ∈ F and v ∈ V . f is extractive under v if there exists x ∗ ∈ E fv such that P i ∈ N x ∗ i = m v . f is strictly extractive under v if f is extractive under v and for all x ∗ ∈ E fv , P i ∈ N x ∗ i = m v . We consider the case in which the contest designer can observe contestants’ valuesand thus, can design CSFs dependent on the values.Under any value tuple, if the number of contestants is greater than or equal to 3,or all contestants have a common value, some CSF is strictly extractive; otherwise,some CSF is extractive, but any CSF is not strictly extractive. Thus, the aggregate ffort equivalence across Nash equilibria holds in the former case but does not in thelatter case. Proposition 2.
Let v ∈ V . Suppose that n ≥ . Let f ∈ F such that for somedistinct i, j, k ∈ N such that i ∈ M v , for any x ∈ X , (i) if x i = m v , then f i ( x ) = 1 ,(ii) if x i = m v and x j > , then f j ( x ) = 1 , and (iii) if x i = m v and x j = 0 , then f k ( x ) = 1 . Then, f is strictly extractive under v . Proposition 3.
Let v ∈ V . Suppose that n = 2 and v ∈ ˆ V . Let f ∈ F suchthat for any i ∈ N , any j ∈ N \ { i } and any x ∈ X , if x k > for any k ∈ N , f i ( x ) = xi = mv − xj = mv +12 ; otherwise, f i ( x ) = xi> − xj> +12 . Then, f is strictlyextractive under v . Proposition 4.
Let v ∈ V . Suppose that n = 2 and v / ∈ ˆ V . (i) Let f ∈ F such thatfor some i ∈ M v , for any x ∈ X , f i ( x ) = x i = m v . Then, f is extractive under v .(ii) Let f ∈ F . f is not strictly extractive under v . The above CSFs make the contestant with the maximum value win with certainty(or with probability ) if their effort is equal to the maximum value (or of themaximum value), in order that the aggregate effort is equal to the maximum value.In the 3-or-more contestant or common-value case, the above CSFs give contestantsan incentive to make a positive effort, to exclude the Nash equilibrium such thatevery contestant’s effort is zero. In the other case, under any extractive CSF, thereexists a Nash equilibrium such that every contestant’s effort is zero. We consider the case in which the contest designer cannot observe contestants’ values,and design CSFs independent of the values.For any CSF f , for some value tuple v , f is not extractive under v . Proposition 5.
Let f ∈ F . Then, for some v ∈ V , f is not extractive under v . We focus on the common-value case and define candidates of CSFs (strictly)extractive under any common value. Let f ∈ F such that for some a ∈ N such that ≤ a ≤ n , for any i ∈ N and any x ∈ X , f i ( x ) = x aa − i P j ∈ N x aa − j if ∃ j ∈ N ( x j > n otherwise .f is extractive under any common value; f with a = 2 is strictly extractive underany common value; under any common value, f with a ≥ a = 2 butdoes not under a ≥ Proposition 6.
Let v ∈ ˆ V . Then, f is extractive under v .Remark . As seen in the proof, for any x ∈ X , x ∈ E fv if for some A ∈ N suchthat | A | = a , for any i ∈ A , x i = m v a and for any i ∈ N \ A , x i = 0. Proposition 7.
Suppose that a = 2 . Let v ∈ ˆ V . Then, f is strictly extractive under v .Remark . As seen in the proof, for any x ∈ X , x ∈ E fv if and only if for some A ∈ N such that | A | = 2, for any i ∈ A , x i = m v and for any i ∈ N \ A , x i = 0. Proposition 8.
Suppose that ≤ a ≤ n . Let v ∈ ˆ V . Then, f is not strictlyextractive under v .Remark . As seen in the proof, for any x ∈ X , x ∈ E fv if for some A ∈ N suchthat | A | = a −
1, for any i ∈ A , x i = m v a ( a − a − and for any i ∈ N \ A , x i = 0. Theaggregate effort in the strategy tuples is ( a − va ( a − a − = va ( a − a − < v . Let a ∈ N N such that 3 ≤ a n ≤ n and lim n →∞ a n = ∞ . Then, lim n →∞ va n ( a n − a n − = v .Under f , each contestant’s effort x is transformed to x aa − , and the winningprobabilities are determined proportionally to the transformed efforts. As a is larger,the elasticity of transformed effort x aa − to effort x , dx aa − /dxx aa − /x = aa − , is smaller; thus,in the Nash equilibrium such that the aggregate effort is equal to the maximumvalue, each active contestant’s effort is smaller, but the number of active contestantsis larger. Discussion
In the unobservable-value case, unless a common value is assumed, there does notexist extractive CSF. In such case, it is necessary to derive CSFs that maximize theexpectation of the aggregate effort under some belief on value tuples. For example,this problem is formalized asmax ( f,x ) ∈ F × X V Z v ∈ V X i ∈ N x i ( v ) dP ( v )s.t. ∀ v ∈ V (cid:16) x ( v ) ∈ E fv (cid:17) ∧ x is measurable , where P is a cumulative distribution function on V (the designer’s belief on valuetuples). This problem remains for future research. ppendix Lemma 1.
Let f ∈ F , v ∈ V , x ∗ ∈ E fv and i ∈ N . Then, u fvi ( x ∗ ) ≥ , and f i ( x ∗ ) v i ≥ x ∗ i .Proof. Because x ∗ ∈ E fv , u fvi ( x ∗ ) ≥ u fvi (cid:0) , x ∗− i (cid:1) = f i (cid:0) , x ∗− i (cid:1) v i ≥
0. Thus, f i ( x ∗ ) v i ≥ x ∗ i . Lemma 2.
Let v ∈ R > and a, b ∈ N such that ≤ b ≤ a . Let x ∗ := va ( b − b ( a − . Let u : R ≥ → R such that for any x ∈ R ≥ , u ( x ) = x aa − x aa − +( b − x ∗ ) aa − v − x . Then, x ∗ ∈ arg max x ∈ R ≥ u ( x ) .Proof. Let φ : R ≥ → R such that for any x ∈ R ≥ , φ ( x ) = − (cid:16) (2 b −
1) ( x ∗ ) aa − + x aa − (cid:17) a − X i =0 ( x ∗ ) ia − x a − − ia − + b ( x ∗ ) a − a − . For any x ∈ R ≥ , du ( x ) dx = φ ( x ) (cid:16) x a − − ( x ∗ ) a − (cid:17)(cid:16) x aa − + ( b −
1) ( x ∗ ) aa − (cid:17) . Note that φ (0) = ( b − ( x ∗ ) a − a − >
0, and φ ( x ∗ ) = − b (2 a − b ) ( x ∗ ) a − a − <
0. Then,by the intermediate value theorem, there exists ¯ x ∈ (0 , x ∗ ) such that φ (¯ x ) = 0. Let x ∈ R ≥ . Because φ is strictly decreasing, φ ( x ) R x ⋚ ¯ x . Thus, du ( x ) dx ≤ x ≤ ¯ x ≥ x < x ≤ x ∗ < x > x ∗ . Note that u (0) = 0 ≤ u ( x ∗ ). Then, for any x ∈ R ≥ , u ( x ∗ ) ≥ u ( x ). Proof of Proposition 1
By Lemma 1, for any i ∈ N , x ∗ i ≤ v i f i ( x ∗ ) ≤ m v f i ( x ∗ ).Hence, P i ∈ N x ∗ i ≤ m v P i ∈ N f i ( x ∗ ) = m v . Proof of Proposition 2
Let x ∗ ∈ X such that x ∗ i = m v and for any l ∈ N \ { i } , x ∗ l = 0. Then, P l ∈ N x ∗ l = m v . It suffices to show that E fv = { x ∗ i } . or any x i ∈ R ≥ \ { x ∗ i } , u fvi ( x ∗ ) = 0 ≥ − x i = u fvi (cid:0) x i , x ∗− i (cid:1) . For any l ∈ N \ { i } and any x l ∈ R ≥ \ { x ∗ l } , u fvl ( x ∗ ) = 0 ≥ − x l = u fvl (cid:0) x l , x ∗− l (cid:1) . Thus, x ∗ ∈ E fv .Let x ∈ X \ { x ∗ } . If x i = m v , then for some l ∈ N \ { i } , x l >
0, and u fvl ( x ) = − x l < u fvl (0 , x − l ). If x i = m v and x j >
0, then u fvj ( x ) = v j − x j < v j − x j = u fvj (cid:0) x j , x − j (cid:1) . If x i = m v and x j = 0, then u fvj ( x ) = 0 < v j = u fvj (cid:0) v j , x − j (cid:1) . Thus, x / ∈ E fv . Proof of Proposition 3
Let x ∗ ∈ X such that for any i ∈ N , x ∗ i = m v . Then, P i ∈ N x ∗ i = m v . It suffices to show that E fv = { x ∗ } .For any i ∈ N and any x i ∈ R ≥ \ { x ∗ i } , u fvi ( x ∗ ) = 0 ≥ − x i = u fvi (cid:0) x i , x ∗− i (cid:1) .Thus, x ∗ ∈ E fv .Let x ∈ X \ { x ∗ } . If for any i ∈ N , x i >
0, for some i, j ∈ N , x i = m v and x j = m v , then u fvj ( x ) = − x j < u fj (0 , x − j ). If for any i ∈ N , x i > x i = m v , then for some i ∈ N , u fvi ( x ) = v i − x i < v i = u fvi (cid:0) m v , x − i (cid:1) . If for some i, j ∈ N , x i > x j = 0, then u fvi ( x ) = v i − x i < v i − x i = u fvi (cid:0) x i , x − i (cid:1) . If forany i ∈ N , x i = 0, then for some i ∈ N , u fvi ( x ) = v i < v i = u fvi (cid:0) v i , x − i (cid:1) . Thus, x / ∈ E fv . Proof of Proposition 4 (i) Let j ∈ N \ { i } . Let x ∗ ∈ X such that x ∗ i = m v ,and x ∗ j = 0. For any x i ∈ R ≥ \ { x ∗ i } , u fvi ( x ∗ ) = 0 ≥ − x i = u fvi (cid:0) x i , x ∗− i (cid:1) . Forany x j ∈ R ≥ \ n x ∗ j o , u fvj ( x ∗ ) = 0 ≥ − x j = u fvi (cid:16) x j , x ∗− j (cid:17) . Thus, x ∗ ∈ E fv . P k ∈ N x ∗ k = m v .(ii) Let i, j ∈ N such that v i > v j . Suppose that f is extractive. Then, there exists x ∗ ∈ E fv such that x ∗ i + x ∗ j = m v . By Lemma 1, m v = x ∗ i + x ∗ j ≤ f i ( x ∗ ) m v + f j ( x ∗ ) v j .Thus, f j ( x ∗ ) ( m v − v j ) ≤
0. Hence, f j ( x ∗ ) = 0. Thus, by Lemma 1, x ∗ j = 0. Thus, x ∗ i = m v . Hence, 0 = u fvi ( x ∗ ) ≥ u fvi (cid:0) , x ∗− i (cid:1) = f i (cid:0) , x ∗− i (cid:1) v i . Thus, f i (cid:0) , x ∗− i (cid:1) = 0.Hence, f i (0 ,
0) = 0. Let y ∗ ∈ X such that y ∗ i = y ∗ j = 0. Because x ∗ ∈ E fv and x ∗ j = y ∗ j , for any y i ∈ R ≥ , u fvi ( y ∗ ) = f i (0 , v i = 0 = u fvi ( x ∗ ) ≥ u fvi (cid:0) y i , x ∗− i (cid:1) = u fvi (cid:0) y i , y ∗− i (cid:1) ; for any y j ∈ R ≥ , u fvj ( y ∗ ) = f j (0 , v j = v j ≥ f j (cid:16) y j , y ∗− j (cid:17) v j − y j = u fvj (cid:16) y j , y ∗− j (cid:17) . Thus, y ∗ ∈ E fv . y ∗ i + y ∗ j = 0 = m v . Thus, f is not strictly extractiveunder v . roof of Proposition 5 Suppose that for any v ∈ V , there exists x ∗ ∈ E fv suchthat P i ∈ N x ∗ i = m v (assumption to a contradiction).Let v ∈ V such that for some i ∈ N , for any j ∈ N \ { i } , v i > v j . Let x ∗ ∈ E fv such that P j ∈ N x ∗ j = v i . By Lemma 1, v i = P j ∈ N x ∗ j ≤ P j ∈ N v j f j ( x ∗ ). Thus, f i ( x ∗ ) = 1, and for any j ∈ N \ { i } , f j ( x ∗ ) = 0. Hence, by Lemma 1, for any j ∈ N \ { i } , x ∗ j = 0, and x ∗ i = v i .Let v, w ∈ R N> such that for some i ∈ N , v i = 1 and w i = 2, and v j < v i and w j < w i for any j ∈ N \ { i } . Then, by the assumption to a contradiction, there exist x ∗ ∈ E fv and y ∗ ∈ E fw such that P j ∈ N x ∗ j = v i and P j ∈ N y ∗ j = w i . Thus, x ∗ i = 1and y ∗ i = 2, and for any j ∈ N \ { i } . x ∗ j = y ∗ j = 0. Moreover, f i ( x ∗ ) = f i ( y ∗ ) = 1.Thus, u fwi (cid:0) , y ∗− i (cid:1) = 2 f i (cid:0) , y ∗− i (cid:1) − f i ( x ∗ ) − > u fwi ( y ∗ ), whichcontradicts that y ∗ ∈ E fw . Proof of Proposition 6
Abuse v as the common value ( v = m v = v i for any i ∈ N ). Let x ∗ ∈ X such that for some A ∈ N such that | A | = a , for any i ∈ A , x ∗ i = va and for any j ∈ N \ A , x ∗ j = 0. Let i ∈ A . By Lemma 2 with b = a , for any x i ∈ R ≥ , u fvi ( x ∗ ) ≥ u fvi (cid:0) x i , x ∗− i (cid:1) . Let j ∈ N \ A . For any x j ∈ R > , u fvj (cid:0) x j , x ∗− j (cid:1) = x aa − j x aa − j + a (cid:0) va (cid:1) aa − v − x j < x aa − j x aa − j + ( a − (cid:0) va (cid:1) aa − v − x j = u fvi (cid:0) x j , x ∗− i (cid:1) ≤ u fvi ( x ∗ ) = 0 = u fvj ( x ∗ ) . Thus, x ∗ ∈ E fv . P i ∈ N x ∗ i = a · va = v . Proof of Proposition 7
By Proposition 6, f is extractive.Abuse v as the common value ( v = m v = v i for any i ∈ N ). Let x ∗ ∈ E fv . Let A := { i ∈ N | x ∗ i > } and α := | A | . If α = 0, for some i ∈ N , u fvi ( x ∗ ) = vn < (2 n − v n = u fvi (cid:0) v n , x ∗− i (cid:1) , which contradicts that x ∗ ∈ E fv . If α = 1, for some i ∈ A , u fvi ( x ∗ ) = v − x ∗ i < v − x ∗ i = u fvi (cid:16) x ∗ i , x ∗− i (cid:17) , which contradicts that x ∗ ∈ E fv . Thus, α ≥
2. Let i ∈ arg max j ∈ A x ∗ j . For any k ∈ A ,0 = ∂u fvk ∂x k ( x ∗ ) = 2 vx ∗ k (cid:16)P l ∈ A ( x ∗ l ) − ( x ∗ k ) (cid:17)(cid:16)P l ∈ A (cid:0) x ∗ l (cid:1) (cid:17) − . hus, for any k ∈ A \{ i } such that x ∗ k < x ∗ i , x ∗ i (cid:16)P l ∈ A ( x ∗ l ) − ( x ∗ i ) (cid:17) = x ∗ k (cid:16)P l ∈ A ( x ∗ l ) − ( x ∗ k ) (cid:17) ,( x ∗ i − x ∗ k ) (cid:16)P l ∈ A \{ i,k } ( x ∗ l ) − x ∗ i x ∗ k (cid:17) = 0, P l ∈ A \{ i,k } ( x ∗ l ) = x ∗ i x ∗ k . Suppose that forsome j ∈ A \ { i } , x ∗ j < x ∗ i (assumption to a contradiction). Then, P l ∈ A \{ i,j } ( x ∗ l ) = x ∗ i x ∗ j . For any k ∈ A \ { i, j } , ( x ∗ k ) ≤ P l ∈ A \{ i,j } ( x ∗ l ) = x ∗ i x ∗ j < ( x ∗ i ) , and thus, x ∗ k < x ∗ i . Hence, for any k ∈ A \ { i } , x ∗ k < x ∗ i . Thus, for any k ∈ A \ { i } , x ∗ i x ∗ j + (cid:16) x ∗ j (cid:17) = P l ∈ A \{ i } ( x ∗ l ) = x ∗ i x ∗ k + ( x ∗ k ) , (cid:16) x ∗ j − x ∗ k (cid:17) (cid:16) x ∗ j + x ∗ k + x ∗ i (cid:17) = 0,and x ∗ j = x ∗ k . Thus, ( α − (cid:16) x ∗ j (cid:17) = x ∗ i x ∗ j , and x ∗ i = ( α − x ∗ j . Hence, 0 = ∂u fvj ∂x j ( x ∗ ) = v ( α − α − ( x ∗ j ) (cid:16) ( α − α +3) ( x ∗ j ) (cid:17) − x ∗ j = v ( α − α − α − α +3) , and α ≥
3. Thus, u fvj ( x ∗ ) =( x ∗ j ) ( ( α − x ∗ j ) +( α − ( x ∗ j ) v − x ∗ j = − v ( α ( α − α − α +3) < u fvj (cid:16) , x ∗− j (cid:17) , which contradictsthat x ∗ ∈ E fv . Hence, for any j ∈ A , x ∗ j = x ∗ i . Thus, 0 = ∂u fvi ∂x i ( x ∗ ) = v ( α − α x ∗ i − x ∗ i = v ( α − α . Hence, v (2 − α ) α = u fvi ( x ∗ ) ≥ u fvi (cid:0) , x ∗− i (cid:1) = 0. Thus, α = 2. Hence, x ∗ i = v . Thus, for any j ∈ A , x ∗ j = v . Hence, P i ∈ N x ∗ i = 2 · v = v . Proof of Proposition 8
Abuse v as the common value ( v = m v = v i for any i ∈ N ). Let x ∗ be a strategy tuple such that for some A ∈ N such that | A | = a − i ∈ A , x ∗ i = va ( a − a − and for any j ∈ N \ A , x ∗ j = 0. Let i ∈ A . By Lemma 2with b = a −
1, for any x i ∈ R ≥ , u fvi ( x ∗ ) ≥ u fvi (cid:0) x i , x ∗− i (cid:1) . Let j ∈ N \ A . For any x j ∈ R > , u fvj (cid:0) x j , x ∗− j (cid:1) = x j (cid:18) x a − j v − x aa − j − ( a − (cid:16) va ( a − a − (cid:17) aa − (cid:19) x aa − j + ( a − (cid:16) va ( a − a − (cid:17) aa − d (cid:18) vx a − j − x aa − j (cid:19) dx j = aa − x − aa − j (cid:16) va − x j (cid:17) . Thus, for any x j ∈ R ≥ , u fvj (cid:0) x j , x ∗− j (cid:1) ≤ x j (cid:18)(cid:0) va (cid:1) a − v − (cid:0) va (cid:1) aa − − ( a − (cid:16) va ( a − a − (cid:17) aa − (cid:19) x aa − j + ( a − (cid:16) va ( a − a − (cid:17) aa − = − ( a − x j (cid:0) va (cid:1) aa − (cid:18)(cid:16) a ( a − a − (cid:17) aa − − (cid:19) x aa − j + ( a − (cid:16) va ( a − a − (cid:17) aa − ≤ u fvj ( x ∗ ) . hus, x ∗ ∈ E fv . P i ∈ N x ∗ i = ( a − · va ( a − a − = ( a − − a − v = v . Thus, f is not strictlyextractive under v . eferences J. Alcalde and M. Dahm. Rent seeking and rent dissipation: A neutrality result.
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