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Conflict and Competition over Multi-Issues

António Osório

Abstract

Real life disputes, negotiations and competitive situations involve multi-issue considerations in which the final outcome depends on the aggregated effort over several dimensions. We consider two allocation systems, the I-system, in which each issue is disputed and award independently, and the A-system, in which all issues are aggregate in a single prize award. In the A-system, we propose a contest success function that aggregates the individuals’ multi-issue efforts in a single outcome. Among other results, we found that the A-system tends to induce higher total effort than the I-system. The model is also able to reproduce a large set of strategic behaviors. For instance, under decreasing returns to effort, individuals maximize their payoffs by distributing effort over all issues, while under increasing returns to effort, individuals focus on a single issue. Hybrid equilibria, in which one individual focus in a single issue while the other individual diversifies effort over all issues, may also emerge when individuals hold different returns to effort. Strategic behavior is simultaneously influenced by the weight of each issue on the final outcome and by comparative advantages. Throughout the manuscript, we link our results with strategic behavior observed in electoral competition, i.e., “issue ownership”,“issue divergence/convergence”and“common value issues”. We expect that our findings will help researchers and practitioners to better understand the process of endogenous selection of issues in competitive contexts and to provide guidance in the implementation of the optimal allocation mechanism.

JEL Classification: C72; D72; D74; D81

A Appendix

In the main text of this paper, in order to simplify the analysis and to provide better intuition, we have focused in the two individuals and two issues case. In this Appendix, we briefly consider the general model with an arbitrary number of issues. In this context, we should introduce extra notation to distinguish between individuals and issues.

A.1 The general model

Let xij(s) denotes the effort of individual i=1,...,n in issue j=1,...,m under the system s{I,A}. Let λij>0 denote the ability of individual i=1,...,n in issue j=1,...,m. Recall that vj denotes the prize of issue j=1,...,m and v=j=1mvj denotes the aggregated prize.

In this context, under the multi-issue A-system, the probability that individual i=1,...,n wins by providing effort in all (or some) issues j=1,...,m, is given by:

(15)pi(A)=(j=1mλijxijαij(A))/(k=1nj=1mλkjxkjαij(A)),

with pi(A)=j=1mλij/n if xij(A)=0 for all i=1,...,n and j=1,...,m. Expression eq. (15) is the generalized version of expression eq. (6).

Under the multi-issue A-system, each individual i=1,...,n simultaneously chooses a profile of efforts, xij(A)0 for j=1,...,m, that maximizes the expected payoff net of the cost of effort, which is given by:

(16)πi(A)=pi(A)vj=1mxij(A),

subject to the participation constraint πi(A)0. Expression eq. (16) is the generalized version of expression eq. (7).

In order to have a sufficiently tractable A-system model, we do the following simplifying assumptions: αij=α and λij(0,1) with i=1nλij=1 for all i=1,...,n and j=1,...,m and n=2.

A.2 General results and proofs

In the case 0<α<1, the solution to the general problem eq. (16) is framed in the following result.

Proposition 9.

For0<α<1,the individualsi=1,2equilibrium efforts in each issuej=1,...,mare given by:

(17)xij(A)=αϕγijv,
whereϕ=siα1siα1/(siα1+siα1)2andγij=λij1/(1α)/siwithsi=j=1mλij1/(1α)fori=1,2andj=1,...,m.

Proof

[Proof of Proposition 10 (and Propositions 3 and 4)] The proof of Proposition 3 is just a particular case of this proof. From the problem 16, the associated set of n×m first order conditions is given by:

(18)αijλijxijαij1(A)(k=1mλikxikαik(A))v/(l=1nk=1mλlkxlkαlk(A))2=1,

for i=1,...,n and j=1,...,m. Under αij=α for all i=1,...,n and j=1,...,m and n=2, after some algebra on the system of 2m first order conditions we obtain that:

(19)λi1xi1α1(A)=λi2xi2α1(A)=...=λimximα1(A),

for i=1,2, which corresponds to 2m2 independent equations. Consequently, we need two additional equations in order to solve the system. One such equation is obtained by noticing that since individuals are not budget constrained, and the total prize and the unit cost of effort are the same, in equilibrium we must have: j=1mxij=j=1mxij. This equality relation can be easily shown by rewriting the system of first order conditions eq. (18) in the form:

xij(A)=αλijxijα(A)(k=1mλikxikα(A))v/(l=1nk=1mλlkxlkα(A))2,

and, then summing over all issues j=1,...,m for each i=1,2. The last independent equation is obtained by any of the first order conditions in the system eq. (18). After some algebra, on this system of 2m equations, we obtain the unique solution given by expression eq. (17), from where expression eq. (8) is a particular case. Now, we need to verify under which conditions such solution corresponds to a Nash equilibrium (Pérez-Castrillo and Verdier 1992; Szidarovszky and Okuguchi 1997). Since the second derivative of each first order condition is strictly negative, i.e.:

(1+α)λijxijα(A)+(1α)(l=1nk=1mλlkxlkα(A)λijxijα(A))(l=1nk=1mλlkxlkα(A))3/(αλijxijα1(A)(k=1mλikxikα(A)))v<0,

then, πi(A) is strictly concave for α[0,1). In addition, since the vector (xi1(A),...,xim(A))R+m is defined on a convex space, then the first order condition is simultaneously necessary and sufficient for a maximum. Finally, since the effort in expression eq. (17) is positive we are left to show that in equilibrium πi(A)0. Note that the winning probability of individual i is given by: pi(A)=siα1/(siα1+siα1), and the individual i total effort is given by: αsiα+1siα+1v/(siαsi+sisiα)2 (i.e., T(A)/2). Consequently, after replacing these expressions into πi(A)=pi(A)vT(A)/2 we obtain the expected payoff:

πi(A)=(siα1/(siα1+siα1)αsiα1siα1/(siα1+siα1)2)v

for i=1,2. Since effort does not create value, the minimum payoff is obtained when effort is maximal, i.e., at si=si (see Proposition 4). Then, it is easy to show that participation is guaranteed for α<2, which is always true for α<1.

The proof of Proposition 4 is just a particular case of the following more general proof. Since in equilibrium each individual provides the same aggregate effort, i.e., j=1mxij=j=1mxij, and T(A)=2j=1mxij(A). Then, after some algebra we obtain 2αsiα1siα1v/(siα1+siα1)2 which is equivalent to expression eq. (9). In order to study whether the aggregate effort T(A) increases with the ability λij simply differentiate T(A) with respect to λij to obtain that:

T(A)/λij=2αλijα/(1α)ϕ(siα1siα1)v/(si(siα1+siα1)),

is strictly positive if si<si because α<1 for all i and j, and the opposite otherwise. Since T(A) is twice continuously differentiable, the maximal effort occurs at si=si. At this point, the Hermitian matrix of i is negative-semidefinite, i.e., with non-positive eigenvalues, which are either 0 or αvj=1mλij2a/(1α)/(4si2).

In the case α1, individuals place all effort in a single issue - the one in which they have highest ability. Let j[i]=argmaxj{λi1,...,λim} denotes the issue in which individual i has highest ability.

Proposition 10.

For

(20)1αmin{(λ1j[1]+λ2j[2])/λ1j[1],(λ1j[1]+λ2j[2])/λ2j[2]},

the individuals equilibrium efforts in each issuej=1,2are given by:

(21)xij[i](A)=αλ1j[1]λ2j[2]v/(λ1j[1]+λ2j[2])2,
wherej[i]=argmaxj{λi1,...,λim},andxij(A)=0forjj[i].

Proof of Proposition 11 (and Proposition 5). The proof of Proposition 5 is just a particular case of this proof. Note that for α1, in order for the sequence of equalities eq. (19) obtained from the first order condition eq. (18) to be satisfied, the issues with larger ability λij must receive lower effort. However, such behavior cannot be an equilibrium because since α1 the effort output j=1mλijxijα(A) is convex in xij(A), and the marginal utility from effort increases. Consequently, optimal behavior requires that each individual places effort in a single issue, i.e., the one with largest abilities (see the discussion in Section 5.1). Let j[i]=argmaxj{λi1,...,λim} denotes the issue in which individual i has the highest ability. Under αij=α for all i=1,...,n and j=1,...,m and n=2, individual i=1,2 chooses xij[i](A)>0 to maximize eq. (16) and set xij(A)=0 for all jj[i]. The associated system of two first order conditions is given by:

αλij[i]xij[i]α1(A)λij[i]xij[i]α(A)v/(λij[i]xij[i]α(A)+λij[i]xij[i]α(A))2=1,

for i=1,2. The solution returns the unique, symmetric and strictly positive equilibrium effort given in eq. (21), from where expression eq. (11) is a particular case. The expected payoff is non-negative if α(λij[i]+λij[i])/λij[i] for i=1,2.[13] The individual i=1,2 second order condition for a maximum is given by:

(λij[i]+λij[i])((1+α)λij[i](α1)λij[i])/(αλij[i]λij[i]v),

which is negative if α<((1+α)λij[i]+λij[i])/λij[i]. This condition is implied by the more restrictive non-negative expected payoff condition α(λij[i]+λij[i])/λij[i] for i=1,2. This inequality corresponds to the existence condition eq. (20), from where the condition eq. (10) is a particular case.

Acknowledgements

Financial support from the Barcelona Graduate School of Economics and Univ. Rovira i Virgili are gratefully acknowledged. I would like to thank Matthias Dahm, Sabine Flamand, David Pérez-Castrillo, Ricardo Ribeiro and Santiago Sanchez-Pages as well as several seminars and congresses participants for helpful comments and discussions. All remaining errors are mine.

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Published Online: 2018-01-20

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