Efficiency in Contests Between Groups

Abstract

We study collective contests in which contestants lobby as groups. Our main observation is that group lobbying effort is efficiently produced in equilibrium, apparently contradicting conventional interpretations of results in the literature. This observation also leads to a substantial simplification of the analysis of equilibria through the application of a decomposition theorem, which shows that, under standard conditions a collective contest can be analyzed by reducing it to a conventional Tullock contest between groups in which each group has a group-specific cost function derived from the individual members’ cost functions. The theorem also allows us to transfer results from standard contest theory to collective contests. For example, it can be used to establish conditions for the existence and uniqueness of equilibria in collective contests. Similarly, we can apply well-known results in comparative statics to study how group composition affects the lobbying effectiveness of groups and to investigate rent dissipation.

Appendix

Proof of Lemma 1

Since Assumptions A1 and A2 imply that we are minimizing a continuous and strictly increasing function on a closed set, the minimum is achieved, though not necessarily uniquely. We will write $\widehat{\mathbf{x}}_{i}\left (y\right )$ for a minimizing x _i in (3).

To show that C _i is strictly increasing we first observe that, since f _i is increasing (by Assumption A1 ), $f_{i}\left (\widehat{\mathbf{x}}_{i}\left (y\right )\right ) = y$. Now suppose that $y^{0} \in \left (0,y\right )$, which entails $f_{i}\left (\widehat{\mathbf{x}}_{i}\left (y\right )\right ) > y^{0}$. Since f _i is continuous, there must be an x _i ⁰ satisfying $\mathbf{x}_{i}^{0} \leq \widehat{\mathbf{x}}_{i}\left (y\right )$ and $f_{i}\left (\mathbf{x}_{i}^{0}\right ) \geq y^{0}$. Since each c _{i ℓ} is strictly increasing, we have

$$\displaystyle{C_{i}\left (y^{0}\right ) \leq \sum _{\ell =1}^{N_{i} }c_{i\ell}\left (x_{i\ell}^{0}\right ) <\sum _{ \ell =1}^{N_{i} }c_{i\ell}\left (\widehat{x}_{i\ell}\left (y\right )\right ) = C_{i}\left (y\right ).}$$

To prove convexity, observe that, if $\mu \in \left (0,1\right )$, concavity of f _i implies that

$$\displaystyle{f_{i}\left (\mu \widehat{\mathbf{x}}_{i}\left (y\right ) +\mu ^{0}\widehat{\mathbf{x}}_{ i}\left (y^{0}\right )\right ) \geq \mu y +\mu ^{0}y^{0},}$$

where μ ⁰ = 1 −μ. Convexity of c _{i ℓ} implies that:

$$\displaystyle\begin{array}{rcl} C_{i}\left (\mu y +\mu ^{0}y^{0}\right )& \leq & \sum _{\ell =1}^{N_{i} }c_{i\ell}\left (\mu \widehat{x}_{i\ell}\left (y\right ) +\mu ^{0}\widehat{x}_{ i\ell}\left (y^{0}\right )\right ) {}\\ & \leq & \sum _{\ell=1}^{N_{i} }\mu c_{i\ell}\left (\widehat{x}_{i\ell}\left (y\right )\right ) +\mu ^{0}c_{ i\ell}\left (\widehat{x}_{i\ell}\left (y^{0}\right )\right ) {}\\ & =& \mu C_{i}\left (y\right ) +\mu ^{0}C_{ i}\left (y^{0}\right ). {}\\ \end{array}$$

The assertion that C _i satisfies $C_{i}\left (0\right ) = 0$ is an immediate consequence of Assumptions A1 and A2. ■

In the proof of Theorem 2, it will prove convenient to use the following lemma.

Lemma 9

Suppose Assumptions A1 and A2 hold and y > 0. If i = 1,…,n and $\widehat{\mathbf{x}}_{i}\left (y\right )$ achieves the minimum in ( 3 ) and k = 1,…,N _i , we have

$$\displaystyle{C_{i}^{{\prime}}\left (y\right ) \leq c_{ ik}^{{\prime}}\left (\widehat{x}_{ ik}\left (y\right )\right )/ \frac{\partial f_{i}} {\partial x_{ik}}\left (\widehat{\mathbf{x}}_{i}\left (y\right )\right ),}$$

with equality if $\widehat{x}_{ik}\left (y\right ) > 0$ . Furthermore,

$$\displaystyle{C_{i}^{{\prime}}\left (0\right ) \geq \min _{ k=1,\ldots,N_{i}}\left \{c_{ik}^{{\prime}}\left (0\right )/ \frac{\partial f_{i}} {\partial x_{ik}}\left (\mathbf{0}\right )\right \}.}$$

Proof

For any y > 0, the assumption that f _i is unbounded above means that there is some x _i ⁰ for which $f_{i}\left (\mathbf{x}_{i}^{0}\right ) > y$. This says that the Slater constraint qualification for the optimization problem in (3) holds (cf. Rockafellar 1972), which means that there is a Lagrange multiplier λ ≥ 0 such that the (necessary) first-order conditions for this optimization problem read:

$$\displaystyle{c_{ik}^{{\prime}}\left (\widehat{x}_{ ik}\left (y\right )\right ) \geq \lambda \frac{\partial f_{i}} {\partial x_{ik}}\left (\widehat{\mathbf{x}}_{i}\left (y\right )\right ),}$$

with equality if $\widehat{x}_{ik} > 0$. Furthermore, marginal group cost is the slope of the minimum function and is therefore equal to the Lagrange multiplier: $C_{i}^{{\prime}}\left (y\right ) =\lambda$. This observation completes the proof that the first displayed inequality is necessary and sufficient.

To complete the proof, note that Assumption A2 (specifically convexity and zero cost of zero input) applied to c _ik implies that $c_{ik}\left (x_{ik}\right ) \geq x_{ik}c_{ik}^{{\prime}}\left (0\right )$ for all x _ik  ≥ 0. Similarly, from A1 we have

$$\displaystyle{f_{i}\left (\mathbf{x}_{i}\right ) \leq \sum _{\ell=1}^{N_{i} }x_{i\ell}\frac{\partial f_{i}} {\partial x_{i\ell}}\left (\mathbf{0}\right ).}$$

Hence, for any y > 0, we have

$$\displaystyle\begin{array}{rcl} C_{i}\left (y\right )& \geq & \min _{\mathbf{x}_{i}}\left \{\sum _{\ell=1}^{N_{i} }x_{i\ell}c_{i\ell}^{{\prime}}\left (0\right ):\sum _{ \ell =1}^{N_{i} }x_{i\ell}\frac{\partial f_{i}} {\partial x_{i\ell}}\left (\mathbf{0}\right ) \geq y\right \} {}\\ & =& y\min _{k}\left \{c_{ik}^{{\prime}}\left (0\right )/ \frac{\partial f_{i}} {\partial x_{ik}}\left (\mathbf{0}\right )\right \}. {}\\ \end{array}$$

(Note that Assumption A1 implies $\left (\partial f_{i}/\partial x_{ik}\right )\left (\mathbf{0}\right ) > 0$.) Since $C_{i}\left (0\right ) = 0$, we can divide by y and let y ⟶ 0 to obtain the second inequality in the statement of the lemma. ■

Proof of Theorem 2

Throughout the proof, we use the fact that the convexity/concavity conditions from Assumptions A1 and A2 and Lemma 1 mean that first order conditions are necessary and sufficient to characterize best responses.

To prove sufficiency in the Separation Theorem, suppose $\widetilde{\mathbf{x}}$ satisfies requirements 1 and 2. Since the first of these says that $\left (\,f_{1}\left (\,\widetilde{\mathbf{x}}_{1}\right ),\ldots,f_{n}\left (\,\widetilde{\mathbf{x}}_{n}\right )\right )$ is an equilibrium of $\mathcal{D}$, the first-order conditions for best responses give

$$\displaystyle{C_{i}^{{\prime}}\left (\,\widetilde{y}_{ i}\right ) \geq \sum \limits _{j\neq i}\widetilde{y}_{j}\left [\sum \limits _{j=1}^{n}\widetilde{y}_{ j}\right ]^{-2},}$$

with equality if $\widetilde{y}_{i} > 0$, where $\widetilde{y}_{i} = f_{i}\left (\,\widetilde{\mathbf{x}}_{i}\right )$ for each i = 1, …, n. Requirement 2 says that $\widetilde{\mathbf{x}}_{i}$ achieves the minimum in the definition of $C_{i}\left (\,\widetilde{y}_{i}\right )$ and Lemma 1 implies that, for any k,

$$\displaystyle{C_{i}^{{\prime}}\left (\,\widetilde{y}_{ i}\right ) \leq c_{ik}^{{\prime}}\left (\,\widetilde{x}_{ ik}\right )/ \frac{\partial f_{i}} {\partial x_{ik}}\left (\,\widetilde{\mathbf{x}}_{i}\right )}$$

with equality if $\widetilde{x}_{ik} > 0$. Combining these inequalities gives

$$\displaystyle{ \sum \limits _{j\neq i}f_{j}\left (\,\widetilde{\mathbf{x}}_{j}\right )\left [\sum \limits _{j=1}^{n}f_{ j}\left (\,\widetilde{\mathbf{x}}_{j}\right )\right ]^{-2} \frac{\partial f_{i}} {\partial x_{ik}}\left (\,\widetilde{\mathbf{x}}_{i}\right ) \leq c_{ik}^{{\prime}}\left (\,\widetilde{x}_{ ik}\right ), }$$

(6)

with equality if $\widetilde{x}_{ik} > 0$. These are the first-order conditions for best responses in $\mathcal{C}$ and show that x ^∗ is a Nash equilibrium.

To prove necessity, let $\widetilde{\mathbf{x}}$ be a Nash equilibrium of $\mathcal{C}$ and write $\widetilde{y}_{i} = f_{i}\left (\,\widetilde{\mathbf{x}}_{i}\right )$ for i = 1, …, n. If $\widetilde{y}_{i} > 0$, the first order conditions for best responses for members of group i in $\mathcal{C}$ are (6) and

$$\displaystyle{\lambda =\sum \limits _{j\neq i}f_{j}\left (\,\widetilde{\mathbf{x}}_{j}\right )\left [\sum \limits _{j=1}^{n}f_{ j}\left (\,\widetilde{\mathbf{x}}_{j}\right )\right ]^{-2},}$$

these can be expressed as

$$\displaystyle{c_{ik}^{{\prime}}\left (\,\widetilde{x}_{ ik}\left (y\right )\right ) \geq \lambda \frac{\partial f_{i}} {\partial x_{ik}}\left (\,\widetilde{\mathbf{x}}_{i}\left (y\right )\right ),}$$

with equality if $\widehat{x}_{ik} > 0$. Since λ ≥ 0, these are the first order conditions for the minimization problem in (3). These conditions are necessary and sufficient by Assumptions A1 and A2, which means $\widetilde{\mathbf{x}}_{i}$ achieves the minimum in the definition of $C_{i}\left (\,f_{i}\left (\,\widetilde{\mathbf{x}}_{i}\right )\right )$. It follows from Lemma 1 that

$$\displaystyle{C_{i}^{{\prime}}\left (\,\widetilde{y}_{ i}\right ) = c_{ik}^{{\prime}}\left (\,\widetilde{x}_{ i\ell}\right )/ \frac{\partial f_{i}} {\partial x_{ik}}\left (\,\widetilde{\mathbf{x}}_{i}\right ).}$$

Combining these observations, we get

$$\displaystyle{C_{i}^{{\prime}}\left (\,\widetilde{y}_{ i}\right ) =\sum \limits _{j\neq i}\widetilde{y}_{j}\left [\sum \limits _{j=1}^{n}\widetilde{y}_{ j}\right ]^{-2}.}$$

These are the first order conditions for $\widetilde{y}_{i}$ being a best response in $\mathcal{D}$.

In the case $\widetilde{y}_{i} = 0$, we must have $\widetilde{\mathbf{x}}_{i} = \mathbf{0}$ (which achieves the minimum in (3)) and the first-order conditions imply that, for all k,

$$\displaystyle{\sum \limits _{j\neq i}f_{j}\left (\,\widetilde{\mathbf{x}}_{j}\right )\left [\sum \limits _{j=1}^{n}f_{ j}\left (\,\widetilde{\mathbf{x}}_{j}\right )\right ]^{-2} \frac{\partial f_{i}} {\partial x_{ik}}\left (\mathbf{0}\right ) \leq c_{ik}^{{\prime}}\left (0\right ).}$$

Combined with the second inequality in Lemma 1, we deduce

$$\displaystyle{\sum \limits _{j\neq i}\widetilde{y}_{j}\left [\sum \limits _{j=1}^{n}\widetilde{y}_{ j}\right ]^{-2} \leq C_{ i}^{{\prime}}\left (0\right ),}$$

which implies that 0 is a best response by group i in $\mathcal{D}$. We have demonstrated that $\widetilde{\mathbf{y}}$ is an equilibrium of $\mathcal{D}$. ■

Proof of Lemma 5

To see that $c_{ik}\left (\widehat{x}_{ik}\left (y\right )\right )$ is increasing in y when Assumption A1* holds, we start by noting that the group cost function can be rewritten:

$$\displaystyle{ C_{i}\left (y\right ) =\min _{\mathbf{x}_{i}}\left \{\sum _{\ell=1}^{N_{i} }c_{i\ell}\left (x_{i\ell}\right ):\sum _{ \ell=1}^{N_{i} }g_{i\ell}\left (x_{i\ell}\right ) \geq \phi _{i}^{-1}\left (y\right )\right \}, }$$

(7)

where ϕ _i ⁻¹ is the inverse function of ϕ _i . Convexity of the cost functions and concavity of the g _{i ℓ} imply the existence of a Lagrange multiplier $\lambda _{i}\left (y\right ) \geq 0$ such that

$$\displaystyle{ \sum _{\ell=1}^{N_{i} }c_{i\ell}\left (\widehat{x}_{i\ell}\left (y\right )\right ) -\lambda _{i}\left (y\right )\sum _{\ell=1}^{N_{i} }g_{i\ell}\left (\widehat{x}_{i\ell}\left (y\right )\right ) <\sum _{ \ell=1}^{N_{i} }c_{i\ell}\left (x_{i\ell}\right ) -\lambda _{i}\left (y\right )\sum _{\ell=1}^{N_{i} }g_{i\ell}\left (x_{i\ell}\right ) }$$

(8)

for any x _i  ≥ 0 satisfying $\mathbf{x}_{i}\neq \widehat{\mathbf{x}}_{i}\left (y\right )$ the constraint in (7). Now suppose y ^∗ > y.

It is possible that $\widehat{\mathbf{x}}_{i}\left (y\right ) =\widehat{ \mathbf{x}}_{i}\left (y^{{\ast}}\right )$. If not, combining (8) with the constraint in (7), we have

$$\displaystyle\begin{array}{rcl} \sum _{\ell=1}^{N_{i} }c_{i\ell}\left (\widehat{x}_{i\ell}\left (y\right )\right ) -\lambda _{i}\left (y\right )\phi _{i}^{-1}\left (y\right )& <& \sum _{\ell =1}^{N_{i} }c_{i\ell}\left (\widehat{x}_{i\ell}\left (y^{{\ast}}\right )\right ) -\lambda _{ i}\left (y\right )\phi _{i}^{-1}\left (y^{{\ast}}\right ), {}\\ \sum _{\ell=1}^{N_{i} }c_{i\ell}\left (\widehat{x}_{i\ell}\left (y^{{\ast}}\right )\right ) -\lambda _{ i}\left (y^{{\ast}}\right )\phi _{ i}^{-1}\left (y^{{\ast}}\right )& <& \sum _{\ell =1}^{N_{i} }c_{i\ell}\left (\widehat{x}_{i\ell}\left (y\right )\right ) -\lambda _{i}\left (y^{{\ast}}\right )\phi _{ i}^{-1}\left (y\right ), {}\\ \end{array}$$

where we have used the fact that $\sum _{l}g_{il}\left (\widehat{x}_{il}\left (y\right )\right ) =\phi _{ i}^{-1}\left (y\right )$ (since g _i is increasing). Adding these inequalities shows that

$$\displaystyle{\left [\phi _{i}^{-1}\left (y^{{\ast}}\right ) -\phi _{ i}^{-1}\left (y\right )\right ]\left [\lambda _{ i}\left (y^{{\ast}}\right ) -\lambda _{ i}\left (y\right )\right ] > 0.}$$

Since ϕ _i is strictly increasing, we have $\phi _{i}^{-1}\left (y^{{\ast}}\right ) >\phi _{ i}^{-1}\left (y\right )$ and may conclude that $\lambda _{i}\left (y^{{\ast}}\right ) >\lambda _{i}\left (y\right )$.

It follows from (4) and (8) that $\widehat{x}_{ik}\left (y\right )$ minimizes $c_{ik}\left (x_{i\ell}\right ) -\lambda _{i}\left (y\right )g_{ik}\left (x_{ik}\right )$ subject to x _ik  ≥ 0 and therefore, for any k for which $\widehat{x}_{ik}\left (y\right )\neq \widehat{x}_{ik}\left (y^{{\ast}}\right )$, we have

$$\displaystyle\begin{array}{rcl} c_{ik}\left (\widehat{x}_{ik}\left (y\right )\right ) -\lambda _{i}\left (y\right )g_{ik}\left (\widehat{x}_{ik}\left (y\right )\right )& <& c_{ik}\left (\widehat{x}_{ik}\left (y^{{\ast}}\right )\right ) -\lambda _{ i}\left (y\right )g_{ik}\left (\widehat{x}_{ik}\left (y^{{\ast}}\right )\right ), {}\\ c_{ik}\left (\widehat{x}_{ik}\left (y^{{\ast}}\right )\right ) -\lambda _{ i}\left (y^{{\ast}}\right )g_{ ik}\left (\widehat{x}_{ik}\left (y^{{\ast}}\right )\right )& <& c_{ ik}\left (\widehat{x}_{ik}\left (y\right )\right ) -\lambda _{i}\left (y^{{\ast}}\right )g_{ ik}\left (\widehat{x}_{ik}\left (y\right )\right ). {}\\ \end{array}$$

In the case $\lambda _{i}\left (y\right ) > 0$, dividing the first inequality by $\lambda _{i}\left (y\right )$, the second by $\lambda _{i}\left (y^{{\ast}}\right )$, adding the results and rearranging gives

$$\displaystyle{\left [ \frac{1} {\lambda _{i}\left (y\right )} - \frac{1} {\lambda _{i}\left (y^{{\ast}}\right )}\right ]\left [c_{ik}\left (\widehat{x}_{ik}\left (y^{{\ast}}\right )\right ) - c_{ ik}\left (\widehat{x}_{ik}\left (y\right )\right )\right ] > 0.}$$

Since we have already shown that $\lambda _{i}\left (y^{{\ast}}\right ) >\lambda _{i}\left (y\right )$, we can deduce that $c_{ik}\left (\widehat{x}_{ik}\left (y^{{\ast}}\right )\right ) > c_{ik}\left (\widehat{x}_{ik}\left (y\right )\right )$. Since c _ik is a strictly increasing function, this implies $\widehat{x}_{ik}\left (y^{{\ast}}\right ) >\widehat{ x}_{ik}\left (y\right )$. In the case $\lambda \left (y\right ) = 0$, we have $\widehat{x}_{ik}\left (y\right ) = 0$ and $c_{ik}\left (\widehat{x}_{ik}\left (y^{{\ast}}\right )\right ) > 0 = c_{ik}\left (\widehat{x}_{ik}\left (y\right )\right )$. Hence, $\widehat{x}_{ik}\left (y^{{\ast}}\right ) >\widehat{ x}_{ik}\left (y\right )$. ■

Proof of Lemma 7

By definition,

$$\displaystyle{C_{i}\left (y\right ) = c_{i1}\left (\widehat{x}_{i1}\left (y\right )\right ) + D_{i}\left (y - g_{i1}\left (\widehat{x}_{i1}\left (y\right )\right )\right )}$$

for any y ≥ 0, where $\widehat{x}_{ik}\left (y\right )$ is the optimal solution of (3). For any y ^′ > y, the definition of C _i implies

$$\displaystyle\begin{array}{rcl} C_{i}\left (y^{{\prime}}\right )& =& c_{ i1}\left (\widehat{x}_{i1}\left (y^{{\prime}}\right )\right ) + D_{ i}\left (y^{{\prime}}- g_{ i1}\left (\widehat{x}_{i1}\left (y^{{\prime}}\right )\right )\right ) {}\\ & \leq & c_{i1}\left (\widehat{x}_{i1}\left (y\right )\right ) + D_{i}\left (y^{{\prime}}- g_{ i1}\left (\widehat{x}_{i1}\left (y\right )\right )\right ). {}\\ \end{array}$$

Hence,

$$\displaystyle\begin{array}{rcl} C_{i}\left (y^{{\prime}}\right ) - C_{ i}\left (y\right )& \leq & D_{i}\left (y^{{\prime}}- g_{ i1}\left (\widehat{x}_{i1}\left (y\right )\right )\right ) - D_{i}\left (y - g_{i1}\left (\widehat{x}_{i1}\left (y\right )\right )\right ) {}\\ & \leq & D_{i}\left (y^{{\prime}}\right ) - D_{ i}\left (y\right ), {}\\ \end{array}$$

using convexity of D _i (Lemma 1) and the fact that $g_{i1}\left (\widehat{x}_{i1}\left (y\right )\right ) > 0$. Dividing by y ^′− y and letting y ^′ ⟶ y gives the result.■