Group size and group success in conflicts

Abstract

This paper analyzes the occurrence of the group-size paradox in situations in which groups compete for rents, allowing for degrees of rivalry of the rent among group members. We provide two intuitive criteria for the group-impact function which for groups with within-group symmetric valuations of the rent determine whether there are advantages or disadvantages for larger groups: social-interactions effects and returns to scale. For groups with within-group asymmetric valuations, the complementarity of group members’ efforts and the composition of valuations are shown to play a role as further factors.

Appendices

Appendix A: Proof of Theorem 1

For ease of notation, define $Q_{-i}= \sum _{j' \in N' \backslash \{i'\}} q_{j'}(\vec{x}_{j'})$ and $Q= \sum _{j'\in N'} q_{j'}(\vec{x}_{j'})$. For notational convenience, in the following proofs we will often omit the arguments of functions, i.e., write $q_{i'}$ instead of $q_{i'}(\vec{x}_{i'})$.

The first order conditions (FOCs) are:

$$\begin{aligned}&\frac{Q_{-i}}{Q^2} \cdot \frac{\partial q_{i'}}{\partial x_k} \cdot v_i(i') - 1 \le 0 \quad \wedge \quad x_k \ge 0 \nonumber \\&\quad \wedge \quad \left( \frac{Q_{-i}}{Q^2} \cdot \frac{\partial q_{i'}}{\partial x_k} \cdot v_i(i') - 1\right) \cdot x_k = 0 \; \forall \; i,k. \end{aligned}$$

(A.1)

We start by showing that the first order conditions are sufficient conditions for an equilibrium. The second order conditions for a local maximum are:

$$\begin{aligned} \frac{Q_{-i}\cdot v_i(i')}{Q^2} \left( \frac{\partial ^2 q_{i'}(\vec{x}_{i'})}{\partial (x_k)^2} - \frac{\partial q_{i'}(\vec{x}_{i'})}{\partial x_k}\frac{2}{Q} \right) < 0 \end{aligned}$$

which holds for $Q_{-i} \ge 0$ and $\frac{\partial ^2 q_{i'}(\vec{x}_{i'})}{\partial (x_k)^2} \le 0$, which holds by Assumption 3. Since the above concavity condition holds for all $x_k \in [0,\infty )$ we only need to verify that $\pi (\infty , \vec{x}_{-k}) \le \pi (x_k^*, \vec{x}_{-k})$ and $\pi (0, \vec{x}_{-k}) \le \pi (x_k^*, \vec{x}_{-k})$. Since the FOC is strictly decreasing in $x_k$, we must have for all $x_k \in [0,x_{i'}^*)$: $\frac{\partial \pi _i^k(x_k,\vec{x}_{-k})}{\partial x_k}>\frac{\partial \pi _i^k(x_{i'}^*,\vec{x}_{-k})}{\partial x_k}$. This means profits are strictly increasing in $x_k$ over the interval $[0,x_{i'}^*)$ and thus $\pi (0, \vec{x}_{-k}) < \pi (x_k^*, \vec{x}_{-k})$. Further, since $\pi (\infty , \vec{x}_{-k})=-\infty < 0 \le \pi (0, \vec{x}_{-k})$ the solution to the FOCs indeed yields a global maximum of the expected payoff for each player.

What is left to show is that there exists a unique solution to the system of FOCs given that $\forall i,k: x_{k}^*=x_{i'}^*$.²³ By Assumption 4 we have for all k, l: $\frac{\partial q_{i'}(x_i,\dots ,x_i)}{\partial x_k} = \frac{\partial q_{i'}(x_i,\dots ,x_i)}{\partial x_l}$. Therefore, if $\forall i,k: x_k^*=x_{i'}^*$, then the system of FOCs can be reduced to:

$$\begin{aligned}&\frac{Q_{-i}}{Q^2} \cdot \frac{\partial q_{i'}(x_{i'}^*,\dots ,x_{i'}^*)}{\partial x_k} \cdot v_i(i') - 1 \le 0 \quad \wedge \quad x_{i'}^* \ge 0 \quad \nonumber \\&\quad \wedge \quad \left( \frac{Q_{-i}}{Q^2} \cdot \frac{\partial q_{i'}(x_{i'}^*,\dots ,x_{i'}^*)}{\partial x_l} \cdot v_i(i') - 1\right) \cdot x_{i'}^* = 0 \; \forall \; i. \end{aligned}$$

(A.2)

where k is an arbitrary member of $i'$.

We furthermore have the following relation between Q, $p_{i'}$ and $x_i$:

$$\begin{aligned} p_{i'} = \frac{q_{i'}(x_i,\dots ,x_i)}{Q} \quad \wedge \quad{{\underline{p}}_{i'}}(Q)= \frac{q_{i'}(0,\dots ,0)}{Q} \end{aligned}$$

where ${{\underline{p}}_{i'}}(Q)$ is the lower bound on the winning probability given a specific Q. Since $q_{i'}(x_i,\dots ,x_i)$ is strictly increasing in $x_i$, we can solve this for $x_i(Q,p_{i'})$ as long as $Q>0$ and $p_{i'} \ge{{\underline{p}}_{i'}}(Q)$. Finally, we can rewrite

$$\begin{aligned} \frac{\partial q_{i'}(x_i(Q,p_{i'}),\dots ,x_i(Q,p_{i'}))}{\partial x_k}=\rho (Q,p_{i'}), \quad \forall p_{i'} \ge{\underline{p}_{i'}}(Q) \end{aligned}$$

which by Assumption 3 is weakly decreasing in Q and $p_{i'}$. The left equation of the Kuhn–Tucker conditions then becomes:

$$\begin{aligned} \frac{1-p_{i'}}{Q} \cdot v_i(i') \cdot \rho (Q,p_{i'}) - 1 \le 0, \quad \forall i. \end{aligned}$$

(A.3)

Since $\rho$ is weakly decreasing in Q and $p_{i'}$, the LHS of (A.3) is strictly decreasing in $p_{i'}$ and Q. Further, for $p_{i'}\rightarrow \infty$, the LHS is negative while for $p_{i'}={q_{i'}(0,\dots ,0)/Q}$ it can be negative or positive. Therefore, for each strictly positive Q there exists a unique $p_{i'} \in [0,\infty )$ which solves the Kuhn–Tucker conditions where $p_{i'}=q_{i'}(0,\dots ,0)/Q$ if $v_i(i')/Q \cdot \rho (Q,0)\le 1$ and $p_{i'}=1$ if $Q=0$. We can therefore form the function $p_{i'}(Q)$ as the solution to the FOC of each group i.

What remains to be shown is that a unique strictly positive ${Q'}^*$ exists such that the winning probabilities $p_{i'}({Q'}^*)$ sum to one. Notice that $p_{i'}(Q)$ has the following properties: It is continuous, $\lim _{Q\rightarrow 0} p_{i'}(Q)=1$ and $\lim _{Q\rightarrow \infty } p_{i'}(Q)=0$ and it is strictly decreasing. Therefore, $\sum _{i'} p_{i'}(Q)$ is also strictly decreasing, continuous and has $\lim _{Q\rightarrow 0} \sum _{i'} p_{i'}(Q)>1$ as well as $\lim _{Q\rightarrow \infty } \sum _{i'} p_{i'}(Q)=0$. It follows by the intermediate value theorem that a ${Q'}^* \in (0,\infty )$ exists such that $\sum _{i'} p_{i'}({Q'}^*)=1$. Since $\sum _{i'} p_{i'}(Q)$ is strictly decreasing, this ${Q'}^*$ is unique. Given a unique ${Q'}^*$, we can obtain unique solutions for $p_{i'}({Q'}^*)$ and thus $x_{i'}^*$ and via $\forall i,k: x_k^{*}=x_{i'}^*$ also for all $x_k^{*}$.

Appendix B: Proof of Theorem 2

The first order condition for an equilibrium is:

$$\begin{aligned} \frac{Q_{-i}}{Q^2} \cdot \frac{\partial q_{i'}}{\partial x_k} \cdot v_i(i') - 1 \le 0 \quad \wedge \quad x_k \ge 0 \quad \wedge \quad \left( \frac{Q_{-i}}{Q^2} \cdot \frac{\partial q_{i'}}{\partial x_k} \cdot v_i(i') - 1\right) \cdot x_k = 0 \; \forall \; i,k. \end{aligned}$$

(B.1)

First notice that Assumption 5 implies Assumption 4, i.e., we are only considering a subset of the impact functions, therefore the results from the proof of Theorem 1 carry over. Since we thus know that the equilibrium is unique given $\forall i,k: x_k^*$, we only need to show that under the more strict Assumption 5, any equilibrium must fulfill $\forall i,k: x_k^*$.

From Assumption 5 we have that

$$\begin{aligned} x_k > x_l \quad \Leftrightarrow \quad \frac{\partial q_{i'}(\vec{x}_{i'})}{\partial x_k} < \frac{\partial q_{i'}(\vec{x}_{i'})}{\partial x_l}. \end{aligned}$$

Therefore, in equilibrium it can never be that case that $x_l^{*} =0$ if $x_k^* >0$ since then the above FOC (B.1) does not hold for at least one group member. Thus, either $\forall k: x_k^*=0$ or $\forall k: x_k^*>0$. If $x_k^*>0$ then inserting the FOC for player k into the FOC for player l yields $x_k^*=x_l^*$ and thus the desired condition.

Inserting this into the FOCs reduces the equilibrium conditions to (A.2). We therefore obtain as shown in the existence proof for each group i a best response function $p_{i'}(Q)$ which is continuous and strictly decreasing for $p_{i'}(Q)>0$. The necessary condition for an equilibrium is now $\sum _{i'} p_{i'} (Q) =1$. Since $\sum _{i'} p_{i'} (Q)$ is strictly decreasing in Q, if there exists a ${Q'}^*$ which solves the necessary condition (as has been shown in the existence proof), it is unique. Since $p_{i'}(Q)$ is monotonic in Q and $x_i(p_{i'}, Q)$ is monotonic in $p_{i'}$, the strategies $x_k^*=x_i(p_{i'}({Q'}^*),{Q'}^*)$ are unique solutions to the FOCs.

We can now write $x_{i''}(p_{i'},Q)$ as the solution of $p_{i'}=\frac{q_{i'}(x_i,\dots ,x_i)}{Q}$ for $x_i$ when $Q>0$. Since $q_{i'}$ is strictly increasing, $x_{i''}$ is a function and strictly increasing in Q for $p_{i'}>0$ and strictly increasing in $p_{i'}$.

$$\begin{aligned} \frac{1-p_{i'}}{Q} \cdot \frac{\partial q_{i'}(x_i(p_{i'},Q),\dots ,x_i(p_{i'},Q))}{\partial x_k} \cdot v_i(i') - 1 \le 0 \end{aligned}$$

It follows from non-increasing RTS and Assumptions 3, and 5 that $\frac{\partial q_{i'}(x_i,\dots ,x_i)}{\partial x_k}$ is non-increasing in $x_i$. Therefore, the LHS of the left equation in (A.2) is strictly decreasing in $p_{i'}$ and Q, which, together with Continuity of $q_{i'}$ from Assumption 3 implies that for each group i there exists a unique best response function $p_{i''}(Q)$ which solves the above FOC. Notice that since the LHS is strictly decreasing in both Q and $p_{i'}$, $p_{i''}(Q)$ must be strictly decreasing whenever the FOC holds with equality, i.e., whenever $p_{i'} \ge 0$. This reduces all equilibrium conditions to $\sum _{i'} p_{i''} (Q) =1$. Since all $p_{i''}$ are decreasing in Q and for $\sum _{i'} p_{i''}(Q)>0$ at least for some i the function $p_{i''}(Q)$ is strictly decreasing, the LHS is strictly decreasing in Q. Thus, if there exists a ${Q'}^*$ which solves the equilibrium condition, it is unique.

Appendix C: A useful Lemma

Lemma 1

Suppose a contest fulfills Assumptions 1, 2, 3, and 4for all groups. Further, $\forall j,k: v_k(m_j)=v_j(m_j)$. Consider two within-group symmetric equilibria, which only differ by the group sizes $i' \ne i'$ and/or the impact functions, $q_{i'}(\cdot ) \ne q_{i'}(\cdot )$. Suppose group i participates under group size $i'$ and impact function $q_{i'}(\cdot )$ with winning probability $p_{i'}^*$. Let the winning probability under group composition $i''$ and impact function $q_{i''}(\cdot )$ be $p_{i''}^*$. Then the following equivalence holds:

$$\begin{aligned} p_{i'}^* \gtreqless p_{i''}^* \quad \Leftrightarrow \quad v_i(i') \frac{\partial q_{i'}(x_{i'}^*,\dots ,x_{i'}^*)}{\partial x_k} \gtreqless v_i(i'')\frac{\partial q_{i''}(x_{i''},\dots ,x_{i''})}{\partial x_k} \end{aligned}$$

(C.1)

where $x_{i''}$ is defined such that

$$\begin{aligned} q_{i'}(x_{i'}^*,\dots ,x_{i'}^*)=q_{i''}(x_{i''},\dots ,x_{i''}). \end{aligned}$$

This Lemma has a very intuitive explanation: If and only if for a switch from $i'$ to $i''$ (while holding the winning probability constant) the LHS of the FOC is too low, the group will respond by increasing the effort from which a higher winning probability results. The only complication in the proof is that one has to address the possibility of a response by other groups which overcompensates this effect.

Proof

The first-order condition for an interior solution, evaluated at the solution, becomes after rearranging terms:

$$\begin{aligned} \forall i,k:\quad v_i(i') \frac{\partial q_{i'} (x_{i'}^*,\dots ,x_{i'}^*)}{\partial x_k} = \frac{{Q'}^*}{(1-p_{i'}^*)}. \end{aligned}$$

(C.2)

We first show that

$$\begin{aligned} p_{i'}^* \gtreqless p_{i''}^* \quad \Rightarrow \quad v_i(i')\frac{\partial q_{i'}(x_{i'}^*,\dots ,x_{i'}^*)}{\partial x_k} \gtreqless v_i(i'')\frac{\partial q_{i''}(x_{i''},\dots ,x_{i''})}{\partial x_k} \end{aligned}$$

(C.3)

and since the cases are exhaustive, the reverse implication is then automatically proven.

$p_{i'}^* > p_{i''}^*$: This implies that either ${Q'}^* <{Q''}^*$ or $q_{i'}(x_{i'}^*,\dots ,x_{i'}^*)>q_{i'}(x_{i''}^*,\dots ,x_{i''}^*)$. We first show that ${Q'}^* <{Q''}^*$ yields a contradiction: If $p_{i'}^* > p_{i''}^*$ then there exists a group j: $p_{j'}^*<p_{j''}^*$. Together with ${Q'}^* <{Q''}^*$, this implies that

$$\begin{aligned} \frac{{Q'}^*}{1-p_{j'}^*}<\frac{{Q''}^*}{1-p_{j''}^*} . \end{aligned}$$

By (C.2) this is equivalent with:

$$\begin{aligned} v_j \frac{\partial q_{j'}(x_{j'}^*,\dots ,x_{j'}^*)}{\partial x_k} < v_j \frac{\partial q_{j''}(x_{j''}^*,\dots ,x_{j''}^*)}{\partial x_k} \end{aligned}$$

(C.4)

Since $q_{j'}(\ldots )$ has constant or decreasing RTS, this implies $q_{j'}(x_{j'}^*,\dots ,x_{j'}^*) \ge q_{j'}(x_{j''}^*,\dots ,x_{j''}^*)$. But since ${Q'}^*<{Q''}^*$, we have $p_{j'}^* > p_{j''}^*$ and thus a contradiction. From this follows that if $p_{i'}^* > p_{i''}^*$, then ${Q'}^* \ge{Q''}^*$ and $q_{i'}(x_{i'}^*,\dots ,x_{i'}^*)>q_{i'}(x_{i''}^*,\dots ,x_{i''}^*)$. The latter implies $x_{i''}> x_{i''}^*$ via the definition of $x_{i''}$. Since $q_{i'}$ has constant or decreasing RTS, we have:

$$\begin{aligned} \frac{\partial q_{i''}(x_{i''},\dots ,x_{i''})}{\partial x_k} \le \frac{\partial q_{i''}(x_{i''}^*,\dots ,x_{i''}^*)}{\partial x_k} \end{aligned}$$

(C.5)

From $p_{i'}^* > p_{i''}^*$ and ${Q'}^* \ge{Q''}^*$ follows ${Q'}^*/(1-p_{i'}^*) >{Q''}^*/(1-p_{i''}^*)$. Using the FOCs, we have:

$$\begin{aligned} v_i(i')\frac{\partial q_{i'}(x_{i'}^*,\dots ,x_{i'}^*)}{\partial x_k} > v_i(i'')\frac{\partial q_{i''}(x_{i''}^*,\dots ,x_{i''}^*)}{\partial x_k}. \end{aligned}$$

Combining this equation with (C.5), immediately yields the $p_{i'}^*>p_{i''}^*$ part of (C.3).

The proof for $p_{i'}^* < p_{i''}^*$ follows the same steps with reverse inequalities and is therefore omitted.

$p_{i'}^* = p_{i''}^*$: This implies that either ${Q'}^* ={Q''}^*$ and $q_{i'}(x_{i'}^*,\dots ,x_{i'}^*)=q_{i'}(x_{i''}^*,\dots ,x_{i''}^*)$ or ${Q'}^* \lessgtr{Q''}^*$ $q_{i'}(x_{i'}^*,\dots ,x_{i'}^*) \lessgtr q_{i''}(x_{i''}^*,\dots ,x_{i''}^*)$.

Suppose ${Q'}^* ={Q''}^*$ and $q_{i'}(x_{i'}^*,\dots ,x_{i'}^*)=q_{i'}(x_{i''}^*,\dots ,x_{i''}^*)$ hold. Then it immediately follows from the FOCs that

$$\begin{aligned} v_i(i') \frac{\partial q_{i'}(x_{i'}^*,\dots ,x_{i'}^*)}{\partial x_k} = v_i(i'') \frac{\partial q_{i'' }(x_{i''}^*,\dots ,x_{i''}^*)}{\partial x_k}. \end{aligned}$$

By definition of $x_{i''}$ we then have the symmetric part of (C.1).

For the case ${Q'}^* \lessgtr{Q''}^*$, $q_{i'}(x_{i'}^*,\dots ,x_{i'}^*) \lessgtr q_{i''}(x_{i''}^*,\dots ,x_{i''}^*)$ we can show that this yields a contradiction. It follows from these assumptions that there exists a group j with $p_{j'}^*\gtreqqless p_{j''}^*$ such that:

$$\begin{aligned} q_{j'}(x_{j'}^*,\dots ,x_{j'}^*)\gtrless q_{j''}(x_{j''}^*,\dots ,x_{j''}^*). \end{aligned}$$

(C.6)

Furthermore,

$$\begin{aligned} \frac{{Q'}^*}{1-p_{j'}^*} \gtrless \frac{{Q''}^*}{1-p_{j''}^*}. \end{aligned}$$

Applying the FOCs gives us:

$$\begin{aligned} \frac{\partial q_{j'}(x_{j'}^*,\dots ,x_{j'}^*)}{\partial x_k} \gtrless \frac{\partial q_{j''}(x_{j''}^*,\dots ,x_{j''}^*)}{\partial x_k}. \end{aligned}$$

But from this follows $x_{j'}^*<x_{j''}^*$ since the group composition does not change between $j'$ and $j''$ and thus

$$\begin{aligned} q_{j'}(x_{j'}^*,\dots ,x_{j'}^*) \lessgtr q_{j''}(x_{j''}^*,\dots ,x_{j''}^*) \end{aligned}$$

(C.7)

which contradicts (C.6).

Suppose ${Q'}^*<{Q''}^*$ and $q_{i'}(x_{i'}^*,\dots ,x_{i'}^*) > q_{i'}(x_{i''}^*,\dots ,x_{i''}^*)$. Then there exists a group j such that: $q_{j'}(x_{j'}^*,\dots ,x_{j'}^*)<q_{j''}(x_{j''}^*,\dots ,x_{j''}^*)$ and by $q_{j'}=q_{j''}$ being a strictly increasing function, $x_{j'}^*<x_{j''}^*$. From this follows by Assumption (3) that

$$\begin{aligned} \frac{\partial q_{j'}(x_{j'}^*,\dots ,x_{j'}^*)}{\partial x_k} > \frac{\partial q_{j''}(x_{j''}^*,\dots ,x_{j''}^*)}{\partial x_k} \end{aligned}$$

By the FOCs we then have:

$$\begin{aligned} \frac{{Q'}^*}{1-p_{j'}^*} > \frac{{Q''}^*}{1-p_{j''}^*}. \end{aligned}$$

Since ${Q'}^*<{Q''}^*$ this implies $p_{j'}^* > p_{j''}^*$. But then there must exist a group g such that $p_{g'}^*< p_{g''}^*$ since $p_{i'}^* \ge p_{i''}^*$. It follows that $q_{g'}(x_{g'}^*,\dots ,x_{g'}^*)<q_{g''}(x_{g''}^*,\dots ,x_{g''}^*)$ and thus $x_{g'}^*< x_{g''}^*$. Due to Assumption (3) we then have for some l in $g'$ that

$$\begin{aligned} \frac{\partial q_g(x_{g'}^*,\dots ,x_{g'}^*)}{\partial x_l} \ge \frac{\partial q_g(x_{g''}^*,\dots , x_{g''}^*)}{\partial x_l} \end{aligned}$$

from which follows

$$\begin{aligned} \frac{{Q'}^*}{1-p_{g'}^*} > \frac{{Q''}^*}{1- p_{g''}^*}. \end{aligned}$$

But then $p_{g'}^* > p_{g''}^*$ which yields the contradiction. $\square$

Appendix D: Proof of Proposition 1

Proof

We employ the total differential:

$$\begin{aligned} \Delta q_{i'}(\vec{x}_{i}) = \sum _k \Delta x_k \frac{\partial{q_{i'}(\vec{x}_{i'})}}{{\partial x_k}} \end{aligned}$$

(D.1)

For equal inputs we can write $q_{i'}(x_i,\dots ,x_i)=g(x_i, i')$. The total differential then becomes for symmetric efforts:

$$\begin{aligned} \Delta g(x_i, i') = \sharp i' \cdot \Delta x_i \frac{\partial{q_{i'}(x_i,\dots ,x_i)}}{{\partial x_k}}, \end{aligned}$$

(D.2)

and thus

$$\begin{aligned} \frac{\partial{q(x_i,\dots ,x_i)}}{{\partial x_k}} = \frac{g({x_i}, i')-g({x'_i}, i')}{\sharp i' \cdot (x_i-x_i')} \end{aligned}$$

(D.3)

for $x_i-x_i' \rightarrow 0$. Similarly, we have:

$$\begin{aligned} \frac{\partial{q_{i''}(x_i\frac{\sharp i'}{\sharp i''},\dots ,x_i\frac{\sharp i'}{\sharp i''})}}{{\partial x_k}} = \frac{g({x_i}\frac{\sharp i'}{\sharp i''}, i'')-g({x_i'}\frac{\sharp i'}{\sharp i''}, i'')}{\sharp i'' \cdot (x_i-x_i') \cdot \frac{\sharp i'}{\sharp i''}} \end{aligned}$$

(D.4)

for $x_i-x_i' \rightarrow 0$. It follows from (D.3) and (D.4) that:

$$\begin{aligned} \frac{\partial{q_{i'}(x_i,\dots ,x_i)}}{{\partial x_k}} = \frac{\partial{q_{i''}(x_i\frac{\sharp i'}{\sharp i''},\dots ,x_i\frac{\sharp i'}{\sharp i''})}}{{\partial x_k}} \end{aligned}$$

(D.5)

since by absent SSIE it holds that $g({x_i}, i')=g({x_i}\frac{\sharp i'}{\sharp i'}, i'')$ and $g({x_i'}, i')=g({x_i'}\frac{\sharp i'}{\sharp i'}, i'')$.

To apply Lemma 1, we need to know what the symmetric effort level $x_{i''}$ of the group after the increase in size would need to be in order to obtain $q_{i'}(x_{i'}^*,\dots ,x_{i'}^*) = q_{i''}(x_{i''},\dots ,x_{i''})$. With absent SSIE we have:

$$\begin{aligned} x_{i''} = \frac{x_{i'}^* \sharp i'}{\sharp i''}. \end{aligned}$$

Lemma 1 then yields:

$$\begin{aligned} p_{i'}^* \gtreqless p_{i''}^* \quad \Leftrightarrow \quad v_i(i')\frac{\partial q_{i'}(x_{i'}^*,\dots ,x_{i'}^*)}{\partial x_k} \gtreqless v_i(i'')\frac{\partial q_{i''}(\frac{x_{i'}^* \sharp i'}{\sharp i''}, \dots ,\frac{x_{i'}^*\sharp i'}{\sharp i''})}{\partial x_k} \end{aligned}$$

which given (D.5) reduces to the desired condition:

$$\begin{aligned} p_{i'}^* \gtreqless p_{i''}^* \quad \Leftrightarrow \quad v_i(i') \gtreqless v_i(i'') \end{aligned}$$

Therefore, if the rent is a public good, the winning probability is independent of group size for an impact function with absent SSIE at the equilibrium effort $x_{i'}^*$ for a change in group size from $\sharp i'$ to $\sharp i''$. If the rent is partly private, it is strictly decreasing in group size.

For the welfare effects, we have:

$v_i(i')=v_i(i'')$: It follows from $p_{i'}^*=p_{i''}^*$ and $S_i(x_{i'}^*,i',i')$ that $x_{i'}^*>x_{i''}^*$. Inserting this into $\pi _i^*-\pi _{i''}^*$ we get: $(p_{i'}^*-p_{i''}^*)(v_i(i')) - (x_{i'}^*-x_{i''}^*)$. Since the first term is equal to zero and the second term negative, we get that ${\pi '_k}^{*}<{\pi _k''}{*}$ from which the statements for average and total utility follow for $v_i(i')=v_i(i'')$.

$v_i(i')>v_i(i'')$: Let $v_i(i'')\rightarrow 0$. Then ${\pi _{k}'}^{*} \rightarrow 0$. Since ${\pi _{k}'}^{*}>0$, and ${\pi _{k}''}^{*}$ is continuous in $v_i(i'')$ the existence of $\underline{v_i(i'')}$ follows.

We cannot specify $\underline{v_i(i'')}$ more precisely under the very general assumptions. Especially, we cannot know, whether $\underline{v_i(i'')}\lesseqgtr \sharp i'v_i(i')/\sharp i''$ which is the private good case. $\square$

Appendix E: Proof of Proposition 2

Proof

We assume throughout that we are in a symmetric, interior equilibrium. By homogeneity of degree $r_i$, we have from Euler’s theorem

$$\begin{aligned} r_i \cdot q_{i'}(x_i,\dots ,x_i) = \sharp i' \cdot x_i \cdot \frac{\partial q_{i'}(x_i,\dots ,x_i)}{\partial x_k}, \end{aligned}$$

(E.1)

and further

$$\begin{aligned} \frac{\partial q_{i'}(x_i,\dots ,x_i)}{\partial x_k} = \frac{r_i \cdot q_{i'}(x_i,\dots ,x_i)}{ \sharp i' \cdot x_i} = \frac{r_i \cdot q_{i'}(1, \dots ,1)}{ \sharp i' \cdot (x_i)^{1-r_i}}. \end{aligned}$$

(E.2)

Using homogeneity of the impact function and the above expression for the partial derivative, we get for the measure of SSIE:

$$\begin{aligned} S_i(i', i'') = \frac{q_{i''}(1,\dots ,1) \cdot \sharp i'^{r_i}}{q_{i'}(1,\dots ,1) \cdot \sharp i''^{r_i}}. \end{aligned}$$

(E.3)

Lemma 1 now tells us that

$$\begin{aligned} p_{i'}^* \gtreqless p_{i''}^* \quad \Leftrightarrow \quad v_i(i')\frac{\partial q_{i'}(x_{i'}^*,\dots ,x_{i'}^*)}{\partial x_k} \gtreqless v_i(i'')\frac{\partial q_{i''}(x_{i''}, \dots ,x_{i''})}{\partial x_k} \end{aligned}$$

(E.4)

where $x_{i''}$ is defined such that

$$\begin{aligned} q_{i'}(x_{i'}^*, \dots ,x_{i'}^*)=q_{i''}(x_{i''},\dots , x_{i''}). \end{aligned}$$

We can make use of homogeneity of degree $r_i$ and solve for $x_{i''}$:

$$\begin{aligned} x_{i''}=\frac{x_{i'}^*\sharp i'}{\sharp i''} \left( \frac{q_{i'}(1, \dots ,1) \cdot \sharp i''^{r_i}}{q_{i''}(1,\dots ,1) \cdot \sharp i'^{r_i}} \right) ^{1/r_i} = \frac{x_{i'}^*\cdot \sharp i'}{ S_i(i', i'')^{1/r_i} \cdot \sharp i''} \end{aligned}$$

(E.5)

where the last step follows from (E.3). Plugging this definition back into (E.4), we get using (E.2):

$$\begin{aligned} p_{i'}^* \gtreqless p_{i''}^* \quad \Leftrightarrow \quad v_i(i')\frac{r_i \cdot q_{i'}(x_{i'}^*, \dots ,x_{i'}^*)}{ \sharp i' \cdot x_{i'}^*} \gtreqless v_i(i'')\frac{r_i \cdot q_{i'}(x_{i''}, \dots ,x_{i''}) S_i(i', i'')^{1/r_i}}{ \sharp i' \cdot x_{i'}^*} \end{aligned}$$

(E.6)

By canceling terms, this simplifies to:

$$\begin{aligned} p_{i'}^* \gtreqless p_{i''}^* \quad \Leftrightarrow \quad \frac{v_i(i')}{v_i(i'')} \gtreqless S_i(i', i'')^{1/r_i} \end{aligned}$$

(E.7)

For the welfare effects, we have:

$\pi _{i'}^{A}\gtreqless \pi _{i''}^{A} \quad \Leftrightarrow \quad p_{i'}^*v_i(i')-x_{i'}^* \gtreqless p_{i''}^*v_i(i'')- x_{i''}^*.$ We furthermore have:

$$\begin{aligned} x_{i'}^*=\frac{r_i v_i(i')}{\sharp i'} p_{i'}^*(1-p_{i'}^*) \end{aligned}$$

from inserting (E.2) into (B.1). Inserting this into the above equation for $x_{i'}^*$ and $x_{i''}^*$ and rearranging terms yields the result. The total welfare effects follow by multiplying with $\sharp i'$ and $\sharp i''$ on the LHS and RHS, respectively.

For public goods $v_i(i')=v_i(i'')$ it immediately follows that for $\pi _i^{k*}\ge \pi _{i''}^{k*}$ to hold, we must have $p_{i'}^* > p_{i''}^*$ and thus the group-size paradox must hold strictly for the welfare group-size paradox to hold. For private goods this is not the case. The average utility criterion yields: $p_{i'}^*/\sharp i' \left( 1-(1-p_{i'}^*)\frac{r_i}{\sharp i'}\right) \ge p_{i''}^*/\sharp i''\left( 1-(1-p_{i''}^*)\frac{r_i}{\sharp i''}\right)$. The total utility criterion yields: $p_{i'}^* \left( 1-(1-p_{i'}^*)\frac{r_i}{\sharp i'}\right) \ge p_{i''}^*\left( 1-(1-p_{i''}^*)\frac{r_i}{\sharp i''}\right)$. For the total utility criterion we therefore again need the group-size paradox to hold strictly for the welfare group-size paradox to hold. $\square$

Appendix F: Proof of Theorem 3

Proof

We prove that the only class of impact functions fulfilling the assumptions is given by:

$$\begin{aligned} q_{i'}(\vec{x}_{i'}) = \zeta _i(\sharp i')(\sharp i')^{r_i} \cdot \left( \sum _{k\in i'} \frac{(x_k)^{\gamma _i}}{\sharp i'}\right) ^{r_i/\gamma _i} \end{aligned}$$

(F.1)

$\gamma _i \in (0,1)$, $r_i \in (0,1]$.

From independence of the class of impact functions, the continuity of the impact functions, strict monotonicity of the impact functions, and Theorem 1 of Gorman (1968), we have that for group sizes larger or equal to 3:

$$\begin{aligned} q_{i'}(\vec{x}_{i'}) = T_{i'} \left[\sum _{k\in i'} f_{k,i'}(x_k) \right] \end{aligned}$$

(F.2)

where $T_{i'}$ and $f_{k,i'}$ are continuous strictly monotone functions and without loss of generality $f_{k,i'}(0)=0$.

By Homogeneity of degree $r_i$, we have after setting $x_l=0$ for all $l\ne k$ and taking the inverse of f,

$$\begin{aligned} T_{i'}[y]&= f_{k,i'}^{-1}(y)^{r_i} T_{i'}[f_{k,i'}(1)] \end{aligned}$$

(F.3)

$$\begin{aligned} T_{i'}'[y]&= f_{k,i'}^{-1}(y)^{r_i-1}{f_{k,i'}^{-1}}'(y) r_i T_{i'}[f_{k,i'}(1)] \end{aligned}$$

(F.4)

Moreover, we can derive from homogeneity and Euler’s homgeneous function theorem the functional equation:

$$\begin{aligned} \frac{T_{i'}[\sum _k y_k]}{T_{i'}'[\sum _k y_k]} r_i =&\sum _k f_{k,i'}'(f_{k,i'}^{-1}(y_k))f_{k,i'}^{-1}(y_k) \end{aligned}$$

(F.5)

to which we can apply Pexider’s functional equation Aczél and Dhombres (1989) $g_0(\sum _k s_k) = \sum _k g_k(s_k)$ with the continuous solution being that all functions are affine. The solution to the functional equation yields therefore

$$\begin{aligned} Ay + B =&\frac{T_{i'}[y]}{T_{i'}'[y]} r_i =\frac{f_{k,i'}^{-1}(y)}{{f_{k,i'}^{-1}}'(y) } \end{aligned}$$

(F.6)

This is an ordinary differential equation with solution:

$$\begin{aligned} f_{k,i'}^{-1}(y)&= (C_{k,i'} y + B_{k,i'})^{1/A_{k,i'}} \end{aligned}$$

(F.7)

$$\begin{aligned} f_{k,i'}(x_k)&= \frac{(x_k)^{A_{k,i'}}-B_{k,i'}}{C_{k,i'}} \end{aligned}$$

(F.8)

Since k was arbitrarily chosen (or by solving the corresponding ordinary differential equations induced by the solutions to RHS of the Pexider equation), this solution must hold for all k. By our normalization assumption, we have that $B_k=0$. It is straightforward to use Assumption 5 to obtain that $A_{k,i'}$ and $C_{k,i'}$ do not depend on k. Moreover, using Assumption 7 we can establish that $A_{k,i'}$ is even consistent across group compositions, i.e., $A_{k,i'}=\gamma _i$. We have therefore derived

$$\begin{aligned} q_{i'}(\vec{x}_{i'}) = \left( \sum _k (x_k)^{\gamma _{i}} \right) ^{r_i/\gamma _i} T_{i'}[f_{i'}(1)] \end{aligned}$$

(F.9)

The only parameter that remains to be chosen is $T_{i'}[f_{k,i'}(1)]$ which may depend on composition. Letting $T_{i'}[f_{k,i'}(1)] = \frac{\zeta _i(i')}{(\sharp i')^{r_i/a_i}-r_i}$ yields our desired functional form. The impact function of $\sharp i' = 2$ follows from independence by substituting $x_k=0$ into an impact function of a composition with $\sharp i''=3$. Order equivalence on the remaining two dimensions and identical returns to scale guarantee that the impact function of two dimensions must be equal to a linear transformation of the impact function with 3 group members and some $x_k=0$.

The cases of $r_i<0$, $r_i> 1$, $\gamma _{i'}>1$, and $\gamma _i\le 0$ are excluded by monotonicity in $x_k$, concavity (both from Assumption 3), and nonnegativity of the range of $q_{i'}$. $\square$

Appendix G: Proof of Theorem 4

Before beginning the main proof, it will be useful to derive the following two lemmas.

Lemma 2

If $a_1 \ge a_2 > a_3$ or $a_3 > a_2 \ge a_1$ and f is an integrable convex function, then

$$\begin{aligned} f(a_1 + a_2 - a_3) > f(a_1) + f(a_2) - f(a_3) \end{aligned}$$

(G.1)

Proof

Case $a_3<a_1$: From convexity of f, we have for all h

$$\begin{aligned} f'(a_1 - a_3 + h) > f'(h) \quad \Leftrightarrow \quad a_3 < a_1 \end{aligned}$$

(G.2)

Integrating both sides gives:

$$\begin{aligned} \int _{a_3}^{a_2} f'(a_1 - a_3 + h)dh > \int _{a_3}^{a_2} f'(h)dh \end{aligned}$$

(G.3)

which yields the desired condition:

$$\begin{aligned} f(a_1 + a_2 - a_3) - f(a_1) > f(a_2) - f(a_3). \end{aligned}$$

(G.4)

Case $a_3>a_1$: From convexity of f, we have for all h

$$\begin{aligned} f'(a_1 - a_3 + h) < f'(h) \quad \Leftrightarrow \quad a_3 > a_1 \end{aligned}$$

(G.5)

Integrating both sides gives:

$$\begin{aligned} \int _{a_2}^{a_3} f'(a_1 - a_3 + h)dh > \int _{a_2}^{a_3} f'(h)dh \end{aligned}$$

(G.6)

which yields the desired condition:

$$\begin{aligned} f(a_1) - f(a_1 + a_2 - a_3) < f(a_3) - f(a_2). \end{aligned}$$

(G.7)

$\square$

The above Lemma 2 can be used to derive the following result:

Lemma 3

Suppose $\vec{v}^{(n)}$ is obtained from a sequence of $\theta$-power mean preserving spreads of $\vec{v}^{(1)}$. Then

$$\begin{aligned} \theta \gtreqless \phi \quad \Leftrightarrow \quad{{\mathcal{M}}}(\vec{v}^{(1)},\phi ) \gtreqless{{\mathcal{M}}}(\vec{v}^{(n)},\phi ) \end{aligned}$$

(G.8)

Proof

Suppose $\vec{v}^{(1)},\dots \vec{v}^{(n)}$ is a sequence of vectors generated by a sequence of $\theta$-power mean preserving spreads. If for all i it holds that

$$\begin{aligned} \theta \gtreqless \phi \quad \Leftrightarrow \quad{{\mathcal{M}}}(\vec{v}^{(i)},\phi ) \gtreqless{{\mathcal{M}}}(\vec{v}^{(i+1)},\phi ) \end{aligned}$$

(G.9)

then it clearly also holds that:

$$\begin{aligned} \theta \gtreqless \phi \quad \Leftrightarrow \quad{{\mathcal{M}}}(\vec{v}^{(1)},\phi ) \gtreqless{{\mathcal{M}}}(\vec{v}^{(n)},\phi ) \end{aligned}$$

(G.10)

We therefore only need to show this property for vectors which differ by a single power mean preserving spread. Notice that for any $\phi$:

$$\begin{aligned}{{\mathcal{M}}}(\vec{v}^{(i)},\phi ) \gtreqless{{\mathcal{M}}}(\vec{v}^{(i+1)},\phi ) \quad \Leftrightarrow \quad{{\mathcal{M}}}((v^{(i)}_H, v^{(i)}_L),\phi ) \gtreqless{\mathcal{M}}((v^{(i+1)}_H,v^{(i+1)}_L),\phi ) \end{aligned}$$

(G.11)

where $(v^{(i)}_H,v^{(i)}_L)$ refers to the vector of the two elements that are changed by the spreading operation and $(v^{(i+1)}_H,v^{(i+1)}_L)$ to the vector of these two elements after application of the spreading operation. Let w.l.o.g. $v^{(i)}_H \ge v^{(i)}_L$ from which immediately follows $v^{(i+1)}_H>v^{(i)}_H \ge v^{(i)}_L>v^{(i+1)}_L$ by the properties of the power mean preserving spread. That is, $v_H^{(i)}$ refers to the element of $\vec{v}^{(i)}$ which is increased to $v_H^{(i+1)}$ and $v_L^{(i)}$ to the decreased element of $\vec{v}^{(i)}$.

From the power mean preserving spread also follows via evaluating (G.11) at equality:

$$\begin{aligned} \left( \frac{1}{2} (v^{(i+1)}_H)^{\theta } + \frac{1}{2} (v^{(i+1)}_L)^{\theta }\right) ^{1/\theta } = \left( \frac{1}{2} (v^{(i)}_H)^{\theta } + \frac{1}{2} (v^{(i)}_L)^{\theta }\right) ^{1/\theta }, \end{aligned}$$

(G.12)

which – after solving for $v^{(i+1)}_H$ – yields:

$$\begin{aligned} (v^{(i+1)}_H)^{\theta } = \left( (v^{(i)}_H)^{\theta } + (v^{(i)}_L)^{\theta } -(v^{(i+1)}_L)^{\theta }\right) ^{1/\theta }. \end{aligned}$$

(G.13)

Combining this condition with (G.11) and (G.9), what is left to show is:

$$\begin{aligned}&\theta \gtreqless \phi \quad \Leftrightarrow \quad \left( (v^{(i)}_H)^{\phi } + (v^{(i)}_L)^{\phi } \right) ^{1/\phi } \nonumber \\&\quad \gtreqless \left( \left( (v^{(i)}_H)^{\theta } + (v^{(i)}_L)^{\theta } -(v^{(i+1)}_L)^{\theta }\right) ^{\phi /\theta } +(v^{(i+1)}_L)^{\phi }\right) ^{1/\phi } \end{aligned}$$

(G.14)

which is implied by the following, more general condition:

$$\begin{aligned} \forall \; \psi> \eta : \quad \left( (v_H^{(i)})^{\eta }+(v_L^{(i)})^{\eta }-(v_L^{(i+1)})^{\eta } \right) ^{1/\eta } > \left( (v_H^{(i)})^{\psi }+(v_L^{(i)})^{\psi }-(v_L^{(i+1)})^{\psi } \right) ^{1/\psi } \end{aligned}$$

(G.15)

Notice that standard mean inequalities or the reverse Jensen inequality from the previous appendix do not apply to prove (G.15). This would also be counterintuitive as then the proof would not rely on $v^{(i)}_H \ge v^{(i)}_L > v^{(i+1)}_L$. We have to distinguish the cases $\psi >0$ and $\psi < 0$.

$\psi >0$: Define $f(a)=a^{\psi /\eta }$, which is strictly convex and $a_1=(v_H^{(i)})^{\eta }$, $a_2=(v_L^{(i)})^{\eta }$, and $a_3=(v_L^{(i+1)})^{\eta }$. If $\eta >0$, we have $a_1\ge a_2 >a_3$, while if $\eta <0$, we have $a_3>a_2 \ge a_1$. In both cases Lemma 2 applies. Employing these definitions in Lemma 2 gives:

$$\begin{aligned} ((v_H^{(i)})^{\eta } + (v_L^{(i)})^{\eta } - (v_L^{(i+1)})^{\eta })^{\psi /\eta } > ((v_H^{(i)})^{\eta })^{\psi /\eta } + ((v_L^{(i)})^{\eta })^{\psi /\eta } - ((v_L^{(i+1)})^{\eta })^{\psi /\eta }, \end{aligned}$$

(G.16)

which simplifies to (G.15).

$\psi <0$: Define $f(a)=a^{\eta /\psi }$, which is strictly convex and $a_1=(v_H^{(i)})^{\psi }$, $a_2=(v_L^{(i)})^{\psi }$, and $a_3=(v_L^{(i+1)})^{\psi }$. Since $\eta <0$, we have $a_3>a_2 \ge a_1$. Employing these definitions in Lemma 2 gives us:

$$\begin{aligned} ((v_H^{(i)})^{\psi } + (v_L^{(i)})^{\psi } - (v_L^{(i+1)})^{\psi })^{\eta /\psi } > ((v_H^{(i)})^{\psi })^{\eta /\psi } + ((v_L^{(i)})^{\psi })^{\eta /\psi } - ((v_L^{(i+1)})^{\psi })^{\eta /\psi } \end{aligned}$$

(G.17)

which is equivalent with (G.15) since $\eta$ is negative and the inequality sign thus changes direction once we exponentiate both sides with $\eta$. $\square$

We now turn to the main proof of the theorem.

Proof

Suppose $\vec{v}'$ is more asymmetric than $\vec{v}$ at mean parameter $\theta$. Then $\exists \omega : \omega \cdot \vec{v}'' = \vec{v}'$ and $\vec{v}''$ is obtained from a sequence of $\theta$-power mean preserving spreads of $\vec{v}$. Thus by Lemma 2,

$$\begin{aligned} \theta \gtreqless \phi \quad \Leftrightarrow \quad{{\mathcal{M}}}(\vec{v}, \phi ) \gtreqless{{\mathcal{M}}}(\vec{v}'', \phi ) \end{aligned}$$

(G.18)

By definition of a $\theta$-power mean preserving spread we have ${{\mathcal{M}}}(\vec{v}, \theta )={{\mathcal{M}}}(\vec{v}'', \theta )$. We therefore obtain:

$$\begin{aligned} \theta \gtreqless \phi \quad \Leftrightarrow \quad \frac{\omega{{\mathcal{M}}}(\vec{v}'', \theta )}{{{\mathcal{M}}}(\vec{v}, \theta )} \gtreqless \frac{\omega{{\mathcal{M}}}(\vec{v}'', \phi )}{{{\mathcal{M}}}(\vec{v}, \phi )} \end{aligned}$$

(G.19)

Making use of the homogeneity of degree 1 of ${\mathcal{M}}$ and the definition of $\omega \vec{v}''$ we have:

$$\begin{aligned} \theta \gtreqless \phi \quad \Leftrightarrow \quad \frac{{{\mathcal{M}}}(\vec{v}', \theta )}{{{\mathcal{M}}}(\vec{v}, \theta )} \gtreqless \frac{{{\mathcal{M}}}(\vec{v}', \phi )}{{{\mathcal{M}}}(\vec{v}, \phi )} \end{aligned}$$

(G.20)

$\square$

Appendix H: Proof of Theorem 5

Proof

The proof proceeds similarly to the one for single-player contests in Cornes and Hartley (2005), with the main difference that first one has to obtain equilibrium conditions that fix relative efforts within each group. The optimality condition for individual k in composition $i'$ yields:

$$\begin{aligned} \frac{\partial q_{i'}}{\partial x_k} \frac{Q_{-i} \cdot v_k}{Q^2} \le 1 \end{aligned}$$

(H.1)

with equality if $x_k > 0$. Notice first that in equilibrium it can never be the case that $Q_{-i}=0$, since then some individual will have an incentive to provide effort $x_k=\epsilon$ with $\epsilon \rightarrow 0$ to win the rent with probability 1. The expression for the partial derivative of the impact function becomes after rearranging terms:

$$\begin{aligned} \frac{\partial q_{i'}}{\partial x_k} = \frac{r_i \cdot q_{i'}^{1-\gamma _i/r_i} \cdot (x_k)^{\gamma _i-1}}{\zeta _i(i')^{-\gamma _i/r_i} (\sharp i')^{1-\gamma _i}} \end{aligned}$$

(H.2)

From this expression can already be derived that if one group member participates in equilibrium, all group members do: Notice that if some group member l of group i participates in equilibrium, $q_{i'} > 0$ and therefore at $x_k=0$, $\frac{\partial q_{i'}}{\partial x_k} = \infty$. But then the first order condition cannot hold for individual k in that equilibrium, since we have $Q>0$ and $Q_{-i}>0$ and thus the LHS of the optimality condition (H.1) is infinite, which is greater than the RHS.

Since either all group members participate or none, we can express the following relationship among efforts within a group:

$$\begin{aligned} (x_k)^{\gamma _i-1} \cdot v_k = (x_l)^{\gamma _i-1} \cdot v_l \quad \forall l,k \end{aligned}$$

(H.3)

Notice that this relation trivially also holds for groups that do not participate. Rearranging and summing over all l yields:

$$\begin{aligned} \left( \frac{1}{\sharp i'}\sum (x_l)^{\gamma _i}\right) ^{1/\gamma _i} = \frac{x_k}{(v_k)^{\frac{1}{1-\gamma _i}}} \left( \frac{1}{\sharp i'}\sum (v_l)^{\frac{\gamma _i}{1-\gamma _i}}\right) ^{1/\gamma _i} \end{aligned}$$

(H.4)

Substituting this relation into the optimality condition yields:

$$\begin{aligned} q_{i'}^{1-\frac{1}{r_i}} \cdot Q_{-i} \cdot V_i \le Q^2 \end{aligned}$$

(H.5)

where $V_i=r_i \zeta _i(i')^{1/r_i} \cdot \left( \frac{1}{\sharp i'} \sum _l (v_l)^{\frac{\gamma _i}{1-\gamma _i}}\right) ^{\frac{1-\gamma _i}{\gamma _i}}$.

We now show that for every group i, there exists a function $p_{i'}(Q)$ such that at $p_{i'}(Q)=\frac{q_{i'}}{Q}$, the optimality condition is fulfilled. $p_{i'}(Q)$ corresponds to a share function of a single player contest (Cornes and Hartley 2005), only with the change of interpretation that it is the share of the whole group on which the within-group equilibrium condition (H.3) has been imposed.

Notice that if $r_i < 1$ we have that $q_{i'}=0$ can never be a best response to any positive $Q_{-i}$, since then the LHS of the above optimality condition (H.4) is infinite. We rewrite the optimality condition therefore in terms of winning probabilities $p_{i'} = q_{i'}/Q$:

$$\begin{aligned} V_i \cdot (1-p_{i'}) = Q^{1/r_i} \cdot p_{i'}^{1/r_i - 1} \end{aligned}$$

(H.6)

It is now easy to see that for all Q there exists a $p_{i'}(Q)$ such that the optimality condition is fulfilled: Notice that for $p_{i'}=1$, the LHS is strictly smaller than the RHS, while for $p_{i'}=0$, the RHS is strictly smaller than the LHS. Since both are continuous functions of $p_{i'}$, by the intermediate value theorem there then exists at least one $p_{i'}(Q)$ such that the optimality condition holds with equality. Further, this point is unique, since the LHS is strictly decreasing in $p_{i'}$, while the RHS is strictly increasing in $p_{i'}$. This proves that there exists a unique best response $p_{i'}(Q)$ to any level of Q. Moreover, $p_{i'}(Q)$ is strictly decreasing in Q as can be verified from the following argument: Suppose Q increases, then the RHS is larger than the LHS of the optimality condition. Since the RHS is strictly increasing in $p_{i'}$ and the LHS strictly decreasing, $p_{i'}$ must decrease in order to maintain equality.

Suppose $r_i=1$, then the optimality condition becomes

$$\begin{aligned} Q_{-i} \cdot V_i \le Q^2 \end{aligned}$$

(H.7)

We can therefore directly solve for the best response winning probability:

$$\begin{aligned} p_{i'}(Q) = \max [0,1-Q/V_i] \end{aligned}$$

(H.8)

which has the properties $p_{i'}(0)=1$ and $p_{i'}(V_i)=0$ and is obviously weakly decreasing in Q.

A Nash equilibrium is now given by a ${Q'}^*$ such that $\sum _{i'} p_{i'}({Q'}^*) =1$. Without loss of generality, assume that the groups are ordered such that first all groups with $r_i<1$ appear in any order and then the groups with $r_i=1$ appear with $V_{i} \ge V_{i+1}$. Notice that for $Q=0$, we have $p_{i'}(0)=1$ for all groups, while for $Q\rightarrow \infty$, we have that $p_{i'}(\infty )\rightarrow 0$. Therefore, $\sum _{i'} p_{i'}(0)> 1 > \sum _{i'} p_{i'}(\infty )$ and a solution exists. For uniqueness, We distinguish two cases. In the first case, at least one group has $r_i<1$ and therefore $\sum _{i'} p_{i'}(Q)$ is strictly decreasing for all $Q\in{\mathbb{R}}$ since $p_1$ is strictly decreasing and all other $p_{i'}$ are at least weakly decreasing. In the second case, all groups have $r_i=1$. In this case, $p_1$ is strictly decreasing on $[0,V_1]$ and $p_2$ is strictly decreasing on $[0,V_2]$. Since for all $i\ge 2$, $p_{i'}(V_2) = 0$, we have that $\sum _{i'} p_{i'}(V_2)<1$. Therefore, all solutions must lie in the interval $[0,V_2]$ in which $\sum _{i'} p_{i'}(Q)$ is strictly decreasing and therefore a unique ${Q'}^*$ exists. Define $n^*$ as the highest index of a group with $r_i=1$ such that $0> 1-{Q'}^*/V_i$, or if no such group exists as the number of groups with $r_i<1$. This completes the proof. $\square$

Appendix I: Proof of Theorem 6

Proof

The proof goes by contradiction. Suppose we have that $V_i\ge V_{i''}$ and $p_{i'}^* < p_{i''}^*$. Then it follows that $V_i\cdot (1-p_{i'}^*)>V_{i''}\cdot (1-p_{i''}^*)$. By (9) this is equivalent to:

$$\begin{aligned} ({Q'}^*)^{1/r_i}\cdot (p_{i'}^*)^{1/r_i-1} > ({Q''}^*)^{1/r_i}\cdot (p_{i''}^*)^{1/r_i-1} \end{aligned}$$

and thus ${Q'}^*>{Q''}^*$.

Since $p_{i'}^* < p_{i''}^*$ there must exist at least one group j such that: $p_{j'}^* >p_{j''}^*$. Therefore, $V_j\cdot (1-p_{j'}^*)>V_j\cdot (1-p_{j''}^*)$, since $V_j$ does not differ between both equilibria. Using (9) for group j gives us:

$$\begin{aligned} ({Q'}^*)^{1/r_j}\cdot (p_{j'}^*)^{1/r_j-1} < ({Q''}^*)^{1/r_j}\cdot (p_{j''}^*)^{1/r_j-1}. \end{aligned}$$

and thus ${Q'}^*<{Q''}^*$ which yields a contradiction. By an analogous proof for $V_{i'}\le V_{i''}$ and $p_{i'}^* > p_{i''}^*$ then follows the theorem. $\square$

Appendix J: Proof of Proposition 3

Proof

From Theorem 6, we have that:

$$\begin{aligned} p_{i'}^* \gtreqless p_{i''}^* \quad \Leftrightarrow \quad r_i \zeta _i(\sharp i') \cdot{{\mathcal{M}}}(\vec{v}_{i'}(i'),{\textstyle \frac{\gamma _i}{1-\gamma _i}}) \gtreqless r_i \zeta _i(\sharp i'') \cdot{{\mathcal{M}}}(\vec{v}_{i''}(i''),{\textstyle \frac{\gamma _i}{1-\gamma _i}}). \end{aligned}$$

(J.1)

Rearranging terms and using $S_i(i',i'') = \zeta _i(i'')/\zeta (i')$ and writing out ${{\mathcal{M}}}(\vec{v}_{i,i''}(i''), \frac{\gamma _i}{1-\gamma _i})$ gives us:

$$\begin{aligned} p_{i'}^* \gtreqless p_{i''}^* \quad \Leftrightarrow \quad \frac{{{\mathcal{M}}}(\vec{v}_{i'}(i'),\frac{\gamma _i}{1-\gamma _i})}{\left( \frac{\sum _{k\in i'} \left( v_k(i'')\right) ^{\frac{\gamma _i}{1-\gamma _i}}+\sum _{k\in i'''}\left( v_k(i'')\right) ^{\frac{\gamma _i}{1-\gamma _i}}}{\sharp i''} \right) ^{\frac{1-\gamma _i}{\gamma _i}}} \gtreqless S_i(i',i'')^{1/r_i}, \end{aligned}$$

(J.2)

which yields:

$$\begin{aligned} p_{i'}^*&\gtreqless p_{i''}^* \; \Leftrightarrow \; \frac{{{\mathcal{M}}}(\vec{v}_{i'}(i'),\frac{\gamma _i}{1-\gamma _i})}{\left( \frac{\sharp i'}{\sharp i''}\frac{1}{ \sharp i'}\underset{k\in i'}{\sum }\left( v_k(i'')\right) ^{\frac{\gamma _i}{1-\gamma _i}} +(1-\frac{\sharp i'}{\sharp i''}) \frac{1}{(\sharp i''-\sharp i')}\underset{k\in i'''}{\sum }\left( v_k(i'')\right) ^{\frac{\gamma _i}{1-\gamma _i}} \right) ^{\frac{1-\gamma _i}{\gamma _i}}} \nonumber \\&\gtreqless S_i(i',i'')^{1/r_i} \end{aligned}$$

(J.3)

from which follows (10), since $\sharp i''-\sharp i' = \sharp i'''$ . $\square$

Appendix K: Proof of Proposition 4

Proof

Part (a) follows from using the definitions of $R_i$ and $v_k$ in (10):

$$\begin{aligned} R_i(i',i'') \frac{\left( \frac{\sum _{k\in i'} (w_k)^{\frac{\gamma _i}{1-\gamma _i}}}{\sharp i'}\right) ^{\frac{1-\gamma _i}{\gamma _i}}}{\left( \frac{\sum _{k\in i'} (w_k)^{\frac{\gamma _i}{1-\gamma _i}} + (\sharp i''/\sharp i'-1)\cdot i' \cdot (w_i^{\mu })^{\frac{\gamma _i}{1-\gamma _i}}}{\sharp i''}\right) ^{\frac{1-\gamma _i}{\gamma _i}}} = S_i(i', i'')^{1/r_i} \end{aligned}$$

(K.1)

Employing the definition of ${{\mathcal{M}}}(\vec{w}_{i,i'},{\textstyle \frac{\gamma _i}{1-\gamma _i}})$ gives us:

$$\begin{aligned} R_i(i',i'') \frac{{{\mathcal{M}}}(\vec{w}_{i'},{\frac{\gamma _i}{1-\gamma _i}})}{\left( \frac{\sharp i'}{\sharp i''}{{\mathcal{M}}}(\vec{w}_{i'},{\frac{\gamma _i}{1-\gamma _i}})^{\frac{\gamma _i}{1-\gamma _i}} +\frac{\sharp i'''}{\sharp i''}{{\mathcal{M}}}(\vec{w}_{i'''},{\frac{\gamma _i}{1-\gamma _i}})^{\frac{\gamma _i}{1-\gamma _i}}\right) ^{\frac{1-\gamma _i}{\gamma _i}}} = S_i(i', i'')^{1/r_i} \end{aligned}$$

(K.2)

which after solving for ${{\mathcal{M}}}(\vec{w}_{i'''},{\textstyle \frac{\gamma _i}{1-\gamma _i}})$ yields the desired condition.

Part (a) follows from ${{\mathcal{M}}}(\vec{w}_{i,i'},{\textstyle \frac{\gamma _i}{1-\gamma _i}})$ being a power mean with $\frac{\gamma _i}{1-\gamma _i}$ as the exponent. Since any power mean is weakly increasing in its exponent, and the exponent is increasing in $\gamma _i$, ${{\mathcal{M}}}(\vec{w}_{i,i'},{\textstyle \frac{\gamma _i}{1-\gamma _i}})$ is increasing in $\gamma _i$. Furthermore, any power mean is strictly increasing in its exponent, whenever there exist two unequal elements over which the mean is formed. Inserting $R_i(i',i'')$ into (10) and simplifying gives the desired result.

(b) Follows directly from Theorem 4 and the assumption that $\vec{w}_{i'''}$ is more asymmetric than $\vec{w}_{i'}$ at $\gamma _i/(1-\gamma _i)$. Note that we only consider cases with $\gamma _i<1$ and thus $\gamma _i/(1-\gamma _i)<\infty$.

(c) It is immediately obvious that $\Gamma (\gamma _i,i',i',R_i,S_i,r_i)$ is increasing in $R_i$ and decreasing in $s_i$. The only difficulty is thus the proof of the behavior of $\Gamma (\gamma _i,i',i',R_i,S_i,r_i)$ with changes in $\gamma _i$. A useful result on which the proof is based is the reverse Jensen inequality (for a more general version and its proof, see Bullen 2003, p. 43):

Lemma 4

If f is convex, $\varrho _1 > 0$ and $\varrho _i < 0$ for all $2 \ge i \ge n$ and $\sum _{j=1}^n \varrho _j = 1$, then $f(\sum _{j=1}^n \varrho _j a_j) \ge \sum _{j=1}^n \varrho _j f(a_j)$ for all $\sum _{j=1}^n \varrho _j a_j \in I$, and the inequality holds strictly, if f is strictly convex and $\exists i,j: a_i \ne a_j$.

We now show that the following term is decreasing in $\gamma _i$:

$$\begin{aligned} \Gamma (\gamma _i,i',i'',R_i,s_i,r_i)=\left( \frac{i''}{i''-i'} \cdot \left( \frac{R_i(i',i'')}{S_i(i',i'')^{1/r_i}}\right) ^{\frac{\gamma _i}{1-\gamma _i}}- \frac{i'}{i''-i'} \right) ^{\frac{1-\gamma _i}{\gamma _i}}. \end{aligned}$$

(K.3)

Define $\theta _i = \frac{\gamma _i}{1-\gamma _i}$, which is increasing in $\gamma _i$. What is to be shown is that the above term is decreasing in $\theta _i$, thus:

$$\begin{aligned} \left( \frac{i''}{i''-i'} \Psi ^{\theta _i}- \frac{i'}{i''-i'} \right) ^{1/\theta _i} > \left( \frac{i''}{i''-i'} \Psi ^{\phi _i}- \frac{i'}{i''-i'} \right) ^{1/\phi _i} \end{aligned}$$

(K.4)

whenever $\phi _i > \theta _i$ and where $\Psi = \frac{R_i(i',i')}{S_i(i',i'')^{1/r_i}}$. Note that $\phi _i,\theta _i \in ( -1 , \infty )$ and therefore we will distinguish the cases $0 < \phi _i$, and $\phi _i < 0$. We will furthermore assume that $\theta _i$ and $\phi _i$ are not zero (which is equivalent to assuming that $\gamma _i \ne 0$.

$0 < \phi _i$: Since $\phi _i$ is positive, we can rewrite condition (K.4) to:

$$\begin{aligned} \left( \frac{i''}{i''-i'} \Psi ^{\theta _i}- \frac{i'}{i''-i'} \left( 1^{\theta _i}\right) \right) ^{\phi _i/\theta _i} > \left( \frac{i''}{i''-i'} \left( \Psi ^{\theta _i}\right) ^{\phi _i/\theta _i}- \frac{i'}{i''-i'} \left( 1^{\theta _i}\right) ^{\phi _i/\theta _i} \right) . \end{aligned}$$

(K.5)

Setting $f(a)=a^{\phi _i/\theta _i}$ (which is strictly convex also for negative $\theta _i$) and $\varrho _1 = \frac{i''}{i''-i'}$, $\varrho _2 = - \frac{i'}{i''-i'}$, and $a_1 = (\Psi )^{\theta _i}$, $a_2 = 1^{\theta _i}$ in the above reverse Jensen inequality directly yields equation (K.5).

$\phi _i < 0$: Since $\theta _i$ is negative, we can rewrite condition (K.4) to:

$$\begin{aligned} \left( \frac{i''}{i''-i'} \Psi ^{\phi _i}- \frac{i'}{i''-i'} \left( 1^{\phi _i}\right) \right) ^{\theta _i/\phi _i} > \left( \frac{i''}{i''-i'} \left( \Psi ^{\phi _i}\right) ^{\theta _i/\phi _i}- \frac{i'}{i''-i'} \left( 1^{\phi _i}\right) ^{\theta _i/\phi _i} \right) . \end{aligned}$$

(K.6)

Setting $f(a) = a^{\theta _i / \phi _i}$ (which is strictly convex since the absolute value of $\phi _i$ is smaller than that of $\theta _i$) and $\varrho _1 =\frac{i''}{i''-i'}$, $\varrho _2 = - \frac{i'}{i''-i'}$, and $a_1 = (\Psi )^{\phi _i}$, $a_2 = 1^{\phi _i}$ in the above reverse Jensen inequality directly yields equation (K.6). Therefore, condition (K.4) follows, which concludes the proof of part c) of the proposition. $\square$