Heterogeneity and Number of Players in Rent-Seeking, Innovation, and Patent-Race Games

Abstract

Many economists have studied rent-seeking contests, innovation tournaments, and patent-race games independently. These three seemingly different games are known to be strategically equivalent under some reasonable assumptions. In these classic games, it is assumed that the value of a prize, i.e. the gain from rent-seeking, achieving an innovation, or obtaining a patent, is exogenously given and does not depend on the number of players, so that an increase in the number of players decreases the winning rate of each player. However, if players engage in R&D and then set quantities à la Cournot, the value of the prize in general depends on the number of players. In this paper, we set up a model with one efficient player and identical inefficient players to analyze how an increase in heterogeneity among players or the number of players changes the wining rate of the efficient player. One of the main results is that if the number of players is larger than some critical value, which can be less than two, an increase in the number of inefficient players always increases the winning rate of the efficient player.

Appendix

Proof of Lemma 1

Define $A \equiv 2(a - c) - 2\delta + (n - 1)k\delta$ and $B \equiv 2(a - c) + 2n\delta + (n - 1)k\delta$. Simple calculations lead to

$$\displaystyle\begin{array}{rcl} \left.\frac{\partial \theta (\delta,k,n)} {\partial \delta } \right \vert _{\varDelta c=k\delta =0}& =& - \frac{2} {B^{2}}(nA + B) <0, {}\\ \frac{\partial \theta \left (\delta,k,n\right )} {\partial k} & =& \frac{2\left (n^{2} - 1\right )\delta ^{2}} {B^{2}} \geq 0, {}\\ \frac{\partial \theta \left (\delta,k,n\right )} {\partial n} & =& -\frac{4\delta \left (\left (a - c\right ) -\delta \left (1 + k\right )\right )} {B^{2}} \leq 0. {}\\ \end{array}$$

The last inequality holds from (7 $^{{\prime}}$).  □ 

Proof of Lemma 2

The relation (16) implies

$$\displaystyle{\left ( \frac{y} {y_{1}}\right )^{R-1} = \frac{Y _{-i}} {Y _{-1}}\theta = \frac{y_{1} + \left (n - 2\right )y} {\left (n - 1\right )y} \theta> \frac{n - 2} {n - 1}\theta,}$$

note that $\varDelta \pi /\varDelta \pi _{1} =\theta$. Because of this inequality and R ≥ 1, the partial derivative in (20) is positive for any n ≥ 2.

As already mentioned, ϕ(1, R, n) = 1 for any R ≥ 1 and

$$\displaystyle{ \frac{y} {y_{1}} =\phi \left (\theta,1,n\right ) = \frac{\theta } {\left (n - 1\right ) -\left (n - 2\right )\theta }.}$$

Since

$$\displaystyle{\frac{\partial } {\partial \theta }\phi \left (\theta,1,n\right ) = \frac{n - 1} {\left \{\left (n - 1\right ) -\left (n - 2\right )\theta \right \}^{2}}> 0,}$$

for $\theta <1$,

$$\displaystyle{ \frac{y} {y_{1}} =\phi \left (\theta,1,n\right ) = \frac{\theta } {\left (n - 1\right ) -\left (n - 2\right )\theta } <\frac{1} {\left (n - 1\right ) -\left (n - 2\right )} = 1.}$$

Since $\phi (\theta,R,n)$ is continuous in R for R > 0, $\phi (\theta,R,n) <1$ for any R close enough to 1. By (21), if $\phi (\theta,R,n) <1$ for some R ₀ which is close to 1, $\frac{\partial \phi (\theta,R,n)} {\partial R}> 0$ at R = R ₀ because y∕y ₁ > 0 and $\ln (y/y_{1}) <0$. Hence, as long as $\phi (\theta,R,n) <1$, an increase in R increases $\phi (\theta,R,n)$. If $\phi (\theta,R,n)$ reaches to 1 at some R = R ₁, that is, $y/y_{1} = 1$ at R = R ₁. From (21), $\partial \phi /\partial R = 0$, thus $\phi (\theta,R,n) = 1$ for any R ≥ R ₁. Therefore, in any case, $\phi (\theta,R,n) \leq 1$ and $\frac{\partial \phi (\theta,R,n)} {\partial R} \geq 0$ for any R ≥ 1.

From (22), $\frac{\partial \phi (\theta,R,n)} {\partial n} \leq 0$ for any R ≥ 1 and n ≥ 2.  □ 

Proof of Lemma 3

Assumption (7 $^{{\prime}}$) implies

$$\displaystyle{ \frac{\partial } {\partial n}\theta \left (\delta,k,n\right ) = \frac{-4\delta \left \{\left (a - c\right ) -\left (1 + k\right )\delta \right \}} {\left \{2\left (a - c\right ) - 2\delta + \left (n - 1\right )k\delta \right \}^{2}} <0.}$$

This fact together with Lemma 2 implies

$$\displaystyle{ \frac{d\phi } {dn} = \frac{\partial \phi } {\partial \theta } \frac{\partial \theta } {\partial n} + \frac{\partial \phi } {\partial n} <0.}$$

This proves Lemma 3.  □