Two-stage elimination contests with optimal head starts

Abstract

We study two-stage elimination Tullock contests. In the first stage all the players compete against each other; then some advance to the second stage while the others are removed. The finalists compete against each other in the second stage, and one of them wins the prize. To maximize the expected total effort, the designer can give a head start to the winner of the first stage when he competes against the other finalists in the second stage. We show that the optimal head start, independent of the number of finalists, always increases the players’ expected total effort. We also show how the number of players and finalists affect the value of the optimal head start.

Appendix

1.1 Proof of Proposition 1

The F.O.C. of the maximization problem (5) is

$$\begin{aligned}&\frac{\alpha ^{2}}{(1+\alpha )^{2}}\frac{ {\displaystyle \sum \nolimits _{\begin{array}{c} j=1 \\ j\ne i \end{array}}^{n}} x_{j}}{\left( {\displaystyle \sum \nolimits _{j=1}^{n}} x_{j}\right) ^{2}}-\frac{1}{(1+\alpha )^{2}} {\displaystyle \sum \limits _{\begin{array}{c} k=1 \\ k\ne i \end{array}}^{n}} \frac{x_{k}}{\left( {\displaystyle \sum \nolimits _{j=1}^{n}} x_{j}\right) ^{2}}\frac{x_{i}}{ {\displaystyle \sum \nolimits _{\begin{array}{c} j=1 \\ j\ne k \end{array}}^{n}} x_{j}}\\&\quad \quad +\,\frac{1}{(1+\alpha )^{2}} {\displaystyle \sum \limits _{\begin{array}{c} k=1 \\ j\ne i \end{array}}^{n}} \frac{x_{k}}{ {\displaystyle \sum \nolimits _{j=1}^{n}} x_{j}}\frac{ {\displaystyle \sum \nolimits _{\begin{array}{c} j=1 \\ j\ne k,i \end{array}}^{n}} x_{j}}{\left( {\displaystyle \sum \nolimits _{\begin{array}{c} j=1 \\ j\ne k \end{array}}^{n}} x_{j}\right) ^{2}}\\&\quad =1 \end{aligned}$$

By symmetry, we obtain

$$\begin{aligned} \frac{\alpha ^{2}}{(1+\alpha )^{2}}\frac{n-1}{n^{2}x}+\frac{1}{(1+\alpha )^{2} }\frac{n^{2}-3n+1}{n^{2}(n-1)x}=1 \end{aligned}$$

Thus, the subgame perfect equilibrium strategy in the first stage is

$$\begin{aligned} x=\frac{\alpha ^{2}(n^{2}-2n+1)+(n^{2}-3n+1)}{(1+\alpha )^{2}n^{2}(n-1)}. \end{aligned}$$

1.2 Proof of Proposition 3

By symmetry, the F.O.C. of the maximization problem (12) yields

$$\begin{aligned}&\frac{(\alpha (k-1)-k+2))^{2}}{(\alpha (k-1)+1)^{2}}\frac{n-1}{n^{2}x} +\frac{1}{(\alpha (k-1)+1)^{2}} {\displaystyle \sum \limits _{s=2}^{k}}\\&\quad \quad \times \left[ x^{s-1}\frac{(n-1)!}{(n-s)!}\frac{(n-s+1)!}{n!x^{s-1}}\frac{n-s}{(n-s+1)^{2}x}\right] \\&\quad \quad -\frac{1}{(\alpha (k-1)+1)^{2}} {\displaystyle \sum \limits _{s=2}^{k}} \left[ x^{s-1}\frac{(n-1)!}{(n-s)!} {\displaystyle \sum \limits _{i=2}^{s}} \frac{1}{(n-i+2)x}\frac{(n-s+1)!}{n!x^{s-1}}\frac{1}{(n-s+1)}\right] \\&\quad =\frac{(\alpha (k-1)-k+2))^{2}}{(\alpha (k-1)+1)^{2}}\frac{n-1}{n^{2}x} +\frac{1}{(\alpha (k-1)+1)^{2}} {\displaystyle \sum \limits _{s=2}^{k}}\\&\quad \quad \times \left( \frac{n-s}{n(n-s+1)x}- {\displaystyle \sum \limits _{i=2}^{s}} \frac{1}{n(n-i+2)x}\right) =1 \end{aligned}$$

Thus, the subgame perfect equilibrium strategy in the first stage is

$$\begin{aligned} x= & {} \frac{(\alpha (k-1)-k+2))^{2}}{(\alpha (k-1)+1)^{2}}\frac{n-1}{n^{2}}+\frac{1}{(\alpha (k-1)+1)^{2}} {\displaystyle \sum \limits _{s=2}^{k}} \\&\times \left( \frac{n-s}{n(n-s+1)}- {\displaystyle \sum \limits _{i=2}^{s}} \frac{1}{n(n-i+2)}\right) . \end{aligned}$$

1.3 Proof of Proposition 4

The F.O.C. of the maximization of the total effort given by (13) is

$$\begin{aligned} \frac{dTE}{d\alpha }&= \frac{-2(k-1)}{\left( k\alpha ^{*}-\alpha ^{*}+1\right) ^{3}}\left( \frac{(k-1)(n-1)(k-2-\alpha ^{*}(1-k))}{n}\right. \\&\quad +\,\left. \left( {\displaystyle \sum \limits _{s=2}^{k}} \left( \frac{n-s}{(n-s+1)}- {\displaystyle \sum \limits _{i=2}^{s}} \frac{1}{(n-i+2)}\right) \right) +(\alpha ^{*}-1)(k-1)^{2}\right) =0 \end{aligned}$$

By some calculations we have

$$\begin{aligned} \alpha ^{*}&=\frac{-n}{(k-1)^{2}}\left( {\displaystyle \sum \limits _{s=2}^{k}} \left( \frac{n-s}{(n-s+1)}- {\displaystyle \sum \limits _{i=2}^{s}} \frac{1}{(n-i+2)}\right) \right) +\frac{n+k-2}{(k-1)}\\&=\frac{-n}{(k-1)^{2}}\left( (k-1)+ {\displaystyle \sum \limits _{s=2}^{k}} \left( \frac{-1}{(n-s+1)}- {\displaystyle \sum \limits _{i=2}^{s}} \frac{1}{(n-i+2)}\right) \right) +\frac{n+k-2}{(k-1)}\\&=\frac{n}{(k-1)^{2}}\left( {\displaystyle \sum \limits _{s=3}^{k+1}} \left( \frac{1}{(n-s+2)}+ {\displaystyle \sum \limits _{i=2}^{s}} \frac{1}{(n-i+2)}\right) \right) +\frac{k-2}{k-1}\\&=\frac{1}{(k-1)^{2}}\left( -k+n {\displaystyle \sum \limits _{s=2}^{k+1}} {\displaystyle \sum \limits _{i=2}^{s}} \frac{1}{(n-i+2)}\right) +1\\&=\frac{1}{(k-1)^{2}}\left( -k+n {\displaystyle \sum \limits _{i=2}^{k+1}} \frac{k-i+2}{(n-i+2)}\right) +1\\&=\frac{1}{(k-1)^{2}} {\displaystyle \sum \limits _{i=2}^{k+1}} \left( n\frac{k-i+2}{n-i+2}-1\right) +1 \end{aligned}$$

When $\alpha$ approaches one we obtain that

$$\begin{aligned} \lim _{\alpha \rightarrow 1}\frac{dTE}{d\alpha }=2\frac{k-1}{k^{3}}\left( \frac{(k-1)(n-1)}{n}- {\displaystyle \sum \limits _{s=2}^{k}} \left( \frac{n-s}{(n-s+1)}- {\displaystyle \sum \limits _{i=2}^{s}} \frac{1}{(n-i+2)}\right) \right) \end{aligned}$$

Since

$$\begin{aligned} {\displaystyle \sum \limits _{s=2}^{k}} \frac{n-s}{(n-s+1)}\le (k-1)\frac{n-2}{n-1} \end{aligned}$$

We obtain,

$$\begin{aligned} \lim _{\alpha \rightarrow 1}\frac{dTE}{d\alpha }\ge 2\frac{(k-1)^{2}}{k^{3}} \left( \frac{n-1}{n}-\frac{n-2}{n-1}\right) >0 \end{aligned}$$

Therefore, the head start $\alpha >1$, independent of the number of the finalists k, increases the players’ total effort.