Optimal allocations of prizes and punishments in Tullock contests

Abstract

We study Tullock contests with n symmetric players. We show that in a contest without an exit option, if prizes and punishments (negative prizes) have the same cost, it is optimal for the designer who wants to maximize the players’ total effort to allocate the entire prize sum to a single punishment without any prize. On the other hand, in a contest with an exit option, it is optimal to allocate the entire prize sum to a single prize and a single punishment, where independent of the costs of the prize and the punishment, the optimal value of the prize is larger than the optimal value of the punishment. We also show that allocating a prize and a punishment in a two-stage contest yields a higher expected total effort than in a one-stage contest.

Appendices

Appendix

1.1 Proof of Proposition 1

The F.O.C. of (1) and (2) yields

$$\begin{aligned}&v\sum \limits _{i=1}^{k}\frac{y^{i-1}\left( \frac{n-1)!}{(n-i)!}\right) \Pi _{j=1}^{i}\left( (n-j)y+x\right) }{\left[ \Pi _{j=1}^{i}\left( (n-j)y+x\right) \right] ^{2}} \nonumber \\&\qquad -\frac{\Pi _{j=1}^{i}\left( (n-j)y+x\right) \Sigma _{j=1}^{i}\frac{1}{ \left( (n-j)y+x\right) }y^{i-1}x\left( \frac{n-1)!}{(n-i)!}\right) }{\left[ \Pi _{j=1}^{i}\left( (n-j)y+x\right) \right] ^{2}} \nonumber \\&\qquad -p\left( \frac{-\Pi _{i=1}^{n-1}(iy+x)\Sigma _{i=1}^{n-1}\frac{1}{(iy+x)} y^{n-1}(n-1)!}{\left( \Pi _{i=1}^{n-1}(iy+x)\right) ^{2}}\right) \nonumber \\&\qquad -p\left( \sum \limits _{i=2}^{l}\frac{y^{n-i}(\frac{(n-1)!}{(i-1)!})\Pi _{j=i-1}^{n-1}(jy+x)-\Pi _{j=i-1}^{n-1}(jy+x)\Sigma _{j=i-1}^{n-1}\frac{1}{ jy+x}y^{n-i}x\left( \frac{(n-1)!}{(i-1)!}\right) }{\left( \Pi _{j=i-1}^{n-1}(iy+x)\right) ^{2}}\right) \nonumber \\&\quad =v\sum \limits _{i=1}^{k}\frac{y^{i-1}\left( \frac{n-1)!}{(n-i)!}\right) -\Sigma _{j=1}^{i}\frac{1}{\left( (n-j)y+x\right) }y^{i-1}x\left( \frac{n-1)! }{(n-i)!}\right) }{\Pi _{j=1}^{i}\left( (n-j)y+x\right) } \nonumber \\&\qquad +p\left( \frac{\Sigma _{i=1}^{n-1}\frac{1}{(iy+x)}y^{n-1}(n-1)!}{\Pi _{i=1}^{n-1}(iy+x)}\right) -p\left( \sum \limits _{i=2}^{l}\frac{y^{n-i}( \frac{(n-1)!}{(i-1)!})-\Sigma _{j=i-1}^{n-1}\frac{1}{jy+x}y^{n-i}x\left( \frac{ (n-1)!}{(i-1)!}\right) }{\Pi _{j=i-1}^{n-1}(iy+x)}\right) \nonumber \\&\quad =1. \end{aligned}$$

(17)

By symmetry, $x=y,$ and then we obtain

$$\begin{aligned}&v\sum \limits _{i=1}^{k}\frac{x^{i-1}\left( \frac{n-1)!}{(n-i)!}\right) -\Sigma _{j=1}^{i}\frac{1}{(n-j+1)x}x^{i}\left( \frac{n-1)!}{(n-i)!}\right) }{\Pi _{j=1}^{i}\left( (n-j+1)x\right) } \nonumber \\&\quad =v\sum \limits _{i=1}^{k}\frac{x^{i-1}\left( \frac{n-1)!}{(n-i)!}\right) - \frac{1}{x}\Sigma _{j=1}^{i}\frac{1}{(n-j+1)}x^{i}\left( \frac{n-1)!}{(n-i)!} \right) }{\frac{n!}{(n-i)!}x^{i}} \nonumber \\&\quad =v\sum \limits _{i=1}^{k}\frac{x^{i-1}(n-1)!-\frac{1}{x}\Sigma _{j=1}^{i} \frac{1}{(n-j+1)}x^{i}(n-1)!}{n!x^{i}} \nonumber \\&\quad =v\sum \limits _{i=1}^{k}\frac{x^{i}x^{-1}(n-1)!-x^{-1}\Sigma _{j=1}^{i} \frac{1}{(n-j+1)}x^{i}(n-1)!}{n(n-1)!x^{i}} \nonumber \\&\quad =v\sum \limits _{i=1}^{k}\frac{1-\Sigma _{j=1}^{i}\frac{1}{(n-j+1)}}{nx} =v\sum \limits _{i=1}^{k}\frac{1-(H_{n}-H_{n-i})}{nx}, \end{aligned}$$

(18)

where $H_{n}=\sum \nolimits _{i=1}^{n}\frac{1}{i}$ is the harmonic series. Similarly, by symmetry, $x=y,$ and then

$$\begin{aligned}&p\left( \frac{\Sigma _{i=1}^{n-1}\frac{1}{(i+1)x}x^{n-1}(n-1)!}{\Pi _{i=1}^{n-1}(i+1)x}\right) -p\sum \limits _{i=2}^{l}\frac{x^{n-i}\left( \frac{(n-1)! }{(i-1)!}\right) -\Sigma _{j=i-1}^{n-1}\frac{1}{(j+1)x}x^{n-i}x\left( \frac{(n-1)!}{(i-1)! }\right) }{\Pi _{j=i-1}^{n-1}(i+1)x} \nonumber \\&\quad =p\left( \frac{\frac{1}{x}\Sigma _{i=2}^{n}\frac{1}{i}x^{n-1}(n-1)!}{ n(n-1)!x^{n-1}}\right) -p\left( \sum \limits _{i=2}^{l}\frac{x^{n-i}\left( \frac{ (n-1)!}{(i-1)!}\right) -\frac{1}{x}\Sigma _{j=i}^{n}\frac{1}{j}x^{n-i}x\left( \frac{(n-1)! }{(i-1)!}\right) }{\frac{n!}{(i-1)!}x^{n-i+1}}\right) . \end{aligned}$$

This implies that

$$\begin{aligned}&p\left( \frac{H_{n}-1}{nx}\right) -p\left( \sum \limits _{i=2}^{l}\frac{ \frac{(n-1)!}{(i-1)!}-\Sigma _{j=i}^{n}\frac{1}{j}\frac{(n-1)!}{(i-1)!}}{ \frac{n(n-1)!}{(i-1)!}x}\right) \nonumber \\&\quad =p\left( \frac{H_{n}-1}{nx}\right) -p\left( \sum \limits _{i=2}^{l}\frac{ 1-(H_{n}-H_{i-1})}{nx}\right) . \end{aligned}$$

(19)

Inserting (18) and (19) in (17) yields

$$\begin{aligned} v\sum \limits _{i=1}^{k}\frac{1-(H_{n}-H_{n-1})}{nx}+p\left( \frac{H_{n}-1}{nx }\right) -p\left( \sum \limits _{i=2}^{l}\frac{1-(H_{n}-H_{i-1})}{nx}\right) =1. \end{aligned}$$

Hence, the players’ symmetric equilibrium effort in a contest with k identical prizes of value v and l identical punishments of value p is

$$\begin{aligned} x=v\sum \limits _{i=1}^{k}\frac{1-(H_{n}-H_{n-1})}{n}+p\left( \frac{ (H_{n}-1)+\Sigma _{i=2}^{l}(H_{n}-H_{i-1}-1)}{n}\right) . \end{aligned}$$

Q.E.D.

Proof of Proposition 2

By (3), the symmetric equilibrium effort with k identical prizes of value v and without punishments is

$$\begin{aligned} x=v\sum \limits _{i=1}^{k}\frac{1-(H_{n}-H_{n-i})}{n}, \end{aligned}$$

and the symmetric equilibrium effort with l identical punishments of value v and without prizes is

$$\begin{aligned} x=v\left( \frac{(H_{n}-1)+\Sigma _{i=2}^{l}(H_{n}-H_{i-1})-1}{n}\right) . \end{aligned}$$

Thus, in order to show the equivalence between a contest with k identical prizes and a contest with $n-k$ identical punishments we need to show that

$$\begin{aligned} \sum \limits _{i=1}^{k}1-(H_{n}-H_{i-1})=(H_{n}-1)-\sum \limits _{i=2}^{n-k}\left( (H_{n}-H_{i-1})-1\right) , \end{aligned}$$

or, alternatively, that

$$\begin{aligned} H_{n}+\sum \limits _{i=2}^{k}(H_{n}-H_{i-1})=n-\sum \limits _{i=1}^{n-k}(H_{n}-H_{n-i}). \end{aligned}$$

This equation is equivalent to

$$\begin{aligned} n-H_{n}= & {} \sum \limits _{i=2}^{k}(H_{n}-H_{i-1})+\sum \limits _{i=1}^{n-k}\left( H_{n}-H_{n-i}\right) \\= & {} (n-1)H_{n}-\left( \sum \limits _{i=2}^{k}H_{i-1}+\sum \limits _{i=1}^{n-k}H_{n-i}\right) \\= & {} (n-1)H_{n}-\left( \sum \limits _{i=1}^{k-1}H_{i}+\sum \limits _{i=1}^{n-k}H_{n-i}\right) . \end{aligned}$$

Using the identity

$$\begin{aligned} \sum \limits _{i=1}^{k-1}H_{i}+\sum \limits _{i=1}^{n-k}H_{n-i}=\sum \limits _{i=1}^{n-1}H_{i}, \end{aligned}$$

we obtain that we need to show that

$$\begin{aligned} n=nH_{n}-\sum \limits _{i=1}^{n-1}H_{i}. \end{aligned}$$

Since

$$\begin{aligned} \sum \limits _{i=1}^{n-1}H_{i}=(n-1)1+(n-2)\frac{1}{2}+\cdots +(n-(n-1))\frac{1}{ n-1}, \end{aligned}$$

we indeed obtain that

$$\begin{aligned} nH_{n}-\sum \limits _{i=1}^{n-1}H_{i}= & {} \left( n+\frac{n}{2}+\frac{n}{3}+\cdots +\frac{ n}{n-1}+\frac{n}{n}\right) -\left( \frac{n-1}{1}+\frac{n-2}{2}+\cdots +\frac{1}{n-1}\right) \\= & {} (n-1)1+\frac{n}{n}=n. \end{aligned}$$

Q.E.D.

1.1 Proof of Proposition 3

We first assume that $k<\frac{n}{2}$ (a similar proof holds when $k>\frac{n}{ 2}$) and show that the symmetric equilibrium effort in a Tullock contest with $k<\frac{n}{2}$ punishments of value $-v$ is higher than in a Tullock contest with k prizes of value v. By (3), we need to show that

$$\begin{aligned} v\left( \frac{(H_{n}-1)+\Sigma _{i=2}^{k}(H_{n}-H_{i-1}-1)}{n}\right) \ge v\sum \limits _{i=1}^{k}\frac{1-(H_{n}-H_{n-i})}{n}. \end{aligned}$$

Thus, it is sufficient to show that

$$\begin{aligned} (H_{n}-1)-\sum \limits _{i=2}^{k}(1-(H_{n}-H_{i-1}))\ge \sum \limits _{i=1}^{k}(1-(H_{n}-H_{n-i})). \end{aligned}$$

This is equivalent to

$$\begin{aligned} (H_{n}-1)-(k-1)+\sum \limits _{i=2}^{k}H_{n}-\sum \limits _{i=2}^{k}H_{i-1} \ge k-\sum \limits _{i=1}^{k}H_{n}-\sum \limits _{i=1}^{k}H_{n-i}, \end{aligned}$$

or

$$\begin{aligned} 2k(H_{n}-1)\ge \sum \limits _{i=1}^{k}H_{n-i}+\sum \limits _{i=2}^{k}H_{i-1}. \end{aligned}$$

Note that

$$\begin{aligned} \sum \limits _{i=1}^{k}H_{n-i}=\sum \limits _{i=n-k}^{n-1}H_{i}. \end{aligned}$$

Thus, we need to show that

$$\begin{aligned} 2k(H_{n}-1)\ge \sum \limits _{i=n-k}^{n-1}H_{i}+\sum \limits _{i=1}^{k-1}H_{i}, \end{aligned}$$

or, alternatively, that

$$\begin{aligned}&2k\left( \frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{k-1}+\frac{1}{k}+\cdots +\frac{1}{(n-k) }+\cdots +\frac{1}{n-1}+\frac{1}{n}\right) \\&\quad \ge (2k-1)+(2k-2)\frac{1}{2}+\cdots +(k+1)\frac{1}{k-1}+ \\&\qquad +k\frac{1}{k}+\cdots +k\frac{1}{n-k}+(k-1)\frac{1}{n-k+1}+\cdots +2\frac{1}{n-2}+ \frac{1}{n-1}. \end{aligned}$$

By moving elements from the RHS to the LHS we have

$$\begin{aligned} \frac{k}{k+1}+\frac{k}{k+2}+\cdots +\frac{k}{n-k}+\frac{k+1}{n-k+1}+\cdots +\frac{ 2k-1}{n-1}+\frac{2k}{n}\ge k. \end{aligned}$$

(20)

Note that for all $1\le j<n-k,$

$$\begin{aligned} \frac{k}{n-k}<\frac{k}{k+j}, \end{aligned}$$

and for all $1\le j\le k$

$$\begin{aligned} \frac{k}{n-k}<\frac{k+j}{n-k+j}. \end{aligned}$$

Then, we have

$$\begin{aligned}&\frac{k}{k+1}+\frac{k}{k+2}+\cdots +\frac{k}{n-k}+\frac{k+1}{n-k+1}+\cdots +\frac{ 2k-1}{n-1}+\frac{2k}{n} \\&\quad \ge (n-k)\frac{k}{n-k}=k. \end{aligned}$$

That is, the inequality (21) holds. Q.E.D.

1.2 Proof of Proposition 4

The F.O.C. of (6) is

$$\begin{aligned}&-p_{1}\left( \frac{\Pi _{i=1}^{n-1}(iy+x)\Sigma _{i=1}^{n-1}\frac{1}{(iy+x) }y^{n-1}(n-1)!}{(\Pi _{i=1}^{n-1}(iy+x))^{2}}\right) \\&\quad -\sum \limits _{i=2}^{n-1}p_{i}\left( \frac{y^{n-i}\frac{(n-1)!}{(i-1)!}\Pi _{j=i-1}^{n-1}(jy+x)-\Pi _{j=i-1}^{n-1}(jy+x)\Sigma _{j=i-1}^{n-1}\frac{1}{ (jy+x)}y^{n-i}x\frac{(n-1)!}{(i-1)!}}{(\Pi _{j=i-1}^{n-1}(jy+x))^{2}}\right) . \end{aligned}$$

By some simple calculations, we obtain that the symmetric equilibrium effort in a contest with punishments of absolute values $p_{i},i=1,\ldots ,n$ is

$$\begin{aligned} x=p_{1}\frac{H_{n-1}}{n}+\sum \limits _{i=2}^{n-1}p_{i}\frac{H_{n}-H_{i-1}-1}{ n}. \end{aligned}$$

Thus, the designer who wishes to maximize the symmetric equilibrium effort has the following maximization problem:

$$\begin{aligned}&\max _{p_{1,\ldots ,}p_{n}}p_{1}\frac{H_{n}-1}{n}+\sum \limits _{i=2}^{n-1}p_{i} \frac{H_{n}-H_{i-1}-1}{n} \\&\quad s.t.\text { \ }\sum \limits _{i=1}^{n-1}p_{i}=1. \end{aligned}$$

Since for all $1<i\le n-1$

$$\begin{aligned} H_{n}-1>H_{n}-H_{i-1}-1, \end{aligned}$$

we obtain that the symmetric equilibrium effort x is maximized for $p_{1}=p$ and $p_{j}=0$ , $2\le j\le n-1.$$\ Q.E.D.$

1.3 Proof of Proposition 5

By inserting the optimal values of the prize and punishment given by (10) into the equilibrium effort given by (8), the optimal effort of a player in a contest with one prize and one punishment is

$$\begin{aligned} x_{\max 1}=\frac{m}{n}\frac{nH_{n}-1}{\alpha nH_{n}+1}. \end{aligned}$$

Similarly, by inserting the optimal values of the prize and punishment given by (11) into the equilibrium effort given by (8), the optimal effort of a player in a contest with one prize and two punishments is

$$\begin{aligned} x_{\max l}=\frac{m}{n}\frac{n\left( H_{n}-\frac{\sum _{i=1}^{l-1}H_{i}}{l}\right) -1}{ \alpha n\left( H_{n}-\frac{\sum _{i=1}^{l-1}H_{i}}{l}\right) +1}. \end{aligned}$$

Now, let $x_{l}=H_{n}-\frac{\sum _{i=1}^{l-1}H_{i}}{l}.$ Note that $x_{l}>x_{l+1}$ for $l=1,2,\ldots$ Since $\frac{d}{dx_{l}}$ ($\frac{nx_{l}-1}{ \alpha nx_{l}+1})=$$n\frac{\alpha +1}{\left( nx_{l}\alpha +1\right) ^{2}}>0$ , we obtain that $x_{\max l}>x_{\max l+1}$ for all $l\ge 1.$ Thus, an allocation of one prize and l punishments yields a higher total effort than an allocation of one prize and $l+1$ punishments. In particular, the optimal allocation of one prize and one punishment yields a higher total effort than the optimal allocation of one prize and any larger number of punishments. Q.E.D.

1.4 Proof of Proposition 6

By (12), the optimal values of the prize and the punishment in a contest with $n-1$ players are

$$\begin{aligned} v_{n-1}= & {} \frac{m(n-1)H_{n}}{1+\alpha (n-1)H_{n-1}} \\ p_{n-1}= & {} \frac{m}{1+\alpha (n-1)H_{n-1}}. \end{aligned}$$

Then, the total effort in a contest with $n-1$ players is

$$\begin{aligned} TE_{n-1}=\frac{m((n-1)H_{n-1}-1)}{1+\alpha (n-1)H_{n-1}}. \end{aligned}$$

(21)

Comparing (12) and (21) yields

$$\begin{aligned} TE_{n}-TE_{n-1}=m\left( \frac{nH_{n}-1}{\alpha nH_{n}+1}-\frac{(n-1)H_{n-1}-1}{ \alpha (n-1)H_{n-1}+1}\right) , \end{aligned}$$

or, alternatively,

$$\begin{aligned}&\frac{TE_{n}-TE_{n-1}}{m} \\&\quad =\frac{nH_{n}-1}{\alpha nH_{n}+1}-\frac{ (n-1)H_{n-1}-1}{\alpha (n-1)H_{n-1}+1} \\&\quad =\frac{nH_{n}-1}{\alpha nH_{n}+1}-\frac{(n-1)(H_{n}-\frac{1}{n})-1}{\alpha (n-1)(H_{n}-\frac{1}{n})+1} \\&\quad =\left( \alpha +1\right) \frac{n+nH_{n}-1}{n^{3}\alpha ^{2}H_{n}^{2}-n^{2}\alpha ^{2}H_{n}^{2}-n^{2}\alpha ^{2}H_{n}+2n^{2}\alpha H_{n}+n\alpha ^{2}H_{n}-n\alpha H_{n}-n\alpha +n+\alpha }. \end{aligned}$$

Denote

$$\begin{aligned} S= & {} n+nH_{n}-1 \\ R= & {} n^{3}\alpha ^{2}H_{n}^{2}-n^{2}\alpha ^{2}H_{n}^{2}-n^{2}\alpha ^{2}H_{n}+2n^{2}\alpha H_{n}+n\alpha ^{2}H_{n}-n\alpha H_{n}-n\alpha +n+\alpha . \end{aligned}$$

When $\alpha =1$, it can be easily verified that

$$\begin{aligned} R(\alpha =1)=(n^{3}H_{n}^{2}-n^{2}H_{n}^{2})+(2n^{2}H_{n}-n^{2}H_{n})+(nH_{n}-nH_{n})+(n-n+1)>0. \end{aligned}$$

Since $S>0,$ we obtain that if $\alpha =1,$$TE_{n}-TE_{n-1}>0$. Note that,

$$\begin{aligned} \frac{dR}{d\alpha }=2n^{2}H_{n}-nH_{n}-n-2n^{2}\alpha H_{n}^{2}+2n^{3}\alpha H_{n}^{2}+2n\alpha H_{n}-2n^{2}\alpha H_{n}+1, \end{aligned}$$

and

$$\begin{aligned} \frac{d^{2}R}{d\alpha ^{2}}=-2n^{2}H_{n}^{2}+2n^{3}H_{n}^{2}+2nH_{n}-2n^{2}H_{n}. \end{aligned}$$

If $\alpha =0,$ we obtain that

$$\begin{aligned} \frac{dR}{d\alpha }(\alpha =0)=n((2n-1)H_{n}-1)+1>0. \end{aligned}$$

Since $H_{n}^{2}>H_{n}$, we have

$$\begin{aligned} \frac{d^{2}R}{d\alpha ^{2}}\ge H_{n}(2n^{3}-4n^{2}+2n). \end{aligned}$$

Thus, $\frac{d^{2}R}{d\alpha ^{2}}\ge 0$ for all $n>1$, and since $\frac{dR }{d\alpha }(\alpha =0)>0$, we obtain that $\frac{dR}{d\alpha }>0$ for all $\alpha >0.$ Hence, we obtain that $\frac{TE_{n}(\alpha )-TE_{n-1}(\alpha )}{m }>0$ for all $\alpha >0$. Q.E.D.

1.5 Proof of Proposition 7

The total effort in the one-stage contest with n players is

$$\begin{aligned} TE_{one}=\frac{v(n-1)+pn(H_{n}-1)}{n}. \end{aligned}$$

By (16), the total effort in the two-stage contest with one player who wins the prize in the first stage and $n-1$ players who compete so as not to pay the punishment cost in the second stage is

$$\begin{aligned} TE_{1,n}=\frac{v(n-1)+p(H_{n-1}+n(H_{n-1}-1))}{n}. \end{aligned}$$

Then we have

$$\begin{aligned} TE_{1,n}-TE_{one}= & {} \frac{pH_{n-1}}{n}+p(H_{n-1}-H_{n}) \\= & {} \frac{p}{n}(H_{n-1}-1)>0. \end{aligned}$$

Similarly, by (16), the total effort in the two-stage contest with one player who pays the punishment cost in the first stage and $n-1$ players who compete for the prize in the second stage is

$$\begin{aligned} TE_{n-1,n}=\left[ \frac{v}{(n-1)^{2}}+p\right] \left[ \sum \limits _{i=1}^{n-1}1-(H_{n}-H_{n-i})\right] +\frac{v(n-2)}{n-1}, \end{aligned}$$

or, alternatively,

$$\begin{aligned} TE_{n-1,n}= & {} \left[ \frac{v}{(n-1)^{2}}+p\right] \left[ (n-1)-\sum \limits _{i=1}^{n-1}\frac{n-i}{n-i+1}\right] +\frac{v(n-2)}{n-1} \\= & {} \left[ \frac{v}{(n-1)^{2}}+p\right] (H_{n}-1)+\frac{v(n-2)}{n-1}. \end{aligned}$$

Then we have

$$\begin{aligned} TE_{n-1,n}-TE_{one}=v(\frac{H_{n}-1}{(n-1)^{2}}+\frac{n-2}{n-1}-\frac{n-1}{n} )=\frac{1}{n^{3}-2n^{2}+n}\left( (H_{n}-2)n+1\right) . \end{aligned}$$

Since $(H_{n}-2)n+1\ge 0$, we obtain that for all $n\ge 2$, $TE_{n-1,n}-TE_{one}\ge 0$. Q.E.D.