Equity and effectiveness of optimal taxation in contests under an all-pay auction

Abstract

The means of contest design may include differential taxation of the prize. This paper considers all-pay auctions under complete information and establishes that, given a revenue-maximizing contest designer who faces a balanced-budget constraint, the optimal taxation scheme corresponding to an all-pay auction is appealing in two senses. First, it ensures extreme equitable final prize valuations. Second, it is effective; it yields total contestants’ efforts that are larger than those obtained under almost any pure-strategy equilibrium of a Tullock-type lottery contest.

Appendix

Proposition 1

The optimal taxation scheme under the APA equalizes the contestants final stakes, that is, $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {-0.5\left( {n_1 -n_2 } \right) ,0.5\left( {n_1 -n_2 } \right) } \right)$.

Proof

The proof includes three parts. We first clarify the properties of the feasible (potentially equilibrium) strategies $\left( {\varepsilon _1 ,\varepsilon _2 } \right)$. For these strategies, we describe in Part 2 the properties of the balanced-budget constraint. We then present in the third part the properties of an equi-effort curve (ee-curve) and by comparing its slope to that of the balanced-budget curve (bb-curve) complete the proof. $\square$

Part 1

Let us show that in equilibrium, if $k=1$, then $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {0,0} \right)$ and if $k>1$, then $-0.5\left( {n_1 -n_2 } \right) \le \varepsilon _1 <0<\varepsilon _2$.

Since $\frac{\partial G_A }{\partial \varepsilon _1 }=-0.5\left( {\frac{n_2 +\varepsilon _2 }{n_1 +\varepsilon _1 }} \right) ^{2}<0,\frac{\partial G_A }{\partial \varepsilon _2 }=0.5+\frac{n_2 +\varepsilon _2 }{n_1 +\varepsilon _1 }>0$, and, by the bb-curve, $\varepsilon _1 \varepsilon _2 \le 0$, considering a deviation from $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {0,0} \right)$, the designer does not have an incentive to reduce $\varepsilon _2$ (so $\varepsilon _2 \ge 0)$ or increase $\varepsilon _1$ (so $\varepsilon _1 \le 0)$. That is, in any equilibrium, $\varepsilon _1 \le 0\le \varepsilon _2$ must be satisfied.

When $k=1$ ($n=n_1 =n_2 )$, by (6), for $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {0,0} \right)$, the total efforts are equal to $\frac{n}{2}+\frac{n^{2}}{2n}=n$. It can be easily verified, by (6), that any alternative feasible taxation scheme attains smaller efforts. Henceforth we therefore assume that $k>1$.

When $k>1$, by (6), for $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {0,0} \right)$, the total efforts are equal to $\frac{n_1 +n_2 }{2k}$ and for the feasible stake-equalizing scheme $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {-0.5\left( {n_1 -n_2 } \right) ,0.5\left( {n_1 -n_2 } \right) } \right)$, the total efforts are equal to $0.5\left( {n_1 +n_2 } \right)$. Since, for $k>1$, $0.5\left( {n_1 +n_2 } \right) >\frac{n_1 +n_2 }{2k}$, we can conclude that $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {0,0} \right)$ does not maximize the contestants’ efforts. Hence, for $k>1$, in equilibrium, $\varepsilon _1 <0<\varepsilon _2$.

Let us complete the proof of Part 1 (establish that, in equilibrium, $-0.5\left( {n_1 -n_2 } \right) \le \varepsilon _1 )$ by showing that when $-n_1 <\varepsilon _1 <-0.5\left( {n_1 -n_2 } \right)$, the corresponding efforts cannot be maximal.

Let $\left( {\varepsilon _1^E ,\varepsilon _2^E } \right) =\left( {-0.5\left( {n_1 -n_2 } \right) ,0.5\left( {n_1 -n_2 } \right) } \right)$ be our benchmark taxation scheme, where $n_1 +\varepsilon _1^E =n_2 +\varepsilon _2^E =0.5\left( {n_1 +n_2 } \right)$, efforts are equal to $G_A^E =\frac{n_2 +\varepsilon _2^E }{2}+\left( {\frac{n_2 +\varepsilon _2^E }{2}} \right) \left( {\frac{n_2 +\varepsilon _2^E }{n_1 +\varepsilon _1^E }} \right) =0.5\left( {n_1 +n_2 } \right)$ and $\frac{n_2 +\varepsilon _2^E }{n_1 +\varepsilon _1^E }=\frac{n_1 +\varepsilon _1^E }{n_2 +\varepsilon _2^E }=1$. Starting from this scheme, let us reduce $\varepsilon _1$ below $-0.5\left( {n_1 -n_2 } \right)$ and show that such a change reduces the efforts, independent of the balanced-budget constraint:¹⁶

• (i) If, after the reduction in $\varepsilon _1$, $n_1 +\varepsilon _1 \ge n_2 +\varepsilon _2$, then efforts are still given by (6), $G_A =\frac{n_2 +\varepsilon _2 }{2}+\left( {\frac{n_2 +\varepsilon _2 }{2}} \right) \left( {\frac{n_2 +\varepsilon _2 }{n_1 +\varepsilon _1 }} \right)$, and $n_2 +\varepsilon _2^E =n_1 +\varepsilon _1^E >n_1 +\varepsilon _1 \ge n_2 +\varepsilon _2$. Therefore, $\frac{n_2 +\varepsilon _2^E }{2}>\frac{n_2 +\varepsilon _2 }{2}$ and $\frac{n_2 +\varepsilon _2^E }{n_1 +\varepsilon _1^E }=1\ge \frac{n_2 +\varepsilon _2 }{n_1 +\varepsilon _1 }$ and, consequently, the move from $G_A^E$ to the new $G_A$ reduces the efforts.

• (ii) If, after the reduction in $\varepsilon _1$, $n_1 +\varepsilon _1 <n_2 +\varepsilon _2$, then efforts are given by $G_A =\frac{n_1 +\varepsilon _1 }{2}+\left( {\frac{n_1 +\varepsilon _1 }{2}} \right) \left( {\frac{n_1 +\varepsilon _1 }{n_2 +\varepsilon _2 }} \right)$ and $n_2 +\varepsilon _2^E =n_1 +\varepsilon _1^E >n_1 +\varepsilon _1$. Therefore, $\frac{n_2 +\varepsilon _2^E }{2}>\frac{n_1 +\varepsilon _1 }{2}$ and $\frac{n_2 +\varepsilon _2^E }{n_1 +\varepsilon _1^E }=1>\frac{n_1 +\varepsilon _1 }{n_2 +\varepsilon _2 }$ and consequently the move from $G_A^E$ to the new $G_A$ reduces the total efforts.

Part 2

In this part we examine the properties of the bb-curve for the relevant schemes satisfying $-0.5\left( {n_1 -n_2 } \right) \le \varepsilon _1 <0<\varepsilon _2$. The bb-curve can be written as $\varepsilon _2^2 +\varepsilon _2 \left( {n_2 -\varepsilon _1 } \right) +\varepsilon _1 \left[ {2\left( {n_1 +\varepsilon _1 } \right) -n_2 } \right] =0$. Since $\varepsilon _1 \le 0\le \varepsilon _2$, the solution of this equation must be the positive root. That is:

$$\begin{aligned} \varepsilon _2 (\varepsilon _1 )=\frac{\varepsilon _1 -n_2 +\left\{ {\left( {n_2 -\varepsilon _1 } \right) ^{2}-4\varepsilon _1 \left[ {2\left( {n_1 +\varepsilon _1 } \right) -n_2 } \right] } \right\} ^{0.5}}{2}. \end{aligned}$$

(17)

A taxation scheme $\left( {\varepsilon _1 ,\varepsilon _2 } \right)$ that satisfies the bb-curve has the following properties:

• a. $n_1 +\varepsilon _1 =n_2 +\varepsilon _2 \left( {>0} \right)$ iff $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {-0.5\left( {n_1 -n_2 } \right) ,0.5\left( {n_1 -n_2 } \right) } \right)$.

• b. If $-0.5\left( {n_1 -n_2 } \right) <\varepsilon _1 <0$, then $n_1 +\varepsilon _1 >n_2 +\varepsilon _2 \left( {>0} \right)$.

• c. For $n_1 +\varepsilon _1 \ge n_2 +\varepsilon _2 >0$, the function (17) that defines the bb-curve is (i) concave, $\frac{\partial ^{2}\varepsilon _2 }{\partial \varepsilon _1^2 }<0$ and (ii) at $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {0,0} \right)$, $\frac{\partial \varepsilon _2 }{\partial \varepsilon _1 }<0$.

The proof of property a is straightforward and therefore omitted.

To prove property b, note that in the range $-0.5\left( {n_1 -n_2 } \right) \le \varepsilon _1 \le 0$ on the bb-curve, only in the extreme point where $\varepsilon _1 =-0.5\left( {n_1 -n_2 } \right)$, the contestants’ stakes are equal, that is, $n_1 +\varepsilon _1 =n_2 +\varepsilon _2$. In the other extreme point where $\varepsilon _1 =0$, $n_1 +\varepsilon _1 >n_2 +\varepsilon _2$. Hence, by the continuity of the bb-curve (17), for every $\varepsilon _1$, $-0.5\left( {n_1 -n_2 } \right) <\varepsilon _1 \le 0$, we get that $n_1 +\varepsilon _1 >n_2 +\varepsilon _2$.

To prove property c, notice that according to (17), the slope of the bb-curve is:

$$\begin{aligned} \frac{\partial \varepsilon _2 }{\partial \varepsilon _1 }=0.5+0.5\left( {n_2 -4n_1 -7\varepsilon _1 } \right) \left\{ {\left( {n_2 -\varepsilon _1 } \right) ^{2}-4\varepsilon _1 \left[ {2\left( {n_1 +\varepsilon _1 } \right) -n_2 } \right] } \right\} ^{-0.5} \end{aligned}$$

(18)

and, therefore,

$$\begin{aligned} \frac{\partial ^{2}\varepsilon _2 }{\partial \varepsilon _1^2 }&= -0.5\left( {n_2 -4n_1 -7\varepsilon _1 } \right) ^{2}\left\{ {\left( {n_2 -\varepsilon _1 } \right) ^{2}-4\varepsilon _1 \left[ {2\left( {n_1 +\varepsilon _1 } \right) -n_2 } \right] } \right\} ^{-1.5} \nonumber \\&-3.5\left\{ {\left( {n_2 -\varepsilon _1 } \right) ^{2}-4\varepsilon _1 \left[ {2\left( {n_1 +\varepsilon _1 } \right) -n_2 } \right] } \right\} ^{-0.5}<0 \end{aligned}$$

(19)

Substituting $\varepsilon _1 =0$ in (18), we get that $\frac{\partial \varepsilon _2 }{\partial \varepsilon _1 }=1-2k<0$.

Let us, finally, find out what is the value of $\varepsilon _1$ that yields the maximal value of $\varepsilon _2 (\varepsilon _1 )$ on the bb-curve. By (18), for $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {0,0} \right)$, we get that $\frac{\partial \varepsilon _2 }{\partial \varepsilon _1 }<0$ and for $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {-0.5\left( {n_1 -n_2 } \right) ,0.5\left( {\text{ n}_1 -\text{ n}_2 } \right) } \right)$, we get that $\frac{\partial \varepsilon _2 }{\partial \varepsilon _1 }=\frac{k-3}{3k-1}$. By (19), the function $\varepsilon _2 (\varepsilon _1 )$ defining the bb-curve is concave in the relevant range $-0.5\left( {n_1 -n_2 } \right) \le \varepsilon _1 <0$. Therefore, for $k>3$, $\varepsilon _2 (\varepsilon _1 )$ has a negative slope at $\varepsilon _1 =0$, a positive slope at $\varepsilon _1 =-0.5\left( {n_1 -n_2 } \right)$ and a zero slope at some intermediate value $\varepsilon _1$, $-0.5\left( {n_1 -n_2 } \right) <\varepsilon _1 <0$, that yields the maximal value of $\varepsilon _2$. Notice that when $k\le 3$, $\frac{\partial \varepsilon _2 }{\partial \varepsilon _1 }<0$ for any $\varepsilon _1$, $-0.5\left( {n_1 -n_2 } \right) \le \varepsilon _1 <0$. That is, the bb-curve is declining in the relevant domain and the maximal value of $\varepsilon _2 =0.5\left( {n_1 -n_2 } \right)$ is obtained at $\varepsilon _1 =-0.5\left( {n_1 -n_2 } \right)$.

For $k\le 3$ and $-0.5\left( {n_1 -n_2 } \right) \le \varepsilon _1 <0$, an increase in $\varepsilon _1$ is always associated with a decrease in $\varepsilon _2$. Since $\frac{\partial G_A }{\partial \varepsilon _1 }<0$ and $\frac{\partial G_A }{\partial \varepsilon _2 }>0$, we directly get that the maximal efforts are obtained in the egalitarian scheme $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {-0.5\left( {n_1 -n_2 } \right) ,0.5\left( {n_1 -n_2 } \right) } \right)$. But when $k>3$ and $-0.5\left( {n_1 -n_2 } \right) \le \varepsilon _1 <0$, an increase in $\varepsilon _1$ can be associated with an increase in $\varepsilon _2$. The optimality of the egalitarian tax scheme needs therefore to be proved taking into account also this possibility (an increasing bb-curve). The third part of the proof establishes the optimality of $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {-0.5\left( {n_1 -n_2 } \right) ,0.5\left( {n_1 -n_2 } \right) } \right)$ also in this case.

Part 3

An ee-curve $\overline{G} _A$ is defined for $n_1 +\varepsilon _1 \ge n_2 +\varepsilon _2 >0$ as follows $\overline{G} _A =\frac{n_2 +\varepsilon _2 }{2}+\frac{\left( {n_2 +\varepsilon _2 } \right) ^{2}}{2\left( {n_1 +\varepsilon _1 } \right) }$. Let us show that the function $\varepsilon _2 \left( {\varepsilon _1 } \right)$ that defines $\overline{G} _A$ is positively sloped, $\frac{\partial \varepsilon _2 }{\partial \varepsilon _1 }>0$, and concave, $\frac{\partial ^{2}\varepsilon _2 }{\partial \varepsilon _1^2 }<0$. Differentiating the function $\varepsilon _2 \left( {\varepsilon _1 } \right)$ we get $\frac{\partial \varepsilon _2 }{\partial \varepsilon _1 }=-\frac{-0.5\left( {\frac{n_2 +\varepsilon _2 }{n_1 +\varepsilon _1 }} \right) ^{2}}{0.5+\frac{n_2 +\varepsilon _2 }{n_1 +\varepsilon _1 }}=\frac{\left( {\frac{n_2 +\varepsilon _2 }{n_1 +\varepsilon _1 }} \right) ^{2}}{1+2\left( {\frac{n_2 +\varepsilon _2 }{n_1 +\varepsilon _1 }} \right) }>0$. Letting $b=\frac{n_2 +\varepsilon _2 }{n_1 +\varepsilon _1 }$, we get that $\frac{\partial \varepsilon _2 }{\partial \varepsilon _1 }=\frac{b^{2}}{1+2b}>0$. Since $b$ is a function of $\varepsilon _1$, we get that $\frac{\partial ^{2}\varepsilon _2 }{\partial \varepsilon _1^2 }=\frac{2b\frac{\partial b}{\partial \varepsilon _1 }\left( {1+2b} \right) -2\frac{\partial b}{\partial \varepsilon _1 }b^{2}}{\left( {1+2b} \right) ^{2}}=\frac{2b\frac{\partial b}{\partial \varepsilon _1 }\left( {b+1} \right) }{\left( {1+2b} \right) ^{2}}$. Substituting in this latter expression $\frac{\partial b}{\partial \varepsilon _1 }=\frac{\frac{\partial \varepsilon _2 }{\partial \varepsilon _1 }\left( {n_1 +\varepsilon _1 } \right) -\left( {n_2 +\varepsilon _2 } \right) }{\left( {n_1 +\varepsilon _1 } \right) ^{2}}=\frac{\frac{\partial \varepsilon _2 }{\partial \varepsilon _1 }-b}{n_1 +\varepsilon _1 }$ (note that in this expression $\varepsilon _2$ is a function of $\varepsilon _1 )$ we get that $\frac{\partial ^{2}\varepsilon _2 }{\partial \varepsilon _1^2 }=\frac{2b\left( {\frac{\partial \varepsilon _2 }{\partial \varepsilon _1 }-b} \right) \left( {b+1} \right) }{\left( {n_1 +\varepsilon _1 } \right) \left( {1+2b} \right) ^{2}}$. Substituting $\frac{\partial \varepsilon _2 }{\partial \varepsilon _1 }=\frac{b^{2}}{1+2b}$, we get that $\frac{\partial ^{2}\varepsilon _2 }{\partial \varepsilon _1^2 }=-\frac{2b^{2}\left( {1+b} \right) ^{2}}{\left( {n_1 +\varepsilon _1 } \right) \left( {1+2b} \right) ^{3}}<0$. Given the properties of the bb-curve and the ee-curve, we will complete the proof by showing that, for $-0.5\left( {n_1 -n_2 } \right) \le \varepsilon _1 <0$, where, $n_1 +\varepsilon _1 \ge n_2 +\varepsilon _2$ (see Part 2), at every point on the bb-curve, the slope of the ee-curve is larger than the slope of the bb-curve.

Substituting $b=\frac{n_2 +\varepsilon _2 }{n_1 +\varepsilon _1 }$ in the bb-curve, we get that $\left[ {1-\frac{n_2 +\varepsilon _2 }{2\left( {n_1 +\varepsilon _1 } \right) }} \right] \varepsilon _1 +\frac{n_2 +\varepsilon _2 }{2\left( {n_1 +\varepsilon _1 } \right) }\varepsilon _2 =0$, or $\left( {1-0.5b} \right) \varepsilon _1 +0.5b\varepsilon _2 =0$ or $b=\frac{2\varepsilon _1 }{\varepsilon _1 -\varepsilon _2 }$.¹⁷ We have to show that $\frac{b^{2}}{1+2b}>\frac{2\left( {n_1 +2\varepsilon _1 } \right) -\left( {n_2 +\varepsilon _2 } \right) }{\varepsilon _1 -\left( {n_2 +2\varepsilon _2 } \right) }$, where the LHS (RHS) expression is the slope of the ee-curve (bb-curve). This inequality can be equivalently written as $\frac{b^{2}}{1+2b}>\frac{2\left( {n_1 +\varepsilon _1 } \right) -\left( {n_2 +\varepsilon _2 } \right) +2\varepsilon _1 }{\varepsilon _1 -\varepsilon _2 -\left( {n_2 +\varepsilon _2 } \right) }$. Dividing the nominator and denominator of the RHS by $\left( {n_1 +\varepsilon _1 } \right)$ we get $\frac{b^{2}}{1+2b}>\frac{2-\frac{n_2 +\varepsilon _2 }{n_1 +\varepsilon _1 }+\frac{2\varepsilon _1 }{n_1 +\varepsilon _1 }}{\frac{\varepsilon _1 -\varepsilon _2 }{n_1 +\varepsilon _1 }-\frac{n_2 +\varepsilon _2 }{n_1 +\varepsilon _1 }}$ and, in terms of $b$, we get $\frac{b^{2}}{1+2b}>\frac{2-b+\frac{2\varepsilon _1 }{n_1 +\varepsilon _1 }}{\frac{\varepsilon _1 -\varepsilon _2 }{n_1 +\varepsilon _1 }-b}$. Since the denominator of the RHS expression is negative (because $\varepsilon _1 <0<\varepsilon _2$ and $b>0)$, $b^{2}\left( {\frac{\varepsilon _1 -\varepsilon _2 }{n_1 +\varepsilon _1 }-b} \right) <\left( {2-b+\frac{2\varepsilon _1 }{n_1 +\varepsilon _1 }} \right) \left( {1+2b} \right)$. After some algebraic manipulations, this inequality takes the form:

$$\begin{aligned} b\left( {1-b} \right) ^{2}+2\left( {b+1} \right) +\frac{1}{n_1 +\varepsilon _1 }\left[ {2\varepsilon _1 \left( {1+2b} \right) -b^{2}\left( {\varepsilon _1 -\varepsilon _2 } \right) } \right] >0 \end{aligned}$$

Substituting $b=\frac{2\varepsilon _1 }{\varepsilon _1 -\varepsilon _2 }$ (which has been calculated above) in all the terms in the above inequality with the exception of $b\left( {1-b} \right) ^{2}$, we get after some algebraic manipulations that the above inequality is equivalent to $b\left( {1-b} \right) ^{2}+2\left( {\frac{2\varepsilon _1 }{\varepsilon _1 -\varepsilon _2 }+1} \right) +\frac{1}{n_1 +\varepsilon _1 }\left[ {2\varepsilon _1 \left( {1+2\frac{2\varepsilon _1 }{\varepsilon _1 -\varepsilon _2 }} \right) -\left( {\frac{2\varepsilon _1 }{\varepsilon _1 -\varepsilon _2 }} \right) ^{2}\left( {\varepsilon _1 -\varepsilon _2 } \right) } \right] >0$ or $b\left( {1-b} \right) ^{2}+2\left( {\frac{3\varepsilon _1 -\varepsilon _2 }{\varepsilon _1 -\varepsilon _2 }} \right) \left( {\frac{n_1 +2\varepsilon _1 }{n_1 +\varepsilon _1 }} \right) >0$. Since $-0.5\left( {n_1 -n_2 } \right) \le \varepsilon _1 <0$, $n_1 +2\varepsilon _1 \ge n_1 +2\left[ {-0.5\left( {n_1 -n_2 } \right) } \right] =n_2 >0$ and, therefore, $\frac{n_1 +2\varepsilon _1 }{n_1 +\varepsilon _1 }>0$. Since $\frac{3\varepsilon _1 -\varepsilon _2 }{\varepsilon _1 -\varepsilon _2 }>0$ (because $\varepsilon _1 <0<\varepsilon _2 )$ and $b>0$, we get that, for $-0.5\left( {n_1 -n_2 } \right) \le \varepsilon _1 <0$, $b\left( {1-b} \right) ^{2}+2\left( {\frac{3\varepsilon _1 -\varepsilon _2 }{\varepsilon _1 -\varepsilon _2 }} \right) \left( {\frac{n_1 +2\varepsilon _1 }{n_1 +\varepsilon _1 }} \right) >0$. $\square$

Proposition 2

When $k>1$ the optimal taxation scheme under any Tullock-type lottery with $0<\alpha <2$ does not equalize the contestants’ final stakes, but preserves their relative magnitude. That is, $\left( {\varepsilon _1 ,\varepsilon _2 } \right) \ne \left( {-0.5\left( {n_1 -n_2 } \right) ,0.5\left( {n_1 -n_2 } \right) } \right)$ and $\varepsilon _1 +\varepsilon _2 >0$.

Proof

The proof is based on three lemmas. $\square$

Lemma 1

For $0<\alpha \le 2$, the total efforts obtained in the designer’s problem, (16), satisfy:

• a. For any $\left( {\varepsilon _1 ,\varepsilon _2 } \right) , \frac{\alpha d^{\alpha }\left( {n_1 +\varepsilon _1 +n_2 +\varepsilon _2 } \right) }{\left( {d^{\alpha }+1} \right) ^{2}}\le 0.25\alpha \left( {n_1 +\varepsilon _1 +n_2 +\varepsilon _2 } \right)$.

• b. For the equilibrium taxation scheme $\left( {\varepsilon _1^E ,\varepsilon _2^E } \right)$, $0.25\alpha \left( {n_1 \!+\! n_2 } \right) \!\le \! \frac{\alpha d^{\alpha }\left( {n_1 +\varepsilon _1^E +n_2 +\varepsilon _2^E } \right) }{\left( {d^{\alpha }+1} \right) ^{2}}$.

Proof of Lemma 1

a. We have to prove that $\frac{d^{\alpha }}{\left( {d^{\alpha }+1} \right) ^{2}}\le 0.25$ or $\left( {d^{\alpha }-1} \right) ^{2}\ge 0$ which is always satisfied.

Hence, for any $\left( {\varepsilon _1,\varepsilon _2 } \right)$:

$$\begin{aligned} \frac{\alpha d^{\alpha }\left( {n_1 +\varepsilon _1 +n_2 +\varepsilon _2 } \right) }{\left( {d^{\alpha }+1} \right) ^{2}}\le 0.25\alpha \left( {n_1 +\varepsilon _1 +n_2 +\varepsilon _2 } \right) . \end{aligned}$$

• b. The selection of $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {-0.5\left( {n_1 -n_2 } \right) ,0.5\left( {n_1 -n_2 } \right) } \right)$, which satisfied all the constraints in the designer’s problem (16), yields total efforts that are equal to $0.25\alpha \left( {n_1 +n_2 } \right)$. Hence, $0.25\alpha \left( {n_1 +n_2 } \right) \le \frac{\alpha d^{\alpha }\left( {n_1 +\varepsilon _1^E +n_2 +\varepsilon _2^E } \right) }{\left( {d^{\alpha }+1} \right) ^{2}}$. $\square$

Lemma 2

In equilibrium, for $0<\alpha <2$ and $k>1$, $\varepsilon _1 <0<\varepsilon _2$.

Proof of Lemma 2

The proof consists of two steps.

• Step 1 Since $p_i >0$, $i=1,2$, the balanced-budget constraint (14) implies that $\varepsilon _1 \varepsilon _2 \le 0$. Let us first show that $\varepsilon _1 \le 0\le \varepsilon _2$. Suppose to the contrary that, in equilibrium, the inequalities $\varepsilon _1 \le 0\le \varepsilon _2$ are not satisfied. Since $\varepsilon _1 \varepsilon _2 \le 0$, this implies that $\varepsilon _2 <0<\varepsilon _1$ and, therefore, after the change in the contestants’ stakes, the stake of contestant 1 (2) is increased (decreased), $d=\frac{n_1 +\varepsilon _1 }{n_2 +\varepsilon _2 }>1$. By the balanced-budget constraint (15), $\varepsilon _2 =-d^{\alpha }\varepsilon _1 <-\varepsilon _1$ and using the result $d>1$, we get that $\varepsilon _2 =-d^{\alpha }\varepsilon _1 <-\varepsilon _1$ or $\varepsilon _1 +\varepsilon _2 <0$. The total efforts, even if constraints 2 and 3 in the designer’s problem (16) are disregarded, are not larger than $0.25\alpha \left( {n_1 +\varepsilon _1 +n_2 +\varepsilon _2 } \right)$. Clearly, under the constraints 2 and 3, the total contestants’ efforts cannot be larger than this amount. Since $\varepsilon _1 +\varepsilon _2 <0$, the equilibrium total efforts are smaller than $0.25\alpha \left( {n_1 +n_2 } \right)$. But this contradicts part (b) of Lemma 1, which implies that the assumption $\varepsilon _2 <0<\varepsilon _1$ cannot be true. Hence, $\varepsilon _1 \le 0\le \varepsilon _2$.

• Step 2 Let us prove that for $k>1,\; \left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {0,0} \right)$ is not optimal. Together with the conditions established in step 1, $\varepsilon _1 \le 0\le \varepsilon _2$, this will complete the proof establishing that, in equilibrium, $\varepsilon _1 <0<\varepsilon _2$. Let us then show that the selection of $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {-0.5\left( {n_1 -n_2 } \right) ,0.5\left( {n_1 -n_2 } \right) } \right)$ is superior to the selection of $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {0,0} \right)$, that is, $\frac{\alpha k^{\alpha }\left( {n_1 +n_2 } \right) }{\left( {k^{\alpha }+1} \right) ^{2}}<0.25\alpha \left( {n_1 +n_2 } \right)$. This latter inequality is equivalent to $0<\left( {k^{\alpha }-1} \right) ^{2}$, which is always satisfied since $k>1$.

$\square$

Lemma 3

An ee-curve $\overline{G} _L$ is defined by $\overline{G} _L \!=\!\frac{\alpha \left[ {\left( {n_1 +\varepsilon _1 } \right) \left( {n_2 +\varepsilon _2 } \right) } \right] ^{\alpha }\left( {n_1 +\varepsilon _1 +n_2 +\varepsilon _2 } \right) }{\left[ {\left( {n_1 +\varepsilon _1 } \right) ^{\alpha }+\left( {n_2 +\varepsilon _2 } \right) ^{\alpha }} \right] ^{2}}$.

• a. If $k>1$, then at the point representing $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {-0.5\left( {n_1 -n_2 } \right) ,0.5\left( {n_1 -n_2 } \right) } \right)$,

  • (1) The slope of the ee-curve $\frac{\partial \varepsilon _2 }{\partial \varepsilon _1 }$ is equal to $-1$.

  • (2) In the margin, an increase in $\varepsilon _i$, $i=\left( {1,2} \right)$, increases total efforts.

  • (3) The slope of the bb-curve $\frac{\partial \varepsilon _2 }{\partial \varepsilon _1 }$ is larger than $-1$.

• b. For $0<\alpha <2$ and $k>1$, in equilibrium $d=\frac{n_1 +\varepsilon _1 }{n_2 +\varepsilon _2 }>1$ (the stake of contestant 1 (2) is reduced (increased), but the final stake of 1 is still larger than that of 2) and therefore $\varepsilon _1 +\varepsilon _2 >0$.

Proof of Lemma 3

a.(1) and a.(2). Given an ee-curve $\overline{G} _L$.

$$\begin{aligned} \begin{array}{l} \frac{\partial \overline{G} _L }{\partial \varepsilon _1 }=\alpha \left( {n_2 +\varepsilon _2 } \right) ^{\alpha } \\ \frac{\left\{ {\begin{array}{l} \left[ {\alpha \left( {n_1 +\varepsilon _1 } \right) ^{\alpha -1}\left( {n_1 +\varepsilon _1 +n_2 +\varepsilon _2 } \right) +\left( {n_1 +\varepsilon _1 } \right) ^{\alpha }} \right] \left[ {\left( {n_1 +\varepsilon _1 } \right) ^{\alpha }+\left( {n_2 +\varepsilon _2 } \right) ^{\alpha }} \right] ^{2} \\ -2\left[ {\left( {n_1 +\varepsilon _1 } \right) ^{\alpha }+\left( {n_2 +\varepsilon _2 } \right) ^{\alpha }} \right] \alpha \left( {n_1 +\varepsilon _1 } \right) ^{\alpha -1}\left( {n_1 +\varepsilon _1 } \right) ^{\alpha }\left( {n_1 +\varepsilon _1 +n_2 +\varepsilon _2 } \right) \\ \end{array}} \right\} }{\left[ {\left( {n_1 +\varepsilon _1 } \right) ^{\alpha }+\left( {n_2 +\varepsilon _2 } \right) ^{\alpha }} \right] ^{4}} \\ \end{array} \end{aligned}$$

or, after some simplification,

$$\begin{aligned} \frac{\partial \overline{G} _L }{\partial \varepsilon _1 }=\frac{\alpha d^{\alpha -1}\left\{ {\alpha \left( {d+1} \right) +d+d^{\alpha }\left[ {d-\alpha \left( {d+1} \right) } \right] } \right\} }{\left( {d^{\alpha }+1} \right) ^{3}} \end{aligned}$$

In a similar way we get that

$$\begin{aligned} \frac{\partial \overline{G} _L }{\partial \varepsilon _2 }=\frac{\alpha d^{\alpha }\left\{ {d^{\alpha }\left[ {\alpha \left( {d+1} \right) +1} \right] +1-\alpha \left( {d+1} \right) } \right\} }{\left( {d^{\alpha }+1} \right) ^{3}} \end{aligned}$$

The slope of an ee-curve is therefore equal to

$$\begin{aligned} \frac{\partial \varepsilon _2 }{\partial \varepsilon _1 }=-\frac{\frac{\partial \overline{G} _L }{\partial \varepsilon _1 }}{\frac{\partial \overline{G} _L }{\partial \varepsilon _2 }}=-\frac{\left\{ {\alpha \left( {d+1} \right) +d+d^{\alpha }\left[ {d-\alpha \left( {d+1} \right) } \right] } \right\} }{d\left\{ {d^{\alpha }\left[ {\alpha \left( {d+1} \right) +1} \right] +1-\alpha \left( {d+1} \right) } \right\} } \end{aligned}$$

For $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {-0.5\left( {n_1 -n_2 } \right) ,0.5\left( {n_1 -n_2 } \right) } \right)$, $d=1$. For $d=1$ we get that $\frac{\partial \overline{G} _L }{\partial \varepsilon _1 }=\frac{\partial \overline{G} _L }{\partial \varepsilon _2 }=0.25\alpha$ and $\frac{\partial \varepsilon _2 }{\partial \varepsilon _1 }=-1$. This means that in the neighborhood of $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {-0.5\left( {n_1 -n_2 } \right) ,0.5\left( {n_1 -n_2 } \right) } \right)$, an ee-curve is negatively sloped and an increase in $\varepsilon _1$ or in $\varepsilon _2$ increases the total efforts [an increase in total efforts shifts an ee-curve upward in the $(\varepsilon _1 ,\varepsilon _2 )$ plane].

a.(3) From the implicit form of the balanced-budget constraint (15) we get that $\left( {n_1 +\varepsilon _1 } \right) ^{\alpha }\varepsilon _1 +\left( {n_2 +\varepsilon _2 } \right) ^{\alpha }\varepsilon _2 =0$. Therefore, the slope of the bb-curve is equal to $\frac{\partial \varepsilon _2 }{\partial \varepsilon _1 }=-\frac{\alpha \left( {n_1 +\varepsilon _1 } \right) ^{\alpha -1}\varepsilon _1 +\left( {n_1 +\varepsilon _1 } \right) ^{\alpha }}{\alpha \left( {n_2 +\varepsilon _2 } \right) ^{\alpha -1}\varepsilon _2 +\left( {n_2 +\varepsilon _2 } \right) ^{\alpha }}$ or $\frac{\partial \varepsilon _2 }{\partial \varepsilon _1 }=-\frac{d^{\alpha -1}\left[ {n_1 +\varepsilon _1 \left( {1+\alpha } \right) } \right] }{n_2 +\varepsilon _2 \left( {1+\alpha } \right) }$ and, therefore, the slope at $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {-0.5\left( {n_1 -n_2 } \right) ,0.5\left( {n_1 -n_2 } \right) } \right)$ is $\frac{\partial \varepsilon _2 }{\partial \varepsilon _1 }=-\frac{n_1 +n_2 -\alpha \left( {n_1 -n_2 } \right) }{n_1 +n_2 +\alpha \left( {n_1 -n_2 } \right) }$. Since $n_1 +n_2 -\alpha \left( {n_1 -n_2 } \right) <n_1 +n_2 +\alpha \left( {n_1 -n_2 } \right)$, the slope of the bb-curve at $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {-0.5\left( {n_1 -n_2 } \right) ,0.5\left( {n_1 -n_2 } \right) } \right)$ satisfies $\frac{\partial \varepsilon _2 }{\partial \varepsilon _1 }>-1$.¹⁸

b. Let us show that a move from $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {-0.5\left( {n_1 -n_2 } \right) ,0.5\left( {n_1 -n_2 } \right) } \right)$, where $d=1$, that involves a marginal increase in $\varepsilon _1$, which preserves the balanced-budget constraint, increases total efforts. This will prove that, in equilibrium, $d\ne 1$.

A marginal change in $d=1$, which is due to a marginal increase in $\varepsilon _1$ that preserves the balanced-budget constraint, still satisfies constraints 2 and 3 in the designer’s problem (16), because at $d=1$ these constraints are satisfied as strict inequalities ($2-\alpha >0)$. By Lemma 3 part (a), at the point which represents $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {-0.5\left( {n_1 -n_2 } \right) ,0.5\left( {n_1 -n_2 } \right) } \right)$, the slope of the bb-curve ($-1<\frac{\partial \varepsilon _2 }{\partial \varepsilon _1 })$ is larger than the slope of the ee-curve ($\frac{\partial \varepsilon _2 }{\partial \varepsilon _1 }=-1)$. Therefore, starting from this point, a marginal increase in $\varepsilon _1$ accompanied by the required change in $\varepsilon _2$, such that the bb-curve is still satisfied, increases the total efforts. Note that if the slope of the bb-curve is positive (not positive) an increase (a decrease) in $\varepsilon _2$ is required. We have shown then that $d\ne 1$. Let us show that, in equilibrium, $(0<)d<1$ is impossible. Suppose to the contrary that $d<1$. By Lemma 2, $\varepsilon _1 <0<\varepsilon _2$, the balanced-budget constraint, $d^{\alpha }\varepsilon _1 +\varepsilon _2 =0$ and the assumption $d<1$, we get that $\varepsilon _2 =-d^{\alpha }\varepsilon _1 <-\varepsilon _1$ or $\varepsilon _1 +\varepsilon _2 <0$. By Lemma 1 part (a), $\frac{\alpha d^{\alpha }\left( {n_1 +\varepsilon _1 +n_2 +\varepsilon _2 } \right) }{\left( {d^{\alpha }+1} \right) ^{2}}\le 0.25\alpha \left( {n_1 +\varepsilon _1 +n_2 +\varepsilon _2 } \right)$. Since $\varepsilon _1 +\varepsilon _2 <0$, $\frac{\alpha d^{\alpha }\left( {n_1 +\varepsilon _1 +n_2 +\varepsilon _2 } \right) }{\left( {d^{\alpha }+1} \right) ^{2}}<0.25\alpha \left( {n_1 +n_2 } \right)$. But this contradicts the second part of Lemma 1. We therefore obtain that the assumption $d<1$ cannot be satisfied. In equilibrium, then $d=1$ and $d<1$ cannot be satisfied. That is, $d>1$. By the balanced-budget constraint we get that $\varepsilon _2 =-d^{\alpha }\varepsilon _1 >-\varepsilon _1$ and this proves the second part of Proposition 2, that is, $\varepsilon _1 +\varepsilon _2 >0$. $\square$

Proposition 3

The total efforts of the contestants corresponding to the optimal taxation scheme under the APA are equal to the average prize valuation, $G_A =0.5\left( {n_1 +n_2 } \right)$. These total efforts are larger than or equal to those obtained under any Tullock-type lottery with $0<\alpha \le 2$.

Proof

The proof will use the following lemma and its consequences.

Lemma 4

In equilibrium, for $0<\alpha \le 2$,

• a. $\varepsilon _1 \le 0\le \varepsilon _2.$

• b. $d\ge 1$ and, therefore, $\varepsilon _1 +\varepsilon _2 \ge 0$.

Proof of Lemma 4

• a. See step 1 in the proof of Lemma 2.

• b. The proof that, in equilibrium, $(0<)d<1$ is impossible is similar to the one presented in the proof of Lemma 3.b for this case. Notice that the difference between the case dealt with in this section and the case described in Lemma 3.b is that here we also allow $\alpha$ to be equal to 2 and we also allow $d$ to be equal to 1.

$\square$

Recall that by Lemma 1 part (a), for $0<\alpha \le 2$,

$$\begin{aligned} \frac{\alpha d^{\alpha }\left( {n_1 +\varepsilon _1 +n_2 +\varepsilon _2 } \right) }{\left( {d^{\alpha }+1} \right) ^{2}}\le 0.25\alpha \left( {n_1 +\varepsilon _1 +n_2 +\varepsilon _2 } \right) \end{aligned}$$

(20)

Suppose now that the designer wishes to maximize the total efforts $\text{0.25}\alpha (n_1 +\varepsilon _1 + n_2 +\varepsilon _2 )$. That is, he faces the problem:

$$\begin{aligned} \begin{array}{l} \mathop {Max}\limits _{\varepsilon _1 ,\varepsilon _2 ,\alpha } \left( {x_1^*+x_2^*} \right) =0.25\alpha \left( {n_1 +\varepsilon _1 +n_2 +\varepsilon _2 } \right) \\ s.t. \\ 1.d^{\alpha }\varepsilon _1 +\varepsilon _2 =0 \\ 2.\alpha -1-d^{\alpha }\le 0 \\ 3.\left( {\alpha -1} \right) d^{\alpha }-1\le 0 \\ 4.\alpha -2\le 0 \\ 5.n_1 +\varepsilon _1 >0 \\ 6.n_2 +\varepsilon _2 >0 \\ \end{array} \end{aligned}$$

(21)

Let us show that the maximal efforts for this problem are equal to $\text{0.}5\left( {n_1 +n_2 } \right)$. Since this amount can be attained by a Tullock-type lottery with $\alpha =2$ [see Lemma 1 part (b)], inequality (20) implies that the maximal efforts under a Tullock-type lottery is also $\text{0.}5\left( {n_1 +n_2 } \right)$, which will complete the proof of Proposition 3.

Consider problem (21) and let $k>1$.¹⁹ Since, by Lemma 4 part (b), in equilibrium, $d\ge 1$. The fulfillment of the constraint $d\ge 1$ and constraint 4 imply that constraint 2 is also satisfied. We can therefore omit constraint 2 and add the constraint $1-d\le 0$ to obtain the following equivalent designer’s problem:

$$\begin{aligned} \begin{array}{l} \mathop {Max}\limits _{\varepsilon _1 ,\varepsilon _2 ,\alpha } \left( {x_1^*+x_2^*} \right) =0.25\alpha \left( {n_1 +\varepsilon _1 +n_2 +\varepsilon _2 } \right) \\ s.t. \\ 1.d^{\alpha }\varepsilon _1 +\varepsilon _2 =0 \\ 2.\left( {\alpha -1} \right) d^{\alpha }-1\le 0 \\ 3.\alpha -2\le 0 \\ 4.1-d\le 0 \\ 5.n_1 +\varepsilon _1 >0 \\ 6.n_2 +\varepsilon _2 >0 \\ \end{array} \end{aligned}$$

(22)

The Lagrangian function is:

$$\begin{aligned} L&= 0.25\alpha \left( {n_1 +\varepsilon _1 +n_2 +\varepsilon _2 } \right) +\nu _1 \left( {d^{\alpha }\varepsilon _1 +\varepsilon _2 } \right) +\mu _1 \left[ {\left( {1-\alpha } \right) d^{\alpha }+1} \right] \\&+\mu _2 \left( {2-\alpha } \right) +\mu _3 \left( {d-1} \right) \end{aligned}$$

and, in addition, by Lemma 4 part (a), in equilibrium, $\varepsilon _1 \le 0\le \varepsilon _2$ and $\varepsilon _1 +\varepsilon _2 \ge 0$. The following Kuhn–Tucker conditions must therefore be satisfied:

$$\begin{aligned} d^{\alpha }\varepsilon _1 +\varepsilon _2&= 0 \\ \mu _1 \left[ {\left( {1-\alpha } \right) d^{\alpha }+1} \right]&= 0 \quad \left( {1-\alpha } \right) d^{\alpha }+1\ge 0 \quad \mu _1 \ge 0 \\ \mu _2 \left( {2-\alpha } \right)&= 0 \quad 2-\alpha \ge 0 \quad \mu _2 \ge 0 \\ \mu _3 \left( {d-1} \right)&= 0 \quad d-1 \ge 0 \quad \mu _3 \ge 0 \end{aligned}$$

Hence,

$$\begin{aligned} \frac{\partial L}{\partial \varepsilon _1 }&= 0.25\alpha +\nu _1 \left( {\frac{\alpha d^{\alpha -1}}{n_2 +\varepsilon _2 }\varepsilon _1 +d^{\alpha }} \right) +\mu _1 \frac{\alpha d^{\alpha -1}\left( {1-\alpha } \right) }{n_2 +\varepsilon _2 }+\frac{\mu _3 }{n_2 +\varepsilon _2 }=0 \\ \frac{\partial L}{\partial \varepsilon _2 }&= 0.25\alpha +\nu _1 \left( {-\frac{\alpha d^{\alpha }}{n_2 +\varepsilon _2 }\varepsilon _1 +1} \right) +\mu _1 \frac{\alpha d^{\alpha }\left( {\alpha -1} \right) }{n_2 +\varepsilon _2 }-\mu _3 \frac{d}{n_2 +\varepsilon _2 }=0 \\ \frac{\partial L}{\partial \alpha }&= 0.25\left( {n_1 \!+\!\varepsilon _1 \!+\! n_2 \!+\! \varepsilon _2 } \right) \!+\!\nu _1 d^{\alpha }\varepsilon _1 \ln d\!+\!\mu _1 \left[ {\left( {1\!-\!\alpha } \right) d^{\alpha }\ln d\!-\!d^{\alpha }} \right] \!-\!\mu _2 =0 \\ \frac{\partial L}{\partial \nu _1 }&= d^{\alpha }\varepsilon _1 +\varepsilon _2 =0 \end{aligned}$$

Suppose that, in equilibrium, $\alpha <2$, so $\mu _2 =0$. In this case, in equilibrium, $d>1$ [see Lemma 3 part (b)] and, therefore, $\mu _3 =0$. Given these requirements, let us consider the following two possibilities:

Possibility 1

$\left( {1-\alpha } \right) d^{\alpha }+1>0$. In this case, $\mu _1 =0$ and we therefore get that:

$$\begin{aligned} \frac{\partial L}{\partial \varepsilon _2 }&= 0.25\alpha +\nu _1 \left( {-\frac{\alpha d^{\alpha }}{n_2 +\varepsilon _2 }\varepsilon _1 +1} \right) =0 \end{aligned}$$

(23)

$$\begin{aligned} \frac{\partial L}{\partial \alpha }&= 0.25\left( {n_1 +\varepsilon _1 +n_2 +\varepsilon _2 } \right) +\nu _1 d^{\alpha }\varepsilon _1 \ln d=0 \end{aligned}$$

(24)

By (23), since $\varepsilon _1 \le 0\le \varepsilon _2$, we get that:

$$\begin{aligned} \nu _1 =\frac{0.25\alpha }{\frac{\alpha d^{\alpha }}{n_2 +\varepsilon _2 }\varepsilon _1 -1}<0 \end{aligned}$$

(25)

and, by (24), we get that:

$$\begin{aligned} 0.25\left( {n_1 +\varepsilon _1 +n_2 +\varepsilon _2 } \right) =-\nu _1 d^{\alpha }\varepsilon _1 \ln d \end{aligned}$$

(26)

Since, in equilibrium, $\varepsilon _1 +\varepsilon _2 \ge 0$ [Lemma 4 part (b)], the LHS expression in (26) is positive, so the RHS expression must be positive. Since $\varepsilon _1 \le 0\le \varepsilon _2$ and $d>1$ [Lemma 4 part (a) and Lemma 3 part (b)], $\varepsilon _1 <0$, because otherwise the RHS in (26) equals zero. But this implies that if $\left( {-\nu _1 d^{\alpha }\varepsilon _1 \ln d} \right)$ is positive, then $\nu _1 >0$ (since $\varepsilon _1 <0$ and $\ln d>0)$, which contradicts inequality (25). We have thus obtained that, in equilibrium, $\alpha <2$ and $\left( {1-\alpha } \right) d^{\alpha }+1>0$ cannot hold.

Possibility 2

$\left( {1-\alpha } \right) d^{\alpha }+1=0$ (recall that we have assumed that $\alpha <2$ and, therefore, $d>1)$. Note that this possibility requires that $\alpha >1$ and that

$$\begin{aligned} \frac{n_1 +\varepsilon _1 }{n_2 +\varepsilon _2 }=\left( {\alpha -1} \right) ^{-\frac{1}{\alpha }} \end{aligned}$$

(27)

Since, by the balanced-budget constraint, $\left( {d^{\alpha }=} \right) \left( {\frac{n_1 +\varepsilon _1 }{n_2 +\varepsilon _2 }} \right) ^{\alpha }=-\frac{\varepsilon _2 }{\varepsilon _1 }$, in equilibrium, $-\frac{\varepsilon _2 }{\varepsilon _1 }=\left( {\alpha -1} \right) ^{-1}$ or:

$$\begin{aligned} \varepsilon _2 =-\varepsilon _1 \left( {\alpha -1} \right) ^{-1} \end{aligned}$$

(28)

By (27) and (28), we get the taxation scheme $\left( {\varepsilon _1 ,\varepsilon _2 } \right)$:

$$\begin{aligned} \varepsilon _1 =\frac{n_2 \left( {\alpha -1} \right) -n_1 \left( {\alpha -1} \right) ^{\frac{1}{\alpha }+1}}{\left( {\alpha -1} \right) ^{\frac{1}{\alpha }+1}+1},\varepsilon _2 =\frac{n_1 \left( {\alpha -1} \right) ^{\frac{1}{\alpha }}-n_2 }{\left( {\alpha -1} \right) ^{\frac{1}{\alpha }+1}+1} \end{aligned}$$

(29)

Let us show that, in equilibrium, this taxation scheme is impossible. From (28) we get that $\varepsilon _2 \left( {\alpha -1} \right) +\varepsilon _1 =0$ and, therefore, $\varepsilon _2 \left( {\alpha -2} \right) +\varepsilon _1 +\varepsilon _2 =0$ or $\varepsilon _1 +\varepsilon _2 =\varepsilon _2 \left( {2-\alpha } \right)$. We therefore get that:

$$\begin{aligned} 0.25\alpha \left( {n_1 +\varepsilon _1 +n_2 +\varepsilon _2 } \right) =0.25\alpha \left[ {n_1 +n_2 +\varepsilon _2 \left( {2-\alpha } \right) } \right] \end{aligned}$$

(30)

Notice that the selection of $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {-0.5\left( {n_1 -n_2 } \right) ,0.5\left( {n_1 -n_2 } \right) } \right)$ and $\alpha =2$ satisfies all the constraints in (22) and yield total efforts that are equal to $0.5\left( {n_1 +n_2 } \right)$. Let us show that the total efforts in our case are smaller than this amount and this contradiction would imply that the assumption that, in equilibrium, $\left( {1-\alpha } \right) d^{\alpha }+1=0$, together with $\alpha <2$, is impossible. Given (30), we have to show that $0.25\alpha \left[ {n_1 +n_2 +\varepsilon _2 \left( {2-\alpha } \right) } \right] <0.5\left( {n_1 +n_2 } \right)$. This latter inequality can be written as $\alpha \varepsilon _2 \left( {2-\alpha } \right) <\left( {n_1 +n_2 } \right) \left( {2-\alpha } \right)$ and since $1<\alpha <2$, we have to prove that, in equilibrium, $\alpha \varepsilon _2 <n_1 +n_2$.

By substituting $\varepsilon _2$, see (29), in the last inequality we get that we have to prove the following inequality:

$$\begin{aligned} \frac{\alpha \left[ {n_1 \left( {\alpha -1} \right) ^{\frac{1}{\alpha }}-n_2 } \right] }{\left( {\alpha -1} \right) ^{\frac{1}{\alpha }+1}+1}<n_1 +n_2 \end{aligned}$$

or

$$\begin{aligned} n_1 \frac{\alpha \left( {\alpha -1} \right) ^{\frac{1}{\alpha }}}{\left( {\alpha -1} \right) ^{\frac{1}{\alpha }+1}+1}-n_2 \frac{\alpha }{\left( {\alpha -1} \right) ^{\frac{1}{\alpha }+1}+1}<n_1 +n_2 \end{aligned}$$

(31)

Finally, let us prove inequality (31) by showing that the coefficient of $n_1$ in the LHS of (31) is smaller than 1. We have to show then that $\frac{\alpha \left( {\alpha -1} \right) ^{\frac{1}{\alpha }}}{\left( {\alpha -1} \right) ^{\frac{1}{\alpha }+1}+1}<1$ or $\alpha \left( {\alpha -1} \right) ^{\frac{1}{\alpha }}<\left( {\alpha -1} \right) ^{\frac{1}{\alpha }+1}+1$ or $\alpha \left( {\alpha -1} \right) ^{\frac{1}{\alpha }}<\left( {\alpha -1} \right) \left( {\alpha -1} \right) ^{\frac{1}{\alpha }}+1$ or $\left( {\alpha -1} \right) ^{\frac{1}{\alpha }}<1$. Since $1<\alpha <2$, the last inequality is satisfied and, therefore, inequality (31) is also satisfied.

The conclusion from the two possibilities is that in the range $0<\alpha \le 2$, the maximal efforts are obtained when $\alpha =2$. By constraints 2 and 4 in problem (22), this implies that $d=1$. Therefore, by the balanced-budget constraint, we get that $\varepsilon _1 +\varepsilon _2 =0$ and, since $d=1$, we get that, in equilibrium, $\left( {\varepsilon _1 ,\varepsilon _2 } \right) =\left( {-0.5\left( {n_1 -n_2 } \right) ,0.5\left( {n_1 -n_2 } \right) } \right)$. The maximal total efforts are therefore equal to $0.5\left( {n_1 +n_2 } \right)$. That is, for any $\alpha$, $1<\alpha <2$, the maximal efforts are smaller than $0.5\left( {n_1 +n_2 } \right)$. $\square$