Effort complementarity and sharing rules in group contests

Abstract

In this paper, we consider a prize-sharing rule design problem in a group contest with effort complementarities within groups by employing a CES effort aggregator function. We derive the conditions for a monopolization rule that dominates an egalitarian rule if the objective of the rule design is to maximize the group’s winning probability. We find conditions under which the monopolization rule maximizes the group’s winning probability, while the egalitarian rule is strictly preferred by all members of the group. Without effort complementarity, there cannot be such a conflict of interest.

Appendices

Appendix 1

Here, we collect all proofs.

Proof of Lemma 1

Recalling $X_{i}=( \sum _{j=1}^{n_{i}}e_{ij}^{r})^{\frac{1}{r}}$ and given $X_{-i}$, maximizing the winning probability of group i means that $X_{i}$ becomes as large as possible at Nash equilibrium in group i. If $X_{i}$ is a strictly increasing function of $A_{i}$, we can maximize $X_{i}$ by maximizing $A_{i}$ subject to $\sum _{j=1}^{n_{i}}a_{ij}=1$.

From (2), let $\phi (X_{i},A_{i})=X_{i}^{\beta }-P_{i}(1-P_{i})A_{i}=0$. Recalling that $\beta \ge 1$, $0<r\le 1$ and that $P_{i}$ is a function of $X_{i}$ through $P_{i}=\frac{X_{i}}{ X_{i}+X_{-i}}$, and by differentiating $\phi$ with respect to $A_i$, we get

$$\begin{aligned} \frac{dX_{i}}{dA_{i}}=-\frac{\phi _{A_{i}}}{\phi _{X_{i}}}=\frac{ P_{i}(1-P_{i})}{X_{i}^{\beta -1}(\beta -1+2P_{i})}>0 \end{aligned}$$

for $\beta >r$ by using $A_{i}=\frac{X_{i}^{\beta }}{P_{i}(1-P_{i})}$ from (2). Thus, $X_{i}$ is a strictly increasing function in $A_{i}$ for $\beta >r$. $\square$

Proof of Proposition 1

From Lemma 1, it is enough to maximize $A_{i}$. It is also enough to maximize the contents in parentheses in $A_{i}$ because $\frac{\beta -r}{r}>0$. Note that $A_{i}^{\frac{r}{\beta -r}}=\sum _{j=1}^{n_{i}}a_{ij}^{\frac{r}{\beta -r}}$ is an additively separable function. Since $r>0$, our maximization problem boils down to

$$\begin{aligned} \max \sum _{j=1}^{n_{i}}a_{ij}^{\frac{r}{\beta -r}}\text { subject to (i) } \sum _{j=1}^{n_{i}}a_{ij}=1\text { and (ii) }a_{ij}\ge 0 \quad \text { for all } \quad j=1,\ldots ,n_{i}. \end{aligned}$$

Thus, it is easy to see that $\frac{r}{\beta -r}\lesseqgtr 1$ dictates the optimal sharing rule. We obtain three cases:

• Case 1: If $2r<\beta$, $A_{i}$ is maximized when $a_{1}=a_{2}=\ldots =a_{n_{i}}=1/n_{i}$.

• Case 2: If $2r=\beta$, $A_{i}$ is constant for any sharing rule.

• Case 3: If $2r>\beta$, $A_{i}$ is maximized when $a_{ij}=1$ for a single j, and $a_{i\ell }=0$ for all other $\ell$.

$\square$

Proof of Lemma 2

Let $\Delta \equiv n_{i}^{-\frac{\beta -2r}{r\beta }}$ in relation to $n_{i}$ in (4). Rewriting (4), we have

$$\begin{aligned} \left( 1-P_{iE}\right) \left( P_{iE}(1-P_{iE})\right) ^{\frac{1}{\beta } }=P_{iE}X_{-i}\Delta . \end{aligned}$$

(10)

By totally differentiating the above, we obtain

$$\begin{aligned}&\left[ -\left( P_{iE}(1-P_{iE})\right) ^{\frac{1}{\beta }}-X_{-i}\Delta + \frac{1}{\beta }\left( 1-P_{iE}\right) \left( P_{iE}(1-P_{iE})\right) ^{ \frac{1}{\beta }-1}\left( 1-2P_{iE}\right) \right] dP_{iE}\\&\quad =P_{iE}X_{-i}d \Delta . \end{aligned}$$

After solving (10) for $X_{-i}$, we substitute it into the above and obtain

$$\begin{aligned}&\frac{1}{P_{iE}}\left( P_{iE}(1-P_{iE})\right) ^{\frac{1}{\beta }}\left[ -1+ \frac{1}{\beta }\left( 1-2P_{iE}\right) \right] dP_{iE}\\&\quad =\left( 1-P_{iE}\right) \left( P_{iE}(1-P_{iE})\right) ^{\frac{1}{\beta }}\frac{ d\Delta }{\Delta } \end{aligned}$$

or

$$\begin{aligned} \frac{dP_{iE}}{d\Delta }=\frac{P_{iE}\left( 1-P_{iE}\right) }{\Delta \left[ -1+\frac{1}{\beta }\left( 1-2P_{iE}\right) \right] }. \end{aligned}$$

Since $-1+\frac{1}{\beta }\left( 1-2P_{iE}\right) <0$ from $\beta \ge 1$, we have $\frac{dP_{iE}}{d\Delta }<0$. By differentiating $\Delta$ with respect to $n_{i}$ and r, we have

$$\begin{aligned} \frac{d\Delta }{dn_{i}}=-\frac{\beta -2r}{r\beta }\times \frac{\Delta }{n_{i} } \end{aligned}$$

and

$$\begin{aligned} \frac{d\Delta }{dr}=\frac{1}{r^{2}}(\log n_{i})\Delta >0, \end{aligned}$$

respectively. We obtain the results using the chain rule. $\square$

Proof of Lemma 3

Recall $U_{iE}(n_{i},r)=\frac{P_{iE}(n_{i},r)}{n_{i}}\left( 1-\frac{1}{\beta }\left( 1-P_{iE}(n_{i},r)\right) \frac{1}{n_{i}}\right) \equiv {\tilde{U}} (n_{i},P_{iE}(n_{i},r))$. This implies

$$\begin{aligned} \frac{\partial {\tilde{U}}}{\partial P_{iE}}=\frac{1}{n_{i}}-\frac{1}{\beta n_{i}^{2}}(1-2P_{iE})=\frac{1}{n_{i}^{2}\beta }(n_{i}\beta +2P_{iE}-1). \end{aligned}$$

(11)

Thus, by totally differentiating ${\tilde{U}}(n_{i},P_{iE}(n_{i},1))$ with respect to $n_{i}$ using (6), we obtain

$$\begin{aligned}&\frac{dU_{iE}(n_{i},1)}{dn_{i}} \\&\quad =\frac{\partial {\tilde{U}}}{\partial n_{i}}+\frac{\partial {\tilde{U}}}{ \partial P_{iE}}\frac{dP_{iE}}{dn_{i}} \\&\quad =\frac{P_{iE}}{n_{i}^{2}}\left[ -1+\frac{2}{n_{i}\beta }(1-P_{iE})\right] + \frac{\left( 2-\beta \right) P_{iE}\left( 1-P_{iE}\right) }{n_{i}\left( 1-2P_{iE}-\beta \right) }\left[ \frac{1}{n_{i}^{2}\beta }(n_{i}\beta +2P_{iE}-1)\right] \\&\quad =\frac{P_{iE}}{n_{i}^{3}}\left[ -n_{i}+\frac{2}{\beta }\left( 1-P_{iE}\right) +\left( 1-P_{iE}\right) \frac{2-\beta }{1-2P_{iE}-\beta } \left( n_{i}+\frac{1}{\beta }\left( 2P_{iE}-1\right) \right) \right] \\&\quad =\frac{P_{iE}}{n_{i}^{3}\beta \left( 1-2P_{iE}-\beta \right) } \\&\qquad \times \Big [ -n_{i}\beta \left( 1-2P_{iE}-\beta \right) +2\left( 1-P_{iE}\right) \left( 1-2P_{iE}-\beta \right) \\&\qquad + \,\left( 1-P_{iE}\Big ) \left( 2-\beta \right) \left( \beta n_{i}-\left( 1-2P_{iE}\right) \right) \right] . \end{aligned}$$

Since $1-2P_{iE}-\beta <0$, we can focus on the sign of the contents of the brackets:

$$\begin{aligned} \left[ \ \cdot \ \right]= & {} -n_{i}\beta \left( 1-2P_{iE}-\beta \right) +n_{i}\beta \left( 2-\beta -2P_{iE}+P_{iE}\beta \right) \\&+2\left( 1-2P_{iE}-\beta \right) \left( 1-P_{iE}\right) -\left( 1-P_{iE}\right) \left( 2-\beta \right) \left( 1-2P_{iE}\right) \\= & {} \beta [(n_{i}-1)+P_{iE}(n_{i}\beta -1+2P_{iE})]>0 \end{aligned}$$

for any $n_{i}\ge 1$. Thus, we conclude that $\frac{dU_{iE}}{dn_{i}}<0$ for any $n_{i}\ge 1$. This implies that $U_{iE}(n_{i},1)<U_{iM}$ for any $n_{i}\ge 2$ when $r=1$. We have completed the proof. $\square$

Proof of Lemma 4

First note that the assumptions $n_{i}\ge 2$ and $1+\frac{1}{n_{i}}>\beta$ imply $2>\beta$. Consider the case of $r=\frac{\beta }{2}$. Since $P_{iE}(n_{i},\frac{\beta }{ 2})=P_{iE}(1,\frac{\beta }{2})=P_{iM}$ by Proposition 1, we have

$$\begin{aligned} U_{iE}\left( n_{i},\frac{\beta }{2}\right)= & {} \frac{P_{iE}\left( n_{i},\frac{\beta }{2}\right) }{n_{i} }\left( 1-\frac{1}{\beta }\left( 1-P_{iE}\left( n_{i},\frac{\beta }{2}\right) \right) \frac{1}{n_{i}} \right) \\= & {} \frac{P_{iM}}{n_{i}}\left( 1-\frac{1}{\beta }(1-P_{iM})\frac{1}{n_{i}} \right) . \end{aligned}$$

By subtracting $U_{iM}=P_{iM}\left[ 1-\frac{1}{\beta }(1-P_{iM})\right]$ from $U_{iE}(n_{i},\frac{\beta }{2})$, we obtain

$$\begin{aligned} U_{iE}\left( n_{i},\frac{\beta }{2}\right) -U_{iM}= & {} P_{iM}\left[ -1+\frac{1}{\beta } (1-P_{iM})+\frac{1}{n_{i}}\left( 1-\frac{1}{\beta }(1-P_{iM})\frac{1}{n_{i}} \right) \right] \\= & {} P_{iM}\left[ -\left( 1-\frac{1}{n_{i}}-\frac{1}{\beta }\left( 1-\frac{1}{ n_{i}^{2}}\right) \right) -\frac{1}{\beta }P_{iM}\left( 1-\frac{1}{n_{i}^{2}} \right) \right] \\= & {} P_{iM}\left( 1-\frac{1}{n_{i}^{2}}\right) \left[ -\frac{n_{i}}{n_{i}+1}+ \frac{1}{\beta }-\frac{1}{\beta }P_{iM}\right] . \end{aligned}$$

Then, the condition of $U_{iE}(n_{i},\frac{\beta }{2})>U_{iM}$ is

$$\begin{aligned} 1-\frac{n_{i}}{n_{i}+1}\beta >P_{iM}. \end{aligned}$$

(12)

That is, if (12) is satisfied, $U_{iE}(n_{i},\frac{\beta }{2} )>U_{iM}$ holds, while $U_{iE}(n_{i},1)<U_{iM}$. Since $\frac{dP_{iE}}{dr}<0$ holds by (5) in Lemma 2 and from (11), $\frac{\partial {\tilde{U}}}{\partial P_{iE}}=\frac{1}{n_{i}^{2}\beta }(n_{i}\beta +2P_{iE}-1)>0$, we have $\frac{dU_{iE}(P_{iE}(r))}{dr}=\frac{\partial \tilde{ U}_{iE}}{\partial P_{iE}}\frac{dP_{iE}}{dr}<0$, which is $U_{iE}$ monotonically decreasing in r. Considering the above facts and given that $U_{iE}$ is continuous in r, there is a unique ${\hat{r}}\in (\frac{\beta }{2} ,1)$, such that $U_{iE}(n_{i},r)<U_{iM}$ holds for all $r\in ({\hat{r}},1]$ and $U_{iE}(n_{i},r)>U_{iM}$ holds for all $r\in [\frac{\beta }{2}, {\hat{r}})$. $\square$

Fig. 1

Share function $P_i(X;A_i)$ of group i and aggregate share function f(X; A)

Proof of Lemma 5

First, focus on the $P_{i}(X;A_{i})$ function. Starting from the original $A_{i}$ and equilibrium $X^{*}$, $A_{i}$ is increased by $\Delta A_{i}>0$. Since $\frac{\partial P_{i}}{\partial A_{i}}>0$ for all X from (9 ), the $P_{i}$ function shifts up vertically. Let ${\tilde{X}}$ be such that $P_{i}(X^{*};A_{i})=P_{i}({\tilde{X}};A_{i}+\Delta A_{i})$ (see Fig. 1). Since $\frac{\partial P_{i}}{\partial X}<0$ from (8), ${\tilde{X}} >X^{*}$ holds, and for any $X\in (X^{*},{\tilde{X}})$, we have $P_{i}(X;A_{i}+\Delta A_{i})>P_{i}(X^{*};A_{i})$. Recall that the equilibrium $X^{*}$ is described by the aggregate share function

$$\begin{aligned} f(X^{*};A)=\sum _{i^{\prime }\ne i}P_{i^{\prime }}(X^{*};A_{i^{\prime }})+P_{i}(X^{*};A_{i})=1. \end{aligned}$$

Let $A_{-i}$ be a vector that removes $A_{i}$ from A. By increasing $A_{i}$ by $\Delta A_{i}$, the equilibrium aggregate effort $X^{**}$ satisfies

$$\begin{aligned} f(X^{**};A_{i}+\Delta A_{i},A_{-i})=\sum _{i^{\prime }\ne i}P_{i^{\prime }}(X^{**};A_{i^{\prime }})+P_{i}(X^{**};A_{i}+\Delta A_{i})=1. \end{aligned}$$

Since $\frac{\partial P_{i^{\prime }}}{\partial X}<0$ for all $i^{\prime }=1,\ldots ,m$, we have $X^{**}>X^{*}$ and

$$\begin{aligned} f({\tilde{X}};A_{i}+\Delta A_{i},A_{-i})= & {} \sum _{i^{\prime }\ne i}P_{i^{\prime }}({\tilde{X}};A_{i^{\prime }})+P_{i}({\tilde{X}};A_{i}+\Delta A_{i}) \\= & {} \sum _{i^{\prime }\ne i}P_{i^{\prime }}({\tilde{X}};A_{i^{\prime }})+P_{i}(X^{*};A_{i})<1. \end{aligned}$$

By the intermediate value theorem, $X^{**}\in (X^{*},{\tilde{X}})$ holds. We conclude $P_{i}(X^{**};A_{i}+\Delta A_{i})>P_{i}(X^{*};A_{i})$.

This implies that as $A_{i}$ increases, $P_{i}(X^*; A_{i})$ increases. That is, maximizing $A_{i}$ achieves the maximum winning probability for group i. $\square$

Appendix 2

Here, we repeat our analysis by using the Epstein and Mealem’s generalized Tullock contest, and show that Lemma 1 and Proposition 1 hold. We confirm this first. The expected payoff of member j in group i is $U_{ij}=\frac{\sum _{j=1}^{n_{i}}e_{ij}^{r}}{ \sum _{j=1}^{n_{i}}e_{ij}^{r}+X_{-i}}a_{ij}-\frac{1}{\beta }e_{ij}^{\beta }$. The first order condition is

$$\begin{aligned} \frac{\partial U_{ij}}{\partial e_{ij}}=\frac{re_{ij}^{r-1}X_{-i}}{ \left( \sum _{j=1}^{n_{i}}e_{ij}^{r}+X_{-i}\right) ^{2}}a_{ij}-e_{ij}^{\beta -1}=0. \end{aligned}$$

This can be rewritten as

$$\begin{aligned} P_{i}(1-P_{i})\frac{re_{ij}^{r}}{\sum _{j=1}^{n_{i}}e_{ij}^{r}} a_{ij}-e_{ij}^{\beta }=0. \end{aligned}$$

(13)

We process a procedure similar to the one at the end of Sect. 2 and get $e_{ij}^{r}=\left( \frac{rP_{i}(1-P_{i})}{X_{i}}\right) ^{\frac{r}{\beta -r} }a_{ij}^{\frac{r}{\beta -r}}$ from (13). By summing up each $e_{ij}^{r}$, we have $\sum _{j=1}^{n_{i}}e_{ij}^{r}=X_{i}=\left( \frac{ rP_{i}(1-P_{i})}{X_{i}}\right) ^{\frac{r}{\beta -r}}\hat{A_i}$ where $\hat{ A_i} = \sum _{j=1}^{n_i}a_{ij}^{\frac{r}{\beta -r}}$. Let ${\hat{\phi }}(X_{i}, \hat{A_i})=X_{i}-\left( \frac{rP_{i}(1-P_{i})}{X_{i}}\right) ^{\frac{r}{ \beta -r}}\hat{A_i}=0.$ By differentiating ${\hat{\phi }}$ with respect to $\hat{A_i}$ and noting that $P_{i}$ is a function of $X_{i}$, we have

$$\begin{aligned} \frac{dX_{i}}{d\hat{A_i}}=-\frac{{\hat{\phi }}_{\hat{A_i}}}{{\hat{\phi }}_{X_{i}}}= \frac{(rP_{i}(1-P_{i})/X_{i})^{\frac{r}{\beta -r}}}{(\beta -r+2rP_{i})/(\beta -r)}>0 \end{aligned}$$

for $\beta >r$ by using $\hat{A_i}=X_{i}\left( \frac{rP_{i}(1-P_{i})}{X_{i}} \right) ^{-\frac{r}{\beta -r}}$. Therefore, since Lemma 1 holds, Proposition 1 also holds in this case. Proposition 2 holds as well. However, Proposition 3 does not hold. We check this second point. Under the egalitarian rule, since $X_{i}= \sum _{j=1}^{n_{i}}e_{ij}^{r}=n_{i}e_{i}^{r}$, we have $e_{i}=n_{i}^{-\frac{2 }{\beta }}r^{\frac{1}{\beta }}P_{iE}^{\frac{1}{\beta }}(1-P_{iE})^{\frac{1}{ \beta }}$ from (13). Using this, we have

$$\begin{aligned} P_{iE}=\frac{n_{i}e_{i}^{r}}{n_{i}e_{i}^{r}+X_{-i}}=\frac{n_{i}^{\frac{\beta -2r}{\beta }}r^{\frac{r}{\beta }}P_{iE}^{\frac{r}{\beta }}(1-P_{iE})^{\frac{r }{\beta }}}{n_{i}^{\frac{\beta -2r}{\beta }}r^{\frac{r}{\beta }}P_{iE}^{ \frac{r}{\beta }}(1-P_{iE})^{\frac{r}{\beta }}+X_{-i}} \end{aligned}$$

(14)

and

$$\begin{aligned} U_{iE}=P_{iE}\frac{1}{n_{i}}-\frac{1}{\beta }e_{i}^{\beta }=\frac{P_{iE}}{ n_{i}}\left( 1-\frac{1}{\beta }(1-P_{iE})\frac{r}{n_{i}}\right) . \end{aligned}$$

Let ${\hat{\Delta }} \equiv n_i^{-\frac{\beta -2r}{\beta }}$ in relation to $n_i$ in (14). We process the same procedure as in the proof of Lemma 2. Rewriting (14), we have

$$\begin{aligned} r^{\frac{r}{\beta }} \left( 1-P_{iE}\right) \left( P_{iE}(1-P_{iE})\right) ^{ \frac{r}{\beta }}=P_{iE}X_{-i} {\hat{\Delta }}. \end{aligned}$$

(15)

By totally differentiating the above expression and conducting the same operations as the proof of Lemma 2, we obtain

$$\begin{aligned} \frac{dP_{iE}}{d {\hat{\Delta }} }=\frac{P_{iE}\left( 1-P_{iE}\right) }{ \hat{ \Delta } \left[ -1+\frac{r}{\beta }\left( 1-2P_{iE}\right) \right] }<0 \end{aligned}$$

because of $\frac{r}{\beta }(1-2P_{iE}) < 1$. Differentiating ${\hat{\Delta }}$ with respect to $n_i$, we have

$$\begin{aligned} \frac{d {\hat{\Delta }} }{d n_{i}}=- \frac{\beta -2r}{\beta }\times \frac{ \hat{ \Delta } }{n_{i}}. \end{aligned}$$

We then obtain $\frac{dP_{iE}}{dn_{i}}=\frac{2r-\beta }{n_i} \frac{ P_{iE}(1-P_{iE})}{r(1-2P_{iE}) -\beta }.$ The sign of this formula depends only on the sign of $2r - \beta$, as well as (6) from Lemma 2. Therefore, Proposition 2 holds in this case.

By processing the same procedure as the proof of Lemma 4, we obtain

$$\begin{aligned} U_{iE}\left( n_{i},\frac{\beta }{2}\right) -U_{iM}= & {} P_{iM}\left[ -1+\frac{1}{\beta } (1-P_{iM})r+\frac{1}{n_{i}}\left( 1-\frac{1}{\beta }(1-P_{iM})\frac{r}{n_{i}} \right) \right] \\= & {} P_{iM}\left( 1-\frac{1}{n_{i}^{2}}\right) \left[ -\frac{n_{i}}{n_{i}+1}+ \frac{r}{\beta }-\frac{r}{\beta }P_{iM}\right] \\= & {} P_{iM}\left( 1-\frac{1}{n_{i}^{2}}\right) \left[ -\frac{n_{i}}{n_{i}+1}+ \frac{1}{2}-\frac{1}{2}P_{iM}\right] \end{aligned}$$

at $r=\frac{\beta }{2}$. For the above expression to be positive, the sign in the brackets needs to be positive. Thus,

$$\begin{aligned} P_{iM}<\frac{1-n_{i}}{n_{i}+1}<0. \end{aligned}$$

However, this condition contradicts the definition of the probability. Lemma 4 does not hold. Therefore, Proposition 3 also fails to hold in the generalized Tullock contest. $\square$