All-pay matching contests

Abstract

We study two-sided matching contests with two sets, each of which includes two heterogeneous players with commonly known types. The players in each set compete in all-pay contests where they simultaneously send their costly efforts and then are assortatively matched. A player has a value function that depends on his type as well as his matched one. This model always has a corner equilibrium in which the players do not exert efforts and are randomly matched. We characterize the interior equilibrium and show that although players exert costly (wasted) efforts, this equilibrium might be welfare superior to the corner equilibrium. We analyze the cross effects of the players’ types on the expected payoffs of the other players as well as on their effect on the players’ expected total effort, and demonstrate the complexity of these cross effects.

Appendix

1.1 Proof of Proposition 2

By (5), player $a_{1}$’s probability of winning in set A is

$$\begin{aligned} p_{1}^{a}=1-\frac{f(a_{2},b_{1})-f(a_{2},b_{2})}{ 2(f(a_{1},b_{1})-f(a_{1},b_{2}))}. \end{aligned}$$

By Condition (1), $(f(a_{1},b_{1})-f(a_{1},b_{2}))$ strictly increases in $a_{1}$ , which implies that $\frac{dp_{1}^{a}}{da_{1}} >0$, or, alternatively, $\frac{dp_{2}^{a}}{da_{1}}<0$. Similarly, $f(a_{2},b_{1})-f(a_{2},b_{2})$ strictly decreases in $a_{2}$, which implies that $\frac{dp_{1}^{a}}{da_{2}}<0$ or, alternatively, $\frac{dp_{2}^{a}}{ da_{2}}>0.$

1) By (5),

$$\begin{aligned} \frac{dp_{1}^{a}}{db_{1}}=\frac{ -f_{b}(a_{2},b_{1})(f(a_{1},b_{1})-f(a_{1},b_{2}))+f_{b}(a_{1},b_{1})(f(a_{2},b_{1})-f(a_{2},b_{2})) }{2((f(a_{1},b_{1})-f(a_{1},b_{2})))^{2}}. \end{aligned}$$

Thus, $\frac{dp_{1}^{a}}{db_{1}}\ge 0$, or $\frac{dp_{2}^{a}}{db_{1}}\le 0$ iff $f_{b}(a_{1}b_{1})(f(a_{2},b_{1})-f(a_{2},b_{2}))-f_{b}(a_{2},b_{1})(f(a_{1},b_{1})-f(a_{1},b_{2}))\ge 0.$

2) Similarly,

$$\begin{aligned} \frac{dp_{1}^{a}}{db_{2}}=-\frac{ -2f_{b}(a_{2},b_{2})(f(a_{1},b_{1})-f(a_{1},b_{2}))+2f_{b}(a_{1}b_{2})(f(a_{2},b_{1})-f(a_{2},b_{2})) }{(2(f(a_{1},b_{1})-f(a_{1},b_{2})))^{2}}. \end{aligned}$$

Thus, $\frac{dp_{1}^{a}}{db_{2}}\ge 0$, or $\frac{dp_{2}^{a}}{db_{2}}\le 0$ iff $f_{b}(a_{2},b_{2})(f(a_{1},b_{1})-f(a_{1},b_{2}))-f_{b}(a_{1}b_{2})(f(a_{2},b_{1})-f(a_{2},b_{2}))\ge 0.$ Likewise, we can obtain similar results for the players’ probabilities in set B. $\square$

1.2 Proof of Proposition 3

By (4), player $a_{1}$’s expected payoff is

$$\begin{aligned} u_{1}^{a}=w_{1}^{a}-w_{2}^{a}+l_{2}^{a}=p_{1}^{b}(f(a_{1},b_{1})-f(a_{2},b_{1}) +f(a_{2},b_{2}))+p_{2}^{b}(f(a_{1},b_{2})-f(a_{2},b_{2})+f(a_{2},b_{1})). \end{aligned}$$

Thus,

$$\begin{aligned} \frac{du_{1}^{a}}{da_{1}}&= f_{a}(a_{1},b_{1})p_{1}^{b}+\frac{dp_{1}^{b}}{ da_{1}}(f(a_{1},b_{1})-f(a_{2},b_{1})+f(a_{2},b_{2})) \\&\quad +f_{a}(a_{1},b_{2})p_{2}^{b}+\frac{dp_{2}^{b}}{da_{1}} (f(a_{1},b_{2})-f(a_{2},b_{2})+f(a_{2},b_{1})). \end{aligned}$$

Since $\frac{dp_{1}^{b}}{da_{1}}=-\frac{dp_{2}^{b}}{da_{1}}$ we have

$$\begin{aligned} \frac{du_{1}^{a}}{da_{1}} &= f_{a}(a_{1},b_{1})p_{1}^{b}+f_{a}(a_{1},b_{2})p_{2}^{b} \\&\quad +\frac{dp_{1}^{b}}{da_{1}} ((f(a_{1},b_{1})-f(a_{1},b_{2}))-2(f(a_{2},b_{1})-f(a_{2},b_{2}))). \end{aligned}$$

We obtain that if $\frac{dp_{1}^{b}}{da_{1}}=0$, then $\frac{du_{1}^{a}}{ da_{1}}>0.$

(2) Likewise,

$$\begin{aligned} \frac{du_{1}^{a}}{da_{2}}&= (f_{a}(a_{2},b_{1})-f_{a}(a_{2},b_{2})+f_{a}(a_{2},b_{2}))(p_{2}^{b}-p_{1}^{b}) \\&\quad + \frac{dp_{1}^{b}}{da_{2}}(f(a_{1},b_{1})-f(a_{2},b_{1})+f(a_{2},b_{2})) \\&\quad + \frac{dp_{2}^{b}}{da_{2}}(f(a_{1},b_{2})-f(a_{2},b_{2})+f(a_{2},b_{1})) \\&= f_{a}(a_{2},b_{1})-f_{a}(a_{2},b_{2}))(p_{2}^{b}-p_{1}^{b}) \\&\quad +\frac{dp_{1}^{b}}{da_{2}} (f(a_{1},b_{1})-f(a_{1},b_{2})-2f(a_{2},b_{1})+2f(a_{2},b_{2})). \end{aligned}$$

Since $(p_{2}^{b}-p_{1}^{b})<0$, by Condition (1) we obtain that $(f_{a}(a_{2},b_{1})-f_{a}(a_{2},b_{2})+f_{a}(a_{2},b_{2}))(p_{2}^{b}-p_{1}^{b})<0.$ Thus, if $\frac{dp_{1}^{b}}{da_{2}}=0$, then $\frac{du_{1}^{a}}{da_{2}}<0.$

(3) By (3), player $a_{2}$’s expected payoffs

$$\begin{aligned} u_{2}^{a}=l_{2}^{a}=f(a_{2},b_{1})p_{2}^{b}+f(a_{2},b_{2})p_{1}^{b}. \end{aligned}$$

Then,

$$\begin{aligned} \frac{du_{2}^{a}}{da_{1}} &= f(a_{2},b_{1})\frac{dp_{2}^{b}}{da_{1}} +f(a_{2},b_{2})\frac{dp_{1}^{b}}{da_{1}} \\ &= (f(a_{2},b_{1})-f(a_{2},b_{2}))\frac{dp_{2}^{b}}{da_{1}}. \end{aligned}$$

Thus, if $\frac{dp_{2}^{b}}{da_{1}}=0$, then $\frac{du_{2}^{a}}{da_{1}}=0.$

(4) Last,

$$\begin{aligned} \frac{du_{2}^{a}}{da_{2}} =(f_{a}(a_{2},b_{1})-f_{a}(a_{2},b_{2}))p_{2}^{b}+(f(a_{2},b_{1})-f(a_{2},b_{2})) \frac{dp_{2}^{b}}{da_{2}}. \end{aligned}$$

Since by Condition 1$(f_{a}(a_{2},b_{1})-f_{a}(a_{2},b_{2}))>0$, we obtain that if $\frac{ dp_{2}^{b}}{da_{2}}=0_{,}$ then $\frac{du_{2}^{a}}{da_{2}}>0.$ Likewise, we can obtain similar results on the players’ expected payoffs in set B. $\square$

1.3 Proof of Proposition 4

(1) By (13), the players’ expected total effort in set A is

$$\begin{aligned} TE_{A}=(f(a_{2},b_{1})-f(a_{2},b_{2}))\frac{p_{1}^{b}-p_{2}^{b}}{2}\left( 1+\frac{ f(a_{2},b_{1})-f(a_{2},b_{2})}{f(a_{1},b_{1})-f(a_{1},b_{2})}\right) . \end{aligned}$$

Then,

$$\begin{aligned} \frac{dTE_{A}}{da_{1}} &= \frac{dp_{1}^{b}}{da_{1}} (f(a_{2},b_{1})-f(a_{2},b_{2}))\left( 1+\frac{f(a_{2},b_{1})-f(a_{2},b_{2})}{ f(a_{1},b_{1})-f(a_{1},b_{2})}\right) \\&\quad +(f(a_{2},b_{1})-f(a_{2},b_{2}))\left( \frac{p_{1}^{b}-p_{2}^{b}}{2}\right) \frac{ -(f_{a}(a_{1},b_{1})-f_{a}(a_{1},b_{2}))(f(a_{2},b_{1})-f(a_{2},b_{2}))}{ (f(a_{1},b_{1})-f(a_{1},b_{2}))^{2}}. \end{aligned}$$

By (17), we have that $\frac{dp_{1}^{b}}{da_{1}}\le 0$ iff

$$\begin{aligned} g_{a}(a_{1},b_{1})(g(a_{1},b_{2})-g(a_{2},b_{2}))-g_{a}(a_{1},b_{2})(g(a_{1},b_{1})-g(a_{2},b_{1}))\le 0, \end{aligned}$$

and by Condition 1 we have

$$\begin{aligned} f_{a}(a_{1},b_{1})-f_{a}(a_{1},b_{2})>0. \end{aligned}$$

Thus, if Condition (17) is satisfied, we obtain that $\frac{ dTE_{A}}{da_{1}}<0.$

(2) By (13),

$$\begin{aligned} \frac{dTE_{A}}{da_{2}} &= \left( \frac{-dp_{2}^{b}}{da_{2}} \right) (f(a_{2},b_{1})-f(a_{2},b_{2}))(1+\frac{f(a_{2},b_{1})-f(a_{2},b_{2})}{ f(a_{1},b_{1})-f(a_{1},b_{2})}) \\ & \quad +(f_{a}(a_{2},b_{1})-f_{a}(a_{2},b_{2}))\frac{p_{1}^{b}-p_{2}^{b}}{2}\left( 1+ \frac{f(a_{2},b_{1})-f(a_{2},b_{2})}{f(a_{1},b_{1})-f(a_{1},b_{2})}\right) \\ & \quad +(f(a_{2},b_{1})-f(a_{2},b_{2}))\left( \frac{p_{1}^{b}-p_{2}^{b}}{2}\right) \frac{ (f_{a}(a_{2},b_{1})-f_{a}(a_{2},b_{2}))f(a_{1},b_{1})-f(a_{1},b_{2})}{ f(a_{1},b_{1})-f(a_{1},b_{2})}. \end{aligned}$$

By (18), $\frac{dp_{2}^{b}}{da_{2}}\le 0$ iff

$$\begin{aligned} g_{a}(a_{2},b_{2})(g(a_{1},b_{1})-g(a_{2},b_{1}))-g_{a}(a_{2},b_{1})(g(a_{1},b_{2})-g(a_{2},b_{2}))\ge 0, \end{aligned}$$

and by Condition 1 we have

$$\begin{aligned} f_{a}(a_{1},b_{1})-f_{a}(a_{1},b_{2})>0. \end{aligned}$$

Thus, if Condition (18) is satisfied, we obtain that $\frac{ dTE_{A}}{da_{2}}>0$. $\square$

1.4 Proof of Proposition 5

By (20), the welfare in the assortative matching contest is

$$\begin{aligned} W_{all} &= (f(a_{1},b_{1})+f(a_{2},b_{2}))(p_{1}^{a}p_{1}^{b}+p_{2}^{a}p_{2}^{b}) \\& \quad +(f(a_{1}b_{2})+f(a_{2},b_{1})(p_{1}^{a}p_{2}^{b}+p_{2}^{a}p_{1}^{b}) \\& \quad -TE_{A}-TE_{B}. \end{aligned}$$

Assume that the lower types in both sets approach zero. Then, by (5) and (8)

$$\begin{aligned} \lim _{a_{2}\rightarrow 0}p_{1}^{a}= & {} \lim _{a_{2}\rightarrow 0}1-\frac{ f(a_{2},b_{1})-f(a_{2},b_{2})}{2(f(a_{1},b_{1})-f(a_{1},b_{2}))}=1 \\ \lim _{b_{2}\rightarrow 0}p_{1}^{b}= & {} \lim _{b_{2}\rightarrow 0}1-\frac{ g(a_{1},b_{2})-g(a_{2},b_{2})}{2(g(a_{1},b_{1})-g(a_{2},b_{1}))}=1. \end{aligned}$$

Likewise, by (13) and (14),

$$\begin{aligned} \lim _{a_{2}\rightarrow 0}TE_{A}= & {} \lim _{a_{2}\rightarrow 0}(f(a_{2},b_{1})-f(a_{2},b_{2}))\frac{p_{1}^{b}-p_{2}^{b}}{2}\left( 1+\frac{ f(a_{2},b_{1})-f(a_{2},b_{2})}{f(a_{1},b_{1})-f(a_{1},b_{2})}\right) =0 \\ \lim _{b_{2}\rightarrow 0}TE_{B}= & {} \lim _{b_{2}\rightarrow 0}(g(a_{1},b_{2})-g(a_{2},b_{2}))\frac{p_{1}^{a}-p_{2}^{a}}{2}\left( 1+\frac{ g(a_{1},b_{2})-g(a_{2},b_{2})}{g(a_{1},b_{1})-g(a_{2},b_{1})}\right) =0. \end{aligned}$$

Then,

$$\begin{aligned} \lim _{a_{2},b_{2}\rightarrow 0}W_{all}=f(a_{1},b_{1}) \end{aligned}$$

On the other hand, by (21),

$$\begin{aligned} \lim _{a_{2},b_{2}\rightarrow 0}W_{r}=\frac{f(a_{1},b_{1})}{2}. \end{aligned}$$

Thus,

$$\begin{aligned} \lim _{a_{2},b_{2}\rightarrow 0}W_{all}-W_{r}>0. \end{aligned}$$

Now, assume that the difference between the lower type and the higher type in each set approaches zero. Then by (5) and (8) we have

$$\begin{aligned} \lim _{a_{2}\rightarrow 0}p_{1}^{a}= & {} \lim _{a_{2}\rightarrow a_{1}}1-\frac{ f(a_{2},b_{1})-f(a_{2},b_{2})}{2(f(a_{1},b_{1})-f(a_{1},b_{2}))}=0.5 \\ \lim _{b_{2}\rightarrow 0}p_{1}^{b}= & {} \lim _{b_{2}\rightarrow b_{1}}1-\frac{ g(a_{1},b_{2})-g(a_{2},b_{2})}{2(g(a_{1},b_{1})-g(a_{2},b_{1}))}=0.5, \end{aligned}$$

and therefore, by (13) and (14),

$$\begin{aligned} \lim _{a_{2}\rightarrow a_{1}}TE_{A}= & {} \lim _{a_{2}\rightarrow 0}(f(a_{2},b_{1})-f(a_{2},b_{2}))\frac{p_{1}^{b}-p_{2}^{b}}{2}\left( 1+\frac{ f(a_{2},b_{1})-f(a_{2},b_{2})}{f(a_{1},b_{1})-f(a_{1},b_{2})}\right) =0 \\ \lim _{b_{2}\rightarrow b_{1}}TE_{B}= & {} \lim _{b_{2}\rightarrow 0}(g(a_{1},b_{2})-g(a_{2},b_{2}))\frac{p_{1}^{a}-p_{2}^{a}}{2}\left( 1+\frac{ g(a_{1},b_{2})-g(a_{2},b_{2})}{g(a_{1},b_{1})-g(a_{2},b_{1})}\right) =0. \end{aligned}$$

Thus, we obtain that

$$\begin{aligned} \lim _{\begin{array}{c} a_{1}\rightarrow a_{_{2}} \\ b_{1}\rightarrow b_{2} \end{array}} W_{all}-W_{r}=2f(a_{1},b_{1}). \end{aligned}$$

$\square$