Two-stage contests with effort-dependent values of winning

Abstract

We study two-stage all-pay contests in which synergy exists between the stages. The value of winning for each contestant is fixed in the first stage while it is effort-dependent in the second one. We assume that a player’s effort in the first stage either increases (positive synergy) or decreases (negative synergy) his value of winning in the second stage. The subgame perfect equilibrium of this contest is analyzed with either positive or negative synergy. We show, in particular, that whether the contestants are symmetric or asymmetric their expected payoffs may be higher under negative synergy than under positive synergy. Consequently, they prefer smaller values of winning (negative synergy) over higher ones (positive synergy).

Appendix

1.1 Proof of Proposition 1

We can see that the functions $F_{p}^{i}(x)$, $i=1,2$, are well defined, strictly increasing on $\left[ 0,x_{\max -p}\right]$, continuous, and that $F_{p}^{1}(0)=0$, $F_{p}^{2}(0)=\frac{1}{\alpha }-\frac{(1-\alpha )^{\frac{ v_{1}^{1}-v_{1}^{2}}{v_{1}^{1}}}}{\alpha }>0$ and $F_{p}^{1}(x_{\max -p})=F_{p}^{2}(x_{\max -p})=1$. Thus, $F_{p}^{i}(x)$, $i=1,2$ are cumulative distribution functions of continuous probability distributions supported on $\left[ 0,x_{\max -p}\right]$. In order to see that the above strategies are an equilibrium, note that when contestant 2 uses the mixed strategy $F_{p}^{2}(x)$, contestant 1’s expected payoff is

$$\begin{aligned} \pi _{p}^{1}=v_{1}^{1}\left( \frac{1}{\alpha }-\frac{(1-\alpha )^{\frac{ v_{1}^{1}-v_{1}^{2}}{v_{1}^{1}}}}{\alpha }\right) \end{aligned}$$

for any effort $x\in [0,x_{\max -p}]$. Since it can be easily shown that for contestant 1, efforts above $x_{\max -p_{a}}$ would lead to a lower expected payoff than $v_{1}^{1}(\frac{1}{\alpha }-\frac{(1-\alpha )^{\frac{ v_{1}^{1}-v_{1}^{2}}{v_{1}^{1}}}}{\alpha })$, any effort in $[0,x_{\max -p}]$ is a best response of contestant 1 to $F_{p}^{2}(x)$. Similarly, when contestant 1 uses the mixed strategy $F_{p}^{1}(x)$, contestant 2’s expected payoff is

$$\begin{aligned} \pi _{p}^{2}=0 \end{aligned}$$

for any effort $x\in [0,x_{\max -p_{a}}]$. Since it can be easily shown that for contestant 2, efforts above $x_{\max -p}$ would result in a non-positive expected payoff, any effort in $[0,x_{\max -p}]$ is a best response of contestant 2 to $F_{p}^{1}(x)$. Hence, the pair $\left( F_{p}^{1}(x),F_{p}^{2}(x)\right)$ is a mixed strategy equilibrium.

1.2 Proof of Proposition 2

It is clear that the functions $F_{n}^{i}(x)$, $i=1,2$, are well defined, strictly increasing on $\left[ 0,x_{\max -n}\right]$, continuous, and that $F_{n}^{1}(0)=0$, $F_{n}^{2}(0)=\frac{1+\alpha }{\alpha }-\frac{(1+\alpha )^{ \frac{v_{1}^{2}}{v_{1}^{1}}}}{\alpha }$ and $F_{n}^{1}(x_{\max -n})=F_{n}^{2}(x_{\max -n})=1$. Thus, $F_{n}^{i}(x)$, $i=1,2$ are cumulative distribution functions of continuous probability distributions supported on $\left[ 0,x_{\max -n}\right]$.

In order to see that the above strategies are an equilibrium, note that when contestant 2 uses the mixed strategy $F_{n}^{2}(x)$, contestant 1’s expected payoff is

$$\begin{aligned} \pi _{n}^{1}=v_{1}^{1}-\frac{v_{1}^{2}}{\alpha }\ln (1+\alpha ) \end{aligned}$$

for any effort $x\in [0,x_{\max -n}]$. Since it can be easily shown that for contestant 1, efforts above $x_{\max -n}$ would lead to a lower expected payoff than $v_{1}^{1}-\frac{v_{1}^{2}}{\alpha }\ln (1+\alpha )$, any effort in $[0,x_{\max -n}]$ is a best response of contestant 1 to $F_{n}^{2}(x)$. Similarly, when contestant 1 uses the mixed strategy $F_{n}^{1}(x)$, contestant 2’s expected payoff is

$$\begin{aligned} \pi _{n}^{2}=v_{1}^{2}-\frac{v_{1}^{2}}{\alpha }\ln (1+\alpha ) \end{aligned}$$

for any effort $x\in [0,x_{\max -n}]$. Since it can be easily shown that for contestant 2, efforts above $x_{\max -n}$ would result in a lower expected payoff than $v_{1}^{2}-\frac{v_{1}^{2}}{\alpha }\ln (1+\alpha )$, any effort in $[0,x_{\max -n}]$ is a best response of contestant 2 to $F_{n}^{1}(x)$. Hence, the pair $\left( F_{n}^{1}(x),F_{n}^{2}(x)\right)$ is a mixed strategy equilibrium.

1.3 Proof of Proposition 4

By (11), when the synergy is negative we have that $\frac{d\pi _{n}^{1}}{d\alpha }\ge 0$, and

$$\begin{aligned} \frac{d\pi _{n}^{1}}{d\alpha }= & {} \frac{1}{\alpha ^{2}}\frac{v_{1}^{2}}{ \alpha +1}(\ln \left( \alpha +1\right) (1+\alpha )-\alpha ) \\\le & {} \frac{1}{\alpha ^{2}}\frac{v_{1}^{2}}{\alpha +1}(\alpha (1+\alpha )-\alpha )=\frac{v_{1}^{2}}{\alpha +1}\le v_{1}^{2} \end{aligned}$$

On the other hand, by (5), when the synergy is positive we have that $\frac{d\pi _{p}^{1}}{d\alpha }\ge 0$ and

$$\begin{aligned} \lim _{\alpha \rightarrow 1}\frac{d\pi _{p}^{1}}{d\alpha }=\lim _{\alpha \rightarrow 1}\frac{1}{\alpha ^{2}}\left( \frac{v_{1}^{1}-\alpha v_{1}^{2}}{ \left( 1-\alpha \right) ^{\frac{v_{1}^{2}}{v_{1}^{1}}}}-v_{1}^{1}\right) =\infty \end{aligned}$$

Thus, when $\alpha$ is sufficiently close to 1, we obtain that $\pi _{p}^{1}-\pi _{n}^{1}>0$.

1.4 Proof of Proposition 5

By (6), if the contestants are symmetric, a contestant’s expected effort in the first stage of the contest under positive synergy is

$$\begin{aligned} E_{p}=\mathop {\displaystyle \int }\limits _{0}^{\frac{-v_{1}}{\alpha }\ln (1-\alpha )}\frac{x}{v_{1}} e^{-\frac{\alpha x}{v_{1}}}dx=\frac{v_{1}}{\alpha ^{2}}(\alpha +(1-\alpha )\ln (1-\alpha )) \end{aligned}$$

And, by (12), if the contestants are symmetric, a contestant’s expected effort in the first stage of the contest under negative synergy is

$$\begin{aligned} E_{n}=\mathop {\displaystyle \int }\limits _{0}^{\frac{v_{1}}{\alpha }\ln (1+\alpha )}\frac{1+\alpha }{v_{1}}xe^{-\frac{\alpha x}{v_{1}}}dx=\frac{v_{1}}{\alpha ^{2}}(\alpha -\ln (1+\alpha )) \end{aligned}$$

Thus we have

$$\begin{aligned} E_{p}-E_{n}=\frac{v_{1}}{\alpha ^{2}}((1-\alpha )\ln (1-\alpha )+\ln (1+\alpha )) \end{aligned}$$

In order to show that $E_{p}-E_{n}>0$, we define

$$\begin{aligned} g(\alpha )=(1-\alpha )\ln (1-\alpha )+\ln (1+\alpha ) \end{aligned}$$

Then, we have

$$\begin{aligned} g^{\prime }(\alpha )=-\frac{1}{\alpha +1}\left( \alpha +\ln \left( 1-\alpha \right) +\alpha \ln \left( 1-\alpha \right) \right) \end{aligned}$$

Since $g(0)=g^{\prime }(0)=0$ and

$$\begin{aligned} g^{\prime \prime }(\alpha )=\frac{\alpha }{\left( 1-\alpha \right) \left( \alpha +1\right) ^{2}}\left( \alpha +3\right) >0 \end{aligned}$$

we obtain that $g^{\prime }(\alpha )>0$ and therefore $g(\alpha )>0$ for all $0<\alpha <1$. Thus, $E_{p}-E_{n}>0$.