Sequential all-pay auctions with head starts

Abstract

We study a sequential all-pay auction with two contestants who are privately informed about a parameter (ability) that affects their cost of effort. We characterize the unique perfect Bayesian equilibrium of this sequential all-pay auction and analyze if giving a head start, i.e., an exogenously determined mechanism that increases the winning probability of the first mover for any level of effort she exerts, improves the contestants’ performance. In particular, we analyze the difference between a multiplicative head start and an additive head start with respect to the effect on the contestants’ performance.

Appendix

1.1 Proof of Proposition 1

The expected highest effort in the two-player model without a head start is equal to contestant 1’s expected effort, while the expected highest effort in the two-player model with a head start is larger than or equal to contestant 1’s expected effort. Thus, in order to prove that a head start increases the expected highest effort it is sufficient to show that a head start increases contestant 1’s expected effort. However, what we actually show is even stronger. We show that the effort of every type of contestant 1 who made a positive effort when there was no head start increases when there is a head start. Therefore we show that

$$\begin{aligned} \beta _{1}(a_{1})\ge b_{1}(a_{1})\text { for all }0\le a_{1}\le 1\text { and }1\le t\le \frac{1}{f_{2}\left( 1\right) } \end{aligned}$$

Note that if Condition 1 holds then since $b_{1}\left( a_{1}\right)$ is increasing in $a_{1}$ and $\tilde{a}\ge 0$ then for all $t>1,$

$$\begin{aligned} \beta _{1}\left( a_{1}\right) =\frac{1}{t}f_{2}^{-1}\left( \frac{1}{ta_{1}} \right) >f_{2}^{-1}\left( \frac{1}{a_{1}}\right) =b_{1}\left( a_{1}\right) \end{aligned}$$

Moreover the lowest type of contestant 1 who is active in the two-player model with a head start is lower than the lowest active type of contestant 1 in the two-player model without any head start. Formally, $\widehat{a}=\frac{ 1}{tf_{2}\left( 0\right) }\le \frac{1}{f_{2}\left( 0\right) }=\widetilde{a}$ for any $t\ge 1.$ Thus, we have

$$\begin{aligned} \beta _{1}\left( a_{1}\right)&= \frac{1}{t}f_{2}^{-1}\left( \frac{1}{ta_{1}} \right) >f_{2}^{-1}\left( \frac{1}{a_{1}}\right) =b_{1}\left( a_{1}\right) \quad \text { for all }\quad \widetilde{a}\le a_{1}\le 1 \\ \beta _{1}\left( a_{1}\right)&= \frac{1}{t}f_{2}^{-1}\left( \frac{1}{ta_{1}} \right) >b_{1}\left( a_{1}\right) =0\quad \text { for all }\widehat{a}\le a_{1}\le \widetilde{a} \\ \beta _{1}\left( a_{1}\right)&= b_{1}(a_{1})=0\quad \text { for all }\quad 0\le a_{1}\le \text { }\widehat{a} \end{aligned}$$

such that the expected effort of contestant 1 with a head start $t$ is higher than her expected effort without any head start. $Q.E.D.$

1.2 Proof of Proposition 2

The expected effort of contestant 2 given an effort $\beta _{1}\left( a_{1},t\right) >0$ of contestant 1 is

$$\begin{aligned} E_{2}(t,a_{1})=\left( 1-F_{2}\left( t\beta _{1}\left( a_{1},t\right) \right) \right) t\beta _{1}\left( a_{1},t\right) \end{aligned}$$

The expected effort of contestant 2 is then

$$\begin{aligned} TE_{2}(t)=\int \limits _{0}^{1}E_{2}(t,a_{1})f_{1}(a_{1})da_{1}=\int \limits _{\widehat{a} }^{1}E_{2}(t,a_{1})f_{1}(a_{1})da_{1}>0 \end{aligned}$$

The function $t\beta _{1}\left( a_{1},t\right) =f_{2}^{-1}\left( \frac{1}{ a_{1}t}\right)$ is increasing in $a_{1}$ as well as in $t$. By Condition 3, we know that $f_{2}^{-1}\left( 1\right) <x^{*}$. Therefore we obtain that, for $t>1$ close enough to $1$ and for all $a_{1}\le 1,$

$$\begin{aligned} t\beta _{1}\left( a_{1},t\right) \le t\beta _{1}\left( a_{1}=1,t\right) =f_{2}^{-1}\left( \frac{1}{t}\right) <x^{*} \end{aligned}$$

Thus, by Condition 2 we have

$$\begin{aligned} \frac{dE_{2}(t,a_{1})}{dt}>0 \end{aligned}$$

So far we have shown that given a type $\widehat{a}\le a_{1}\le 1$ of contestant 1 that exerts a positive effort, the expected effort of contestant 2 increases in $t$ as long as $t$ is sufficiently close to $1$. By Condition 1, the interval of types of contestant 1 who exert a positive effort increases in $t$, i.e., $\frac{d\widehat{a}}{dt }=\frac{d}{dt}\left( \frac{1}{tf_{2}\left( 0\right) }\right) \le 0$ and therefore, if $t$ is sufficiently close to $1$ we established that

$$\begin{aligned} \frac{d}{dt}TE_{2}(t)=\frac{d}{dt} \int \limits _{0}^{1}E_{2}(t,a_{1})f_{1}(a_{1})da_{1}=\frac{d}{dt}\int \limits _{\widehat{a} }^{1}E_{2}(t,a_{1})f_{1}(a_{1})da_{1}>0 \end{aligned}$$

$Q.E.D.$

1.3 Proof of Proposition 5

The expected highest effort for $0<t\le f_{2}^{-1}\left( 1\right)$ is given by

$$\begin{aligned} HE(t)&= \int \limits _{0}^{\overline{a}}\left( \int \limits _{t}^{1}tf_{2}\left( a_{2}\right) da_{2}\right) f_{1}\left( a_{1}\right) da_{1} \\&+\int \limits _{\overline{a}}^{1}\left( \int \limits _{0}^{\gamma _{1}\left( a_{1}\right) +t}\gamma _{1}\left( a_{1}\right) f_{2}(a_{2})da_{2} \right. \\&\left. +\int \limits _{\gamma _{1}\left( a_{1}\right) +t}^{1}\left( \gamma _{1}\left( a_{1}\right) +t\right) f_{2}(a_{2})da_{2}\right) f_{1}(a_{1})da_{1} \\&= t\left( 1-F_{2}\left( t\right) \right) F_{1}\left( \frac{1}{f_{2}\left( t\right) }\right) \\&+\int \limits _{\overline{a}}^{1}\left( \left( f_{2}{}^{-1}\left( \frac{1}{a_{1}} \right) -t\right) F_{2}\left( f_{2}{}^{-1}\left( \frac{1}{a_{1}}\right) \right) \right. \\&\left. +f_{2}{}^{-1}\left( \frac{1}{a_{1}}\right) \left( 1-F_{2}\left( f_{2}{}^{-1}\left( \frac{1}{a_{1}}\right) \right) \right) \right) f_{1}(a_{1})da_{1} \\&= t\left( 1-F_{2}\left( t\right) \right) F_{1}\left( \frac{1}{f_{2}\left( t\right) }\right) \\&+\int \limits _{\frac{1}{f_{2}\left( t\right) }}^{1}\left( f_{2}{}^{-1}\left( \frac{1}{a_{1}}\right) -tF_{2}\left( f_{2}{}^{-1}\left( \frac{1}{a_{1}}\right) \right) \right) f_{1}(a_{1})da_{1} \end{aligned}$$

Therefore

$$\begin{aligned} \frac{dHE\left( t\right) }{dt}&= \left( 1-F_{2}\left( t\right) -tf_{2}\left( t\right) \right) F_{1}\left( \frac{1}{f_{2}\left( t\right) } \right) -t\left( 1-F_{2}\left( t\right) \right) \frac{f_{2}^{\prime }\left( t\right) }{\left( f_{2}\left( t\right) \right) ^{2}} \\&+t\frac{f_{2}^{\prime }\left( t\right) }{\left( f_{2}\left( t\right) \right) ^{2}}\left( 1-F_{2}\left( t\right) \right) f_{1}\left( \frac{1}{f_{2}\left( t\right) } \right) \\&-\int \limits _{\frac{1}{f_{2}\left( t\right) }}^{1}F_{2}\left( f_{2}{}^{-1}\left( \frac{1}{a_{1}}\right) \right) f_{1}(a_{1})da_{1} \\&= \left( 1-F_{2}\left( t\right) -tf_{2}\left( t\right) \right) F_{1}\left( \frac{1}{f_{2}\left( t\right) }\right) -\int \limits _{\frac{1}{f_{2}\left( t\right) } }^{1}F_{2}\left( f_{2}{}^{-1}\left( \frac{1}{a_{1}}\right) \right) f_{1}(a_{1})da_{1} \end{aligned}$$

Now, since $f_{2}\left( t\right) \rightarrow \infty$ and $tf_{2}\left( t\right) \rightarrow 0$ when $t\rightarrow 0$ we have

$$\begin{aligned} \lim _{t\rightarrow 0}\frac{dHE\left( t\right) }{dt}=-\int \limits _{0}^{1}F_{2}\left( f_{2}{}^{-1}\left( \frac{1}{a_{1}}\right) \right) f_{1}(a_{1})da_{1}<0 \end{aligned}$$

Moreover, since the expected total effort for $0<t\le f_{2}^{-1}\left( 1\right)$ is given by

$$\begin{aligned} TE(t)&= t\left( 1-F_{2}\left( t\right) \right) F_{1}\left( \frac{1}{ f_{2}\left( t\right) }\right) -t\left( 1-F_{1}\left( \frac{1}{f_{2}\left( t\right) }\right) \right) \\&+\int \limits _{\overline{a}}^{1}\left( \left[ 2-F_{2}\left( f_{2}{}^{-1}\left( \frac{1}{a_{1}}\right) \right) \right] f_{2}{}^{-1}\left( \frac{1}{a_{1}} \right) \right) f_{1}(a_{1})da_{1} \\&= t\left( \left( 2-F_{2}\left( t\right) \right) F_{1}\left( \frac{1}{ f_{2}\left( t\right) }\right) -1\right) \\&+\int \limits _{\overline{a}}^{1}\left( \left[ 2-F_{2}\left( f_{2}{}^{-1}\left( \frac{1}{a_{1}}\right) \right) \right] f_{2}{}^{-1}\left( \frac{1}{a_{1}}\right) \right) f_{1}(a_{1})da_{1} \end{aligned}$$

we have

$$\begin{aligned} \frac{dTE}{dt}&= \left( 2-F_{2}\left( t\right) \right) F_{1}\left( \frac{1}{ f_{2}\left( t\right) }\right) -1\\&+t\left( -f_{2}\left( t\right) F_{1}\left( \frac{1}{f_{2}\left( t\right) }\right) -\left( 2-F_{2}\left( t\right) \right) f_{1}\left( \frac{1}{f_{2}\left( t\right) }\right) \frac{ f_{2}^{\prime }\left( t\right) }{\left( f_{2}\left( t\right) \right) ^{2}} \right) \\&+\frac{f_{2}^{\prime }\left( t\right) }{\left( f_{2}\left( t\right) \right) ^{2}}\left( 2-F_{2}(t)\right) tf_{1}\left( \frac{1}{f_{2}\left( t\right) }\right) \\&= \left( 2-F_{2}\left( t\right) -tf_{2}\left( t\right) \right) F_{1}\left( \frac{1}{f_{2}\left( t\right) }\right) -1 \end{aligned}$$

And since $f_{2}\left( t\right) \rightarrow \infty$ and $tf_{2}\left( t\right) \rightarrow 0$ when $t\rightarrow 0$ we obtain that

$$\begin{aligned} \lim _{t\rightarrow 0}\frac{dTE}{dt}=-1 \end{aligned}$$

$Q.E.D.$