Information disclosure in all-pay contests with costly entry

Abstract

In this paper, we accommodate the costly entry of contestants and examine an information design problem when the organizer can decide how to generate contestants’ private information. The information designer should take into account both ex ante entry incentives and post-entry effort elicitation. We show that no transparency (full transparency) induces greater expected aggregate effort if the entry cost is lower (higher) than a threshold. We further consider randomized disclosure policies and identify the optimal degree of transparency, which increases with the entry cost to attract entry. In particular, depending on the entry cost, diverse randomized disclosure policies could be optimal. Our results indicate that endogenous participation plays a crucial role in the design of information revelation.

Appendices

Appendix

Proof of Lemma 3

For a contestant i, suppose that his opponent participates with probability $q^{\prime \prime }$, $i^{\prime }s$ ex ante expected payoff from participation equals

$$\begin{aligned} U_{i} &= (1-q^{\prime \prime })v+q^{\prime \prime }\left[ \Pr (v_{i}=v_{H})u_{i}(v_{i}=v_{H})+\Pr (v_{i}=v_{L})u_{i}(v_{i}=v_{L})\right] \\ &= (1-q^{\prime \prime })v+q^{\prime \prime }[p(1-p)(v_{H}-v_{L})], \end{aligned}$$

where $u_{i}(v_{i}=v_{H})$ is the interim payoff of a $v_{H}$-type, which equals $(1-p)(v_{H}-v_{L})$ by (3), and $u_{i}(v_{i}=v_{L})$ is the interim payoff of a $v_{L}$-type, which equals 0, by Lemma 1.

A contestant would be indifferent between participating and staying out:

$$\begin{aligned} c=(1-q^{\prime \prime })v+q^{\prime \prime }p(1-p)(v_{H}-v_{L}), \end{aligned}$$

whereby we solve for the equilibrium entry probability $q^{\prime \prime }=\min \left\{\frac{v-c}{v-p(1-p)(v_{H}-v_{L})},1\right\}\in [0,1]$.

Under full transparency, after $i^{\prime }s$ entry, i observes his valuation $v_{i}$, which could be $v_{H}$ or $v_{L}$. Note that i will earn $(1-q^{\prime \prime })v_{i}$ from participation, regardless of effort, since his opponent will be absent with probability $(1-q^{\prime \prime })$. In fact, under full transparency, this contest game is equivalent to an incomplete-information all-pay auction with the adjusted valuations $v_{H}^{\prime \prime }=q^{\prime \prime }v_{H}$ and $v_{L}^{\prime \prime }=q^{\prime \prime }v_{L}$ and probability p. Therefore, by Lemma 1, the equilibrium strategy is given by

$$\begin{aligned} F(x|v_{i} &= v_{H})=\frac{x-(1-p)v_{L}^{\prime \prime }}{pv_{H}^{\prime \prime }},\quad \forall x\in [(1-p)v_{L}^{\prime \prime },(1-p)v_{L}^{\prime \prime }+pv_{H}^{\prime \prime }]; \\ F(x|v_{i} &= v_{L})=\frac{x}{(1-p)v_{L}^{\prime \prime }}, \quad \forall x\in [0,(1-p)v_{L}^{\prime \prime }]. \end{aligned}$$

Based on the equilibrium characterization, we calculate the sum of contestants’ equilibrium efforts as follows.

$$\begin{aligned} TE_{Full}(c ) & = 2q^{\prime \prime }\left[ p((1-p)v_{L}^{\prime \prime }+\frac{1}{2} pv_{H}^{\prime \prime })+(1-p)\frac{1}{2}(1-p)v_{L}^{\prime \prime }\right] \\ &= 2(q^{\prime \prime })^{2}\left[ p((1-p)v_{L}+\frac{1}{2}pv_{H})+(1-p)\frac{1}{2} (1-p)v_{L}\right] \\ &= (q^{\prime \prime })^{2}[p^{2}v_{H}+(1-p^{2})v_{L}]. \end{aligned}$$

$\square$

Proof of Proposition 2

By Lemma 3, $TE_{Full}(c)=(q^{\prime \prime })^{2}\left[ p^{2}v_{H}+(1-p^{2})v_{L}\right]$, where $q^{\prime \prime }=\min \{\frac{ v-c}{v-p(1-p)(v_{H}-v_{L})},1\}$. To show the result, we consider $[v-p(1-p)(v_{H}-v_{L})]^{2}\le v[p^{2}v_{H}+(1-p^{2})v_{L}]$, where

$$\begin{aligned}{}[v-p(1-p)(v_{H}-v_{L})]^{2}=[p^{2}v_{H}+(1-p^{2})v_{L}]^{2}\le v[p^{2}v_{H}+(1-p^{2})v_{L}]\text {,} \end{aligned}$$

since $p^{2}v_{H}+(1-p^{2})v_{L}<v=pv_{H}+(1-p)v_{L}$.

Case 1: If $c\in [v-\{v[p^{2}v_{H}+(1-p^{2})v_{L}]\}^{\frac{1}{2}}$, $p(1-p)(v_{H}-v_{L}))$, we have $\frac{v-c}{v-p(1-p)(v_{H}-v_{L})}>1$, $q^{\prime \prime }=1$, and $TE_{Full}(c)=[p^{2}v_{H}+(1-p^{2})v_{L}]$. By Lemma 2, $TE_{No}(c)=(1-\frac{c}{v})^{2}v$. Therefore, $TE_{Full}(c)\ge TE_{No}(c)$ holds if and only if

$$\begin{aligned}{}[p^{2}v_{H}+(1-p^{2})v_{L}]\ge \left( \frac{v-c}{v}\right) ^{2}v, \end{aligned}$$

i.e.,

$$\begin{aligned} c\ge v-\{v[p^{2}v_{H}+(1-p^{2})v_{L}]\}^{\frac{1}{2}}\text {.} \end{aligned}$$

Case 2: If $c\in [p(1-p)(v_{H}-v_{L})$, v], we have $\frac{v-c}{ v-p(1-p)(v_{H}-v_{L})}\le 1$, $q^{\prime \prime }=\frac{v-c}{ v-p(1-p)(v_{H}-v_{L})}$, and $TE_{Full}(c)=(\frac{v-c}{v-p(1-p)(v_{H}-v_{L})} )^{2}[p^{2}v_{H}+(1-p^{2})v_{L}]$. By Lemma 2, $TE_{No}(c)=(1-\frac{c}{v})^{2}v$. $TE_{Full}(c)\ge TE_{No}(c)$ holds if and only if

$$\begin{aligned} \left( \frac{v-c}{v-p(1-p)(v_{H}-v_{L})}\right) ^{2}[p^{2}v_{H}+(1-p^{2})v_{L}]\ge \left( \frac{v-c}{v}\right) ^{2}v, \end{aligned}$$

i.e.,

$$\begin{aligned} \left( \frac{1}{v-p(1-p)(v_{H}-v_{L})}\right) ^{2}[p^{2}v_{H}+(1-p^{2})v_{L}]\ge \left( \frac{1 }{v}\right) ^{2}v. \end{aligned}$$

Since $v-p(1-p)(v_{H}-v_{L})=p^{2}v_{H}+(1-p^{2})v_{L}$, the above inequality holds if and only if

$$\begin{aligned} v\ge v-p(1-p)(v_{H}-v_{L}), \end{aligned}$$

which holds, whenever $p\in (0,1)$. $\square$

Proof of Lemma 4

For a randomized disclosure policy with degree of transparency $\alpha \in [0,1]$, a contestant’s ex ante expected payoff is simply a linear combination of the two payoffs resulting from full transparency and no transparency, which is $\alpha p(1-p)(v_{H}-v_{L})$. Recall that with probability $\alpha$, contestants are fully informed, and with probability $1-\alpha$, contestants are fully uninformed.

A contestant would be indifferent between participating and staying out:

$$\begin{aligned} c=(1-q_{\alpha })v+q_{\alpha }[\alpha p(1-p)(v_{H}-v_{L})], \end{aligned}$$

where $q_{\alpha }$ is the entry probability of the other contestant. Therefore, the equilibrium entry probability $q_{\alpha }=\min \left\{\frac{v-c}{ v-\alpha p(1-p)(v_{H}-v_{L})},1\right\}$, as $q_{\alpha }\in [0,1]$.

After the entry stage, information about the prize will be disclosed according to the $\alpha$-degree randomized policy. When contestants are perfectly uninformed, analogous to the proof of Lemma 2, a participating contestant will exert effort following the uniform distribution over $[0,q_{\alpha }v]$, i.e., $F(x|v)=\frac{x}{q_{\alpha }v} ,\forall x\in [0,q_{\alpha }v]$. The resulting aggregate effort equals $2q_{\alpha }[\frac{1}{2}q_{\alpha }v]=(q_{\alpha })^{2}v$. Analogously, when contestants are perfectly informed, depending on his/her valuation, a participating contestant will exert effort following the distributions below:

$$\begin{aligned} F(x|v_{i} &= v_{H})=\frac{x-(1-p)v_{L}(\alpha )}{pv_{H}(\alpha )},\quad \forall x\in [(1-p)v_{L}(\alpha ),(1-p)v_{L}(\alpha )+pv_{H}(\alpha )]; \\ F(x|v_{i} &= v_{L})=\frac{x}{(1-p)v_{L}(\alpha )},\quad \forall x\in [0,(1-p)v_{L}(\alpha )], \end{aligned}$$

where $v_{H}(\alpha )=q_{\alpha }v_{H}$ and $v_{L}(\alpha )=q_{\alpha }v_{L}$ . The resulting aggregate effort equals $q_{\alpha }[p^{2}v_{H}(\alpha )+(1-p^{2})v_{L}(\alpha )]=(q_{\alpha })^{2}[p^{2}v_{H}+(1-p^{2})v_{L}]$.

To summarize, for an $\alpha$-degree randomized policy, with probability $\alpha$, contestants are perfectly informed, and the resulting aggregate effort equals $(q_{\alpha })^{2}[p^{2}v_{H}+(1-p^{2})v_{L}]$; and with probability $1-\alpha$, contestants are perfectly uninformed, and the resulting aggregate effort equals $(q_{\alpha })^{2}v$. Therefore, the ex ante expected aggregate effort resulting from an $\alpha$-degree disclosure policy equals

$$\begin{aligned} TE^{\alpha }(c) &= (1-\alpha )(q_{\alpha })^{2}v+\alpha (q_{\alpha })^{2}[p^{2}v_{H}+(1-p^{2})v_{L}] \\ &= (q_{\alpha })^{2}[(1-\alpha )v+\alpha (p^{2}v_{H}+(1-p^{2})v_{L})]. \end{aligned}$$

$\square$

Proof of Theorem 1

For each $c\in [0,v)$, to determine the optimal $\alpha \in [0,1]$, we consider two cases: $c\ge {\widehat{c}}(\alpha )$ and $c\le {\widehat{c}}(\alpha )$, where ${\widehat{c}}(\alpha )=\alpha p(1-p)(v_{H}-v_{L})$. Note that $c\ge {\widehat{c}}(\alpha )$ if and only if $\alpha \le \frac{c}{p(1-p)(v_{H}-v_{L})}$, we thus consider two cases: Case 1 where $\alpha \le \frac{c}{p(1-p)(v_{H}-v_{L})}$ and Case 2 where $\alpha \ge \frac{c}{p(1-p)(v_{H}-v_{L})}$ as follows.

Case 1: For $\alpha \le \frac{c}{p(1-p)(v_{H}-v_{L})}$, $TE^{\alpha }(c)= \frac{(v-c)^{2}}{v-\alpha p(1-p)(v_{H}-v_{L})}$, which increases with $\alpha$. Therefore, the constrained optimum $\alpha ^{*}=\min \{\frac{c }{p(1-p)(v_{H}-v_{L})},1\}$ in Case 1.

Case 2: For $\alpha \ge \frac{c}{p(1-p)(v_{H}-v_{L})}$, since $v=pv_{H}+(1-p)v_{L}$,

$$\begin{aligned} TE^{\alpha }(c)=(1-\alpha )v+\alpha \left[ p^{2}v_{H}+(1-p^{2})v_{L}\right] \end{aligned}$$

which decreases with $\alpha$, as $v:=pv_{H}+(1-p)v_{L}>p^{2}v_{H}+(1-p^{2})v_{L}$. Therefore, the constrained optimum $\alpha ^{*}=\frac{c}{p(1-p)(v_{H}-v_{L})}$ in Case 2.

Combining the two cases, $\alpha ^{*}(c)=\min \left\{\frac{c}{ p(1-p)(v_{H}-v_{L})},1\right\}$ for each $c\in [0,v)$. $\square$