Correlated equilibria and communication equilibria in all-pay auctions

Abstract

We study cheap-talk pre-play communication in static all-pay auctions. For the complete information case of two bidders, all correlated equilibria are payoff equivalent to the Nash equilibrium if there is no reserve price, or if it is commonly known that one bidder has a strictly higher value. Similarly, for the independent private values case of two bidders with no reserve price, all communication equilibria are payoff equivalent to the Bayesian Nash equilibrium. Hence, in such environments the Nash equilibrium and the Bayesian Nash equilibrium predictions are robust to pre-play communication between the bidders. On the other hand, if there are three or more symmetric bidders, or two symmetric bidders and a positive reserve price, then with complete information there may exist correlated equilibria such that the bidders’ payoffs are higher than in any Nash equilibrium, and with independent private values there may exist communication equilibria such that the bidders’ payoffs are higher than in any Bayesian Nash equilibrium. In these cases, pre-play cheap talk may affect the outcomes of the game, since the bidders have an incentive to coordinate on such equilibria.

Appendix

1.1 Proofs of Section 3

Proof of Proposition 3

Denote $u_{i}=\frac{1}{v-r}U_{i}$ for $i=1,2$, and fix $\left( u_{1},u_{2}\right) \in {\mathbb {R}} _{+}^{2}$ such that $v^{2}u_{i}+r^{2}u_{j}\le r^{2}$. The bidders are given recommendations according to the following probability distribution, where “bid above r” means “bid uniformly on $\left( r,v\right]$”.³⁶

Suppose bidder 1 is suggested to bid 0 and bids $b\in \left( r,v\right]$ instead.³⁷ Then his payoff is

$$\begin{aligned}{} & {} \left( \frac{u_{2}}{u_{2}+\frac{r}{v+r}\left( 1-u_{1}-u_{2}\right) }+\frac{ \frac{r}{v+r}\left( 1-u_{1}-u_{2}\right) }{u_{2}+\frac{r}{v+r}\left( 1-u_{1}-u_{2}\right) }\left( \frac{b-r}{v-r}\right) \right) v-b \\{} & {} \quad =\frac{r^{2}u_{1}+v^{2}u_{2}-r^{2}}{\left( v^{2}-r^{2}\right) \left( u_{2}+ \frac{r}{v+r}\left( 1-u_{1}-u_{2}\right) \right) }\left( v-b\right) \le 0 \end{aligned}$$

where the inequality follows from $r^{2}u_{1}+v^{2}u_{2}\le r^{2}$. If bidder 1 is suggested to bid r, then it is clearly optimal to comply, since the opponent bids 0 in such case. If bidder 1 is suggested to bid above r, then he is indifferent between all bids since the payoff from bidding $b\in \left[ r,v\right]$ is

$$\begin{aligned} \left( \frac{r}{v}+\left( 1-\frac{r}{v}\right) \left( \frac{b-r}{v-r}\right) \right) v-b=0 \end{aligned}$$

To summarize, bidder 1 gets a payoff of $v-r$ when he is suggested to bid r, and zero payoff otherwise. Hence, his ex ante payoff is $u_{1}\left( v-r\right) =U_{1}$. Using a similar argument for bidder 2, we conclude that the considered correlation rule is a correlated equilibrium, and it achieves the desired payoffs. $\square$

Proof of Proposition 4

Fix $U\in \left[ 0,{\overline{U}}\right]$, where ${\overline{U}}=\frac{2\left( v-r\right) }{n}\frac{\left( n-2\right) v^{2}+\left( n-2\right) vr+2nr^{2}}{ \left( 9n-14\right) v^{2}+\left( 6n-8\right) vr+\left( n+6\right) r^{2}}$. Consider the following symmetric correlation rule. First, a pair of bidders is randomly chosen, with each pair being equally likely to be chosen. Next, the bidders receive private bid recommendations without being told whether they have been chosen. The bidders who are not chosen are recommended to bid 0, and the chosen bidders are given recommendations according to the following probability distribution, where “bid low” means “bid uniformly on $\left( r, \frac{1}{2}\left( v+r\right) \right]$”, and “bid high” means “bid uniformly on $\left( \frac{1}{2}\left( v+r\right) ,v\right]$ ”, and where $x=\frac{1}{v+r}\left( \frac{v-r}{4}-\frac{ 3v+r}{v}\frac{n}{8}U\right)$.³⁸

If a bidder is suggested to bid 0, then he knows that either he was not chosen (which happens with probability $\frac{n-2}{n}$), or that he was chosen but only his opponent was suggested to bid above 0 (which happens with probability $\frac{2}{n}\left( \pi _{l}+\pi _{h}\right)$).

If this bidder bids $b\in \left( r,\frac{1}{2}\left( v+r\right) \right]$ instead, then he has a chance to win only if none of his opponents bid high. In particular, bidder i could win if (i) he was not chosen, and one chosen bidder bids low (which happens with probability $\frac{n-2}{n}\left( 2\pi _{l}\right)$); (ii) he was not chosen, and two chosen bidders bid low (which happens with probability $\frac{n-2}{n}\pi _{ll}$); (iii) he was chosen, and his opponent bids low (which happens with probability $\frac{2}{n }\pi _{l}$). The expected payoff this bidder is then

$$\begin{aligned}{} & {} \left( \frac{\frac{n-2}{n}\left( 2\pi _{l}\right) +\frac{2}{n}\pi _{l}}{ \frac{n-2}{n}+\frac{2}{n}\left( \pi _{l}+\pi _{h}\right) }\left( \frac{b-r}{ \frac{1}{2}\left( v+r\right) -r}\right) +\frac{\frac{n-2}{n}\pi _{ll}}{\frac{ n-2}{n}+\frac{2}{n}\left( \pi _{l}+\pi _{h}\right) }\right. \nonumber \\ {}{} & {} \left. \left( \frac{b-r}{\frac{1 }{2}\left( v+r\right) -r}\right) ^{2}\right) v-b \end{aligned}$$

(2)

Note that (2) is equal to $-r$ if $b=r$. If $b=\frac{1}{2}\left( v+r\right)$, then (2) becomes

$$\begin{aligned} \frac{\frac{n-2}{n}\left( 2\pi _{l}+\pi _{ll}\right) +\frac{2}{n}\pi _{l}}{ \frac{n-2}{n}+\frac{2}{n}\left( \pi _{l}+\pi _{h}\right) }v-\frac{1}{2} \left( v+r\right)= & {} \frac{n}{8}\frac{\left( 9n-14\right) v^{2}+\left( 6n-8\right) vr+\left( n+6\right) r^{2}}{\left( v-r\right) \left( \left( n-2\right) v+nr+nU\right) }\nonumber \\ {}{} & {} \left( U-{\overline{U}}\right) \le 0 \end{aligned}$$

(3)

where the inequality holds since $U\le {\overline{U}}$. Since (2) is convex in b, this implies that it is nonpositive for every $b\in \left[ r,\frac{1}{2}\left( v+r\right) \right]$.

If this bidder bids $b\in \left( \frac{1}{2}\left( v+r\right) ,v\right]$ instead, then he wins for sure if none of his opponents bid high, and has a chance to win otherwise. In particular, bidder i wins for sure if (i) he was not chosen, and none of the chosen bidders bid high (which happens with probability $\frac{n-2}{n}\left( 2\pi _{l}+\pi _{ll}\right)$); (ii) he was chosen, and his opponent does not bid high (which happens with probability $\frac{2}{n}\pi _{l}$). Also bidder i could win if (i) he was not chosen, and one chosen bidder bids high (which happens with probability $\frac{n-2}{n }\left( 2\pi _{h}+2\pi _{hl}\right)$); (ii) he was not chosen, and two chosen bidders bid high (which happens with probability $\frac{n-2}{n}\pi _{hh}$); (iii) he was chosen, and his opponent bids high (which happens with probability $\frac{2}{n}\pi _{h}$). The expected payoff of this bidder is then

$$\begin{aligned}{} & {} \left( \frac{\frac{n-2}{n}\left( 2\pi _{l}+\pi _{ll}\right) +\frac{2}{n} \pi _{l}}{\frac{n-2}{n}+\frac{2}{n}\left( \pi _{l}+\pi _{h}\right) }+\frac{ \frac{n-2}{n}\left( 2\pi _{h}+2\pi _{hl}\right) +\frac{2}{n}\pi _{h}}{\frac{ n-2}{n}+\frac{2}{n}\left( \pi _{l}+\pi _{h}\right) }\left( \frac{b-\frac{1}{2 }\left( v+r\right) }{v-\frac{1}{2}\left( v+r\right) }\right) \right. \nonumber \\{} & {} \quad \left. +\frac{\frac{n-2}{n}\pi _{hh}}{\frac{n-2}{n}+\frac{2}{n}\left( \pi _{l}+\pi _{h}\right) }\left( \frac{b-\frac{1}{2}\left( v+r\right) }{v-\frac{1 }{2}\left( v+r\right) }\right) ^{2}\right) v-b \end{aligned}$$

(4)

Note that (4) is equal to (3) if $b=\frac{1}{2} \left( v+r\right)$, and (4) is equal to zero if $b=v$. Since (4) is convex in b, this implies that it is nonpositive for every $b\in \left( \frac{1}{2}\left( v+r\right) ,v\right]$.

If a bidder is suggested to bid low, then he knows that he is chosen, and faces exactly one chosen opponent. This opponent bids 0, low, or high with probabilities $\frac{\pi _{l}}{\pi _{l}+\pi _{ll}+\pi _{hl}}$, $\frac{\pi _{ll}}{\pi _{l}+\pi _{ll}+\pi _{hl}}$, and $\frac{\pi _{hl}}{\pi _{l}+\pi _{ll}+\pi _{hl}}$, respectively. The expected payoff of this bidder from bidding any $b\in \left( r,\frac{1}{2}\left( v+r\right) \right]$ is

$$\begin{aligned} \left( \frac{\pi _{l}}{\pi _{l}+\pi _{ll}+\pi _{hl}}+\frac{\pi _{ll}}{\pi _{l}+\pi _{ll}+\pi _{hl}}\left( \frac{b-r}{\frac{1}{2}\left( v+r\right) -r} \right) \right) v-b= & {} \frac{U}{\frac{2}{n}\left( \pi _{l}+\pi _{ll}+\pi _{hl}\right) }\\ {}\ge & {} 0 \end{aligned}$$

If he bids $b\in \left( \frac{1}{2}\left( v+r\right) ,v\right]$ instead, then his payoff is

$$\begin{aligned}{} & {} \left( \frac{\pi _{l}+\pi _{ll}}{\pi _{l}+\pi _{ll}+\pi _{hl}}+\frac{\pi _{hl}}{\pi _{l}+\pi _{ll}+\pi _{hl}}\left( \frac{b-\frac{1}{2}\left( v+r\right) }{v-\frac{1}{2}\left( v+r\right) }\right) \right) v-b \\{} & {} \quad =\frac{U}{\frac{2}{n}\left( \pi _{l}+\pi _{ll}+\pi _{hl}\right) }\frac{ 2\left( v-b\right) }{v-r}<\frac{U}{\frac{2}{n}\left( \pi _{l}+\pi _{ll}+\pi _{hl}\right) } \end{aligned}$$

If a bidder is suggested to bid high, then he knows that he is chosen, and faces exactly one chosen opponent. This opponent bids 0, low, or high with probabilities $\frac{\pi _{h}}{\pi _{h}+\pi _{hl}+\pi _{hh}}$, $\frac{\pi _{hl}}{\pi _{h}+\pi _{hl}+\pi _{hh}}$, and $\frac{\pi _{hh}}{\pi _{h}+\pi _{hl}+\pi _{hh}}$, respectively. The expected payoff of this bidder from bidding any $b\in \left( \frac{1}{2}\left( v+r\right) ,\frac{1}{2}v\right]$ is

$$\begin{aligned}{} & {} \left( \frac{\pi _{h}+\pi _{hl}}{\pi _{h}+\pi _{hl}+\pi _{hh}}+\frac{\pi _{hh}}{\pi _{h}+\pi _{hl}+\pi _{hh}}\left( \frac{b-\frac{1}{2}\left( v+r\right) }{v-\frac{1}{2}\left( v+r\right) }\right) \right) v-b \\{} & {} \quad =\left( \frac{v+r}{2v}+\frac{v-r}{v}\left( \frac{b-\frac{1}{2}\left( v+r\right) }{v-r}\right) \right) v-b=0. \end{aligned}$$

If he bids $b\in \left( r,\frac{1}{2}\left( v+r\right) \right]$ instead, then his payoff is

$$\begin{aligned}{} & {} \left( \frac{\pi _{h}}{\pi _{h}+\pi _{hl}+\pi _{hh}}+\frac{\pi _{hl}}{\pi _{h}+\pi _{hl}+\pi _{hh}}\left( \frac{b-r}{\frac{1}{2}\left( v+r\right) -r} \right) \right) v-b\\ {}{} & {} \quad =\left( \frac{r}{v}+\frac{v-r}{v}\left( \frac{b-r}{v-r} \right) \right) v-b=0 \end{aligned}$$

Each bidder gets a payoff of $\frac{U}{\frac{2}{n}\left( \pi _{l}+\pi _{ll}+\pi _{hl}\right) }$ when he is suggested to bid low, and zero payoff otherwise. Hence, his ex ante payoff is U. Thus the considered correlation rule is a correlated equilibrium, and it achieves the desired payoffs. $\square$

1.2 Proofs of Section 4

Proof of Proposition 5

(ii) In every Bayesian Nash equilibrium, type 0 of each bidder bids 0 and gets a zero payoff. Let us represent the equilibrium strategy of bidder i of type v as a distribution function $G_{i}:\left\{ 0\right\} \cup \left[ r,\infty \right) \rightarrow \left[ 0,1\right]$, let ${\underline{b}} _{i}$ and ${\overline{b}}_{i}$ be the infimum and the supremum of the support of his equilibrium bids, and let $U_{i}\ge 0$ be his equilibrium payoff.

Note that $U_{i}\le v-{\overline{b}}_{i}$. Also note that bidder j of type v can secure a payoff arbitrarily close to $v-{\overline{b}}_{i}$ by bidding slightly above ${\overline{b}}_{i}$. Thus $U_{j}\ge v-{\overline{b}}_{i}$. Reversing the roles of i and j, and rearranging, we get $U_{i}=U_{j}=U$.

Next, note that bidder i of type v can secure a payoff arbitrarily close to $p_{j}v-r$ by bidding slightly above r, and thus winning when the opponent is of type 0. Hence, $U\ge \max \left\{ p_{1}v-r,p_{2}v-r,0\right\}$. If this inequality is strict, then neither bidder of type v bids 0 with positive probability, and thus $\underline{b }_{i},{\underline{b}}_{j}\ge r$. Moreover, $U>\max \left\{ p_{1}v-r,p_{2}v-r,0\right\}$ implies that each bidder of type v must be winning with positive probability against the opponent of type v. Then ${\underline{b}}_{i}<{\underline{b}}_{j}$ is impossible, since bidder i who bids below ${\underline{b}}_{j}$ always loses against the opponent of type v. But ${\underline{b}}_{i}={\underline{b}}_{j}={\underline{b}}$ is impossible either: the requirement of winning with positive probability against the opponent of type v implies that both bidders bid ${\underline{b}}$ with positive probability, which cannot happen in equilibrium since each bidder could profitably deviate to a slightly higher bid. Thus $U=\max \left\{ p_{1}v-r,p_{2}v-r,0\right\}$.

Let $p_{1}\le p_{2}$. It is straightforward to check that the following is a Bayesian Nash equilibrium. Types 0 of both bidders bid 0. Type v of bidder 1 bids 0 and r with probabilities $\frac{\min \left\{ r-p_{2}v,0\right\} }{\left( 1-p_{1}\right) v}$ and $\frac{p_{2}-p_{1}}{ 1-p_{1}}$, respectively, and bids uniformly on $\left( r,\min \left\{ \left( 1-p_{2}\right) v+r,v\right\} \right]$ otherwise; type v of bidder 2 bids 0 with probability $\frac{\min \left\{ r-p_{2}v,0\right\} }{ \left( 1-p_{2}\right) v}$, and bids uniformly on $\left( r,\min \left\{ \left( 1-p_{2}\right) v+r,v\right\} \right]$ otherwise. $\square$

Proof of Proposition 6

Denote by $U_{i}^{*}\left( v_{i}\right)$ the interim expected payoff of player i of type $v_{i}$ in the Bayesian Nash equilibrium $\left( \mu _{1}^{*},\mu _{2}^{*}\right)$, and by $U_{i}\left( v_{i}\right)$ the interim expected payoff of player i of type $v_{i}$ in the communication equilibrium $\mu$. Let $\mu _{i}\left( \cdot |v_{i}\right)$ be the marginal probability measure of $\mu$ on $A_{i}$ conditional on $v_{i}$ (defined in Sect. 2).

By the definition of Bayesian Nash equilibrium, for every i and $P_{i}$ -a.e. $v_{i}$, it is unprofitable to deviate to any strategy $\widetilde{\mu }_{i}(\cdot |v_{i})$:

$$\begin{aligned} U_{i}^{*}\left( v_{i}\right)= & {} v_{i}\int \nolimits _{T_{j}}\left( \int \nolimits _{A_{j}}\int \nolimits _{A_{i}}\rho _{i}\left( b\right) \mu _{i}^{*}(db_{i}|v_{i})\mu _{j}^{*}(db_{j}|v_{j})\right) P_{j}\left( dv_{j}\right) -\int \nolimits _{A_{i}}b_{i}\mu _{i}^{*}(db_{i}|v_{i}) \nonumber \\\ge & {} v_{i}\int \nolimits _{T_{j}}\left( \int \nolimits _{A_{j}}\int \nolimits _{A_{i}}\rho _{i}\left( b\right) {\widetilde{\mu }} _{i}(db_{i}|v_{i})\mu _{j}^{*}(db_{j}|v_{j})\right) P_{j}\left( dv_{j}\right) -\int \nolimits _{A_{i}}b_{i}{\widetilde{\mu }}_{i}(db_{i}|v_{i}) \end{aligned}$$

(5)

By the definition of communication equilibrium, for every i and $P_{i}$ -a.e. $v_{i}$, it is unprofitable to deviate to the following strategy: first, randomize over the type reports according to $P_{i}$, and then, regardless of the mediator’s recommendation, choose bids according to some bidding strategy ${\widetilde{\mu }}_{i}(\cdot |v_{i})$:

$$\begin{aligned} U_{i}\left( v_{i}\right)= & {} v_{i}\int \nolimits _{T_{j}}\left( \int \nolimits _{A}\rho _{i}\left( b\right) \mu (db|v_{i},v_{j})\right) P_{j}\left( dv_{j}\right) -\int \nolimits _{A_{i}}b_{i}\mu _{i}(db_{i}|v_{i}) \nonumber \\\ge & {} v_{i}\int \nolimits _{T_{j}}\left( \int \nolimits _{A}\left( \int \nolimits _{A_{i}}\rho _{i}\left( {\widehat{b}}_{i},b_{j}\right) \widetilde{\mu }_{i}(d{\widehat{b}}_{i}|v_{i})\right) \int \nolimits _{T_{j}}\mu (db|\widehat{ v}_{i},v_{j})dP_{i}\left( d{\widehat{v}}_{i}\right) \right) P_{j}\left( dv_{j}\right) \nonumber \\{} & {} -\int \nolimits _{A_{i}}{\widehat{b}}_{i}{\widetilde{\mu }}_{i}(d {\widehat{b}}_{i}|v_{i}) \nonumber \\= & {} v_{i}\int \nolimits _{T_{j}}\left( \int \nolimits _{A_{j}}\int \nolimits _{A_{i}}\rho _{i}\left( {\widehat{b}}_{i},b_{j}\right) \widetilde{\mu }_{i}(d{\widehat{b}}_{i}|v_{i})\mu _{j}(db_{j}|v_{j})\right) P_{j}\left( dv_{j}\right) -\int \nolimits _{A_{i}}{\widehat{b}}_{i}{\widetilde{\mu }}_{i}(d {\widehat{b}}_{i}|v_{i})\nonumber \\ \end{aligned}$$

(6)

For every i and $v_{i}\ne 0$, such that both (5) and (6) hold, perform the following operations. Take (5) with ${\widetilde{\mu }}_{i}=\mu _{i}$ and (6) with ${\widetilde{\mu }} _{i}=\mu _{i}^{*}$, add the two resulting inequalities, and divide by $v_{i}$. Then for every i and $P_{i}$-a.e. $v_{i}$ we get

$$\begin{aligned}{} & {} \int \nolimits _{T_{j}}\left( \int \nolimits _{A_{j}}\int \nolimits _{A_{i}}\rho _{i}\left( b\right) \mu _{i}^{*}(db_{i}|v_{i})\mu _{j}^{*}(db_{j}|v_{j})+\int \nolimits _{A}\rho _{i}\left( b\right) \mu (db|v_{i},v_{j})\right) P_{j}\left( dv_{j}\right) \nonumber \\{} & {} \quad \ge \int \nolimits _{T_{j}}\left( \int \nolimits _{A_{j}}\int \nolimits _{A_{i}}\rho _{i}\left( b\right) \mu _{i}(db_{i}|v_{i})\mu _{j}^{*}(db_{j}|v_{j})+\int \nolimits _{A_{j}}\int \nolimits _{A_{i}}\rho _{i}\left( b\right) \mu _{i}^{*}(db_{i}|v_{i})\mu _{j}(db_{j}|v_{j})\right) P_{j}\left( dv_{j}\right) \nonumber \\ \end{aligned}$$

(7)

Next, integrate (7) with respect to $P_{i}$ over the set of types of bidder i for which inequality (7) holds. Note that the resulting inequality continues to hold even if we integrate over $T_{i}$ because (7) is satisfied for $P_{i}$-a.e. $v_{i}$, and because we have assumed that $P_{i}\left( \left\{ 0\right\} \right) =0$. Sum up the resulting inequalities over i:

$$\begin{aligned}{} & {} \int \nolimits _{T}\left( \int \nolimits _{A_{j}}\int \nolimits _{A_{i}}\sum \limits _{k=1,2}\rho _{k}\left( b\right) \mu _{i}^{*}(db_{i}|v_{i})\mu _{j}^{*}(db_{j}|v_{j})+\int \nolimits _{A}\sum \limits _{k=1,2}\rho _{k}\left( b\right) \mu (db|v_{i},v_{j})\right) P\left( dv\right) \nonumber \\{} & {} \quad \ge \int \nolimits _{T}\left( \int \nolimits _{A_{j}}\int \nolimits _{A_{i}}\sum \limits _{k=1,2}\rho _{k}\left( b\right) \mu _{i}(db_{i}|v_{i})\mu _{j}^{*}(db_{j}|v_{j})+\int \nolimits _{A_{j}}\int \nolimits _{A_{i}}\sum \limits _{k=1,2}\rho _{k}\left( b\right) \mu _{i}^{*}(db_{i}|v_{i})\mu _{j}(db_{j}|v_{j})\right) P\left( dv\right) \nonumber \\ \end{aligned}$$

(8)

Since $\sum \nolimits _{k=1,2}\rho _{k}\left( b\right) =1$ for every b, inequality (8) holds as an equality. This implies that the following inequalities hold as equalities as well for $P_{i}$-a.e. $v_{i}$: inequality (5) when ${\widetilde{\mu }}_{i}=\mu _{i}$, and inequality (6) when ${\widetilde{\mu }}_{i}=\mu _{i}^{*}$.

Hence, for every i, $P_{i}$-a.e. $v_{i}$, the following is true for any ${\widetilde{\mu }}_{i}(\cdot |v_{i})$:

$$\begin{aligned} U_{i}^{*}\left( v_{i}\right)= & {} v_{i}\int \nolimits _{T_{j}}\left( \int \nolimits _{A_{j}}\int \nolimits _{A_{i}}\rho _{i}\left( b\right) \mu _{i}(db_{i}|v_{i})\mu _{j}^{*}(db_{j}|v_{j})\right) P_{j}\left( dv_{j}\right) -\int \nolimits _{A_{i}}b_{i}\mu _{i}(db_{i}|v_{i}) \\\ge & {} v_{i}\int \nolimits _{T_{j}}\left( \int \nolimits _{A_{j}}\int \nolimits _{A_{i}}\rho _{i}\left( b\right) {\widetilde{\mu }} _{i}(db_{i}|v_{i})\mu _{j}^{*}(db_{j}|v_{j})\right) P_{j}\left( dv_{j}\right) -\int \nolimits _{A_{i}}b_{i}{\widetilde{\mu }}_{i}(db_{i}|v_{i}) \end{aligned}$$

and

$$\begin{aligned} U_{i}\left( v_{i}\right)= & {} v_{i}\int \nolimits _{T_{j}}\left( \int \nolimits _{A_{j}}\int \nolimits _{A_{i}}\rho _{i}\left( b_{i},b_{j}\right) \mu _{i}^{*}(db_{i}|v_{i})\mu _{j}(db_{j}|v_{j})\right) P_{j}\left( dv_{j}\right) -\int \nolimits _{A_{i}}b_{i}\mu _{i}^{*}(db_{i}|v_{i}) \\\ge & {} v_{i}\int \nolimits _{T_{j}}\left( \int \nolimits _{A_{j}}\int \nolimits _{A_{i}}\rho _{i}\left( b_{i},b_{j}\right) {\widetilde{\mu }} _{i}(db_{i}|v_{i})\mu _{j}(db_{j}|v_{j})\right) P_{j}\left( dv_{j}\right) -\int \nolimits _{A_{i}}b_{i}{\widetilde{\mu }}_{i}(db_{i}|v_{i}) \end{aligned}$$

This implies that $\left( \mu _{i}^{*},\mu _{j}\right)$ is a Bayesian Nash equilibrium. In this equilibrium bidder i of type $v_{i}$ gets payoff $U_{i}\left( v_{i}\right)$, and bidder j of type $v_{j}$ gets payoff $U_{j}^{*}\left( v_{j}\right)$. Since by assumption the Bayesian Nash equilibrium is unique, for every i we have $\mu _{i}=\mu _{i}^{*}$, and thus $U_{i}^{*}\left( v_{i}\right) =U_{i}\left( v_{i}\right)$ for every $P_{i}$-a.e. $v_{i}$.³⁹$\square$

Proof of Proposition 7

We show that for $p\in \left[ 0,\frac{r}{v}\right)$ there exists a communication equilibrium such that each bidder of type v gets a payoff of $\frac{\left( r^{2}-p^{2}v^{2}\right) \left( v-r\right) }{v^{2}+r^{2}-2pv^{2} }$. Consider the following symmetric communication rule. If a bidder reports type$\ 0$, then he is suggested to bid 0. If a bidder reports type v and his opponent reports type 0, then this bidder is suggested to bid r or “bid above r” (which means “bid uniformly on $\left( r,v\right]$”), with probabilities ${\widehat{\pi }}=\frac{\left( r-pv\right) \left( v+r\right) }{v^{2}+r^{2}-2pv^{2}}$ and $1-{\widehat{\pi }}$, respectively. If both bidders report type v, then they are given recommendations according to the following probability distribution.⁴⁰

We need to check the incentives to tell the truth and to comply with the recommendations only for the bidders of type v, since the bidders of type 0 have no incentive to lie or to disobey.

If a bidder of type v has reported v and is suggested to bid 0, then he knows that his opponent must be of type v and bids r or above r, with probabilities $\frac{\pi _{r}}{\pi _{r}+\pi _{h}}$ and $\frac{\pi _{h} }{\pi _{r}+\pi _{h}}$, respectively. If this bidder bids $b\in \left( r,v \right]$ instead, then his payoff is⁴¹

$$\begin{aligned} \left( \frac{\pi _{r}}{\pi _{r}+\pi _{h}}+\frac{\pi _{h}}{\pi _{r}+\pi _{h}} \left( \frac{b-r}{v-r}\right) \right) v-b=\left( \frac{r}{v}+\frac{v-r}{v} \left( \frac{b-r}{v-r}\right) \right) v-b=0 \end{aligned}$$

If a bidder of type v has reported v and is suggested to bid r, then it is clearly optimal to comply, since the opponent bids 0, regardless of the type, in such case.

If a bidder of type v has reported v and is suggested to bid above r, then he knows that either his opponent is of type 0 and thus bids 0, or his opponent is of type v and bids 0 or high, with probabilities $\frac{ p\left( 1-{\widehat{\pi }}\right) }{p\left( 1-{\widehat{\pi }}\right) +\left( 1-p\right) \left( \pi _{h}+\pi _{hh}\right) }$, $\frac{\left( 1-p\right) \pi _{h}}{p\left( 1-{\widehat{\pi }}\right) +\left( 1-p\right) \left( \pi _{h}+\pi _{hh}\right) }$, and $\frac{\left( 1-p\right) \pi _{hh}}{p\left( 1-\widehat{ \pi }\right) +\left( 1-p\right) \left( \pi _{h}+\pi _{hh}\right) }$, respectively. The expected payoff of this bidder from bidding any $b\in \left[ r,v\right]$ is

$$\begin{aligned}{} & {} \left( \frac{p\left( 1-{\widehat{\pi }}\right) +\left( 1-p\right) \pi _{h}}{ p\left( 1-{\widehat{\pi }}\right) +\left( 1-p\right) \left( \pi _{h}+\pi _{hh}\right) }+\frac{\left( 1-p\right) \pi _{hh}}{p\left( 1-{\widehat{\pi }} \right) +\left( 1-p\right) \left( \pi _{h}+\pi _{hh}\right) }\left( \frac{b-r }{v-r}\right) \right) v-b \\{} & {} \quad =\left( \frac{r}{v}+\frac{v-r}{v}\left( \frac{b-r}{v-r}\right) \right) v-b=0 \end{aligned}$$

To summarize, if a bidder of type v truthfully reports his type and follows the recommendations, then he gets a payoff of $v-r$ when he is suggested to bid r, and zero payoff otherwise. Hence, his ex ante payoff is $\left( p{\widehat{\pi }}+\left( 1-p\right) \pi _{r}\right) \left( v-r\right) =\frac{\left( r^{2}-p^{2}v^{2}\right) \left( v-r\right) }{ v^{2}+r^{2}-2pv^{2}}$.

If a bidder of type v has reported 0, then he is suggested to bid 0. He knows that either his opponent is of type 0 and thus bids 0, or his opponent is of type v and bids r or above r, with probabilities p, $\left( 1-p\right) {\widehat{\pi }}$, and $\left( 1-p\right) \left( 1-\widehat{ \pi }\right)$, respectively. If this bidder bids $b\in \left( r,v\right]$, then his payoff is

$$\begin{aligned}{} & {} \left( \left( p+\left( 1-p\right) {\widehat{\pi }}\right) +\left( 1-p\right) \left( 1-{\widehat{\pi }}\right) \left( \frac{b-r}{v-r}\right) \right) v-b\\ {}{} & {} \qquad \le \max \left\{ \left( p+\left( 1-p\right) {\widehat{\pi }}\right) v-r,0\right\} =\frac{\left( r^{2}-p^{2}v^{2}\right) \left( v-r\right) }{ v^{2}+r^{2}-2pv^{2}} \end{aligned}$$

where the inequality follows from the fact that payoff is a linear function of b and is thus maximized either at $b=r$ or at $b=v$.

Thus the considered communication rule is a communication equilibrium, and it achieves the desired payoffs.

$\square$

Proof of Proposition 8

The proof is similar to the proof of part (ii) of Proposition 5. In every Bayesian Nash equilibrium, type 0 of each bidder bids 0 and gets a zero payoff. Let the equilibrium strategy of bidder i of type v be represented by distribution function $G_{i}:\left\{ 0\right\} \cup \left[ r,\infty \right) \rightarrow \left[ 0,1\right]$, ${\underline{b}}_{i}$ and ${\overline{b}}_{i}$ be the infimum and the supremum of the support of his equilibrium bids, and $U_{i}\ge 0$ be his equilibrium payoff.

Note that $U_{i}\le \prod \nolimits _{k\ne i}\left( p+\left( 1-p\right) G_{k}\left( {\overline{b}}_{i}\right) \right) v-{\overline{b}}_{i}$. Also note that bidder j of type v can secure a payoff arbitrarily close to $\prod \nolimits _{k\ne i,j}\left( p+\left( 1-p\right) G_{k}\left( {\overline{b}} _{i}\right) \right) v-{\overline{b}}_{i}$ by bidding slightly above $\overline{ b}_{i}$. Thus $U_{j}\ge U_{i}$. Since this is true for every pair of i and j, we have $U_{i}=U$ for every i.

Next, note that bidder i of type v can secure a payoff arbitrarily close to $p^{n-1}v-r$ by bidding slightly above r, and thus winning when the opponent is of type 0. Hence, $U\ge \max \left\{ p^{n-1}v-r,0\right\}$. If this inequality is strict, then neither bidder bids 0 with positive probability, and thus ${\underline{b}}_{i}\ge r$ for every i. Moreover, $U>\max \left\{ p^{n-1}v-r,0\right\}$ implies that each bidder of type v must be winning with positive probability against some opponent of type v. Then it is impossible to have ${\underline{b}}_{i}<{\underline{b}}=\min _{j\ne i} {\underline{b}}_{j}$, since bidder i who bids below ${\underline{b}}$ always loses against the opponents of type v. But ${\underline{b}}_{i}={\underline{b}}$ is impossible either: the requirement of winning with positive probability against some opponent of type v implies that the bidders who bid ${\underline{b}}$ must do so with positive probability, which cannot happen in equilibrium since each bidder could profitably deviate to a slightly higher bid. Thus $U=\max \left\{ p^{n-1}v-r,0\right\}$.

It is straightforward to check that the following is a Bayesian Nash equilibrium. Type 0 of each bidder bids 0. Type v of each bidder bids 0 with probability $x=\frac{1}{1-p}\left( \max \left\{ \left( \frac{r}{v} \right) ^{\frac{1}{n-1}}-p,0\right\} \right)$, and bids according to $G\left( b\right) =\frac{1}{1-p}\left( \left( \max \left\{ p^{n-1}-\frac{r}{v },0\right\} +\frac{b}{v}\right) ^{\frac{1}{n-1}}-p\right)$ on $\left( r,\min \left\{ \left( 1-p^{n-1}\right) v+r,v\right\} \right]$. $\square$

Proof of Proposition 9

We show for sufficiently small $p>0$ there exists a communication equilibrium such that each bidder of type v gets a payoff of $2p^{n-1}v$. Consider the following symmetric communication rule. If a bidder reports type $\ 0$, then he is suggested to bid 0. If exactly one bidder reports v, then this bidder is suggested to“bid low” (which means “bid uniformly on $\left( 0,\frac{1}{2}v \right]$”). If $m>1$ bidders report v, then a pair of bidders out of these m bidders is randomly chosen, with each pair being equally likely to be chosen. The bidders receive private bid recommendations without being told whether they have been chosen. The bidders who are not chosen are recommended to bid 0, and the chosen bidders are given recommendations according to the following probability distribution where “bid high” means “bid uniformly on $\left( \frac{1}{2}v,v\right]$”, and where

$$\begin{aligned} g=\sum \limits _{k=1}^{n-1}\frac{\left( n-1\right) !}{k!\left( n-1-k\right) ! }\left( 1-p\right) ^{k}p^{n-1-k}\left( \frac{2}{k+1}\right) =2\left( \frac{1 }{n}\frac{1-p^{n}}{1-p}-p^{n-1}\right) \end{aligned}$$

is the probability that a bidder who submitted report v was chosen and that he is not the only one who submitted v.⁴²

We need to check the incentives to tell the truth and to comply with the recommendations only for the bidders of type v, since the bidders of type 0 have no incentive to lie or to disobey.

If a bidder of type v has reported v and is suggested to bid 0, then he knows that either he was not chosen (which happens with probability $1-p^{n-1}-g$), or that he was chosen but only his opponent is suggested to bid above 0 (which happens with probability $g\pi _{l}$).

If this bidder bids $b\in \left( 0,\frac{1}{2}v\right]$ instead, then he has a chance to win only if none of his opponents bid high. In particular, bidder i could win if (i) he was not chosen, and one chosen bidder bids low (which happens with probability $\left( 1-p^{n-1}-g\right) 2\pi _{l}$); (ii) he was not chosen, and two chosen bidders bid low (which happens with probability $\left( 1-p^{n-1}-g\right) \pi _{ll}$); (iii) he was chosen, and his opponent bids low (which happens with probability $g\pi _{l}$). The expected payoff of this bidder is then

$$\begin{aligned} \left( \frac{\left( 1-p^{n-1}-g\right) 2\pi _{l}+g\pi _{l}}{1-g}\left( \frac{ b}{\frac{1}{2}v}\right) +\frac{\left( 1-p^{n-1}-g\right) \pi _{ll}}{1-g} \left( \frac{b}{\frac{1}{2}v}\right) ^{2}\right) v-b \end{aligned}$$

(9)

Note that (9) is equal to 0 if $b=0$. If $b=\frac{1}{2}v$, then (9) becomes

$$\begin{aligned} \left( \frac{\left( 1-p^{n-1}-g\right) \left( 2\pi _{l}+\pi _{ll}\right) +g\pi _{l}}{1-g}\right) v-\frac{1}{2}v=\left( \frac{\left( 3\frac{1-p^{n-1}}{ g}-\frac{9}{4}\right) p^{n-1}}{1-g}-\frac{1}{4}\right) v \nonumber \\ \end{aligned}$$

(10)

Note that (10) is equal to $-\frac{1}{4}v$ if $p=0$, and that ( 10) is continuous in p. Thus for p small enough (9) is nonpositive at $b=0$ and at $b=\frac{1}{2}v$, and it is convex in b on $\left( 0,\frac{1}{2}v\right]$, which implies that (9) is nonpositive for every $b\in \left( 0,\frac{1}{2}v\right]$.

If this bidder bids $b\in \left( \frac{1}{2}v,v\right]$ instead, then he wins for sure if none of his opponents bids high, and has a chance to win otherwise. In particular, bidder i wins for sure if (i) he was not chosen, and none of the chosen bidders bid high (which happens with probability $\left( 1-p^{n-1}-g\right) \left( 2\pi _{l}+\pi _{ll}\right)$); (ii) he was chosen, and his opponent does not bid high (which happens with probability $g\pi _{l}$). Also bidder i could win if (i) he was not chosen, and one chosen bidder bids high (which happens with probability $\left( 1-p^{n-1}-g\right) 2\pi _{hl}$); (ii) he was not chosen, and two chosen bidders bid high (which happens with probability $\left( 1-p^{n-1}-g\right) \pi _{hh}$). The expected payoff of this bidder is then

$$\begin{aligned}{} & {} \left( \frac{\left( 1-p^{n-1}-g\right) \left( 2\pi _{l}+\pi _{ll}\right) +g\pi _{l}}{1-g}+\frac{\left( 1-p^{n-1}-g\right) 2\pi _{hl}}{1-g}\left( \frac{b-\frac{1}{2}v}{\frac{1}{2}v}\right) \right. \nonumber \\{} & {} \quad \left. +\frac{\left( 1-p^{n-1}-g\right) \pi _{hh}}{1-g}\left( \frac{b- \frac{1}{2}v}{\frac{1}{2}v}\right) ^{2}\right) v-b \end{aligned}$$

(11)

Note that (11) is equal to (10) if $b=\frac{1}{2}v$, and (11) is equal to zero if $b=v$. Thus for p small enough ( 11) is nonpositive at $b=\frac{1}{2}v$ and at $b=v$, and it is convex in b on $\left( \frac{1}{2}v,v\right]$, which implies that (11) is nonpositive for every $b\in \left( \frac{1}{2}v,v\right]$.

If a bidder of type v has reported v and is suggested to bid low, then he knows that either he faces no opponents (with probability $p^{n-1}$), or that he was chosen and faces one chosen opponent who bids 0, low, or high with probabilities $g\pi _{l}$, $g\pi _{ll}$, and $g\pi _{hl}$, respectively. The expected payoff of this bidder from bidding any $b\in \left( 0,\frac{1}{2}v\right]$ is

$$\begin{aligned}{} & {} \left( \frac{p^{n-1}+g\pi _{l}}{p^{n-1}+g\left( \pi _{l}+\pi _{ll}+\pi _{hl}\right) }+\frac{g\pi _{ll}}{p^{n-1}+g\left( \pi _{l}+\pi _{ll}+\pi _{hl}\right) }\frac{b}{\frac{1}{2}v}\right) v-b\\ {}{} & {} \quad =\frac{2p^{n-1}v}{2p^{n-1}+ \frac{1}{2}g}\ge 0 \end{aligned}$$

If he bids $b\in \left( \frac{1}{2}v,v\right]$ instead, then his payoff is

$$\begin{aligned}{} & {} \left( \frac{p^{n-1}+g\left( \pi _{l}+\pi _{ll}\right) }{p^{n-1}+g\left( \pi _{l}+\pi _{ll}+\pi _{hl}\right) }+\frac{g\pi _{hl}}{p^{n-1}+g\left( \pi _{l}+\pi _{ll}+\pi _{hl}\right) }\left( \frac{b-\frac{1}{2}v}{\frac{1}{2}v} \right) \right) v-b\\{} & {} \quad =\frac{2p^{n-1}v}{2p^{n-1}+\frac{1}{2}g}\left( \frac{ v-b}{\frac{1}{2}v}\right) \\{} & {} \quad <\frac{2p^{n-1}v}{2p^{n-1}+\frac{1}{2}g} \end{aligned}$$

If a bidder of type v has reported v and is suggested to bid high, then he knows that he was chosen and faces one chosen opponent who bids low or high, with equal probabilities. Thus the expected payoff of this bidder from bidding any $b\in \left( 0,v\right]$ is equal to zero.

To summarize, if a bidder of type v truthfully reports his type and follows the recommendations, then he gets a payoff of $\frac{2p^{n-1}v}{ 2p^{n-1}+\frac{1}{2}g}$ when he is suggested to bid low, and zero payoff otherwise. Hence, his ex ante payoff is $2p^{n-1}v$.

If a bidder of type v has reported 0, then he is suggested to bid 0. He knows that he faces no active opponents with probability $p^{n-1}$; one active opponent who bids low with probability $\left( n-1\right) \left( 1-p\right) p^{n-2}+2d\pi _{l}$, where $d=\left( 1-p^{n-1}-\left( n-1\right) \left( 1-p\right) p^{n-2}\right)$; two active opponents who both bid low, both bid high, or one bids low and another high with probabilities $d\pi _{ll}$, $d\pi _{hh}$, and $2d\pi _{hl}$, respectively.

If this bidder bids $b\in \left( 0,\frac{1}{2}v\right]$, then his payoff is

$$\begin{aligned} \left( p^{n-1}+\left( \left( n-1\right) \left( 1-p\right) p^{n-2}+d2\pi _{l}\right) \left( \frac{b}{\frac{1}{2}v}\right) +d\pi _{ll}\left( \frac{b}{ \frac{1}{2}v}\right) ^{2}\right) v-b \end{aligned}$$

(12)

Note that if $b=0$, then (12) is equal to $p^{n-1}v$ which is smaller than the payoff from truthtelling $2p^{n-1}v$. If $b=\frac{1}{2}v$ then (12) becomes

$$\begin{aligned} \left( p^{n-1}+\left( n-1\right) \left( 1-p\right) p^{n-2}+d\left( 2\pi _{l}+\pi _{ll}\right) \right) v-\frac{1}{2}v \end{aligned}$$

(13)

Note that (13) is equal to $-\frac{1}{4}v$ if $p=0$, and that ( 13) is continuous in p. Thus for p small enough (12) is smaller than $2p^{n-1}v$ at $b=0$ and at $b=\frac{1}{2}v$, and it is convex in b on $\left( 0,\frac{1}{2}v\right]$, which implies that (13) is smaller than the payoff from truthtelling $2p^{n-1}v$ for every $b\in \left( 0,\frac{1}{2}v\right]$.

If this bidder bids $b\in \left( \frac{1}{2}v,v\right]$, then his payoff is

$$\begin{aligned} \left( p^{n-1}+\left( n-1\right) \left( 1-p\right) p^{n-2}+d\left( 2\pi _{l}+\pi _{ll}\right) +2d\pi _{hl}\left( \frac{b-\frac{1}{2}v}{\frac{1}{2}v} \right) +d\pi _{hh}\left( \frac{b-\frac{1}{2}v}{\frac{1}{2}v}\right) ^{2}\right) v-b \nonumber \\ \end{aligned}$$

(14)

Note that (14) is equal to (13) if $b=\frac{1}{2}v$, and (14) is equal to zero if $b=v$. Thus for p small enough ( 14) is smaller than $2p^{n-1}v$ at $b=\frac{1}{2}v$ and at $b=v$, and it is convex in b on $\left( \frac{1}{2}v,v\right]$, which implies that (14) is smaller than the payoff from truthtelling $2p^{n-1}v$ for every $b\in \left( \frac{1}{2}v,v\right]$.

Thus the considered communication rule is a communication equilibrium, and it achieves the desired payoffs. $\square$