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Participation constraints in first-price auctions

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Abstract

We study the endogenous participation problem when bidders are characterized by a two-dimensional private information on valuations and participation costs in first-price auctions. Bidders participate whenever their private costs are less than or equal to the expected revenue from participating. We show that there always exists an equilibrium in this general setting with two-dimensional types of ex-ante heterogeneous bidders. When bidders are ex-ante homogeneous, there is a unique symmetric equilibrium, but asymmetric equilibria may also exist. We provide conditions under which the equilibrium is unique (not only among symmetric ones). In the symmetric equilibrium, we show that the equilibrium cutoff of participation costs described above which bidders never participate, is lower when the distribution of participation costs is first-order stochastically dominated.

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Notes

  1. Related terminology includes participation cost, participation fee, entry cost, or opportunity costs of participating in the auction. Participating costs differs from entry fee in that entry fee is part of the seller’s revenue while participation cost is more general. As we only study the bidder’s problem in this paper, we do not distinguish these two terminology.

  2. Hendricks et al. (2003) find only around 25 percent of potential bidders participate in the auctions that took place within 15 years in the united states. Bajari and Hortacsu (2003), Li and Zheng (2009), Athey S et al. (2011), Li and Zhang (2010, 2015), Krasnokutskaya and Seim (2011), Roberts and Sweeting (2013) for various auctions. See Feng, Lu and Sun (2016) for details.

  3. Another example is Menezes and Monteiro (2000).

  4. The support for valuations is normalized to be \(\left[ 0,1\right]\). Bidders with participation costs higher than 1 will not participate in the auction and such bidders are of no practical interest. If the upper bounds of the supports for the participation costs are higher than 1, the above distributions on the participation costs should be interpreted as the truncated distributions of the original distributions on \(\left[ 0,1\right]\).

  5. We will study the special case where \(v_{i}\) and \(c_{i}\) are independently distributed in Sects. 4 and 5. When there are atoms in the distribution, \(k_{i}\left( v_{i},c_{i}\right)\) can incorporate Dirac delta functions to handle the infinite density.

  6. The description of the equilibria can be slightly different under different informational structures on \(K_{i}\left( v_{i},c_{i}\right)\). For example, when \(v_{i}\) is private information and \(c_{i}\) is exogenously fixed for all bidders, \(K_{i}\left( v_{i},c_{i}\right) =K_{i}\left( v_{i}\right)\) (See Cao and Tian 2010), the equilibrium is described by a valuation cutoff \(v_{i}^{*}\) for bidder i such that bidder i submits a bid whenever \(v_{i}\ge v_{i}^{*}\).

  7. Subscripts A and B are used to indicate variables associated with different distributions.

References

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Acknowledgements

We are grateful to two anonymous referees and the editor of this journal for their valuable comments. Wei Wang is the corresponding author.

Funding

Xiaoyong Cao: Financial supports from the National Natural Science Foundation of China (NSFC-72273023 NSFC-72273030) are gratefully acknowledged. Shao-Chieh Hsueh: Financial support from the Social Science Foundation in Xiamen University of Technology (YSK22024R), Natural Science Foundation of Xiamen, China (Grant No. 3502Z202373068), and the Education Reform Foundation in Xiamen University of Technology (JG202232) are gratefully acknowledged. Wei Wang: The financial support from the National Natural Science Foundation of China (Grant No. 72173019), Beijing Social Science Foundation (Grant No. 20JJA004), the Chinese Ministry of Education Research Funds on Humanities and Social Sciences (Grant No. 21YJA790056), and the Fundamental Research Funds for the Central Universities in UIBE (Grant No. CXTD14-04) are gratefully acknowledged.

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Appendix

Appendix

1.1 Proof of Proposition 1

First, notice that it follows from the inverse function theorem that all inverse bidding functions are also differentiable. Hence, \(\Phi _{i}\left( v\right)\) and thus \(R_{i}\left( v\right)\) are differentiable for all \(i\in N\). It follows that \(c_{i}^{*}\left( v\right)\) is differentiable for all \(i\in N\). Then, we can apply the envelope theorem in Eq. (5). Differentiating Eq. (5) derives:

$$\begin{aligned} c_{i}^{*\prime }\left( v\right) =\Phi _{i}\left( v\right) , \end{aligned}$$

for all \(i\in N\). Since \(\Phi _{i}\left( v\right)\) is differentiable, \(c_{i}^{*\prime }\left( v\right)\) is also differentiable. Last, integrating Eq. (8) from 0 to 1 and noticing \(c_{i}^{*}\left( 0\right) =0\) for all \(i\in N\), we derive Eq. (7). To see why \(c_{i}^{*}\left( 0\right) =0\), notice that since \(0\le b_{i}^{k_{l}}\left( v\right) \le 0\), we have \(b_{i}^{k_{l}}\left( 0\right) =0\), for all \(i\in N\), \(k=\left\{ 1,2,\ldots ,n-1\right\}\), and \(l\in \left\{ 1,,2,\ldots ,C_{n-1}^{k}\right\}\). It follows that \(\Phi _{i}\left( 0\right) =0\), for all \(i\in N\). Hence, evaluating Eq. (5) at \(v=0\) yields \(c_{i}^{*}\left( 0\right) =0\), for all \(i\in N\). This completes the proof.

1.2 Proof of Proposition 2

First, (i) Eq. (8) implies \(c_{i}^{*\prime }\left( v\right) \ge 0\) since \(K_{i}\left( \cdot ,\cdot \right)\) is a distribution function defined on \(\left[ 0,1\right] \times \left[ 0,1\right]\). Since \(c_{i}^{*\prime }\left( v\right)\) is differentiable for all \(i\in N\), it is clear that Eq. (8) implies \(c^{*\prime }\left( v\right)\) is increasing:

$$\begin{aligned} c_{i}^{*\prime \prime }\left( v\right) =\Phi _{i}^{\prime }\left( v\right) \ge 0, \end{aligned}$$

for all \(i\in N\). Second, Eq. (7) implies \(c_{i}^{*}\left( v\right) \ge 0\). Notice that \(\Phi _{i}\left( \cdot \right) \in \left[ 0,1\right]\) since it is the probability of winning for bidder i. Equation (7) implies:

$$\begin{aligned} c_{i}^{*}\left( v\right) =\int _{0}^{v}\Phi _{i}\left( s\right) ds\le \int _{0}^{v}ds=v. \end{aligned}$$

Last, notice that:

$$\begin{aligned} c_{i}^{*\prime }\left( v\right) =\Phi _{i}\left( v\right) =1-\sum _{k=0}^{n-1}\sum _{l=1}^{C_{n-1}^{k}} \left( {\bar{P}}\left( b_{i}^{k_{l}}\left( v\right) ;O_{i}^{k_{l}}\right) Q\left( N_{-i}/O_{i}^{k_{l}}\right) \right) , \end{aligned}$$

where

$$\begin{aligned} {\bar{P}}\left( b_{i}^{k_{l}}\left( v\right) ;O_{i}^{k_{l}}\right) :=\sum _{p=1}^{k} \left( \left( \int _{v_{p}^{k_{l^{\prime }}}\left( b_{i}^{k_{l}}\left( v\right) \right) } ^{1}\int _{0}^{c_{p}^{*}\left( \tau \right) }k_{p}\left( c,\tau \right) dcd\tau \right) {\underline{Q}}_{p}^{k_{l}}{\bar{Q}}_{p}^{k_{l}}\right) . \end{aligned}$$

Notice that Maskin and Riley (2003) show that if the upper endpoint of the support of the valuation distributions is the same for all bidders in first-price auctions, then the upper endpoints of the supports of all bidders’ equilibrium bid distributions are the same. Since \(v_{i}\) have the same upper endpoint of their distributions, we have \(b_{i}\left( 1\right) =b^{\star }\) and \(v_{j}\left( b_{i}\left( 1\right) \right) =1\), where \(j\ne i\) and \(b^{\star }\) is a constant. Evaluating \(c_{i}^{*\prime }\left( v\right)\) at \(v=1\) derives \(c_{i}^{*\prime }\left( 1\right) =1\) since \(v_{j}^{k_{l^{\prime }}}\left( b_{i}^{k_{l}}\left( 1\right) \right) =1\) for all \(i\in N\). This completes the proof.

1.3 Proof of Theorem 1

We establish the existence by appying the Schauder-Tychonoff fixed-point theorem, which states that any continuous mapping from a compact convex non-empty subset of a locally convex topological space to itself has a fixed point (Burton 2005). For all \(i\in N\), define a mapping \(h_{i}:\left[ 0,1\right] \rightarrow \left[ 0,1\right]\) as follows:

$$\begin{aligned} h_{i}\left( t_{i},{\textbf{c}}^{*}\left( \cdot \right) \right) :=\Phi \left( t_{i}; {\textbf{c}}^{*}\left( \cdot \right) \right) . \end{aligned}$$

Since \(K\left( \cdot ,\cdot \right)\) is integrable over both arguments and bidding functions are differentiable (hence continuous), \(h_{i}\left( t_{i},{\textbf{c}}^{*}\left( \cdot \right) \right)\) is a continuous mapping from \(\left[ 0,1\right] \times \left[ 0,1\right] ^{n}\rightarrow \left[ 0,1\right]\), for all \(i\in N\). Define \(H\left( t,{\textbf{c}}^{*}\left( \cdot \right) \right) :\left[ 0,1\right] ^{n}\times \left[ 0,1 \right] ^{n}\rightarrow \left[ 0,1\right] ^{n}\) as follows:

$$\begin{aligned} H\left( t,{\textbf{c}}^{*}\left( \cdot \right) \right) =\left( h_{1}\left( t_{1}, {\textbf{c}}^{*}\left( \cdot \right) \right) ,h_{2}\left( t_{2},{\textbf{c}}^{*} \left( \cdot \right) \right) ,\ldots ,h_{n}\left( t_{n},{\textbf{c}}^{*}\left( \cdot \right) \right) \right) ^{\prime }. \end{aligned}$$

Since \(h_{i}\left( t_{i},{\textbf{c}}^{*}\left( \cdot \right) \right)\) is continuous for all \(i\in N\), \(H\left( t,{\textbf{c}}^{*}\left( \cdot \right) \right)\) is also a continuous mapping. Next, let \(C\left( \left[ 0,1\right] \right)\) be the sapce of continuous functions \(\psi\) mapping from \(\left[ 0,1\right]\) to \(\left[ 0,1\right] ^{n}\) with the sup norm on \(\left[ 0,1\right]\):

$$\begin{aligned} \left\| \psi \right\| =\text {sup}_{v\in \left[ 0,1\right] }\left| \psi \left( v\right) \right| , \end{aligned}$$

where \(\left| \cdot \right|\) is a norm in \(R^{n}\) and for any \(x\in R^{n}\), \(\left| \cdot \right|\) is defined as \(\left| x\right| =\underset{i\in N}{\text {max}}\left| x_{i}\right|\). Define

$$\begin{aligned} M=\left\{ \psi \in C\left( \left[ 0,1\right] \right) :\left\| \psi \right\| \le 1\right\} . \end{aligned}$$

Notice that \({\textbf{c}}^{*}\left( \cdot \right) \in M\). By definition, M is equicontinuous and equibounded. It follows from the Arzela-Ascoli theorem (Burton 2005) that M is compact. In addition, M is certainly convex by definition.

Define a mapping \(T:M\rightarrow M\) as follows:

$$\begin{aligned} \left( T{\textbf{c}}^{*}\right) \left( v\right) :=\int _{0}^{v}H\left( t,{\textbf{c}} ^{*}\left( \cdot \right) \right) dt. \end{aligned}$$

We next establish that T is a continuous mapping. To this end, let \(\phi ,\varphi \in M\). It suffices to show that for any \(\epsilon >0\), there is a \(\delta\) such that \(\left\| \phi -\varphi \right\| <\delta\) implies \(\left\| T\phi -T\varphi \right\| <\epsilon\). Since \(h_{i}\left( t_{i},{\textbf{c}}^{*}\left( \cdot \right) \right)\) is a continuous mapping from \(\left[ 0,1\right] \times \left[ 0,1\right] ^{n}\rightarrow \left[ 0,1\right]\), for any \(\epsilon >0\), and \(t_{i}\in \left[ 0,1\right] ^{n}\), there exists an \(\delta >0\) such that \(\left| h_{i}\left( t_{i},\phi \left( \cdot \right) \right) -h_{i}\left( t_{i},\varphi \left( \cdot \right) \right) \right| <\epsilon\) when \(\left\| \phi -\varphi \right\| <\delta\), for all \(i\in N\). To establish the continuity of T, notice that for \(\delta\) such that \(\left\| \phi -\varphi \right\| <\delta\), we have:

$$\begin{aligned} \left\| T\phi -T\varphi \right\|&=\underset{v\in \left[ 0,1\right] }{\text {sup}}\left| \int _{0}^{v}H\left( t,\phi \left( \cdot \right) \right) dt-\int _{0}^{v}H\left( t,\varphi \left( \cdot \right) \right) dt\right| \\&=\underset{v\in \left[ 0,1\right] }{\text {sup}}\left( \underset{i\in N}{\text {max}}\left| \int _{0}^{v}h_{i}\left( t_{i},\phi \left( \cdot \right) \right) dt_{i}-\int _{0} ^{v}h_{i}\left( t_{i},\varphi \left( \cdot \right) \right) dt_{i}\right| \right) \\&\le \underset{v\in \left[ 0,1\right] }{\text {sup}}\left( \underset{i\in N}{\text {max}}\int _{0}^{v}\left| h_{i}\left( t_{i},\phi \left( s\right) \right) -h_{i}\left( t_{i}, \varphi \left( s\right) \right) \right| dt_{i}\right) \\&<\underset{v\in \left[ 0,1\right] }{\text {sup}}\left( \underset{i\in N}{\text {max}}\int _{0}^{v}\epsilon dt_{i}\right) \\&=\underset{v\in \left[ 0,1\right] }{\text {sup}}\left( v\epsilon \right) =\epsilon . \end{aligned}$$

It follows from the Schauder-Tychonoff fixed-point theorem that T has a fixed point in M. Hence, there exists at least an equlibrium cutoff \({\textbf{c}}^{*}\left( \cdot \right)\). This completes the proof.

1.4 Proof of Proposition 3

First, notice that \(c^{*}\left( v\right)\) is differentiable. Then, we can applying the envelope theorem in Eq. (5). Notice that \(v^{k}\left( b^{k}\left( v\right) \right) =v\) in a symmetric equilibrium. Differentiating Eq. (5) derives:

$$\begin{aligned}&c^{*\prime }\left( v\right) =\left[ \int _{0}^{1}\int _{c^{*}\left( \tau \right) }^{1}k\left( c,\tau \right) dcd\tau \right] ^{n-1} \\&\quad +\sum _{k=1}^{n-1}\left( C_{n-1}^{k}\left( \int _{0}^{v}\int _{0}^{c^{*}\left( \tau \right) }k \left( c,\tau \right) dcd\tau \right) ^{k}\left( \int _{0}^{1}\int _{c^{*}\left( \tau \right) }^{1} k\left( c,\tau \right) dcd\tau \right) ^{n-1-k}\right) ,\\&\quad =\sum _{k=0}^{n-1}\left( C_{n-1}^{k}\left( \int _{0}^{v}\int _{0}^{c^{*}\left( \tau \right) }k \left( c,\tau \right) dcd\tau \right) ^{k}\left( \int _{0}^{1}\int _{c^{*}\left( \tau \right) }^{1} k\left( c,\tau \right) dcd\tau \right) ^{n-1-k}\right) \\&\quad =\left( \int _{0}^{v}\int _{0}^{c^{*}\left( \tau \right) }k\left( c,\tau \right) dcd\tau +\int _{0}^{1} \int _{c^{*}\left( \tau \right) }^{1}k\left( c,\tau \right) dcd\tau \right) ^{n-1}\\&\quad =\left( \int _{0}^{v}\int _{0}^{c^{*}\left( \tau \right) }k\left( c,\tau \right) dcd\tau +\int _{v}^{1} \int _{c^{*}\left( \tau \right) }^{1}k\left( c,\tau \right) dcd\tau +\int _{0}^{v}\int _{c^{*} \left( \tau \right) }^{1}k\left( c,\tau \right) dcd\tau \right) ^{n-1}\\&\quad =\left( \int _{0}^{v}\int _{0}^{1}k\left( c,\tau \right) dcd\tau +\int _{v}^{1}\int _{c^{*} \left( \tau \right) }^{1}k\left( c,\tau \right) dcd\tau \right) ^{n-1}\\&\quad =\left( 1-\int _{v}^{1}\int _{0}^{1}k\left( c,\tau \right) dcd\tau +\int _{v}^{1}\int _{c^{*} \left( \tau \right) }^{1}k\left( c,\tau \right) dcd\tau \right) ^{n-1}\\&\quad =\left( 1-\int _{v}^{1}\int _{0}^{c^{*}\left( \tau \right) }k\left( c,\tau \right) dcd\tau \right) ^{n-1}. \end{aligned}$$

It follows that \(c_{i}^{*\prime }\left( v\right)\) is also differentiable and \(c^{*\prime }\left( 1\right) =1\). Last, integrating Eq. (8) from 0 to 1 and noticing \(c^{*}\left( 0\right) =0\) for all \(i\in N\), we derive Eq. (7). To see why \(c_{i}^{*}\left( 0\right) =0\), notice that \(b^{k}\left( 0\right) =0\) and \(v^{k}\left( 0\right) =0\), since \(v^{k}\left( \cdot \right)\) is the inverse bidding function. It follows that \(c^{*}\left( 0\right) =0\). This completes the proof.

1.5 Proof of Theorem 2

The existence of the symmetric equilibrium can be established by the Schauder-Tychonoff fixed-point theorem. Here we only need to prove the uniqueness of the symmetric equilibrium. Suppose, by contradiction, that we have two different symmetric equilibria \(x\left( v\right)\) and \(y\left( v\right)\). Then we have:

$$\begin{aligned} x^{\prime }\left( v\right) =\left[ 1-\int _{v}^{1}\int _{0}^{x\left( \tau \right) } k\left( \tau ,c\right) dcd\tau \right] ^{n-1}, \\ y^{\prime }\left( v\right) =\left[ 1-\int _{v}^{1}\int _{0}^{y\left( \tau \right) } k\left( \tau ,c\right) dcd\tau \right] ^{n-1}. \end{aligned}$$

Without loss of generality, suppose \(x\left( 1\right) >y\left( 1\right)\), then by the continuity of \(x\left( v\right)\) and \(y\left( v\right)\), there exists a \(v^{*}\in \left[ 0,1\right)\) such that \(x\left( v^{*}\right) =y\left( v^{*}\right) :=c\left( v^{*}\right)\) and \(x\left( \tau \right) >y\left( \tau \right)\) for all \(\tau \in \left( v^{*},1\right]\) by noting that \(x\left( 0\right) =y\left( 0\right)\). Consider two mutually exclusive cases. First, if \(k\left( v,c\right) >0\) with positive probability measure on \(\left( v^{*},1\right) \times \left( c\left( v^{*}\right) ,1\right)\), then \(x\left( \tau \right) >y\left( \tau \right)\) for all \(\tau \in \left( v^{*},1\right]\) implies that

$$\begin{aligned} \int _{0}^{x\left( \tau \right) }k\left( \tau ,c\right) dc>\int _{0}^{y\left( \tau \right) }k\left( \tau ,c\right) dc, \end{aligned}$$

for all \(\tau \in \left( v^{*},1\right)\). It follows that \(x^{\prime }\left( v^{*}\right) <y^{\prime }\left( v^{*}\right)\), which is a contradiction to \(x\left( \tau \right) >y\left( \tau \right)\) for all \(\tau \in \left( v^{*},1\right]\). Hence, we must have \(x\left( \tau \right) =y\left( \tau \right)\) for all \(\tau \in \left( v^{*},1\right]\) in this case. Second, if \(k\left( v,c\right) >0\) with zero probability measure on \(\left( v^{*},1\right) \times \left( c\left( v^{*}\right) ,1\right)\), then, by the same argument, \(x^{\prime }\left( \tau \right) =y^{\prime }\left( \tau \right)\) for all \(\tau \in \left( v^{*},1\right]\). It follows from \(x\left( v^{*}\right) =y\left( v^{*}\right)\) that \(x\left( \tau \right) =y\left( \tau \right)\) for all \(\tau \in \left( v^{*},1\right]\) in this case. Hence, both cases lead to a contradiction to \(x\left( 1\right) >y\left( 1\right)\). Therefore, we have established that there exists a \(v^{*}\in \left[ 0,1\right)\), such that \(x\left( \tau \right) =y\left( \tau \right)\) for all \(\tau \in \left[ v^{*},1\right]\). If \(v^{*}=0\), the proof is complete.

Now, consider an arbitrary closed interval \(\left[ \alpha ,\beta \right] \subset \left[ 0,1\right]\). Suppose \(x\left( \alpha \right) =y\left( \alpha \right)\) and \(x\left( \beta \right) =y\left( \beta \right)\); and, without loss of generality, \(x\left( \tau \right) >y\left( \tau \right)\) for all \(\tau \in \left( \alpha ,\beta \right) .\) It follows from the argument above that \(x^{\prime }\left( \tau \right) <y^{\prime }\left( \tau \right)\) for \(\tau \in \left( \alpha ,\beta \right)\), which is in contradiction to \(x\left( \tau \right) >y\left( \tau \right)\) for all \(\tau \in \left( \alpha ,\beta \right)\). Hence, we have established that \(x\left( v\right) =y\left( v\right)\) for all \(v\in \left[ 0,1\right]\). Therefore, the symmetric equilibrium is unique. This completes the proof.

1.6 Proof of Corollary 1

We compare the expected payoffs for n and \(n+1\) bidders as follows:

$$\begin{aligned} c_{n}^{*}\left( v\right) =\int _{0}^{v}\left[ 1-\int _{t}^{1}\int _{0}^{c_{n}^{*}\left( \tau \right) } k\left( c,\tau \right) dcd\tau \right] ^{n-1}dt,\\c_{n+1}^{*}\left( v\right) =\int _{0}^{v} \left[ 1-\int _{t}^{1}\int _{0}^{c_{n+1}^{*}\left( \tau \right) }k\left( c,\tau \right) dcd\tau \right] ^{n}dt, \end{aligned}$$

If \(c_{n+1}^*(v)\ge c_{n}^*(v)\), then

$$\begin{aligned} \int _{0}^{v}\left[ 1-\int _{t}^{1}\int _{0}^{c_{n+1}^{*}\left( \tau \right) } k\left( c,\tau \right) dcd\tau \right] ^{n}dt<\int _{0}^{v}\left[ 1-\int _{t}^{1}\int _{0}^{c_{n} ^{*}\left( \tau \right) }k\left( c,\tau \right) dcd\tau \right] ^{n-1}dt, \end{aligned}$$

which contradicts the assumption that \(c_{n+1}^*(v)\ge c_{n}^*(v)\), and thus \(c_{n+1}^*(v)<c_{n}^*(v)\).

1.7 Proof of Proposition 4

First, notice that \(F_{A}=F_{B}\) implies \(f_{A}=f_{B}\). Next, we establish that \(c_{A}^{*}\left( 1\right) >c_{B}^{*}\left( 1\right)\). Suppose, by contradiction, that \(c_{A}^{*}\left( 1\right) \le c_{B}^{*}\left( 1\right)\). Then, there exists a \(v^{*}\in \left[ 0,1\right)\) such that \(c_{A}^{*}\left( v\right) \le c_{B}^{*}\left( v\right)\) for all \(v\in \left( v^{*},1\right]\). This implies that

$$\begin{aligned} c_{A}^{*\prime }\left( v^{*}\right) =\left[ 1-\int _{v^{*}}^{1}f\left( \tau \right) G_{A}\left( c_{A}^{*}\left( \tau \right) \right) d\tau \right] ^{n-1}, \\ c_{B}^{*\prime }\left( v^{*}\right) =\left[ 1-\int _{v^{*}}^{1}f\left( \tau \right) G_{B}\left( c_{B}^{*}\left( \tau \right) \right) d\tau \right] ^{n-1}. \end{aligned}$$

Since \(G_{A}<G_{B}\) and \(c_{A}^{*}\left( v\right) \le c_{B}^{*}\left( v\right)\) for all \(v\in \left( v^{*},1\right]\), we have \(G_{A}\left( c_{A}^{*}\left( \cdot \right) \right) <G_{B}\left( c_{B}^{*}\left( \cdot \right) \right)\). It follows from that \(c_{A}^{*\prime }\left( v^{*}\right) >c_{B}^{*\prime }\left( v^{*}\right)\). It follows that there exists a \({\bar{v}}^{*}>v^{*}\) such that \(c_{A}^{*}\left( v\right) >c_{B}^{*}\left( v\right)\) for all \(v\in \left( v^{*},{\bar{v}}^{*}\right)\), leading to a contradiction.

Then, we show \(c_{A}^{*}\left( v\right) >c_{B}^{*}\left( v\right)\) for all \(v\in \left( 0,1\right]\). Without loss of generality, assume there exists a \(v^{*}\in \left[ 0,1\right)\) such that \(c_{A}^{*}\left( v^{*}\right) =c_{B}^{*}\left( v^{*}\right)\) and \(c_{A}^{*}\left( v\right) >c_{B}^{*}\left( v\right)\) for all \(v\in \left( v^{*},1\right]\). This implies that \(c_{A}^{*\prime }\left( v^{*}\right) >c_{B}^{*\prime }\left( v^{*}\right)\), which means:

$$\begin{aligned} \int _{v^{*}}^{1}f\left( \tau \right) G_{A}\left( c_{A}^{*}\left( \tau \right) \right) d\tau <\int _{v^{*}}^{1}f\left( \tau \right) G_{B}\left( c_{B}^{*}\left( \tau \right) \right) d\tau . \end{aligned}$$

In addition, since \(c_{A}^{*\prime }\left( v^{*}\right) >c_{B}^{*\prime }\left( v^{*}\right)\), there exists a \({\underline{v}}^{*}<v^{*}\) such that \(c_{A}^{*}\left( v\right) >c_{B}^{*}\left( v\right)\) for all \(v\in \left( {\underline{v}}^{*},v^{*}\right)\) and \(c_{A}^{*}\left( {\underline{v}}^{*}\right) =c_{B}^{*}\left( {\underline{v}}^{*}\right)\) by noticing \(c_{A}^{*}\left( 0\right) =c_{B}^{*}\left( 0\right)\). This implies that \(c_{A}^{*\prime }\left( {\underline{v}}^{*}\right) >c_{B}^{*\prime }\left( {\underline{v}}^{*}\right)\). To see this, notice that Since \(G_{A}<G_{B}\) and \(c_{A}^{*}\left( v\right) <c_{B}^{*}\left( v\right)\) for all \(v\in \left( {\underline{v}}^{*},v^{*}\right)\), we have \(G_{A}\left( c_{A}^{*}\left( \cdot \right) \right) <G_{B}\left( c_{B}^{*}\left( \cdot \right) \right)\) for all \(v\in \left( {\underline{v}}^{*},v^{*}\right)\), meaning

$$\begin{aligned} \int _{{\underline{v}}^{*}}^{v^{*}}f\left( \tau \right) G_{A}\left( c_{A}^{*} \left( \tau \right) \right) d\tau <\int _{{\underline{v}}^{*}}^{v^{*}}f \left( \tau \right) G_{B}\left( c_{B}^{*}\left( \tau \right) \right) d\tau . \end{aligned}$$

Hence, we have

$$\begin{aligned} c_{A}^{*\prime }\left( {\underline{v}}^{*}\right)&=\left[ 1-\int _{{\underline{v}}^{*}}^{1}f\left( \tau \right) G_{A}\left( c_{A}^{*} \left( \tau \right) \right) d\tau \right] ^{n-1}\\&=\left[ 1-\int _{{\underline{v}}^{*}}^{v^{*}}f\left( \tau \right) G_{A}\left( c_{A}^{*} \left( \tau \right) \right) d\tau -\int _{v^{*}}^{1}f\left( \tau \right) G_{A}\left( c_{A}^{*} \left( \tau \right) \right) d\tau \right] ^{n-1}\\&>\left[ 1-\int _{{\underline{v}}^{*}}^{v^{*}}f\left( \tau \right) G_{B}\left( c_{B}^{*} \left( \tau \right) \right) d\tau -\int _{v^{*}}^{1}f\left( \tau \right) G_{B}\left( c_{B}^{*} \left( \tau \right) \right) d\tau \right] ^{n-1}\\&=\left[ 1-\int _{v^{*}}^{1}f\left( \tau \right) G_{B}\left( c_{B}^{*}\left( \tau \right) \right) d\tau \right] ^{n-1}=c_{B}^{*\prime }\left( {\underline{v}}^{*}\right) . \end{aligned}$$

This then implies that there exists a \({\tilde{v}}\in \left( {\underline{v}}^{*},v^{*}\right)\) such that \(c_{A}^{*}\left( v\right) >c_{B}^{*}\left( v\right)\) for all \(v\in \left( {\underline{v}}^{*},{\tilde{v}}\right) \subset \left( {\underline{v}}^{*},v^{*}\right)\), resulting in a contradiction. Hence, \(c_{A}^{*}\left( v\right) >c_{B}^{*}\left( v\right)\) for all \(v\in \left( 0,1\right]\). This completes the proof.

1.8 Proof of Proposition 5

We prove this theorem by using the contraction mapping theorem. First, let \(C\left( \left[ 0,1\right] \right)\) be the sapce of continuous functions \(\psi\) mapping from \(\left[ 0,1\right]\) to \(\left[ 0,1\right] ^{2}\) with the sup norm on \(\left[ 0,1\right]\):

$$\begin{aligned} \left\| \psi \right\| =\text {sup}_{v\in \left[ 0,1\right] }\left| \psi \left( v\right) \right| , \end{aligned}$$

where where \(\left| \cdot \right|\) is a norm in \(R^{n}\) and for any \(x\in R^{n}\), \(\left| \cdot \right|\) is defined as \(\left| x\right| =\underset{i\in N}{\text {max}}\left| x_{i}\right|\). Define

$$\begin{aligned} M=\left\{ \psi \in C\left( \left[ 0,1\right] \right) :\left\| \psi \right\| \le 1\right\} . \end{aligned}$$

It is certainly that the normed space \(\left( M,\left\| \cdot \right\| \right)\) is a Banach space. Notice that \({\textbf{c}}^{*}\left( \cdot \right) =\left( c_{1}\left( \cdot \right) ,c_{2}\left( \cdot \right) \right) \in M\).

Next, define a mapping \(h_{i}\left( t_{i},{\textbf{c}}^{*}\left( \cdot \right) \right)\) from \(\left[ 0,1\right] \times \left[ 0,1\right] ^{2}\rightarrow \left[ 0,1\right]\), for all \(i\in \left\{ 1,2\right\}\) as follows:

$$\begin{aligned} h_{1}\left( t_{1},{\textbf{c}}^{*}\left( \cdot \right) \right) =1-\int _{v_{2}\left( b_{1} \left( t_{1}\right) \right) }^{1}\int _{0}^{c_{2}^{*}\left( \tau \right) }k_{2}\left( c,\tau \right) dcd\tau \\ h_{2}\left( t_{2},{\textbf{c}}^{*}\left( \cdot \right) \right) =1-\int _{v_{1}\left( b_{2} \left( t_{2}\right) \right) }^{1}\int _{0}^{c_{1}^{*}\left( \tau \right) }k_{1}\left( c,\tau \right) dcd\tau \end{aligned}$$

Define \(H\left( t,{\textbf{c}}^{*}\left( \cdot \right) \right) :\left[ 0,1\right] ^{2}\times \left[ 0,1\right] ^{2}\rightarrow \left[ 0,1\right] ^{2}\) as follows:

$$\begin{aligned} H\left( t,{\textbf{c}}^{*}\left( \cdot \right) \right) =\left( h_{1}\left( t_{1},{\textbf{c}} ^{*}\left( \cdot \right) \right) ,h_{2}\left( t_{2},{\textbf{c}}^{*}\left( \cdot \right) \right) \right) ^{\prime }. \end{aligned}$$

Then, we define a mapping \(T:M\rightarrow M\) as follows:

$$\begin{aligned} \left( T{\textbf{c}}^{*}\right) \left( v\right)&=\int _{0}^{v}H\left( t,{\textbf{c}}^{*} \left( \cdot \right) \right) dt\\&=\left[ \begin{array}{c} \int _{0}^{v}\left( 1-\int _{v_{2}\left( b_{1}\left( t_{1}\right) \right) }^{1}\int _{0}^{c_{2}^{*} \left( \tau \right) }k_{2}\left( c,\tau \right) dcd\tau \right) dt_{1}\\ \int _{0}^{v}\left( 1-\int _{v_{1}\left( b_{2}\left( t_{2}\right) \right) }^{1}\int _{0}^{c_{1}^{*} \left( \tau \right) }k_{1}\left( c,\tau \right) dcd\tau \right) dt_{2} \end{array}\right] \\&=\left[ \begin{array}{c} v-\int _{0}^{v}\int _{v_{2}\left( b_{1}\left( t_{1}\right) \right) }^{1}\int _{0}^{c_{2}^{*} \left( \tau \right) }k_{2}\left( c,\tau \right) dcd\tau dt_{1}\\ v-\int _{0}^{v}\int _{v_{1}\left( b_{2}\left( t_{2}\right) \right) }^{1}\int _{0}^{c_{1}^{*} \left( \tau \right) }k_{1}\left( c,\tau \right) dcd\tau dt_{2} \end{array}\right] \end{aligned}$$

Since \(g_{i}\left( \cdot \right)\) and \(f_{i}\left( \cdot \right)\), \(i\in \left\{ 1,2\right\}\), are integrable over both arguments and bidding functions are differentiable (hence continuous), \(h_{i}\left( t_{i},{\textbf{c}}^{*}\left( \cdot \right) \right)\) and \(H\left( t,{\textbf{c}}^{*}\left( \cdot \right) \right)\) are continuous, for all \(i\in N\).

It suffices to show that \(Tc^{*}\) is a contraction mapping. To show this, for any \(x,y\in M\), where \(x=\left( x_{1},x_{2}\right) ^{\prime }\) and \(y=\left( y_{1},y_{2}\right) ^{\prime }\), we have:

$$\begin{aligned} \left( Tx\right) \left( v\right) -\left( Ty\right) \left( v\right)&=\left[ \begin{array}{c} \int _{0}^{v}\int _{v_{2}\left( b_{1}\left( t_{1}\right) \right) }^{1}\int _{0}^{y_{2} \left( \tau \right) }k_{2}\left( c,\tau \right) dcd\tau dt_{1}-\int _{0}^{v}\int _{v_{2} \left( b_{1}\left( t_{1}\right) \right) }^{1}\int _{0}^{x_{2}\left( \tau \right) }k_{2} \left( c,\tau \right) dcd\tau dt_{1}\\ \int _{0}^{v}\int _{v_{1}\left( b_{2}\left( t_{2}\right) \right) }^{1}\int _{0}^{y_{1} \left( \tau \right) }k_{1}\left( c,\tau \right) dcd\tau dt_{2}-\int _{0}^{v}\int _{v_{1} \left( b_{2}\left( t_{2}\right) \right) }^{1}\int _{0}^{x_{1}\left( \tau \right) }k_{1}\left( c,\tau \right) dcd\tau dt_{2} \end{array}\right] \\&=\left[ \begin{array}{c} \int _{0}^{v}\left( \int _{v_{2}\left( b_{1}\left( t_{1}\right) \right) }^{1}\int _{x_{2} \left( \tau \right) }^{y_{2}\left( \tau \right) }k_{2}\left( c,\tau \right) dcd\tau \right) dt_{1}\\ \int _{0}^{v}\left( \int _{v_{1}\left( b_{2}\left( t_{2}\right) \right) }^{1}\int _{x_{1} \left( \tau \right) }^{y_{1}\left( \tau \right) }k_{1}\left( c,\tau \right) dcd\tau \right) dt_{2} \end{array}\right] . \end{aligned}$$

Hence, we have

$$\begin{aligned}&\left\| \left( Tx\right) \left( v\right) -\left( Ty\right) \left( v\right) \right\| =\underset{v\in \left[ 0,1\right] }{\text {sup}}\underset{i\in \left\{ 1,2\right\} }{\text {max}}\left| \int _{0}^{v}\int _{v_{j}\left( b_{i}\left( t_{i}\right) \right) }^{1} \int _{x_{j}\left( \tau \right) }^{y_{j}\left( \tau \right) }k_{j}\left( c,\tau \right) dcd\tau dt_{i}\right| \\&\quad \le \underset{v\in \left[ 0,1\right] }{\text {sup}}\underset{i\in \left\{ 1,2\right\} }{\text {max}}\\&\quad \int _{0}^{v}\left( {\bar{k}}_{j}\left( 1-v_{j}\left( b_{i}\left( t_{i}\right) \right) \right) \underset{s\in \left[ 0,1\right] }{\text {sup}}\underset{j\in \left\{ 1,2\right\} }{\text {max}} \left| x_{j}\left( s\right) -y_{j}\left( s\right) \right| \right) dt_{i}\\&\quad =\left\| x\left( s\right) -y\left( s\right) \right\| \left( \underset{v\in \left[ 0,1\right] }{\text {sup}}\underset{i\in \left\{ 1,2\right\} }{\text {max}}\left( \int _{0}^{v}{\bar{k}}_{j} \left( 1-v_{j}\left( b_{i}\left( t_{i}\right) \right) \right) dt_{i}\right) \right) \\&\quad \le \left\| x\left( s\right) -y\left( s\right) \right\| \left( \underset{i\in \left\{ 1,2\right\} }{\text {max}}\left( \int _{0}^{1}{\bar{k}}_{j}\left( 1-v_{j}\left( b_{i}\left( t_{i}\right) \right) \right) dt_{i}\right) \right) \\&\quad <\left\| x\left( s\right) -y\left( s\right) \right\| , \end{aligned}$$

where \({\bar{k}}_{j}:=\text {sup}_{c,v\in \left[ 0,1\right] }k_{i}\left( c,v\right)\). The last inequality holds because \(\text {sup}_{c,v\in \left[ 0,1\right] }k_{i}\left( c,v\right) <\frac{1}{\int _{0}^{1}\left[ 1-v_{i} \left( b_{j}\left( v\right) \right) \right] dv}\) for \(i,j\in \left\{ 1,2\right\}\) and \(i\ne j\). Therefore, \(T:M\rightarrow M\) is a contraction mapping. Since M is a complete metric space, it follows from the contraction mapping theorem that T has a unique fixed point in M. Hence, functional equation \({\textbf{c}}^{*}=T{\textbf{c}}^{*}\) has a unique solution.

This completes the proof.

1.9 Proof of Corollary 2

Suppose the valuations and participation costs are independently distributed, i.e. \(K_{i}\left( v_{i},c_{i}\right) =F_{i}\left( v_{i}\right) G_{i}\left( c_{i}\right)\) for all \(i\in N\). The most of the argument in the proof of Proposition 5 applies here. However, we could have a weaker sufficient condition. To see this, notice that

$$\begin{aligned}&\left\| \left( Tx\right) \left( v\right) -\left( Ty\right) \left( v\right) \right\| =\underset{v\in \left[ 0,1\right] }{\text {sup}}\underset{i\in \left\{ 1,2\right\} }{\text {max}}\left| \int _{0}^{v}\int _{v_{j}\left( b_{i}\left( t_{i}\right) \right) }^{1} \left( G_{j}\left( y_{j}\left( \tau \right) \right) -G_{j}\left( x_{j}\left( \tau \right) \right) \right) f_{j}\left( \tau \right) d\tau dt_{i}\right| \\&\quad \le \underset{v\in \left[ 0,1\right] }{\text {sup}}\underset{i\in \left\{ 1,2\right\} }{\text {max}}\int _{0}^{v}\int _{v_{j}\left( b_{i}\left( t_{i}\right) \right) }^{1}\left| G_{j} \left( y_{j}\left( \tau \right) \right) -G_{j}\left( x_{j}\left( \tau \right) \right) \right| f_{j} \left( \tau \right) d\tau dt_{i}\\&\quad \le \underset{v\in \left[ 0,1\right] }{\text {sup}}\underset{i\in \left\{ 1,2\right\} }{\text {max}}\left( {\bar{g}}_{j}\int _{0}^{v}\int _{v_{j}\left( b_{i}\left( t_{i}\right) \right) }^{1} \left| y_{j}\left( \tau \right) -x_{j}\left( \tau \right) \right| f_{j}\left( \tau \right) d\tau dt_{i}\right) \\&\quad \le \underset{v\in \left[ 0,1\right] }{\text {sup}}\underset{i\in \left\{ 1,2\right\} }{\text {max}}\left( {\bar{g}}_{j}\int _{0}^{v}\int _{v_{j}\left( b_{i}\left( t_{i}\right) \right) }^{1}\left( \underset{s\in \left[ 0,1\right] }{\text {sup}}\underset{j\in \left\{ 1,2\right\} }{\text {max}}\int _{0}^{s}\left| y_{j}\left( u\right) -x_{j}\left( u\right) \right| du\right) f_{j} \left( \tau \right) d\tau dt_{i}\right) \\&\quad =\left\| x\left( s\right) -y\left( s\right) \right\| \left( \underset{v\in \left[ 0,1\right] }{\text {sup}}\underset{i\in \left\{ 1,2\right\} }{\text {max}}\left( \int _{0}^{v}\int _{v_{j} \left( b_{i}\left( t_{i}\right) \right) }^{1}{\bar{g}}_{j}f_{j}\left( \tau \right) d\tau dt_{i}\right) \right) \\&\quad \le \left\| x\left( s\right) -y\left( s\right) \right\| \left( \underset{i\in \left\{ 1,2\right\} }{\text {max}}\left( \int _{0}^{1}\int _{v_{j}\left( b_{i}\left( t_{i}\right) \right) }^{1}{\bar{g}}_{j}f_{j} \left( \tau \right) d\tau dt_{i}\right) \right) \\&\quad =\left\| x\left( s\right) -y\left( s\right) \right\| \left( \underset{i\in \left\{ 1,2\right\} }{\text {max}}\left( {\bar{g}}_{j}\int _{0}^{1}\left[ 1-F_{j}\left( v_{j}\left( b_{i}\left( t_{i}\right) \right) \right) \right] dt_{i}\right) \right) \\&\quad <\left\| x\left( s\right) -y\left( s\right) \right\| , \end{aligned}$$

where \({\bar{g}}_{j}:=\text {sup}_{c\in \left[ 0,1\right] }g_{i}\left( c\right)\). The last inequality holds because \(\text {sup}_{c\in \left[ 0,1\right] }g_{j}\left( c\right) <\frac{1}{\int _{0}^{1}\left[ 1-F\left( v_{j}\left( b_{i}\left( v\right) \right) \right) \right] dv}\) for \(i,j\in \left\{ 1,2\right\}\) and \(i\ne j\). Last, notice that the sufficient conditions when distribution functions are independent and when the sufficient condition in the general case can be rewritten as follows:

$$\begin{aligned}{} & {} \int _{0}^{1}\int _{v_{j}\left( b_{i}\left( v\right) \right) }^{1}{\bar{g}}_{j}f_{j}\left( \tau \right) d\tau dv<1,\\{} & {} \int _{0}^{1}\int _{v_{j}\left( b_{i}\left( v\right) \right) }^{1}{\bar{k}}_{j}d\tau dv<1. \end{aligned}$$

It follows that

$$\begin{aligned} \int _{0}^{1}\int _{v_{j}\left( b_{i}\left( v\right) \right) }^{1}{\bar{g}}_{j}f_{j}\left( \tau \right) d\tau dv<\int _{0}^{1}\int _{v_{j}\left( b_{i}\left( v\right) \right) }^{1}{\bar{k}}_{j}d\tau dv<1. \end{aligned}$$

The sufficient condition when distribution functions are independent holds when the sufficient condition in the general case holds.

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Cao, X., Hsueh, SC., Tian, G. et al. Participation constraints in first-price auctions. Int J Game Theory 53, 609–634 (2024). https://doi.org/10.1007/s00182-023-00884-x

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