Equilibria in second-price auctions with private participation costs

Abstract

We study equilibria in second-price auctions where bidders are independently and privately informed about both their values and participation costs, and where the joint distributions of these values and costs across bidders are not necessarily identical. 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). We find that the marginal density of participation costs and the concentration of values matter for the uniqueness. The presence of private information on participation costs tends to reduce multiplicity of participation equilibria, although multiplicity still persists.

Appendix

Proof of Lemma 1

If $m_i=0$, none of the other bidders will participate, the probability of which is

$$\begin{aligned} \prod _{j\ne i}F_{c_j^{*}}(0)=\prod _{j\ne i}\int _{0}^{1}\int _{c_j^*(\tau )}^{1}k_j(\tau ,c_j){\mathrm{d}}c_j{\mathrm{d}}\tau . \end{aligned}$$

Otherwise, at least one other bidder submits a bid. Then,

$$\begin{aligned} \prod _{j\ne i}F_{c_j^{*}}(m_i)=\prod _{j\ne i}\left[ 1-\int _{m_i}^1\int _0^{c_j^*(\tau )}k_j(\tau , c_j){\mathrm{d}}c_j{\mathrm{d}}\tau \right] . \end{aligned}$$

Thus, the cutoff for individual i, $i\in {1,2,\ldots n}$, which is $c_i^{*}(v)=\int _{0}^{v}(v_i-m_i){\mathrm{d}}\prod _{j\ne i} F_{c_j^{*}}(m_i)$, can be expressed as

$$\begin{aligned} c_i^{*}(v)= & {} \int _{0}^{v}(v_i-m_i){\mathrm{d}}\prod _{j\ne i}\left[ 1-\int _{m_i} ^1\int _0^{(c_j^*(\tau ))}k_j(\tau , c_j){\mathrm{d}}c_j {\mathrm{d}}\tau \right] \\&+\,v\prod _{j\ne i} \left[ \int _{0}^{1}\int _{(c_j^*(\tau ))}^{1}k_j(\tau ,c_j){\mathrm{d}}c_j{\mathrm{d}}\tau \right] . \end{aligned}$$

Integrating by parts, we have

$$\begin{aligned} c_i^{*}(v)= & {} \int _{0}^{v}(v-m_i){\mathrm{d}}\prod _{j\ne i}\left[ 1-\int _{m_i}^1\int _0^{(c_j^*(\tau ))}k_j(\tau ,c_j){\mathrm{d}}c_j {\mathrm{d}}\tau \right] \\&+\,v\prod _{j\ne i}\left[ \int _{0}^{1}\int _{(c_j^*(\tau ))}^{1}k_j (\tau ,c_j){\mathrm{d}}c_j {\mathrm{d}}\tau \right] \\= & {} (v-m_i)\prod _{j\ne i}\left[ 1-\int _{m_i}^1\int _0^{(c_j^*(\tau ))}k_j(\tau , c_j){\mathrm{d}}c_j {\mathrm{d}}\tau \right] \big |_{0}^{v_i}\\&+\,v\prod _{j\ne i}\left[ \int _{0}^{1}\int _{(c_j^*(\tau ))}^{1}k_j(\tau ,c_j){\mathrm{d}}c_j {\mathrm{d}}\tau \right] \\&+\int _{0}^{v}\prod _{j\ne i}\left[ 1-\int _{m_i}^1\int _0^{(c_j^*(\tau ))}k_j(\tau , c_j){\mathrm{d}}c_j {\mathrm{d}}\tau \right] {\mathrm{d}}m_i\\= & {} -v\prod _{j\ne i}\left[ 1-\int _{0}^1\int _0^{(c_j^*(\tau ))}k_j(\tau , c_j){\mathrm{d}}c_j {\mathrm{d}}\tau \right] \\&+\,v\prod _{j\ne i}\left[ \int _{0}^{1}\int _{(c_j^*(\tau ))}^{1}k_j(\tau ,c_j){\mathrm{d}}c_j {\mathrm{d}}\tau \right] \\&+\int _{0}^{v}\prod _{j\ne i}\left[ 1-\int _{m_i}^1\int _0^{(c_j^*(\tau ))}k_j(\tau , c_j){\mathrm{d}}c_j {\mathrm{d}}\tau \right] {\mathrm{d}}m_i. \end{aligned}$$

Since

$$\begin{aligned} \int _{0}^1\int _0^{(c_j^*(\tau ))}k_j(\tau , c_j){\mathrm{d}}c_j {\mathrm{d}}\tau {+}\int _{0}^{1}\int _{(c_j^*(\tau ))}^{1}k_j(\tau ,c_j){\mathrm{d}}c_j {\mathrm{d}}\tau = \int _0^1\int _0^{1}k_j(\tau , c_j){\mathrm{d}}c_j {\mathrm{d}}\tau {=}1, \end{aligned}$$

thus we have

$$\begin{aligned} c_i^{*}(v)=\int _{0}^{v}\prod _{j\ne i}\left[ 1-\int _{m_i}^1\int _0^{c_j^*(\tau )}k_j(\tau , c_j){\mathrm{d}}c_j {\mathrm{d}}\tau \right] {\mathrm{d}}m_i. \end{aligned}$$

$\square$

Proof of Lemma 2

• (i) Let $v=0$ in the expression of $c_i^{*}(v)$, we have the result.

• (ii) Since

  $$\begin{aligned} c_i^{*}(v)=\int _{0}^{v}\prod _{j\ne i}\left[ 1-\int _{m_i}^1\int _0^{c_j^*(\tau )}k_j(\tau , c_j){\mathrm{d}}c_j {\mathrm{d}}\tau \right] {\mathrm{d}}m_i \leqslant \int _0^{v}{\mathrm{d}}m_i=v \end{aligned}$$

  by $0\le \int _{m_i}^1\int _0^{c_j^*(\tau )}k_j(\tau , c_j){\mathrm{d}}c_j{\mathrm{d}}\tau \le \int _{0}^{1}\int _{0}^{1}k_j(\tau , c_j) {\mathrm{d}}c_j {\mathrm{d}}\tau = 1$, thus $0\le c_i^{*}(v)\le v$.

• (iii) Letting $v=1$ in (3), we have the result.

• (iv)
  $$\begin{aligned} \frac{{\mathrm{d}}c^*_i(v)}{{\mathrm{d}} v}=\prod _{j\ne i}\left[ 1-\int _{v}^1\int _0^{c_j^*(\tau )}k_j(\tau , c_j){\mathrm{d}}c_j {\mathrm{d}}\tau \right] \ge 0 \end{aligned}$$

  by noting that $\int _{v}^1\int _0^{c_j^*(\tau )}k_j(\tau , c_j){\mathrm{d}}c_j {\mathrm{d}}\tau \le 1$. Then,

  $$\begin{aligned} \frac{{\mathrm{d}}^2 c^*_i(v)}{{\mathrm{d}} v^2}{=}\sum _{k\ne i}\prod _{j\ne i, j\ne k}\left[ 1{-}\int _{v}^1\int _0^{c_j^*(\tau )}k_j(\tau , c_j){\mathrm{d}}c_j {\mathrm{d}}\tau \right] \int _0^{c_k^*(v)}k_j(\tau , c_j){\mathrm{d}}c_j {\mathrm{d}}\tau {\ge } 0. \end{aligned}$$

$\square$

Proof of Theorem 1

In the following, we apply the Schauder–Tychonoff fixed-point theorem (see Burton 2005, p. 185) ), 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, to show the existence of the equilibria.

Let $h_i(m_i, \mathbf c ^*)=\prod _{j\ne i}[1-\int _{m_i}^1\int _0^{c_j^*(\tau )}k_j(\tau , c){\mathrm{d}}c{\mathrm{d}}\tau ]$ with $\mathbf c ^*=(c_1^*,\ldots ,c_n^*)$. Since $k_j(\tau , c)$ is integrable over c as it is a density function, there exists a continuous function $\gamma _j(\tau , c)$ with $\frac{\partial \gamma _j(\tau , c)}{\partial c}=k_j(\tau , c)$ such that $\int _0^{c_j^*(\tau )}k_j(\tau , c){\mathrm{d}}c=\gamma _j(\tau , c_j^*(\tau ))-\gamma _j(\tau ,0)$. Thus, $h_i(m_i, \mathbf c ^*)=\prod _{j\ne i}[1-\int _{m_i}^1[\gamma _j(\tau , c_j^*(\tau ))-\gamma _j(\tau ,0)]{\mathrm{d}}\tau ]$, which is a continuous mapping from $[0, 1]\times [0,1]^n\rightarrow [0,1]$.

Let $H(m, \mathbf c ^*))=(h_1(m_1, \mathbf c ^*)), h_2(m_2, \mathbf c ^*)),\ldots , h_n(m_n, \mathbf c ^*)))'$, which is a continuous mapping from $[0,1]^n\times [0,1]^n\rightarrow [0,1]^n$. By Lemma 2, H is bounded above by one. Define

$$\begin{aligned} M=\{ {\mathbf{c}}\in \varphi \mid : \Vert {\mathbf{c}}\Vert \le 1\}, \end{aligned}$$

where $\varphi$ is the space of continuous functions $\phi$ defined on $[0,1]^n\rightarrow [0,1]^n$ with $\Vert c\Vert =\sup _{0\le v\le 1}c(v)$. Then, by Ascoli Theorem, M is compact. M is clearly convex.

Define an operator $P: M\rightarrow M$ by

$$\begin{aligned} (P\mathbf c )(v)=\int _0^vH(s, \mathbf c (\cdot )){\mathrm{d}}s. \end{aligned}$$

To see that P is continuous, let $\phi \in M$ and let $\epsilon >0$ be given. We show that there exists an $\eta >0$ such that $\varphi \in M$ and $\Vert \phi -\varphi \Vert <\eta$ implies $\Vert (P\phi )(v)-(P\varphi )(v)\Vert \le \epsilon$. Now

$$\begin{aligned} \big |(P\phi )(v)-(P\varphi )(v)\big |=\big |\int _0^v[H(s, \phi (\cdot ))-H(s, \varphi (\cdot ))]{\mathrm{d}}s\big | \end{aligned}$$

and $h_i$ is continuous, so for $\epsilon >0$, there is an $\eta$ such that $\big | \phi (\tau )-\varphi (\tau )\big |<\eta$ implies $\big |h_i(m_i, \phi (\tau ))-h_i(m_i, \varphi (\tau ))\big |<\epsilon$. Thus, for $\Vert \phi -\varphi \Vert <\eta$, we have

$$\begin{aligned} \big |(P\phi )(v)-(P\varphi )(v)\big |= & {} \big |\int _0^v[H(s, \phi (\cdot ))-H(s, \varphi (\cdot ))]{\mathrm{d}}s\big |\\= & {} \int _0^1\big |H(s, \phi (\cdot ))-H(s, \varphi (\cdot ))\big |ds\le \epsilon . \end{aligned}$$

Then, by Lemma 2, P is a continuous function from M to itself. Thus, by Schauder–Tychonoff fixed-point theorem, there exists a fixed point, i.e., a solution for the functional differential equation system exists. $\square$

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 way of contradiction, that we have two different symmetric equilibria x(v) and y(v). Then, we have

$$\begin{aligned} x'(v)= & {} \left[ 1-\int _{v}^1\int _0^{x(\tau )}k(\tau , c){\mathrm{d}}c{\mathrm{d}}\tau \right] ^{n-1}\\ y'(v)= & {} \left[ 1-\int _{v}^1\int _0^{y(\tau )}k(\tau , c){\mathrm{d}}c{\mathrm{d}}\tau \right] ^{n-1}. \end{aligned}$$

Suppose $x(1)>y(1)$, then by the continuity of x(v) and y(v), we can find a $v^*$ such that $x(v^*)=y(v^*)=c(v^*)$ and $x(v)>y(v)$ for all $v\in (v^*, 1]$ by noting that $x(0)=y(0)$.

Case 1: If $k(v,c)>0$ with positive probability measure on $(v^*,1)\times (c(v^*),1)$, then $x(\tau )>y(\tau )$ for all $\tau \in (v^*, 1]$ implies that

$$\begin{aligned} \int _0^{x(\tau )}k(\tau , c){\mathrm{d}}c>\int _0^{y(\tau )}k(\tau , c){\mathrm{d}}c \end{aligned}$$

for $\tau \in (v^*,1)$. Then, we have $x'(v^*)<y'(v^*)$ which is a contradiction to $x(v)>y(v)$ for $v>v^*$. So we have $x(1)=y(1)$.

Now suppose there exists an interval $[\alpha ,\beta ]\subset [0,1]$ such that $x(\alpha )=y(\alpha )$ and $x(\beta )=y(\beta )$ while for all $v\in (\alpha ,\beta )$, $x(v)>y(v)$ and for all $v\in [\beta ,1]$, $x(v)=y(v)$, by the same logic above, we have $x(\beta )=y(\beta )$ and $x'(v)<y'(v)$ for $v\in (\alpha ,\beta )$, which is inconsistent with $x(v)>y(v)$ for all $v\in (\alpha ,\beta )$. Thus, we can prove that $x(v)=y(v)$ for all $v\in [0,1]$ and so the symmetric equilibrium is unique.

Case 2: If $k(v,c)>0$ with zero probability measure on $(v^*,1)\times (c(v^*),1)$, then we have $x'(v)=y'(v)$ for all $v\in (v^*,1]$. By $x(v^*)=y(v^*)$, we have $x(v)=y(v)$ for all $v>v^*$, which is a contradiction to $x(v)>y(v)$. Thus, there is a unique symmetric equilibrium.

Then, in both cases, we prove that there is a unique symmetric equilibrium. $\square$

Proof of Proposition 1

Based on (P2), define a mapping

$$\begin{aligned} (Pc)(v)=\int _0^v{\mathrm{d}}s-\int _0^v\int _s^1\left( \begin{array}{cc} 0 &{}\quad f_2(\tau )\\ f_1(\tau ) &{}\quad 0 \end{array}\right) \left( \begin{array}{c } G_1(c_1(\tau ))\\ G_2(c_2(\tau )) \end{array}\right) {\mathrm{d}}\tau {\mathrm{d}}s, \end{aligned}$$

where $c=(c_1, c_2)'$.

Take any $x(v)=(x_1(v), x_2(v))'$ and $y(v)=(y_1(v), y_2(v))'$ with x(v), $y(v)\in \varphi$ where $\varphi$ is the space of monotonic increasing continuous functions defined on $[0,1]\rightarrow [0,1]$. First note that by mean value theorem, $\forall i=1, 2$,

$$\begin{aligned} G_i(x_i(\tau ))-G_i(y_i(\tau ))=g_i(\widehat{x}_i(\tau )) (x_i(\tau )-y_i(\tau )), \end{aligned}$$

where $\widehat{x}_i(\tau )$ is some number between $x_i(\tau )$ and $y_i(\tau )$. In the following, for presentation convenience, denote

$$\begin{aligned} h(\widehat{x}_1(\tau ),\widehat{x}_2(\tau ),\tau )=\left( \begin{array}{cc} 0 &{}\quad g_2(\widehat{x}_2(\tau ))f_2(\tau )\\ g_1(\widehat{x}_1(\tau ))f_1(\tau ) &{}\quad 0 \end{array}\right) . \end{aligned}$$

Then, we have

$$\begin{aligned} |(Px)(v)- & {} (Py)(v)| = |\int _0^v\int _s^1h(\widehat{x}_1(\tau ),\widehat{x}_2(\tau ),\tau )\left( \begin{array}{c } x_1(\tau )-y_1(\tau )\\ x_2(\tau )-y_2(\tau ) \end{array}\right) {\mathrm{d}}\tau {\mathrm{d}}s|\\\le & {} \int _0^v\int _s^1h(\widehat{x}_1(\tau ),\widehat{x}_2(\tau ),\tau ){\mathrm{d}}\tau {\mathrm{d}}s \sup _{0<v\le 1}|x(v)-y(v)|\\\le & {} \int _0^1\int _s^1h(\widehat{x}_1(\tau ),\widehat{x}_2(\tau ),\tau ){\mathrm{d}}\tau {\mathrm{d}}s \sup _{0<v\le 1}|x(v)-y(v)|\\\le & {} \int _0^1\left( \begin{array}{cc} 0 &{} \sup _{[0,1]} g_2(c)(1-F_2(s))\\ \sup _{[0,1]} g_1(c)(1-F_1(s))&{} 0 \end{array}\right) {\mathrm{d}}s \\&\quad \sup _{0<v\le 1}|x(v)-y(v)|. \end{aligned}$$

Thus, when $\sup _{[0,1]} g_i(c)\int _0^1(1-F_i(s)){\mathrm{d}}s=\sup _{[0,1]} g_i(c)E(v_i)<1$, or $\sup _{[0,1]} g_i(c)<\frac{1}{E(v_i)}$ by noting that $E(v_i)=\int _0^1sf_i(s){\mathrm{d}}s=\int _0^1(1-F_i(s)){\mathrm{d}}s$, the above mapping is a contraction, so there exists a unique equilibrium.

We further show Proposition 1 also holds when the support of $c_i$ is a subset of [0, 1] and the proof is slightly modified. Suppose that the support of $G_i(c_i)$ is $[c_l,c_h]\subset [0,1]$ and $G_i(c_i)$ is differentiable on $(c_l,c_h)$. Then, for any $c_1(\tau )$ and $c_2(\tau )$, if $c_l\le c_i(\tau )\le c_h$ for $i=1,2$, we can follow the proof above to apply the Mean Value Theorem. Otherwise, if one of them is not in $[c_l,c_h]$, an extra treatment is needed. For instance, if $c_l\le c_1(\tau )\le c_h$ and $c_2(\tau )<c_l$, then $|G_i(c_1(\tau ))-G_i(c_2(\tau ))|=|G_i(c_1(\tau ))-G_i(c_l)|\le \sup _{[c_l,c_h]}g_i(c_i)|(c_1(\tau )-c_l)|<\sup _{[c_l,c_h]}g_i(c_i)|(c_1(\tau )-c_2(\tau ))|$. Similar inequalities hold for other possible cases. Thus, if $\sup _{[c_l,c_h]} g_i(c_i)<\frac{1}{E(v_i)}$ for $i=1, 2$, the equilibrium is unique. $\square$

Proof of Proposition 2

We first prove necessity. Suppose, in an asymmetric equilibrium, bidder 2 never participates, then bidder 1 participates if and only if $v_1\ge c_1$ and thus we have $c_1^*(v_1)=v_1$. Simplifying R, the expected revenue of bidder 2 with $v_2=v_h$ when he participates in the auction while bidder 1 participates whenever $v_1\ge c_1$, we get (6) and thus necessity holds.

Next we prove sufficiency. Suppose (6) holds. Consider the strategies that bidder 2 never participates and bidder 1 participates whenever $v_1\ge c_1$. Given the strategy of bidder 2, bidder 1’s best response is to participate whenever $v_1\ge c_1$. Given the strategy of bidder 1, since (6) holds, the expected revenue of bidder 2 with $v_2=v_h$ is less than $c_l$; thus, the best response for bidder 2 is never participating for any type. Thus, there exists an asymmetric equilibrium in which one bidder never participates and sufficiency satisfies. $\square$

Proof of Corollary 1

Suppose we have an equilibrium in which bidder 1 always participates and bidder 2 never participates. Then, bidder 1 always participates is a best response to bidder 2’s strategy since $v_l>c_h$. For bidder 2’s strategy to be a best response, we just check that (6) holds with $c_l<c_h<v_l<v_h$. To see this, note that $F(c_h)=0$, $\int _{c_h}^{v_h}(v_h-x)dF(x)=\int _{v_l}^{v_h}(v_h-x){\mathrm{d}}F(x)=v_h-E(v)$ and $\int _{c_l}^{c_h}xG(x){\mathrm{d}}F(x)=0$, thus

$$\begin{aligned} v_hF(c_h)-\int _{c_l}^{c_h}xG(x){\mathrm{d}}F(x)+\int _{c_h}^{v_h}(v_h-x){\mathrm{d}}F(x)-c_l=v_h-E(v)-c_l<0 \end{aligned}$$

by noting that $v_h-E(v)<c_l$. $\square$

Proof of Corollary 2

Let $\lambda (c)=v_hF(c)+\int _{c}^{v_h}(v_h-x){\mathrm{d}}F(x)-c$ with $c\in [0,v_h]$ and notice that (6) can be written as $\lambda (c_h)-\int _{c_l}^{c_h}xG(x){\mathrm{d}}F(x)+c_h-c_l<0$. Note that when v is distributed on [0, 1], $v_h=1$, and we have $\lambda (c)=F(c)+\int _c^1(1-x){\mathrm{d}}F(x)-c$. First we prove for any strictly convex $F(\cdot )$ with support [0, 1], there exists a unique $\overline{c}\in (0,1)$ such that $\lambda (c)<0$ if and only if $c\in (\overline{c},1)$. To see this, note that from $\lambda (c)$, we have $\lambda (0)=\int _0^1(1-x){\mathrm{d}}F(x)>0$, $\lambda (1)=0$ and $\lambda '(c)=-1+cf(c)$. When $F(\cdot )$ is strictly convex on [0, 1], $\lambda (c)$ is a strictly convex function of c with $\lambda (1)=0$ since $\lambda ''(c)=f(c)+cf'(c)>0$. Also note that $\lambda '(1)=f(1)-1>0$, by the strict convexity of $\lambda (c)$ with $\lambda (1)=0$, there exists a unique $\overline{c}$ such that $\lambda (c)<0$ if and only if $c\in (\overline{c},1)$. Thus, we have $\lambda (c_h)>0$ for $c_h\in (\overline{c},1)$. Note that $c_h-c_l-\int _{c_l}^{c_h}xG(x){\mathrm{d}}F(x)=0$ when $c_h=c_l$. By continuity, when $c_h-c_l$ is sufficiently small,

$$\begin{aligned} \lambda (c_h)-\int _{c_l}^{c_h}xG(x){\mathrm{d}}F(x)+c_h-c_l\le 0 \end{aligned}$$

for all $c_h\in (\overline{c},1)$. Thus, (6) holds with $v_h=1$ and an asymmetric equilibrium in which one bidder never participates exists. The other bidder participates whenever $v\ge c$. $\square$