Discriminatory auctions with resale

Abstract

We consider multi-unit discriminatory auctions where ex ante symmetric bidders have single-unit demands and resale is allowed after the bidding stage. When bidders use the optimal auction to sell items in the resale stage, the equilibrium in the auction without resale is no longer an equilibrium in an auction with resale. We find a symmetric and monotone equilibrium when there are two units for sale, and, interestingly, we show that there may not be a symmetric and monotone equilibrium if there are more than two units.

Appendix

Proof of Proposition 1

The proof follows from an argument that is similar to the one made in Example 1. Suppose every bidder other than bidder 1 uses the no resale equilibrium strategy. Consider bidder 1 with value 1 and the alternative strategy $\left( \beta _{1}^{N}\left( 1\right) ,\beta _{1}^{N}\left( 1\right) ,0,\ldots ,0\right)$ (similar strategies can be found for other values in $\left( 0,1\right)$). With this strategy, this bidder will receive two units. She can sell the second unit using a second-price auction with a reserve price $\frac{1}{2} ,$ which gives her an expected revenue that is strictly higher than $\mathbb {E}\left[ Y_{k}^{\left( n-1\right) }\right]$. This is because $\mathbb {E}\left[ Y_{k}^{\left( n-1\right) }\right]$ is the expected revenue of the second-price auction with no reserve price, and the expected revenue of a second-price auction with optimal reserve price is strictly higher than that. Since this bidder has paid $\mathbb {E}\left[ Y_{k}^{\left( n-1\right) }\right]$ for the second unit and gets strictly more than $\mathbb {E}\left[ Y_{k}^{\left( n-1\right) }\right]$ in the resale stage, this deviation strictly increases her utility. $\beta ^{N}\left( x\right)$ is not an equilibrium of discriminatory auctions with resale. $\square$

Proof of Lemma 1

First, by method of contradiction, suppose that $\beta ( x) >\frac{23}{64}x$ for some $x\in \left[ 0,c\right] .$ Consider a bidder with value $y=\theta ^{-1}\left( \beta \left( x\right) \right)$. When this bidder receives three units from the auctioneer, his total payment for second and third object is $\delta \left( y\right) +\theta \left( y\right) \ge 2\theta \left( y\right) =2 \beta \left( x\right) >\frac{23}{32}x,$ whereas his expected revenue from resale for this case is only $\frac{23}{32} x$.¹⁸ Therefore, he makes a loss when he receives 3 units. So, he is better off by deviating to $(\beta ( y) ,0,0)$. $\square$

Proof of Lemma 2

For (i), by method of contradiction, suppose that there exists $t\in \left[ 0,\theta \left( 1\right) \right] \$ such that $\frac{d\beta ^{-1}}{dt}( t) >\frac{32}{23}$ and $\delta ^{-1}\left( t\right) <\theta ^{-1}\left( t\right) .$ Then we can argue that type $\theta ^{-1}\left( t\right) \equiv y$ strictly benefits by deviating to $\left( \beta \left( y\right) ,\delta \left( y\right) ,t+\varepsilon \right)$ for a small enough $\varepsilon$. This is because, by deviating to $t+\varepsilon$ from t for his third bid (and this is feasible because $\delta ^{-1}\left( t\right) <\theta ^{-1}\left( t\right)$), this bidder (i) increases the probability of getting three units and (ii) increases his net utility when he sells two units to unassigned bidders (his payment increases by $\varepsilon$ and his expected revenue increases by strictly more than $\frac{23}{32}\times \frac{32}{23}\times \varepsilon =\varepsilon .$)

Next, because $\beta \left( 0\right) =0$ and $\beta \left( x\right) \le \frac{ 23}{64}x$ for all $x\in \left[ 0,c\right] ,$ we have $\beta ^{\prime }\left( 0\right) \le \frac{23}{64}$ or $\frac{\mathrm{d}(\beta ^{-1} ) }{\mathrm{d}x} (0)\ge \frac{64}{23}>\frac{32}{23}.$ Because $\beta$ is continuously differentiable, there exists $d\le c$ such that $\frac{\mathrm{d}(\beta ^{-1} ) }{\mathrm{d}x} (x)>\frac{32}{23}$ for all $x\in \left[ 0,d\right]$; part (i) implies that $\delta \left( x\right) =\theta \left( x\right)$ for all $x\in \left[ 0,d\right] .$$\square$

Proof of Lemma 3

By Lemma 2, we know that for all $x\in [ 0,d] ,$ we have $\beta ( x) \in \left[ 0,\frac{23}{64}x\right]$ and $\delta \left( x\right) =\theta \left( x\right)$. Let us define the net utility of a buyer with value $x\in [ 0,d]$ when he is a seller in the resale stage:¹⁹

$$\begin{aligned} s\left( x\right)&= \beta ^{-1}\left( \theta \left( x\right) \right) ^{3}\left( \frac{23}{32}\beta ^{-1}\left( \theta \left( x\right) \right) -2\theta \left( x\right) \right) \\&\quad + 3 \beta ^{-1}\left( \theta \left( x\right) \right) ^{2}\left( x-\beta ^{-1}\left( \theta \left( x\right) \right) \right) \left( \frac{5}{12}\beta ^{-1}\left( \theta \left( x\right) \right) -\theta \left( x\right) \right) . \end{aligned}$$

First, if $s'( x) >0$ then we have $\theta ( x) =\beta ( x)$. This is because whenever $s'(x)$ is positive, a bidder with value x becomes strictly better off by increasing his second and third bids by $\varepsilon$, and doing this would be feasible if $\beta \left( x\right) >\theta \left( x\right)$.

Next, by method of contradiction, suppose that there exists no $e\in (0,1]$ such that for all $x\in \left[ 0,e\right] ,$$\beta \left( x\right) =\delta \left( x\right) =\theta \left( x\right) .$ This means there exists $f>0$ such that we have $\beta \left( x\right) >\theta \left( x\right)$ for all $x\in (0,f]$. Note that Lemma 1 implies $\frac{23}{32}\beta ^{-1}(\theta (x))>2\theta (x)$ and $\frac{5}{12}\beta ^{-1}(\theta (x))>\theta (x)$. As a result, $s(x)>0$. This implies that $s'(y)>0$ for some $y\in (0,f]$ since $s(0)=0$. Therefore, we have $\theta (y) =\beta ( y)$. $\square$

Proof of Lemma 4

Consider a bidder with value $x\in \left( 0,e\right)$ who bids as if his value is z (which is very close to x.) His expected utility is given by

$$\begin{aligned} u\left( x,z\right) =z^{3}\left( x-3\beta \left( z\right) +\frac{23}{32} z\right) +R\left( x,z\right) , \end{aligned}$$

where $R\left( x,z\right)$, first, is his expected utility from the resale stage when he is a buyer and, second, is given by

$$\begin{aligned} R\left( x,z\right)&= 6\int _{z}^{\min \{1,2x\}}\int _{\frac{k}{2}}^{x}\int _{ \frac{k}{2}}^{l}\left( x-m\right) \mathrm{d}m\mathrm{d}l\mathrm{d}k\\&\quad +6\int _{z}^{\min \{1,2x\}}\int _{x}^{k}\int _{\frac{k}{2}}^{x}\left( x-m\right) \mathrm{d}m\mathrm{d}l\mathrm{d}k \\&\quad +6\int _{z}^{\min \{1,2x\}}\int _{\frac{k}{2}}^{k}\int _{0}^{\frac{k}{2} }\left( x-\frac{k}{2}\right) \mathrm{d}m\mathrm{d}l\mathrm{d}k\\&\quad +6\int _{z}^{\min \{1,2x\}}\int _{0}^{\frac{ k}{2}}\int _{0}^{l}\left( x-\frac{k}{2}\right) \mathrm{d}m\mathrm{d}l\mathrm{d}k, \end{aligned}$$

where k, l and m denote the realizations for the highest, the second highest, and the third highest values among the competitors, respectively; the first two terms in the summation represent the cases in which the bidder with value x pays the third highest value; and the last two terms in the summation represent the cases in which the bidder with value x pays the reserve price.

A necessary condition for an equilibrium is $\left. \frac{\partial u\left( x,z\right) }{\partial z}\right| _{z=x}=0$. Note that

$$\begin{aligned} \frac{\partial R\left( x,z\right) }{\partial z}&= -6\left( \int _{\frac{z}{2} }^{x}\int _{\frac{z}{2}}^{l}\left( x-m\right) \mathrm{d}m\mathrm{d}l+\int _{x}^{z}\int _{\frac{z}{ 2}}^{x}\left( x-m\right) \mathrm{d}m\mathrm{d}l\right. \\&\quad +\int _{\frac{z}{2}}^{z}\int _{0}^{\frac{z}{2} }\left( x-\frac{z}{2}\right) \mathrm{d}m\mathrm{d}l \\&\quad \left. +\int _{0}^{\frac{z}{2}}\int _{0}^{l}\left( x- \frac{z}{2}\right) \mathrm{d}m\mathrm{d}l\right) = x^{3}+\frac{5}{8}z^{3}-3x^{2}z. \end{aligned}$$

The necessary condition can be rewritten as

$$\begin{aligned} \left. \frac{\partial }{\partial z}\left( z^{3}\left( x-3\beta \left( z\right) +\frac{23}{32}z\right) \right) + \left( x^{3}+\frac{5}{8}z^{3}-3x^{2}z\right) \right| _{z=x} = 0. \end{aligned}$$

This differential equation will have a unique solution, which is $\beta \left( x\right) =\frac{3}{8}x$. $\square$