All-pay auctions with endogenous bid timing: an experimental study

Abstract

Using a controlled laboratory experiment, we examine individuals’ bid timing decisions in complete information all-pay auctions and find that homogeneous bidders are more likely to enter the early bidding stage under a favor-early tie-breaking rule. Furthermore, revenue in an endogenous-entry treatment, in which both sequential and simultaneous all-pay auctions exist, is either lower than or equal to that in an exogenous-entry treatment with only simultaneous all-pay auctions. Additionally, in simultaneous all-pay auctions, individuals do not always employ a mixed strategy as predicted by the risk-neutral model. Instead, our data is better rationalized by a risk- and loss-aversion model.

Appendices

A Proofs

1.1 Equilibrium characterization for all-pay auctions with n players

The analysis for heterogeneous bidders follows Konrad and Leininger (2007), and we focus on equilibrium characterization for homogeneous players. Assuming a favor-late tie-breaking rule, we first characterize the SPNE in sequential all-pay auctions.

Following Konrad and Leininger (2007), we first consider the subgame at the late stage (L). The maximum bids from the early stage (E) are characterized by $\bar{x}_{E}\equiv \max _{i\in E} \{x_{i}\}$. $x_{L}^{i}$ represents the best response in stage L for bidder i. Given $\bar{x}_{E}$, we have:

• 1. If $c \bar{x}_{E}> v$, $x_{L}^{i}=0$.

• 2. If $c \bar{x}_{E}= v$, if there is only one player in the late stage, she is indifferent between bidding v and 0. If there are more than one bidder, either all of them bid 0 or one of them bids v.

• 3. If $c \bar{x}_{E}<v$, if there is only one player in the late stage, she would bid $\bar{x}_{E}$. If there are more than one bidder, it becomes a simultaneous all-pay auction with lower bound $\bar{x}_{E}$, and the expected payoff for all late player is always 0.

Now we consider stage E. As $\bar{x}_{L}\equiv \max _{i\in L} \{\bar{x}_{i}\}$ and $\bar{x}_{i}=\bar{x}=\frac{v}{c}$ for homogeneous bidders, the equilibrium bids in stage E for bidder i: $x_{E}^{i}$, is characterized below.

• 1. If there is more than one bidder in stage E, there are two types of equilibrium strategies.

  • (a) One player bids $\bar{x}$ and all others bid 0.

  • (b) All players bid 0.

• 2. If there is one and only one bidder i in stage E, similarly, $x_{E}^{i}=\{\bar{x},0\}$.

When all bidders enter either stage E or stage L together, the game becomes a simultaneous all-pay auction and the unique symmetric Nash equilibrium is that bidders randomize continuously on $[0,\bar{x}]$. The expected payoff for everyone is always 0.

Furthermore, when the bid timing decision is endogenous, conditional on the subgame where all early players bid 0 in the equilibrium, $q^{*}_{i}=0$ is a best reply if $\prod q_{j\ne i}>0$. Otherwise, $q^{*}_{i} \in [0, 1]$. $\square$

Proof of Proposition 1

The proof for the unique symmetric NE follows the line of the argument in Baye et al. (1996), and we focus on the analysis of $G_{1}(x)$ and $G_{2} (x)$. The proof for homogeneous bidders is similar to $G_{1}(x)$, so we omit it.

As the expected payoff for bidder 2 is always 0 in the equilibrium (Siegel 2009), we obtain:

$$\begin{aligned} U_{2}(x)=G_{1}(x)u(v-x)+(1-G_{1}(x))u(-x)=0. \end{aligned}$$

Consequently,

$$\begin{aligned} G_{1}(x)=\frac{-u(-x)}{u(v-x)-u(-x)}. \end{aligned}$$

The utility function is defined below:

$$\begin{aligned} u(x) = \left\{ \begin{array}{ll} x^{\alpha } &{} \text{ if } x\ge 0 \\ -\lambda (-x)^{\alpha } &{} \text{ otherwise. } \end{array} \right. \end{aligned}$$

As $G_{1}(\frac{v}{2})u(\frac{v}{2})+(1-G_{1}(\frac{v}{2}))u(-\frac{v}{2})=0$, $G_{1}(\frac{v}{2})=\frac{\lambda }{1+\lambda }\ge \frac{1}{2}$.

As $\forall x \in (0, v)$, $U_{2}(x)=0$ in the equilibrium, $U'_{2}(x)=0$ and $U''_{2}(x)=0$. Defining $g_{1}(x)=G'_{1}(x)$, we obtain:

$$\begin{aligned} g_{1}(x)=\frac{(1-G_{1}(x))u'(-x)+G_{1}(x)u'(v-x)}{u(v-x)-u(-x)}>0. \end{aligned}$$

Furthermore, $U_{2}''(x)=0$ yields the following equation:

$$\begin{aligned} g'_{1}(x)(u(v-x)-u(-x))= & {} 2g_{1}(x)(u'(v-x)-u'(-x))-G_{1}(x)u''(v-x)\\&-(1-G_{1}(x))u''(-x). \end{aligned}$$

Defining $Z(x)=2g_{1}(x)(u'(v-x)-u'(-x))-G_{1}(x)u''(v-x)-(1-G_{1}(x))u''(-x)$, we show that $Z(x)\le 0$ when $0\le x \ll v$, and that $Z(x)\ge 0$ when $0\le v-x\ll v$.

First, $\exists$ $\frac{v}{2}\le x_{1}=\frac{v}{1+\lambda ^{\frac{1}{\alpha -1}}}<v$, $u'(v-x_{1})=u'(-x_{1})$. As u(x) is concave with $x\ge 0$, $\forall$ $x\ge x_{1}$, $u'(v-x)\ge u'(-x)>0$, and by explicit calculation, $u''(v-x)\le -u''(-x)<0$. Together with $G_{1}(x)\ge \frac{1}{2}$, $Z(x)\ge 0$.

Second, as $G_{1}(\frac{v}{2})\ge \frac{1}{2}$, then $\exists$ $x_{0}\le \frac{v}{2}$, $G(x_{0})=\frac{1}{2}$. Furthermore, $\forall$ $x \le x_{0}<x_{1}$, $0<u'(v-x)\le u'(-x)$, and by explicit calculation, $-u''(-x) \le u''(v-x)<0$. Together with $G_{1}(x)\le \frac{1}{2}$, $Z(x)\le 0$.

Because of the continuity of Z(x), $\exists$ y where $x_{0}\le y \le x_{1}$, $Z(y)=0$ and $g_{1}'(y)=0$. Now we show that y is unique.

If $\exists$ $Z(y_{1})=Z(y_{2})=0$, by the continuity of Z(x) and the intermediate value theorem, one of them, e.g., $y_{1}$, must have $Z'(y_{1})\le 0$. However, we know that

$$\begin{aligned} Z'(x)= & {} 2g'_{1}(x)(u'(v-x)-u'(-x))+2g_{1}(x)(-u''(v-x)+u''(-x))\\&-g_{1}(x)u''(v-x)+G_{1}(x)u'''(v-x)+g_{1}(x)u''(-x)\\&+(1-G_{1}(x))u'''(-x) \end{aligned}$$

When $x=y_{1}$, $Z'(y_{1})=3g_{1}(y_{1})(u''(-y_{1})-u''(v-y_{1}))+G_{1}(y_{1})u'''(v-y_{1})+(1-G_{1}(y_{1}))u'''(-y_{1})>0$, which is a contradiction.

Along the same line of proof for $G_{1}(x)$, $G''_{2}(x)<0$ with $0\le x \ll v$, and $G''_{2}(x)>0$ with $0\le v-x \ll v$. Additionally, as $G_{2}(0)=\frac{u((1-c_{1})v)}{u(v)}$ and $G^{rn}_{2}(0)=\frac{(1-c_{1})v}{v}$, by the concavity of u(x) with $x\ge 0$, $G_{2}(0)\ge G^{rn}_{2}(0)$. $\square$

B Experimental instruction

This is an experiment in decision-making. The experiment will proceed in two parts and you will make a series of decisions in each part. At the end, you will fill out a post-experiment questionnaire.

This experiment has 12 participants. Each of you has been randomly assigned an experiment ID at the beginning of the experiment. The experimenter will use this ID to pay you at the end of the experiment.

Rounds: The experiment consists of 30 rounds of two-person auctions.

Endowment: Each of you has 125 tokens as an endowment at the beginning of each round.

Prize Values: At the beginning of each round, an object with a value of 100 tokens will be auctioned within each two-person group.

Matching: At the beginning of each round, you will be randomly matched with another person. You are equally likely to be matched with any other person in the room.

Decisions: In each round, you must make two decisions. First, both you and your match choose independently and simultaneously which bidding stage you want to enter. The entry decisions are then announced to both of you. Second, you and your match each choose a bid in your respective bidding stage.

Bids: There are two bidding stages: the early stage and the late stage.

• 1. If you enter early and your match enters late, you will choose a bid first. After observing your bid, your match will choose his or her bid.

• 2. If you enter late and your match enters early, your match will choose a bid first. After observing his or her bid, you will choose your own bid.

• 3. If both of you choose the same stage, you will bid simultaneously.

Cost of the Bid: The cost of the bid captures the idea that it is sometimes more or less costly to submit a bid. In the experiment, it is determined by a random number generator at the beginning of each round. For each round, with 50% chance, the cost of the bid is 1 token for you and 0.8 tokens for your match. With 50% chance, the cost of the bid is 0.8 tokens for you and 1 token for your match. Here is a numerical example:

• 1. If the cost of your bid is 1 token and you bid 50, then you will pay 50 tokens.

• 2. If the cost of your bid is 0.8 tokens and you bid 50, then you will pay 40 tokens.

Bid Range: Your bid can be any integer between 0 and 125, inclusive.

Profits: In each round, your profits will be determined by (1) your bid; (2) your match’s bid; (3) the cost of your bid; and (4) the entry decisions in the event of a tie.

Profits$=$ Your Endowment- the cost of your bid*your bid + the value of the object if you win $=$ 125—the cost of your bid*your bid+ 100 if you win

For example, in a given round, if you bid 40 and the cost of your bid is 0.8 in this round, then

• 1. If you win the auction, then your profit is $125-0.8*40+100 = 193$ tokens

• 2. If you lose the auction, then your profit is $125-0.8*40 = 93$ tokens

Note: You will always pay for your bid, which is equal to the cost of your bid* your bid, no matter whether you win or lose.

The tie-breaking rule: If you and your match bid exactly the same amount, and

• 1. Both of you enter in the same stage, we will randomly choose one as the winner.

• 2. If one and only one of you enter in the early stage, then the early bidder will be the winner.

Cumulative Profits: Your cumulative profits will be the sum of your profits in all rounds.

Feedback: At the end of each round, you will get the following feedback on your screen:

• 1. Your entry decision

• 2. Your match’s entry decision

• 3. Your bid

• 4. Your match’s bid

• 5. Your profits

• 6. Your match’s profits

• 7. Your cumulative profits

History: In each round, your and your matches’ bids and entry decisions in each previous round, your and your matches’ profits in each previous round, as well as your cumulative profits up until the last round will be displayed in a history box.

Review Questions: To help you understand the experiment, we will go over nine review questions before we start the auction. You can also find these review questions in the appendix for your reference. You will get 25 tokens for answering each of the review questions correctly.

Exchange Rate: $1 = 250 tokens. Please do not communicate with each other during the experiment. If you have a question, feel free to raise your hand, and an experimenter will come to help you.