Appendix A: Instructions
Below we include a translation from German of the instructions that we used in the experiment.
Welcome to our experiment! In this experiment you can earn money. How much you will get depends on your own decisions and on the decisions of the other participants. From now on please do not communicate with other participants. If there are any questions, please raise your hand! We will come to you and answer your question. If you break this rule, we will have to exclude you from this experiment and all payments.
In this experiment, ECU is used as the currency. At the end of the experiment, your ECU-payoffs will be converted to Euro and paid out in cash. The conversion rate is 100 ECU = 1 Euro.
Procedures
The experiment consists of 20 auctions in total. In each auction, you have the opportunity to sell a fictitious good.
At the beginning of the experiment, four sellers are randomly chosen and randomly assigned to each other. These sellers will interact in each of the 20 auctions.
At most three goods are sold in each auction. Thus, not every seller will be able to sell her good. An auction proceeds as follows:
The price starts at 20 ECU. The sellers then have 5 s to decide whether they want to sell their goods at this price by clicking on the button “Sell the good at this price”. The remaining time is displayed in the upper right corner of the screen.
Sellers who do not want to sell their good at this price do not have to do anything. After 5 s, the experiment proceeds automatically.
When all sellers have made their decisions and not all three goods were sold, the price will be raised by 5 ECU to 25 ECU and all remaining sellers in the auction have 5 s to decide whether they want to sell their goods at the new price.
When not all three goods are sold at this price, the price will again be raised by 5 ECU to 30 ECU and all remaining sellers in the auction decide again.
The price will be raised by 5 ECU-steps until either the three goods are all sold
or the price reaches the upper limit of 100 ECU, without three sellers having sold at this price.
If more sellers want to sell their goods at a certain price than goods are demanded in the auction, it will be randomly determined which seller is allowed to sell her good.
The sellers have no costs. This means that sellers who sell their goods in the auction receive a payoff equally to their selling prices. Sellers who did not sell their good in an auction do not receive a payoff from this auction.
After each auction, the sellers will be informed about all prices that were realized in this auction.
Concluding remarks
At the end of the experiment, the sum of payoffs from all 20 auctions will be converted into Euro and paid to you. In addition, you will receive an amount of 7.50 Euro for your participation irrespective of the decisions in the experiment.
Appendix B: A more elaborate approach to modeling first-round bidding
We consider the simplest case in which a bidder following the sophisticated bidding strategy expects the three other bidders to pursue the myopic strategy. In the first auction, as there is no information on bidding behavior from earlier periods available, a sophisticated bidder optimizes his bidding behavior under the assumption that myopic bids follow a discrete uniform distribution on the interval \(\{20, 25, \ldots ,100\}\).Footnote 24 In our case, this assumption reflects that in the very first auction, there is no anchor to which players can adjust their bids. To determine the optimal sophisticated bidding behavior at the current price b, we need to distinguish between three cases depending on the number of products that have already been sold during the auction.
Suppose none of the competing bidders has submitted a bid at price \(b-5\). Now let \(p{:}=(b-(b-5))/(100-(b-5))=5/(105-b)\) be the probability that a competing bidder bids at the current price \(b>20\). Then, the expected payoff of a player with the sophisticated strategy from submitting a bid at b is given by
$$\begin{aligned} {\mathbb {E}}\left[ \pi _s^{\mathrm{bid}}\right] =b\left( 1-p^3+\dfrac{3}{4}p^3\right) =b\left( 1-\dfrac{p^3}{4}\right) . \end{aligned}$$
(6)
Note that \(1-p^3\) is the probability that at most two other competitors bid at the same time, which means that the bidder sells his product with certainty. Similarly, \(p^3\) represents the probability that all other bidders simultaneously submit bids, which means that the bidder has a winning probability of only 3 / 4.
Let \({\tilde{p}} \mathrel {\mathop :}=(b+5-b)/(100-b)=5/(100-b)\) be the probability that the myopic bid is \(b+5\). Similar to the case in which the bidder submits a bid, the expected payoff from waiting at the current price b then amounts to
$$\begin{aligned} {\mathbb {E}}\left[ \pi _s^{\mathrm{wait}}\right]= & {} (b+5)\left( \left( 1-{\tilde{p}}+\dfrac{1}{2} {\tilde{p}}\right) 3p^2\left( 1-p\right) +\left( 1-{\tilde{p}}^2+\dfrac{2}{3} {\tilde{p}}^2\right) 3p\left( 1-p\right) ^2\right. \nonumber \\&\left. +\left( 1-{\tilde{p}}^3+ \dfrac{3}{4}{\tilde{p}}^3\right) \left( 1-p^3-3p^2(1-p) -3p(1-p)^2\right) \right) . \end{aligned}$$
(7)
Solving \(\mathbb {E}\left[ \pi _s^{\mathrm{bid}}\right] =\mathbb {E}\left[ \pi _s^{\mathrm{wait}}\right] \) gives \(b\approx 85.07\) as the (relevant) solution. Hence, if no product has been sold, the sophisticated strategy consists of accepting a clock price of 90 (where \(\mathbb {E}\left[ \pi _s^{\mathrm{bid}}\right] >\mathbb {E}\left[ \pi _s^{\mathrm{wait}}\right] \)).
Suppose next that one of the competing bidders has already sold his product at a price lower than \(b-5\). Then, the expected payoff for a bidder following the sophisticated strategy and submitting a bid at b is given by
$$\begin{aligned} {\mathbb {E}}\left[ \pi _s^{\mathrm{bid}}\right] =b\left( 1-p^2+\dfrac{2}{3}p^2\right) =b\left( 1-\dfrac{p^2}{3}\right) . \end{aligned}$$
Analogously, waiting for another tick of the price clock yields an expected payoff of
$$\begin{aligned}&{\mathbb {E}}[\pi _s^{\mathrm{wait}}]=(b+5)\\&\quad \times \left( \left( 1-{\tilde{p}}+\dfrac{1}{2} {\tilde{p}}\right) 2p\left( 1-p\right) + \left( 1-{\tilde{p}}^2+\dfrac{2}{3}{\tilde{p}}^2\right) (1-p^2-2p(1-p))\right) . \end{aligned}$$
Again, solving \(\mathbb {E}\left[ \pi _s^{\mathrm{bid}}\right] =\mathbb {E}\left[ \pi _s^{\mathrm{wait}}\right] \) gives \(b\approx 76.33\) as the (relevant) solution. Hence, if one product has been sold, the sophisticated bid is 80.
Last, consider the case where only one more product can be sold to the buyer. Then, the expected payoffs amount to
$$\begin{aligned} {\mathbb {E}}\left[ \pi _s^{\mathrm{bid}}\right] =b\left( 1-p+\dfrac{1}{2}p\right) =b\left( 1-\dfrac{p}{2}\right) . \end{aligned}$$
Analogously, waiting for another tick of the price clock results in an expected payoff of
$$\begin{aligned} {\mathbb {E}}\left[ \pi _s^{\mathrm{wait}}\right] =(b+5)\left( 1-{\tilde{p}}+\dfrac{1}{2} {\tilde{p}}\right) \left( 1-p\right) . \end{aligned}$$
From \(\mathbb {E}\left[ \pi _s^{\mathrm{bid}}\right] =\mathbb {E}\left[ \pi _s^{\mathrm{bid}}\right] \Leftrightarrow b=48.75\), it follows that the sophisticated bid is 50.
The bidding behavior in the first round can thus be summarized as follows:
$$\begin{aligned} b_m\left( g,{\bar{b}}_{t-1},x\right) \big \vert _{t=1}\sim U[20,100] \end{aligned}$$
and
$$\begin{aligned} b_s\left( g,{\bar{b}}_{t-1},x,\epsilon \right) \big \vert _{t=1}= {\left\{ \begin{array}{ll} 90 &{} \text {if } g=3 \\ 80 &{} \text {if } g=2 \\ 20 &{} \text {if } g=1. \end{array}\right. } \end{aligned}$$
This bidding behavior implies that actual bids in the first period of the reverse multi-unit Dutch auction are significantly higher than the one predicted by standard economic theory, which equals 20 ECU.
Note that although we only covered the case where one player applies a sophisticated strategy, the above argument also holds for the case where two to four players follow the sophisticated bidding strategy. This is due to the fact that if the bidder applying the sophisticated bidding strategy expects three other players following a sophisticated strategy as well, he is going to enter the auction at a price of 20 for \(g\in \{1,2,3\}\). As the bidding strategy continuously depends on the distribution of bid types and given the intermediate value theorem, sophisticated bids higher than 20 can be supported for certain ranges of shares x.
Appendix C: Trembling-hand equilibrium
Assume a share of x bidders follow the sophisticated bidding strategy of bidding \({\bar{b}}_{t-1}-5\) but might err in doing so. Assume the errors are distributed according to the Poisson distribution \(P(\lambda ,k)\), where k is the number of steps of deviation and \(\lambda \) is the variance/expected value of the distribution. The share \(1-x\) follows the myopic strategy. A best response to this setup might be to bid \({\bar{b}}_{t-1}-5\) depending on x and \(\lambda \). This is the case if the following holds:
$$\begin{aligned} {\bar{b}}_{t-1} -10\le & {} \left( 1-\left( P(\lambda ,0)x\right) ^3\right) \left( {\bar{b}}_{t-1}-5\right) + \frac{3}{4}\left( P(\lambda ,0)x\right) ^3\left( {\bar{b}}_{t-1}-5\right) \\&+\,\left( 1-(1-x)^3\right) \sum ^{\frac{{\bar{b}}_{t-1}}{5}-5}_{i=1}P(\lambda ,i) \frac{2}{3}\left( {\bar{b}}_{t-1}-5i\right) \\&+\,\left( 1-(1-x)^2\right) \sum ^{\frac{{\bar{b}}_{t-1}}{5}-5}_{i=1}P(\lambda ,i)^2 \frac{1}{2}\left( {\bar{b}}_{t-1}-5i\right) \end{aligned}$$
The left-hand side is the reduction in profits if one were to bid \({\bar{b}}_{t-1}-10\) instead of \({\bar{b}}_{t-1}-5\). The first line of the right-hand side is the probability that there is at least one bidder following the myopic bidding heuristic plus the probability that all other bids are sophisticated and non-erring, which leads to a winning probability of 3 / 4. The second line is the probability that there is one trembling bidder who follows the sophisticated bidding strategy and submits an initial bid strictly smaller than one step below \({\bar{b}}_{t-1}\) times the resulting winning probability and profit when everyone enters at the next step. The third line follows the same logic given that there are two sophisticated trembling bids at the same bid step strictly smaller than one step below \({\bar{b}}_{t-1}\). This can be rearranged to
$$\begin{aligned} 5\ge & {} \frac{1}{4}\left( P(\lambda ,0)x\right) ^3\left( {\bar{b}}_{t-1}-5\right) \\&-\,\left( 1-(1-x)^3\right) \sum ^{\frac{{\bar{b}}_{t-1}}{5}-5}_{i=1}P(\lambda ,i)\frac{2}{3}\left( {\bar{b}}_{t-1}-5i\right) \\&-\,\left( 1-(1-x)^2\right) \sum ^{\frac{{\bar{b}}_{t-1}}{5}-5}_{i=1}P(\lambda ,i)^2\frac{1}{2}\left( {\bar{b}}_{t-1}-5i\right) . \end{aligned}$$
This is satisfied for large ranges of \(\lambda \) and x. In particular, it is fulfilled for all combinations of \(\lambda \), x, and \({\bar{b}}_{t-1}\) that we find empirically in Sect. 4. The only exception here are the periods 16–20 in which x and \(\lambda \) are such that bidding \({\bar{b}}_{t-1}-5 \) is not trembling-hand perfect for the entire support of \({\bar{b}}_{t-1}\). However, in these five periods, the observed values for \({\bar{b}}_{t-1}\) are low enough such that together with \(x\approx 0.65\) and \(\lambda \approx 0.208\), they form a trembling-hand perfect equilibrium again.
Appendix D: Estimation procedure
In what follows, we first characterize the estimation procedure (subsection “Procedure and intuition”) and then show that it is a valid approach (subsections “Choice of norm”, “Existence of a minimum”, and “Uniqueness of a minimum”).
Procedure and intuition
We can derive the ex ante probabilities of observing between one and four bids (as four bidders interact) at a given number of steps below the highest previous bid in each auction. Table 4 lists all corresponding ex ante probabilities that n bids are of either myopic or sophisticated type. By comparing the ex ante probabilities with the observed number of auctions with a given number of initial bids, we can estimate the distribution of sophisticated and myopic strategies that fits our data best.
We capture the errors in the sophisticated bidding strategy by assuming that they are distributed according to a discrete probability mass function. High price undercuts are rare and our model shows that underbidding by exactly one price step is a trembling-hand perfect equilibrium for large parameter spaces. Transferring this idea to our formulation of the error term, we need a distribution function that allows for relatively high probabilities for small errors and vice versa. For this reason, we assume that the errors are Poisson distributed, \(P(y;\lambda )=\lambda ^y{\mathrm {e}}^{-\lambda }/y!\), where y is the number of steps below \({\bar{b}}_{t-1}\), and estimate \(\lambda \), the parameter that measures the expected probability of placing a bid based on erroneous beliefs in our case.
Given the share x of sophisticated bids, the functions in Table 4 minus the observed shares define the distance between the ex ante and the observed probabilities of n initial bids of either type. Let us denote these functions \(f_j^{\{z\}}\), where z is the number of initial bids of either type and \(j\in \{m,s,se\}\) stands for the three strategies: myopic, sophisticated, or sophisticated with error. Take the example where exactly \(z=2\) initial bids are classified as type \(j=s\) in 55 of all 325 auctions. This means that according to the above definition, we have a function
$$\begin{aligned} f_s^{\{2\}}=6\left( P(0;\lambda )x\right) ^2(1-x)^2-\dfrac{55}{325}. \end{aligned}$$
From the table it becomes clear that we are looking at an over-determined system of equations. To estimate the probabilities given the number of observations per case (i.e., the values for x and \(\lambda \)), we transform the system into a minimization problem by defining a function \(f: \mathbb {R}^{2} \rightarrow \mathbb {R}^{9}\) with equations \(f_j^{\{z\}}\) as components.Footnote 25 For a classical solution \(\tilde{\mathbf {x}}=(\tilde{x},\tilde{\lambda })\), it holds that
$$\begin{aligned} \left\| f(\tilde{\mathbf {x}})\right\| _2 = 0. \end{aligned}$$
In combination with the positive homogeneity of a norm, one can minimize the norm of f under the constraints
$$\begin{aligned} 0\overset{!}{\le } x\overset{!}{\le } 1. \end{aligned}$$
With \(\mathbf {x}=\left( x,\lambda \right) \), this problem can be written as
$$\begin{aligned} \underset{{\mathbf {x}} \in [0,1]\times {\mathbb {R}}_+}{\min }\left\{ \left\| f({\mathbf {x}}) \right\| _2 \right\} . \end{aligned}$$
Thus, we look for those ex post probabilities \(\mathbf {x}\) in the model functions, such that the distance between x and the observed share is minimized.
We use our approach—rather than a maximum likelihood estimation conducted with individual data—because based on our model assumption concerning bidding behavior, each initial bid in an auction below the highest price in the previous auction is unambiguously assigned to the sophisticated bidding strategy, which means that the observed distribution of initial bids is deterministic. Hence, our method finds the best estimate for the underlying distribution of sophisticated and myopic individual bids. In addition, our approach provides us with an estimate of the distribution of errors among the share of sophisticated bids.
Choice of norm
In the minimization problem, we use the Euclidean (or 2-)norm. Since one could also minimize the \(L^1\) or even an \(L^p\) norm, this choice may not be obvious, but actually follows naturally from the problem. One can only minimize a function \(f: \mathbb {K} \mapsto \mathbb {R}\) with \(\mathbb {K}\subset \mathbb {R}^n\), but in our case, the function maps to \(\mathbb {R}^{9}\). Thus, we minimize
$$\begin{aligned} \langle f^\intercal , f \rangle , \end{aligned}$$
where \(\langle \cdot ,\cdot \rangle \) is the standard inner product in \(\mathbb {R}\). The norm induced by \(\langle \cdot ,\cdot \rangle \) is the 2-norm
$$\begin{aligned} \left\| \mathbf {x} \right\| _2 = \sqrt{\langle \mathbf {x} ,\mathbf {x}\rangle } \end{aligned}$$
and, therefore, the choice of the norm follows directly from the problem.
Existence of a minimum
The existence of a minimum (and not just an infimum) is guaranteed, because \(\left\| f(\mathbf {x})\right\| _2\) is continuous. The constraints require x to be in the compact space [0, 1]. \(\lambda \) represents the expected value of the Poisson distribution and is, therefore, bounded by \(\lambda _{\text {max}}=15\), because that is the maximum number of steps a bidder can erroneously deviate (given \({\bar{b}}_{t-1}=100\)). Continuous functions attain a minimum on compact spaces (Theorem of Weyerstraß).
Uniqueness of a minimum
Every norm is a convex function, because by the triangle inequality and the positive homogeneity, it holds that
$$\begin{aligned}&\forall \Theta \in (0,1)\quad \forall x,y \in \mathbb {K}\\&\left\| \Theta x+(1-\Theta )y\right\| \le \Theta \left\| x\right\| + (1-\Theta )\left\| y\right\| . \end{aligned}$$
Therefore, the worst-case scenario is that \(\left\| f(\mathbf {x)}\right\| \) is constant for a small space \(\mathbb {S}\subset [0,1]\) around a critical point. However, it can be easily check that for every critical point \(\mathbf {x}\) found, the Hessian matrix is strictly positive definite. It follows that the minimum is unique.