# Going… going… wrong: a test of the level-k (and cognitive hierarchy) models
of bidding behaviour††thanks: I would especially like to thank my supervisor
Miguel Ballester for his insightful suggestions and constant support. For
useful comments and discussions, I would like to thank Johannes Abeler, Jesper
Åkesson, Elizabeth Baldwin, Adam Brzezinski, Riccardo Camboni, Vince Crawford,
Jasmin Droege, Caspar Jacobs, Nat Levine, Luke Milsom, Frieder Neunhoeffer,
Rossa O’Keeffe-O’Donovan, Bernhard Kasberger, Paul Klemperer, Alex Teytelboym
and Jasmine Theilgaard. I would also like to thank audiences at the
Transatlantic Theory Workshop, SasCa PhD Conference, Oligo Workshop, and
various seminars and workshops in Oxford. Finally, I am grateful to Darija
Halatova and Tommaso Batistoni for coding assistance and to Exeter College and
the OEP Fund for their generous financial support. Links: algorithm; pre-
registration; experiment.

Itzhak Rasooly Department of Economics, University of Oxford.

(November 1, 2021)

In this paper, we design and implement an experiment aimed at testing the
level-$k$ model of auctions. We begin by asking which (simple) environments
can best disentangle the level-$k$ model from its leading rival, Bayes-Nash
equilibrium. We find two environments that are particularly suited to this
purpose: an all-pay auction with uniformly distributed values, and a first-
price auction with the possibility of cancelled bids. We then implement both
of these environments in a virtual laboratory in order to see which theory can
best explain observed bidding behaviour. We find that, when plausibly
calibrated, the level-$k$ model substantially under-predicts the observed bids
and is clearly out-performed by equilibrium. Moreover, attempting to fit the
level-$k$ model to the observed data results in implausibly high estimated
levels, which in turn bear no relation to the levels inferred from a game
known to trigger level-$k$ reasoning. Finally, subjects almost never appeal to
iterated reasoning when asked to explain how they bid. Overall, these findings
suggest that, despite its notable success in predicting behaviour in other
strategic settings, the level-$k$ model (and its close cousin cognitive
hierarchy) cannot explain behaviour in auctions.

Keywords: auction, behavioral game theory, experimental design, level-$k$
models
JEL Codes: C72, C90, D44, D83, D90

##  1 Introduction

Although the study of auctions remains dominated by equilibrium based
approaches, in recent years more ‘behavioural’ alternatives have emerged. Most
prominent among these is Crawford and Iriberri,’s (2007) level-$k$ model,
which posits that individuals iteratively best respond to a naive ‘first
thought’ about how one might bid in an auction. If accurate, the level-$k$
model would substantially revise our understanding of how auctions ought to
designed (Crawford et al.,, 2009; De Clippel et al.,, 2019); and would
similarly revise how we think about individual reasoning in incomplete
information settings. Understanding whether the level-$k$ model is accurate is
thus a critical issue.

Surprisingly, however, the existing literature provides little insight into
the predictive performance of level-$k$ and whether it can outperform the
natural benchmark of Bayes-Nash equilibrium. There is some evidence that the
level-$k$ model might be able to explain the ‘winner’s curse’ in common value
auctions (Crawford and Iriberri,, 2007; Costa-Gomes and Shimoji,, 2015); but
also evidence that it might do so for the wrong reasons (Ivanov et al.,,
2010). In addition, there does not appear to be any serious comparison of
level-$k$ and equilibrium in the rather simpler independent private setting
(IPV) that forms the starting point for most auctions research. For example,
while Crawford and Iriberri, (2007) do compare level-$k$ and equilibrium using
an IPV dataset, the auction structure that generated this dataset leads
level-$k$ and equilibrium to make near identical predictions — making a formal
comparison of the models almost impossible.111Indeed, Crawford and Iriberri,
(2007) restrict themselves to comparing level-$k$ with a rival behavioural
theory, quantal response equilibrium.222See also Kirchkamp and Reiß, (2011)
for a brief experimental comparison of level-$k$ and equilibrium, albeit in a
paper not devoted to level-$k$ reasoning. Since the goal of their paper was
not to assess the performance of level-$k$, it is perhaps not surprising that
their experiments do not cleanly disentangle level-$k$ and equilibrium: for
example, in their first experiment, the predictions of the models co-incide
entirely. See also Gillen, (2009), An, (2017), Hortaçsu et al., (2019) and
Galavotti et al., (2018) for work on the level-$k$ model in the context of
observational data. These papers do not attempt to test the accuracy of the
level-$k$ model or compare it with equilibrium; and indeed doing so using
observational data (when individual valuations are unknown) would obviously be
much more challenging than doing so with an experimental approach.

It is also difficult to learn very much about the relative performance of
level-$k$ and equilibrium using existing experimental data sets since these
have often been generated by first-price auctions with uniformly distributed
valuations (Kagel and Levin,, 1995). Unfortunately, the predictions of
level-$k$ and equilibrium entirely coincide in such settings (for any $k\geq
1)$. Moreover, while it is easy to disentangle level-$k$ and equilibrium using
‘exotic’ value distributions, it is doubtful that such distributions would be
understood by all subjects in any actual experiment. The challenge is
therefore to find an experimental design that is both sufficiently simple for
subjects to understand and yet dramatically disentangles the models.

We begin by proposing an environment that does exactly this: a discrete333We
work with discrete models throughout both for the sake of realism (any actual
auction must be discrete) but also because the level-$k$ model would be ill-
defined in a continuous action setting (we elaborate more upon these points
later on). all-pay auction with uniformly distributed values. We first analyse
the equilibrium predictions in this environment, proving that it possesses
exactly one symmetric equilibrium. We thus extend a well known result from
continuous auction theory to the discrete auction model – and do so using
completely novel (and elementary) arguments.444Our argument holds for
arbitrary value distributions and can be straightforwardly modified to allow
for factors like risk aversion. It also extends immediately to the first-price
auction; indeed, we suspect that it could be extended to fairly arbitrary
auction structures (although we have not formally verified this). In the the
course of our proof, we construct an algorithm555To aid future experimenters,
we have implemented the algorithm and made it available online. See:
https://auctionsolver.herokuapp.com/ that is guaranteed to identify this
equilibrium, finding that it closely approximates the continuous equilibrium
of traditional auction theory given the parameters chosen in our experiment.
In this equilibrium, bids increase (quadratically) in valuations, rising to
around half of valuations at the maximum possible valuation that players can
draw.

We then conduct a similar analysis of the level-$k$ model but uncover a
surprising contrast. Solving the level-$1$ player’s optimisation problem
reveals a corner solution: such types bid zero for all valuations. As a
result, level-$2$ and level-$3$ types also bid close to zero; the former
‘overcutting’ the level-$1$s by (almost always) bidding $1$ and the latter
‘overcutting’ the level-$2$s by (almost always) bidding $2$. Clearly, this is
very different from equilibrium; and so we have found an environment that
dramatically separates the models. In addition, this environment leads the
models to possess very different comparative statics. For example, in the
level-$k$ model making the bid discretisation more coarse substantially
inflates the predicted bids; whereas in the equilibrium model, it has very
little effect.

Next, we conduct a similar exercise in the first-price auction, again assuming
uniformly distributed values to make the environment experimentally
implementable (and to force the two level-0 specifications proposed by
Crawford and Iriberri, (2007) to coincide – we elaborate on this point later
on). To break the equivalence of the level-$k$ and equilibrium models, we
introduce the possibility that subject bids are ‘cancelled’ (in which case
they lose the auction automatically). We then solve for the cancellation
probability that maximises the ‘distance’ between the two theories. This turns
out to separate the models in a similar way: level-$k$ bids are once again
close to zero, whereas equilibrium bids are roughly a quarter of valuations.
Thus, we once again manage to disentangle the models in an environment simple
enough to be implemented in an experiment.

We then conduct an experiment designed to put these insights into practice.
The basic idea of the experiment is straightforward. We create the
aforementioned environments in a virtual laboratory, allowing us to see which
of the models under consideration better predict the observed bids. However,
the experiment also possesses various other features aimed at testing the
predictive success of the level-$k$ model. For example, we elicit individuals’
levels using a variant on Arad and Rubinstein, (2012)’s 11-20 game; study
whether behaviour varies with the bid discretisation in the manner predicted
by level-$k$ theory; and finally ask individuals to explain why they bid in
the way that they did to see if the level-$k$ model accurately characterises
their conscious reasoning processes.

The experiment yields four central findings. First, if individual levels are
restricted to the $1-3$ range (to capture an intuitive limit on the number of
thinking steps that individuals can be expected to perform), then the
level-$k$ predictions are far lower than the observed bids. In addition, the
level-$k$ predictions are substantially less accurate than those of
equilibrium. These conclusions hold across all the auction structures and
treatments that we consider and under a variety of approaches to model testing
and comparison. Moreover, this conclusion continues to hold if one allows
individuals to vary across auction rounds but retains the requirement that
individual levels are ‘plausible’ (i.e. in the 1–3 range).

Second, if one fits the model without any such plausibility restrictions, one
finds levels typically estimated in the 30–35 range. We view this almost as
reductio ad absurdum of the model since it conflicts with almost all prior
experimental work on this issue (Crawford et al.,, 2013), clashes with basic
intuition about the number of thinking steps that individuals can plausibly be
expected to conduct, and sits uneasily beside findings from linguistics that
individuals struggle to even comprehend higher order statements whose order
exceeds three (Arad and Rubinstein,, 2012). Moreover, methods for penalising
model flexibility (e.g. the Bayesian information criterion) generally do not
support the inclusion of these very high levels.

Third, individuals’ levels as estimated from the auctions bear almost no
resemblance to their levels as inferred from a game known to trigger level-$k$
reasoning (the 11-20 game). This is hard to square within the context of the
level-$k$ model: for if levels reflect either cognitive sophistication or
beliefs about the cognitive sophistication of one’s opponents, then one would
expect to find a positive correlation between estimated levels across games.
Conversely, the absence of positive correlation across any two games can be
taken as prima facie evidence that level-$k$ reasoning is not operative in at
least one of the games under consideration.

Finally, we find that subjects are unlikely to cite iterated reasoning when
asked to explain why they bid in the way that they did. In other words,
subjects are unlikely to offer conjectures about the bidding strategies of
their opponents – and even less likely to ground these conjectures in what
they take to be their opponents’ conjectures about their own bidding strategy.
Admittedly, we do see evidence of iterated reasoning in the explanations given
by at least one subject – and this subject was (perhaps not coincidentally)
likely to bid in line with the predictions of the level-$k$ model. However,
such reports are very rare, suggesting that the level-$k$ model captures at
best a small fraction of observed bidding behaviour.

We then conduct a series of extensions and robustness checks. For example, we
consider the impact of incorporating risk aversion into the models, allowing
subjects to best respond to a distribution over levels (as in Camerer et al.,
(2004)), changing the $L_{0}$ specification, as well as conducting various
tweaks to our empirical analyses (e.g. including subjects who submit dominated
bids). We do not find that any of these exercises alter our substantive
conclusions.

The remainder of this article is structured as follows. Section 2 outlines an
equilibrium and level-$k$ analysis of the discrete all-pay and first-price
auctions. Section 3 presents our experimental design. Section 4 contains our
empirical results and Section 5 our robustness checks. Finally Section 6
concludes with a discussion of why the level-$k$ model does so poorly in the
auction setting despite its well documented successes in other domains of
strategic behaviour.

##  2 Separating level-$k$ from equilibrium

###  2.1 The all-pay auction

To begin, let us describe our first environment which sharply disentangles the
level-k and equilibrium models: a (discrete) all-pay auction with uniformly
and independently distributed values. There are $n\geq 2$ bidders, each
assumed to be risk neutral (we relax this assumption in Section 5). As in our
experiment, each player’s valuation is drawn from the set
$\mathbb{X}=\\{0,1...,x\\}$ for some $x\in\mathbb{N^{+}}$; and the bids that
they may submit are restricted to the very same set $\mathbb{X}$. Let
$\beta\colon\mathbb{X}\rightarrow\mathbb{X}$ denote a generic pure strategy
(i.e. ‘bidding function’). We study the case of discrete values and bids since
(i) this is inevitably the set-up in the experiment that follows, as indeed it
must be in any auction experiment; (ii) if we had instead assumed that the
bids were continuous, as is more standard in the literature, then the
predictions of the level-k model would be ill-defined.666In particular, the
level-$2$ player would not possess a best response: for any number
$\epsilon>0$ that they might bid, there exists a feasible bid
$\epsilon^{\prime}<\epsilon$ that would yield them a higher expected pay-off.

Since this is an all-pay auction, individuals need to pay their bid no matter
whether they win the object. However, they win the object if and only if their
bid is strictly higher than all bids submitted by their opponents. Note that
we assume that individuals cannot win the object in the event of a tie. While
this is a slightly unusual assumption, it matches the tie-breaking rule that
we adopt in the experiment. This tie-breaking rule, in turn, was chosen
because it tremendously simplifies the equilibrium (and to a lesser extent the
level-k) analysis.

To begin with, we consider what Bayes-Nash Equilibrium (henceforth
‘equilibrium’) predicts that individuals will do in this situation. In
particular, we will focus on symmetric equilibria. We do this for three
reasons. First of all, it is unclear how individuals could be expected to
coordinate on asymmetric equilibria without any communication in a one-shot
game. Second, symmetric equilibria are analytically more tractable. Third,
based on computational experiments reported in Rasooly and Gavidia-Calderon,
(2020), it does not seem as though discrete auction games like those described
above possess asymmetric equilibria which are substantially different from
their symmetric equilibrium. Therefore, one might hope that the restriction to
symmetric equilibria is ‘almost’ without loss of generality.

Unfortunately, since our auction is discrete we cannot directly apply results
from ‘standard’ auction theory. We can, however, establish the following very
useful result (which holds entirely independently of the distribution from
which each player’s valuation is drawn).

###### Proposition 1

The discrete all-pay auction has exactly one symmetric equilibrium.

Proposition 1 contains two claims: that a symmetric equilibrium exists and
that it is unique. Establishing the existence part is straightforward: since
the game is finite, one can essentially just apply Harsanyi, (1967). However,
establishing uniqueness is considerably more involved: see Appendix A for the
details. Importantly, the uniqueness proof is based on an algorithm that
explicitly constructs the (candidate) equilibrium (see Appendix B for examples
of the algorithm in action). Thus, our result does not merely tell us that
there is exactly one symmetric equilibrium, but also shows us exactly how to
compute it.777This is invaluable from an experimental perspective since ‘brute
force’ approaches to equilibrium computation take several hours even for
auctions with a handful of possible values and essentially intractable for
auctions of the size considered in our experiment (Rasooly and Gavidia-
Calderon,, 2020). In contrast, our algorithm is able to compute the symmetric
equilibria of our experimental games in less than a second.

Having developed this algorithm, we now use it compute the symmetric
equilibrium in the case of the distribution (uniform) used in our experiment.
Figure 1 displays the result for the case of $x=15$ and $n=3$: a dot
represents a bid that is submitted with positive probability. As can be seen,
the equilibrium is in (properly) mixed strategies; but it is closely
approximated, quantitatively speaking, by the continuous equilibrium888See,
for instance, Klemperer, (1999) for a derivation. of ‘textbook’ theory:

$\beta(v)=\left(\frac{n-1}{n}\right)\frac{v^{n}}{x^{n-1}}$ (1)

As in the continuous model, equilibrium bids increase slowly when valuations
are low but quickly when valuations are high. When values equal the maximum
$x$, bids are roughly equal to the equilibrium bid in the first-price auction,
i.e. $x(n-1)/n$.

Figure 1: Equilibrium in the all-pay auction

Having discussed the predictions of equilibrium (our main benchmark), we now
turn to the predictions of the level-k model. As formulated by Crawford and
Iriberri, (2007), the model begins with a ‘naive’ type of player (the
level-$0$) who either bids their valuation (the ‘truthful’ specification) or
either bids any element of the strategy space $\mathbb{X}$ with equal
probability (the ‘random’ specification). This is meant to capture a player’s
‘first thought’ about how their opponents will play the game. Higher types are
then defined inductively: for $k\geq 1$, a level-$k$ player submits the bid
that maximises their expected payoff given that their opponents are all level
$k-1$ (see Crawford and Iriberri, (2007) for interpretation and elaboration).
In the following, we will write
$\beta^{k}\colon\mathbb{X}\rightarrow\mathbb{X}$ to denote a level-$k$
player’s bidding strategy.

We now derive the behaviour of the types. Assume, as in our experiment, that
values are uniformly distributed on $\mathbb{X}$ (retreating from the
generality of Proposition 1). Given the ‘truthful’ specification ($b=v$),
level-$0$ bids are then uniform on $\mathbb{X}$. Note, however, that this is
also the ‘random’ specification. We thus see that, quite aside from making the
experimental rules easier for subjects to grasp, imposing uniform values
ensures that the two level-$0$ specifications lead to identical predictions, a
great advantage when it comes to testing the model.

Given that level-$0$ bids are uniform, the level-$1$ player solves the problem

$\max_{b\in\mathbb{X}}\quad\pi(v,b)\equiv
v\mathbb{P\text{(win}}|b)-b=v\left(\frac{b}{x+1}\right)^{n-1}-b.$ (2)

This leads to a corner solution: bidding 0 for all values $v\in\mathbb{X}$. To
see why this is true, note that the benefit to such a player of increasing
their bid by 1 is the gain in probability $\Delta\mathbb{P}$ multiplied by
their valuation $v$. If $n=2$, then $\Delta\mathbb{P}=1/(x+1)$; so the benefit
of bidding one more is $v/(x+1)$. Meanwhile, the cost of bidding one more is
simply $1$. Since $v/(x+1)<1$ for all $v\in\mathbb{X}$, the benefit of
increasing one’s bid is always lower than the cost – which means that the
optimal bid is 0. Moreover, if it is optimal to bid 0 when $n=2$, it must be
optimal for any $n\geq 2$: adding more bidders decreases the expected payoff
from submitting any positive bid, while leaving the pay-off from bidding 0
fixed.

It is then straightforward to derive the strategies of higher levels. For now,
assume that a level-k player breaks ties by choosing the lowest optimal bid
(while this assumption is natural since it allows us to avoid dominated bids,
we relax it in Section 5). Given this assumption, a level-2 player sets
$\beta(v)=1$ for every value $v\geq 2$ (‘over-cutting’ their level-1
opponents); and sets $\beta(v)=0$ otherwise. Similarly, a level-$3$ player
sets $\beta(v)=2$ for all $v\geq 3$; and sets $\beta(v)=0$ otherwise. Thus,
for all ‘reasonably low’ levels, a level-k type player bids

$\beta_{LK}(v)=\begin{cases}k-1&\mbox{if }v\geq k\\\
0&\mbox{otherwise}\end{cases}$ (3)

The strategy $\beta_{LK}$ describes level-$k$ behaviour at ‘reasonably low’
levels. At higher levels, however, this simple pattern breaks down. The
strategy $\beta_{LK}$ calls for a level-$k$ player with a value of $k-1$ to
bid $0$, thereby earning a (certain) payoff of $0$. There is another option,
however: bid $1$ in the hope of winning when one’s opponents all bid $0$. As
$k$ rises, this second option becomes more attractive; and eventually it pays
to deviate from $\beta_{LK}$. It is easy to check, however, that level-$k$
players must bid $\beta_{LK}$ provided that

$k\leq(x+1)^{\frac{n-1}{n}}+1;$ (4)

an inequality that gives concrete meaning to the phrase ‘reasonably low’ in
the previous paragraph.

Finally, we argue that level-$k$ can never coincide with equilibrium; with the
implication that level-$k$ bidding functions must cycle as
$k\rightarrow\infty$. To prove this, note that, by Proposition 4 of Rasooly
and Gavidia-Calderon, (2020), the discrete all-pay auction with uniform values
does not possess a symmetric equilibrium in pure strategies. Of course, it
does have a symmetric equilibrium (see our Proposition 1), so this must be in
mixed strategies. Now, given our assumption about level-k tie breaking, it is
clear that a level-k player could never choose a mixed strategy. As a result,
for any $k\in\mathbb{N}^{+}$, level-k can never coincide with the symmetric
equilibrium. Moreover, this means that the level-k predictions must cycle. To
see this, suppose that they did not cycle. Since the set of bidding functions
is finite, there would need to exist some $k\in\mathbb{N}$ such that
$\beta^{k}$ coincides with $\beta^{k^{\prime}}$ for all $k^{\prime}>k$.
(Otherwise, a strategy would need to re-appear at some stage – which would
necessarily trigger a cycle.) But then $\beta^{k}$ would be a symmetric, pure
strategy equilibrium, contradicting Rasooly and Gavidia-Calderon, (2020)’s
Proposition 4.

The following proposition summarises the findings of the previous paragraphs.

###### Proposition 2

For every level $k$ that satisfies (4) and every value $v\in\mathbb{X}$, a
level-$k$ bidder sets $\beta^{k}(v)=\beta_{LK}(v)$. Moreover, there is no
$k\in\mathbb{N}$ at which $\beta^{k}$ coincides with equilibrium; with the
implication that $\beta^{k}$ cycles as $k\rightarrow\infty$.

The figure below plots both the equilibrium predictions (or technically the
continuous approximation) and level-k predictions (for levels 1–3). As can be
seen, the auction structure has done an excellent job disentangling the
predictions of the models. First, given any reasonable calibration of the
levels, the level-k model predicts bids that are substantially lower than
those predicted by equilibrium. Second, while the equilibrium bidding function
is generally increasing, the level-k bidding functions are flat at all but one
value, providing a second contrast which one can test. Finally, our non-
convergence result means that the theories remain separated at arbitrarily
high levels. It follows then, that we have found a powerful way to disentangle
the models, and to do so without abandoning the uniform values specification
which is so useful when implementing any actual experiment.

Figure 2: Comparing level-$k$ with equilibrium

###  2.2 The first-price auction

It is clearly desirable to find a second auction structure which sharply
separates the models – that way, we can check the relative performance of the
models in two different strategic settings. To this end, we now examine the
possibility of doing this within the first-price auction. As noted by Crawford
and Iriberri, (2007), the level-k and equilibrium models coincide (for every
$k\geq 1$) in the first-price auction with uniformly distributed values. To
break this equivalence, we now suppose that each player’s bid only goes
through with probability $p\in(0,1)$; otherwise their bid is cancelled and
they cannot win the auction. We now consider what probability $p$ maximally
disentangles the theories.

Formally, the model structure is much as before: for example, values and bids
are restricted to the same set of consecutive integers $\mathbb{X}$. However,
there are two differences. First, since this is now a first-price auction, a
player pays their bid if and only if they win the object. Second, as just
mentioned, each player’s bid is now cancelled with probability $1-p\in(0,1)$;
and a player wins if and only if their (non-cancelled) bid is strictly higher
than all the non-cancelled bids submitted by their opponents.

As before, we begin by demonstrating that a symmetric equilibrium exists and
is unique999While our uniqueness result still holds under arbitrary value
distributions, it now requires the restriction that $p<1$; the role of this
assumption is to ensure that all bids have positive probability of winning and
thus to rule out equilibria involving dominated bids. Since this restriction
holds in our experiment, this technical point need not concern us here.:

###### Proposition 3

If $p\in(0,1)$, then the discrete first-price auction has exactly one
symmetric equilibrium.

The arguments used to establish this result are very similar to those used to
prove Proposition 1. In particular, the argument is based on the very same
algorithm that constructs the equilibrium. As before, using this algorithm
reveals that the equilibrium bids are well approximated by the continuous
equilibrium, which is now (under uniform values)101010Equation (5) may be
derived by noting the formal equivalence of our model with a model with an
uncertain number of bidders. One can then apply Theorem 1 of Harstad et al.,
(1990). Alternately, one can note that our model is also equivalent (at least
as far as undominated equilibria are concerned) with a model in which each
player’s value distribution has a mass of probability on zero. One can then
use the ‘standard formula’ (see, e.g., Krishna, (2009))
$\beta(v)=v-\int_{0}^{v}\frac{G(y)}{G(v)}dy,$ and insert
$G(v)=\left(1-p+pv/x\right)^{n-1}$ (recalling that $G(v)$ is the probability
that the maximum valuation is $v$ or lower).

$\beta(v)=\left(\frac{n-1}{n}\right)v-\frac{x(1-p)}{np}\left[1-\left(\frac{1-p}{1-p+p(v/x)}\right)^{n-1}\right]$
(5)

As can be seen from Figure 2, introducing the possibility that bids are
cancelled makes the bidding less aggressive. If $p=1$, i.e. bids are never
cancelled, our expression reduces to $\beta(v)=v(n-1)/n$. In other words, we
recover the standard formula for equilibrium with uniformly distributed
values. Moreover, it is easy to check that for any $v\in\mathbb{X}$,
$\beta(v)$ is strictly decreasing in $p$. Finally it can be shown (using
L’Hôpital’s rule) that $\beta(v)\rightarrow 0$ as $p\rightarrow 0$ for all
$v\in\mathbb{X}$. This is also not surprising: if one is almost certain to win
the auction anyway (in the event that one’s own bid is not cancelled), then
one may as well bid $b\approx 0$.

We now turn to the predictions of the level-k model. Given that level-$0$ bids
are either cancelled or uniform on $\mathbb{X}$, a level-$1$ player solves the
problem

$\max_{b\in\mathbb{X}}\quad(v-b)\mathbb{P\text{(win}}|b)=(v-b)\left(1-p+\frac{pb}{x+1}\right)^{n-1}$
(6)

Since the objective function in (6) is single peaked, it is obvious that the
integer that maximises this function can be obtained by finding the
$b\in[0,x]$ that maximises the function and rounding up or down. In other
words, we have a good continuous approximation of the exact level-$1$ bidding
function. To simplify formulae, we will therefore proceed as if level-$1$
types could choose any $b\in[0,x]$ (although we will always use the exact
model predictions when analysing the data from our experiment). Routine
optimisation subject to the inequality constraint then reveals that

$\beta^{1}(v)=\max\left\\{\left(\frac{n-1}{n}\right)v-\left(\frac{1-p}{p}\right)\frac{x}{n},\hskip
3.00003pt0\right\\}$ (7)

As in equilibrium, level-$1$ bids are decreasing in the probability $p$.
Moreover, for all $p\leq 1/n$, the level-$1$ player sets $\beta(v)=0$ for all
$v\in\mathbb{X}$, i.e. we recover the same corner solution as in the all-pay
auction. Intuitively, if $p$ is quite low, then there is a large chance that
the player’s opponents’ bids are all cancelled – so the player may as well
just bid $0$ and receive their entire valuation whenever they ‘get lucky’.

Our goal is to disentangle the models. To this end, we define the distance
between the theories as

$d(p)\equiv\int_{0}^{x}\left|\beta(v)-\beta^{1}(v)\right|dv.$ (8)

Of course, the distance $d(p)$ is simply the total area between the
(approximated) level-$1$ and equilibrium bidding functions. We now study the
probability $p^{*}$ that maximises this distance. Recall from above that the
level-$1$ player bids $\beta(v)=0$ for all $v\in\mathbb{X}$ provided that
$p\leq 1/n$. Meanwhile, every equilibrium bid $\beta(v)$ is strictly
decreasing in $p$ (moving closer to 0). From this, we see that we could never
maximise (8) by choosing some $p\in[0,1/n)$. The reason is that increasing $p$
(by a sufficiently small $\epsilon>0$) would increase every equilibrium bid
$\beta(v)$ while leaving the level-k bids unchanged, thus increasing the
distance between the two theories. In other words, we have the following easy
result.

###### Proposition 4

$p^{*}\geq 1/n$.

Unfortunately, $p^{*}$ cannot be solved analytically. Nonetheless, we can
compute the optimal $p$ numerically given various values of $n$ (see Appendix
C). We find, for example, that $p^{*}\approx 0.54$ if $n=2$, $p^{*}\approx
0.34$ if $n=3$ and $p^{*}\approx 0.26$ if $n=4$. Overall, then, choosing
$p=1/n$ seems like a good approximation of the optimal probability (the loss
in terms of distance is small). This has the further advantage of making the
probabilities easier for subjects to grasp than the exact solutions would be;
so we ultimately opt for $p=1/n$ in our experiment.

Assuming that $p=1/n$, it is then straightforward to derive the exact level-k
bidding strategies. As noted, the level-$1$ player sets $\beta^{1}(v)=0$ for
all $v\in\mathbb{X}$. As a result, the level-$2$ player bids $1$ (when their
value is high) or bids $0$ (when their value is low). Similarly, the level-$3$
player bids $2$ for high values and $0$ otherwise. This simple pattern
characterises the behaviour of level-$k$ bidders when $k$ is ‘reasonably low’
relative to the number of valuations $x+1$. More formally, we can say the
following:

###### Proposition 5

Fix a set of types $\mathbb{K}=\\{1,2,...\bar{k}\\}$, let $p=1/n$, and define

$v^{*}(k)=\frac{k-1}{1-\left(1-p\right)^{n-1}}$

Then if $x$ is sufficiently large, every level-$k$ type in $\mathbb{K}$ bids
$\beta^{k}(v)=k-1$ for $v>v^{*}(k)$ and $\beta^{k}(v)=0$ otherwise.

Given the parameters ultimately chosen for our experiment ($n=2$, $p=1/2$,
$x=100$), level-$k$ behaviour once again cycles at very high levels. To
illustrate this, Figure 3 plots the maximum bid submitted by the first $100$
levels (the full level-$k$ bidding functions are available in the
supplementary materials). As can be seen, the model produces cycles of length
$29$ (so level-$7$ coincides with level-$36$, which in turn coincides with
level $65$, etc.) Moreover, far from cycling locally around the equilibrium
predictions, level-$k$ cycles are quite dramatic: the maximum bid ranges from
0 to 20. This is all the more surprising once we realise that all bids above
50 are strictly dominated (by bidding zero), so the highest level-$k$ bid
traces out around 40% of the non-dominated strategy space.

Figure 3: Level-$k$ cycling

Finally, Figure 4 compares the models in the case of $n=2$ and $p=1/2$. As can
be seen, we have once again managed to disentangle the theories – and in a
dramatic way. First, given any plausible calibration of the levels – recall
that these are supposed to be in the 1–3 region – level-k bids are
significantly lower than equilibrium bids. Second, while equilibrium bids are
increasing in values, level-k bids are almost always non-increasing. Finally,
the non-convergence of level-$k$ to equilibrium means that the models remain
separated even if one were to ascribe implausibly high levels to the
experimental participants.

Figure 4: Comparing level-$k$ with equilibrium

##  3 Experimental design

To test the insights of the previous section, we conducted an experiment with
84 participants in collaboration with the Nuffield Centre for Experimental
Social Sciences. All participants were students at the University of Oxford
and no participant took part in more than one session. The experiment was
conducted online and written in oTree (Chen et al.,, 2016).111111The full
experimental instructions can be viewed here: https://auction-
experiment.herokuapp.com/

The essential idea of the experiment is straightforward – to implement the two
aforementioned auction structures in a virtual laboratory and ask subjects how
much they wanted to bid. It is then a relatively straightforward matter to
check which of the two theories under consideration can better predict (and
rationalise) the observed bidding behaviour.

More precisely, each subject in the experiment participated in a first-price
auction (as described previously) with $n=2$, $p=1/2$ and an all-pay auction
(as described in the previous section) with $n=2$. Following the auctions,
subjects played a modified version of Arad and Rubinstein, (2012)’s 11-20 game
that is proposed in Alaoui and Penta, (2016). The purpose of this game was to
allow for a rough calibration of each participant’s level.121212We use Alaoui
and Penta, (2016)’s modification of the original game since it prevents
cycling (level-$k$ behavior coincides for all $k\geq 9$). As Alaoui and Penta,
(2016) note, this is then useful if one wishes to calibrate individual levels
without the potentially controversial assumption that each individual’s level
is the smallest possible level that rationalises their observed behaviour.
Finally, we measured each subject’s risk aversion using the ‘bomb’ risk
elicitation task (Crosetto and Filippin,, 2013).

Before participating in either of the auctions, subjects were given extensive
quizzes to check their understanding of the rules. They were not permitted to
proceed past the quiz until they had answered a full set of questions
correctly. We view this as an important feature of the study: subjects are
unlikely to make meaningful choices if they cannot even understand the rules
of the game that they are playing.

Each subject bid in two rounds of each of the auction types.131313They were
also asked to explain the general considerations determining how they chose to
bid after every second round. Moreover, we did not inform subjects of the bids
submitted by their opponents in previous rounds. We made these two choices –
only two rounds coupled with very little feedback – since the level-$k$ model
is supposed to be a model of initial play, as opposed to a model of the
outcomes of a long-run learning process (Crawford and Iriberri,,
2007).141414Indeed, it is for this reason that Crawford and Iriberri, (2007)
only test their model using data from the first five rounds of the experiments
they examine. In other words, we sought to test the level-$k$ model within its
intended domain of applicability.

Subjects were informed (and regularly reminded) that they would never bid
against the same opponent twice across the rounds of an auction. That is, we
used the ‘perfect stranger matching’ protocol. We opted for this protocol
since the two theories under consideration are static theories – in the sense
that they treat the game as a one-shot interaction – and we did not want to
complicate the analysis by introducing learning or reputational
considerations.

Bidding was incentivised through a procedure similar to that employed in
Filiz-Ozbay and Ozbay, (2007). In every round, individuals were asked how much
they would want to bid given ten possible valuations that they might have.
They were informed that they should bid carefully since one of these
valuations would be their actual valuation and they would be committed to
bidding the amount they had entered. In other words, one in ten bids were
selected to ‘count’. We opted for this procedure, as opposed to incentivising
every bid, since it allowed us to collect more data given a fixed number of
rounds, thereby further reducing feedback or learning effects (and making the
environment more hospitable to the level-$k$ model).

Finally, we randomly allocated the participants into two treatments: an
‘integer bid’ treatment (to which 2/3 of participants are allocated), and a
treatment in which bids were constrained to be multiples of 5. The idea here
was to check whether varying the bid treatment has the strong effects
predicted by level-$k$ theory or the relatively weak effects predicted by the
equilibrium model. In the level-$k$ model, increasing the discretisation by a
factor of 5 increases the predicted bids by roughly a factor of 5, assuming
that levels are sufficiently low for individual behavior to be characterised
by Propositions 2 and 5 (see the supplementary materials for an exact
description of the effects of the bid discretisation on the model’s
predictions). In contrast, varying the bid discretisation does not have large
quantitative effects on equilibrium bids: for example in the first-price
auction the average equilibrium bid increases by just 4.7% (again, see the
supplementary materials).

The experiment was pre-registered along with a full analysis plan.151515See
https://www.socialscienceregistry.org/trials/8011. The experiment also
received Ethics Approval from both the University of Oxford’s and Centre for
Experimental Social Sciences’ institutional review boards.

##  4 Results

###  4.1 Pooling the data

We begin by asking which theory best explains the overall patterns observed in
the bidding data. To this end, we pool all the valuations and bids, compute
the mean observed bid for every valuation and plot this against the
corresponding predictions of the two models. For the moment, we assume that
levels cannot exceed three161616We impose this constraint because all of the
evidence of which we are aware suggests that insofar as individuals use
level-$k$ reasoning, it rarely goes beyond the third level. Indeed, this is
what we find when examining the responses of our subjects in the 11-20 game:
out of those who mentioned level-$k$ reasoning when asked to explain their
choices, not one made a choice consistent with more than three levels of
reasoning. and drop the handful of subjects who submitted at least one
dominated bid in one of the auctions.

Figure 5 displays the results for the case of the first price auction with
integer bids (see also Figure 8 for the treatment in which bids are multiples
of 5). As found in many prior studies of first-price auctions, we observe
substantial overbidding relative to the (risk-neutral) equilibrium benchmark
(Kagel and Levin,, 1995). Indeed, the average bid is around 46% of the average
value, whereas equilibrium predicts that it should be around 20%. However,
while the equilibrium model struggles to match the data, it is clear that the
level-$k$ model performs considerably worse, predicting bids that are at least
an order of magnitude lower than those commonly observed. As a result, it is
clear that equilibrium provides a substantially better (albeit highly
imperfect) description of the average bidding function observed in the
experimental data.

Figure 5: Predictions and data in the first-price auction

We then conduct a similar analysis of the all-pay auction. As shown in Figure
6, equilibrium does a somewhat better job of tracking mean bids in this
environment (see also Figure 9 for the case where bids are multiples of 5).
However, over-bidding relative to equilibrium is again observed on average at
low to medium values.171717The overbidding that we observe is consistent with
results from the only other experimental study of IPV all-pay auctions of
which we are aware (Noussair and Silver,, 2006). Unfortunately, however, we
were unable to obtain the data from this experiment and so were unable to use
it to complement our analyses. As before, the level-$k$ model predicts bids
that are orders of magnitude below those that are observed on average, at
least if one imposes the restriction that levels cannot exceed 3. Thus, our
initial examination of the data suggests a very clear ranking of the two
models under consideration.

Figure 6: Predictions and data in the all-pay auction

To verify these conclusions more formally, we also compute the root-mean-
square prediction errors of both of the theories. In the case of equilibrium,
this is straightforward since the model is ‘parameter free’. In the case of
level-$k$, we obtain a prediction by assigning each datapoint the level from
the $1-3$ range that minimises the model’s prediction error (this is a
generous assumption that will tend to overstate the model’s predictive
performance). Table 1 displays the results. As can be seen, both models make
size-able prediction errors. Consistent with the graphical evidence, however,
the prediction errors of level-$k$ are substantially greater than those of
equilibrium.

Table 1: Prediction errors

| T1 FP | T1 AP | T2 FP | T2 AP
---|---|---|---|---
Equilibrium | 15.1 | 13.3 | 14.1 | 14.3
Level-$k$ | 21.3 | 19.8 | 21.4 | 24.0

  * •

Notes: T1 and T2 denote the integer bid and multiples of five treatments; FP
and AP abbreviate ‘first-price’ and ‘all-pay’.

Next, we examine the effect of varying the bid discretisation. Recall that the
level-$k$ model predicts a large effect from changing the discretisation from
integers to multiples of 5: bids should increase by roughly a factor of 5. In
contrast, the equilibrium model predicts much smaller effects. Given these
very different predictions, we now conduct a comparison of the integer bid and
multiples of 5 bid treatments.

As a preliminary, we confirm that the treatments are balanced, confirming that
the randomisation worked as expected (see Table 3). Next, we inspect the
average bids across the two treatments and auction structures (see Table 4).
As can be seen, changing the bid discretisation makes essentially no
difference to the first-price auction (in both cases, the average bid is about
46% of the average valuation). In the all-pay auction, restricting bids to
multiples of five does appear to slightly increase the average bids (which
rise from 45% to 50% of the average valuation), but clearly not in the
dramatic way predicted by the level-$k$ model. In neither auction, then, do
the data support the surprising comparative static prediction of level-$k$
theory.

###  4.2 Structural estimates

In light of the evidence above, it is clear that the level-$k$ model cannot
count for observed average behaviour, at least when calibrated with a
plausible distribution of types. However, this still leaves at least two
issues unresolved. First, even if level-$k$ cannot describe the average
bidder, it is still possible that it accounts for a substantial minority of
the observed behaviour. Second, it remains to be seen whether the model might
be rescued by allowing for a greater number of possible types. We now address
both of these issues.

To do this, we follow the literature in estimating a mixture of types model
(see, e.g., Stahl and Wilson, (1994), Bosch-Domenech et al., (2002) and
Crawford and Iriberri, (2007)). To obtain stochastic versions of the relevant
theories, we assume that subjects submit bids with probabilities that are
proportional to a normal distribution whose mean is the predicted bid that
they are ‘supposed’ to submit (according to the relevant theory). More
formally, let $\mathbb{K}\subset\mathbb{N}^{+}$ denote the set of types,
indexing equilibrium by $k=0$ and level-$k$ types in the obvious way. Given
that a player $i$ has value $v_{i}\in\mathbb{X}$ and type $k\in\mathbb{K}$, we
assume that the chance they submit a bid $b_{i}\in\mathbb{X}$ is given by:

$\mathbb{P}(b_{i}|v_{i},k)\propto\exp\left[-\frac{1}{2}\left(\frac{b_{i}-b_{p}}{\sigma_{k}}\right)^{2}\right]$
(9)

where $b_{p}$ is the bid predicted by their type (given their valuation),
$\sigma_{k}$ is a noise parameter that controls the variance of their bids,
and the constant of proportionality is chosen to ensure that the probabilities
sum to $1$.181818We also experimented with a model containing logit errors.
However, these had the unfortunate tendency of preferring level-$1$ to all
higher levels even though level-$1$ predictions were the furthest from the
observed bids (recall that level-$1$ types bid zero regardless of their
valuation). This produces a single peaked distribution whose mode is the
predicted bid $b_{p}$. As $\sigma_{k}\rightarrow\infty$, this distribution
converges to uniform noise; and as $\sigma_{k}\rightarrow 0$, it converges to
a degenerate distribution which places all probability mass on the predicted
bid $b_{p}$.

For now, assume that each individual’s type is fixed (though we relax this in
many of our analyses). Then the probability that an individual submits a
vector of (twenty) bids $\bm{b_{i}}$ given their vector of values $\bm{v_{i}}$
is (assuming independence of errors):

$\mathbb{P}(\bm{b_{i}}|\bm{v_{i}},k)=\prod_{i=1}^{20}\mathbb{P}(b_{i}|v_{i},k)$
(10)

We assume that the individual we observe is drawn randomly from the
population; and write $p_{k}\in[0,1]$ to denote the population proportion of
type $k$ (obviously, $\sum_{k}p_{k}=1)$. By the law of total probability, the
unconditional probability of their bids is

$\mathbb{P}(\bm{b_{i}}|\bm{v_{i}})=\sum_{k\in\mathbb{K}}\mathbb{P}(\bm{b_{i}}|\bm{v_{i}},k)p_{k}$
(11)

and so the ‘likelihood’ of all the observed decisions is (combining Equations
9 \- 11)

$L=\prod_{i=1}^{n}\mathbb{P}(\bm{b_{i}}|\bm{v_{i}})\propto\prod_{i=1}^{n}\left(\sum_{k\in\mathbb{K}}\left[\prod_{i=1}^{20}\left(\exp\left[-\frac{1}{2}\left(\frac{b_{i}-b_{p}}{\sigma_{k}}\right)^{2}\right]\right)\right]p_{k}\right)$
(12)

In every estimation exercise, we compute the parameters
$\bm{p}=(p_{k})_{k\in\mathbb{K}}$ and
$\bm{\sigma}=(\sigma_{k})_{k\in\mathbb{K}}$ that maximise the likelihood (12).
We then calculate standard errors with the jack-knife. While we consider the
case where individual levels are fixed (as above), we also consider the case
where individual levels are permitted to vary by round – although we do insist
throughout that levels are fixed within a round.191919Recall that all bids
within a round must be made simultaneously, so there is no possibility for
learning within rounds. In contrast, individuals could conceivably learn
between rounds – although the scope for such learning is limited by the near-
total absence of feedback.

As a preliminary exercise, we compare an equilibrium only model with the
level-$k$ model in which levels are restricted from $1-3$. To avoid
introducing too great an imbalance in the number of parameters between the two
respective models, we restrict both to a single noise parameter $\sigma$.
Table 5 displays the results. As one might have expected from Figures 5 \- 6,
equilibrium does a better job than any of the level-$k$ models at explaining
the data, attaining a higher log-likelihood in all specifications. This
advantage becomes even greater if we punish the level-$k$ models for their
extra flexibility; for example, using the Bayesian information criterion. This
simply reinforces the conclusion that we reached earlier: equilibrium does a
better job than the level-$k$ model at accounting for the overall patterns in
the data.

We now examine the extent to which the level-$k$ model might be salvaged by
adding higher levels. To this end, we now estimate a series of models,
gradually adding levels until we reach the point where the model cycles. Three
main points emerge from this exercise.202020Since the resulting tables are too
large to be contained within the appendix, they have been made available here.
First, while the level-$k$ model can achieve a higher likelihood than
equilibrium once the full suite of levels is added, it can do this only by
placing most probability on very high levels: for example, in the first price
auction, most probability mass is placed on level-$32$. As argued previously,
is doubtful whether we can plausibly ascribe so many thinking steps to
participants. Second, the level-$k$ model fits the data with a very unusual
distribution, placing some probability mass on level-1, none on intermediate
levels, and the rest on the very high levels just discussed. Of course, the
idea that individuals do either one thinking step or 32 thinking steps might
appear to be rather strange. Finally, the inclusion of these very high levels
is almost never recommended by the Bayes-Information Criterion. For all these
reasons, it seems difficult to rescue the level-$k$ model through a strategy
of adding a very large number of levels.

Although a pure level-$k$ model appears to be outperformed by equilibrium, a
final question we investigate is whether a hybrid model might be able to
outperform both. To this end, we now re-estimate the model for all auction
structures, allowing for both equilibrium and level-$k$ types and no longer
restricting the model to a single noise parameter $\sigma$. We also return to
our previous assumption that levels are in the plausible range ($1$-$3$).
Table 6 displays the results. As can be seen, most probability mass is placed
on equilibrium – suggesting again that the level-$k$ model fits the data
relatively poorly. However, there is still some probability mass placed on
level-$k$ types. Thus, the results from the hybrid model are consistent with
the possibility that the level-$k$ model characterises the behaviour of a
small minority of subjects, even if it struggles to capture the general
patterns in the bidding data.

###  4.3 Correlating levels

We now examine whether individual levels as estimated from the two auctions
bear any resemblance to the levels as inferred from a game known to trigger
level-$k$ reasoning: a variant on the 11-20 game proposed by Alaoui and Penta,
(2016). To this end, we calculate the level $k$ and precision $\sigma_{k}$
that maximises the likelihood of each individual’s bids (i.e. Equation 10).
Thus, we now assign each individual a definite level, as opposed to a
distribution of levels; and no longer allow for equilibrium types. After doing
this for both the first-price and all-pay auction, we then compute each
individual’s level as inferred from the 11-20 game, discarding those whose
implied level exceeds 4.212121We do this partly because such levels are at
odds with prior experimental evidence but also because not one of such
subjects invoke level-$k$ reasoning when asked to explain their choice.
Finally, we compute the correlation between the levels inferred from each
auction and the level as inferred from the 11-20 game.

Figure 10 plots the levels estimated from the first-price and all-pay auctions
against the levels as inferred from the 11-20 game. As can be seen, there is
no discernible relationship between the levels; and indeed the correlations
are estimated to be $0.00$ and $-0.09$ in the case of the first-price and all-
pay auctions respectively.222222This result reflects those of Georganas et
al., (2015), who find no correlation between estimated individual levels
across two families of games (albeit in a non-auction setting). From the
perspective of level-$k$ theory, this is rather unexpected. For if levels
reflect either cognitive sophistication or beliefs about the cognitive
sophistication of one’s opponents – the two leading interpretations of
individual levels – then one would expect to find a positive correlation
between individual levels across games. Conversely, the absence of such a
positive correlation suggests that individuals are not using level-$k$
reasoning in at least one of the games.

###  4.4 Subject reports

Finally, we examine the explanations given by participants as to why they
chose to bid in the way that they did.232323Historically, there seems to have
been a widespread belief in economics that one cannot learn anything from
individuals from asking them why they behaved in the way they did (see e.g.
Becker ??). As far as we can see, however, this belief was not adopted on the
basis of any kind of empirical evidence. More to the point, it is obviously
false: for example, macroeconomists appear to learn a great deal from
unemployment statistics, which in turn are almost always based on individual
self-reports (Thaler,, 2015). In particular, we consider whether they appealed
to iterated reasoning.242424A strict definition of iterated reasoning would be
reports of the form: “I bid in way X because I expected my opponent to bid in
way Y; and I expected my opponent to bid in way Y because I thought they would
think I would bid in way Z.” In other words, this would need to involve (1) a
conjecture about the bidding strategies followed by the player’s opponents (2)
an attempt to ground this conjecture using a conjecture about the conjectures
of the opponents about the player’s bidding strategy. On a weaker definition,
we would simply require some kind of conjecture about the opponents’ bidding
strategy (along with an attempt to optimise given this conjecture). Since the
level-$k$ model allows for level-$1$ players, we used this weaker (and more
permissive) definition when categorising the reports. Assuming that subjects’
conscious reasoning was indeed captured by the level-$k$ model and further
assuming that subjects did not for some reason wish to lie to their
experimenter about their reasoning, one would expect to see iterated reasoning
cited in their explanations. Conversely, one can take the absence of such
reports as evidence that the level-$k$ model did not in fact characterise
their conscious reasoning.

In their explanations, the majority of subjects (around 55%) explicitly
indicated an awareness of the basic bidding trade-off: bidding more increases
one’s probability of winning, but decreases one’s pay-off if one wins the
auction. However, the vast majority of subjects did not appear to use iterated
reasoning. Indeed, we categorise responses as invoking iterated reasoning in
just two cases.252525All categorisations were independently checked by another
researcher who was not involved in the study. The reader is invited to examine
the responses themselves which are provided (along with all other non-
demographic data generated by the experiment) in the supplementary materials.

To be clear, we are not claiming that the level-$k$ model did not characterise
the reasoning given by any of the subjects. For example, one subject (who also
bid in line with the model) wrote that

  * Since my opponent has a 50% chance of their bid being cancelled, it’s better to keep bids low in hopes theirs [sic] gets cancelled but mine does not. I raised the bid to more than 1 because in the event my opponent is using the same strategy and bids very low, I still have a chance of winning the auction.

Thus, this subject not only considered the level-$1$ strategy (bidding zero)
but then went beyond it – a clear case of level-$k$ reasoning.262626While the
bids submitted by the subjects did not exactly conform to any of the levels
they were very close, with the subjects bidding a mix of 2’s and 3’s at
relatively low valuations. What is evident, however, is that reasoning was
provided by only a small minority of subjects, a finding that is also in line
with the choice data outlined in the previous section.

##  5 Robustness checks

We now report on the results of a variety of extensions and robustness checks.

Risk aversion. While the previous analysis assumed that individuals are risk-
neutral, we now consider whether the level-$k$ model might be salvaged by
allowing individuals to be risk-averse.272727Of course, expected utility
maximisers should be approximately risk-neutral over such small stakes
(Arrow,, 1970; Rabin,, 2000). Thus, the ‘risk aversion’ discussed here should
really be thought of as capturing utility curvature within a reference
dependent model (Bleichrodt et al.,, 2019). To begin, consider the all-pay
auction. As noted in Section 2.1, level-1 players (if assumed to be risk-
neutral) are predicted to bid 0 (for all values), thereby earning a certain
payoff of zero. If they submitted a positive bid, then they would be exposed
to risk: either their pay-off would be positive (if they win the auction) or
negative (if they lose the auction). Now, given that level-1 players choose to
bid 0 when they are risk neutral, it is obvious that they must also choose to
bid 0 if they are assumed to be risk averse. Thus, introducing risk aversion
makes no difference to the model’s predictions – and also (for this reason)
makes no difference to the predictions for high levels (provided that they are
sufficiently low to be characterised by bidding strategy $\beta_{LK}$).

While risk aversion cannot improve the model’s performance in the all-pay
auction, matters are more complicated in the first-price auction. In general,
holding the bidding strategies of one’s opponents fixed, introducing risk
aversion must (weakly) increase one’s optimal bid. In our setting, this means
that introducing risk aversion can make positive bidding by a level-1 type
optimal, thereby bringing the model’s predictions closer in line with the
observed data. Note, however, that introducing risk aversion also makes
equilibrium bids more aggressive (Holt,, 1980), which might be expected to
increase the predicted accuracy of our benchmark. It is therefore unclear at
the onset whether introducing risk aversion improves the relative performance
of the level-$k$ model.

To study this issue more carefully, we assume that every individual $i$ has
constant relative risk aversion (i.e. $u_{i}(x)=x^{\alpha}_{i}$ where $x_{i}$
is the individual’s pay-off and $\alpha_{i}>0$ determines their level of risk
aversion.) We then estimate every individual’s parameter $\alpha_{i}$ using
their choice in the “bomb” risk elicitation task (Crosetto and Filippin,,
2013). We then re-calculate the level-$k$ model’s predictions for every player
using their estimated parameter $\alpha_{i}$ (along with their estimated level
as in Section 4.1). We also re-calculate the equilibrium predictions on the
assumption that every individual has the average estimated risk aversion
parameter of the subject pool ($\alpha=0.72$).282828The code can be viewed
here: https://github.com/Itzhak95/risk_aversion/tree/master. The $\alpha=0.72$
is obtained after dropping the two outliers who collected $61$ or more out of
the $62$ boxes. If such subjects are included, we obtain $\alpha=1.8$ (risk-
seeking), in which case allowing non-linear utility is unable to improve the
performance of the level-$k$ model.

Figure 7 displays the results for the main treatment (see also 11 for the case
where bids are multiples of five). As can be seen, both the level-$k$ and
equilibrium predictions are now somewhat improved and considerably closer to
the actual data. Overall, however, it is clear that equilibrium still
outperforms level-$k$ dramatically – a conclusion confirmed by the prediction
error rates reported in Table 7.

Figure 7: Risk aversion in the first-price auction

Cognitive hierarchy. We now ask if any conclusions might change if individuals
are assumed to best respond to a distribution over levels, as in Camerer et
al., (2004)’s cognitive hierarchy model. To this end, we assume (following
Camerer et al., (2004)) that levels are Poisson distributed with mean $\tau$.
Using standard maximum likelihood methods, we then estimate $\tau$ for the
participants in our experiments using their choices in the 11-20 game, finding
that $\tau\approx 1.9$. Finally, we compute the predictive errors of the
cognitive hierarchy model and compare those with the associated errors for
equilibrium. As shown in Table 7, moving from level-$k$ to cognitive hierarchy
makes essentially no difference to the prediction errors. This is not a
surprise: obviously, best responding to a mix of levels between 1 and 3 yields
(in our context) very similar optimal bids to best responding to level-2.

Including dominated bids. As noted earlier, our main analysis excluded the
small number of subjects who submitted at least one dominated bid on the
grounds that such subjects may have misunderstood the rules of the auction.
However, excluding such subjects is a debatable choice; and we re-ran all
analyses after including such subjects. This did not change any of our
substantive conclusions (see Table 7).

Changing the tie-breaking rule. In the previous analysis, we assumed that
level-$k$ players always chose the lowest optimal bid if multiple bids were
optimal. While this allowed the model to avoid predicting that individuals
choose dominated strategies, it is natural to ask whether any of our
conclusions hinge on this assumption. To investigate this, we now make the
opposite assumption – that level-$k$ players always choose the highest bid out
of the bids that they believe are optimal.292929Obviously, this does not
exhaust the universe of possible tie-breaking rules: one could also assume
that players sometimes choose the highest optimal bid and sometimes choose the
lowest optimal bid. Re-running our main analyses, we see that this slightly
worsens the model’s fit with the data (although the change is small). This is
not a surprise: the main change is that the model now predicts some dominated
bids, and such bids are very rare in the data.

Dropping the second round. As noted in Section 3, the level-$k$ model is
supposed to be a model of initial play. In order to test the model in its
natural domain, we therefore blunted subject feedback and restricted the
number of rounds to two. Nonetheless, one might worry that even two rounds is
too many to count as ‘initial play’. To test this possibility, we re-ran the
analyses after dropping the second round. While this does somewhat improve
level-$k$’s performance, the model remains clearly outperformed by equilibrium
(see Table 5).

Changing the $L_{0}$ specification. The level-$k$ auction model, as formulated
by Crawford and Iriberri, (2007), assumes that level-$0$ players either bid
their value (the ‘truthful’ specification) or randomise uniformly over
$\mathbb{X}$. Since the predictions of such a model seem strongly disconfirmed
by the data, one might wonder whether the model might be improved by choosing
an alternative level-0 specification. While our data do not allow us to
absolutely rule this possibility out, there are various reasons to think that
this is not likely to be a fruitful theoretical exercise.

First, as stressed by Crawford et al., (2013), one must be careful to ensure
that any L0 specification represents a ‘strategically naive’ assessment of how
others will play the game. This is most obviously important if one wishes to
connect the model to its standard interpretation as a model of iterated
reasoning given some naive but initially plausible starting point. In
addition, it is vital to restrict the range of possible level-$0$ strategies
if one wishes the model to make anything like definite predictions in
strategic situations; and to be clearly disentangled from rival models. For
instance, if an analyst were allowed to specify that level-$0$ play coincided
with the symmetric equilibrium, it would be impossible to disentangle the
equilibrium from the predictions of the level-$k$ model: for every $k\geq 1$,
the level-$k$ prediction would simply be the symmetric equilibrium strategy.

Thus, if one did want to change the $L_{0}$ specification, one would need to
make sure that it represents a psychologically plausible way in which a player
might naively expect others to play. This brings us to our second point,
however: there simply do not appear to be a large number of such $L_{0}$
specifications. Other than the specifications proposed by Crawford and
Iriberri, (2007), we are only able to think of one alternative: bidding
uniformly between 0 and one’s value $v$ (modifying their ‘random’
specification so as to remove dominated bids). However, opting for this L0
specification would render the model unable to match even the general trend of
bidding in the most experimentally studied of all auction formats – the first-
price auction with independent, uniform values (and no possibility of
cancelled bids).303030In this setting, a large number of auction experiments
report overbidding relative to the risk-neutral equilibrium
$\beta(v)=(n-1)v/n$ (see Kagel and Levin, (1995) for a review). However, it is
easy to see the level-$0$ specification suggested here would predict bids that
are substantially lower than the risk-neutral equilibrium. As a result, it
would result in level-$k$ predictions that are far too low (at least, given a
plausible calibration of the levels). Note also that other apparently
plausible $L_{0}$ specifications, such as thinking that one’s opponents bid
half their value, are excluded on the grounds that $L_{0}$ reasoning is
supposed to be non-strategic (whereas bidding a fraction of one’s value quite
clearly reflects an attempt to strategically balance one’s chance of winning
with one’s payoff if one does win). Moreover, such specifications imply that
level-$k$ bidders should submit a large mass of bids just above the maximum
value bid by $L_{0}$ (for instance $L_{1}$ players in the all-pay auction
should bid $50$ for all $v\geq 51$). We do not observe such masses in this
experiment nor in the other auctions experiments of which we are aware.

Finally, we note that the level-$k$ predictions as formulated in Section 2 are
robust to a very wide range of level-0 specifications. For example, in the
all-pay auction, level-$k$ bids will be characterised by Proposition 2 for any
$L_{0}$ specification that satisfies

$F(b)\leq\bigg{(}\frac{b}{x+1}\bigg{)}^{\frac{1}{n-1}}$ (13)

where $F(b)$ is the probability that the level-$0$ type bids $b$ or lower
(i.e. the CDF of $b$). In particular, this means that Proposition (5) will
continue to hold provided that the CDF $F(b)$ that is bounded by the uniform
CDF from above. Similarly, in the first-price auction, level-$1$ players will
bid 0 (as in Proposition 5) for any level-$0$ specification that satisfies

$F(b)\leq\left(\frac{n-1}{n}\right)\left(\left(\frac{x}{x-b}\right)^{n-1}-1\right)$
(14)

for all $b\in\mathbb{X}$.313131This inequality comes from checking that a
player with $v=x$ would rather bid $0$ than any $b\in\mathbb{X}$. Moreover,
while these inequalities are sufficient for our level-$k$ characterisation to
hold, they are absolutely not necessary. For example, in the all-pay auction,
the level-1 bidding strategy turns out to be a best response to the empirical
bid distribution observed in the experiment, even though that distribution
does not satisfy inequality (13).

In light of the reasons above, we think that it is unlikely that one can find
a level-0 specification that is simultaneously non-strategic, psychologically
plausible and generates predictions that match the data in both this and other
auction experiments. Ultimately though, this conclusion is somewhat more
speculative than the central conclusion of this study: that the level-$k$
model as specified by Crawford and Iriberri, (2007) cannot predict or
rationalise the observed bidding behaviour.

##  6 Concluding remarks

In this paper, we have designed and implemented an experiment aimed at testing
the level-$k$ model of auctions. Overall, the evidence would appear to
strongly reject the model. When plausibly calibrated, the model produces bids
that are at least an order of magnitude too low. Moreover, fitting the model
without such a plausibility constraint results in estimated levels in the
30–35 range; and these levels in turn fail to bear any relation to levels
inferred from the 11-20 game. In addition, changing the bid discretisation has
little effect on the observed bids (in contrast to the model’s predictions);
and subjects almost never cite iterated reasoning when asked to explain why
they bid in the way they did. Thus, while one might be able to disagree with
any of the pieces of evidence separately, it seems that the evidence
collectively casts strong doubt on the model’s ability to predict or even
rationalise the observed data.

This finding might be seen as surprising given that the experimental
environment was carefully chosen to be the level-$k$ model’s natural domain of
applicability. Individual feedback was severely restricted and (following
Crawford and Iriberri, (2007)) the number of rounds was restricted to just
two. However, the level-$k$ model fared poorly despite the initial play
setting.

The findings also might be seen as surprising in light of evidence that the
level-$k$ model very accurately captures facets of human behaviour in other
settings, whether ‘beauty contests’ (Nagel,, 1995), co-ordination games
(Costa-Gomes et al.,, 2009), strategic communication (Crawford,, 2003; Cai and
Wang,, 2006) or zero sum betting (Brocas et al.,, 2014). We thus close with
some speculations as to why the model appears to do so badly at predicting
bids in auctions despite its successes in predicting behaviour in other areas.

First, recall that the level-$k$ model requires a level-0 anchor, i.e. some
sort of naive starting point that forms the basis of the iterated reasoning.
In many settings, such an anchor is easy to provide. For example, one’s
natural first thought (in a strategic communication setting) is that one’s
opponents will tell the truth; and a natural first thought (in the context of
guessing games) is that all possible guesses are equally likely. In contrast,
it is not at all clear how the level-0 ought to be specified in the setting of
auctions. Crawford and Iriberri, (2007) select what appear to be the two best
proposals – that subjects either bid their valuation or randomise uniformly
over the strategy space. However, since both of these proposals involve
dominated bids, it is not obvious whether either represents a natural first
thought for how one’s opponents will play. More to the point, it does not seem
that there exists more plausible L0 specifications than those proposed by
Crawford and Iriberri, (2007) (see section 5 for discussion). Thus, the
failure of the level-$k$ model in this context may well be down to the lack of
a salient and psychologically plausible anchor.

Second, iterated reasoning is cognitively much more taxing in Bayesian games
like auctions than the simpler settings to which the level-$k$ model is
normally applied. To calculate one’s optimal action, one must combine one’s
conjecture about one’s opponent’s strategy with their distribution over types
to arrive at their distribution over actions. One must then solve an (often
non-trivial) optimisation problem to find the optimal action given this
distribution. Thus, even if a plausible level 0 anchor were available, it is
unclear whether individuals would be able to correctly solve for the L1
strategy – and even less clear whether they would be able to correctly
identify the higher order strategies.

If this is right, it follows that the level-$k$ model should be applied only
in situations which possess both an intuitively appealing ‘first thought’
along with easily computable best response dynamics. Rigorously testing
whether these features are indeed necessary for the predictive success of the
model – along with identifying strategic settings which possess these
apparently necessary features – would seem to be an important task for future
research.

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##  Appendix A Proofs

Proof of Proposition 1. Since our game has a symmetric and finite normal form,
it must have a symmetric equilibrium (see Harsanyi, (1967) and Cheng et al.,
(2004) for details). Turning to uniqueness, we now generalise three lemmas
from Rasooly and Gavidia-Calderon, (2020) to the case of mixed strategies. Let
$\sigma$ denote an arbitrary symmetric equilibrium strategy; and for any
$v\in\mathbb{X}$, let $s(v)\subseteq\mathbb{X}$ denote the set of bids that
are submitted with positive probability at value $v$ (i.e. the support at
$v$).

###### Lemma 1

In any symmetric equilibrium $\sigma$, $s(0)=\\{0\\}$.

Proof. Consider a player with $v=0$. If they bid $0$, they get (regardless of
their opponents’ strategies) a payoff
$\pi(v=0,b=0)=0\times\mathbb{P\text{(win}}|b=0)-0=0$. On the other hand, if
they submit any bid $b\geq 1$, they get (given any opponent strategy profile)
$\pi(v=0,b)=0\times\mathbb{P\text{(win}}|b)-b=-b<0$. Thus, bidding $0$
strictly dominates bidding any $b\geq 1$; which means that
$\mathbb{P}(b=0|v=0)=1$ in any equilibrium (and in any symmetric equilibrium).
$\square$

######  Lemma 2 (Monotonicity)

Consider any symmetric equilibrium $\sigma$ and consider any two values
$v,v^{\prime}\in\mathbb{X}$ with $v>v^{\prime}$. Then for any $b\in s(v)$ and
$b^{\prime}\in s(v^{\prime})$, $b\geq b^{\prime}$.

Proof. For contradiction, let us suppose that $b<b^{\prime}$. Since $b$ is
submitted with positive probability when the value is $v$, bidding $b$ must be
weakly optimal. In particular, it must be weakly better than bidding
$b^{\prime}$:

$\begin{split}&\pi(v,b)\geq\pi(v,b^{\prime})\\\
\iff&v\mathbb{P\text{(win}}|b)-b\geq
v\mathbb{P\text{(win}}|b^{\prime})-b^{\prime}\\\ \iff&b^{\prime}-b\geq
v\left(\mathbb{P\text{(win}}|b^{\prime})-\mathbb{P\text{(win}}|b)\right).\end{split}$
(15)

Similarly, since $b^{\prime}$ is submitted with positive probability when the
value is $v^{\prime}$, it must be weakly better than $b$:

$\begin{split}&\pi(v^{\prime},b^{\prime})\geq\pi(v^{\prime},b)\\\
\iff&v^{\prime}\mathbb{P\text{(win}}|b^{\prime})-b^{\prime}\geq
v^{\prime}\mathbb{P\text{(win}}|b)-b\\\
\iff&v^{\prime}\left(\mathbb{P\text{(win}}|b^{\prime})-\mathbb{P\text{(win}}|b)\right)\geq
b^{\prime}-b.\end{split}$ (16)

Inequalities (15) and (16) jointly imply that

$v^{\prime}\left(\mathbb{P\text{(win}}|b^{\prime})-\mathbb{P\text{(win}}|b)\right)\geq
v\left(\mathbb{P\text{(win}}|b^{\prime})-\mathbb{P\text{(win}}|b)\right).$
(17)

Since $b^{\prime}>b$, and $b$ is a symmetric equilibrium bid, it must be that
$\mathbb{P\text{(win}}|b^{\prime})>\mathbb{P\text{(win}}|b)$ (for example, one
wins with a bid of $b^{\prime}$, but not with a bid of $b$, if all of one’s
opponents bid $b$ – and this happens with positive probability). Since
$\mathbb{P\text{(win}}|b^{\prime})-\mathbb{P\text{(win}}|b)>0$, inequality 17
then yields $v^{\prime}\geq v$, which contradicts our initial assumption that
$v^{\prime}<v$. This establishes that $b\geq b^{\prime}$ as claimed. $\square$

######  Lemma 3 (No gaps)

In any symmetric equilibrium $\sigma$, the bids that are submitted with
positive probability are a set of consecutive integers.

Proof. Suppose for contradiction that this were false. Then there must exist
two bids $b_{high}\geq b_{low}+2$ that are submitted with positive probability
even though no bid in between $b_{high}$ and $b_{low}$ is submitted with
positive probability. Now if types were to deviate from bidding $b_{high}$ to
bidding $b_{low}+1$, their payment would fall. However, their probability of
winning would remain unchanged: both before and after the deviation, they win
the auction if and only if all opponents bid $b_{low}$ or lower. So this
deviation would be strictly profitable, implying that this could not have been
an SE. $\square$

We now introduce the concept of jump form.

###### Definition A.1

For every bid $i\in B^{\sigma}$, we define the $i$th jump $j_{i}$ as

$j_{i}=v_{i}+\mathbb{P}(b<i|v_{i})$

where $v_{i}$ is the minimum value $v\in\mathbb{X}$ such that
$\mathbb{P}(b=i|v)>0$. We refer to the vector of jumps
$j=(j_{0},j_{1},...,j_{m})$ as a jump vector and say that it is increasing if
$j_{i+1}>j_{i}$ for every $i\in\\{0,1,...,m-1\\}$.

Under the lemmas outlined previously, this is an equivalent representation of
behavioural strategies.

###### Lemma 4

There is a bijection between the set of gapless and monotone strategies that
satisfy $s(0)=0$ and the set of increasing jump vectors.

Proof. To establish the first claim, take any gapless and monotone strategy
that specifies $\mathbb{P}(b=0|v=0)=1$ and convert it into a jump vector $j$.
To show that $j$ is increasing, consider any two consecutive jump points
$j_{i},j_{i+1}\in j$. If the strategy is monotone, every value that submits a
bid $b=i+1$ must be weakly larger than the every value that submits a bid
$b=i$. So in particular, $v_{i+1}\geq v_{i}$. This yields two possibilities:
either $v_{i+1}>v_{i}$ or $v_{i+1}=v_{i}$. In the first instance
($v_{i+1}>v_{i}$),

$\begin{split}j_{i+1}&\equiv v_{i+1}+\mathbb{P}(b<i+1|v_{i+1})\\\ &\geq
v_{i}+1+\mathbb{P}(b<i+1|v_{i+1})\\\ &\geq v_{i}+1\\\
&>v_{i}+\mathbb{P}(b<i|v_{i})\equiv j_{i}\end{split}$ (18)

where the final inequality holds since $\mathbb{P}(b<i|v_{i})<1$ (which in
turn holds since necessarily $\mathbb{P}(b=i|v_{i})>0$). So in that case,
$j_{i+1}>j_{i}$. In the second instance ($v_{i+1}=v_{i}$), we have

$\begin{split}j_{i+1}&\equiv v_{i+1}+\mathbb{P}(b<i+1|v_{i+1})\\\
&=v_{i}+\mathbb{P}(b<i+1|v_{i})\\\
&=v_{i}+\mathbb{P}(b<i|v_{i})+\mathbb{P}(b=i|v_{i})\\\
&>v_{i}+\mathbb{P}(b<i|v_{i})\equiv j_{i}\end{split}$ (19)

where the inequality again holds since $\mathbb{P}(b=i|v_{i})>0$. Either way,
then, we see that monotonicity implies that $j_{1+1}>j_{i}$. Since this true
for all $i=2,3,...,m$, this then implies that that vector of jumps is
increasing. So every monotone behavioural strategy generates an increasing
vector of jumps.

We now argue that any increasing vector of jumps $j=(j_{1},...,j_{m})$ is
consistent with a unique monotone and gapless behavioural strategy that
satisfies $\mathbb{P}(b=0|v=0)$. First, use $j$ to compute the minimum values
that submit each of the possible bids $v_{i}$ using the fact that
$v_{i}=\lfloor j_{i}\rfloor$ for all $i=1,...,m$. (Notice that, since
$j_{i+1}>j_{i}$ for every $i$, $v_{i+1}=\lfloor j_{i+1}\rfloor\geq
v_{i}=\lfloor j_{i}\rfloor$, i.e. the $v_{i}$ are weakly increasing.) We will
begin by showing how this uniquely determines the behavioural strategy for all
values other than the $v_{i}$.

First, consider all $v\in\mathbb{X}$ with $v<v_{1}$. Recall that $v_{1}$ is
the minimum $v\in\mathbb{X}$ such that $\mathbb{P}(b=1|v)>0$. So
$\mathbb{P}(b=1|v_{1})>0$; and $\mathbb{P}(b=1|v)=0$ for any value $v<v_{1}$.
Since $\mathbb{P}(b=1|v_{1})>0$, monotonicity implies that
$\mathbb{P}(b>1|v)=0$ for any $v<v_{1}$. Since $\mathbb{P}(b=1|v)=0$ and
$\mathbb{P}(b>1|v)=0$ for all $v<v_{1}$, $\mathbb{P}(b=0|v)=1$ for all
$v<v_{1}$, i.e. all such values bid $0$ with probability $1$.

Next, consider now all $v\in\mathbb{X}$ with $v>v_{m}$ (where $m$ is the
largest bid submitted). By definition, $v_{m}$ is the minimum value $v$ such
that $\mathbb{P}(b=m|v)>0$. So $\mathbb{P}(b=m|v_{m})>0$. By monotonicity,
this means that $\mathbb{P}(b\geq m|v)=1$ for all $v>v_{m}$. But $m$ is the
largest bid submitted with positive probability. Hence, $\mathbb{P}(b=m|v)=1$
for all $v>v_{m}$, i.e. all such values bid $m$ with probability $1$.

Finally, consider all remaining values $v$ such that $v\neq v_{i}$ for all
$i=1,...,m$. For any such value, we can find a bid $i$ such that
$v_{i}<v<v_{i+1}$. Using the previous two arguments, one can show that
$\mathbb{P}(b\geq i|v)=1$ and $\mathbb{P}(b\leq i|v)=1$. But then
$\mathbb{P}(b=i|v)=1$, i.e. all such values bid $i$ with probability $1$.

It remains to consider the values $v\in\mathbb{X}$ such that $v=v_{i}$ for
some $i=1,2,...,m$. Consider then any such value and let $B^{v}$ denote the
set of bids $i$ such that $v_{i}=v$. Since the $v_{i}$ are weakly increasing
(in the bid $i$), $B^{v}$ must be a set of consecutive integers (or a set
containing a single integer). Label these integers
$b^{\prime},b^{\prime}+1,...,b^{\prime}+k$. For all such bids, $v$ is the
minimum value such that $\mathbb{P}(b|v)>0$. So all such bids must belong to
the support of the strategy at value $v$.

We now argue that $\mathbb{P}(b\geq b^{\prime}-1|v)=1$, i.e. bids strictly
lower than $b^{\prime}-1$ cannot belong to the support. To see this, note
that, since the $v_{i}$ are increasing, $v_{b^{\prime}}\geq v_{b^{\prime}-1}$.
However, $v_{b^{\prime}}=v\neq v_{b^{\prime}-1}$: otherwise, the bid
$b^{\prime}-1$ would belong to $B^{v}$. Hence,
$v_{b^{\prime}}=v>v_{b^{\prime}-1}$. So the minimum value that bids
$b^{\prime}-1$ is strictly smaller than $v$. By monotonicity, this means that
the value $v$ must bid at least $b^{\prime}-1$ with probability $1$.

Next, we argue that $\mathbb{P}(b>b^{\prime}+k|v)=0$. The argument is similar.
Since the $v_{i}$ are increasing, $v_{b^{\prime\prime}}\geq v_{b^{\prime}+k}$
for any higher bid $b^{\prime\prime}>b^{\prime}+k$. However,
$v_{b^{\prime\prime}}\neq v_{b^{\prime}+k}=v$: otherwise, the bid $b^{\prime}$
would be in $B^{v}$. So $v_{b^{\prime\prime}}>v_{b^{\prime}+k}$. So the
smallest value that bids $b^{\prime\prime}$ is strictly larger than $v$. As a
result, $\mathbb{P}(b=b^{\prime\prime}|v)=0$.

Given the previous arguments, we know that the support at $v$ must contain
$B^{v}=\\{b^{\prime},b^{\prime}+1,...,b^{\prime}+k\\}$. In addition, the only
other bid that may belong to the support is $b^{\prime}-1$. So, for every bid
$i\in B^{v}$, we can write

$\begin{split}j_{i}&=v_{i}+\mathbb{P}(b<i|v)=v+\sum_{j=b^{\prime}-1}^{i-1}\mathbb{P}(b=j|v)\\\
\end{split}$ (20)

Thus, we have the system of equations

$\begin{split}j_{b^{\prime}}-v&=\mathbb{P}(b=b^{\prime}-1|v)\\\
j_{b^{\prime}+1}-v&=\mathbb{P}(b=b^{\prime}-1|v)+\mathbb{P}(b^{\prime}|v)\\\
.\\\ .\\\
j_{b^{\prime}+k}-v&=\mathbb{P}(b=b^{\prime}-1|v)+\mathbb{P}(b=b^{\prime}|v)+...+\mathbb{P}(b=b^{\prime}+k-1)\end{split}$
(21)

Given the vector $j$, the first equation uniquely determines
$\mathbb{P}(b=b^{\prime}-1|v)$ (which may equal zero). Using the value
obtained for $\mathbb{P}(b=b^{\prime}-1|v)$, the second equation uniquely
determines $\mathbb{P}(b=b^{\prime\prime}-1|v)$. Continuing in this manner,
the system of equations uniquely pins down $\mathbb{P}(b=i|v)$ for
$i=b^{\prime}-1,b^{\prime},...,b^{\prime}+k-1$. Given the previous arguments,
the only other bid in the support is $b=b^{\prime}+k$ (the maximum bid). This
is then also uniquely determined using

$\mathbb{P}(b=b^{\prime}+k|v)=1-\sum_{i=b^{\prime\prime}-1}^{b^{\prime}-1}\mathbb{P}(b=i|v)$
(22)

The argument above shows how a vector of jumps uniquely pins down a
probability distribution over bids for any value $v\in\mathbb{X}$. Repeating
the argument for every such value, we see that it is consistent with a unique
behavioural strategy. $\square$

Having recast strategies in jump form, we can now define our algorithm:

###### Definition A.2

Let $\hat{j}=(\hat{j_{1}},...\hat{j}_{m})$ denote the output of the following
algorithm:

  1. 1.

Impose $j_{0}$ = 0.

  2. 2.

Starting at $i=1$, find the minimum $j_{i}\in(j_{i-1},S]$ such that
$\pi(v=\lfloor j_{i}\rfloor,b=i)-\pi(v=\lfloor j_{i}\rfloor,b=i-1)\geq 0$.

  3. 3.

If there is no such $j_{i}$, the algorithm terminates.

  4. 4.

There there does exist such a $j_{i}$, repeat (from step 2) for $i+1$.

We then observe the following.

###### Lemma 5

The vector $\hat{j}$ is the only possible symmetric equilibrium of the all-pay
auction.

Proof. Suppose for contradiction that there were some symmetric equilibrium
$j\neq\hat{j}$. If $j\neq\hat{j}$, there must be some jump at which they
differ. Let $j_{i}$ denote the first such jump (i.e. $i$ is the smallest
number such that $j_{i}\neq\hat{j_{i}})$. There are two possibilities:
$j_{i}<\hat{j_{i}}$ or instead $j_{i}>\hat{j_{i}}$. We will argue that each
leads to a contradiction.

To see why the first case is impossible, first define $\pi^{\hat{j}}(v,b)$ and
$\pi^{j}(v,b)$ as the payoff of a player with a value $v$ who bids $b$ given
that their opponents all bid according to $\hat{j}$ and $j$ respectively. Now
recall that, by construction, $\hat{j_{i}}$ is the minimum jump such that

$\pi^{\hat{j}}(v=\lfloor\hat{j_{i}}\rfloor,b=i)\geq\pi^{\hat{j}}(v=\lfloor\hat{j_{i}}\rfloor,b=i-1)$
(23)

Since $j_{i}<\hat{j_{i}}$, this means that

$\pi^{j}(v=\lfloor j_{i}\rfloor,b=i)<\pi^{j}(v=\lfloor j_{i}\rfloor,b=i-1)$
(24)

But then, given that one’s opponents bid according to $j$, one should never
bid $b=i$ with a value $v=\lfloor j_{i}\rfloor$. That is,
$\mathbb{P}(b=i|v=\lfloor j_{i}\rfloor)=0$. But then $v=\lfloor j_{i}\rfloor$
cannot be the smallest value that bids $i$ (contradiction).

To see why the second case is impossible, recall again that every
$\hat{j_{i}}$ satisfies

$\begin{split}\pi^{\hat{j}}(v=\lfloor\hat{j_{i}}\rfloor,b=i)&\geq\pi^{\hat{j}}(v=\lfloor\hat{j_{i}}\rfloor,b=i-1)\iff\\\
\lfloor\hat{j_{i}}\rfloor\mathbb{P}^{\hat{j}}(\text{win}|b=i)-1&\geq\lfloor\hat{j}_{i}\rfloor\mathbb{P}^{\hat{j}}(\text{win}|b=i-1)\end{split}$
(25)

As a preliminary, note that since the RHS is non-negative,
$\lfloor\hat{j_{i}}\rfloor\mathbb{P}^{\hat{j}}(\text{win}|b=i)-1\geq 0$ and so
$\lfloor\hat{j_{i}}\rfloor\geq 1$ for every $i$. Let us record this fact.

Turning to the main argument, fix the value at $\hat{j_{i}}$ but consider now
changing opponent bidding from $\hat{j}$ to $j$. If $j_{i}>\hat{j_{i}}$, then
$\mathbb{P}^{j}(\text{win}|b=i)>\mathbb{P}^{\hat{j}}(\text{win}|b=i)$.
However, since $j$ and $\hat{j}$ agree for all previous jumps,
$\mathbb{P}^{j}(\text{win}|b=i-1)=\mathbb{P}^{\hat{j}}(\text{win}|b=i-1)$.
Since $\hat{\lfloor j_{i}\rfloor}\geq 1$ (established earlier), this all means
that

$\begin{split}\lfloor\hat{j_{i}}\rfloor\mathbb{P}^{j}(\text{win}|b=i)-1&>\lfloor\hat{j}_{i}\rfloor\mathbb{P}^{j}(\text{win}|b=i-1)\iff\\\
\pi^{j}(v=\lfloor\hat{j_{i}}\rfloor,b=i)&>\pi^{j}(v=\lfloor\hat{j_{i}}\rfloor,b=i-1)\end{split}$
(26)

If $j$ is an SE, this implies that
$\mathbb{P}^{j}(b=i-1|v=\lfloor\hat{j_{i}}\rfloor)=0$. However, this is
impossible: since $j_{i}>\hat{j}_{i}$,
$\mathbb{P}^{j}(b=i-1|v=\lfloor\hat{j_{i}}\rfloor)=\min\\{j_{i},\lfloor\hat{j}_{i}\rfloor+1\\}-\max\\{\hat{j}_{i-1},\lfloor\hat{j}_{i}\rfloor\\}>\hat{j}_{i}-\max\\{\hat{j}_{i-1},\lfloor\hat{j}_{i}\rfloor\\}=\mathbb{P}^{\hat{j}}(b=i-1|v=\lfloor\hat{j_{i}}\rfloor)\geq
0$ and so $\mathbb{P}^{j}(b=i-1|v=\lfloor\hat{j_{i}}\rfloor)>0$. $\square$

By the argument above, there is no possible symmetric equilibrium except for
$\hat{j}$. At we noted earlier, however, our game must possess some symmetric
equilibrium. From this, it follows that it has exactly one symmetric
equilibrium (namely, $\hat{j}$). This concludes the proof. $\square$

Proof of Proposition 2. See main text. $\square$

Proof of Proposition 3. A proof of this result can be easily reconstructed
from the proof of Proposition 1 (and is available from the author upon
request). $\square$

Proof of Proposition 4. We argue by induction. First, we show that the desired
pattern of bidding holds when $k=1$. Next, we show that if it holds for any
level $k-1\in\mathbb{K}$, then it also holds (provided that $x$ is
sufficiently large) for any level $k\in\mathbb{K}$.

Starting with the base case ($k=1$), note that level-$1$ bidders choose a bid
$b\in\mathbb{X}$ to maximise

$\pi^{1}(v,b)=(v-b)\left(\frac{n-1}{n}+\frac{b}{n(x+1)}\right)^{n-1}$ (27)

(inserting $p=1/n$ into 5). Considering this function’s continuous extension,
observe that

$\frac{\partial\ln[\pi^{1}(v,b)]}{\partial
b}=\frac{(n-1)(v-(x+1))-nb}{(v-b)(b+(n-1)(x+1))}$ (28)

Since are searching for the optimal bid, we can just consider the range $b\leq
v$. In this range, the denominator of (28) is non-negative. Moreover, since
$v\leq x$ for all $v\in\mathbb{X}$, the numerator is negative. Thus,
$\partial\ln[\pi^{1}(v,b)]/\partial b\leq 0$ and so
$\partial\pi^{1}(v,b)/\partial b\leq 0$. Since this is true of the function’s
continuous extension, it follows that $\pi(v,0)\geq\pi(v,b)$ for every value
$v\in\mathbb{X}$ and every positive integer bid $b\in\mathbb{X}$. That is,
bidding zero is optimal (as claimed).

Next, we show that, if the desired pattern of bidding holds for level
$k-1\in\mathbb{K}$, then it also holds (provided that $x$ is sufficiently
large) for level $k\in\mathbb{K}$. Suppose then that all players of level
$k-1$ bid as proposed. Then they bid $0$ when $v\leq v^{*}(k-1)$, i.e. at
$\lfloor v^{*}(k-1)+1\rfloor$ distinct values; and otherwise, they bid $k-2$.
Hence, for a level-$k$ player,

$\mathbb{P}^{k}(\text{win}|b)=\begin{cases}1&\text{if}\hskip 5.0ptb>k-2\\\
\left(1-p+p\left(\frac{\lfloor
v^{*}(k-1)+1\rfloor}{x+1}\right)\right)^{n-1}&\text{if}\hskip
5.0ptb\in\\{1,...,k-2\\}\\\ (1-p)^{n-1}&\text{if}\hskip 5.0ptb=0\\\
\end{cases}$ (29)

Given this, it is immediate that $b^{*}\in\\{0,1,k-1\\}$, i.e. there are at
most three optimal bids. Moreover, $\pi^{k}(v,b=0)-\pi^{k}(v,b=1)\rightarrow
v(1-p)^{n-1}-(v-1)(1-p)^{n-1}<0$ as $x\rightarrow\infty$, and so
$b^{*}\in\\{0,k-1\\}$ if $x$ is sufficiently large. Finally, observe that the
latter is optimal iff

$\begin{split}\pi(v,b=k-1)>\pi(v,b=0)&\iff v-(k-1)>v(1-p)^{n-1}\\\ &\iff
v>\frac{k-1}{1-(1-p)^{n-1}}\end{split}$ (30)

i.e. iff $v>v^{*}(k)$ (as claimed). $\square$

##  Appendix B Illustrating the algorithm

In this section, we illustrate the algorithm using two examples. In both,
values are uniform so (in any symmetric strategy profile)

$\begin{split}\mathbb{P}(b<i)&=\mathbb{P}(v<\lfloor
j_{i}\rfloor)+\mathbb{P}(v=\lfloor j_{i}\rfloor)(j_{i}-\lfloor
j_{i}\rfloor)\\\ &=\frac{\lfloor j_{i}\rfloor}{S}+\frac{1}{S}(j_{i}-\lfloor
j_{i}\rfloor)\\\ &=\frac{j_{i}}{S}\end{split}$ (31)

where $S$ is the number of possible valuations.

Example 1. Consider an all-pay auction with uniform values and $n=2$. To find
the first jump point, we look for the minimum $j_{1}\in[2,S]$ such that

$\pi(v=\lfloor j_{1}\rfloor,b=1)-\pi(v=\lfloor j_{1}\rfloor,b=0)\geq 0$ (32)

Clearly, $\pi(v=\lfloor j_{1}\rfloor,b=0)=0$ since
$\mathbb{P\text{(win}}|b=0)=0$. In addition,
$\mathbb{P\text{(win}}|b=1)=\mathbb{P}(b=0)=j_{1}/S$ and so $\pi(v=\lfloor
j_{1}\rfloor,b=1)=\lfloor j_{1}\rfloor j_{1}/S-1$. So we look for the minimum
$j_{1}$ such that

$\frac{\lfloor j_{1}\rfloor j_{1}}{S}-1\geq 0$ (33)

Of course, the solution will depend on $S$. If $S=100$, $j_{1}=10$ (and
similarly $j_{1}=\sqrt{S}$ for any square $S$). If $S=99$, the solution
remains at $j_{1}=10$. However, if $S=101$, the solution becomes $j_{1}=10.1$,
implying randomisation. (One can prove that, as $S\rightarrow\infty$, the
proportion of solutions that are integer solutions converges to $1/2$.) Let us
suppose that $S=101$ so that the first jump is $j_{1}=10.1$.

To find the next jump, we look for the smallest $j_{2}$ that satisfies

$\begin{split}&\pi(v=\lfloor j_{2}\rfloor,b=2)-\pi(v=\lfloor
j_{2}\rfloor,b=1)\geq 0\\\ \iff&\lfloor
j_{2}\rfloor\left(\mathbb{P\text{(win}}|b=2)-\mathbb{P\text{(win}}|b=1)\right)\geq
1\\\ \iff&\lfloor j_{2}\rfloor\left(\mathbb{P}(b\leq
1)-\mathbb{P}(b=0)\right)\geq 1\\\ \iff&\lfloor
j_{2}\rfloor\left(\frac{j_{2}}{101}-\frac{10.1}{101}\right)\geq 1\\\
\end{split}$ (34)

One can check that the solution is $j_{2}=16.4125$, so the equilibrium vector
of jumps starts with $(0,10.1,16.4125,...)$. Returning to behavioural
strategies, this means that all values $v\in\\{0,1,...,9\\}$ bid $0$ (with
probability $1$), value $v=10$ randomises between $b=0$ and $b=1$ with
probabilities $0.1$ and $0.9$, values $v\in\\{11,...,15\\}$ bid $1$ (with
probability $1$), and finally the value $v=16$ bids $b=1$ with probability
$0.4125$. (Since we haven’t computed $j_{3}$, we cannot technically determine
the probability with which value $v=16$ bids $v=2$; but in fact $j_{3}\geq 17$
so this probability is $1-0.4125$.)

Example 2. Consider now the first price auction with cancelled bids. To find
the first jump $j_{1}$, we look for the smallest $j_{1}\in(0,S]$ such that

$\begin{split}&\pi(v=\lfloor j_{1}\rfloor,b=1)\geq\pi(v=\lfloor
j_{1}\rfloor,b=0)\\\ \iff&(\lfloor
j_{1}\rfloor-1)\mathbb{P\text{(win}}|b=1)\geq\lfloor
j_{1}\rfloor\mathbb{P\text{(win}}|b=0)\\\ \iff&(\lfloor
j_{1}\rfloor-1)(1-p+p\frac{j_{1}}{S})^{n-1}p\geq\lfloor
j_{1}\rfloor(1-p)^{n-1}p\\\ \iff&(\lfloor
j_{1}\rfloor-1)(1-p+p\frac{j_{1}}{S})^{n-1}\geq\lfloor
j_{1}\rfloor(1-p)^{n-1}\\\ \end{split}$ (35)

For instance, suppose that $p=1/2$, $n=2$ and $S=101$. Then (35) reduces to

$\lfloor j_{1}\rfloor j_{1}-101-j_{1}\geq 0$ (36)

which is almost quadratic in $j_{1}$. One can check that this inequality is
satisfied by $j_{1}=11$: then $\lfloor j_{1}\rfloor
j_{1}-101-j_{1}=11^{2}-101-11=9>0$. On the other hand, it won’t hold if
$j_{1}<11$: for then $\lfloor j_{1}\rfloor j_{1}-101-j_{1}\leq 10\times
11-101-11=-2<0$. So the minimum $j_{1}$ that satisfies the inequality is $11$;
this is the first jump. This means that all values smaller than $11$ must bid
zero (with probability $1$); but the value $v=11$ never bids zero with any
probability.

##  Appendix C Optimising $p$

To find the probability $p^{*}\in[0,1]$ that maximises $d(p)$, recall that
$p^{*}>0$ (from Proposition 5). Moreover, it is obvious that $p^{*}<1$: if
$p=1$, then $d(p)=0$ (whereas $d(p)>0$ is plainly possible). Hence,
$p^{*}\in(0,1)$, which means that the optimal $p$ must satisfy the first-order
condition

$\frac{\partial}{\partial
p}\int_{0}^{x}\left|\beta(v)-\beta^{1}(v)\right|dv=0.$ (37)

One can check that $\beta(v)\geq\beta^{1}(v)$ for all $v\in[0,x]$. So (37) is
equivalent to

$\frac{\partial}{\partial p}\int_{0}^{x}\beta(v)dv=\frac{\partial}{\partial
p}\int_{0}^{x}\beta^{1}(v)dv.$ (38)

When $n=2$, standard calculations reveal that

$\displaystyle\frac{\partial}{\partial p}\int_{0}^{x}\beta(v)dv$
$\displaystyle=\frac{\partial}{\partial
p}\int_{0}^{x}\frac{pv^{2}}{2(pv-p+1)}dv$
$\displaystyle=\frac{\partial}{\partial
p}\frac{x^{2}\left(p(3p-2)-2(p-1)^{2}\ln(1-p)\right)}{4p^{2}}$
$\displaystyle=\frac{x^{2}((2-p)p-2(p-1)\ln(1-p))}{2p^{3}}$ (39)

Similarly, when $n\geq 3$, we find that

$\displaystyle\frac{\partial}{\partial p}\int_{0}^{x}\beta(v)dv$
$\displaystyle=\frac{\partial}{\partial
p}\int_{0}^{x}\left(\frac{n-1}{n}\right)v-\frac{x(1-p)}{np}\left[1-\left(\frac{1-p}{1-p+p(v/x)}\right)^{n-1}\right]dv$
$\displaystyle=\frac{\partial}{\partial
p}\frac{x^{2}\left(2+np((n-1)p-2)\right)-2(1-p)^{n}}{2(n-2)np^{2}}$
$\displaystyle=\frac{x^{2}\left(p\left(n(1-p)^{n}-2(1-p)^{n}+n+2\right)+2\left((1-p)^{n}-1\right)-np^{2}\right)}{(n-2)n(p-1)p^{3}}$
(40)

Turning to the level-$1$ bidding function, we find that

$\displaystyle\frac{\partial}{\partial p}\int_{0}^{x}\beta^{1}(v)dv$
$\displaystyle=\frac{\partial}{\partial
p}\int_{v^{*}}^{x}\left(\frac{n-1}{n}\right)v-\left(\frac{1-p}{p}\right)\frac{x}{n}\hskip
3.00003ptdv$ $\displaystyle=\frac{\partial}{\partial
p}\frac{x^{2}(np-1)^{2}}{2(n-1)np^{2}}$
$\displaystyle=\frac{x^{2}(np-1)}{(n-1)np^{3}}$ (41)

where $v^{*}=(1-p)x/(n-1)p$.

To find the optimal probability when $n=2$, we now substitute (C) and (C) into
the first-order condition (38), obtaining

$\frac{(2-p)p-2(p-1)\ln(1-p)}{2p^{3}}=\frac{np-1}{(n-1)np^{3}}$ (42)

Similarly, when $n\geq 3$, we substitute (C) and (C) and into the first-order
condition, yielding

$\frac{p\left(n(1-p)^{n}-2(1-p)^{n}+n+2\right)+2\left((1-p)^{n}-1\right)-np^{2}}{(n-2)n(p-1)p^{3}}=\frac{np-1}{(n-1)np^{3}}$
(43)

Either way, we obtain an equation that can be solved numerically to reveal
candidates for $p^{*}$. It is then a straightforward matter to check which is
of these candidates generate a larger distance. Since our problem must have a
global maximiser (by the Weierstrass theorem), and the global maximiser must
satisfy the first-order condition, the $p$ satisfying the first-order
condition that generates the largest distance must be the global maximiser.

Table 2: Optimal cancellation probabilities (rounded) $n$ | 2 | 3 | 4 | 5 | 6 | 7 | 8
---|---|---|---|---|---|---|---
$p^{*}$ | 0.536 | 0.343 | 0.256 | 0.204 | 0.170 | 0.145 | 0.127

Table 2 outlines for the solutions for $n\in\\{1,...,8\\}$. As can be seen,
the solution is generally quite close to $1/n$ — so the bound identified in
Proposition 5 comes close to identifying the exact solution. Note also that,
as might be expected, the optimal probability $p$ does not depend on the scale
parameter $x$.

##  Appendix D Tables and figures

Figure 8: The first-price auction when bids are multiples of 5 Figure 9: The
all-pay auction when bids are multiples of 5 Table 3: Balance Table

| Integer bids | Multiples of five |  $p$-value
---|---|---|---
Mean age | 31.4 | 32.0 | 0.29
Share male | 0.54 | 0.43 | 0.33
Share maths | 0.39 | 0.38 | 0.95
Share humanities | 0.24 | 0.14 | 0.36
Share social-sciences | 0.30 | 0.29 | 0.93

  * •

Notes: This table displays how demographics (age, sex and subject) vary
between the two treatment groups. The column ‘$p$-value’ reports $t$-tests of
the hypothesis that the relevant variable is on average equal across the
groups.

Table 4: Average bids

| First-price | All-pay
---|---|---
| T1 | T2 |  $p$-value | T1 | T2 |  $p$-value
Average bid | 23.6 | 23.0 | 0.85 | 22.2 | 25.4 | 0.12
Average value | 50.9 | 49.6 | 0.27 | 49.4 | 50.8 | 0.89
Ratio | 0.46 | 0.46 |  | 0.45 | 0.50 |

  * •

Notes: This table reports the average bids across two treatments and auction
structures. The column ‘$p$-value’ reports $t$-tests of the hypothesis that
the relevant variable is on average equal across the groups.

Table 5: Comparing equilibrium and level-$k$

|  | T1FP | T1AP | T2FP | T2AP
---|---|---|---|---|---
BNE | LL | -4407.9 | -4332.7 | -1024.7 | -1042.2
| BIC | 8820.5 | 8670.2 | 2053.1 | 2088.1
L1 | LL | -4530.4 | -4501.2 | -1088.8 | -1128.6
| BIC | 9065.4 | 9007.1 | 2181.3 | 2260.9
L1-L2 | LL | -4530.4 | -4501.2 | -1085 | -1126.9
| BIC | 9070.1 | 9011.7 | 2177.4 | 2261.2
L1-L3 | LL | -4530.1 | -4501.2 | -1071.1 | -1120.7
| BIC | 9074.3 | 9016.4 | 2153.4 | 2252.6

  * •

Notes: This table reports the log-likelihoods and associated Bayes-Information
Criteria values for four different structural models. In the first model,
equilibrium is the only type. The subsequent models are populated by level-1
types, level-1 and 2 types, and level-1, 2 and 3 types respectively.

Table 6: The hybrid model

| T1FP | T1AP | T2FP | T2AP
---|---|---|---|---
| Subject | Round | Subject | Round | Subject | Round | Subject | Round
$p_{0}$ | 0.758 | 0.760 | 0.815 | 0.757 | 0.905 | 0.858 | 0.741 | 0.788
| (0.059) | (0.046) | (0.053) | (0.06) | (0.066) | (0.054) | (0.149) | (0.083)
$p_{1}$ | 0.036 | 0.046 | 0.000 | 0.010 | 0.095 | 0.044 | 0.048 | 0.071
| (0.025) | (0.02) | (0.0) | (0.01) | (0.066) | (0.034) | (0.048) | (0.04)
$p_{2}$ | 0.133 | 0.027 | 0.019 | 0.055 | 0.000 | 0.000 | 0.048 | 0.024
| (0.051) | (0.016) | (0.019) | (0.022) | (0.0) | (0.0) | (0.048) | (0.024)
$p_{3}$ | 0.073 | 0.167 | 0.167 | 0.178 | 0.000 | 0.098 | 0.164 | 0.117
| (0.041) | (0.041) | (0.051) | (0.057) | (0.0) | (0.048) | (0.141) | (0.073)
$\sigma_{0}$ | 27.3 | 27.3 | 19.9 | 20.2 | 23.2 | 23.7 | 21.7 | 21.3
| (1.9) | (1.8) | (1.6) | (1.5) | (3.0) | (2.4) | (2.8) | (2.0)
$\sigma_{1}$ | 3.8 | 1.8 | - | 2.3 | 5.4 | 6.3 | 2.7 | 2.4
| (2.0) | (0.3) | - | (0.4) | (2.2) | (0.7) | (0.0) | (0.4)
$\sigma_{2}$ | 9.2 | 4.2 | 1.1 | 1.0 | - | - | 2.7 | 1.0
| (0.6) | (0.5) | (0.0) | (0.0) | - | - | (0.0) | (0.0)
$\sigma_{3}$ | 13.3 | 10.5 | 5.3 | 7.9 | - | 4.1 | 14.5 | 6.5
| (1.1) | (1.2) | (0.6) | (2.1) | - | (0.9) | (7.0) | (4.1)
LL | -4241.5 | -4217.1 | -4043.8 | -4004.1 | -987.9 | -982.6 | -962.9 | -940.9

  * •

Notes: The ‘subject’ columns report analyses in which subjects’ levels are
fixed; the ‘round’ columns report analyses in which levels are allowed to vary
between rounds. The symbols $p_{i}$ and $\sigma_{i}$ respectively denote the
probability and noise parameter of type $i\in\mathbb{K}$ (equilibrium is
indexed by $i=0$). LL abbreviates ‘log-likelihood’ and jack-knife standard
errors are in parentheses. When $p_{i}=0$, the associated parameter
$\sigma_{i}$ is unidentified and therefore omitted from the table.

Figure 10: Comparing estimated levels

Notes: The first panel plots levels as estimated from the first-price auction
against levels as inferred from the 11-20 game. The estimated correlation
between these variables is 0 (95% confidence interval: -0.37 to 0.37). The
second panel plots levels as estimated from the all-pay auction against levels
as inferred from the 11-20 game. The estimated correlation between these
variables is -0.09 (95% confidence interval: -0.44 to 0.30).

Table 7: Robustness checks

|  | T1FP | T1AP | T2FP | T2AP
---|---|---|---|---|---
Risk aversion | BNE | 19.4 | - | 17.9 | -
Level-$k$ | 26.9 | - | 22.7 | -
Cognitive hierarchy | BNE | 21.7 | 18.2 | 19.7 | 18.8
Level-$k$ | 31.3 | 23.8 | 26.6 | 29.6
Dominated bids | BNE | 22.6 | 19.4 | 19.5 | 19.0
Level-$k$ | 30.8 | 29.9 | 22.7 | 26.9
Tie-breaking | BNE | 21.7 | 18.2 | 19.7 | 18.8
Level-$k$ | 30.3 | 29.5 | 23.4 | 26.6
Dropping round 2 | BNE | 18.4 | 18.2 | 18.3 | 17.4
Level-$k$ | 26.5 | 29.0 | 22.2 | 24.0

  * •

Notes: This table reports the root-mean-square prediction errors of the
relevant theories following a series of robustness checks. In every case,
level-$k$ predictions are obtained by assigning each datapoint the level from
the $1-3$ range that minimises the model’s prediction error (a procedure that
will tend to overstate the model’s predictive performance).

Figure 11: Risk aversion (multiples of 5 treatment)