Discussion Paper No. 874

THE MINIMUM APPROVAL MECHANISM
IMPLEMENTS
THE EFFICIENT PUBLIC GOOD ALLOCATIONTHEORETICALLY AND EXPERIMENTALLY

Takehito Masuda
Yoshitaka Okano
Tatsuyoshi Saijo

May 2013
Revised September 2013

The Institute of Social and Economic Research
Osaka University
6-1 Mihogaoka, Ibaraki, Osaka 567-0047, Japan

The Minimum Approval Mechanism Implements the Efficient Public Good Allocation
Theoretically and Experimentally
September 2013
Takehito Masuda a,b,*, Yoshitaka Okano c, and Tatsuyoshi Saijo c, d
a Graduate School of Economics, Osaka University 1-7, Machikaneyama, Toyonaka, Osaka 560-0043, Japanb Research Fellow of the Japan Society for the Promotion of Science, Kojimachi Business Center Building,

5-3-1 Kojimachi, Chiyoda, Tokyo 102-0083, Japanc Research Center for Social Design Engineering, Kochi University of Technology, Tosayamada, Kami, Kochi

782-8502, Japand Center for Environmental Innovation Design for Sustainability, Osaka University, 2-1 Yamada-oka, Suita, Osaka

565-0871, Japan

Abstract
We propose the minimum approval mechanism (MAM) for a standard linear public goodenvironment with two players. Players simultaneously and privately choose theircontributions to the public good in the first stage. In the second stage, they simultaneously
decide whether to approve the other’s choice. Both contribute what they choose in the firststage if both players approve; otherwise, both contribute the minimum of the two choicesin the first stage. The MAM multiply implements the Pareto-efficient allocation in
backward elimination of weakly dominated strategies (BEWDS), limit logit agent quantal
response equilibrium, subgame perfect minimax regret equilibrium, level-k thinking, anddiagonalization heuristics. Moreover, the MAM is unique under plausible conditions.
Overall, contributions in the MAM experiment averaged 94.9%. Quantifying subjects’
responses to the questionnaire revealed the heterogeneity of reasoning processes to beconsistent with the model.
JEL Classification: C72; C92; D74; H41; P43
Keywords: Public good experiment; Approval mechanism; Multiple implementation

Corresponding author. Postal address: Research Center for Social Design Engineering, Kochi Universityof Technology, Tosayamada, Kami, Kochi 782-8502, Japan. Tel.: +81-887-57-2795; Fax: +81-887-57-2366.
E-mail addresses: tmasuda@iser.osaka-u.ac.jp (T. Masuda), okano.yoshitaka@kochi-tech.ac.jp (Y. Okano),and tatsuyoshisaijo@gmail.com (T. Saijo).
Abbreviations: MAM = minimum approval mechanism, BEWDS = backward elimination of weakly
dominated strategies, MCM = mate choice mechanism.
*11. Introduction
The free-rider problem lies at the heart of public good provision. To solve this problem,
theorists have traditionally concentrated on designing mechanisms yielding efficient
outcomes, given that players are self-interested. In this mechanism design or
implementation approach, individual theorists have discretion over which equilibrium
concept or behavioral rule to use, such as dominant strategy equilibrium, Nash equilibrium,or subgame perfect Nash equilibrium (SPNE). However, recent experimental results haveraised questions concerning this fundamental assumption. 1 It is also widely observed thatexperimental subjects are heterogeneous in their motive to choose their contributions to the
public good.
We offer an alternative approach, multiple implementations, in order to bridge
gaps between the theory and experiments. That is, public good mechanisms should berobust to as many as possible equilibrium concepts or behavioral rules. More precisely, we
design the minimum approval mechanism (MAM) that multiply implements thesymmetric Pareto-efficient allocation for linear public good environments with two players.
The MAM consists of two stages. Each player chooses his/her contribution in thefirst stage, following which each player observes the other’s choice and approves ordisapproves the same. If both players approve, their final contributions are what they chosein the first stage. If either disapproves, both finally contribute the minimum of theirfirst-stage choices.
Under the MAM, both players fully contribute in all five equilibrium concepts or
heuristics: (i) backward elimination of weakly dominated strategies (BEWDS; Saijo et al.,2013) 2, (ii) limit logit agent quantal response equilibrium (limit LAQRE; Mckelvey and
Palfrey, 1995; 1998), (iii) subgame perfect minimax regret equilibrium (SPMRE; Renou andSchlag, 2011), (iv) level-k thinking (Costa-Gomes and Crawford, 2006), and (v)diagonalization heuristics.
Moreover, we show that the MAM is a unique BEWDS-implementing approvalmechanism, satisfying the following natural properties. Forthrightness requires that theoutcome must be what players choose whenever both approve the other's first-stage
choices. Voluntariness mandates that no player is forced to finally contribute more thanwhat he/she chooses in the first stage. 3 Monotonicity states that final contributions increaseas their first-stage choices increase.
In experiments under the perfect stranger matching protocol, we observe, aspredicted, a rapid convergence to efficient allocation in the MAM. Subjects successfully
See Chen (2008) for an excellent survey.
As Saijo et al. (2013) noted, Kalai (1981) used BEWDS.
3 In other words, we only provide refunds to players, rather than using coercive power.
122

sustained high average contribution rates, beginning with 76.9% in the first period andending with 89.7% in the final period. The overall average contribution rate is 94.9%. 4 Our
experimental results of the MAM are in sharp contrast with those of existing mechanismsthat need dozens of iterations to converge to the efficient Nash equilibrium in experiments.
We also conduct experiments for the mate choice mechanism (MCM) of Saijo et al. (2013),
which theoretically does not work in our public good environment and observe that the
average contribution fluctuates within a middle range.
In order to capture subjects’ reasoning processes, we apply a coding scheme
similar to the one of Cooper and Kagel (2005). Two coders independently quantified
subjects’ responses to open questions during and after the session. The coding confirmsthat subjects are heterogeneous in their motive. In particular, 20.0% of subjects in the MAM
are deemed to have used diagonalization heuristics. Moreover, 16.7% and 11.7% of subjects
in the MAM and MCM, respectively are deemed to minimize regret. Finally, 21.7% and45.0% of subjects in the MCM are deemed to have followed level-1 and level-2 thinking,respectively.
Our approach differs from much of the experimental research on the public goodmechanism, which thus far has mainly focused on evaluating mechanism designs based on
dominant strategy equilibrium, Nash equilibrium, or SPNE. For dominant strategy
mechanisms, Attiyeh et al. (2000) found that subjects in the pivotal mechanism frequently
misrepresent their private information and become stuck at a weakly dominated Nashequilibrium. To solve this problem, Saijo et al. (2007) introduced secure implementation (ordouble implementation in dominant strategy equilibrium and Nash equilibrium). Further,
Cason et al. (2006) experimentally showed that the frequency of dominant strategy play
was significantly increased in a secure Groves–Clarke mechanism relative to a nonsecure
pivotal mechanism, although the former’s frequency was only slightly over 80%.
The Groves–Ledyard (Groves and Ledyard, 1977) and Walker (1981) mechanisms
have difficulty in attaining efficient Nash equilibria in one-shot play, because of subjects’bounded rationalities. 5 Andreoni and Varian (1999) reported similar results for Varian’s(1994) extensive form mechanism with the unique SPNE. Chen and Plott (1996), Chen andTang (1998), Chen and Gazzale (2004), Healy (2006), and Healy and Mathevet (2012)
characterized the stable mechanisms to achieve efficient Nash equilibria as rest points ofvarious learning dynamics. Nonetheless, these studies still required dozens of repetitions,which may not be feasible in practical situations. A notable exception is Falkinger et al.
(2000), who observed that under Falkinger’s (1996) mechanism, contributions approached
4 We obtained a similar result in the simplified MAM (SMAM) treatment, in which the player with thehigher first-stage choice alone can proceed to the second stage.
5 See Chen and Plott (1996) and Chen and Tang (1998).

3the efficient Nash equilibrium in only a few periods. 6
The remainder of this paper is organized as follows. Section 2 overviews the
MCM proposed by Saijo et al. (2013), introduces the MAM with examples, and presents theimplementation results. Section 3 describes the proof of the uniqueness of the MAM.
Section 4 argues in favor of the MAM’s robustness. Section 5 describes the experimental
design. Section 6 discusses the experimental results. Section 7 provides an analysis ofsubjects’ responses to the open-ended questionnaire. Section 8 discusses further
implications and concludes.
2. The model and the MAM
Consider a voluntary contribution mechanism (VCM) in the provision of a public goodwith two players. Each player i has an initial endowment w > 0, and he/she must decide
the contribution si ∈ S =[0, w] . The sum of the contribution is multiplied by α ∈ (0.5, 1) ,and the benefit passes to every player, which expresses the non-rivalness of the public) ( w − si ) + α(si + s j ), j ≠ i . Then, s=
good. That is, player i’s payoff is ui (si , s j=s=
0 is the12
dominant strategy equilibrium, and hence, no public good is provided.
When there are only two levels of the contribution (i.e., 0 or w) when players face
the prisoner’s dilemma, Saijo et al. (2013) introduced the notion of the MCM following thedilemma decision. By knowing the other player’s choice between cooperation (contributing
w and abbreviated by C) and defection (contributing 0 and abbreviated by D) in thedilemma or the first stage, each player must approve (y) or disapprove (n) it. If both players
approve the choices of their counterparts in the first stage, the outcome or payoff vector iswhat they choose. If either player disapproves, however, the outcome is that when bothchoose D.
<Figure 1 about here.>
Let us explain how BEWDS works. Figure 1 shows an example where w = 10 and
α =0.7. Note that this is an unusual extensive form game tree. Consider the case where

player 1 chooses D and player 2 chooses C (termed subgame (D,C) in Figure 1). If bothchoose y in the second stage, players 1 and 2 obtain 17 and 7, respectively. Otherwise, bothobtain 10. Then, player 1 compares (17,10) with (10,10) and chooses y, since y weakly
dominates n. A vector x weakly dominates z if every component in x is at least greater thanor equal to that in z, and there is at least one component with strict inequality. Similarly,
player 2 chooses n since (10,10) weakly dominates (7,10). Therefore, the outcome insubgame (D,C) is (10,10). By repeating the same procedure for every subgame, players canContributions averaged 90.5% of subjects’ endowments in experiments with linear payoffs. However,this mechanism proportionally taxes or subsidizes to the extent of the deviation of a player’s contribution
from the mean contribution, and hence, some players are forced to contribute more than they choose.
64construct the normal form game. Then, by comparing (14,10) with (10,10), the playerchooses C since the former weakly dominates the latter. Thus, BEWDS achieves (C,C) in theprisoner’s dilemma game with the MCM.
Saijo et al. (2013) showed that subjects achieve nearly full cooperation in the MCM
experiment, where each subject was randomly paired with another, and the same two
subjects were not paired more than once. With the help of a related experiment, the authors
also showed that BEWDS is most consistent with the data among the behavioral principles
based upon Nash equilibrium, SPNE, neutrally stable strategies, and BEWDS. 7 Following
Saijo et al. (2013), we use BEWDS as a basic behavioral principle and restrict ourselves tothe class of mechanisms where each player chooses si ∈ S in the first stage and y or n in thesecond stage. We call this mechanism an approval mechanism. In other words, the MCM isan approval mechanism.
However, the simple extension of the MCM with two strategies 0 and w to a casewith many strategies cannot implement the symmetric Pareto-efficient outcome. For
simplicity, hereafter, we call the extended MCM with many strategies as the MCM. In the
MCM, if both players approve the other player’s choice in the first stage, then bothcontribute what they choose, whereas if either one disapproves, then both contribute 0. To
see why the MCM breaks down, consider the case with w = 10 and α =0.7 once more. For
simplicity, each player can choose a contribution from {0,2,4,6,8,10}. Table 1 shows player
1’s (the row player) payoff table. The symmetric Pareto-efficient outcome is (14,14) whenboth players contribute 10. Consider subgame (10,8). Then, both choose y, since for player 1(12.6,10) weakly dominates (10,10), and for player 2 (14.6,10) weakly dominates (10,10).
Thinking in reverse, player 2 will find choosing 8 a profitable deviation, which shows that
the MCM cannot achieve efficiency.
<Table 1 about here.>
In order to implement the symmetric Pareto-efficient outcome, we introduce theMAM. In this mechanism, if both players approve the contribution of the other in the firststage, then both contribute what they choose, whereas if either one disapproves, then bothcontribute the minimum of the two contributions in the first stage.
Using Table 1, let us consider two examples of subgames under the MAM. In
subgame (6,2) on the left-hand side of Table 2, if either disapproves the other’s choice, themechanism returns 4 (= 6 − 2) to player 1, and both players contribute 2 and obtain 10.8.
Therefore, player 1 chooses n since (10.8,10.8) weakly dominates (9.6,10.8), and player 2chooses y since (13.6,10.8) weakly dominates (10.8,10.8). Therefore, both obtain 10.8.
7 Neutrally stable strategies also attain (C,C), though the outcomes of the Nash equilibrium and the SPNE
show every possibility, namely (C,C), (C,D), (D,C), and (D,D). As shown in Saijo et al. (2013), the strategies
of BEWDS are a subset of neutrally stable strategies even though both achieve (C,C).

5

<Table 2 about here.>
Next, consider subgame (8,10) on the right-hand side of Table 2. Player 1 chooses
y and player 2 chooses n; hence, the payoff is (13.2,13.2). In general, player i chooses y (n) if
his/her first-stage choice is smaller (larger) than that of player j ≠ i : si < ( >)s j . Table 3
shows player 1’s payoff matrix in the reduced normal form game.
Clearly, choice 10 weakly dominates the rest, and hence, the MAM implements
the symmetric Pareto-efficient outcome in BEWDS. Table 3 presents a special structure: thepayoff in subgame (m,m) is the same as that in (m,n) and (n,m) for all n > m, whereas thepayoff in (n,n) is greater than that in (m,m) for all n > m. If the payoff matrix satisfies thesetwo conditions, we say that it has an echelon structure.
<Table 3 about here.>
Proposition 1. The MAM implements the symmetric Pareto-efficient outcome in BEWDS.
Proof. First, we show that player 1’s payoff matrix in the reduced normal form game under
the MAM has an echelon structure. Fix a ∈ S , and choose any b ∈ S with b > a. Consider
subgame (a,b). Since { w − a + a( a + a)} − { w − b + a( a + b )} = (b − a)(1 − a ) > 0 , player 2chooses n in the second stage. Then, player 1’s payoff becomes w − a + a( a + a). In otherwords, for all b > a, the payoff in subgame (a,b) is the same as that in (a,a) under the MAM.
Similarly, we find that for all b > a, the payoff in subgame (b,a) is the same as that in (a,a)
under the MAM. Choose any a, b ∈ S with b > a. Then, the payoff in subgame (b,b) is greaterthan that in (a,a) since w − b + 2 ab − ( w − a + 2 aa) =(2 a − 1)(b − a) > 0 .
We further show that choice w weakly dominates s1 ∈ [0, w). Consider the casewith s2 ∈ [0, s1 ]. From the first part of the echelon structure, we have u1 (s1 , s2 ) = u1 (s2 , s2 )= u1 ( w , s2 ). 8 Consider the case with s2 ∈ (s1 , w]. Using the two properties of the echelon

structure, we have u1 (s1 , s2 ) = u1 (s1 , s1 ) < u1 (s2 , s2 ) = u1 ( w , s2 ). In other words, choice w
weakly dominates s1 ∈ [0, w). █3. The MAM is unique
There could be various ways to implement the symmetric Pareto outcome using approvalmechanisms under BEWDS. For example, consider the average approval mechanism(AAM): if either player disapproves, then both contribute the average of the two choices in

8

Let t = ( t 1 , t 2 ) be the decision in the second stage. Then, we must write ui ((s1 , t1 ),(s2 , t2 )) . Thus,ui (s1 , s2 ) should be ui (s1 , s2 ) = ui ((s1 , y ),(s2 , n)) . In other words, we must specify the decision in the second
stage in order to obtain the payoff of player i. By slight abuse of notation, we can also use it both ways
whenever the decision in the second stage is clear.

6the first stage. The symmetric Pareto outcome is attained by the AAM under BEWDS. 9
Under the AAM, however, the player who chooses a smaller number must
contribute more. In order to exclude this coercion, we state that an approval mechanism isvoluntary if each player’s contribution does not exceed his/her first-stage choice. Clearly,
the MAM satisfies voluntariness and the AAM does not. 10
Moreover, we show that the MAM is a unique mechanism for satisfying
voluntariness and several other conditions. For this purpose, let us introduce the following
definitions. We refer to the players’ choices in each stage ((s1 , t1 ),(s2 ,t2 )) as a path. An
approval mechanism g is a function that associates a contribution g((s1 , t1 ),(s2 ,t2 )) =( g1 ((s1 , t1 ),(s2 ,t2 )),g2 ((s1 , t1 ),(s2 ,t2 ))) for every path ((s1 , t1 ),(s2 ,t2 )) . First, we require thatapproval mechanisms be forthright in the following manner: if both make the samefirst-stage choice or if both choose y in the second stage, then they contribute what theychoose in the first stage; moreover, given the first-stage choices, the outcomes when either
chooses n are the same.11 Formally, we have:
Definition 1. An approval mechanism g is forthright if:(s1 , s2 ) if s1 = s2 , where “･” indicates either y or n;(i) g((s1 , ⋅),(s2 ,⋅)) =(ii) g((s1 , y ),(s2 ,y )) = (s1 , s2 ) for every (s1 , s2 ) ; and((s1 , n),(s2 ,y )) g=((s1 , y ),(s2 ,n)) g((s1 , n),(s2 ,n)) for every (s1 , s2 ) .
(iii) g=In the following, vector inequality is taken component-wise as usual. The next
property is voluntariness, as argued above.
Definition 2. An approval mechanism g is voluntary if g((s1 , t1 ),(s2 ,t2 )) ≤ (s1 , s2 ) for every
path ((s1 , t1 ),(s2 ,t2 )) .
The third property, monotonicity, is a plausible condition that reflects players’
willingness to pay more. It states that if players weakly increase their first-stage choices,then their contributions weakly increase regardless of their decisions in the second stage.
Definition 3. An approval mechanism g is monotonic if (s1′ , s2′ ) ≥ (s1 , s2 ) implies
It is easy to check that the AAM implements a Pareto-efficient allocation in BEWDS. More generally, forany λ ∈ [0 , 1] , λ − weighted AAM, where if either says no then both players finally contribute9

λs1 + (1 − λ )s2 , also implements a Pareto-efficient allocation in BEWDS.

10

The budget-balanced mechanism proposed by Falkinger et al. (2000), under which subjects contributeover 90% of endowments, is similar to the AAM, because it taxes (subsidizes) players with a contribution
lower (higher) than the group mean.
11 This definition is slightly different from forthrightness introduced by Saijo et al. (1996).

7g((s1′ , t1 ),(s2′ ,t2 )) ≥ g((s1 , t1 ),(s2 ,t2 )) for every (t1 , t2 ) .

The above properties are sufficient to prove the uniqueness of the MAM. 12
Proposition 2. Suppose that a forthright approval mechanism implements the Pareto-efficient
allocation in BEWDS. If it is voluntary and monotonic, then it must be the MAM.
Proof. See Appendix. █4. Multiple implementations of the MAM
In this section, we explore other equilibrium properties and the outcome of heuristics
under the MAM. First, we present the main result of implementation in several equilibrium
concepts and heuristics. 13 Then we derive the result step by step.
Proposition 3. The MAM multiply implements the symmetric Pareto-efficient outcome in limit
LAQRE, SPMRE, level-k thinking, and diagonalization heuristics.
Proof. It immediately follows from Lemmata 2-5. █The concept of SPNE is often used to analyze multistage games. We denote player
i’s strategy by=pi (si ,ti (⋅)) , where si ∈ S , and ti (⋅) is a function that associates y or n toeach first-stage choice (s1 ,s2 ) . Let P=P=
P be the pure strategy set. With a slight abuse
1
2
of notation, we write i’s payoff function induced by the MAM as ui . A strategy profilep = (p1 , p2 ) is a Nash equilibrium if for each i, j with j ≠ i , ui (p) ≥ ui (pi′ ,p j ) for all pi′ ∈ P .

A Nash equilibrium p is subgame perfect if it constitutes a Nash equilibrium in every
subgame.
Lemma 1. In the MAM, the set of SPNE contributions is the set of symmetric contributions.
Proof. See Appendix. █Although it is artificial, the following forthright mechanism is voluntary, but not monotonic. Consider aforthright rule in which a player who announces a bigger contribution than the other in the first stage
must contribute 0, and the other player must contribute what he/she chooses in the first stage if either one
chooses n in the second stage. Apparently, this mechanism is voluntary. However, it is not monotonic,
0 < g1 ((2, n), (2, y )) =2. It is easy to check that the mechanism implements asince g1 ((6, n), (2, y )) =Pareto-efficient allocation in BEWDS.
13 We thank an anonymous referee for pointing out this possibility.
128Mckelvey and Palfrey (1995, 1998) introduced logit quantal response equilibrium(QRE) concept for the normal and extensive form games. They modeled the players whomake noisy best response to other players. The probability of choosing an action is
proportional to the exponential function of the observed payoff of the action, divided by aparameter µ > 0 . Even in such a case, the MAM can implement the efficient outcome inlimiting equilibrium, which survives as players become fully sensitive to their payoffs:µ → 0 . Assume µ is common to both players and all their information sets. To ensure theexistence of the limiting equilibrium by Theorem 1 of Mckelvey and Palfrey (1998), we
focus on the case where players contribute in integer.
1

2

Let σi = (σi , σi ) be player i’s behavioral strategy with the superscript denoting
stage, and let σ = (σ1 ,σ2 ) be a strategy profile. Let πi (si ) be i’s expected payoff fromwhen he/she chooses si ∈ Sd ={1,2,2 ,w} , while all the other actions of the players are

πi (ti|si ,s j ) , ti = y ,n be i’s expected payoff whenprescribed by σ . Similarly, let πi (ti ) =he/she chooses ti at subgame ( si ,s j ), while j ≠ i is following σ in the subgame, wherethe vertical “|” indicates “conditional on.” A strategy profile σ is an LAQRE if it solves
the following equations: for each i = 1,2, j ≠ i ,exp(πi (y )/µ )exp(π (s )/µ )for each si ∈ Sd , and σi2 (y|s1 ,s2 ) =σ1i (si ) = i iexp(πi (y )/µ )+exp(πi (n)/µ )∑ exp(πi (x)/µ)x∈Sdfor each si ,s j ∈ Sd . A strategy profile σ is a limit LAQRE, if it is an LAQRE as µ → 0 .
In the limit, since the player with a higher first-stage choice surely chooses n, theother player randomizes y and n by indifference. Still, the reduced normal form gameapproaches the echelon form. 14 Then, analogous to Proposition 5 of Anderson et al. (2001)
for minimum effort games, the full contribution is the unique equilibrium. Therefore, wehave Lemma 2.
Lemma 2. The MAM implements the symmetric Pareto-efficient outcome in limit LAQRE.
Proof. See Appendix. █The third alternative equilibrium concept is SPMRE. Some subjects imagine
foregone payoffs from unchosen strategies. This is close to the simplest version of εminimax regret equilibrium introduced by Renou and Schlag (2011), who incorporated intogame-theoretic models experimental evidence that people avoid ambiguity and thedecision-theoretic formulation of such behavior. The basic idea is that a player imagines
14

For a more general discussion of the QRE, see Mckelvey and Palfrey (1995, 1998).

9foregone payoffs from unchosen strategies. For every pure strategy profile p = ( p1 , p2 ) ,=by Ri ( p) max p′i ui ( pi′ , p j ) − ui ( pi , p j ) , j ≠ i . This is theplayer i’s regret at p is defineddifference between his/her maximal payoff when player i best responds to j’s strategy p j
and his/her payoff when he/she plays pi . A strategy profile p is a minimax regret
equilibrium if for each i, j with j ≠ i and for all pi′ , max p′ Ri ( pi , p ′j ) ≤ max p′ Ri ( pi′ , p ′j ) . Injjother words, each player chooses the strategy that minimizes the largest possible regret in
the equilibrium. 15 We say that a minimax regret equilibrium p is subgame perfect if theaction profile induced by p constitutes a minimax regret equilibrium in every subgame.
Lemma 3. The MAM implements the symmetric Pareto-efficient outcome in SPMRE.
Proof. See Appendix. █We now propose two alternative heuristics that predict that players make the fullcontribution. The first one is level-k thinking. We apply such heuristics to the reduced
normal form game induced by elimination of weakly dominated strategies. Consider Table
3 as an illustration. Assume that player 1 expects player 2 to choose each number withequal probability (level-0) and to eliminate weakly dominated strategies in the second
stage. In the case of the MAM, player 1’s best response to level-0 is to choose 10. Thus,
efficiency can be achieved by any pair of players of level-1 or above. 16
Lemma 4. The MAM implements the symmetric Pareto-efficient outcome in level-1 or above.
Proof. Straightforward. █The fifth possibility is a heuristic called diagonalization, of which we find notable
descriptions given by the subjects in the questionnaires during and after the experiment.
Some subjects wrote that the player who chooses the higher first-stage choice will
disapprove the other’s choice, and both would obtain the payoffs on the diagonal line in
the payoff table. Then, subjects can easily find that full contributions attain the maximum
payoff among the payoffs on the diagonal line. Thus, we have the following simple
proposition.

15
16

Here, we assume that each player believes that his/her opponent may choose any strategy.
Lemma 4 also holds if we consider the level-0 player who chooses w.

10Lemma 5. If both players follow diagonalization in the MAM, then the outcome is symmetric
Pareto-efficient.
Proof. Straightforward. █Now, Proposition 3 is verified for the MAM. Proposition 4 for the MCM provides
the counterparts to Propositions 1 and 3 for the MAM. Moreover, the results are quite
contrasting. Consider Table 1 again. Let us begin with BEWDS. Since both obtain 10 ifeither chooses n, player 1 will know that player 2 disapproves if and only if (s1 ,s2 ) = (0,x)
with x ≥ 2 , (2,x) with x ≥ 6 , or (4,10). Then, the choices of 0, 8, and 10 are eliminated, but
any choice of 2, 4, and 6 survives under BEWDS. Every pair of contributions such that bothplayers get at least 10 can be supported as an SPNE. 17 Consider level-k thinking. Given that
player 2 randomly chooses any number, player 1 can see that the best response is to choose
6. This also minimizes regret. Another step of iterated reasoning leads player 1 tocontribute 4, which is the best response to player 2’s 6. The third iteration result uniquely
selects the choice of 2. 18 (2,2) is also a limit LAQRE contributions. Let β= (1 − α ) / α .
Proposition 4. Under the MCM, the following holds.
i) A player following BEWDS chooses any contribution in (0, βw] .
ii) There exists an SPNE where players contribute ( g1 , g2 ) if and only if βg2 ≤ g1 ≤ β−1 g2 .
iii) A limit LAQRE supports contributions (1,1) if players choose their contributions in integers.
iv) The unique SPMRE contribution is βw .
v) Level-k player’s contribution is βk w , which decreases as k increases.
Proof. See Appendix. █5. Experimental design
We conducted four treatments in order to test the performances of the approvalmechanisms. The VCM was the control treatment, while the other three treatments were
the MCM, MAM, and simplified MAM (SMAM). Under the SMAM, if both players makethe same first-stage choice, the game ends. Otherwise, only the player with the higher
first-stage choice proceeds to the second stage. If he/she approves, they contribute what
they choose in the first stage. If he/she disapproves, both players contribute the minimumof their choices in the first stage. The SMAM also implements the efficient allocation in
17
18

That is, (x,x) with x ∈ {0, 2, 4, 6, 8, 10} , (4,2), (6,4), (8,4), (8,6), (10,6), (10,8), and their permutations.
Proposition 4-iv) also holds if we consider the level-0 player who chooses w.

11

BEWDS. 19 Each of the above-mentioned four treatments had three experimental sessions.
These 12 sessions were conducted at Osaka University in March, April, September, andNovember 2011. Subjects were recruited from Osaka University through campus-wide
advertisements and were inexperienced in this particular type of experiment. No
individual participated in more than one session. The experiment was computerized using
the experimental software z-Tree (Fischbacher, 2007). Twenty subjects participated in eachsession. Each was seated at a computer terminal assigned by lottery. All terminals wereseparated by partitions. No communication was allowed between subjects.
Each subject had a set of printed instructions, a record sheet, and a payoff table(called the points table). 20 In the distributed payoff table, each subject was considered to beplayer 1. Player i’s payoff is given by =ui 300{(24 − gi ) + 0.7( g1 + g2 )} for each pair ofcontribution ( g1 , g2 ) . Since we consider two players with identical utility functions andendowments, the payoff table is identical for all subjects. Subjects were informed that they
had identical payoff tables.
Instructions were read aloud by an experimenter for approximately 10 minutes.
After that, subjects were given another 10 minutes to ask questions. Then, subjects
proceeded to the payment periods. There was no practice period. Each session consisted of
19 periods under the perfect stranger matching protocol. Subjects were informed that eachof them would meet any other subject just once.
The approval mechanism treatments continued as follows. At the beginning ofeach period, all subjects endowed with 24 tokens were anonymously paired. In the firststage, subjects were asked to enter their contributions as nonnegative integers into a box in
the display and to write down their choices along with their reasoning in thecorresponding row in the record sheet. Once all subjects had finished their tasks, theyclicked the OK button. In the second stage, the first-stage choices of both players and thepayoff matrix in the second stage were displayed. After all players wrote down thefirst-stage choices of their counterparts, they chose to approve or disapprove by clicking
the relevant radio button and recording the decision along with the reason. If a subject in
the SMAM did not proceed to the second stage, he/she simply wrote down the first-stage
choice of his/her counterpart before the other subjects finished the second stage and circled“none” while facing the waiting screen. Once all the subjects in the second stage hadfinished the above tasks and clicked the OK button, they proceeded to the results screen.
The results screen included the first-stage choices of both players, their “approve” or
19 Recall that under the MAM, the decision by the player with the higher first-stage choice alone
determines the outcome in every second stage. The SPNE, SPMRE, limit LAQRE, and level-k thinking
outcomes of the SMAM are the same as those of the MAM.
20 See Online Appendix.

12

“disapprove” decisions, and their payoffs (points earned) in the period. No information on
the choices of the other nine groups was provided to the subjects. Finally, the subjects
wrote down the decisions of their counterparts and the points they had earned and clicked
the Next button to begin the next period. The treatment without the second stage became
the VCM treatment.
After playing for 19 periods, the subjects completed a questionnaire and wereimmediately paid privately in cash. Each subject was paid an amount proportional to thesum of the points that he/she had earned for the 19 periods. Individual payments ranged
from $49.91 to $87.97.
6. Experimental results
Figure 2 shows the time path of average contributions over the 19 periods arranged bymechanism. Individual contributions were evaluated after the second stage in the MAM,
MCM, and SMAM.
<Figure 2 about here.>
We first assess the results of the VCM. The time path of average contribution in
the VCM shows a similar pattern to that of previous VCM experiments (e.g., Ledyard,
1995). In the first period, subjects contributed 15.3% of endowments (3.68 tokens), with
contribution rates gradually decreasing to 5.7% (1.37 tokens) in the last period. The
hypothesis test based on Spearman’s rank correlation coefficient shows that the downward
trend in average contributions was statistically significant=(ρ -0.916, p < 0.001) . Datapooled across all 19 periods and 3 sessions show that the subjects contributed 10.2% ofendowments (2.44 tokens). Out of 570 outcomes (10 pairs × 19 periods × 3 sessions), there isno case where both players contributed all endowments.
Introducing the MAM facilitates almost full contributions among subjects, even in
the earlier period. Contributions in the first period average 76.9% of endowments (18.45
tokens). They then rise repeatedly for the first five periods, achieving 97.8% (23.47 tokens)in period 5. Then, cooperation is sustained with an average contribution rate of over 95% inevery period except the final one. Moreover, the convergence to the efficient allocation
under the MAM is statistically supported by Spearman’s rank correlation test, which helps
examine the upward trend in average contributions=(ρ 0.557,=
p 0.009) . When the data
are pooled across all 19 periods and 3 sessions, subjects in the MAM contribute on average
94.9% of endowments (22.78 tokens). Out of 570 outcomes, there are 475 outcomes where
both players contributed all endowments. Table 4 compares the contributions for the four
treatments using a two-tailed Mann–Whitney test. This test shows that subjects in the
MAM contributed significantly more than those in the VCM, with the test statistic z = 9.445
13(p < 0.001). 21<Table 4 about here.>
The contributions in the SMAM show similar patterns to those in the MAM. It
took five periods to achieve a contribution rate of over 90%, and almost all subjects
maintained full contributions until the end of the session. Spearman’s rank correlation testshows that the upward trend in average contributions is statistically significant=(ρ 0.625,=
p 0.004) . Overall, the average contribution rate of the SMAM is 89.9%. Of the

570 outcomes, there were 424 outcomes where both players contributed all endowments.
The contribution rate of the SMAM was significantly lower than that of the MAM (z = 3.799,
p < 0.001). 22 There is no significant difference, however, in the first-stage choices between
the SMAM and MAM (z = 0.456, p = 0.649). The contribution rate of the SMAM issignificantly higher than that of the VCM (z = 9.275, p < 0.001).
In contrast to the MAM and the SMAM, only 7 of the 540 pairs of the MCM
achieved full contributions and the approval of both players. The average contribution
under the MCM fluctuated within a relatively narrow range, but decreased repeatedly, andit remained higher than that of the VCM in every period. Spearman’s rank correlation testshows that the downward trend in average contributions is statistically significant=(ρ -0.525, p < 0.001) . The overall average contribution rate of the MCM is 28.4% (6.83

tokens), which is significantly higher than that of the VCM (z = 7.284, p < 0.001). However,the contribution rate was significantly lower than that of the MAM and the SMAM (z =9.450 and p < 0.001 for the MCM vs. the MAM and z = 9.133 and p < 0.001 for the MCM vs.
the SMAM).
In reality, the average contributions by subjects in one session are mutuallydependent, although in a two-tailed Mann–Whitney test we assume that they are
independent. In order to account for this dependence, we also run a regression to predict
contribution after the approval stage, using three approval mechanism dummies andclustering standard errors by session. The estimation results are similar to those of a
two-tailed Mann–Whitney test, indicating that all three approval mechanisms significantly
increase contribution relative to the VCM. In particular, the MAM increases contribution by
20.34 tokens with a standard error of 0.633, followed by the SMAM which increases

The null hypothesis is that the contribution rates are the same between the MAM and the VCM. We use
the method presented by Andreoni and Miller (1993). We first calculate the average contribution rate ofeach subject across periods and then calculate the test statistic using these averages, in order to eliminate
cross-period correlation.
22 The statistical difference exists mainly because the lower choices of two of the subjects in the SMAM
were disapproved. One of them mistakenly thought that she had to allocate 24 tokens across 19 periods.
The other subject intentionally chose low contributions, expecting that his/her matched subjects might
approve by giving up a small number of their points.
21

14

contribution by 19.1 tokens and a standard error of 0.886, and the MCM which increases
contribution by 4.38 tokens and a standard error of 0.659.
7. Heterogeneous reasoning processes coexist
We report a third-party assessment and classification of descriptions in the questionnaireduring and after the experiment, called coding, in order to investigate which type of
reasoning process (including diagonalization, regret minimization, and level-k thinking)was used by the subjects. 23 The coding proceeded as follows. First, we read the responses
in the MAM and the MCM separately. Then, we created 30 sentence-based categories fromtypical responses (see Online Appendix). Two research assistants (hereafter coders) wereindependently instructed, and each coder created the coding for all 120 subjects in the
MAM and the MCM. For each subject, the coders were required to read subjects’descriptions and determine the categories to which his/her descriptions belong and tocheck all relevant categories (0 = description does not fall into the category, 1 = description
falls into the category). We refer to a coder’s binary decision as the rating. Coders never met
face-to-face, and no efforts were made to reconcile the differences in individual coding.
For each category of our interest, we regard a subject rated positively by at least
one coder as a follower of the equilibrium strategy or heuristics in the category. In
particular, Category 10 is designed to capture subjects in the MAM who directly mentioned
the diagonal line of the payoff table. Such subjects are deemed to follow diagonalization
heuristics. Category 12 is designed to capture regret minimization in the first stage in the
MAM and MCM. Categories 16 and 17 are set to capture level-1 and level-2 in the MCM,
respectively. 24,25
Let us summarize the coding result. In the MAM, 20.0% (= 12/60) of the subjects
are deemed to follow diagonalization heuristics, and 16.7% (= 10/60) subjects fall in the
regret minimization category. In the MCM, on the other hand, 21.7% (= 13/60) and 45.0%(= 27/60) of the subjects are deemed to follow level-1 and level-2 thinking, respectively. In
addition, 11.7% (= 7/60) of the subjects are deemed to follow regret minimization, whichappears commonly in both treatments. Although no category of our interest records anabsolute majority, these results suggest that subjects are heterogeneous in reasoning
23 This method is similar to that presented by Cooper and Kagel (2005). Note that our experiments did notallow communication among subjects, while Cooper and Kagel (2005) investigated specific words andattitudes in informal communication that facilitated cooperation among group members.
24 We did not make level-k categories for the MAM because we could not find such responses in thequestionnaires. The same applies to regret minimization in the second stage for both the MAM and theMCM.
25 At the start of the coding process, Category 12 and 13 were classified as regret minimization. However,
we found that the sentence of Category 13 was too imprecise to capture regret minimization. Therefore, we
do not consider Category 13 to describe regret minimization in the analysis.

15

processes, and the appearance frequency of each reasoning process may depend on themechanisms. 26
8. Discussion and conclusion
Prior theoretical work on public good mechanisms has given theorists broad latitude thechoice of equilibrium concept or behavioral rule, such as dominant strategy equilibrium,
Nash equilibrium, or SPNE. However, recent experimental studies reveal that there is awide gap between the theory and subjects’ behavior.
This paper, in contrast, theoretically showed that the MAM multiply implements
the Pareto-efficient provision of the public good in five equilibrium concepts and heuristics,
namely BEWDS, limit LAQRE, SPMRE, level-k thinking, and diagonalization heuristics.
Consistent with our predictions, we observed rapid convergence towards efficient
allocation in the MAM sessions. Further, our inclusive coding of subjects’ responses helped
pinpoint their reasoning processes. The coding results suggest that the subjects areheterogeneous in reasoning processes. Although our public good environment is simple,
our result is in stark contrast to those of recent experimental studies, which have often
needed dozens of repetitions to attain efficiency in the laboratory.
To our knowledge, this is the first theoretical and experimental study tosuccessfully design a public good mechanism robust to several behavioral rules. This paper
should at least stimulate constructive discussion on mechanism design, to emphasize thealignment of players with various behavioral rules towards a social optimum.
One direction for further research is to evaluate experimental approvalmechanisms in more general public good environments. First, if we consider asymmetriccases, Proposition 1 still holds if two players have moderately different preferences
towards the public good. 27 Further, if the endowment is heterogeneous, the problem
reduces to a symmetric case, through the definition of a new strategy space, where eachplayer/subject chooses a contribution rate relative to his/her endowment in the first stage.
If the public good is harmful for some player, the MAM with transfer implements thesymmetric Pareto-efficient allocation in BEWDS (see Masuda, 2013). The second extension
is toward the n-person case. In public good environments with binary choices, Huang et al.
We did not obtain sufficiently high agreement in ratings between any two coders for the following two
reasons. First, even though our coders had excellent and comparable game theoretic skills, the sensitivityof coders to track subjects’ reasoning processes, or to pick up any subtle nuance in Japanese, isheterogeneous. Second, we made coders work independently throughout the coding process to prevent
coordination among them, though experiments in psychology permit communication between two coders
before rating each subject.
27 If we write player i’s benefit per public good as α , the conditions are α ≠ α , α , α ∈ (0, 1) , and
i
1
2
1
2
26

α1 + α2 > 1 .

16

(2013) succeeded in constructing a variant of an n-person approval mechanism. Masuda(2013) also generalized the mechanism of Huang et al. (2013) to homogeneous n-person
public good environments with continuous contributions and linear payoffs in the spirit of
the SMAM, where players other than those with the minimum choice can proceed to thesecond stage. Third, in a more realistic setting, we may consider the ratification process forinternational agreements involving countries; nature determines the order of the player
who may decide to approve or otherwise. Our results hold in such cases also.
Appendix
This appendix gives the proofs for Propositions 2 and 3.
Proof of Proposition 2. Let g be a forthright approval mechanism that implements thePareto-efficient allocation in BEWDS and that satisfies voluntariness and monotonicity. Let(s1 , s2 ) according to condition(s1 , s 2 ) be any first-stage choice. If s1 = s2 , g((s1 , ⋅),(s2 , ⋅)) =(i) of forthrightness, which shows the outcome of the MAM. Consider the case wheres1 ≠ s2 . Without loss of generality, assume that s1 > s2 , and consider two cases:=
s1 w and s1 < w. First, consider subgame ( w , s 2 ) with w > s2 . In order to implement

the efficient allocation in BEWDS, w must weakly dominate s 2 for player 2 in the reduced
normal form game. In other words,gg(A.1) u2 (( w , ⋅),( w , ⋅)) ≥ u2 (( w , t1 ),(s2 , t2 ))gwhere ui is i’s payoff function induced by g, and ti is player i’s decision under BEWDS.

t=y . Then, by conditions (i) and (ii) of forthrightness, we have
Suppose that t=
12(A.2) u2g (( w , y ),(s2 , y ))) u2g (( w , ⋅),( w , ⋅)) ,= ( w − s2 ) + α( w + s2 ) > ( w − w ) + α( w + w=
which contradicts to (A.1). Thus,=t1 n=
or t2 n. Together, condition (iii) of forthrightnessggand (A.2) show that u2g (( w , y ),(s2 , y )) > u=u=
2 (( w , y ),( s2 , n ))2 (( w , n ),( s2 , y ))u2g (( w , n),(s2 , n)) . Hence, y weakly dominates n for player 2 in the second stage, namely

t2 = y , which implies t1 = n .
Consider now the reduced normal form game under g, and compare the outcomesin subgames (s2 , s2 ) and (w, s2 ). Since player 1 chooses w by assumption, we haveu1g (( w , n),(s2 , y )) ≥ u1g ((s2 , ⋅),(s2 , ⋅)) . Let g(( w , n),(s2 , y )) = ( g1 , g2 ). Then,gu1g (( w , n),(s2 , y ))= w − g1 + α( g1 + g2 ) ≥ w − s2 + α(s2 + s2 ) = u1 ((s2 , ⋅),(s2 , ⋅)) . Since g is

voluntary, ( g1 , g2 ) ≤ ( w , s2 ). Because of the monotonicity of g, ( g1 , g2 ) ≥ (s2 , s2 ) . Theintersection of these three inequalities shows ( g1 , g2 ) = (s2 , s2 ). Thus, according to17

condition (iii) of forthrightness, we have(( w , n),(s2 , y )) g=(( w , y ),(s2 , n)) g=(( w , n),(s2 , n)) (s2 , s2 ).
(A.3) g=In other words, both players contribute s2 , which is the minimum of ( w , s2 ) ,when at least one player chooses n.
Now take any s1 ∈ (s2 , w) . Together with ( w , s2 ) ≥ (s1 , s2 ) ≥ (s2 , s2 ) and (A.3), sinceg is monotonic, we have (s2 , s2 ) = g((w , n),(s2 ,y )) ≥ g((s1 , n),(s2 ,y )) ≥ g((s2 , n),(s2 ,y )) = (s2 , s2 ).
Hence, we have g((s1 , n),(s2 ,y )) = (s2 , s2 ). In other words, g is the MAM. █Proof of Lemma 1. Suppose, on the contrary, that for some SPNE p =((s1 , t1 (⋅)),(s2 , t2 (⋅))) ,players contribute differently, i.e., g1 ≠ g2 . Then, by construction of the MAM, s1 ≠ s2
and ti (s1 , s2 ) = y for i = 1,2. Define p′ = ( p1′ , p2 ) with p1′ = (s1′ , t1(⋅)) as the strategy profile
where player 1 deviates from p only in the first stage. Without loss of generality, assume

s1 > s2 . Assume s1′ = s2 , then u1 ( p′)= w − s2 + α(s2 + s2 ) > w − s1 + α(s1 + s2 )= u1 ( p) by
s1 > s2 , which is a contradiction. Thus, g1 = g2 . Next, we show that any symmetric pair ofcontributions can be supported as an SPNE. Let s ∈ S . Consider the strategy profileq =((s , t(⋅)),(s , t(⋅))) , where t(⋅) is a function such that t(s1 , s2 ) = n for all s1 , s2 ∈ S .
Clearly, p constitutes a Nash equilibrium (n,n) for every second-stage subgame. Now,
consider the reduced normal form game. Suppose that player 1 unilaterally deviates tos1′ > s in the first stage. Let q1′ = (s1′ , t(⋅)) and let q ′ = (q1′ , q 2 ) . Since both players choose nat (s1′ , s ) and by s1′ > s , we have u1 (q ′) = w − s + α(s + s ) = u1 (q ) . Suppose s1′ < s . Then,
we have u1 (q ′) = w − s1′ + α(s1′ + s1′ ) < w − s + α(s + s ) = u1 (q ) . The above argument also

applies to player 2. Hence, q is an SPNE. █Proof of Lemma 2. First, consider the second stage. Take any (s1 ,s2 ) . Note that σ2j (n)=1-σ2j (y ) ,ui ((si , y ),(s j , y ))=(w − si )+α(si + s j ) and if (ti , t j ) ≠ (y , y ) , ui ((si , ti ),(s j , t j ))=(w − s )+α(s + s )=πi (n) where s = min{si , s j } . Then,
22(A.4) πi (y )= ∑ s2j (t j )ui ((si , y ),(s j , t j )) =sj (y ){(w si )+α(si + s j )} + {1-s j (y )}πi ( n ).
t j ∈{y ,n }By substituting (A.4) into the second equation satisfied by the LAQRE, we have
 1+exp{α(s -s )s2 (y )/µ}  -1
if si <s j ,
i
j
j


1

(A.5) si2=(y ) [ 1+exp{(pp}]if s1 =s2 ,
0.5
i (n )- i (y ))/µ=-12
 1+exp{(1-α )(si -s j )s j (y )/µ}  if si >s j .

18

2
2
ss
For each si ∈ Sd , πi (si )= ∑ s1j (x )πi (si , x ), where πi (si , x ) =i (y ) j (y )ui ((si , y ),(x , y )) +
s∈Sd

2
2
(1 − ssi (y ) j (y ))ui ((si , n ),(x , n )) . Since2222πi (si , s j ) =ss
i (y ) j (y ){(w − si )+α(si + s j )} + (1 − ssi (y ) j (y )){(w − s )+α( s + s )}22αssi (y ) j (y )( s j − si ) + (2α − 1)s + w if si <s j ,= (2α − 1)s + w
if s1 =s j ,22i (y ) j (y )( s j − si ) + (2α − 1)s + w if si >s j , (1 − α )ss

we have

πi (si )= ∑ s1j (x )πi (si , x ) = ∑ s1j (x )πi (si , x ) + ∑ s1j (x )πi (si , x )
x∈Sd

(A.6)

x ≤ si∑x > si∑122
s1j (x ){(2α − 1)min{si , x} + w} +(1 − α ) sss
j (x ) i (y ) j (y )( x − si )x∈Sdx ≤ si=122+ α ∑ sssj (x ) i (y ) j (y )( x − si ).
x > si22Now, let us consider limiting equilibrium. If s1 =s2 , by (A.5), (σ1 (y ) , σ2 (y ))=(0.5,0.5) for any µ . Consider next that s1 <s2 . Consider player 2’s equilibrium strategy.

Note by s1 <s2 , and α(s1 -s2 )s22 (y )/µ < 0 , we have 0.5 < σ12 (y ) < 1 for any µ . Define afunction of x , f (x ; s1 ,s2 )= [ 1+exp{(1-α )(s2 -s1 )x} ] . Then, by (A.5), 0.5 < σ12 (y ) < 1 for any-12
f (s12 (y )/µ ; s1 ,s2 ) > 0 for any µ .
µ , and since f is decreasing, f (0.5/µ ; s1 ,s2 ) > s2 (y ) =Moreover, lim f (0.5/m ; s1 ,s2 )=0 . Then, lim σ22 (y )=0. Consider player 1’s equilibrium
m→ 0

m→ 0

2
2

strategy. Note f (0.5/µ ; s1 ,s2 )/µ > s (y )/µ= f (s12 (y )/µ ; s1 ,s2 )/µ > 0 . By letting y =0.5/µ ,lim=;s1 ,s2 )/lim 2 y / {1+exp((1-α )(s2 -s1 )y )}= lim 2 / {1/y +exp((1-α )(s2 -s1 )y )/y }=0 .
f (0.5/mm
y →∞

m→ 0

y →∞

Then, lim σ (y )/m =0. By substituting this into (A.5), we have
m→ 0

2
2

-1

-1
2
2
2

lim s (=y ) 1+exp{α(s1 -s2 )lim (sm= 0.5. Therefore, (σ1 (y ),σ2 (y ))2 (y )/ )} = [ 1+exp(0)]

m→ 0
m→ 0

2
2
=(0.5 , 0) in limiting equilibrium. Similarly, if s1 >s2 , (σ1 (y ),σ2 (y )) = (0 , 0.5) in limiting
21equilibrium.
Next, we will show that for each i, σ1i (w) → 1 as µ → 0 . For each i, define a122function of x , hi ( x ; si , µ ) =sss
j (x ) i (y ) j (y )( x − si ) . By the above argument, for any

si , x ∈ Sd with x ≠ si , we have

lim hi ( x ; si , µ ) = lim σ1j (x ) ⋅ lim σm2 (y ) ⋅ lim σ2k (y ) ⋅ lim ( x − si )= lim σ1j (x ) ⋅ 0.5 ⋅ 0 ⋅ lim( x − si )=0
µ→0

µ→0

µ→0

µ→0

µ→0

µ→0

µ→0

where (m , k ) = (i , j ) if x > si and (m , k ) = ( j , i ) otherwise. Finally, hi (si ; si , µ ) =0. Hence,hi ( x ; si , µ ) converges to the constant function 0 on S as µ → 0 . Then, the second and thirdterms in (A.6) also converge to 0. Then, for each i and any si ∈ Sd , we haveπi (si ) → ∑ σ1j (x ){(2α − 1)min{si , x} + w} as µ → 0.
x∈Sd

19

Now we assume on the contrary that there exists i and s < w withs=min{x ∈ Sd |s1i (x )>0} in the limiting equilibrium. Let s′ ∈ {s + 1 ,...,w} . Then , we have1
πi (s′) − πi (s ) → ∑ saj (x )(2 − 1)(min{ s′, x} − min{ s , x})

(A.6)
=

∑

s ≤ x ≤ s′ − 1

x∈Sd

1
sa
j (x )(2 − 1)( x − s ) +1∑ sa
j (x )(2 − 1)( s′ − s )as m → 0.

s′≤ x ≤ w

Together with α ∈ (0.5 , 1) , s < s′ , and σ1i (x ) ≥ 0 for each i and any x ∈ Sd ,(A.6) is nonnegative. By the definition of limit LAQRE and the assumption of s1i (s )>0 ,(A.6) is nonpositive. Then, (A.7) must be 0. Then, s1i (s′) =s1i (s )>0 in the limiting
equilibrium. Together with s=min{x ∈ Sd |s1i (x )>0} and the definition of limit LAQRE, it
implies that σ1i is the uniform distribution over {s,...,w} , that is, s1i (x )=1/(w-s+1) if
x ∈ {s,...,w} and σ1i (x )=0 otherwise. By substituting s1j (x )=1/(w-s+1) into (A.6),

πi (s′) − πi (s ) converges to some strictly positive value, a contradiction to s1i (s′) =s1i (s )>0 .

Hence, for each i, σ1i (w) → 1 as µ → 0 . Therefore, the unique limit LAQRE outcome isefficient. █, s2 ) y if si < s j ,Proof of Lemma 3. Let p =(( w , t1 (⋅)),( w , t2 (⋅))) with for each i, ti (s1=(s1 , s2 ) , andor n if si s=
ti (s1 , s2 ) = y=j , ti ( s1 , s2 ) n otherwise. If s1 = s2 , g(( s1 , ⋅),( s2 , ⋅)) =hence both players are indifferent between y and n. Next, consider the case s1 ≠ s2 . Since tiprescribes weakly dominant choice in every second stage, for each i, j ≠ i and eacht ′j ∈ { y , n} , ui (ti (s1 , s2 ), t ′j |s1 , s2 ) = maxti′ ui (ti′ , t ′j |s1 , s2 ) . Then, for each i, ti′′ ≠ ti implies
Ri (ti (s1 , s2 ), ⋅|s1 , s2 ) =0 < Ri (ti′′, y|s1 , s2 ) . Hence, p prescribes the SPMRE choice in every
second stage. Then, the reduced normal form game has an echelon structure. Sinceui ( w , w|t1 , t2 ) = 2 αw > w + (2 α − 1)si =ui (si , w|t1 , t2 ) for each i, j ≠ i and any si ≠ w , wehave max Ri ( w , s j |t1 , t2 )= 0 < (2 a − 1)( w − si )= Ri (si , w|t1 , t2 )= max Ri (si , s j |t1 , t2 ) . █s j ∈Ss j ∈S

Proof of Proposition 4. Let g be the mate choice mechanism (MCM).
(i) BEWDS. Consider the second stage. Take any a , b ∈ S . Note that ( w − a) + a(a + b ) ≥ w
if and only if a ≤ b−1 b . Then, t1 ( a , b ) = y if a ≤ b−1 b and t1 ( a , b ) = n otherwise. 28 Then,in the reduced normal form game, u1 ( a , b=
) ( w − a) + a(a + b ) andw otherwise. A similar) ( w − b ) + a(a + b ) if bb ≤ a ≤ b−1 b and u=u2 ( a , b=
1 ( a , b ) u=
2 ( a, b)argument holds for player 2.
28

We assume that a player chooses y if y and n are indifferent.

20

First, we show that for each i, choice βw weakly dominates any choicesi ∈ {0} ∪ (βw , w] . Take any i and si ∈ (βw , w] . If s j ∈ [βsi , w] , then βs j < si ≤ β−1 s j .
) ( w − βw) + α(βw + s j )Together with βs j ≤ βw ≤ β−1 s j , it implies that ui (βw , s j=

> ( w − si ) + α(si + s j )=ui (si , s j ) . If s j ∈ (β2 w , βsi ) , then we have βs j < βw < β−1 s j andsi > β−1 s j . Then, ui (βw , s j=
) ( w − βw) + α(βw + s j ) > w =ui (si , s j ) . If s j ∈ [0, β2 w] , then wew =ui (si , s j ) . Next, assume si = 0 .
have βw ≥ β−1 s j and si > β−1 s j . Then, ui (βw , s j ) =ui (0, s j ) for any s j ≥ 0 ,Then, βs j > si , for any s j > 0 . Then, we have ui (βw , s j ) ≥ w =with strict inequality for s j > β2 w . Second, for any si ∈ (0, βw] , si is the unique best
response toβ−1 si in the reduced normal form game and hence si survives under

BEWDS. Therefore, we get the desired result.
(ii) SPNE. Consider the second stage. Take any SPNE p =((s1 , t1 (⋅)),(s2 , t2 (⋅))) . If
−1s1 > β s2 or s2 > β−1 s1 , some player chooses n in subgame (s1 , s2 ) , otherwise the player
does not provide the best response. Hence, the players finally contribute nothing in p. Next,
take any (s1 , s2 ) such that βs2 ≤ s1 ≤ β−1 s2 . We show that there exists an SPNE where bothplayers approve and hence contribute (s1 , s2 ) . Consider the strategy profileq =((s1 , t(⋅)),(s2 , t(⋅))) with t(s1 , s2 ) = y and n otherwise. Clearly, q constitutes a Nashequilibrium for every second-stage subgame. Now, consider the reduced normal form
game. Assume player 1 deviates to s1′ ≠ s1 . By construction of q, such a deviation results in(n,n). Together with s1 ≤ β−1 s2 , this implies that u1 (q=) ( w − s1 ) + α(s1 + s2 ) ≥ w = u1 (s1′ , q ) .
Hence, choosing s1 is the best response to s2 . This is also true for player 2. Hence, q is an
SPNE with contributions (s1 , s2 ) .
(iii) LAQRE. Let Sd = {0,1, , w} . Take any s1 ,s2 ∈ Sd . If (ti , t j ) ≠ (y , y ) ,ui ((si , ti ),(s j , t j ))=w=πi (n) . Then, πi (y ) = αs2j (y )(s j − βsi ) + πi (n) . Together, this and thedefinition of LAQRE imply that12{1+exp(-αssi2 (y ) {1+exp((pp=µ )} -βsi )/µ )} 1 .
(A.8)=i (n )- i (y ))/j (y )(s j

On the other hand, by the first equation satisfied by LAQRE, for each si ∈ S ,22
ss
we have πi (si )= ∑ s1j (x )πi (si , x ), where πi (si , x ) =i (y ) j (y )ui ((si , y ),(x , y )) +x∈Sd22
(1 − ssi (y ) j (y ))ui ((si , n ),(x , n )) . Since22πi (si , x ) =ssi (y ) j (y ){(w − si )+α(si + s j )} +2222
(1 − ssαss
i (y ) j (y ))w =i (y ) j (y )(s j − β si ) + w , we have

(A.9)

πi (si )= ∑ σ1j (x )πi (si , x ) =α ∑ σ1j (x )σi2 (y )σ2j (y )(x − βsi ) + w ∑ σ1j (x )
x∈Sd

x∈Sd

x∈Sd

Now let us consider limiting equilibrium. Consider first (s1 , s2 ) = (0,0) . Then,21

(σ12 (y ),σ22 (y ))=(0,0) constitutes the unique limiting equilibrium in the second stage.
Consider next (s1 , s2 ) ≠ (0,0) . If si ∈ (βs j , β−1 s j ) for each i = 1,2, j ≠ i , (σ12 (y ),σ22 (y ))=(1,1)
constitutes the unique limiting equilibrium. Similarly, if β−1 si < w and s j ∈ (β−1 si , w] , then(σi2 (y ),σ2j (y ))=(0.5,0) constitutes the limiting equilibrium. If si = β−1 s j , then(σi2 (y ),σ2j (y ))=(0.5,1) constitutes the limiting equilibrium.
In order to get the first-stage behavioral strategy in limiting equilibrium, for each i,22
let us see ssi (y ) j (y )(x − β si ) as a function of x ∈ Sd . Then, by the limiting equilibrium
22
choices in the second stage, we have ss
i (y ) j (y )(x − β si ) → x − β si as µ → 0 if
22−1− β)si as µ → 0 if x = β−1 si , andx ∈ [βsi , β−1 si ) , ssi (y ) j (y )(x − β si ) → 0.5(β22ssi (y ) j (y )(x − β si ) → 0 as µ → 0 otherwise. By substituting these results into (A.9), if

β−1 si is an integer, πi (si ) → α
and otherwise πi (si ) → α∑−1x∈Sd ∩[ βsi ,β si )∑−1x∈S ∩[ βs ,β s )

σ1j (x )(x − βsi ) + 0.5ασ1j (β−1 si )(β−1 − β)si + w as µ → 0

σ1j (x )(x − βsi ) + w as µ → 0. Next, we show that thereiiexists a limiting equilibrium dwhere
bothplayers choose 1 with probability 1. Suppose thatfor each i, σ1i (1) → 1 as µ → 0 . By β < 1 , we have for each i,πi (1) − πi (0) → α∑x∈Sd ∩[ β ,β−1 )

σ1j (x )(x − β) + w − w = α(1 − β) > 0.

Let si > 1 . We have for each i,πi (1) − πi (si ) → α∑x∈Sd ∩[ β ,β−1 )

σ1j (x )(x − β) + w − α∑x∈Sd ∩[ βsi ,β−1 si )σ1j (x )(x − βsi ) − w≥ α ⋅ 1 ⋅ (1 − β) − α ⋅ 1 ⋅ (1 − βsi )= αβ(si − 1) > 0
where the equality holds if βsi ≤ 1 . Therefore, we get the desired result. █(βw , t1 (⋅)),(βw , t2 (⋅))) be a strategy profile with for each i,(iv) SPMRE. Let p =j ≠ i , ti (s1 =, s2 ) y if si ≤ β−1 s j and ti (s1 , s2 ) = n otherwise. If si = β−1 s j , player i is
indifferent between y and n. Next, consider the case si ≠ β−1 s j . Since ti prescribes weakly
dominant choice in every second stage, for each i, j ≠ i and each t ′j ∈ { y , n} , ti′ ≠ ti (s1 , s2 )implies ui (ti (s1 , s2 ), t ′j |s1 , s2 ) = maxti′′ ui (ti′′, t ′j |s1 , s2 ) > ui (ti′ , y|s1 , s2 ) . Hence, p prescribes theunique SPMRE choice in every second stage.
Consider the reduced normal form game, which is identical to the one induced by
BEWDS. Take any s1 , s2 ∈ S . As shown in part (i), player i best responds to player j if

si = βs j . Then, for each i, maxsi ui (si , s j |t1 , t2 )= w − βs j + a(βs j + s j )= w + (1 − β)s j .
Consider player 1’s regret. If s1 > β−1 s2 or s2 > β−1 s1 , either player chooses n and henceu1 (s1 , s2 ) = w . Then, R1 (s1 , s2 )= ( w + (1 − β)s2 ) − w= (1 − β)s2 . If βs2 ≤ s1 ≤ β−1 s2 , bothplayers choose y in the second stage. Then, we have u1 (s1 , s2 )= w − s1 + α(s1 + s2 ) andR1 (s1 , s2 ) = ( w + (1 − β)s2 ) − { w − s1 + α(s1 + s2 )} = (βs1 − β2 s2 ) /(1 + β) , which is maximized at22s2 = βs1 . Hence, player 1’s maximum regret at s1 is given by R (s , w)= (1 − β)w if s1 < βw ,maxs R1 (s1 , s2 ) =  1 1
 R1 (s1 , βs1 ) = β(1 − β) s1 otherwise.
2

By α ∈ (0.5,1) , we have β < 1 . Together with s1 ≤ w , it implies that s1 = βw minimizes
the maximum regret and gives the solution. .The same argument holds for player 2.

Therefore, (βw , βw ) constitutes the unique minimax regret equilibrium of the reduced
normal form game.
(v) Level-k. First, we show that βw is the best response to level-0 player. Given
that the other player uniformly randomizes in the reduced normal form game, by a similarargument in (iii), the expected payoff π′i (si ) when player i chooses si ismin{ w ,β−1 s }i. dx + 1 (min{ w , β−1 si } − βsi )2 / 2 w + 1.
π′i (si )= ∫(x − βsi ) / w=
βsiThen, si =βw maximizes π′i (si ) . Let B(k) be a sequence of the level-k player’s contribution

where k ≥ 1 . Then, B(1) = βw and B(k) = BR(B(k − 1)) until B(k) = B(k - 1), where BR(.) is thebest response to B(k − 1) in the reduced normal form game induced by elimination of
weakly dominated strategies. Since BR(s ) = βs for any s ∈ S , we get the desired result. █Acknowledgements
We would like to thank two anonymous referees for their suggestions. We also thankKanemi Ban, Vincent Crawford, Yukihiko Funaki, Keigo Inukai, Michihiro Kandori,
Tetsuya Kawamura, Shuhei Morimoto, Yuko Morimoto, Masahiro Okuno, Kazuki Onji,
Ariel Rubinstein, Tatsuhiro Shichijo, Masanori Takaoka, Masanori Takezawa, Wataru
Tamura, Toshio Yamagishi, and Takafumi Yamakawa. Keiko Takaoka provided
outstanding research assistance. We are also grateful to the participants at the 11th SAET
Conference; the 2011 Japanese Economic Association Autumn Meeting; the 2011, 2012, and2013 Asia-Pacific Economic Science Association Meeting; and the seminar at Osaka
Prefecture University for their helpful comments. This research was supported by the Joint
Usage/Research Center at ISER, Osaka University and “Experimental Social Sciences:
Toward Experimentally-based New Social Sciences for the 21st Century,” the latter being a
project under the aegis of the Grant-in-Aid for Scientific Research on Priority Areas of the
Ministry of Education, Science, and Culture of Japan.
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25

Figure 1. Prisoner’s dilemma game with the MCM in Saijo et al. (2013).
Figure 2. Average contributions by period, sorted by mechanism.

0

2

4

6

8

10

0

10

11.4

12.8

14.2

15.6

17

2

9.4

10.8

12.2

13.6

15

16.4

4

8.8

10.2

11.6

13

14.4

15.8

6

8.2

9.6

11

12.4

13.8

15.2

8

7.6

9

10.4

11.8

13.2

14.6

10

7

8.4

9.8

11.2

12.6

14

Table 1. Player 1’s payoff table when w = 10 and α =0.7.

1

y

n

y

(9.6,13.6)(10.8,10.8)

n(10.8,10.8)(10.8,10.8)

Subgame (6,2)

y

n

y

(14.6,12.6)(13.2,13.2)

n(13.2,13.2)(13.2,13.2)

Subgame (8,10)

Table 2. Payoff tables in subgames (6,2) and (8,10).

2

0

2

4

6

8

10

0

10

10

10

10

10

10

2

10

10.8

10.8

10.8

10.8

10.8

4

10

10.8

11.6

11.6

11.6

11.6

6

10

10.8

11.6

12.4

12.4

12.4

8

10

10.8

11.6

12.4

13.2

13.2

10

10

10.8

11.6

12.4

13.2

14

Table 3. Player 1’s payoff table in the reduced normal form game under the MAM.

3

Treatment SMAM
MAM
3.799**
SMAM
MCM

MCM
9.450**
9.133**

VCM
9.445**
9.275**
7.284**

Notes: ** p < 0.01.
Table 4. Two-tailed Mann–Whitney test statistics for the equality of average contributions,grouped by treatment.

4