The emergence of norms of cooperation in stag hunt games with production
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Bagnoli, Lidia; Negroni, Giorgio Working Paper The emergence of norms of cooperation in stag hunt games with production Quaderni - Working Paper DSE, No. 626 Provided in Cooperation with: University of Bologna, Department of Economics Suggested Citation: Bagnoli, Lidia; Negroni, Giorgio (2008) : The emergence of norms of cooperation in stag hunt games with production, Quaderni - Working Paper DSE, No. 626, Alma Mater Studiorum - Università di Bologna, Dipartimento di Scienze Economiche (DSE), Bologna, https://doi.org/10.6092/unibo/amsacta/4624 This Version is available at: https://hdl.handle.net/10419/159467 Standard-Nutzungsbedingungen: Die Dokumente auf EconStor dürfen zu eigenen wissenschaftlichen Zwecken und zum Privatgebrauch gespeichert und kopiert werden. Sie dürfen die Dokumente nicht für öffentliche oder kommerzielle Zwecke vervielfältigen, öffentlich ausstellen, öffentlich zugänglich machen, vertreiben oder anderweitig nutzen. Sofern die Verfasser die Dokumente unter Open-Content-Lizenzen (insbesondere CC-Lizenzen) zur Verfügung gestellt haben sollten, gelten abweichend von diesen Nutzungsbedingungen die in der dort genannten Lizenz gewährten Nutzungsrechte. Terms of use: Documents in EconStor may be saved and copied for your personal and scholarly purposes. You are not to copy documents for public or commercial purposes, to exhibit the documents publicly, to make them publicly available on the internet, or to distribute or otherwise use the documents in public. If the documents have been made available under an Open Content Licence (especially Creative Commons Licences), you may exercise further usage rights as specified in the indicated licence. https://creativecommons.org/licenses/by-nc/3.0/
The emergence of norms of cooperation in stag hunt games with production Lidia Bagnoli , Giorgio Negroniy February 6th, 2008 Abstract In this paper we study a two agents asymmetric stag hunt game. The model has an in…nity of strict, Pareto rankable Nash equilibria. The equilibrium selection problem is solved by appealing to the stochastic stability concept put forward by Young (1993). We prove two main results. When the action sets are numerable in…nite sets, then for any value of the distributive parameter we can expect the emergence of a norm involving less than maximal cooperation. When instead the action sets are …nite sets of a particular type (in the sense that each agent can choose his maximum optimal e¤ort and fractions of this), then for some value of the distributive parameter we can expect the emergence of a norm involving maximal cooperation. keywords: asymmetric stag hunt game; stochastic stability; cooperation norms. 1 Introduction According to Skyrms (2004), the stag hunt is a story that became a game. The game is a prototype of the social contract while the story is told by Rousseau. Consider two hunters who have to decide whether to cooperate to hunt a stag. Suppose that, in the case they succeed in hunting the stag, the catch is divided equally. Hunting stags is demanding and it requires the cooperation of both. Suppose that, while waiting for the stag, a hare happens to pass within reach of one of them; hunting hares is much easier: it requires a minimum e¤ort and it can be done successfully without the cooperation of the other hunter. Let us assume that no binding agreement is possible for the two hunters. Although for each of them half a stag is more valuable than a hare, they can not be sure that the other player will provide the required e¤ort. In other terms, although the situation in which both hunt the stag is a Nash equilibrium which Pareto dominates the other equilibrium in which both hunt the hare (the minimax Regione Emilia Romagna, Bologna. yBologna University and DARTT. E-mail: [email protected] 1
solution), the former equilibrium may fail to be risk dominant. See Carlsson and Van Damme (1993). In this paper we follow Bryant (1983) and Cooper (1999) and use the stag hunt game as a model of team production. We assume that agents’e¤ort are complementary inputs so that total output of the team is determined by the least e¤ort. In terms of the stag hunt parable, this means that we focus on the total catch of the hunters rather than on the stag/hare alternative. It is a standard result in games with strategic complementarities that, when the marginal bene…t from coordinated actions is greater than the marginal cost of e¤ort, then any common level of e¤ort is a strict Nash equilibrium; see Cooper and John (1988). Moreover, since all individuals prefer an equilibrium in which all players supply higher e¤ort, all these Nash equilibria can be Pareto ranked. We are thus in a situation in which since the Nash equilibrium concept neither prescribes nor predicts the outcome of the game, it needs to be supplemented with an adequate theory of equilibrium selection. Any traditional re…nement would not help as long as the play is simultaneous; of course if we consider a sequential game, then the only subgame perfect equilibrium is the Pareto optimum one. However, the evidence from experimental economics suggests that the Pareto dominant equilibrium is quite unlikely in the simultaneous stag hunt game; in this case, in fact, life can indeed be "inside the production possibility frontier" (Cooper (1999), pg. 151). See Van Huyck, Battalio and Beil (1990), Van Huyck, Cook and Battalio (1997) and the discussion in Crawford (1991, 1995). We depart from Bryant (1983) in three respects. First we consider a two agents economy with identical (separable) utility functions but di¤erent productivities. Second, although the distributive parameter (x)regulating the distribution of the joint product among the two agents is …xed1, we do not assume from the outset that the resulting distribution is egalitarian.2Third, we assume that our stag hunt game is played by boundedly rational, randomly matched players, along the lines suggested by Young (1993, 1998). In a sense we are considering a stag hunt played by boundedly rational "strangers".3 Since our strangers are engaged in a strategic game, they need to form an expectation on the behavior of their opponent. Following Young we consider the case in which this expectation is shaped by the accumulation of antecedents, according to an inductive process. Suppose that each agent collects a sample of size kfrom the last mpast plays of the game, with k < m. Given khe then extracts the empirical frequencies with which each (pure) strategy was played in the past by other agents. With probability 1each agent then chooses an 1We assume that it is determined by the existing distributional rule (left unexplained) endorsed by the society. 2As pointed out by Cooper (1999), the coordination problems arising in Byant’s (1983) model is partly a consequence of the rule that distributes equally the fruits of the cooperation regardless of individual e¤ort levels. See also Bryant (1994). 3This last assumption makes our contribution departing also from Crawford’s (1991) evolutionary approach to the stag hunt as well as from Crawford’s (1995). 2
action which is a best reply to these empirical frequencies while with probability each agent makes a mistake, i.e. he chooses an action which is not a best reply. The strategies chosen in the current period are recorded and in the next period the game will be played, along the same lines, by another draw of agents from the same population. Following Young (1993) we say that a state is stochastically stable if, in the long run, it can be observed with positive probability when the probability of mistakes is small and the sample is su¢ ciently limited. When there is a unique stochastically stable state this is the equilibrium that, in the long run, will be observed with probability close to one so that it becomes the conventional way of playing the game. In this sense the approach gives us a theory of equilibrium selection. We …rst analyze the case in which agents can choose among a continuum of pure strategies (i.e. e¤ort levels ei2[0; emax i]) Let be(x)denote the Nash equilibrium where bei(x)2[0;bemax i(x)] :For any xwe have a continuum of Nash equilibria which can be Pareto ranked. We say that a Nash equilibrium involves maximal cooperation if at least one agent supplies his maximum optimal e¤ort, that is bei(x) = bemax i(x): Our …rst result says that, for any value of the distributive parameter, interactions of boundedly rational strangers converge to a stochastically stable state eS 1(x); eS 2(x)which, since it involves less that maximal cooperation (i.e. eS i(x)<bemax i(x)) is not Pareto e¢ cient;the precise state to which our economy converges depends on the value of the distributive parameter. However, we also show that at any stochastically stable state each agent supplies an e¤ort which is never smaller than half of his optimal maximum level. Hence, since 1 2bemax i(x)eS i(x)<bemax i(x);we say that in our economy the stochastically stable state involves a minimal cooperation. This is the equilibrium that is easiest to ‡ow into from all other states in the sense that it is more robust to agents’mistakes than all other equilibria; for this reason it tends to persists and becomes a conventional way of playing the stag hunt game for our strangers (or a conventional social contract). This is quite surprising because, since our strangers can choose among a continuum of actions, they are in the case in which the risk of miscoordination is the highest. Nevertheless our …rst result, far from suggesting that the social contract "might degenerate spontaneously into the state of nature" as claimed by Skyrms (2004; pg. 12), tells us that we can quite con…dently expect the emergence of a norm involving minimal (but positive) cooperation. To return to the stag hunt metaphor, it is true that at this stochastically stable state the hunters do not get the maximal catch they could; however, each gets more than the catch he could get alone (i.e. in the state of nature). Our …rst result also suggests that it can e¤ectively be di¢ cult for our agents to improve upon the conventional social contract. In fact, although our strangers may realize (peraphs with the help of an external observer) that, for the given value of the distributive parameter, a larger pie can be achieved if they supply higher optimal e¤ort (thus resulting in an equilibrium which Pareto dominates the stochastically stable one), the resulting equilibrium is not stochastically 3
stable.4 We then analyze the interactions of boundedly rational strangers when each can choose among a …nite number of pure strategies. Our second result says that, although equilibria with minimal cooperation may still be stochastically stable, now it is also possible to observe the emergence of a norm involving maximal cooperation. However the emergence of this norm is not due to any e¢ ciency considerations; in fact, we show that the stochastically stable equilibrium involving maximal cooperation is Pareto e¢ cient for some values of the distributive parameter only and provided that the number of strategies available is not too big and agents have su¢ ciently di¤erent productivities. Lastly we show that when each agent can choose more that two actions, the equilibrium with maximal cooperation corresponding to the value of the distributive parameter suggested by the utilitarian unweighted cooperative solution will never be stochastically stable. We remark that our results are not due to any incentive problem; see Legros and Mathews (1993), Vislie (1994) and Hvide (2001). The remaining of the paper is organized as follow. In Section 2 we present a variant of Bryant (1983) symmetric coordination problem. In Section 3 we introduce our asymmetric coordination game. In Section 4 we brie‡y summarize Young’s (1993) concept of stochastic stabile state. Section 5 then discusses the stochastic stability of our asymmetric coordination game …rst when agents have a continuum of strategies and then when agents can choose only among a …nite number of discrete strategies. Section 7 summarizes our results. 2 A symmetric coordination game In this section we brie‡y present a variant of Bryant’s (1983) coordination game. Consider two equally productive agents engaged in a joint project and let Y=min [e1; e2] be the technology available, where ei2[1; emax]. Suppose that the outcome of the cooperation is divided according to the distributional parameter xso that agent 1 gets Y1=xY and agent 2 gets Y2= (1 x)Y: Let us …rst suppose5 x= 1=2:Denoting by Vi(Yi; ei) = Yibeithe payo¤ of the generic agent i; we get Vi= 2min [e1; e2]bei=amin [e1; e2]bei; 4One possibility open to our strangers would be to agree on conditional contracts, whose enforcement is ensured by an external observer, in which (a) they agree on a particular division of the fruits of social cooperation and (b) they supply their maximal optimal e¤ort, given this value of the distributive parameter:However this alternative is not viable in our economy since there is no external observer. A still di¤erent alternative would be to consider self-enforcing social contracts, as in Binmore (1994). We leave the exploration of this alternative to future research. 5As it can be veri…ed, x= 1=2is also to the cooperative solution of the game. See Cooper (1999). 4
where we assume a > b: This is the game studied by Van Huyck et al. (1990). Because of the technological complementarity, and since e¤ort is costly, no agent has the incentive to choose an e¤ort such that his contribution to the joint project is larger than the contribution of the other agent. Let e2=ee: The payo¤ of agent 1 is V1(ee; ee) = (ab)eeif e1=eeand V1(e1;ee) = (ab)e1if e1<ee: Analogously for agent 2. Since V1(ee; ee)V1(e1;ee) = (eee1) (ab);it follows that all the pro…les (e1; e2)=(e; e)with eemax are Nash equilibria, being a > b: Suppose now that the distributive parameter is x: Then V1=x min [e1; e2]be1 V2=(1 x) min [e1; e2]be2: Let e2=ee: The payo¤ of agent 1 is V1(ee; ee)=(x b)eeif e1=eeand V1(e1;ee) = (x b)e1if e1<ee: Hence V1(ee; ee)V1(e1;ee)if (eee1) (x b) 0;that is if xb=: Consider now player 2 and let e1=ee: Then V2(ee; ee) = ((1 x)b)eeif e2=eeand V2(e2;ee) = ((1 x)b)e2if e2<ee: Hence V2(ee; ee)V2(e2;ee)if (eee2) ((1 x)b)0;that is if x(b)=: It turns out that all the pro…les (e1; e2)=(e; e)with eemax are Nash equilibria if and only if b= x(b)=: In this case the game is a stag hunt. However if the distributive parameter does not satis…es this condition, then the game admits only one Nash equilibrium in which (e1; e2) = (0;0) : 3 An asymmetric coordination game In this section we modify the basic model by allowing some form of heterogeneity; speci…cally we assume (a) that individual e¤orts are not equally productive and (b) that the distribution of the fruits of joint production is not a priori egalitarian. Moreover, since we want to consider a game that maintains the structure of the stag hunt for any value of the distributive parameter, we modify the Van Huyck et al. (1990) model by considering a non linear e¤ort disutility. In the next section we shall use this model to study the problem of equilibrium selection by appealing to the inductive argument put forward by Young (1993). As in Cooper (1999), we consider an economy populated by two individuals engaged in a joint project. The output produced is given by Y= min [e1; e2](1) where e1and e2denote the e¤ort levels chosen by the two agents while and are two real numbers representing the (possibly) heterogeneous individuals’ productivities. We suppose that there is an already established distributional rule which determines how the output is shared by the two individuals; letting xdenote the share of the production going to agent 1, we get Y1=xY and Y2= (1 x)Y: The output received by each individual is entirely consumed. 5
Let6 V1=xmin [e1; e2]e2 1 V2= (1 x) min [e1; e2]e2 2: (2) denote the agents’payo¤s and let emax 1==2and emax 2==2be the maximum feasible level of e¤orts for the two agents. Let Gdenote the game in which agents simultaneously selects their e¤ort levels from the sets Si= [0; emax i]and receive a payo¤ given by (2) :Given the distributional parameter, and given agent’s 2 e¤ort, the problem faced by agent 1 is to choose7e1to maximize xe1e2 1 subjected to e1e2:Let be1be the solution of this problem, where: be1=8 > < > : x 2if e2x2 2; e2if e2x2 2: (3) Analogously, the problem faced by agent 2 is to choose e2to maximize (1 x)e2e2 2subjected to e2e1:Let be2be the solution of this problem, where: be2=8 > < > : (1 x) 2if e1(1 x)2 2; e1if e1(1 x)2 2: (4) We can state the following result.8 Proposition 1 For any x2[0;1] there is an in…nity of strict, pure strategies Pareto rankable Nash equilibria. Let :Then: (a) the set of Nash equilibria is (be1; be1)where be1min bemax 1; 1bemax 2= min x 2;(1x)2 2; (b) for any given x; the equilibrium in which at least one agent o¤ers his maximum optimal e¤ort, i.e. be1=bemax 1=xemax 1and be2=bemax 2= (1 x)emax 2; is Pareto dominant. Proof. See the Appendix. For each player we can write the payo¤ corresponding to any Nash equilibrium as: V1(x) = be1(x) (x be1(x)) V2(x) = be1(x)(1 x)2be1(x)(5) 6The quasi-linearity with respect to the consumption good is essential because in our economy the consumption good is the numeraire; cfr. Ray et al.(2006). 7We restrict our analysis to the case of pure strategies only. 8In a similar model, Anderson et al. (2001) argue that a change in e¤ort cost does not a¤ect the Nash equilibria. This is not necessarily true in our model. 6
In this section we restrict our analysis to the case in which, for any value of the distributive parameter, at least one agent chooses his maximum level of optimal e¤ort9, i.e. bei=bemax i:We say that the corresponding Nash equilibrium involves maximal cooperation:From (1) and Proposition 1 it follows that at any Nash equilibrium the level of production with maximal cooperation is Y=8 > < > : bemax 1=x2 2if x x; bemax 2= (1 x)2 2if x x; (6) where x2 2+2=1 2+1 :Production attains its maximum level Ywhen x=x. The following table shows the income and the utility distributions as functions of x; corresponding to the Nash equilibrium with maximal cooperation. xxxx Y1(x)x22 2x(1 x)2 2 Y2(x)x(1 x)2 2(1 x)22 2 V1(x)x22 4(1 x)2 2hx1 + 2 222 22i V2(x)x2 21x2 22+ 1 (1 x)22 4 (7) The maximum optimal e¤ort is supplied by agent 1 when xxand by agent 2 when xx:Notice that xcorresponds to the unweighted utilitarian cooperative solution of the model. The next Lemma establishes some relevant properties of the equilibrium payo¤s functions V1(x)and V2(x). Lemma 2 Consider the payo¤ functions V1(x)and V2(x)given by (5) :Then: a) V1(x) = V2(x)=0for x= 0 and x= 1:For any 0< x x; V1(x) is an increasing and convex function while for any xx; V1(x)is a concave function with a maximum V1at x=x1:For any 0< x x; V2(x)is a concave function with a maximum V2at x=x2while for any xx; V2(x)is a decreasing and convex function. V1(x)and V2(x)are maximized respectively when x1=2+2 22+2; x2=2 2+ 22;(8) where, for any (; ); x2< x< x1and x2<1 2< x1. 9Notice that since Bryant (1983) considers only an egalitarian distribution, he …nds that there is only one Pareto dominant equilibrium. In our case instead, we have one Pareto dominant equilibrium for any x: 7
b) Let > : Then V1< V 2:Moreover V2(x)> V1(x)for every x<x3 and V2(x)V1(x)for every xx3where x32+2 32+2;(9) and x2< x< x3: c) Let < : Then V1> V 2:Moreover V2(x)> V1(x)for every x<x4 and V2(x)V1(x)for every xx4where x422 2+ 32;(10) and x4< x< x1: Proof. Omitted since it relies on simple algebraic manipulations. The next Lemma shows that the game has two properties that will be useful in the following analysis. Lemma 3 The game Gis acyclic and satis…es the bandwagon property. Moreover, let L(e)denote the length of the shortest best reply path originated in the strategy pro…le e;then L= max L(e) = 2: Proof. See the Appendix. Acyclicity means that the best reply graph contains no directed cycles, a property satis…ed by all coordination games. A su¢ cient condition for the (marginal) bandwagon property to hold for generic (i.e. not necessarily acyclic) symmetric games has been proved by Kandori and Rob (1998):A reformulation which holds for acyclic but not necessarily symmetric two players games is given by Binmore, Samuelson and Young (2003). Let = (2; S1; S2; V1; V2)be a …nite acyclic game with two players and let SNdenote the set of all strict Nash equilibria of the game. exhibits the bandwagon property if for each s2SN and s2Swith si6=sifor any i= (1;2) the following conditions is satis…ed V1(s1; s2)V1(s1; s2)V1(s1; s2)V1(s1; s2);(11) with an analogous condition holding for agent 2. This essentially says that, for both agents, deviations from the equilibrium strategy are more costly when the opponent plays his part of the equilibrium. Example 1. Consider Table (7) and let = 2 and = 1:Figure 1 plots Vias a function of the distributive parameter:The maximum value of V1is 1=9 and it is achieved when x=x1= 5=9; the maximum value of V2is 1=6and it is achieved when x=x2= 1=6:Notice that V1=V2for x=x3= 5=13: Lastly, when x=x= 0:2(that is when the output produced is the maximum possible), V1= 0:04 and V2= 0:16: 8
although agents observe that the Nash equilibrium awas played mtimes in succession, some agent by mistake plays action B: The resistance of the direct path considered is r(a; b) = 1 4k: As shown in the second column of Table 2, when direct paths only are considered, the total resistance of the brooted tree is r(a; b) + r(c; b) = 3 8k: Other two brooted trees are possible, both involving a composite path. In the …rst, the path followed by economy is c!a!b: As before, suppose that the convention cis played mtimes in succession and that one player by mistake plays action A: Speci…cally, we suppose that he plays Afor k0periods and Cfor the remaining kk0periods. It turns out that the minimum number of mistakes that are su¢ cient to shift the economy from cto aare k0=3 8k for agent 1 and k0=3 4kfor agent 2. Since min 3 8;3 4=3 8;it follows that the resistance of the path is r(c; a) = 3 8k: Suppose now that once in afor a su¢ cient long period of time, one player by mistake plays action B: Proceeding as before, it turns out that the minimum number of mistakes that are su¢ cient to shift the economy from ato bare k0=5 8kfor agent 1 and k0=1 4kfor agent 2. Since min 5 8;1 4=1 4;we have r(a; b) = 1 4k: Consider now the complete composite path c!a!b;along this path the total resistance of the tree rooted at bis thus r(c; a) + r(a; b) = 5 8k: In the second, the path followed by the economy is a!c!b: By repeating the same reasoning we …nd that along this path the total resistance of the tree rooted at bis r(a; c) + r(c; b) = 7 8k: Table 2 shows all the possible irooted trees22 for our game (12) : idirect path comp. path 1 comp. path 2 P(i) a b ! 3 8k a ( 1 8k c b ! 3 4k c! 3 8k a c ! 1 8k b! 3 8k a1 2k b a ! 1 4k b ( 1 8k c a ! 3 4k c! 1 8k b c ! 3 8k a! 1 4k b3 8k c a ! 3 4k c ( 3 4k b a ! 1 4k b! 3 4k c b ! 3 8k a! 3 4k c k Table 2 The stochastic potential of state iis the minimum total resistance over all the irooted trees;hence, among all the trees rooted at ithe stochastic potential of this state identi…es the minimal tree. As shown in the last column of Table 2, the stochastic potential of state bis P(b) = min [r(a; b) + r(c; b); r (a; c) + r(c; b); r (c; a) + r(a; b)] = min 1 4+1 8k; 3 4+1 8k; 3 8+1 4k =3 8k 22 For games involving more strict Nash equilibria it is not possible to disentangle between direct and composite paths. 15
The stochastically stable state is the state with minimum stochastic potential. From inspection of the last column of Table 2 we conclude that min [P(a); P (b); P (c)] = P(b) so that the unique stochastically stable state of the above game is b; that is the strict Nash equilibrium pro…le (B;B)in which each agent supply half of his optimal maximum e¤ort.23 5 Stochastically stable states in the stag-hunt game with production In this Section we derive the stochastically stable states for the economy described in Section 3. Let assume that this economy is populated by Nboundedly rational agents and let N1and N2be the sub-populations of agents 1 and 2 respectively; in each period, one agent is randomly selected from each subpopulation to play the stage-game. Since agents are boundedly rational, they are concerned with their stage-game strategies only. Since Young (1993) results hold for a …nite game, in order to apply his approach we have to shift from the game Gto a game G(de…ned below) with a …nite strategy set. Here is a real number su¢ ciently small and we interpret 1= as a degree of precision with which we measure e¤ort. For smaller and smaller values of ; since we can discriminate more …nely between e¤ort levels, the number of possible actions increases. In the limit as !0, agents can choose their e¤ort from a continuum of values. We shall consider two cases: in the …rst e¤ort is a continuous variable (in the sense just speci…ed) while in the second e¤ort is a discrete variable. 5.1 Case 1: e¤ort is (in the limit) a continuous variable Let xxand assume that agents can choose their equilibrium24 e¤ort levels from the …nite and discrete sets S1and S2respectively25 where S1=f0; ; 2; :::; bemax 1; bemax 1g[eS 1 S2=f0; ; 2; :::; (bemax 1); bemax 1g[eS 1: (13) 23 A careful reader should have noticed that in our case there is no composite path with total resistance smaller than the total resistance of the direct path. 24 Notice that the original action set for the generic agent iis Si=0; :::; bemax i; :::; emax i: However, since our game satis…es the bandwagon property, we can exclude all the e¤ort levels greater than the maximum optimal one; these actions will never be a Nash equilibrium and do not alter the resistances of transition between states. Here eS idenotes the stochastically stable e¤ort level; we include this in the set of feasible actions in order to derive exact results. See Binmore, Samuelson and Young (2003). 25 When instead xx;we have to consider the sets S0 1and S0 2;where S0 1=0; 1; :::; 1bemax 2; 1bemax 2[1eS 2 S0 2=0; ; :::; bemax 2; bemax 2[eS 2: 16
In the limit as !0, agents can choose their e¤ort from a continuum of values. In this case, our Proposition 5 establishes the existence of a stochastically stable state. In what follows we denote by Gthe stage game where the two randomly matched players simultaneously choose their e¤ort from the set Sigiven by (13) and receive a payo¤ given by (5) : Consider two Nash equilibria e= (e1; e1)and e= (e1; e1)and let ebe the initial state. Since the game satis…es the bandwagon property (Lemma 3), in order to derive the resistances it is thus su¢ cient to analyze the restrict game where the only strategies available are those corresponding to these two equilibria, that is S1=fe1; e1gand S2=fe1; e1g:Two cases are possible: either e1> e1or e1< e1:The former corresponds to a situation in which we exit from the state eto the right while the latter corresponds to a situation in which we exit from the state eto the left. We show in Claim 11 in the Appendix that the resistance of the path e!e, with e > e; is r+(e; e) = 8 > > > < > > > : 2(e1+e1) (1 x)k if x x e1+e1 x k if x x; (14) while the resistance of the path e!e, with e < e; is r(e; e) = 8 > > > > < > > > > : 1e1+e1 x k if x x 12e1+e1 (1 x)k if x x: (15) From (14) and (15) we notice respectively that r+(e; e)is an increasing function of e1and e1while r(e; e)is a decreasing function of e1and e1:For a given state e; and any value of x; it follows that: (a) the least resistance on an exit path is found on a direct path leading to the adjacent state, i.e. e=e+= (e1+; (e1+)) for (14) and e=e= (e1; (e1)) for (15) ; (b) let (e)denote the following erooted tree 0r(0;0) !r(;2) !::: er(e;e) !e r(e+;e)e+ :::: r(bemax;bemax)bemax (16) where each edge is weighted by the least resistance –given by (14) and (15) – involved in the corresponding transition. Let P(e)denote the minimum stochastic potential associated with the generic state e: Then (e)is the arborescence with minimum stochastic potential P(e): 17
As shown in Claim 12 in the Appendix, we can write the stochastic potential associated with a generic state eas26 P(e) = P(e) + r(e; e)r(e; e ) P(e) = P(e+) + r(e+; e)r(e; e +) (17) Since the game Gis acyclic, we know from Young (1993) that it has at least one stochastically stable state, es =eS 1; eS 2:This is the state which minimizes the stochastic potential over all the possible states, i.e. es = arg mine P(e):Let27 0< es <bemax;then it must be P(es )< P (es +)and P(es )< P(es );conditions satis…ed when28 8 < : r(es +; es )r(es ; es +)<0 r(es ; es )r(es ; es )<0: (18) From (14) ;(15) and (18) it then follows that, for any value of the distributive parameter, es is a stochastically stable state if es x 2 1x 1 + x21< 2: Therefore, as !0the stochastically stable state tends to the equilibrium es=eS 1(x); eS 2(x)= x 2 (1 x) 1 + x21;x 2 (1 x) 1 + x21!: Notice that eS 1=8 > > < > > :bemax 11x2 1+x(21) if x x bemax 2111x 1+x(21) if x x; (19) so that, for any xand for any ; bemax i> eS isince (1x) 1+x(21) <1if x x x2 1+x(21) <1if x x: 26 From (14) and (15) we may also compute the resistances of the path e!e: When e > e; r(e; e) = r(e+; e) ; when e < e; r+(e; e) = r(e; e): 27 The cases es =bemax and es = 0 will be considered below. 28 As shown above, for any value of x; the resistances r(e; e +)and r(e; e )are respectively an increasing and a decreasing function of e: Then, if conditions (18) are satis…ed, it follows that Pes + (k1) < P es +kand Pes (k1) < P es k:This ensures that the stochastic potential has a global minimum at es . 18
From Point (b) of Proposition 1 it then follows that the stochastically stable state esis not Pareto e¢ cient: In previous analysis we assumed 0< es <bemax:We have now to verify that e¤ectively the lower and upper bound of the action set can not be stochastically stable states. Suppose …rst that es =bemax:Since a state involving an e¤ort grater than the maximum optimal one can not be a Nash equilibrium, it follows that bemax is stochastically stable only if P(bemax)< P (bemax );condition satis…ed when r(bemax ; bemax)r(bemax;bemax )<0:(20) From (14) ;(15) we may write (20) as 8 > < > : x 2<x 2 1x 1+x(21) + 2if x x (1x)2 2<x 2 1x 1+x(21) + 2if x x (21) from which we conclude that bemax is stochastically stable when either > 1 and xxor > 2and xx;where 1x22 (2x+(1x)) 2(1x)2 2(2x+(1x)) : (22) Since these conditions can not be satis…ed as !0;it follows that bemax can not be a stochastically stable state. Suppose now that es = 0:Since a state involving an e¤ort smaller that zero is not feasible, it follows that zero e¤ort is a stochastically stable state only if P(0) < P ();condition satis…ed when r(; 0) r(0; )<0:(23) From (14) ;(15) ;since r(; 0) r(0; ) = 8 > > < > > : 1 + 1x(1+2) 2x(1x)if x x 1 + 1+x(1+2) 2x(1x)if x x; it follows that (23) is never satis…ed. We can summarize this discussion in the following Proposition. Proposition 5 Let Gbe the continuous game and let Gbe a discrete approximation of Gwith precision 1= where < 1for xxand < 2for xx: 19
Let xbe given. As !0; Ghas a unique stochastically stable equilibrium given by: eS 1(x) ; eS 2(x)= x 2 (1 x) 1 + x21; x 2 (1 x) 1 + x21!:(24) The stochastically stable equilibrium is not Pareto e¢ cient. Proposition 5 says that when !0and for any value of the distributive parameter, the game played by boundedly rational strangers converges to a stochastically stable but Pareto ine¢ cient state. The precise equilibrium to which our economy converges depends on the value of the distributive parameter. When sampling is su¢ ciently large (although incomplete) and both k=m and the probability of mistakes are su¢ ciently small, in the long run the equilibrium eS 1; eS 2will be observed with the highest positive probability so that it tends to persist and becomes the conventional way of playing the stag hunt game. To get an idea of how much the stochastically stable levels of e¤ort di¤er from the maximum optimal levels, consider (19) and notice that eS 1(x) = 1 2bemax 1= 1 2bemax 21for x=xwhile 1 2bemax 1< eS 1(x)<bemax 1when x < xand 1 2bemax 21< eS 1(x)<bemax 21when x>x:Analogously for player 2. Therefore, for any x; the stochastically stable e¤ort is not smaller that half of the maximum optimal e¤ort.29 Notice that when xx;the lower (i.e. =), the lower the distance between the stochastically stable equilibrium and the equilibrium with maximal e¤ort. The opposite obtains for xx: Substituting (24) into (1) and (2) yields the production and the individual payo¤s at the stochastically stable state: YS(x) = 2x 2 (1x) 1+x(21) VS 1(x) = 2x2 4 (1x) (1+x(21))21x+ 2x2 VS 2(x) = 2x 4 (1x)2 (1+x(21))22 (1 x) + x2 (25) Example 2. Let us reconsider now previous example 1. Let = 2 and = 1:The stochastically stable equilibrium is the strategy pro…le eS 1(x); eS 2(x)=xx2 3x+ 1;2x2x2 3x+ 1 : The associated level of production is YS(x) = 2x1x 3x+1 while agent’s payo¤s are VS 1; V S 2= x21x (3x+ 1)2(7x+ 1) ; x (1 x)22x+ 2 (3x+ 1)2!: 29 However, letting VS(x) = VS 1(x); V S 2(x)and V(x) = (V1(x); V2(x)) ;this does not mean that VS(x)V(x) is maximum when x=x: 20
Figures 4 and 5 plot respectively the total production and the individual utility corresponding to the stochastically stable equilibrium (dotted curves). To facilitate the comparison with previous Example 1, we have also plotted in these Figures the total production and the individual payo¤s when agents supply their maximum optimal e¤ort. 10.750.50.250 0.5 0.375 0.25 0.125 0 Y Y Figure 4 - Total production at the stochastically stable equilibria (dotted curve). 10.750.50.250 0.2 0.15 0.1 0.05 0 V1; V2V1; V2 Figure 5 - V1and V2at the stochastically stable equilibria (dotted curves). It is evident from (24) and (25) that the stochastically stable equilibrium depends on the distributive parameter. By varying xwe obtain a di¤erent stochastically stable equilibrium. We can then de…ne the set of stochastically stable equilibria. 21
De…nition 6 Consider the stochastically stable Nash equilibria in which, for any given x; agents supply the e¤ort levels eS 1; eS 2:The stochastically stable Utility Distribution Frontier (S-UDF) describes how the corresponding utility pair VS 1; V S 2varies with x: Figure 6 plots two utility frontiers, both derived for == 2:The outer locus is the UDF while the S-UDF is represented by the inner locus. In the same Figure we have also plotted three di¤erent equilibrium paths V2(V1);each derived for a speci…c value of x: Let x=xand consider the corresponding linear path. Suppose that agents supply their optimal maximum e¤ort so that the economy is at the point in which the linear path intersects the UDF. Proposition 5 tells us that this Pareto e¢ cient Nash equilibrium is not stochastically stable. To see why, suppose that in all the past mperiods agents played the Nash strategies (bemax 1; bemax 1):For any sample of size kthey consider, the history of the play instructs our agents to continue to select these strategies. Hence, if agents do not make mistakes, we expect to observe in the long run the Pareto e¢ cient equilibrium. Suppose now that agents do make mistakes. Speci…cally, let agent 1 choose by mistake e1=bemax 1 < bemax 1from periods t=m+ 1 to t=m+k00 inclusive, where k00 k: If this number of mistakes is appropriate (see (45) in the Appendix), then it induces agent 2 to choose e1=(bemax 1)as his best reply. This, in turn, is su¢ cient to move the economy from the Pareto e¢ cient equilibrium to the ine¢ cient equilibrium (bemax 1; (bemax 1)). In terms of Figure 6, this corresponds to a move from the point in which the linear path intersects the UDF to a point on the same path, but below the UDF. Of course, this ine¢ cient equilibrium need not be stochastically stable. Proposition 5 says that a stochastically stable state does exist: even if the probability of making a mistake tends to zero, the fact that agents can do a mistake is su¢ cient to make the Pareto e¢ cient equilibrium not a stochastically stable state and to drive the economy away from it. In Figure 6, the stochastically stable equilibrium corresponds to the point in which the linear path intersects the S-UDF. At this particular stochastically stable state, each agent supply exactly one half of his maximum optimal e¤ort. Let (e1; e2) = (0;0) be the state of nature in which no agent provides any e¤ort to the joint project. Proposition 5, far from suggesting that the social contract "might degenerate spontaneously into the state of nature" as claimed by Skyrms (2004; pg. 12), tells us that we can quite con…dently expect the emergence of a minimal level of social cooperation. 22
0.40.30.20.10 0.4 0.3 0.2 0.1 0 V1 V2 V1 V2 Figure 6 - UDF, S-UDF and equilibrium paths. This discussion also suggests that it can e¤ectively be di¢ cult for our agents to improve upon the conventional social contract. To see why, let x=xand consider the corresponding stochastically stable state, eS 1; eS 2. Suppose that, although in the past mperiods agents played these strategies, agent 1 chooses by mistake30 e1=eS 1+ > eS 1from periods t=m+ 1 to t=m+k0inclusive, where k0k;suppose also that these mistakes can induce agent 2 to choose an higher level of e¤ort as his best reply, i.e. e2=eS 1+> eS 1:It can e¤ectively be the case that these mistakes move the economy from eS 1; eS 2 to the Pareto e¢ cient equilibrium. However, Proposition 5 says that – since eS 1; eS 2is the stochastically stable state — the number of mistakes needed to move the economy from eS 1; eS 2to (bemax 1; bemax 1)is bigger than the number of mistakes needed for a move in the opposite direction to occur. In other terms, in the long run the probability of observing (bemax 1; bemax 1)is smaller than the probability of observing eS 1; eS 2:This gives a sense in which to improve upon the conventional social contract can be quite hard for our boundedly rational strangers.31 30 As we have seen, since all the equilibria along the path V2(V1)corresponding to this particular value of xare possible, the choice of supplying the maximum optimal e¤ort is exposed to the strategic risk of ending up with a lower payo¤; this is the case if the other player does not make his part (i.e if he does nor supply his maximal optimal e¤ort). 31 Of course, the above argument does not imply that it is impossible to improve upon the stochastically stable social contract. Suppose that at the beginning of each stage game, an external observer instructs our strangers to play a cooperative solution. The particular cooperative solution is irrelevant. They could agree to implement the utilitarian distribution, resulting in a distributive parameter x=xU. Alternatively, they could agree to implement the Nash bargaining solution (corresponding to x=xN) or the Rawlsian solution (corresponding to x=xR). The only possiblity for our agents to improve upon the conventional social contract is to sign a conditional contract whose enforcement is assured by the external observer. In this contract they agree (a) on a particular division of the fruits of social cooperation (for instance, x=x) and (b) to supply their maximal optimal e¤ort, given this value of x: The enforcement is ensured by the external observer who can punish any detected deviation from this contract. 23
Remark 1. We can use Theorem 2 in Ellison (2000) to show that our economy converges to the stochastically stable state in …nite time. Let the radius of the generic state ebe the minimum number of mistakes needed to leave this state; in our case R(e) = min (r(e; e +); r (e; e )) :Ellison introduces the concept of modi…ed coradius in order to formalize the observation that a large change will occur more rapidly if it involves a gradual change between consecutive states. Let rT(e)denote the minimum total resistance over all possible paths from eto es:De…ne the adjusted total resistance, r T(e);by subtracting from rT(e)the radius of the intermediate states through which the path passes. In our model we have r T(e) = 8 < : r(e; e +)if e < es r(e; e )if e < es: The adjusted coradius CRof the stochastically stable equilibrium is the maximum r T(e)over all possible di¤erent states. In our model, r(e; e +)is increasing in ewhile r(e; e )is decreasing in e;then the maximum value of r T(e)is found when e=esfor e < esand when e=es+for e > es: Hence CR(es) = max (r(es; es); r (es+; es)) :When R(es )> CR(es ); Theorem 2 in Ellison (2000) gives some information on the speed of the adjustment. Since in our economy the conditions (18) are satis…ed, it then necessarily follows that R(es )> CR(es )so that we can apply Ellison’s Theorem 2:Let W(e; es ; )denote the expected wait until a state esis …rst reached from any di¤erent state ein the perturbed model. Then from Theorem 2, point b) in Ellison (2000) it follows that32 W(e; es ; ) = OCR(es ) as !0:Since CR(es) = r(es; es);we have CR(es) = 8 < : 22es (1x)if x < x 2es x if x > x: (26) From (24) and (26) we obtain CR(es)<8 < : 1=2x 1x+x2if x < x 2=1x 1x+x2if x > x where 2= 1 1:Then, for any x; since max 1= max 2=1 2;we get CR(es )<1 2:Hence there exists a positive constant such that W(e; es ; )< p: This solution, however, seems too demanding for our boundedly rational strangers. See also Binmore (1994) for a telling criticism. 32 Following Ellison, we write f(z) = O(g(z)) for z!zas a short-hand for "there exists a constant Csuch that f(z)=g (z) = Cas z!z": 24
for agent 1 and (e1; e2)! G2(e1;bemax 2)! G1(bemax 1;bemax 2)! G2(bemax 1;bemax 2) for agent 2. In order to understand these paths, let (e1; e2)be given and consider agent’s 1 best reply. This leads to the pro…le (bemax 1; e2);where bemax 1=x 2. Given this new pro…le, and since xx;agent’s 2 best reply leads to the Nash equilibrium pro…le (bemax 1; bemax 1). We have thus derived the …rst path. Let now (e1; e2)be given and consider agent’s 2 best reply. This leads to the pro…le (e1;bemax 2)where bemax 2= (1 x) 2. Given this new pro…le, agent’s 1 best reply leads to the pro…le (bemax 1;bemax 2). However, from this last pro…le, agent’s 2 best reply leads to the Nash equilibrium pro…le (bemax 1;bemax 2):We have thus derived the second path. Suppose xxProceeding as above, we can derive the possible best reply paths starting from (e1; e2):These are respectively: (e1; e2)! G1(bemax 1; e2)! G2(bemax 1;bemax 2)! G11bemax 2;bemax 2 for agent 1 and (e1; e2)! G2(e1;bemax 2)! G11bemax 2;bemax 2 for agent 2. Let L(e)denote the length of the shortest path of best reply with origin in e: In the case just analyzed we have L(e) = 2: B) Let e2<2x 2and e1(1 x)2 2 The best reply paths for agent 2 are as for previous case A. For player 1 the possible best reply paths originated in (e1; e2)are respectively (e1; e2)! G11e2; e2= (be1; be1) when xx(with be1bemax 1) and (e1; e2)! G11e2;e2!(if e2bemax 2! G21e2;bemax 2! G11bemax 2;bemax 2 if e2bemax 2! G21be2;be2 when xx: Notice that L(e) = 1: C) Let e2<2x 2and e1<(1 x)2 2 The best reply paths for agent 1 are as for previous case B. For player 2 the possible best reply paths originated in (e1; e2)are respectively 31
(e1; e2)! G2(e1; e1)!(if e1bemax 1! G1(bemax 1; e1)! G2(bemax 1; bemax 1) if e1bemax 1! G1(be1; be1) when xxand (e1; e2)! G2(e1; e1) = 1be2;be2 when xx(where e2bemax 2). Notice that L(e) = 1: D) Let e22x 2and e1<(1 x)2 2 The best reply paths for agent 1 are as for previous case A while those of player 2 are as for previous case C. Notice that L(e)=1:This ends the proof of the …rst part. We observe that, since only a (pure strategy) Nash equilibrium pro…le can be a sink of the best reply graph, the proof of the acyclicity is equivalent to a proof of a necessary condition for the existence of (pure strategies) Nash equilibria. 2) Bandwagon property. Let (be1;be1)be the set of Nash equilibria, where be1min x 2;(1x)2 2. Following Binmore, Samuelson and Young (2003), a su¢ cient condition for the game to exhibit the bandwagon property is that: 1(be; e) = V1(be1; be1)V1(e1; be1)V1(be1; e2) + V1(e1; e2)0 2(be; e) = V2(be1; be1)V2(e1; be1)V2(be1; e2) + V2(e1; e2)0 (32) where be= (be1; be1)is any Nash equilibrium of the game and e= (e1; e2)is any non-equilibrium strategy pro…le. Recall that V1(be1; be1) = be1xbe2 1 V2(be1; be1) = be1(1 x)2be2 1: The following table give informations on the relevant payo¤s: V1(e1; be1)V2(e1; be1) if e1be1:e1xe2 1e1(1 x)2be2 1 if e1be1:be1xe2 1be1(1 x)2be2 1 V1(be1; e2)V2(be1; e2) if be11e2:be1xbe2 1be1(1 x)e2 2 if be11e2:e2xbe2 1e2(1 x)e2 2 32
and lastly V1(e1; e2)V2(e1; e2) if e11e2:e1xe2 1e1(1 x)e2 2 if e11e2:e2xe2 1e2(1 x)e2 2 Next table summarizes all the possible situations: 1(be; e) 2(be; e) 1e2< e1<be1x (be1e1)>0(1 x) (be1e1)>0 e1< 1e2<be1x be11e21>0(1 x)be11e1>0 e1<be1< 1e20 0 1e2<be1< e10 0 be1< 1e2< e1x 1e2be1>0(1 x)1e2be1>0 be1< e1< 1e2x (e1be1)>0(1 x) (e1be1)>0 Since all the entries of this table are non negative, 1(be; e)and 2(be; e)are non negative as well. This ends the proof. Claim 10 Consider an acyclic and …nite game with two players. Let bs= (bs1;bs2)be any strict Nash equilibrium of the game and let s= (s1; s2)be any di¤erent strategy pro…le. Let the bandwagon property (32) be satis…ed for any (strict) Nash equilibrium pro…le bsand for any other pro…le s: Then the transition from the (strict) Nash equilibrium bsto the (strict) Nash equilibrium s= (s1; s2) involves the minimum number of mistakes if the other player by mistakes chooses strategy si. Proof. Since the game is acyclic, we know from Young (1993) that there exists a stochastically stable equilibrium and it coincides with a strict pure strategies Nash equilibrium. Let bsbe an arbitrary Nash equilibrium. We want to …nd the minimum number of mistakes that player 2 must make in order to move the economy from bs= (bs1;bs2)to the other Nash equilibrium s= (s1;s2): Analogous considerations holds when the mistakes are made by agent 1. Let m be the memory size and kbe the sample size used by both agents. Suppose that the economy has been in the state bsfor a long period of time and consider …rst the case in which by mistake agent 2 chooses s2:Speci…cally let us suppose that agents 2 choose s2by mistake from period t=m+ 1 to t=m+k0inclusive, where k0k: If agent 1 draws a sample that includes these k0choices of s2;as well as kk0choices of bs2;then agent 1 deduces that the probability that agent 2 plays bs2is 2= 1 k0 kand that the probability 33
that agent 2 plays s2is 12=k0 k:It then follows that agent 1 is indi¤erent between bs1and s1if the number of mistakes s2made by agent 2 is k0=A1kV1(bs1;bs2)V1(s1;bs2) V1(bs1;bs2)V1(s1;bs2) + V1(s1; s2)V1(bs1; s2)k: (33) Consider now the case in which by mistake agent 2 chooses s 26=s2:As before, let us suppose that agents 2 choose s 2by mistake from period t=m+ 1 to t=m+k0inclusive, where k0k: If agent 1 draws a sample that includes these k0choices of s 2;as well as kk0choices of bs2;then agent 1 deduces that the probability that agent 2 plays bs2is 2= 1 k0 kand that the probability that agent 2 plays s 2is 12=k0 k:It then follows that agent 1 is indi¤erent between bs1and s1if if the number of mistakes s 2made by agent 2 is k0=B1kV1(bs1;bs2)V1(s1;bs2) V1(bs1;bs2)V1(s1;bs2) + V1(s1; s 2)V1(bs1; s 2)k: (34) The number of mistakes involving strategy s2is the minimum if A1< B1: Since bsis a strict Nash equilibrium, the numerators of A1and B1are strictly positive. Suppose now that the bandwagon property (32) is satis…ed; when referred to player 1, this requires 1(bs; s) = V1(bs1;bs2)V1(s1;bs2)V1(bs1; s2) + V1(s1; s2)0: In the case of A1we have bs= (bs1;bs2)and s= (s1; s2):In the case of B1we have bs= (bs1;bs2)and s= (s1; s 2):Since also the denominator37 of A1and B1 are non negative, it then follows that A1< B1if 1(s; s) = V1(s1; s2)V1(bs1; s2)V1(s1; s 2) + V1(bs1; s 2)0:(35) Since s= (s1; s2)is a Nash equilibrium, this condition is satis…ed if the bandwagon property holds for s= (s1; s2)and s= (bs1; s 2): Claim 11 .Consider the game Gand let e= (e1; e1)be an arbitrary initial Nash equilibrium: (a) The path of exit from eto the right and involving the minimum number of mistakes, is the path leading to the adjacent state e+= (e1+; (e1+)) : The resistance of the path e!e+is r(e; e +) = 8 > > > < > > > : 2(2e1+) (1 x)k if x x; 2e1+ x k if x x: (36) 37 Since eand beare two strict Nash equilibria, it follows that the denominator of A1is strictly positive. If the denominator of B1is zero, then the condition A1< B1is always satis…ed. If instead the denominator of B1is positive, the condition A1< B1is satis…ed if (35) holds. 34
(b) The path of exit from eto the left and involving the minimum number of mistakes, is the path leading to the adjacent state e= (e1; (e1)) : The resistance of the path e!eis r(e; e ) = 8 > > > > < > > > > : 12e1 x k if x x; 122e1 (1 x)k if x x: (37) Proof. Consider two Nash equilibria e= (e1; e1)and e= (e1; e1)and let ebe the …xed initial state. Since the game satis…es the bandwagon property (Lemma 3), we know from Claim 10 in the Appendix that the path of least resistance from eto eis the direct path. We now show eto emust be adjacent states. Since the path of least resistance is a direct path, in order to derive the resistance it is thus su¢ cient to analyze the restrict game where the only strategies available are those corresponding to these two equilibria, that is S1=fe1; e1g and S2=fe1; e1g:Two cases are possible: either e1> e1or e1< e1:The former corresponds to a situation in which we exit from the state eto the right (i.e. such that e1> e1) while the latter corresponds to a situation in which we exit from the state eto the left (i.e. such that e1< e1). A) Let e1> e1and consider the following payo¤ matrix e1e1 e1e1(x e1); e1(1 x)2e1e1(x e1);(1 x)e1(e1)2 e1e1x e2 1; e1(1 x)2e1e1(x e1); e1(1 x)2e1 (38) Let (1; 2)be a mixed strategy pro…le where 1(resp. 2) is the probability that agent 1 (resp. 2) plays e1(resp. e1). The best reply correspondence is V1(e1; 2)V1(e1; 2)() 21e1+e1 x V2(1; e1)V2(1; e1)() 112e1+e1 (1 x) (39) Let mbe the memory size and kbe the sample size used by both agents. Suppose that in the past plays agents 2 choose e1by mistake from period t=m+ 1 to t=m+k0inclusive, where k0k: If actual agent 1 draws a sample that includes these k0choices of e1;as well as kk0choices of e1;then agent 1 deduce that 2= 1 k0 kand 12=k0 k:It then follows from (39) that the minimum numbers of mistakes past agents 2 must make in order to induce actual agent 1 to choose e1as best reply is 35
k0(e1+e1) x k: (40) In other words, k0mistakes by agent 2 are su¢ cient to move the economy from eto e: Analogously, suppose in the past agents 1 choose e1by mistake from period t=m+ 1 to t=m+k00 inclusive, where k00 k: If actual agent 2 draws a sample that includes these k00 choices of e1;as well as kk00 choices of e1;then agent 2 deduce that 1= 1k00 kand 11=k00 k:It then follows from (39) that the minimum numbers of mistakes past agents 1 must make in order to induce actual agent 2 to choose e1as best reply is k00 2(e1+e1) (1 x)k: (41) In other words, k00 mistakes by agent 1 are su¢ cient to move the economy from eto e: Since (40) and (41) are both increasing functions of (e1+e1);and since we are considering the case e1> e1;it follows that the number of mistakes that is su¢ cient to displace the economy from eto eis minimized when e1=e1+: Therefore, when the game is in state e; the path of exit (from this state) to the right (i.e. such that e1> e1) with the minimum number of mistakes is the path leading to the state e=e+= (e1+; (e1+)) :The resistance in going from eto e+is the minimum number of mistakes su¢ cient to shift the economy from the …rst equilibrium to the second one; since the minimum between (40) and (41) depends on whether xis greater or smaller than x;we have r(e; e +) = 8 > > > < > > > : 2(2e1+) (1 x)k if x x; 2e1+ x k if x x: B). Let e1< e1and consider the following payo¤ matrix e1e1 e1e1(x e1) ; e1(1 x)2e1xe1e2 1; (1 x)e1(e1)2 e1e1x e2 1;e1(1 x)(e1)2e1(x e1) ; e1(1 x)2e1 (42) Let (1; 2)be a mixed strategy pro…le where 1(resp. 2) is the probability that agent 1 (resp. 2) plays e1(resp. e1). The best reply correspondence is V1(e1; 2)V1(e1; 2)() 2e1+e1 x V2(1; e1)V2(1; e1)() 12e1+e1 (1 x) (43) 36
Proceeding as before, it follows from (43) that the minimum numbers of mistakes past agent 2 must make in order to induce actual agent 1 to choose e1 as best reply is k01(e1+e1) x k(44) while the minimum numbers of mistakes past agent 1 must make in order to induce actual agent 2 to choose e1as best reply is k00 =12(e1+e1) (1 x)k: (45) As before, k0mistakes by agent 2 or k00 mistakes by agent 1 are su¢ cient to move the economy from eto e: Since (44) and (45) are both decreasing functions of (e1+e1);and since we are considering the case e1< e1;it follows that the number of mistakes that is su¢ cient to displace the economy from eto eis minimized when e1=e1: Therefore, when the game is in state e; the path of exit (from this state) to the left (i.e. such that e1< e1) with the minimum number of mistakes is the path leading to the state e=e= (e1; (e1)) :The resistance in going from eto eis the minimum number of mistakes su¢ cient to shift the economy from the …rst equilibrium to the second one; as before, since the minimum between (44) and (45) depends on whether xis greater or smaller than x;we have r(e; e ) = 8 > > > > < > > > > : 12e1 x k if x x; 122e1 (1 x)k if x x: This ends the proof. Claim 12 Consider the game G:Since the erooted tree with minimum stochastic potential P(e)is (e)where 0r(0;0) !r(;2) !::: er(e;e) !e r(e+;e)e+ :::: r(bemax;bemax)bemax; then P(e+) = P(e) + r(e; e +)r(e+; e) P(e) = P(e) + r(e; e )r(e; e): Proof. Consider, without loss of generality, the following erooted tree: e2r1 !er2 !er3 e+r4 e+ 2; 37
where r1r(e2; e ); r2r(e; e); r3r(e+; e)and r4 r(e+ 2; e +):We have P(e) = r1+r2+r3+r4:(46) Consider now the following (e+)rooted tree: e2r1 !er2 !er0 3 !e+r4 e+ 2; where r0 3r(e; e +):We then have P(e+) = r1+r2+r0 3+r4:(47) From (46) and (47) we have P(e+) = r1+r2+r0 3+r4+r3r3 =P(e) + r0 3r3 =P(e) + r(e; e +)r(e+; e): Lastly, consider the following (e)rooted tree: e2r1 !er0 2 er3 e+r4 e+ 2; where r0 2r(e; e ):We then have P(e) = r1+r0 2+r3+r4:(48) From (46) and (48) we have P(e) = r1+r0 2+r3+r4+r2r2 =P(e) + r0 2r2 =P(e) + r(e; e )r(e; e): Proof of Corollary 8 Recall that when xx;the UDF is an increasing function for 0xx2 and a decreasing function for x2< x x;when instead xx;the UDF is a decreasing function for x< x x1and an increasing function for x1< x 1 where x2and x1are both given in (8) : Let n > 1:Then eis the stochastically stable state when either x2(0; xmax n) (0; x)or x2xmin n;1(x;1) :Notice that x2< xmax nwhen n<n 1()while xmin n< x1when n < n 2()where n 1() = int 1 + 1 22 n 2() = int 1 + 2 2: 38
Notice that 2< n 1()if < q1 2and 2< n 2()if > p2: Let suppose q1 2:Then n 2()<2n 1(): (a) When nn 1()then ebelongs to the increasing arm of the UDF for any x2(0; xmax n)and x2xmin n;1; (b) When n < n 1()then ebelongs to the increasing arm of the UDF for any x2(0; x2)and x2xmin n;1;ebelongs to the decreasing arm of the UDF for any x2(x2; xmax n). Let suppose p2:Then n 1()<2n 2(): (a) When nn 2()then ebelongs to the increasing arm of the UDF for any x2(0; xmax n)and x2xmin n;1; (b) When n < n 2()then ebelongs to the increasing arm of the UDF for any x2(0; xmax n)and x2(x1;1);ebelongs to the decreasing arm of the UDF for any x2xmin n; x1. Let suppose q1 2< < p2:Then n 2()<2and n 1()<2:Since n > 1; then ebelongs to the increasing arm of the UDF for any x2(0; xmax n)and x2xmin n;1. Proof of Corollary 9. From Corollary 8, by noting that when n > 1;then for any values of we get xmax n< x< xmin n: 39
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