The overlap between Extensive-Form and

Subgame Perfection.

Emiliano Catoniniy

January 2017 - PRELIMINARY AND INCOMPLETE

Battigalli (1997) …rst proved that in dynamic games with and no relevant ties, the unique backward induction is extensive-form rationalizable (Pearce, 1984). Other authors use the outcome inclusion between extensive-form rationalizability and backward induction to obtain the same result. Allowing for simultaneous moves, perfect equilibrium outcomes are only a subset of the backward induction ones. Thus, showing that at least one subgame perfect equilibrium outcome distribution is extensive-form rationalizable requires a di¤erent . Here I prove that in all …nite dynamic games with observable actions there always exists an outcome distribution which is induced by both a subgame perfect equilibrium in behavioral strategies and a in strongly rationalizable strategies. Moreover, I show that only for a subgame perfect equilibrium path, strategic reasoning under belief in this path may yield as unique prediction the path itself. Keywords: Extensive-form rationalizability, Strong Rationalizabil- ity, Subgame Perfect Equilibrium, Path agreements.

1 Introduction

Battigalli [1] was the …rst to prove that in dynamic games with perfect information and no relevant ties, the unique backward induction outcome is extensive-form rationalizable (Pearce, [16], Battigalli, [1]). Chen and Micali [9] come to the same conclusion as a corollary of the following result: extensive-form rationalizability re…nes backward induction. Heifetz and

A special thanks goes to Pierpaolo Battigalli, this paper would not exist without his mentoring. Thank you also to Adam Brandenburger, Shurojit Chatterji, Yi-Chun Chen, Alfredo Di Tillio, Amanda Friedenberg, Amanda Jakobsson, Atsushi Kajii, Mattia Landoni, Xiao Luo, Elena Manzoni, Andres Perea, Burkhard Schip- per, Madhav Shrihari Aney, Satoru Takahashi, Elias Tsakas, and Dimitrios Tsomocos for precious suggestions. yHigher School of Economics, ICEF, [email protected]

1 Perea [11] also prove Battigalli’s theorem by showing the outcome equivalence of extensive- form-rationalizability and backward induction in the same class of games. Unfortunately, dynamic games with perfect information are a very small class of games. In applications, players are very frequently allowed to move simultaneously — or equivalently, before observing each other’s move. Repeated games with perfect monitoring are a promi- nent example of games with observable actions. In this class of games, the set of backward induction outcomes can expand dramatically. Therefore, games are usually solved with some notion of equilibrium. In absence of payo¤ uncertainty, subgame perfect equilibrium is by far the most widespread and accepted equilibrium concept.1 However, the possible multiplicity calls for further re…nements. A vast literature, stemming from the seminal contribution of Kohlberg and Mertens [12], re…nes subgame perfect equilibrium with forward induction ar- guments.2 However, the forward induction arguments employed in this literature have two main shortcomings. First, they are quite opaque: it is very hard to understand what kind of strategic reasoning they actually capture beyond the simple examples analyzed in the papers. Second, they do not seem to capture all possible steps of forward induction reasoning that players may apply.3 Another stream of literature, stemming from the seminal contribution of Pearce [16], tackles forward induction reasoning from a completely di¤erent perspective. Abstracting away from any equilibrium concept, extensive-form rationalizability aims to capture in a transparent way all possible steps of reasoning that players may perform by forward induction. In absence of …rst-order-belief restrictions (which may throw equilibrium reasoning in the mix), this literature probably culminates in the notion of Strong Rationalizability [4], which is given an epistemic characterization. The orthogonality of the two approaches, makes it very di¢ cult to establish any connec- tion between Strong Rationalizability and subgame perfect equilibrium However, the exis- tence of a common prediction between the two solution concepts would be extremely valuable in terms of robustness and would, so to say, pacify the advocates of the two approaches. Such prediction would be even more valuable if obtainable under a uni…ed approach to strategic reasoning and equilibrium reasoning, as the one o¤ered by Selective Rationalizability [7].4 The main result of this paper a¢ rms that in all …nite dynamic games with observable ac- tions, there always exists a Nash equilibrium in strongly rationalizable strategies that induces a subgame perfect equilibrium outcome distribution. Extending the results of Catonini [8] to

1 (Kreps and Wilson [13]) coincides with subgame perfect equilibrium in this class of games. 2 See, for instance, Govindan and Wilson [10] and Man [14]. 3 See Catonini (2017) for a critical survey. 4 Under Strong--Rationalizability (Battigalli [3], Battigalli and Siniscalchi, [5]), the …rst to introduce …rst-order belief restrictions into the structure of Strong Rationalizability, it is easy to prove that, in games with observable actions, all subgame perfect equilibria can be obtained, hence also those that Strong Rationalizability excludes.

2 mixed strategies in the obvious way, it turns out that the support of the distribution can be induced by Selective Rationalizability under …rst-order belief restrictions on such Nash equi- librium.5 Using the same arguments, I also show that only a subgame perfect equilibrium path can be delivered by Strong--Rationalizability/Selective Rationalizability6 as unique prediction in absence of o¤-the-path …rst-order belief restrictions.

Section 2 provides the preliminary de…nitions. Section 3 states the main results and provides an intuition of their proof. Section 4 proves the key lemma.

2 Preliminaries

7 Primitives of the game. Let I be the …nite set of players. For any pro…le (Xi)i I and 2 any = J I, I write XJ := j J Xj, X := XI , X i := XI i . Let (Ai)i I be the …nite ; 6   2 nf g t 2 sets of actions potentially available to each player. Let H t=1;:::;T A be the set  [ [ f;g of histories, where h0 := H is the root of the game and T is the …nite horizon. For f;g 2 any h = (a1; :::; at) H and l < t, it holds h = (a1; :::; al) H, and I write h h.8 Let 2 0 2 0  Z := z H : h H; z h be the set of terminal histories (henceforth, outcomes or f 2 8 2 6 g paths)9, and H := H Z the set of non-terminal histories (henceforth, just histories). For n each i I, let Ai : H  Ai be the correspondence that assigns to each history h, always 2 10 observed by player i, the set of actions Ai(h) = available at h. Thus, H has the following 6 ; property: For every h H, (h; a) H if and only if a A(h). Note that to simplify notation 2 2 2 every player is required to play an action at every history: when a player is not truly active at a history, her set of feasible actions consists of just one "wait" action. For each i I, let 2 ui : Z R be the payo¤ function. The list = I; H; (ui)i I is a …nite game with complete ! 2 information and observable actions.

Derived objects. A strategy of player i is a function si : h H si(h) Ai(h). 2 7! 2 Let Si denote the set of all strategies of i. The set of mixed strategies of player i is denoted

by (Si). Abusing notation, for a pro…le of mixed strategies  = (j)j I j I (Sj),  2 2  2 will be used also to denote the induced probability distribution over S, and ui() will denote the expected payo¤ of player i under . A strategy pro…le s S naturally induces a unique 2 5 With a caveat: if the equilibrium does not provide strict incentive to remain on path, Selective Rational- izability will yield a superset of the equilibrium outcomes. 6 The two are equivalent under path restrictions: see Catonini (2017). 7 The main notation is almost entirely taken from Osborne and Rubinstein [15]. 8 H endowed with the precedence relation is a tree with root h0. 9 "Path" will be used with emphasis on the moves, and "outcome" with emphasis on the end-point of the game. 10 When player i is not truly active at history h, Ai(h) consists of just one "wait" action.

3 outcome z Z. Let  : S Z be the function that associates each strategy pro…le with the 2 ! induced outcome. For any h H, the set of strategies of i compatible with h is: 2

Si(h) := si Si : z h; s i S i; (si; s i) = z : f 2 9  9 2 g

For any (Sj)j I S, let Si(h) := Si(h) Si. Moreover, let: 2  \

Si[h] = ai Ai(h): si Si(h); si(h) = ai : 2 9 2  For any J I, let H(SJ ) := h H : SJ (h) = denote the set of histories compatible with  2 6 ; SJ . For any h = (h ; a) H, let p(h) denote the immediate predecessor h of h. 0 2  0 Since the game has observable actions, each history h H is the root of a subgame (h). 2 In (h), all the objects de…ned above will be denoted with h as superscript, except for single histories and outcomes, which will be identi…ed with the corresponding history or outcome of h h the whole game, and not rede…ned as shorter lists of action pro…les. For any h H, si Si , h h h h h 2 2 and h h, si h will denote the strategy si Si such that si (h) = si (h) for all h h. For h h j h 2 h h h h  any Si Si , Si h will denote the set of allb strategiesb si Sbi such that si = si h for some h b h b j 2 e e je b si Si . b b b 2 b b

Equilibria. A mixed strategy pro…le  = (i)i I i I (Si) is an equilibrium if, for 2 2  2 all i I and si with i(si) = 0, ui() ui(si;  i). Moreover,  is SPE if it induces an 2  equilibrium in every subgame. See Section 4 for a formal de…nition.

Beliefs. In this dynamic framework, beliefs are modeled as Conditional Probability Systems (Renyi, [17]; henceforth, CPS).

De…nition 1 A Conditional Probability System on (S i; (S i(h))h H ) is a mapping ( ): 2 j S i 2 S i(h) h H [0; 1] satisfying the following axioms:  f g 2 !

CPS-1 for every C (S i(h))h H , ( C) is a probability measure on S i; 2 2 j

CPS-2 for every C (S i(h))h H , (C C) = 1; 2 2 j

S i CPS-3 for every E 2 and C;D (S i(h))h H , if E D C, then (E C)=(E D)(D C). 2 2 2   j j j H The set of all CPS’s on (S i; (S i(h))h H ) is denoted by  (S i). 2 For brevity, the conditioning events will be indicated with just the information set, which represents all the information acquired by players through observation. For each set J  I i of opponents of player i, and for each set of strategy sub-pro…les SJ SJ , I say that n f g H  a CPS i  (S i) strongly believes SJ if, for all h H(SJ ), i(SJ SI (J i ) h) = 1. 2 2  n [f g j

4 Rationality. I consider players who reply rationally to their conjectures. By rationality I mean that players, at every information set, choose an action that maximizes expected utility given the conjecture about how co-players will play and the expectation to reply rationally again in the continuation of the game. This is equivalent (see Battigalli [2]) to playing a sequential best reply to the CPS.

H De…nition 2 Fix i  (S i). A strategy si Si is a sequential best reply to i if for 11 2 2 every h H(si), si is a continuation best reply to  ( h), i.e. for every si Si(h), 2 i j 2

ui((si; s i))i(s i h) ui((si; s i))i(s ei h). j  j s i S i(h) s i S i(h) 2X 2X e H I say that a strategy si is rational if it is a sequential best reply to some i  (S i). The 2 set of sequential best replies to  is denoted by ( ). For each h H, the set of continuation i i 2 best replies to i( h) is denoted by r(i; h). The set of best replies to a conjecture i (S i) j 2 in the normal form of the game is denoted by r(i). b Elimination procedures.I provide a very general notion of elimination procedure for a subgame (h), which encompasses all the procedure I am ultimately interested in, or that will be needed for the proofs.

h De…nition 3 Fix h H. An elimination procedure in (h) is a sequence ((Si;q)i I )q1=0 2 2 where, for every i I, 2 h h RP1 Si;0 = Si ;

h h RP2 Si;n 1 Si;n for all n N;  2 h h h h h RP3 for every si Si; = n NSi;n, there exists i that strongly believes (S i;q)q1=0 such 2 h 2h 1h \ that si (i ) Si; . 2  1 h h Lemma 1 For every elimination procedure ((Si;q)i I )q1=0 and every h h, ((Si;q(h) h)i I )q1=0 2 j 2 is an elimination procedure. b b b Proof. See Catonini [6].

Indeed, elimination procedures have been de…ned purposedly to encompass the implica- tions in the of traditional elimination procedures for the whole game. In a subgame, substrategies can be eliminated "exogenously" and not because they are not sequential best replies to any valid conjecture in the subgame. On the other hand, substrategies can survive

11 It would be totally immaterial to require si to be optimal also at the histories precluded by itself.

5 even if the opponents do not reach the subgame anymore. Note that the elimination can stop for some steps and then resume: for this reason, RP2 allows a weak inclusion at all steps

Now I can de…ne Strong Rationalizability.

De…nition 4 Strong Rationalizability (Battigalli and Siniscalchi, [4]) is an elimination pro- q n cedure ((Si )i I )q1=0 where for every n N and i I, si Si if and only if there exists i 2 q n 1 2 2 2 that strongly believes (S i)q=0 such that si (i). 2 I will talk formally of …rst-order-belief restrictions on a path of play. To do so, I need the de…nition of Strong--Rationalizability.

H h De…nition 5 For each i I, …x i  (S i). Strong--Rationalizability (Battigalli [3], 2  q Battigalli and Siniscalchi [4]) is the elimination procedure ((Si;)i I )q1=0 such that, for every 2 n N, i I, and si Si, si Si;n if and only if si (i) for some i i that strongly 2 2q n 1 2 2 2 2 believes (S i;)q=0 .

3 Main results

Consider an elimination procedure where no SPE is ever eliminated "exogenously". That is, for any belief that guarantees at least the expected payo¤ of a SPE of the game, its sequential best replies always survive the elimination step. Then, a Nash equilibrium that yields the same outcome distribution as a SPE survives the procedure.

h Lemma 2 Fix h H and an elimination procedure ((Si;q)i I )q1=0 such that for each n N, 2 h 2 h h h n 1 2 i I, SPE (j )j I j I (Sj ), and i that strongly believes (S i)q=0 , if 2 2 2  2 u ((s ; s ))h(s h0) u (h), s i S i i i i i i i 2 j 

h P h h h then (i) Si;n. Thus, there exist a SPE  and an equilibrium  (S ) such that  2 1 h(S(z)) = h(S(z)) for all z Z. 2 e e The (very rough) intuition for this result is the following. Since a SPE can only be eliminated "endogeneously", if a SPE is eliminated at some step of the procedure, there must be a unilateral deviation from the SPE path(s) that is pro…table whatever the deviator can expect thereafter. The key is to show that one of the possible post-deviation beliefs corresponds to an outcome distribution of a SPE of the subgame, which can be used to create a SPE of the whole game that substitutes the eliminated one. Note that the opponents may be surprised by the deviation and thus play any continuation best reply thereafter too. This

6 creates a set of continuation plan pro…les where all plans are best replies to some conjecture over the continuation plans of the others. In turn, this guarantees that any equilibrium in this reduced subgame is actually an equilibrium of the subgame. If an equilibrium is not a SPE, there exists a unilateral deviation from its path(s) that the deviator would take whatever SPE of the post-deviation subgame she expects thereafter. Thus, in this subgame, a SPE can only be eliminated "endogeneously", and we can iterate until subgames of depth 1 are reached. There, any equilibrium is a SPE. This intuition is re…ned here through an example, directly in the context of Theorem 1: the entire game , and Strong Rationalizability as elimination procedure.

A B H I C — Ann — D A B S T U n ! n E 1; 1 0; 0 Q 2; 2 0; 0 0; 0 F 0; 0 3; 1 A B N O P R 0; 0 4; 4 0; 0 n G 0; 0 L 6; 3 5; 0 1; 2  ! M 5; 0 6; 3 2; 2

Strong Rationalizability goes as follows. First, Bob eliminates all strategies that prescribe U. Second, Ann eliminates C:E. Third, Bob eliminates all strategies that prescribe H. Fourth, Ann eliminates D:Q. Fifth, Bob eliminates all strategies that prescribe S. Note that at step 4, Ann eliminates a SPE path: (D; (Q; S)), which is sustained by the o¤- path equilibrium (E;H). In the proof of Lemma 9, which is more general than Lemma 2, the external recursive procedure (the one indexed by k) proceeds as follows for every elimination of a SPE path. At step 3, Bob, while planning to play S after D, may play any surviving continuation plan s(C) S3 (C) after C. Then, the elimination of D:Q by Ann at step 4 implies that B 2 Bj for each belief over S3 (C), Ann expects a higher payo¤ than under (D; (Q; S)), thus she Bj may play any sequential best reply to such belief (Lemma 7). The two things together imply that the "reduced subgame" S3 (C) features all the sequential best replies of both players j 3 2 1 i = Ann; Bob to all i’s that strongly believe S i (C);S i (C);S i (C). Then, every Nash j j j equilibrium of the reduced subgame S3 (C) is a Nash equilibrium of the whole subgame. As j in the internal recursive procedure in the proof of Lemma 9 (the one indexed by t), pick one of these Nash, say (F; I:P ). It does not induce a SPE path of ((C)). But then, there must exist a unilateral deviation from (F;I) which is pro…table for any SPE of the subsequent subgame. Consider a deviation of Ann to G. Indeed, Ann prefers all the equilibria of ((C; (G; I))) to (F;I) (for P cannot be played with probability higher than 1=2 in equilibrium). Then, Ann may play any best reply to an equilibrium conjecture of ((C; (G; I))) (Lemma 8). Bob, while planning to play I, may play any surviving continuation plan s(C;(G;I)) S3 (C; (G; I)) B 2 Bj after (C; (G; I)) (Lemma 3). So, since ((C; (G; I))) has depth 1, all its equilibria will survive

7 until step 3. This is the basis step in the proof of Lemma 9; if the subgame had higher depth, the survival of one SPE path of the subgame would have been guaranteed by the Induction Hypothesis. Going back to the example, pick the surviving equilibrium (L; N) and look for a new Nash equilibrium of (C) that prescribes L and N: (G:L; I:N). This time, it induces a SPE path of (C). So we found a Nash s S3 (C) inducing a SPE path of (C), which Ann 2 j prefers to (D; (Q; S)). So, if by backward induction we select (G:L; I:N) after C and (Q; S) after D, we …nd a new SPE of , which induces a di¤erent path than (D; (Q; S)). The proof of Lemma 9 assumes by contradiction that all SPE paths are eliminated, and for each of them, the external recursive procedure …nds as above a SPE subpath that is able to generate, by backward induction, a new SPE path. The external recursive procedure starts from (one of) the last SPE path(s) to be eliminated along the elimination procedure. So, the newly generated SPE path must have been eliminated no later than the one under consideration. Thus, the SPE subpath generated after the deviation that "killed" the old path survives until the step of elimination of the new path. This guarantees two things. First, if the deviation that kills the new path preceeds the deviation that killed the old path, the internal recursive procedure is then able to generate a Nash inducing a SPE subpath that is compatible with the SPE subpath already …xed after the old deviation. So, when the new SPE subpath will be imposed by backward induction, the resulting SPE paths are a subset of those generated by imposing the old SPE subpath, hence they are eliminated no later than the SPE path under consideration. Second, the new deviation cannot follow the old one, for after the old deviation, the …xed SPE subpath is induced by a surviving Nash. Then, the new deviations occur higher and higher in the tree, until the candidate deviations are actually exhausted and a contradiction arises.

By Lemma 2, the overlap between Strong Rationalizability and Subgame Perfect Equi- librium is immediate.

Theorem 1 There exists a SPE (i)i I i I (Si) and an equilibrium in strongly ratio- 2 2  2 nalizable strategies (i)i I i I (Si1) such that for every z Z, (S(z)) = (S(z)). 2 2  2 2

Proof. Lemmae 2 can be applied with strong rationalizability as eliminatione procedure 0 and h := h . 

Fix now a path z Z and, for each i I, let i be set of all CPS’s i such that 2 12 2 i(S i(z) h) = 1 for all h z. Suppose that Strong--Rationalizability yields z as unique j 13 prediction under (i)i I . Then, no unilateral deviation from z has survived. This means 2 that in each of the corresponding subgames, no SPE has survived. But then, by Lemma 2,

12 I show in [6] that these restrictions are equivalent to strong belief in S i(z). 13 Or equivalently, by the results in [6], Selective Rationalizability.

8 there must be at least one SPE against which the deviator would not deviate. With this, I can construct a SPE with path z. Thus in absence of other belief restrictions, subgame perfection is a necessary (but not su¢ cient)14 condition for players to certainly comply with the path they believe all opponents will comply with.

15 Theorem 2 Fix z Z. If (S1) = z , then there exists a SPE (i)i I i I (Si) such 2 f g 2 2  2 that for every i I, i(Si(z)) = 1. 2 Proof. Fix a history h that immediately follows a unilateral deviation by player j I h n 2 from z, i.e. h z, p(h) z, and h (S j(z)). Let (Sn)n1=0 = (S(h) h)n1=0. Fix m N, h 6 h m 1 2 j n m 1 2 i = j, and i t.s.b. (S i;n)n=0 . By (S1) = z there exists i t.s.b. (S i;)n=0 such 6 f g n m 1 that i(S i(h) p(h)) = 0 and (i)(h) = . Hence, there exists i t.s.b. (S i;)n=0 with h h j 6 h ; h h i (s i h) = i( s i S i(h): s i h = s i h) for all h H such that = (i)(h) Si;m. j 2 j j 2 e ; 6  So the hypothesis of Lemma 2 holds for every i = j. By (S1) = z there is m N such that  6  f g 2 h e e e e e Sm = , so Lemma 2 cannot hold. Thus the hypothesis of Lemma 2 must be violated for j and ; h h h v some v m, SPE  , and j t.s.b (S j;v)q=0. Fix s j S1j; S j(z). By (S1) = z ,  h h h 2  f 16g r(s j) Sj(z). For every s j with j (s j h) > 0, by constructing beliefs as above, I obtain h  v h hj h h s(s j) S j;(z) such that s(s j)(h) = s j(h) for all h H and s(s j)(h) = s j(h) for all h2 h v h 0 2 h h h h H . Fix j t.s.b. (S j;v)q=0 such that j(s(s j) h ) = j(s(s j) h) = j (s j h). Then, 62 e e j e j e j e e j. Hence, by the violatione of the hypothesis of Lemma 2,e( )(h) = . Thus, by j 2 j ; e e e h r(s j) Sj(z), (j)(z) = . So, uj(z) is higher than j’sexpected payo¤ against j , hence,  6 ; h by the violation of the hypothesis of Lemma 2, also than uj( ). Repeating for all j and post-deviation h, z is a SPE path. 

4 Proof of the main lemma.

h h h h Additional notation is needed. Fix h H, (sj )j I ; (sj )j I S , H H , i J I, 2 2 h Hh h h 2 h h 2 h  2  i  (S i), h h, (j )j I j I (Sj ), and (j )j I j I (Sj ). 2  2 2  2 2 2  2 b b e b sh =H sh ifb for eache h H, sh(h) = sh(h);  J J 2 J J b b b h h h 17 J ( ) (SJ ) is thee productb ofe the marginale distributions (j )j J ;   2 2 14 In [8] I show that not even when the SPE is unique, its path is necessarily delivered as unique predicition under thee corresponding path restrictions. e 15 SPE in mixed strategies are formally de…ned in the next section, using additional notation. For every SPE in mixed strategy there is a SPE in behavioral strategies that induces the same outcome distribution, thus the theorem holds also with SPE in behavioral strategies. 16 Formally, this is shown by Lemma 3. 17 h This is an exception to the rule of subscripts: J is not a (sub-)pro…le of individual distributions but an h uncorrelated joint distribution. Equilibria (j )j I will be represented as the joint uncorrelated distribution h 2  they induce. e e e 9 H(h) := H(Supph), h[h] := (Supph)[h];  J J J J h h h h 18 Di( ) := h H( i) Hb( ): p(h) H(b ) ;  e f 2 e ne 2e g i h h h h D ( ) :=eh H H( ): p(h) eH( ) h H( ) ;  e f 2 en e 2 e^ 2 i g h h h h h h J h is the producte of (j h)j eJ and i h (e Si ) is def. for every si Si as:  j e e j 2 ej 2 e 2 b b b h h h h e –b ( h)(sh) =  ( seh bSh(h): sheh =b sh )= (Sh(h)) if h H( ), i j i i f i 2 i i j i g i i 2 i h bh h h h h h b – (i h)(si ) = i ( si Si : si h = si ) else; e jb e f 2 b j b g e b b e b b h h h h h h h 19  =e bif for everye z ( ) andbj J, ( h)(S (z)) =  (S (z));  J J 2 2 j j j j j b b b b b h h  =h h if for every z h and j J, ( h)(Sbh(z)) = h(Sh(z));  eJ J 2 j j e j j j b b b b b h h h h h h h h h h i =  i if i ( h) =  bi; i =  i if ib( h) =  i;  e j e j b b b b b b h h h h h h h i( ) is i’sexp. payo¤b under  ; ( i) := maxb h h h h ui((s ; s )) i(s ); si Si s i Supp i i i i  2 2 h h h h h h he P h e  ieh  i if ( i h) (e i); ei  i if i ( h)  i; e e  j  j   j  b b h h  isb a SPE of (h) ifb for every h Hh,  h is an equilibrium of (h);  e e 2 e j e for any set of unordered20 non-terminale historiese H H and any sete of SPE H =  e e  h h H h e ( )h H of the corresponding subgames, E ( ) is the set of SPE of (h)  such that 2 e fore every h H Hh, h h = h. e e e 2 \ j e e e e h h I will use the fact that = and = are transitive and that = implies =. Moreover, when h h i = i : b b b h e for every h h with p(h) H(h) and h h h, ( h)(Sh(h)) = (h h)(Sh(h)); ~ 2   i j i i j i b b h h i h h h h 21 if  h = e hbfor all h e D ( ), thenb = e. e e • i j i j 2 i i e b b b b h h Whene eh is ane equilibriume and  =h eh, I will often use the fact that  h is an j equilibriumb for all h H(h). Moreover: b b 2 e b e e h h h h h h h h  if  i =  ie, ( i) ( i) = i( ) and if  is an equil., ( i) = ( i);  b b b b b b b b 18 Unilateral deviations by player i from the paths induced by h. 19 Notiee that only the outcomese induced with positive probabilitye by h and not alle those compatible with h b J matter. e 20 b For every two histories h; h0 in the set, h h0 and h0 h. 21 h h h h 6h h 6 h h h h By  =  and , ( h)(S (h)) =  (S (h)). For every z h, by  h =  h, ( h)(S (z)) = i i ~ i j i i i i j i j i j i (h h)(Sh(z)). Together,b (h h)(Shb(z)) = h(bSh(bz)). b e i j i i j i i i b ee e b b e b b e e e e e e e e b e 10 h h h h h h h h H(h) h if i =  i, for every si Suppi , there is si (i ) such that si = si . | 2 2 b b b b b b b b b b e e I start with a lemma about the combination of substrategies. Consider a player who may be surprised by history h, in the sense that she may play a strategy that allows h while believing that the co-players do not until h is actually reached. This player can keep the b b same beliefs and the same strategy out of (h), whatever she believes the co-players would b do and hence however she may play after h. The proof is immediate and is here omitted. b However, a formal proof can be found in [6]. b h h h Lemma 3 Fix an elimination procedure ((Si;q)i I )q 0, i I, n N, h H , and i t.s.b. 2  h n 1 h h h 2h h 2 2 h n 1 (S i;q)q=0 such that i (S i(h) p(h)) = 0. Fix si (i ), i t.s.b. (S i;q(h) h)q=0 and j 2 b j h h b si (i ). 2 h hb h b h h h b b b Considerb the unique si = si such that for every h H , si (h) = si (h). h h h h n 1 62 h h h There exists  =  t.s.b.b b (S ) such that for everyb h H ,  ( h) =  ( h), i i i;q q=0 62 i j i j h h e h h e e e e and si (i ) (so thatb b(i )(h) = implies (i )(h) = ). b 2 e 6 ; 6 ; e e e e b b e e e h For the next …ve lemmata, …x n N, h H, an elimination procedure (Sq )q 0 and  h h  h 2 2 h i i h  (Sn 1). Let H := H( ). For every i I, let Di := Di( ) and D := D ( ). 2 2 e This Lemma exploits Lemmae 3 to combine di¤erent reactionse of player i to unexpectede deviations from H.

h h h h Lemma 4 Fix v n and i I such that i (r( i)) and for every i =  i t.s.b. h v 1 h  h 2 i 2h h h h (S i;q)q=0 , (i ) Si;v. For every h D , …x i (Si;v(h) h). There exists i (Si;v)  2 2 j 2 h h i eh e h e e such that  =  and for every h D ,  h =  . i i e2 i j i e e b e b Proof.b e e b e b h h i h h I show that for every i I, si Suppi , and & : h D si Suppi , there exists  2 2 2 7! 2 sh Sh such that sh =H sh and sh =h &(h) for all h D i. Usinge all such esh’s, it is easy i 2 i;v i i i 2 i h e e b to construct the desired i . e b h h h b h v 1 b he h e h b h Fix i =  i t.s.b. (S i;q)q=0 . Since si r( i), by there exists si (i ) such that 2 | 2 for every h H, sh(h)b = sh(h). For every h D i, &(h) Supph Sh (h) h, so there 2 i i 2 2 i  i;v j h e h v 1 e h e exists i t.s.b. (S i;q(h) h)q=0 such that &(h) (i ). Thus, by repeatedly applying Lemma e e j e e2 e  e e e h h h h v 1 h e h hb H h h h 3, I can …nd i =  i t.s.b. (S i;q)q=0 and si (i ) such that si = si and si = &(h) i e e e 2 h h for all h D . By hypothesis of the Lemma, (i ) Si;v.  e 2 e e b e b b e h Fore the next four lemmata suppose that  ise an equilibrium and:

h h h v 1 h h A0 for every v n, i I and i =  i t.s.b.e (S i;q)q=0 , (i ) Si;v,  2  e 11 so that every i I satis…es the hypotheses of Lemma 4. 2 Lemma 5 is a characterization of equilibrium which will turn out to be useful. Since the arguments for it are standard, the proof is omitted.

h h h h Lemma 5 Fix (i )i I i I (Si ):  is an equilibrium if and only if for every h H( ), 2 2 h2  h 2 i I and ai Ai(h) i [h], calling Ha := (h; (ai; a i)) h , 2 2 n i a i  i[h] b b 2 b h b h h hb h h ( i h) ( i h)(S i(h)) + ui(z) ( i h)(S i(z)) i( h): (F) h Hh Z j  j z Hh Z  j  j 2 ai n 2 ai \ P e e P e b b b b Lemma 6 converts a condition on CPS’sinto F for some related conjectures.

h h  h h h Lemma 6 Fix  =  , h H , i I, ai Ai(h)  [h], h H Z, v n, and  t.s.b.  2 2 2 n i 2 ai n  i h v h h h h bh (S i;q(h) h)q=0 such that (i)  i h i , (ii) for every h Hai (Z h ),  i h  i for j j  2 b n [ f g j  h bh e b h bh h v 1 h h ehe some  i (S i;v(h) h), and (iii) for every i =  i t.s.b. (S i;q)q=0 , if i = i , then b b2 j b e b e h b e b b (i )(he) = . Then F holds. b ; e e e e h h h h h h Proof.b Let  i := i ( h). By Lemma 4 there exists  i (Sv ) such that  i =  i, j h 2 h h b b h h h h h h  i h =  i for all h Hai Z, and  i h =  i h for all h Di Hai . Fix i =  i t.s.b. j 2 bn j j 2 n h v e he h h h h h h e (S i;q)q=0 such that i = i (one exists because i ( h) =  i implies i ( h) h =  i h = e e e e j e j j j h b b e i ( h)). j b b b b h e For every h Di Hai and z h, e b 2 n h h e h h h e h h h h h h h  i(S i(z)) =  i(S i(h)) ( i h)(S i(z)) =  i(S i(h)) ( i h)(S i(z)) =  i(S i(z));  j  j e e where the …rst and thee last equalitiese are by thee chaine rule ofe probability,e ande the central h h h h h equality is by and by construction. For every z (Supp ), by  =  ,  i(S i(z)) = h h ~ h h h 2 h h  (S (z)). Hence every s S (h) = h S (h) induces with  and  (1) and with i i i i h Ha i i i 62 [ 2 i e e h h and h h (2) the same distribution over outcomes. By equilibrium, r(h )(h) = ; by i i e i e j j b e e 6 ; 1, r(h )(h) r(h )(h) = ; by h h, r(h )(h) = ; by h H(h ), (h)(h) = ; by i [ i 6 ;  i 6 ; 2 i i 6 ; h e h h h h e (i )(h) = , r(i ; h)(h) = ; by i ( h) h =  i h, r( i h)(h) = ; thus, ; b ; j b j j j ; b h b h h h b h h b( i) ( i h)(S i(h)) + ui(z) ( i h)(S i(z)) i( h): (H) h Hh Z  j z Hh Z  j  j 2 ai n e 2 ai \ P e P e h h h h By equilibrium, r( i h) Si(h) = ; by r( i h)(h) = and 2, ( i h) = ( i h); by h j n h 6 ; h j h ; h h j h jh equilibrium ( i h) = i( h) = i( h) (by  h =  h) By  i =  i =  i and j j j j j eh h b h h hb h he h , for every h Hai , ( i h)(S i(h)) = ( i h)(S i(h)). For each h Hai Z,  i  i h. ~ e2 je b j e b 2 ne b j So H implies F.  e e e b e e b e 12 For the next two Lemmata, …x a set of unordered histories H Hh and a set of SPE H h   = ( )h H such that: 2 b b e A1 for everye b h H, there exists an equilibrium h = h such that h (Sh(h) h). 2  2 n j e e e e b e e e e Lemma 7 claims that if only until step n 1 an equilibrium that mimics a given SPE survives, at step n one deviation is always strictly preferred to continuing as the equilibrium prescribes.

h h H h h Lemma 7 Suppose that there exists  E ( ) such that  =  but there is no equi- h h 2 h h b h librium  (Sn) such that  =  . Then there exist l I and h Dl ( h H H ) 2 2 2 n [ 2 h h 22 eh h h v e such that for every (j )j=l j=l(Sj;n(h) h), v n and l  l t.s.b. (S l;q(h) h)q=0, 6 2  6 j   b e j b b h hb bh v 1 h h h b b h there exists l =  l t.s.b. (S l;q)q=0 such that l = l and (l )(h) = (so by A0 b b 6 ; b b h h b be b (l ) Sl;v). b  e e e e e b h h h eProof. Suppose not. For every i I and h Di ( h H H ) =: Di, …x  i, v(h), and i 2 2 n [ 2 h h e bh h b that violate the statement, and let  i :=  i. By Lemma 4 there exists  (Sn) such b e b 2 b h h h b b i b h h h e that  =  =  and for every i I, h H ( j=iDj), and h D H , i h = i h. h 2 2 [ [ 6 2 \ j j I show that  is an equilibrium. e b b e e  e h b b b b Fixb h eH , i I, and ai Ai(h)  [h]. If there exists h h such thatb h e H, 2 2 2 n i  2 h h hb h h h h h h  =  =  =  , so, by ,  =  ; then  h is an equilibrium, so by Lemma 5 ("only • j h h e h e b e e h e e e e h h if") F holds. If Hai Z H, for every h Hai Z,  i h =  i and by , ( i) = ( i h); n  2 n j j b h h h e h b e h b h b h h b h by  =  , i( h) = i( h) and by , ( i h)(S i(h)) = ( i h)(S i(h)); so since  h j j ~ j j j b b b bh e e b is an equilibrium by Lemma 5 ("only if") holds. If H Di = , …x v := min h v(h) F ai h H Di b\ 6 ; b 2 ai \ b b b h h h h 23 and h := arg min h v(h): for every h H Di,  h  ( h) (S (h) h) = h H Di ai i i i;ve 2 ai \ 2 \ j  j 2 j 6 ;e and for every h Hh (Z D ), h h (Sh (h) h); thereforee by Lemma 6 holds. Thus b e ai ei i e i;v e e F e e 2 nh [ j 2 j b e by Lemma 5 ("if"),  is an equilibrium.  e b e e e Lemma 8 is the "dual"b of Lemma 7: if an equilibrium has survived n steps but it does not mimic a SPE (within a subset), then there is a deviation from one of the equilibrium paths that the deviator could take whenever thereafter she expects at least the payo¤ of a SPE of the subgame (within a subset).

h h h h h Lemma 8 Suppose that  (Sn) and for every h H,  =  , but there is no h 2 2 h h h H h e e h  E ( ) with  =  . Then there exist p I and h Dp( ) ( h H H ) such that 2 2 e b2 n [ 2 b e e e e 22 h h Note that the statement needs not hold for all  l (S j;n(h) h): this is due to thee factb that equilibria e 2 j e are not correlated. b 23 h h h h n 1 h h h For every j = i, there is j =  j t.s.b. (S l;qb)q=0 ; by , (bj )(b h) = ; by A0 (j ) Sj;n. 6 | 6 ;  e e 13 h h H h h h v h h for every  E ( ), v n and p  p t.s.b. (S p;q(h) h)q=0, there exists p =  p 2   j h v 1 b h h h h h h t.s.b. (S p;q)q=0 such that p = p and (p )(h) = (so by A0 (p ) Sp;v). e 6 ;  e e h h h Proof. Suppose not. Fore everye i I ande h Di ( h H H ) =:eDi, …x  , v(h), and i 2 h h 2 n [ 2   that violate the statement. Construct  =  such that for everye h H withb p(h) H b, e b 62 2 h h H b b i eh  h E ( ), and in particular, for every i I, h H ( j I Dj), and h D H , j 2 2 2 [ [ 2 2 \ h h h b e h h h eH e  i h = ie h.b I show that  is an equilibrium with  =  E ( ) for all h H He. je j h e b 2 b h 2 \ Then,b bye Lemma 5 ("only if") for every h H,  h is ane equilibrium,e e b and so  Eh(H ). b b  2 h j e 2 b b Fix h H , i I, andb ai Ai(h)  [h]. Ifb there exists h h such that h bH, 2 2 2 n i  2 h h h h h h eh bh e b  =  =  =  , so by  =  ; then  h is an equilibrium, so by Lemma 5 ("only • j h e e h b e e h e e e e h h h if") F holds. If Hai Z H, for every h Hai Z,  i h =  i and by , ( i) = ( i h); n  2 n j j b h e h e h h b hb h h h h by  =  , i( h) = i( h) and by , ( i h)(S i(h))b = ( i h)(S i(h))b ; so since  h j j ~ j j j b b b bh e b is an equilibrium by Lemma 5 ("only if") holds. If H Di = , …x v := min h v(h) F ai h H Di b\ 6 ; b 2 ai \ b e b e b h h eh h e 24 and h := arg min h v(h): for every h H Di,  h  ( h) (S (h) h) = h H Di ai i i i;ve 2 ai \ 2 \ j  j 2 j 6 ;e e and for every h Hh (Z D ), h h h (Sh (h) h); hence by Lemma 6 holds. b e ai e i i e i;v e e e Fe 2 n [h j  2 b j e Thus by Lemma 5 ("if")  is an equilibrium.e  e b e e e e Now I can prove the fundamentalb Lemma.

h Lemma 9 Fix h H, m N, an elimination procedure (Sq )q 0, a set of unordered histories  1 w 2 h 2 H h H = h ; :::; h H h and a set of SPE  = ( )h H s.t. A1 holds for n = m and: f g  n f g 2 b e h;v e hb v h;v bA2 for every v w, there exists an equilibrium  (Sm 1) such that h i I Di( ) 2  q h;v h;v h2q 2 [ and for every q < v, if h H( ),  =  ; 2 b b h h H h h h n 1 h h A3 for each i I, n m,  E ( ) and i  i t.s.b. (S i)q=0 , (i ) Si;n. 2  2 b b e  b Then there exist h Eh(H ) and an equilibrium h (Sh ) such that h = h. 2 2 m  b e e Proof. The proof is by induction on the depth of (h). Inductive hypothesis (d) The Lemma holds for every h H such that (h) has depth not bigger than d. 2 Basis step (1) For every i I, n m and equilibrium of (h) h such that Supph h h h 2  h h  Sn 1, by A3 r( i) Si;n. Inductively, Supp Sm.   INDUCTIVE STEP (d+1) Suppose not. I will …nd a contradiction through a recursive 0 procedure. Set k = 0 and H := H. Recursive step (k) b 24 See the previous footnote.

14 k k If k > 0, H and H are de…ned in step k 1. Let n m be the greatest q N such k h;k h H h;k  h h;k 2 h;k that there exist  E ( ) and an equilibrium  (Sq 1) with  =  . If 2 k 2 0 h H h H k > 0, by the last remark of the previous steps, E ( ) ::: E ( ). Then by  A3 k  implies A0. Moreover, n weakly decreases with k. Thene H satis…es A1 ande A2 with n in 25 h;k h place of m. Lemma 7 yields l I and h Dl( ) k H . 2 2 n [h H h h 2 De…ne ((Si;q)i I )q1=0 as ((Si;q(h) h))i I )q1=0. By 1, it is an elimination procedure. Fix 2 j b2 b h h v 1 e h;k i = l and v n. For every i t.s.b. (S i;q)q=0 , since h Dl( ), by Lemma 3 there exists 6  b b 2 h h h;k h v 1 b b h h h h h h i =  i t.s.b. (S i;q)q=0 such that i = i . By A0, (i ) Si;v; by , (i )(h) = . b  | 6 ; e b b h e h h h Hence, together with Lemma 7, for every i I, v n, (j )j=i j=i(Si;n) and i  i 2  6 2  6 b e eh v h h e e b e b e b b t.s.b. (S i;q)q=0, (i ) Si;v = (F). k  6 ; b 0 e b Let Hb 0 := H Hbh h:b I show that there exist h Eh(H ) and an equilibrium h = \ 63 2  h eh h b  such that Supp b Sn. Suppose not (G). I will …ndb ab contradictione through a recursive e  b q h;q h;q procedure.b For everyb q bw + k such that hq H0, let h := hq and  :=  h, whiche is  h 2 q q h;q j h;q an equilibrium becausee by h k H , h h , so by h D (b ), h H( ). Set h H i I i 62 [ 2  e 2 [ 2 2 b t = 0. b b b b b b b Recursive step (t) If t > 0, Ht, and Ht are de…ned in step t 1, and satisfy A1 and A2 with n in place of m and h in place of eh.26 Let  := w + k + t. For every i I, let h;t h h e t h h h h 2 i be the set of i (Si;n) such that for every h H H(i ), i = i . Note that for 2 b 2 \ everyb i I, b b b b e e 2 b e e b b e h;t h h h h h h h h h i = h Ht z Zh Hh i (Si ): i (Si (z)) = i (Si (h)) i (Si (z)) (Si;n), \ 2 \ 2 [ f 2  g \ b e e b b b b b b e e b e e b b27 h;tb e e an intersection of convex and compact sets. Hence i is convex and compact. Then, since b h;t expected utility is linear, the reduced game with strategy sets (i )i I , if non-empty, features 2 h;t b h h h;t h n an equilibrium  . For later reference, for each i I, …x i =  i t.s.b. (S i;q)q=0 2 h;+1 b h h;t h;t h h b b bh;+1 hb and i ((i )) (r( i )) such that, by  i =  i and , i = i for all 2  | bt h;te b h;+1 b h b h;te e e b eh;t h H H( i ); by F, i (Si;n). I show that i is non-empty and that  is an 2 b \ e2 e e b e equilibrium ofb the whole (bh). b b b e e e 0 b h;0 h h;+1 h;0 h;t e h;t If t = 0 and H = , i = (Si;n) = (by F) and i i , so i (r( i )) ; b 6 ; 2 2 b b h;+1 h;t b b b b too. Else, for notational convenience let  i :=  i and proceed as follows. e b b b e e 25 q k hq h q q h;q h q For every h H ,  (Sn(h ) h ) and  (Sn 1) come from A1 and A2 if h H, from some 2 2 j b 2 e 2 previous step if hq H. q 26 q 62 t h h q q h;q h b q k For every h H , e (Sn(h ) h ) and b (Sn 1) come from the outer recursive step if h H , 2 2q k j 2 2 from some previous stepb if h H . b b 62 27 Clearly, (Sh )eis convexe and compact. Eachb set of the kind h (Sh):(h)(Sh(z)) = (h)(Sh(h)) c , i;n f i 2 i i i i i  g b b b hb hb b h b h where c is a constant, is clearly convex and compact too. Notice that if h H(i ), i satis…es (i )(Si (z)) = h 62 e ( )(Sh(h)) c as 0 = 0. b b b b b b i i  b b e b b b e b 15 g For every h Hh and q , let Qh := g q : h Ht Hh and Qh := Qh.I 2  q f  2 \ g +1  b h e h;q h e h;qe h;q e show that for every q Q  + 1 , there exists i (Si;n) such that i = i and e 2 g [ fg g 2 e for every g Qh, h;q =hb h . Fix q Qh  + 1b and supposeb to have shownb it alreadyb 2 q i i 2  [ f g bh b h;q b h;q h;q h b for every g Qq q . Since i (r( i )), by  (Sn) and F, Lemma 4 yields 2 nf g e 2 2 h;q (Sh ) suchb that h;q = bh;q and forb every h bD i(h;q)b: i 2 i;n i  i 2 b b b b b b b - if h Ht, h;q h = h; b e b 2 i j i b e h e et eh h;q h;max Qq h q h;q - if h H but Qq = e, i h = i h (where max Qq < q because h j I Dj( ) 62 6 ; j e j 2 [ 2 \ t h;q h b t H and j IeDj( b) H = h H ); e e e [ 2 \e f g 6 e e b t h;q h h;q h;q h h;q h h - if h e h for some h H ,  h =e h,e so that by  =  =  and ,  =  . b2 i j i j i i i • i i b b b b h; h;t 28 h;t h;+1 h;+1 h;t h;+1 Thene   = e. So,e ande e  exist and  b e . Since  e= i 2 i 6 ; i 2 i i h;+1, h;b+1 (br(h;t)). Then bh;t (br(h;t)) too. b b b i i 2 i i 2 i e b t b By ,b F implies A0b with h in placeb of h; Hb sati…es A1 with n in place of m. By the lastb remark of the previouse steps Ehe(Ht ) Eeh(H0 ), so, by G, h;t satis…es the hypotheses b  e h;t 29 e e hb of Lemma 8. Lemma 8 yields p I and h Dp( ) h Ht H . 2 2 n [ 2 h b h e De…ne the elimination procedure ((Si;q)i I )q1=0 := ((Si;q(h) he))i I )q1=0. Fix i = p and 2 e e j 2 6 h h v e h;t b h h h;t v n. For every i t.s.b. (S i;q)q=0, since h Dp( ), by Lemma 3 there exists i =  i  2 h v 1 h h h hb h h b b b t.s.b. (S i;q)q=0 such that i = i . By A0, (i ) Si;v; by , (i )(h) = . Hence,  t | 6 ; b e b h eb h Hb h hb e h ve together with Lemma 8, for every i I, v n,  E ( ) and i  i t.s.b. (S i;q)q=0, e 2e  2e  e (h) Sh . For every q Qh, let h;q := h;q h, whiche is an equilibrium because by i  i;v 2  j h q q h;q b h;q e h h Ht H , h h , so by h i I Di( ), h H( ). Then A3, A2, and A1 are 62 [ 2  2 [ 2 2 satis…ed withe Ht Hh in place of H, nb in placeb b of m, andb h in place of h. So, by the e e \ h h b Ht b h h Induction Hypothesis, there exist  E ( ) and an equilibrium  =  such that h e b 2 Supp Sh = Sh(h) h. e  n n j b h h t+1 t e Since h h Ht H and H is …nite, H := h H : h h h is a set of 62 [ 2 f 2 6 g [ f g unorderede histories thate keep shorteningb with t, until a contradiction is obtained. Before, let e e Ht+1 Ht h h +1 e h;e+1 e h;te30  :=    h Ht Ht+1 , h := h,  :=  . Then, increase t by 1 [ f gnf g 2 n ande run againe noting whate follows: for every h Hbt such thatb h h, h h = h, so that e e e t+1 t 2 b e j Eh(H ) Eh(H ). e  e e e e e e h h k+1 k Since h k H and H is …nite, H := h H : h h h is a set of 62 [h H f 2 6 g [ f g unordered histories2 that keep shortening with k, until a contradiction is obtained. Before, b e e b b 28 By recursive step t 1, max Qh = .  29 b t h;t h;t h h Without loss of generality assume that for every i I and h H H(i ), i = i . 30 2 2 n Overwrite the previous, temporary choice. b b e e e e e e e

16 k+1 k H H h h w+k+1 h;w+k+1 h;k 31 let  :=    k k+1 , h := h,  :=  . Increase k by 1 [ f gnf gh H H 2 n k and run again noting whatb follows: for every h H such that h h, h h = h, so that k+1 k b h H h H 2 b e j E ( ) E ( ).  b  b Proof of Lemma 2. Apply Lemma 9 with empty H and H , and m = . 1  b b References

[1] Battigalli, P., “On rationalizability in extensive games”, Journal of Economic Theory, 74, 1997, 40-61.

[2] Battigalli, P., “Dynamic Consistency and Imperfect Recall”, Games and Economic Be- havior, 20(1), 1997, 31-50.

[3] Battigalli, P., “Rationalizability in In…nite, Dynamic Games of Incomplete Information,” Research in Economics, 57, 2003, 1-38.

[4] Battigalli, P. and M. Siniscalchi, “Strong Belief and Forward Induction Reasoning,” Journal of Economic Theory, 106, 2002, 356-391.

[5] Battigalli P. and M. Siniscalchi, “Rationalization and Incomplete Information,”The B.E. Journal of Theoretical Economics, 3(1), 2003, 1-46.

[6] Catonini, E., “On non-monotonic strategic reasoning”, working paper, 2017.

[7] Catonini, E., “Rationalizability and Epistemic Priority Orderings”,working paper, 2017.

[8] Catonini, E., “Self-Enforcing Agreements and Forward Induction Reasoning”, working paper, 2017.

[9] Chen, J., and S. Micali, “The order independence of iterated dominance in extensive games”, Theoretical Economics, 8, 2013, 125-163.

[10] Govindan, S., and R. Wilson, “On forward induction,”Econometrica, 77, 2009, 1-28.

[11] Heifetz, A., and A. Perea, “On the Outcome Equivalence of Backward Induction and Extensive Form Rationalizability”, International Journal of , 44, 2015, 37–59.

[12] Kohlberg, E. and J.F. Mertens, “On the Strategic Stability of Equilibria”, Econometrica, 54, 1986, 1003-1038.

31 k h;k h;k h;k h h h;k k+1 Note that for each h H H( ),  =  =  =  , so  satis…es A2 with H . 2 \

e e 17 e e [13] Kreps, D.M. and R. Wilson, “Sequential equilibria”, Econometrica, 50, 1982, 863-94.

[14] Man, P. “Forward Induction Equilibrium”, Games and Economic Behavior, 75(1), 2012, 265-276.

[15] Osborne, M. J. and A. Rubinstein, “A Course in Game Theory”, 1994, Cambridge, Mass.: MIT Press.

[16] Pearce, D., “Rational Strategic Behavior and the Problem of Perfection”, Econometrica, 52, 1984, 1029-1050.

[17] Renyi, A., “On a New Axiomatic Theory of Probability”, Acta Mathematica Academiae Scientiarum Hungaricae, 6, 1955, 285-335.

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