Epistemic Game Theory: Online Appendix

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1 Epsemc Game Theory: Onlne Appendx Edde Dekel Lucano Pomao Marcano Snscalch July 18, 2014 Prelmnares Fx a fne ype srucure T I, S, T, β I and a probably µ S T. Le T µ I, S, T µ, βµ I be a ype srucure ha adms µ as a common pror and such ha T µ T for every. Fx a player and a ype profle T µ. Defne E 1 {s,, : µ s,, > 0} Suppose E k has been defned for every 1 < k n and le E n+1 { s, : s, E n, j I s.. j j and µ s, j, j } j > 0 Le E n1 En and E I E. Proposon. Le be n CP µ and ν µ E. Defne T j ν proj Tj E for every j and le T ν I, S j, Tj ν, βν j be he ype srucure generaed by he common pror ν. Then s n CP ν. In parcular, f µ s mnmal for j I Proof. We frs prove ha for all j and all j T µ j, ν S T µ j { j} > 0 marg S j T j ν j marg S j T j µ j. By defnon, f µ S T µ j { j} > 0 hen hen ν µ. µ s j, j j µ s j, j S j { j } s j, j S j { j } s j, j S j T j µ 1

2 For every s j, j S j T j. If ν S T µ j { j} some k. For every s j, j S j T j, f hen µ s j, j Sj { j } > 0 s j, j S j { j } E k+1 E, hus > 0 hen j proj Tj E k for ν s j, j Sj { j } µ s j, j S j { j } µ E Therefore µ s j, j j 1 µe 1 µe µ s j, j S j { j } s j, j S j { j } s j, j S j T j µ ν s j, j S j { j } s j, j S j { j } s j, j S j T j ν ν s j, j j We can conclude ha for every j I, β ν j j β µ j j for each j T ν j. I remans o prove ha ϕ T ν ϕ T µ. For every j, j T ν j and k 0, le ϕ k j T ν j be he k-h order belef of ype j n he ype srucure T ν. Defne ϕ k j T µ analogously. For every j I and j T ν j, we have βv j β µ j, hence ϕ 1 j T ν j ϕ 1 j T µ j. Suppose ϕ k j T ν j ϕ k j T µ j for all j, k K and j T ν j. Then β ν j { s j, j : ϕ K j T ν j h K j} β µ j { s j, j : ϕ K j T µ j h K } j for every h K j X j K 1. Therefore ϕ K+1 j T µ j ϕ K+1 j T ν j. Snce hs s rue for every K, we have ϕ j T µ j ϕ j T ν j for every j Tj ν, n parcular, for. Ths concludes he proof ha s n CP ν. An analogous resul holds for ype profles. We om he proof, whch s an almos exac replca of he proof of Proposon 1. Proposon. Le be n CP µ and defne ν µ E. Defne T ν proj T E for every I and le T ν I, S, T ν, βν I be he ype space generaed by he common pror ν. Then s n CP ν. In parcular, f µ s mnmal for hen ν µ. 2

3 Evens across ype srucures Le R µ, B k R µ and CBR µ be he evens correspondng o, respecvely, raonaly, k-h order belef n raonaly and common belef n raonaly n he ype srucure T µ. In he proofs we wll no formally dsngush beween CBR and CBR µ. Ths s jusfed by he nex resul. Proposon. If s, CP µ CBR, hen s, CBR µ. Proof. Le R, B k R and CBR be he evens correspondng o, respecvely, raonaly, k-h order belef n raonaly and common belef n raonaly n he unversal ype srucure H I, S, H, f I. For every, le ψ T : S T S H be he map defned as ψ T s, s, ϕ T for every s,. As s well known, B k R and R are measurable evens, and ψ T µ and ψ T are measurable maps. Furhermore, for every, every even E S H and every ype T, f ϕ T E β ψ T 1 E where ψ T j ψ j T. Defne analogously he funcons ψ T µ I. Le ψ I ψ. I can be easly checked ha R ψ T 1 R and R µ ψ T µ 1 R. Suppose for every k K we have B k R ψ T 1 B k R and B k R µ ψ T µ 1 B k R. I follows from β B K R β ψ T 1 B K R f ϕ T B K R ha s, B B K R f and only f s, ϕ T B B K R. Equvalenly, B B K R ψ T 1 B B K R for every. Therefore B B K R ψ T 1 B B K R. Hence B K+1 R B K R B B K R ψ T 1 B K R ψ T 1 B B K R ψ T 1 B K+1 R By nducon, we can conclude ha B k R µ ψ T µ 1 B k R for every k. Moreover, CBR k B k R k ψ T 1 B k R ψ T 1 k B k R ψ T 1 CBR The exac same argumens apply o he ype srucure T µ, herefore we have CBR µ 3

4 ψ T µ 1 CBR. Le s, CP µ CBR. By applyng he resuls above and he assumpon ϕ T ϕ T µ, we can conclude herefore s, CBR µ. 1 β CBR β ψ T 1 CBR f ϕ T CBR f ϕ T µ CBR β µ ψ T µ 1 CBR β µ CBR µ Oher evens of neres whch appear n he nex proofs are CB [φ] and CB [n]. The argumen behnd he prevous proposon can be easly adaped o show ha we do no need o dsngush beween hese evens and her counerpars n he ype srucure T µ. Proof of Theorem 4 1 Clam. For every k, E k CBR. Proof. For every profle s,, f µ s, > 0 hen βµ s, > 0 and snce s n CBR B CBR hen s, CBR. Therefore E 1 CBR. Suppose he clam s proved for every k K. If s, E K+1 here exs s, E K and a player j such ha j j and βµ j s j, j > 0. Snce j s n B j CBR hen s j, j CBR j. Therefore s, CBR. Therefore, by nducon, we conclude ha for every k, E k CBR. We now show ha µ S T defnes a correlaed equlbrum. Le µ s j, j > 0 for some player j and par s j, j. Then s j, j, s j, j E k for some k and some s j, j S j T j. Pck l j. Then µ s j, j l β µ l s j, j > 0 4

5 Snce l s n CBR l B l R hen s j, j R j. Therefore s j s a bes response o marg S j β µ j j marg S j µ j marg S j µ s j, j where he las equaly follows from AI ndependence. Therefore µ S T s a correlaed equlbrum. 2 Le ν S be a correlaed equlbrum dsrbuon. Then u s, s ν s s s S s S u s, s ν s s for every s S. Le T µ {s : ν s > 0} and T µ I T µ. µ S T as µ s, ν s Defne he pror f s and µ s, 0 oherwse. Defne β µ o be generaed by µ, ha s β µ s, µ s, for every and every s,. We have a well defned ype srucure T µ I, S, T µ, βµ I admng µ as a common pror. The pror sasfes Condon AI rvally, snce for every s and f µ s, > 0 hen s. If µ s, > 0 hen s and s s a bes response o ν s, hence s, R. Moreover, f β s, > 0 hen µ s, > 0 hence s, R. Therefore, f µ s, > 0 hen s, RCBR. Proof of Theorem 8 I s enough o prove ha f s, CP µ CB [n] and µ s mnmal for hen µ sasfes AI. As before, s mmedae o check ha for every k, E k CB [n]. Le µ s j, j > 0. There exs s j, j such ha s, E k for some k and µ s j, j s j, j > 0. 5

6 Le l j. Then µ s j, j l > 0 and snce s j, j CB [n], hen l s n B [n], hence s j n j j. To conclude, f µ s j, j > 0 hen s j n j j. Therefore µ sasfes AI. Proof of Theorem 7 I s convenen o prove here a slghly sronger resul. Theorem. 7b If here s a probably µ S T, a uple CP µ [φ] CB [φ] B R and ν µ E sasfes AI, hen here exs σ S for all such ha σ σ I s a Nash Equlbrum and φ k σ k. As n he proof of Theorem 2, f CB [φ] hen for every and every k, E k CB [φ]. The res of he proof s based on Aumann and Brandenburger Clam. For every s, f ν s, > 0 hen ν s ν s s, φ s. Proof. For every s,, f ν s, > 0 hen s,, s, E k for some k. Snce E k CB [φ] and E k E k+1 hen s, [φ]. Hence ν s s, ν s β ν s φ s where he frs equaly follows from AI. Therefore ν s ν s s, ν s, φ s ν s, φ s. s, s, Clam. For every s, ν s I 1 ν s. Proof. Suppose for K < I and every s S and I, ν s 1, s 2,.., s K,..., s I K ν s ν s K+1,..., s I We know from he prevous clam ha hs s rue for K 1. Suppose s rue for 1 6

7 some K > 1. Then ν s 1,..., s I marg S K+1 ν s 1,..., s K, s K+2,..., s I s K+1 ν s K+1 marg S K+1 ν s 1,..., s K, s K+2,..., s I ν s K+1 ν s 1,..., s K, s K+1, s K+2,..., s I ν sk+1 s K+1 S K+1 ν s 1 ν s K marg SK+1... S I ν s K+1, s K+2,..., s I ν sk+1 s K+1 S K+1 ν s 1 ν s K ν s K+1 marg SK+1... S I ν s K+1, s K+2,..., s I s K+1 S K+1 ν s 1 ν s K+1 ν s K+2,..., s I Therefore he clam holds for every K I. Clam. If ν s > 0 hen φ s k ν s k Proof. By combnng he prevous wo clams, f ν s, > 0 hen ν s s, φ s ν s ν s k. k Defne σ marg S ν. Le σ s > 0. Fx a player j and he ype j. By assumpon j [φ] j. By he clams above and AI ndependence, n he uple ν s j ν s s j, j φj s σ s hence ν s j > 0. Le be a ype such ha ν s, j > 0. Snce j B R j, hen s s a bes response o he frs order belef of ype. Because j CB [φ] j, hen [φ],.e. he frs order belef of s gven by he conjecure φ k σ k. To conclude, for every player, every sraegy n he suppor of σ s a bes response o he conjecure k σ. Therefore, σ s a Nash Equlbrum. Proof of Theorem 9 Le belong o CP µ [φ] CB [φ] B R CB [n] 7

8 Noce ha µ s no assumed o be mnmal. Le ν µ E. From Proposon 2, we have ha f CP µ hen CP ν. Therefore, s n CP ν [φ] CB [φ] B R CB [n] As before, s mmedae o check ha for every and every k, E k CB [n]. Le ν s j, j > 0. There exs s j, j such ha s j, j, s j, j E k for some k and, and ν s j, j s j, j > 0. Le l j. Then ν s j, j l > 0 and snce s j, j CB [n], hen l s n B [n], hence s j n j j. To conclude, f ν s j, j > 0 hen s j n j j. Therefore ν sasfes AI. We can now apply Theorem 7b. Proof of Theorem 14 By sandard argumens, we can fnd wo ypes 1, 2 [ T ] CB [ T ] CB [ψ] CB R. Le ϕ, for every. Defnon. A ype,1 1,...,,N N such ha:,1 1 1 of player s reachable n N seps f here exss a sequence N and N [ ] For all n N, βn 1 n 1 n > 0 Le RE N be he se of ypes reachable n N seps. Snce he ype srucure T s mnmal, every ype s reachable n a fne number of seps. We need o show ha for every N every player and ype n RE N, f ψ s > 0 hen s s opmal o he conjecure φ defned as φ s T β ψ s for every s S. Le be n RE N. Snce T s mnmal, s whou loss of generaly o assume N > 2. Le 1,...,,N N be a sequence reachng n N-seps. Clam. There exs a sequence 1, 2,..., N n T such ha N, ϕ n, n ϕ n n for all n N and β [ ] n n+1 > 0 for every n N 1. 8

9 Proof. Snce 1 ϕ [ ], 1 and β > 0 hen here mus exs a ype 2 such ha ϕ 2, 2 ϕ 2 [ ] 2 and β > 0. A smple argumen by nducon concludes he proof. Clam. For every 2 < n N, n s n [ψ] R CB [ T ] CB [ψ] CB R. Proof. I can be easly proved by nducon. Suppose ψ s > 0. Snce N 1 s n B [ψ], hen β N 1, s > 0. Snce N 1 s n B R hen s s a bes response o he frs order belef over sraeges of ype, defned as he conjecure For every s S. φ s T β, s Snce s n B [ T ] B [ψ], for every, s such ha β, s > 0 here s a ype T such ha ϕ, ϕ and s ψ. Therefore φ s T β N, s T :ϕ, ϕ T β ψ s β, ψ 9

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