interpreted as Player 1s mixed action, is actually viewed as Player 2' s beliefs about Player 1's actions; and similarly for o2 Then, instead of assuming that players are rational, we assume that they are certain that their opponent is rational. And, instead of assuming that their first-order beliefs are correct we assume that they are certain of their opponent's first-order beliefs Roughly speaking, if Player 1 believes that Player 2 will choose an action a2 with positi probability (i.e. if 2(a2)>0); if she is certain that he is rational; and if she is certain that his belief about her actions is given by 1, then it must be the case that a2 is a best reply to 1. A similar argument holds for Player 2's beliefs about Player 1, and we can conclude that( 1, 2)must be a Nash equilibrium Let us state and prove the result; additional comments will follow Proposition 0.2 Fix a two-player game G=(N, (Ai, lilieN), with N=(1, 2, and a profile of mixed actions=()ieNC△(A1)×△(A2) (1)If there exists a model M=(Q2, (Ti, ai, pilieN) for G and a state w in the model such that B(fB)∩B(a1=]n then o is a Nash equilibrium of (the mixed extension of)G (2)If o is a Nash equilibrium of (the mixed extension of)G, then there exists a model M=(9,(T,a,n)l∈N) for G and a state w in the model such that w∈B(B)∩B(a1 ]∩a2=c2 We begin with two preliminary results Lemma 0.3 For any event E C Q, every Player i N, and every state w E Q2: WE Bi(E) iff supppi ((ti(w))CE Proof of Lemma 0.3: suppose w E B (E). Then pi( Et(w))= l. This requires that D2(u|t(∞)>0→∈E, so supp pi(|t1(u)cE. The converse is obvious.■ Lemma0.4 For every player i∈N, every t;∈T; and every state w∈9,w∈Bl(t;)诳 ti=ti(u); hence, B (a;=ai])=ai=ai and Bi(a-i=q))=[a-i=g, and B (R=Ri Proof: For every w, pi(t:( w))=l if ti=ti(w), and Pi(titi w)=0 otherwise. This proves the first claim. Now ai is T-measurable; hence, if w ai=ail, then t(w)c [ ai ai). and if w E Bi(a;=ail), then necessarily w E ti(w)c ai=ail( because otherwise we could not have pi(la;=aillti ())=1); and similarly for a-i=q]. The second claim follows immediately. Finally, note that r i=w: aiw)Eri(a-iw)) must clearly be Ti -measurable, and the third claim follows by the same argument as above. linterpreted as Player 1’s mixed action, is actually viewed as Player 2’s beliefs about Player 1’s actions; and similarly for φ2. Then, instead of assuming that players are rational, we assume that they are certain that their opponent is rational. And, instead of assuming that their first-order beliefs are correct, we assume that they are certain of their opponent’s first-order beliefs. Roughly speaking, if Player 1 believes that Player 2 will choose an action a2 with positive probability (i.e. if φ2(a2) > 0); if she is certain that he is rational; and if she is certain that his belief about her actions is given by φ1, then it must be the case that a2 is a best reply to φ1. A similar argument holds for Player 2’s beliefs about Player 1, and we can conclude that (φ1, φ2) must be a Nash equilibrium. Let us state and prove the result; additional comments will follow. Proposition 0.2 Fix a two-player game G = (N,(Ai , ui)i∈N ), with N = {1, 2}, and a profile of mixed actions φ = (φi)i∈N ⊂ ∆(A1) × ∆(A2). (1) If there exists a model M = (Ω,(Ti , ai , pi)i∈N ) for G and a state ω in the model such that ω ∈ B(R) ∩ B([α1 = φ1] ∩ [α2 = φ2]) then φ is a Nash equilibrium of (the mixed extension of) G. (2) If φ is a Nash equilibrium of (the mixed extension of) G, then there exists a model M = (Ω,(Ti , ai , pi)i∈N ) for G and a state ω in the model such that ω ∈ B(R) ∩ B([α1 = φ1] ∩ [α2 = φ2]). We begin with two preliminary results. Lemma 0.3 For any event E ⊂ Ω, every Player i ∈ N, and every state ω ∈ Ω: ω ∈ Bi(E) iff supp pi(·|ti(ω)) ⊂ E. Proof of Lemma 0.3: suppose ω ∈ Bi(E). Then pi(E|ti(ω)) = 1. This requires that pi(ω 0 |ti(ω)) > 0 ⇒ ω 0 ∈ E, so supp pi(·|ti(ω)) ⊂ E. The converse is obvious. Lemma 0.4 For every player i ∈ N, every ti ∈ Ti and every state ω ∈ Ω, ω ∈ Bi(ti) iff ti = ti(ω); hence, Bi([ai = ai ]) = [ai = ai ] and Bi([α−i = q]) = [α−i = q], and Bi(Ri) = Ri . Proof: For every ω, pi(ti |ti(ω)) = 1 if ti = ti(ω), and pi(ti |ti(ω)) = 0 otherwise. This proves the first claim. Now ai(·) is Ti-measurable; hence, if ω ∈ [ai = ai ], then ti(ω) ⊂ [ai = ai). and if ω ∈ Bi([ai = ai ]), then necessarily ω ∈ ti(ω) ⊂ [ai = ai ] (because otherwise we could not have pi([ai = ai ]|ti(ω)) = 1); and similarly for [α−i = q]. The second claim follows immediately. Finally, note that Ri = {ω : ai(ω) ∈ ri(α−i(ω))} must clearly be Ti-measurable, and the third claim follows by the same argument as above. 4