The bottom line is that mixed strategies, and indeed perhaps Nash equilibria in general are best interpreted as representing "consistent beliefs Upper and Lower Hemicontinuity of the Nash correspondence I conclude with a technical note. If we fix the number of players and the size of each players strategy set, then we can identify a finite game with the collection of payoff vectorse with a point u in Euclidean space. Thus, denote by N(u) the set of Nash equilibria of the finite game with corresponding to u. Formally, if M=lier Ail, the space of games is just R1M,soN:RM→Iler△A1) It is fairly easy to see that N(u) is uhc. However, it is not lhc, as the game in Fig. 2 demonstrates L R ,10,0 Figure 2 LHC fails at A=0 Let ux E R2- denote the payoffs in the game with parameter A E R. It is easy to see that For A<0, there is a unique NE, (1, 1)=(T, L) For A>0, the NE are(1, 1),(0,0)and (2, 1) For A=0, the NE are(1, 1)and(, 0) for any T E [0, 2]-this includes(0,0)and (3,1)=(3,0) It is easy to see that there is no violation of uhc if we consider payoff sequences uAn generated by any sequence A-0-this much we expected However, there is a violation of lhc: any(a1,0) with an E(0, 2)is in N(uo), but it is mpossible to select a∈N(uxn) so that a→(a,0)The bottom line is that mixed strategies, and indeed perhaps Nash equilibria in general, are best interpreted as representing “consistent beliefs.” Upper and Lower Hemicontinuity of the Nash correspondence I conclude with a technical note. If we fix the number of players and the size of each player’s strategy set, then we can identify a finite game with the collection of payoff vectors—i.e. with a point u in Euclidean space. Thus, denote by N(u) the set of Nash equilibria of the finite game with corresponding to u. Formally, if M = Q i∈I |Ai |, the space of games is just RI·M, so N : RI·M ⇒ Q i∈I ∆(Ai). It is fairly easy to see that N(u) is uhc. However, it is not lhc, as the game in Fig. 2 demonstrates. L R T 1,1 0,0 B 0,0 λ, 2 Figure 2: LHC fails at λ = 0. Let uλ ∈ R2·4 denote the payoffs in the game with parameter λ ∈ R. It is easy to see that: • For λ < 0, there is a unique NE, (1,1) = (T,L). • For λ > 0, the NE are (1,1), (0,0) and ( 2 3 , λ 1+λ ). • For λ = 0, the NE are (1,1) and (x, 0) for any x ∈ [0, 2 3 ]—this includes (0,0) and ( 2 3 , λ 1+λ ) = ( 2 3 , 0) It is easy to see that there is no violation of uhc if we consider payoff sequences uλn generated by any sequence λ n → 0—this much we expected! However, there is a violation of lhc: any (α1, 0) with α1 ∈ (0, 2 3 ) is in N(u0), but it is impossible to select α n ∈ N(uλn ) so that α n → (α, 0). 9