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# Dene x t arg maxx x f x t then we have 1 2 for

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Unformatted text preview: {X , Y } Deﬁne the order Y � X , which induces a lattice structure Given s ∈ S , let X (s ) = {i ∈ I | si = X }, Y (s ) = {i ∈ I | si = Y }. We deﬁne the payoﬀ functions as � Bi (X (s ), X ) ui ( si , s − i ) = Bi (Y (s ), Y ) if si = X , if si = Y It can be veriﬁed that payoﬀ functions satisfy increasing diﬀerences. 24 Game Theory: Lecture 7 Supermodular Games Example II– Cournot As a Supermodular Game with Change of Order Consider Cournot duopoly model. Two ﬁrms choose the quantity they produce qi ∈ [0, ∞). Let P (Q ) with Q = qi + qj denote the inverse demand (price) function. Payoﬀ function of each ﬁrm is ui (qi , qj ) = qi P (qi + qj ) − cqi . Assume P � (Q ) + qi P �� (Q ) ≤ 0 (ﬁrm i ’s marginal revenue decreasing in qj ). We can now verify that the payoﬀ functions of the transformed game deﬁned by s1 = q1 , s2 = −q2 have increasing diﬀerences in (s1 , s2 ). 25 Game Theory: Lecture 7 Supermodular Games Monotonicity of Optimal Solutions Key theorem about monotonicity of optimal solutions: Theorem (Topkis) Let X ⊂ R be a compact set and T be some partially ordered set. Assume that the function f : X × T → R is continuous [or upper semicontinuous] in x for all t ∈ T and has increasing diﬀerences in (x , t ). Deﬁne x (t ) ≡ arg maxx ∈X f (x , t ). Then, we have: 1 2 For all t ∈ T , x (t ) is nonempty and has a greatest and least element, denoted by x (t ) and x(t ) respectively. ¯ ¯ ¯ For all t � ≥ t , we have x (t � ) ≥ x (t ) and x(t � ) ≥ x(t ). Summary: if f has increasing diﬀerences, the set of optimal solutions x (t ) is non-decreasing in the sense that the largest and the smallest selections are non-decreasing. 26 Game Theory: Lecture 7 Supermodular Games Proof By the assumptions that for all t ∈ T , the function f (·, t ) is upper semicontinuous and X is compact, it follows by the Weierstrass’ Theorem that x (t ) is nonempty. For all t ∈ T , x (t ) ⊂ X , therefore is bounded. Since X ⊂ R, to establish that x (t ) has a greatest and lowest element, it suﬃces to show that x (t ) is closed. Let {x k } be a sequence in x (t ). Since X is compact, x k has a limit point x . ¯ By restricting to a subsequence if necessary, we may assume without loss of generality that x k converges to x . ¯ Since x k ∈ x (t ) for all k , we have f (x k , t ) ≥ f (x , t ), ∀ x ∈ X. Taking the limit as k → ∞ in the preceding relation and using the upper semicontinuity of f (·, t ), we obtain f (x , t ) ≥ lim sup f (x k , t ) ≥ f (x , t ), ¯ k →∞ ∀ x ∈ X, thus showing that x belongs to x (t ), and proving the closedness claim. ¯ 27 Game Theory: Lecture 7 Supermodular Games Proof Let t � ≥ t . Let x ∈ x (t ) and x � = x (t � ). ¯ By the fact that x maximizes f (x , t ), we have f (x , t ) − f (min(x , x � ), t ) ≥ 0. This implies (check the two cases: x ≥ x � and x � ≥ x ) that f (max(x , x � ), t ) − f (x � , t ) ≥ 0. By increasing diﬀerences of f...
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## This document was uploaded on 03/19/2014 for the course EECS 6.254 at MIT.

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