lecture7 notes

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 03/19/2014 for the course EECS 6.254 at MIT.

Ask a homework question - tutors are online