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 nondecreasing in the sense that the largest and the smallest
selections are nondecreasing.
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.
 Spring '10
 AsuOzdaglar

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