Game Theory with Applications to Finance and Marketing

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Since players’ payoff functions are linear (hence continuous) in mixedstrategies, and since the simplex Σ i is compact, there exist solutionsto players’ best response problems (Weierstrass theorem). That is, r isnonempty. Moreover, if σ i1 , σ i2 ∈ r i (σ −i ), then clearly for all λ ∈ [0, 1],λσ i1 + (1 − λ)σ i2 ∈ r i (σ −i ),proving that r i is convex, which in turn implies that r is convex. Finally,we claim that r is u.h.c. To see this, suppose instead that r fails tobe upper hemi-continuous at some profile σ ∈ Σ, then there must existsome player i, a sequence {σ −i,n ; n ∈ Z + } converging to σ −i in Σ −i ,and a sequence {σ i,n ′ ; n ∈ Z +} in Σ i converging to some σ i ′ ∈ Σ i withσ i,n ′ ∈ r i(σ −i,n ) for all n ∈ Z + such that σ i ′ is not contained in r i(σ −i ).This would mean that there exists some σ i ′′ ∈ Σ i such that u i (σ i ′′ , σ −i ) >u i (σ i ′, σ −i). Since u i (·) is continuous, and since the sequence {σ n ′ ; n ∈Z + } converges to σ ′ , this would mean that u i (σ i ′′ , σ −i,n ) > u i (σ i,n, ′ σ −i,n )for n sufficiently large, a contradiction to the fact that σ i,n ′ ∈ r i(σ −i,n )for all n ∈ Z + .Now since r(·) is non-empty, convex, and u.h.c., and since Σ is nonempty,convex, and compact in R n , by theorem 3 there must existssome σ ∗ ∈ Σ such that r(σ ∗ ) = σ ∗ .25. From the proof of theorem 2, it is understandable why discontinuouspayoff functions may result in non-existence of NE’s. When payofffunctions fail to be continuous, it may happen that the best responsecorrespondence is not non-empty, or fails to be upper hemi-continuous.For example, consider the single-player game where his payoff functionf : R → R is defined as f(x) = −|x| · 1 [x≠0] − 1 [x=0] , then the playerhas no best response. Even if all best responses are well defined, thebest response correspondence my fail to be upper hemi-continuous, asthe following example shows. Suppose we have a two-player strategicgame where S 1 = S 2 = [0, 1], u 1 (s) = −(s 1 − s 2 ) 2 , andu 2 (s) = −(s 1 − s 2 − 1 3 )2 · 1 [s1 ≥ 1 3 ] − (s 1 − s 2 + 1 3 )2 · 1 [s1 < 1 3 ].Observe that in general lim s1 ↓ 1 u 2 (s) ≠ lim s1 ↑ 1 u 2 (s), proving that u 233is not continuous in s. However, r 2 (·) is a well-defined correspondence;14
in fact, for all s 1 , r 2 (s 1 ) is a singleton, and hence r 2 (·) is in fact a(single-valued) function. Still, r 2 (·) is not upper hemi-continuous (whenthe correspondence is single-valued, upper hemi-continuity becomes thecontinuity of the single-valued function). You can check that r 2 (s 1 ) andr 1 (s 2 ) do not intersect, and hence this two-player game has no NE atall.26. Example 1 (see the section on dominance equilibrium) can be usedto understand the above theorems 1 and 2. Define h i (·) ≡ r i (r j (·)).Apparently, this function can be restricted to the domain of defintion[0, 1 ], which is a non-empty, convex, compact subset of R, and moreover,the functional value of h i is also contained in [0, 1 ]. If this function22intersects with the 45-degree line, then the intersection defines a purestrategy NE. (This game has no mixed strategy NE because a firm’sprofit is a strictly concave function of his own output level given anyoutput choice of its rival.) Now it follows from Brouwer’s fixed pointtheorem (a special version of theorem 3) that if h i (·) is continuous 6then there exists an NE for the game, and moreover, as one can verify,generically a continuous h i (·) will intersect the 45-degree line in an oddnumber of times. 727. Theorem 4: Consider a two-player game where each player i simultaneouslychooses a number s i in the unit interval, and where playeri’s payoff function u i (s 1 , s 2 ) is continuous in (s 1 , s 2 ). This game has amixed strategy NE.Proof Call this game Γ. Consider a sequence of modified games {Γ n }in normal form where the set of players and the payoff functions arethe same as in Γ, but the players’ common pure strategy space isS n = {0, 1 n , 2 n , · · · , n − 1n , 1}.Theorem 2 implies that for all n ∈ Z + , Γ n has an NE (denoted σ n )in mixed strategy. Since the set of probability measures on [0, 1] is6 With the specification P (q 1 + q 2 ) = 1 − q 1 − q 2 , h i (q i ) = 1− 1−q i22, which is indeed acontinuous function of q i .7 Brouwer’s fixed point theorem says that if f : A → A is continuous, where A ⊂ R n isnon-empty, compact and convex, then there exists x ∈ A such that f(x) = x.15
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Game Theory with Applications to Finance and Marketing

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