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  Strategic Game: Nash equilibriumECON6036 1 Strategic Game De fi nition: A strategic game denoted by  h N, ( A i ) , ( u i ) i  consists of the fol-lowing components: ã  a  fi nite set  N   (of players) ã  For all  i ∈ N  , a set of actions  A i  (the set of actions available for  i ) ã  For all  i ∈ N  , a utility function u i  :  A → R where  A  =  ×  j ∈ N  A  j  and  u i ( a )  is said to be  i ’s payo ff  under action pro fi le a . Remark 1  : If for all   i, A i  is   fi  nite, then the game is said to be   fi  nite. 2 Nash Equilibrium De fi nition 2  Nash equilibrium of a strategic game   h N, ( A i ) , ( u i ) i  is a pro  fi  le  a ∗ ∈  A  of actions with the property that for every player   i  ∈  N,  we have  u i ¡ a ∗ i ,a ∗− i ¢ ≥ u i ¡ a i ,a ∗− i ¢  for all   a i  ∈ A i . De fi nition 3  For any   a − i  ∈  A − i , de   fi  ne   B i  ( a − i )  to be the set of player   i ’s best actions given   a − i  : B i  ( a − i ) =  { a i  ∈ A i  :  u i  ( a i ,a − i ) ≥ u i  ( a 0 i ,a − i ) ∀ a 0 i  ∈ A i } B i  ( a − i )  is said to be  i ’s best response function (correspondence).An alternative de fi nition of Nash equilibirum1  De fi nition 4  A Nash equilibrium is a pro  fi  le   a ∗ of actions for which  a ∗ i  ∈ B i ¡ a ∗− i ¢  for all   i ∈ N  . One application is Cournot competition. 3 Fixed Point Theorem A game has a Nash Equilibrium if there exists  a ∗ such that  a ∗ i  ∈ B i ¡ a ∗− i ¢  for all  i ∈ N.  (1)De fi ne  B  :  A → A  by  B ( a ) =  × i ∈ N  B i  ( a − i ) .  Then (1) can be written asthere exists  a ∗ such that  a ∗ ∈ B ( a ∗ ) .  (2)Fixed point theorems give conditions on  B  under which there indeedexists a value of   a ∗ for which  a ∗ ∈ B ( a ∗ ) . Theorem 5  (Kakutani’s   fi  xed point theorem) Let   X   be a compact convex subset of   R ∗ and let   f   :  X   →  X   be a set-valued function. There exists  x ∗ ∈ X   such that   x ∗ ∈ f   ( x ∗ )  if   f  ( . )  satis   fi  es the following two conditions: ã  ∀ x ∈ X  , the set   f   ( x )  is non-empty and convex  ã  the graph of   f   is closed (i.e., for all sequence   { x n }  and   { y n }  such that  y n  ∈ f   ( x n ) ∀ n,x n  → x  and   y n  → y , we have   y ∈ f   ( x ) ). 2  4 Existence of Nash Equilibrium Theorem 6  (Existence of Nash equilibrium) The strategic game   h N, ( A i ) , ( u i ) i has a Nash equilibrium if for all   i ∈ N  ã  A i  is non-empty, compact, convex subset of a Euclidian space  ã  u i  is continuous on   A  and quasi-concave on   A i ,i  = 1 ,...,n. The sketch of the proof is as follows. De fi ne  B  :  A  →  A  by  B ( a ) = × i ∈ N  B i  ( a − i ) . Then  B i  ( a − i )  is non-empty because  u i is continuous and  A i is compact.  B i  ( a − i )  is also convex because  u i is quasi-concave on  A i . Thissuggests that  B  is non-empty and convex ( fi ll in the detail).  B  has a closedgraph because each  u i is continuous. Hence by Kakutani’s theorem,  B  has a fi xed point. As we have noted, any  fi xed point is a Nash equilibrium. Proof.  De fi ne the set-valued function  B  :  A ⇒ A  by B ( a ) = ( B 1  ( a − 1 ) ,B 2  ( a − 2 ) ,...,B n  ( a − n )) . (Recall that  B i  is the best response function of   i .) Notice that if   a  ∈  B ( a ) then  a i ∈ B i ( a − i )  for all  i ; and hence  a  is a Nash equilibrium. Consequently,if we can show that Kakutani’s theorem applies to  B , the proof will becomplete. First notice that  A  is easily checked to be compact and convexsince each  A i  has these two properties. (Verify.) Moreover,  B  is nonempty-valued since, by Weierstrass’ theorem, each  B i  is nonempty-valued. To seethat  B  is convex-valued, consider any  x,y  ∈  B ( a ) . Then, for all  i  ∈  N,u i ( x i ,a − i ) =  u i ( y i ,a − i )  ≥  u i ( a i ,a − i )  for all  a i  ∈  A i , and thus by using thequasi-concavity of   u i  on  A i , we  fi nd u i ( λx i  + (1 − λ ) y i ,a − i ) ≥ u i ( x i ,a − i ) ≥ u i  ( a i ,a − i ) for all  a i ∈ A i  and all  λ ∈ [0 , 1] .  Since this holds for each  i , we conclude that λx +(1 − λ ) y ∈ B ( a ) , and hence that  B  is convex-valued. It remains to check B  has a closed graph. Towards this end, take any sequences  a m and  b m in  A such that  a m → a,b m → b  and  b m ∈ B ( a m ) . We need to show that  b ∈ B ( a ) ,that is,  b i  ∈ B i ( a − i )  for all  i . Suppose, by contradiction, that there exists an i ∈ N   such that  b i  / ∈ B i ( a − i ) . Then there exists an  x i  ∈ A i  such that k ≡ u i ( x i ,a − i ) − u i ( b i ,a − i )  >  0 . 3  By continuity of   u i , for all  ε >  0 ; there exists an integer  M  ε  such that ¯¯ u i ( x i ,a m − i ) − u i ( x i ,a − i ) ¯¯  < ε  and ¯¯ u i ( b mi  ,a m − i ) − u i ( b i ,a − i ) ¯¯  < ε for all  m ≥ M  ε . So, for any  ε >  0  and any  m ≥ M  ε , we have u i ( x i ,a m − i )  > u i ( x i ,a − i ) − ε  =  u i ( b i ,a − i ) + k − ε > u i ( b mi  ,a m − i ) + k − 2 ε. Thus, choosing  ε  =  k/ 2 , we obtain  u i ( x i ,a m − i )  > u i ( b mi  ,a m − i )  for all  m ≥ M  α/ 2 ,contradicting  b mi  ∈ B i ( a m − i )  for all  m . We can then conclude that  b i ∈ B i ( a − i ) for all  i  ∈  N   and hence  b  ∈  B ( a ) . The proof is completed by applyingKakutani’s  fi xed point theorem.Applications: Cournot duopoly; Bertrand duopoly.4
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