1 Bertrand duopoly with incomplete information

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Suzanna Gibbs
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1 Game Theory Solution to Problem Set 5 1 Bertrand duopoly ith incomplete information The game i de ned by I = f1; g ; et of player A i = [0; 1) T i = fb L ; b H g, ith p(b L ) = u i (b i ; p i ; p j ) = (a pi b i p j )p i ; if p i 6 a b i p j 0 ; otherie : In addition, the trategy et S i (8i = 1; ) are the et of all function i : T i i, i.e. mapping from type into action. The ymmetric puretrategy BNE are the BNE i hich i (b i = b L ) = j (b j = b L ) and i (b i = b H ) = j (b j = b H ): No, in order to nd the ymmetric pure trategy BNE, rt uppoe that p i 6 a b i p j ; 8i = 1;. Since e are focuing on ymmetric equilibria, e have that: p i (b i ) = arg max p i (a pi b i p L j )p i + (1 )(a p i b i p H j )p i F OC : (a p i b i p L j ) p i + (1 )(a p i b i p H j ) (1 )p i = 0 a b i p L j p i (1 )b i p H j = 0 p i (b i ) = a b ip L j (1 )b i p H j No, impoing ymmetry and denoting p L and p H the equilibrium price of the lo type and high type repectively, e have that: p L = a b Lp L (1 )b L p H p L( + b L ) = a (1 )b L p H p L = a (1 )b Lp H ( + b L ) 1

2 Similarly, for the high type: p H = a b Hp L ( + (1 )b H ) Combining the expreion for p L and p H, e get: p L = a ( + (1 )(b H b L )) ( + (1 )b H + b L ) p H = a ( (b H b L )) ( + (1 )b H + b L ) (1) () If the equilibrium price and quantitie are all poitive (i.e. if e in fact have an interior olution), p L and p H a de ned above contitute a ymmetric BNE. Note that p L >0 alay hold, o the condition that have to hold for both price to be poitive i: (b H b L ) > 0: The quantitie are poitive if: q(p H; p H) = a p H b H p H > 0 q(p H; p L) = a p H b H p L > 0 q(p L; p H) = a p L b L p H > 0 q(p L; p L) = a p L b L p L > 0 Remember that b H > b L, and notice that p L > p H :Thi implie that if the inequality q(p H ; p L ) > 0 hold, the other three inequalitie ill alo hold. Hence, in order to check that the equilibrium quantitie are poitive, it i enough to check that the folloing hold: q(p H; p L) = a p H b H p L > 0 p H < a b H p L a ( (b H b L )) ( + (1 )b H + b L ) < a b H p L a ( (b H b L )) < a( + b H (b H b L )) b H a ( + (1 )(b H b L )) ( (b H b L )) < ( + b H (b H b L )) b H ( + b H b L (b H b L )) = ( b H )( + b H (b H b L )) + b L b H = ( b H )( (b H b L )) + b H ( b H + b L ) 0 < (1 b H )( (b H b L )) + b H ( b H + b L ) 0 < (1 b H )( + b H (b H b L )) + b H (1 + b L ) (1 b H )A + B if b H < 1 then q(p H; p L) > 0.

3 No, note that the expreion A i alay poitive, and o i B. Hence, if e aume that b H < 1 (hich i u cient but not neceary), e have that q(p H ; p L ) > 0, and expreion (1) and () decribe a ymmetric pure trategy BNE. A ar game T 1 = T = f; g ; p() = p() = 1 A 1 = A = fa; Ng 1 : T 1 1 (pure trategie) : T (pure trategie) Payo : t 1 = ; t = A 8; 8 10; 0 t 1 = ; t = A 8; 4 10; 0 t 1 = ; t = A 4; 8 10; 0 t 1 = ; t = A 6; 6 10; 0 The approach e take here to nding the pure trategy BNE i to look at all poible trategie (AA, AN, NA, NN) for one of the player, and check for each if it can be an equilibrium trategy. Firt, uppoe player follo the folloing trategy: = Then, e can derive the bet-repone of player 1: 9 Eu 1 (A; ; ) = 8 >= Eu 1 (N; ; ) = 0 ) BR Eu 1 (A; ; ) = 1 1 ( ) = >; Eu 1 (N; ; ) = 0 Similarly, if e uppoe that 1 =, the bet-repone for player i =. Hence, by the ymmetry of the game e have found to xed-point. That i, e have the folloing to pure trategy BNE: 3

4 i = ; j =, i f1; g No, uppoe player play according to: = Then, e can derive the bet-repone of player 1: 9 Eu 1 (A; ; ) = 1 >= Eu 1 (N; ; ) = 0 ) BR Eu 1 (A; ; ) = 1 ( ) = >; Eu 1 (N; ; ) = 0 But if 1 =, the bet-repone of player i BR ( 1 ) =. Hence, thi i not a BNE. No, uppoe player play according to: = Then, e can derive the bet-repone of player 1: 9 Eu 1 (A; ) = 1 >= Eu 1 (N; ) = 0 ) BR Eu 1 (A; ) = 7 1 ( ) = >; Eu 1 (N; ) = 0 But if 1 =, the bet-repone of player i BR ( 1 ) =. Hence, thi i not a BNE. To ummarize: the to BNE in pure trategie are: i = ; j =,for i; j f1; g ; i 6= j: 4

5 3 I Information alay bene cial? The game i decribed by: and the payo : T 1 = fa; bg ; p(a) = p(b) = 1 A 1 = fu; Dg ; A = fl; C; Rg t 1 = a t 1 = b L C R L C R U 4; 1 3; 0 3; 3 U 4; 1 3; 3 3; 0 D 5; 4 ; 5 ; 0 D 5; 4 ; 0 ; 5 Finally, a pure trategy for player 1 i a function 1 : T 1 1, herea a pure trategy for player i an A. Firt note that if e had aumed that thee ere to eparate game of complete information, e could have ued repeated deletion of trictly dominated trategie to nd unique pure trategy NE - (U, C) and (U, R) - for the game correponding to "a" and "b" repectively. In both of thee game, the equilibrium payo for player ould have been 0 (and the payo for player 1 ould have been 3). No, let u return to the incomplete information game decribed above. Note rt that no L trictly dominate both C and R for player. Thi implie that in any BNE, it mut be the cae that player play trategy L. Then, if player play L, the bet-repone for player 1 i trategy DD, i.e. to play D he /he i type a, and D hen /he i type b. Hence, e have found a unique BNE: = ( 1 ; ) = (DD; L): Finally, note that the equilibrium payo for player in the incomplete information game i 4 > 0. That i, if e believe that the olution concept ued here (NE and BNE) are appropriate a predictor of real-orld outcome, player ould actually bene t from having le information. 3.1 Alternative approach In order to illutrate di erent ay of looking for BNE, thi ubection preent an approach to quetion 3 that i not baed on the fact that C and R are trictly dominated by L. In thi cae, e need to conider the di erent poible trategie for player. The trategie for player can take the folloing form: L C 5

6 R LC (i.e., a randomization beteen L and C), LR CR LCR We need to check each of thee and ee if e can have a BNE on each particular form. Firt, uppoe S = L D S 1 = D + if t1 = a if t 1 = b Eu (L) = 1 f4 + 4g = 4 Eu (C) = Eu (R) = :5 hich implie that ((D; D) ; L) i a BNE. No, uppoe S = C U S 1 = U + Eu (L) = 1 if t1 = a if t 1 = b Eu (C) = Eu (R) = 1:5 Hence, BR (S 1 ) = L 6= C, o thi i not a BNE. Similarly, e can ho that player cannot play S = R in any BNE. Suppoe player mixe beteen fc; Rg :Since for both S = C and S = R, the bet repone of player 1 (both type) i U, hich mean that thi i te bet-repone alo to a trategy in hich player mixe beteen fc; Rg : But if player 1 play U, the bet repone of player i L: Thu, a mix beteen fc; Rg cannot be played in a BNE. Suppoe player mixe beteen fl; Cg ith probabilitie f; 1 g. For player to mix it mut be the cae that her expected payo of playing L and C are equal and greater than the expected utility of playing R. To compute the expected utility, e need to pecify hat player believe that the to type of player 1 ill do. Suppoe player believe that type a mixe beteen fu; Dg ith probability fp 1 ; 1 p 1 g and type b mixe beteen fu; Dg ith probability fp ; 1 p g :Then: 6

7 Eu (L) = 1 [p 1( 1) + (1 p 1 )4] + 1 [p ( 1) + (1 p )4] = 4 :5(p 1 + p ) Eu (C) = 1 ( p 1 3p ) = :5 :5p 1 1:5p Eu (R) = 1 ( 3p p ) = :5 :5p 1:5p 1 Hoever, then for Eu (L) = Eu (C) e mut have that p = 1:5; hich i impoible. A imilar analyi rule out a mix beteen fl; Rg. Finally, uppoe player i mixing among fl; C; Rg. A e have hoed, for player to mix beteen fl; Cg or fl; Rg, it mut be the cae that ome of the probabilitie de ned above exceed 1. Hence, it i impoible to induce player to mix beteen thee action. To ummarize, the unique BNE i ((D; D) ; L), here: Eu (DD; L) = 4 u 1 (DD; L j a) = 5 u 1 (DD; L j b) = 5: 7

CMSC 474, Introduction to Game Theory Maxmin and Minmax Strategies

CMSC 474, Introduction to Game Theory Maxmin and Minmax Strategie Mohammad T. Hajiaghayi Univerity of Maryland Wort-Cae Expected Utility For agent i, the wort-cae expected utility of a trategy i i the