Université du Maine Théorie des Jeux Yves Zenou Correction de l examen du 16 décembre 2013 (1 heure 30)

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1 Université du Maine Théorie des Jeux Yves Zenou Correction de l examen du 16 décembre 2013 (1 heure 30) Problem (1) (8 points) Consider the following lobbying game between two firms. Each firm may lobby the government in hopes of persuading the government to make a decision that is favorable to the firm. The two firms, and, independently and simultaneously decide whether to lobby () ornot(). Lobbying entails a cost of 15. Not lobbying costs nothing. If both firms lobby or neither firm lobbies then the government takes a neutral decision, which yields 10 to both firms. If firm lobbies and does not lobby, then the government makes a decision that favors firm, yielding zero to firm and 30 to firm. Finally, if firm lobbies and does not, the government s decision yields to firm and zero to firm. Assume that 25. The normal form of this game is (player 1 is firm and player 2 is firm ): \ 5, 5 15, 0 0, 15 10, 10 (1a) Determine the pure-strategy Nash equilibrium of this game (if it has any). (1 point) The BR functions are given by (since 25, then 15 10): \ 5, 5 15, 0 0, 15 10, 10 As a result, there are two pure-strategy Nash equilibria: ( ) and(). 1

2 (1b) Compute the mixed-strategy Nash equilibrium of this game (if it has any). (3 points) Assume that firm believes that firm plays strategy with probability and strategy with probability 1. Assume also that firm believes that firm plays strategy with probability and strategy with probability 1. In that case, the value of that makes firm indifferent between and is given by: 5 +( 15)(1 ) Firm s expected payoff of playing = 0 +10(1 ) Firm s expected payoff of playing Rearranging yields = i.e., this is the probability that firm chooses to play that makes firm indifferent between playing and. Itiseasilyverified that 0 1. Similarly, the value of that makes firm indifferent between and is given by: 5 +15(1 ) Firm s expected payoff of playing = 0 +10(1 ) Firm s expected payoff of playing Rearranging yields = 1 2 i.e., this is the probability that firm chooses to play that makes firm indifferent between playing and. The mixed strategy NE is thus: ½µ µ ¾ 2 (1c) Given the mixed-strategy Nash equilibrium computed in part (1b), what is the probability that the government makes a decision that favors firm? (2 points) We would like to calculate the probability that (, ) occurs, i.e. firm plays and firm plays. This probability is: (1 ) = 1 µ

3 (1d) As rises, does the probability that the government makes a decision favoring firm rise or fall? Is this good from an economic standpoint? (2 points) We need to calculate: [(1 )] = ( 20) 2 0 Thus, as increases, the probability of ( ) decreases. However, as becomes larger, ( ) is a better outcome. Problem (2) (12 points) Consider a Cournot game with two firms where the (inverse) demand function is given by: ( () if b () = 0 otherwise where = is the market demand and b = 1 ( ). It is assumed that 0 () 0. Each firm =1 2 has a total cost function equals to ( )= 2 The two firms simultaneously choose their quantities and we want to determine the Nash equilibrium of this game in quantities, which we denote by ( 1 2). (2a) Show under which conditions there is a unique and interior Nash equilibrium. (2 points) We need to verify that the second-order conditions and the boundary conditions are always satisfied. The profit offirm is given by: = () ( )=[ ()] 2 =[ ( + )] 2 We have for 6= : ( ) = 0 () () 2 =0 The second-order condition for =12 is given by: 2 ( ) 2 = 00 () 2 0 () 2 0 3

4 Asufficient condition for the second-order condition to be satisfied is: 00 () 0. Boundary conditions: Wewanttoshowthat =(00) is not a Nash equilibrium and that ³ = is not a Nash equilibrium. 2 2 () Let us show that =(00) is not a Nash equilibrium. The first boundary condition (fix at =0and show that firm wants to deviate from =0)isgivenby: ( ) =0 =0 = (0) 0 As a result, if (0),thenfirm always wants to deviate from = 0 and thus =(00) cannot be a Nash equilibrium. () Letusnowshowthat ³ = is not a Nash equilibrium. Fix 2 2 at = 2 b and let us show that firm wants to deviate from = b2. The second boundary condition is thus given by: ( ) = 2 = 2 = 0 ( b ) b 2 ( b ) b 0 This is clearly always negative since b =0and 0 ( b ) 0. To summarize, if both 00 () 0 and (0), then there is a unique and interior Nash equilibrium. (2b) Assume that =1and that () =. Calculate the unique Nash equilibrium of this game. Give the Nash equilibrium values of the quantities, 1, 2. (4 points) First-order conditions: ( 1 2 ) = 2 1 (1 + 1 )+ 2 =0 2 ( 1 2 ) = 2 2 (1 + 2 )+ 1 =0 2 which lead to the following best-reply functions: ( 2 ) 1 ( 2 )= ( 2) 2(1+ 1 ) 2 ( 1 ) 2 ( 1 )= ( 1) 4

5 Combining these two equations, we obtain: that is 1 = 2(1+ 1 ) ( 1 ) 4(1+ 1 )(1+ 2 ) 1 = Plugging back this value in 2 ( 1 ), we obtain: 2 = ( ) 3+4( ) ( ) 3+4( ) (2c) Assume that =1and that () =. Determine the social optimum solution in quantities, 1, 2. (2 points) Here The social optimum is defined such that: max {Π( 1 2 )= 1 ( 1 2 )+ 2 ( 1 2 )} 1 2 Π( 1 2 )=( ) First-order conditions: Π = =0 (1) Π = =0 2 (2) If we substract these two equations, we first obtain that: Plugging the value of 2 from (3) into (1), we obtain: 1 = Plugging this value into (3), we finally get: 2 = 2 = (3) 2 2( ) 2( ) 5

6 (2d) Assume that =1and that () =. Assume also that firm 1 decides first its quantity 1 and then firm 2 decides its quantity 2. Determine the subgame-perfect equilibrium of this game where you denote the equilibrium quantities by 1 and 2. Under which condition there is a first-mover advantage in terms of quantities, i.e. 1 2? When is there no first-mover advantage? (4 points) We solve the model backward. The best reply function of firm 2 is (as above) given by: 2 ( 1 ) 2 ( 1 )= 1 Plugging this value into the profit function of firm 1, we obtain: First-order condition yields: This is equivalent to: 1 = ( 1 2 ) = 1 ( 1) = (1 + 1 ) 2 1 = (1+ 1) 1 =0 1 = ( ) 2( ) By plugging this value into the best-reply function of firm 2, we obtain: 2 = ( ) = 4( )(1+ 2 ) Thus 1 2 if and only if: which is equivalent to: (1+ 2 ) 1 1+4(1+ 2) 2 4(1+ 2 ) Observe that 1+4(1+ 2) 2 4(1+ 2 ) is increasing in 2. As a result, there is a first-mover advantage if 1 is low enough or 2 is large enough. 6