Game Theory Solutions to Problem Set 3. S i = q i 2 [0; 1) : 0 Otherwise. P b n q i ) > 0, then each player i maximizes her payo : max. (a b.

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1 Cournot Oligopoly Game Theory Solutions to Problem Set 3 Strategies set for player i S i = q i 2 [0; ) Payo s, < P (a b n P q i )q i cq i if (a b n q i ) > 0 i = i= i= 0 Otherwise Suppose that p = (a P b n q i ) > 0, then each player i maximizes her payo i= F OC a b q i = max q i (a b a nx q i )q i i= cq i nx qj q i c = 0 j6=i P b n qj j6=i Given that we are focusing on symmetric NE, we can impose qi some q satisfying c = q for q + b(n )q = (a c) q (a c) = b(n + ) Q X q n(a c) i = b(n + ) i Here, we need to check that indeed p = (a P b n q i ) > 0 i=

2 Hence, the symmetric NE is nx p = (a b q i ) = i= n(a c) = (a b b(n + ) ) a(n + ) n(a c) = (n + ) = a + nc (n + ) > 0 X q i = q = (a c) b(n + ) i As n! lim n! X i qi = n(a c) (a c) lim = n! b(n + ) b nx ) p = (a b q i ) = c ) i = 0 i= As n!, we approach the competitive equilibrium. 2 Cournot Oligopoly In the same way as in the previous question, we nd q = a bq 2 c q 2 = a bq c 2 = a c = a c 2 2 q 2 2 q This is a system of two equations in two unknows. Solve this system to nd the NE 2

3 q = a c 3 4 q = a ) q = a 3b a 2 c + c 2 c2 2c 3b + c 2 3b = 3b (a 2c + c 2 ) ) q 2 = a c 2 = 3b (a + c 2c 2 ) Knowing that 0 < c < c 2 we have that 2 q 6b (a 2c + c 2 ) q = 3b (a 2c + c 2 ) > 3b (a + c 2c 2 ) = q 2 And since both rms are facing the same prince p = a b (q + q2) = a 3 (a 2c + c 2 + a + c 2c 2 ) = 3 (a + c + c 2 ) > 0 we can show that the pro t is greater for rm 3 Bertrand Oligopoly (p c ) q > (p c 2 ) q 2 Strategies for player i S i = p i 2 [0; ) < (a bp i )(p i c) if p i < p j P ayoffs i (p i ; p j ) = 2 (a bp i)(p i c) if p i = p j 0 Otherwise Notice that, in equilibrium, no rm will suggest a price p < c or p > a b Thus, we can restrict attention to S i = p i 2 [c; a b ] 3

4 Now consider the following cases Case c < p i < p j a b In this case, rm i capture the entire market. However, since c < p i, then rm j has the incentive to decrease it s price such that c < p j < p i. It would then capture the entire market a have a strictly positive pro t. And using the same argument, there could not be a NE where c < p j < p i a b Case 2 c = p i < p j a b In this case, holding p j constant, rm i has the incentive to increase p i Case 3 c < p i = p j = p < a b In this case, both rms share the market demand, as is indicated in i (p i ; p j ) In this case, each rm has incentive to lower p by an arbitrary small amount " > 0 (say to p 0 ) and thus capture the entire market demand. Since the market demand function is continuous, when " is small enough, (a bp 0 )(p 0 c) > 2 (a bp)(p c). Case 4 p i = p j = p = a b Setting p = a, (p) = 0 and thus each rm has the obvious incentive to b decrease its price, as (a bp 0 )(p 0 c) > 0 for c < p 0 < a b Case 5 p i = p j = p = c Here, both rms have pro ts (c) = 0 and there is no incentive to deviate, since increasing the price, holding the other players price constant, will still yield 0 pro ts. Together, this implies that the unique pure strategy NE is (c, c). 4 Bertrand Oligopoly with Fixed Costs 4. Firm has xed costs F>0 Suppose rm 2 is a monopolist. Then rm 2 faces the following optimization problem max (p c) (a bp) p The solution to the above problem is p m = a+bc The pro ts of rm 2 when it charges the price p m are m = (a bc)2 There are two cases to consider. (i) F > (a The pure-strategy Nash equilibria are bc)2 4

5 (ii) F 6 (a (p = p; p 2 = p) where p 2 [c; p m ] (p ; p 2 = p m ) for p > p m bc)2 Let ^p be de ned by (^p c) (a b^p) = F ^p < p m That is, ^p is the smallest price at which the pro ts of rm 2 (when it is a monopolist) are equal to rm s xed cost F It is easy to check that q ^p = a + bc (a bc) 2 F The pure-strategy Nash equilibria are (p = p; p 2 = p) where p 2 [c; ^p] 4.2 Both rms have xed costs F>0 We need to consider three di erent cases. (i) F > (a The pure-strategy Nash equilibria are (ii) F < (a bc)2 (p ; p 2 ) where p > a b ; p 2 > a b bc)2 The pure-strategy Nash equilibrium is (p = ^p; p 2 = ^p) ; where ^p is de ned in part (a). (iii) F = (a bc)2 The pure-strategy Nash equilibria are where p m = a+bc 4.3 Di erent marginal costs (p = p m ; p 2 ) with p 2 > p m (p ; p 2 = p m ) with p > p m (p ; p 2 ) where p > a b ; p 2 > a b ; We have to consider two cases. (i) c 2 < p m = a+bc The pure-strategy Nash equilibria are (p = p; p 2 = p) where p 2 [c ; c 2 ] (ii) c 2 > p m The pure-strategy Nash equilibria are (p = p; p 2 = p) where p 2 [c ; p m ] (p = p m ; p 2 ) with p 2 > p m 5

6 Finally, note that the result in this question (as in parts a and b) depends on the assumption that one of the rms capture the entire market when they charge the same price. Clearly, this is not an entire realistic feature of the model, and a good exercise would be to investigate how the results change when this assumption is changed in some way. 5 Bertrand Oligopoly with "Meeting" Market demand is q = a bp Set of strategies We compute the monopoly price S i = R + fy; Ng max(a bp)(p c) p F OC a p + bc = 0 p M = a + 2 c Suppose S 2 = p 2 Y if p >< 2 > p M ) (p = p M ; fy; Ng) BR (p if p 2; Y ) = 2 = p M ) (p = p M ; fy; Ng); (p > p M ; Y ) if c < p > 2 < p M ) (p 2 [p 2; ); Y ); (p = p 2; N) if p 2 = c ) (p 2 [c; ); fy; Ng) if p >< 2 > p M ) (p = p M ; fy; Ng) BR (p if p 2; N) = 2 = p M ) (p < p M ; fy; Ng) if c < p > 2 < p M ) (p < p 2; fy; Ng) if p 2 = c ) (p 2 [c; ); fy; Ng) Since the game is symmetric, then if p >< > p M ) (p 2 = p M ; fy; Ng) BR 2 (p if p ; Y ) = = p M ) (p 2 = p M ; fy; Ng); (p 2 > p M ; Y ) if c < p > < p M ) (p 2 2 [p 2; ); Y ); (p 2 = p ; N) if p = c ) (p 2 2 [c; ); fy; Ng) 6

7 if p >< > p M ) (p 2 = p M ; fy; Ng) BR 2 (p if p ; N) = = p M ) (p 2 < p M ; fy; Ng) if c < p > < p M ) (p 2 < p ; fy; Ng) if p = c ) (p 2 2 [c; ); fy; Ng) Now, we can nd the set of NE directly by looking for a xed point of these best-responses. Alternatively, we follow the approach used in questions 3-4, i.e. to divide the set of possible strategy pairs into a limited number of di erent "cases," and for each case evaluate whether we can have a NE of that particular form. This is more tedious than in questions 3-4, but in principle not di cult. Either way, we get the following four di erent types of equilibria NE type NE type 2 NE type 3 (p; Y ) ; (p; Y ) such that p 2 c; p M p M ; Y ; p 2 > p M ; Y p > p M ; Y ; (p 2 ; Y ) NE type 4 f(c; fy; Ng) ; (c; fy; Ng)g 6 First-Price Auction Set of players is f; ; ng. Each player i submit a bid b i 2 [0; ) Without loss of generality, we can restrict attention to bids b i 2 [0; v]. To construct the payo matrix, de ne the following bid The payo s are b i = max j6=i b j < v b i, if b i > b u i (b i ; b i i) = k+ (v b i) ; if b i = b i and there are k winners 0; otherwise Suppose that player i is the only winner of the auction, i.e., b i > b i If b i < v, then all other players have incentive to bid either b i or (b i + ") ; so it cannot be a NE. If b i = v; then player i has incentive to lower her bid s.t. it would still be b i > b i But we just argued that such a scenario could not be a NE. Thus, if a NE exists, it has to be that we have more than one winner. Suppose the winning bid is b i = b i, and there are (k + ) winners and (k + ) 2; then the winner s payo is u i (b i ; b i) = k + (v b i) If b i = b i < v; then every player has incentive to bid b i < b < v and thus this could not be NE. 7

8 If b i = b i < v, then every player has incentive to bid b i < b < v. Hence, this could not be a NE. If b i = b i = v; then no player has incentive to deviate. Increasing ones bid s.t. b v would not increase the payo (it would remain at zero). Bidding b > v would yield a negative payo. And if one of the winning players bid less that v, then she does not gain from this deviation. NE b i = b i = v with at least 2 winners.