Problem Set 2 Answers

Transcription

1 Problem Set 2 Answers BPH8- February, 27. Note that the unique Nash Equilibrium of the simultaneous Bertrand duopoly model with a continuous price space has each rm playing a wealy dominated strategy. Consider an alteration of the model in which prices must be named in some discrete unit of account (e.g. pennies) of size. Assume that both rms have identical constant marginal cost of c. a Explain why the unique Nash Equilibrium of the Bertrand duopoly model with a continuous price space is a wealy dominated strategy. Consider charging some price c+" versus charging the NE price of c. Regardless of the price chosen by Firm 2, Firm will receive a payo of zero when choosing p = c. If p 2 < c, then Firm does not sell anything and receives zero. If p 2 = c, then the rms split the maret but price equals cost. If p 2 > c, then Firm captures the entire maret but charges price equal to cost and earns zero. However, when p = c + ", for " >, then there are some price choices by Firm 2 for which Firm earns a positve pro t. When p 2 < c + ", then Firm earns zero. But, when p 2 c + ", Firm will earn a positive pro t. It will either share the maret at price c + " if p 2 = c + " or it will capture the entire maret if p 2 > c + ". Either way, Firm earns a positive pro t. Writing out two abbreviated rows of the normal form of the game we have: Firm 2 p 2 < c p 2 = c c < p 2 < c + " p 2 = c + " p 2 > c + " Firm p = c = = = = = p = c + " = = = > > This shows that choosing p = c is a wealy dominated strategy. b Show that both rms naming prices equal to the smallest multiple of that is strictly greater than c is a PSNE of this game. Argue that it does not involve either rm playing a wealy dominated strategy. Assume that c is a multiple of. Let c + be the smallest price greater than c that the rm can charge (if c = $5, and = $:, then c + = $5:). Suppose both rms charge p = p 2 = c +, so that both rms earn a positive pro t. oes either rm want to change its price? If either rm raises its price then it will earn zero. If either rm lowers its price then it must lower its price to c (since choosing a price below c results in negative pro t), and will also earn zero. Thus, neither rm would choose to deviate from p = p 2 = c +. In this scenario, it is also true that p = p 2 = c is also a PSNE of this game. Now assume that c is not a multiple of (let c = $5: and = $). In this case, p = p 2 = c is NOT a PSNE because choosing price equal to c is not an available action. Let p = p 2 = floor (c)+, where floor (c) rounds c down to the closest multiple of. In this case, choosing p = p 2 = floor (c) + is a PSNE to the game, and both rms earn a positive pro t in this equilibrium. To see this, if one rm lowers price then that rm will capture the entire maret but will have p i < c and earn a negative pro t. If one rm raises price then that rm sells nothing and earns zero pro t. So neither rm wishes to deviate.

2 For completeness, note that it is possible to have a PSNE where both rms charge the same price and that price is greater than floor (c) + (perhaps floor (c) + 2 or floor (c) + 3). Consider the case where c = $5:, Q = 2 (min (p ; p 2 )) is the maret demand function, and each rm s demand follows the demand function from class ( rm i sells if p i > p j, rm i sells 2 Q if p i = p j, and rm i sells Q if p i < p j ). Let = $2, so that prices must be even integers. Each rm charging p = p 2 = $6 is a PSNE note that neither rm wishes to deviate from that strategy given what the other rm is doing. Now, consider p = p 2 = $8. This leads to a pro t of ($8 $5:) 48 = $43:52 for each rm. Neither rm would wish to increase its price because then its pro t would drop to. Would either rm be willing to cut its price to $6, the lowest possible price that yields nonnegative pro t? If one rm dropped its price to $6 it would capture the entire maret and sell 97 units at a pro t of 99 cents apiece, leading to a pro t of $96:3 for the rm. Since that is less than $43:52, neither rm would wish to deviate from p = p 2 = $8. This result hinges upon the slope of the demand function and the size of, as with the same structure in the example and = $ the strategy pro le p = p 2 = $8 is NOT a PSNE (if one rm charges $7 then this rm maes $:99 on 96.5 units and has a pro t of $92:35). Note that choosing a price of c + (or floor (c) + ) is not a wealy dominated strategy. Choosing a price of c + yields a strictly greater payo than ANY other price choice when the other rm also chooses a price of c +. The ey di erence between choosing c + when the price space is discrete and c when the price space is continuous is that choosing c when the price space is continuous does NOT yield a strictly higher payo than other potential actions when the other rm chooses c, it merely yields a payo that is no worse than other payo s. c Argue that as!, this equilibrium converges to both rms charging prices equal to c. As! the price space moves from the discrete case to the continuous case. We have already seen how the PSNE when the price space is continuous is for both rms to charge a price of c. We also now that both rms charging c + (or floor (c) + ) is a NE of the discrete price game. Taing lim! c + we get c. 2. Consider the three-player game below, which starts with P: 2

3 P A P2 A P3 L R L R a Show that the outcome (A; A) that results in payo s (; ; ) is NOT the outcome from a Nash equilbrium. Assume that (A; A) is the outcome from a NE. We need to specify P3 s strategy in order to completely determine the NE. P3 can choose either L or R. If P3 chooses L, then P would choose. If P3 chooses R, then P2 would choose. Regardless of what P3 chooses, either P or P2 would want to switch strategies. b Find all pure strategy Nash equilibrium to this game. Suppose P3 choose L. Then P would choose (to receive 3) while P2 could choose either A or. Can any player switch strategies to receive a strictly higher payo? No. P is receiving his highest payo, P2 receives regardless of whether A or is chosen, and P3 receives regardless of whether L or R is chosen. Thus, ; A; L and ; ; L are PSNE. Alternatively if P3 chooses R we have that ; ; R and A; ; R are PSNE. Alternatively, consider what a 3-player strategic form of the game would be. It would be (in this case because all players have 2 strategies) a cube. It is di cult to visualize a cube, so we brea it apart into two matrices, and let one player choose which matrix to play. P2 P2 A A P A,, 3,, P A,,, 3, 3,, 3,,, 3,, 3, - % L R P3 3

4 3. Consider 2 rms who play a simultaneous Cournot game. Maret demand is given by a bq bq 2, with a > and b >. Firm has constant marginal cost of c and Firm 2 has constant marginal cost of c 2 with c < c 2 and no xed costs for either rm. The pro t to rm i is: i (q i ; q j ) = (a bq i bq j ) q i c i q i These rms, however, are not solely concerned with pro t, but are also concerned with inequity in production. Thus, instead of maximizing pro t they maximize utility by choosing quantity, where utility is given by: U i (q i ; q j ) = i (q i ; q j ) (q i q j ) 2 where > is a constant which measures how much the rm dislies inequity in production. a Find the best response functions for this simultaneous Cournot game. Player wants to maximize: Player 2 maximizes a similar equation: U (q ; q 2 ) = (a bq bq 2 ) q c q (q q 2 = a 2bq bq 2 c 2 (q q 2 ) = a 2bq bq 2 c 2 (q q 2 ) = a 2bq bq 2 c 2q + 2q 2 2bq + 2q = a bq 2 c + 2q 2 q = a bq 2 c + 2q 2 2b + 2 U 2 (q ; q 2 ) = (a bq 2 bq ) q 2 c 2 q 2 (q 2 q ) 2 = a 2bq 2 bq c 2 2 (q 2 q ) = a 2bq 2 bq c 2 2 (q 2 q ) = a 2bq 2 bq c 2 2q 2 + 2q 2bq 2 + 2q 2 = a bq c 2 + 2q q 2 = a bq c 2 + 2q 2b + 2 Technically, the best response functions are: q = Max ; a bq 2 c + 2q 2 2b + 2 q 2 = Max ; a bq c 2 + 2q 2b + 2 b Find the pure strategy Nash equilibrium for these rms which dislie inequity. 4

5 To nd this simply substitute in for either q or q 2 : (2b + 2) q = a c + 2q 2 bq 2 (2b + 2) q = a c + (2 b) q 2 a bq c 2 + 2q (2b + 2) q = a c + (2 b) 2b + 2 (2b + 2) 2 q = 2ba + 2a 2bc 2c + (2 b) (a bq c 2 + 2q ) checaf terthis (2b + 2) 2 q = 2ba + 2a 2bc 2c + 2a 2bq 2c q ba + b 2 q (2b + 2) 2 q + 4bq 4 2 q b 2 q = 2ba + 2a 2bc 2c + 2a 2c 2 ba + bc 2 (2b + 2) 2 q + 4bq 4 2 q b 2 q = ba + 4a 2bc 2c 2c 2 + bc 2 4b 2 q + 8bq q + 4bq 4 2 q b 2 q = ba + 4a 2bc 2c 2c 2 + bc 2 3b 2 q + 2bq = ba + 4a 2bc 2c 2c 2 + bc 2 q = ba + 4a 2bc 2c 2c 2 + bc 2 3b 2 + 2b q = ba 2bc + bc 2 3b 2 + 2b + 4a 2c 2c 2 3b 2 + 2b q = a 2c + c 2 3b a 2c 2c 2 3b 2 + 4b Now you do not need the result in this form, it just maes it easy to see that if = the NE quantity matches that in part c below. For Firm 2 we will have a similar quantity, only replacing the c s and c 2 s in Firm s NE quantity with c 2 s and c s. So: q 2 = a 2c 2 + c 3b a 2c 2 2c 3b 2 + 2b c In the standard asymmetric cost Cournot model without inequity the PSNE is q i = a 2ci+cj 3b for i = ; 2. Let q be Firm s equilibrium production in the standard model and eq be Firm s equilibrium production in the model with inequity. Show that if c < c 2 then q > eq. Now we need to show that: a 2c + c 2 3b > ba + 4a 2bc 2c 2c 2 + bc 2 3b 2 + 2b (a 2c + c 2 ) 3b 2 + 2b > (ba + 4a 2bc 2c 2c 2 + bc 2 ) 3b 3b 2 a 6b 2 c + 3b 2 c 2 + 2ba 24bc + 2bc 2 > 3b 2 a + 2ba 6b 2 c 6bc 6bc 2 + 3b 2 c 2 24bc + 2bc 2 > 6bc 6bc 2 8bc 2 > 8bc c 2 > c Technically I just showed that if q > eq then c 2 > c, but we can simply reverse the steps to show that if c < c 2 then q > eq. 4. Consider a capacity-constrained duopoly pricing game. Firm j s capacity is q j for j = ; 2, and each rm has the same constant cost per unit of output of c up to this capacity limit. Assume that the maret demand function x (p) is continuous and strictly decreasing at all p such that x (p) > and that there exists a price ep such that x (ep) = q + q 2. Suppose also that x (p) is concave. Let p () = x () denote the inverse demand function. Given a pair of prices charged, sales are determined as follows: consumers try to buy at the low-priced rm rst. If demand exceeds this rm s capacity, consumers are served in order of their valuations, 5

6 starting with high-valuation consumers. If prices are the same, demand is split evenly unless one rm s demand exceeds its capacity, in which case the extra demand spills over to the other rm. Formally, the rms sales are given by the functions x (p ; p 2 ) and x 2 (p ; p 2 ) satisfying: x If p j > p i : i (p ; p 2 ) = Min fq i ; x (p i )g x j (p ; p 2 ) = Min fq j ; Max fx (p j ) q i ; gg Ifp 2 = p = p : x (p) x i (p ; p 2 ) = Min q i ; Max 2 ; x (p) q j a Suppose that q < b c (q 2 ) and q 2 < b c (q ), where b c () is the best-response function for a rm with constant marginal costs of c. Show that p = p 2 = p (q + q 2 ) is a Nash Equilibrium of this game. Consider that Firm lowers its price, so that it chooses p < p = p (q + q 2 ). If Firm does this, then Firm still sells q units (since it was selling its capacity at p ), but now sells all the units it can for a lower price. This is not an optimal change by Firm. Now consider that Firm raises its price, so that it chooses p > p = p (q + q 2 ). We now that Firm 2 will produce q 2 units because it produces q 2 units when both Firm and Firm 2 choose p (q + q 2 ), so it will produce q 2 units when it chooses p 2 = p (q + q 2 ) and Firm chooses p > p (q + q 2 ). Firm s best response to Firm 2 producing q 2 is given by b c (q 2 ), but by assumption this is more than Firm can produce. The question then becomes what amount should Firm produce if it cannot produce b c (q 2 ), and this amount is q. How does Firm produce this amount? By choosing p = p (q + q 2 ). This leads us right bac to where we started. A similar pair of arguments can be made for Firm 2 to show that p = p 2 = p (q + q 2 ) is a NE. b Argue that if either q > b c (q 2 ) or q 2 > b c (q ), then no PSNE exists. This is only true if p (q + q 2 ) > c. If p (q + q 2 ) > c then either rm can guarantee itself a positive pro t if it charges a low price (if p (q + q 2 ) = 6 and c = 9, either rm (or both) can charge a price of and guarantee a positive pro t for itself). Thus, any NE would have both rms maing some positive sales since they can guarantee a payo better than by charging a price close to c. Suppose p i < p j. If Firm j is maing positive sales, Firm i must be selling at capacity. Firm i can then raise its price slightly and still sell at capacity. This will increase Firm i s pro t, meaning that the two rms charging di erent prices cannot be a NE. This leaves p i = p j for some level of p as the only potential NE of the game. are possible: There are 3 cases that a If p = p 2 > p (q + q 2 ), then at least one rm sells below capacity and this rm would be better o by slightly lowering its price and either selling at full capacity or stealing all the customers from the other rm. b If p = p 2 < p (q + q 2 ), then both rms sell at full capacity. Each rm could gain by increasing its price to p (q + q 2 ) which would still enable it to sell at full capacity. c If p = p 2 = p (q + q 2 ). This leads us bac to the NE of the rst part of this question. However, now we have either q > b c (q 2 ) or q 2 > b c (q ) by assumption Thus, the best response for Firm to Firm 2 producing q 2 is to produce less than full capacity, which means that it needs to raise its price. We have already seen that both rms charging di erent prices is not a NE, and that both rms charging a price above p (q + q 2 ) is not a NE. 5. Consider the following simultaneous game: 6

7 Player 2 w x y z w 4; 4 5; 3 6; 2 ; Player x 3; 5 6; 6 7; 2; 7 y 2; 6 ; 7 8; 8 2; z ; 7; 2 ; 2 3; 3 a Find all pure strategy Nash equilibria to the stage game. There are two PSNE to this game: () both choose w and (2) both choose z. For parts b-d, assume the game above is in nitely repeated. b Find a strategy pro le that results in an outcome path in which both players choose x in every period and the strategy pro le you have found is a subgame perfect Nash equilibrium (SPNE). Player chooses x in period (or ) and continues to choose x unless a deviation he observes a deviation by player 2. e ne a deviation as any choice by player 2 other than x. If a deviation occurs player then chooses z for the remainder of the game (note that there are two PSNE so two potential punishment strategies, choosing w or choosing z). Player 2 uses the same strategy. If this is the set of strategies, they will be a SPNE if: X X 6 i + 3 i i= i= Note that there are other strategy pro les that would also wor I m sure you all have thought of some that I did not. c Find a strategy pro le that results in an outcome path in which both players choose x in every odd period and y in every even period and the strategy pro le you have found is a subgame perfect Nash equilibrium. Players and 2 choose x in every odd period and y in every even period unless a deviation occurs. A deviation occurs when anything but x is chosen in an odd period and anything but y is chosen in an even period. If a deviation occurs then the punishment will be to play z every period for the remainder 7

8 of the game. This is an SPNE if: X 8 2 i X i i= i= + X 3 i i= ( ) ( + ) You can solve this to show that if :453 then the players will prefer not to deviate. Also, we can chec to mae sure that no deviation occurs in the periods when x is chosen. Then we would essentially have: X 6 2 i X i i= i= + X 3 i i= ( ) ( + ) You can solve this to show that if :479 then the players will prefer not to deviate. Thus, the players would need a :479 in order for this set of strategies to be an SPNE. For :453 :479 the players would cooperate for the rst period (assuming that the rst period is and is even) but then would deviate from the suggested strategies in the second period. Again, there are other SPNE that can be used to support this as an outcome. d Assume that = :4, where is the discount factor. Find a strategy pro le that results in an outcome path in which both players choose y in every period and the strategy pro le you have found is a subgame perfect Nash equilibrium. Similar to part b, but now both players choose y instead of x unless they observe a deviation. A deviation is any choice of strategy other than y by the other player. If a deviation occurs, then the 8

9 other player will punish by choosing z forever. This is an SPNE if: X X 8 i + 3 i i= i= As long as :375 this will be an SPNE to the game. Since = :4 it is an SPNE. What if the players had decided to use w (the other PSNE to the game) as their punishment strategy? Then we would have: X X 8 i + 4 i i= i= This would NOT be an SPNE to the game because = :4 and we would need :42857 in order for it to be an SPNE. So the choice of punishment strategy maes a di erence in this particular setting with = :4. 6. Consider a developer who wishes to purchase parcels of land. If the developer purchases all parcels, the developer receives a payment of. If the developer does not purchase all parcels, the developer receives a payment of. The developer must purchase each parcel of land from the landowner who owns the land. Consider landowners who each own a parcel of land. That parcel has value of v i to the landowner, where v i U[; ]. The individual landowners now their own value for the land but the developer does not. Also, the landowners do NOT now the values of other landowners. The game can be modeled as a sequential game. The developer maes an o er w i to each landowner. Each landowner only observes his own w i and must mae a decision to accept or reject that w i. If all landowners accept their own o er w i, then the landowners each receive w i as a payment from the developer; the developer pays an amount P i= w i, and the developer receives a payment of. If ANY landowner chooses to reject w i, then the developer maes no payment to any landowner and acquires no parcels of land the developer receives but pays. The landowners, who still own their land, receive v i. For simplicity, assume the seller sets w i = w j for all i; j. The developer maximizes expected utility, and receives w if aggregation is successful (which occurs only if all landowners accept the o er) and if not. Note that Pr (ev > v) for the uniform distribution U ; is ev. Assume the developer is ris neutral. Find a subgame perfect Nash equilibrium to this game with landowners. Be sure to set up the developer s expected utility function correctly. Each of the landowners will accept any w v i and reject any o ers w < v i. The developer, nowing 9

10 this, then maximizes expected utility by choosing w: max w U d = ( w)! = w! ( ) + ( w) w! = w! ( ) + ( w) w! = ( w) = ( w) w = w + w w = + The ey to solving the developer s problem is to correctly specify the probability that the o er is accepted. For one individual that probability is w, while for all individuals that probability is. So the SPNE is for all landowners to accept any w vi and to reject any w < v i, while the w developer o ers each landowner w = There are three rational pirates, A, B, and C, who nd gold coins and must decide how to distribute them. The pirates have a strict order of seniority: A is superior to B and C, and B is superior to C. The pirate world s rules of distribution are as follows:. The most senior pirate proposes a distribution of coins among the pirates. 2. The pirates, including the proposer, then vote on whether to accept this distribution. 3. If the proposed allocation is approved by a majority or a tie vote, then the proposal is implemented. 4. If the proposed allocation is NOT approved, the proposer is thrown overboard from the pirate ship and left at sea, and the next most senior pirate maes a new proposal to begin the system again. Pirates base their acceptance of the proposal on three factors. First, each pirate wants to remain on the ship (not be thrown overboard). Second, each pirate wants to maximize the number of gold coins he receives. Third, each pirate would prefer to throw another overboard, if all other results would otherwise be equal (so that if Pirate C is o ered coins he prefers to vote against the proposal as he gets to throw the proposer overboard if the proposal is not approved, and C is guaranteed to get at least coins from any other proposal made). w a (5 points) Suppose that Pirate C is the only remaining pirate. the coins? Hint: This is easy, do not overthin it. How does he propose to distribute Pirate C proposes that he eeps all coins because there are no other pirates with whom to share. b (5 points) Now suppose that Pirate A has been thrown overboard and that Pirates B and C remain. How does Pirate B propose to distribute the coins? Pirate B proposes that he eeps all coins because he (B) will vote to accept the distribution while C will vote to reject. But Pirate B has the tiebreaer so the distribution is accepted.

11 c (5 points) Now suppose all three pirates are still on the ship. equilibrium to this game. Find a subgame perfect Nash This is a little more involved. We can use parts a and b here as we now what Pirates B and C will do if they get to mae a proposal. Now, loo at the outcome if A is tossed overboard: Pirate B will receive all coins and Pirate C will receive. If A o ers B any amount of coins (even ) will Pirate B accept the distribution no, because if B votes against the distribution he nows he will get all the coins PLUS he gets the satisfaction of throwing A overboard. So Pirate A will never o er Pirate B any coins. What about Pirate C? If A is thrown overboard C will get zero coins Pirate A nows that if he o ers C coins C will vote against him and he (A) will be thrown overboard. But what if A o ers C a single coin? Pirate C will accept because he nows that he will get coins if A is thrown overboard actually, Pirate C will accept any amount of coins above, but Pirate A will only o er coin to C because A wants to maximize his own share of gold coins. So the SPNE is: Pirate A o ers Pirate B coins and Pirate C coin (eeping 99 for himself). Pirate B rejects ANY o er by Pirate A and Pirate C votes to accept any o er by Pirate A in which C receives at least coin. Pirate A votes for his own distribution because he wants to stay on the ship. If A is thrown overboard then Pirate B will propose coins for himself and for C. Pirate C will vote against B s proposal and Pirate B will vote for his own proposals. If A and B are thrown overboard then Pirate C will propose that he eep all coins and he will, obviously, vote for his own proposal. And then he will nd another crew.