If V = 4, there are only four times at which trade can happen before the pie vanishes.

Transcription

1 Problem Set 4 1. Imagine an alternating-offer bargain game, but instead of discounting future payoffs by δ and keeping the size of the pie fixed at 1, assume that in each period the value of the pie is reduced by one In the first period, v 1 = V, in the second v 2 = V 1, in the third v 3 = V 2, etc., and there is no discounting. (a). Find the subgame perfect Nash equilibrium f V = 3 and V = 4. If V = 4, there are only four times at which trade can happen befe the pie vanishes. In the fourth offer period, there is once slice left, and player 2 is making an offer to player 1. Player 2 should offer player 1 nothing, and keep the whole pie. Since player 1 is indifferent between accepting and rejecting, he accepts, and the game ends. In the third offer period, there are two slices left, and player 1 is making an offer to player 2. Player 1 should offer player 2 one slice, since we know that player 2 can get a slice by rejecting now and waiting to make an offer next period. So player 1 gets a slice and player 2 gets a slice. Otherwise, player 2 rejects. In the second offer period, there are three slices left, and player 2 is making an offer to player 1. Player 2 should offer player 1 one slice, since we know that player 1 can get a slice by rejecting now and waiting to make an offer next period. So player 2 gets two slices and player 1 gets one slice. Otherwise, player 1 rejects. In the first offer period, there are four slices left, and player 1 is making an offer to player 2. Player 1 should offer player 2 two slices, since we know that player 2 can get two slices by rejecting now and waiting to make an offer next period. So player 1 gets two slices and player 2 gets two slices. Otherwise, player 2 rejects. If V = 3, the wk is the same as the final three offer periods above, but switching the names of player 1 and player 2. (b). Solve f the subgame perfect Nash equilibrium f arbitrary V (you can guess at the solution, then explain why you think it is true). If V is even, t 1 = V 2 and if V is odd, t 1 = V 2 +1 The reason is that as V grows, the players are basically tacking on an extra period to the front of the case studied in (a). If we add a slice to an odd period, the pie becomes even, and the players should split it equally because delays lead to lost slices of pie. If we add a slice to an even period, we get an odd pie, and there is essentially once slice up f grabs. If players split it equally in even periods, a rejections of t 1 = V/2+1 leads to having V/2 pie in the next period f the reject, which is the same as if he had accepted today. (If you really wanted, you can prove this fmally by mathematical induction on V = 3,4,...:

2 Start with V = 3 as the odd base case and V = 4 as the even base case Suppose that it is a SPNE to offer t 1 = V/2 if V is even and t 2 = V/2 +1 if V is odd. Then show it is an SPNE to offer t 1 = V/2 + 1 if V is even and t 1 = V/2 if V is odd. This is easy to do, since we just do one step of the kind of wk we did in part (a). This proves that the induction hypothesis holds. Since the claim holds f V = 3,4, and we can show the claim holds f any V +1 starting from any V, then this is an equilibrium f all V.) 2. There is a firm that produces specialized machines, which require a substantial investment in learning the details of the client s business; think of industrial sewing machines equipment f physics labaties. The firm invests efft e in producing the machine, at cost c 2 e2. The client s profits are increasing in the efft put into the machine, with π(e) = ae. The firm can sell the machine on the market f a payoff be, b < a. If the client has no machine, the client gets a pay-off zero. (a.) Suppose that the timing of the game is that (i) The firm quotes a price p to the client, (ii) The client accepts rejects (after which the players both get payoffs of zero), (iii) The firm decides how much efft to put into the machine. Draw an extensive fm representing this game and solve f a subgame perfect Nash equilibrium. We wk backwards from the end of the game. Since the firm goes last, the price p is already decided, so that when it maximizes p c 2 e2 the solution is e = 0. Then the payoff to the client of accepting is p, while the payoff to rejecting is zero, so the client should reject. Then the firm gets a payoff of zero no matter what it does in the first subgame, so any p 0 is an equilibrium strategy. (All of these moves and counter-moves are the subgame perfect Nash equilibrium of the game) (b.) Suppose that the timing of the game is that (i) The client quotes a price p to the firm, (ii) The firm accepts rejects (after which the players both get payoffs of zero), (iii) The firm decides how much efft to put into the machine. Draw an extensive fm representing this game and solve f a subgame perfect Nash equilibrium. We wk backwards from the end of the game. Since the firm goes last, the price p is already decided, so that when it maximizes p c 2 e2 the solution is e = 0. Then the firm s payoff from accepting is p while its payoff to rejecting is 0 (since it intends to exert no efft anyway), so that it should accept as long as p 0 and reject otherwise. Then the client s payoff is p, so the client should pick p = 0. (Or, you could have the firm accept any p > 0 and reject otherwise, and the client pick a price of zero. This is another subgame perfect Nash equilibrium.) (All of these moves and counter-moves are the subgame perfect Nash equilibrium of the game)

3 (c.) Suppose that the timing of the game is that (i) The firm chooses the level of efft, e, (ii) The client quotes a price p to the firm f the machine, (iii) The firm accepts rejects (in which case the firm sells the machine on the market). Draw an extensive fm representing this game and solve f a subgame perfect Nash equilibrium. We wk backwards from the end of the game. The firm accepts if p ce 2 /e be ce 2 /2, p be, and rejects otherwise. Then the client should offer a price of p = be, since otherwise the firmwill reject, andtheclient wouldget apayoff of zeroinstead ofae p = ae be = (a b)e > 0. Then the firm looks fward, taking the future moves into account, and maximizes its profits, maxbe c e 2 e2 which has solution e = b/c. (All of these moves and counter-moves are the subgame perfect Nash equilibrium of the game) (d.) Suppose that the timing of the game is that (i) The firm chooses the level of efft, e, (ii) The firm quotes a price p to the client f the machine, (iii) The client accepts rejects (in which case the firm is fced to sell the machine on the market). Draw an extensive fm representing this game and solve f a subgame perfect Nash equilibrium. We wk backwards from the end of the game. The client accepts if ae p 0 and rejects otherwise, ae p. Then the firm wants to maximize maxp c p 2 e2 subject to the client accepting, p ae. So the highest price the firm can quote is p = ae. Then the firm chooses efft to maximize maxae c e 2 e2 which has solution e = a/c. (All of these moves and counter-moves are the subgame perfect Nash equilibrium of the game) (e.) Explain your results above. Why do the results change across different bargaining scenarios? Whenever the efft choice is made after the pricing-accept/reject decision, the firm selects zero efft. This is because the firm s compensation doesn t depend on its efft choice, so it simply wants to minimize cost. In c, the efft choice is made befe the price is quoted, but since the client chooses the price, he captures some of the gains from trade, and this reduces the firm s incentives to invest. In d, the firm quotes the price and chooses efft befe the accept/reject decision, and gets all of the games from trade. (f.) Solve the social planner s problem: max(ae p)+(p c e 2 e2 )

4 and thereby find the efficient level of efft. Is it greater less than the subgame perfect Nash level in each of the above contracting scenarios? Explain your results. The solution is e = a/c. This is greater than a and b f sure, and is also greater than c, since a > b. It is the same as the solution in part d. The difference between c and d is that in c, the client gets some of the gains from trade by quoting a price equal to the firm s outside offer; this reduces the firm s incentive to invest, since it is indifferent between accepting the deal and going to find another buyer, and the extra, firm-specific wk is wasted. In d, however, the firm has all the bargaining power and gets all the gains from trade, so that the efficient efft is expended, since the client is the one who is made indifferent between accepting and getting a payoff of zero rejecting and getting a payoff of zero. So the point of the exercise is that timing and control of different features of a contract have imptant implications f whether efficient trade results, and what the payoffs are. You don t want to give people bad incentives, so it is good to think about these things ahead of time and not just hope that reputational legal concerns keep people from making disadvantageous decisions. 3. (a). Consider the infinitely repeated game with discount rate δ, where the strategic fm below is the stage game: B L R U 1,1 2,5 A D 2,0 0,0 Sketch a graph of the payoffs the players can achieve using any set of mixed strategies. Show what utilities could be suppted using the Nash threats folk theem. Show that f sufficiently patient players, it is a subgame perfect Nash equilibrium to play (U,L) in every period. What is the minimum δ that achieves cooperation? The Nash equilibrium of the stage game is (D,R), giving payoffs (0,0). Consider the trigger strategies: If the histy is (U,L),(U,L),...,(U,L), play (U,L) this period. After any other histy, play (D, R) Does either player have a profitable deviation? F the row player, following the proposed strategies gives a payoff of and deviating gives a payoff of So deviating is never profitable. 1+δ1+δ = 1 2+δ0+δ = 2

5 F the column player, following the proposed strategies gives a payoff of 1+δ1+δ = 1 and deviating gives a payoff of 5+δ0+δ = 5 So cooperating is better than deviating if 1 5 δ 4 5 THEREFORE, as long as δ 4/5, the trigger strategies are a Subgame Perfect Nash Equilibrium of the infinitely repeated game. (b). Consider the infinitely repeated game with discount rate δ, where the strategic fm below is the stage game: 2 L C R U 4,4 0,5 5,6 1 M 5,0 1,1 6,0 D 6, 5 0, 6 7, 7 Sketch a graph of the payoffs the players can achieve using any set of mixed strategies. Show what utilities could be suppted using the Nash threats folk theem. Show that f sufficiently patient players, it is a subgame perfect Nash equilibrium to play (U,L) in every period. What is the minimum δ that achieves cooperation? Are there multiple ways to design the punishments f cheating? What is the lowest δ you can achieve?

6 There are three Nash equilibria in the stage game: (M,C), (D,L), and (U,R). Consider the following trigger strategies: If the histy is (U,L),(U,L),...,(U,L), then play (U,L). After any other histy, play (M, C). Cooperating then gives both players a discounted payoff of 4+δ4+δ = 4 Then the most profitable deviation f the row player is to choose D and get a payoff of 6, followed by the punishment, and the most profitable deviation f the column player is to choose R and get a payoff of 6, followed by the punishment. This gives a discounted payoff of deviating to either player of Then cooperating is better than deviating if 6+δ1+δ 2 1+δ = 6+ δ 4 6+ δ Or 4 6()+δ δ 2 5 THEREFORE, if δ 2/5, the above trigger strategies are a subgame perfect Nash equilibrium of the infinitely repeated game. Are there other ways to enfce cooperation? Think about these strategies: If the histy is (U,L),(U,L),...,(U,L), then play (U,L). After any histy in which the row player deviated first, play (U,R) After any histy in which the column player deviated first, play (D,L) After any histy in which both players deviated first at the same time, play (M,C). This punishes the deviat with a 5 payoff fever, and rewards the player who didn t cheat with a 6 fever. The payoff from deviating f either player is Then cooperating is better than deviating if 6 5δ 5δ 2 5δ = 6 5 δ δ 4 6() 5δ

7 δ 2 11 THEREFORE, if δ 2/11, the above trigger strategies are a subgame perfect Nash equilibrium of the infinitely repeated game. This is the lowest you can get, since a payoff of 5 is the wst you can impose on a player, given that the player is trying to maximize his payoff. (c). Suppose the game from part (b) is repeated T < times, and the discount rate δ = 1. Is there an equilibrium of the repeated game in which the agents play (U,L) f the first T 1 periods, and (M,C) in period T? 4. Consider the N-player prisoners dilemma: If all players choose S, they each get a payoff of N If a single player chooses C while all others choose S, that player gets N 2 and the others get 1 If me than one player confesses, all players who confess get zero while any player who remains silent gets 1 i. What is the Nash equilibrium of the N-player prisoners dilemma? ii. Consider the infinitely repeated game with discount fact δ where the N-player prisoners dilemma is the stage game. Show that f sufficiently patient players, it is a subgame perfect Nash equilibrium to play S in every period. What is the minimum δ that achieves cooperation? iii. What happens to the minimum δ as N increases? i. All players confess is the Nash equilibrium of the game. In fact, it is a strictly dominant strategy to confess. Suppose all other players are silent; by confessing, I get a payoff of N 2, and from silent, I get a payoff of N, so confessing is better. Suppose at least one player is

8 confessing; by remaining silent, I get a payoff of 1, while by confessing I get a payoff of zero, so confessing is better. Therefe, no matter what my opponents are doing, I get a higher payoff by confessing, so it is a strictly dominant strategy. ii. Consider the following trigger strategies: After any histy in which all players chose silent in all previous periods, play silent this period. After any other histy, confess. The discounted payoff of cooperating is Deviating and confessing gives a payoff Then cooperating is better than deviating if N +δn +δ 2 N +... = N N 2 +δ0+δ = N 2 N N2 δ 1 1 N THEREFORE, if δ 1 1/N, the trigger strategies are a subgame perfect Nash equilibrium of the infinitely repeated game. iii. As N increases, we get δ(2) = 1 1/2 = 1/1, δ(3) = 2/3, δ(4) = 3/4,..., which tends to 1. Since 1/N 0 as N, when the game is really big, the discount fact will be close to 1, making cooperating me difficult. 5. There s a market with two firms who are contemplating acting collusively. If they cut production so that each is making half the monopoly quantity, they each get π m = 2. If they act as Cournot competits, they both get π c = 1. If one player plays the monopoly quantity, the other player can take advantage of the situation and increase output, getting a larger share of the market f himself at the expense of his partner, who gets 0 if this occurs; call this value π d, f optimal deviation. Here s a strategic fm f the game: FirmB Collude Compete Collude 2,2 0,π d FirmA Compete π d, 0 1,1 i. F what values of π d does this have the fm of a prisoner s dilemma, i.e., each player has a dominant strategy to play competitively? Assume this inequality holds f the rest of the problem. ii. Assume the game is repeatedly infinitely with discount fact δ. Show that f sufficiently patient players, it is a subgame perfect Nash equilibrium to collude in every period. What is the minimum δ that achieves cooperation?

9 iii. How does the minimum δ change as π d increases? Does this make collusion me less likely to succeed? i. If π d > 2, the players have a dominant strategy to compete. ii. The Nash equilibrium of the stage game is (Compete, Compete), giving payoffs (1, 1). Colluding gives payoffs (2, 2). Consider the trigger strategies If the histy is (Collude, Collude), (Collude, Collude),..., (Collude, Collude), play Collude this period. If the histy is anything else, Compete this period. Then the payoff to cooperating is and the payoff to deviating is Then cooperating is better than deviating if Solving f δ gives 2+δ2+δ 2 2+δ = 2 π d +δ +δ 2 +δ = π d + δ 2 π d + δ 2 ()π d +δ δ π d 2 π d 1 THEREFORE, as long as δ π d 2, the above trigger strategies are a subgame perfect Nash π d 1 equilibrium of the infinitely repeated game. iii. The minimum δ is δ = π d 2 π d 1 Differentiating with respect to π d gives so the minimum δ is increasing in π d. δ π d = (π d 1) (π d 2) (π d 1) 2 = 1 (π d 1) 2 > 0 (Optional). We mentioned that if the buyer privately knew his value of the good, the seller would have a me difficult time capturing all the value as happens in the dictat game. This question exples this idea. There is a single buyer and a single seller. The buyer s value is distributed unifmly between zero and 1, so Pr[v b < x] = F(x) = x and Pr[v b > x] = 1 F(x) = 1 x. The buyer knows her value, but the seller does not. The seller has zero value f the good. The buyer s utility is u b = δ k (v t) if they trade at price t at date k, and zero otherwise. The seller s utility is u s = δ k t

10 if they trade at price t at date k, and zero otherwise. There are potentially two trading dates: k = 0,1. (a). If the seller gets to make a single offer to the buyer announces a t which the buyer can accept reject what should it be? What is the probability that trade occurs? Trade occurs only if v t, so the probability trade occurs is Pr[v t] = 1 t. Then the seller s expected profits are maxpr[v t]t = max(1 t)t t t so t = 1/2. Then probability trade occurs is 1/2 as well. (b). If the seller gets to make two offers t 0 and t 1 and both players discount the second trade by δ as described above, what should the offers be? Think of it like this: There is a range of values [ˆv,1] f which the buyer accepts right away, and a range [ṽ,ˆv] where the buyer accepts in the second period, and a range [0,ṽ] f which the buyer rejects both offers (why?). So, there is an agent ˆv f whom ˆv t 0 = δ(ˆv t 1 ), and an agent ṽ f whom δ(ṽ t 1 ) = 0. Solve backwards. and Solving f ˆv and ṽ yields ṽ = t 1 ˆv = t 0 δt 1 You could just write out the seller s discounted profits and maximize: max t 0,t 1 (1 ˆv)t 0 +δ(ˆv ṽ)t 0 ( max 1 t ) ( 0 δt 1 t0 δt 1 t 0 +δ t 1 )t 0 t 0,t 1 But this is actually pretty ugly to solve. If you solve by backwards induction, the seller in the second period is solving ( t0 δt 1 max t 1 )t 1 t 1 where t 0 is a constant that was already determined. Maximizing yields the optimal t 1 : t 1 = t 0/2 I m not kidding, it s really that simple. Then in period one, the seller is maximizing ( max 1 t ) ( 0 δt 0 /2 t0 δt 0 /2 t 0 +δ t 0 Maximizing yields the optimal t 0 : This implies that t 0 = 2 δ t 1 = 2(2 δ) ) t 0 /2 t 0

11 (c). How do your equilibrium t 0 and t 1 depend on δ? As δ 1, what happens to t 0 and t 1? As δ 1, t 0 = 0/2 and t 1 = 0/2. The price goes to zero in both periods ( marginal cost, since marginal cost is zero here). (d). Does the seller make me profits from making two offers as δ 1, a single offer? Coase conjectured that as the time between offers went to zero, a monopolist would be unable to maintain high prices, and would end up with low profits. What do you observe here? That s what we see here. Since the monopolist is competing against his future self, letting δ 1 erodes all of his market power. It becomes like a Bertrand game where we have firm 1 at date 0 and firm 2 at date 0 + (1 δ), and as δ 1, they re essentially competing at the same time. This is a funny result because you d think as δ 1, it would approximate the monopoly outcome with only one period of trade, but instead, you get intra-self/inter-tempal competition that leads to the perfectly competitive outcome. Crazy, right?