Answer Key: Problem Set 4

Answer Key: Problem Set 4 Econ 409 018 Fall A reminder: An equilibrium is characterized by a set of strategies. As emphasized in the class, a strategy is a complete contingency plan (for every hypothetical situation when the player has to move), rather than what the player does in equilibrium; this distinction can be important in dynamic games. As an example, consider the following game L 1 R l r l r (, ) (0, 3) (1, 1) (3, 0) where the thick red lines are drawn using backward induction as in the class. Consider the following three statements which one is a characterization of the equilibrium? (a) They get (1, 1) (or (1, 1) ). (b) Player 1 plays R, and player plays l (or (R, l) ). (c) 1 plays R. plays r if 1 did L, and plays l if 1 did R (or more compactly (R, (r, l)) if you are confident that this representation is unambiguous). 1 The answer is (c). Statement (a) merely lists the outcome payoffs. Statement (b) describes the outcome actions on the equilibrium path. Statement (c) fully describes the action each player chooses at each occasion when the player is to move (in other words, at each of her decision nodes). In this example, the equilibrium strategy of player 1 is choose R, and the equilibrium strategy of player is choose r if player 1 did L, and choose l if player 1 did R. The equilibrium is the list of these two strategies; hence (c). 1 Or a tongue-twister version: Player 1 plays R. Player plays r if player 1 played L, and plays l if player 1 played R. 1

Question 1 The unique SPNE can immediately be found by backward induction. Start with the last subgame, with 1 moving; she will do as it yields 4, more than she would get by playing A. At the previous subgame, which begins with playing, he picks, as it yields more than letting 1 play and choose at the subsequent node. Similarly, going back, you can verify that at every subgame, is chosen by the player in charge, anticipating the optimal behavior by her opponent who would pick if given the chance. The unique SPNE outcome of the game is thus (1, 0), and SPNE strategies are (,, ) for player 1 and (, ) for player. In other words, player 1 chooses regardless of the history and player chooses regardless of the history. The credible threat of the opponent defecting at the next node makes it impossible for the players to attain a mutually preferable payoff. 1 A A 1 A A 1 A (3, 5) (1, 0) (0, ) (3, 1) (, 4) (4, 3) Figure 1: A graphical representation of the subgame perfect equilibrium Question a. In the unique SPE, the equilibrium strategies are play L in every period regardless of the history for both players. We need to show that these strategies form a NE in every subgame of the game; although this was covered in the lecture, we reiterate it here. Start from the last subgames which correspond to the last period (there are many of these, each of which depends on what happened before e.g., if T =, there are 4 such subgames following the histories (L, L), (H, L), (L, H), (H, H)). In these subgames, the players will now play the stage game one final time. The only NE in these subgames is (L, L) (check it in the stage game matrix). Knowing this, let s move to the penultimate subgames and repeat the reasoning. The players are now going to play the stage game once, and after that in the following period (the last one) they will pick L. Therefore, they are facing the following game: Note the use of the phrase regardless of the history. It ensures that the strategy indeed specifies unambiguously what to do at every decision node a player faces.

Player Player 1 H L L (11 + 1, 0 + 1) (1 + 1, 1 + 1) H (10 + 1, 10 + 1) (0 + 1, 15 + 1) Again the unique NE is (L, L). Repeating the reasoning ( backward induction ) we can reach the first period. b. Yes. First consider the whole game (not its proper subgames). We first show that the proposed strategies form a NE in the whole game. Given the proposed strategy of player 1, the best response of player is to choose L regardless of the history. Similarly, player 1 s proposed strategy is the best response to the proposed strategy of player. Therefore the proposed strategies form a NE in the whole game. Now, to show that the proposed strategies form a SPE, we need to show that the proposed strategies induce a NE in every subgame. Since every subgame looks exactly like the whole game, the strategies L in every period regardless of the history induce a NE in each subgame as well, by the same argument in the last paragraph. Therefore the proposed strategies form a SPE. c. We will show that the following strategies constitute a SPE: H if t = 1 s i = H in period t if (H, H) was played in all period s < t L otherwise for i = 1,. Indeed we will show that these strategies form a NE in every subgame. There are an infinite number of subgames but we can divide all subgames in two categories; (i) all subgames in which (H, H) has been played so far and (ii) all other subgames (i.e., all those that some other action profile appeared before). Let s examine the first category (checking the second category is straightforward and replicates the argument made in part b). We will check whether (s 1, s ) is a NE in these subgames. To do this, we fix player 1, strategy at s 1 and check whether s is a BR for against s 1 (definition of NE). Indeed, if sticks to s he gets u (s 1, s ) = 10 + δ10 + = 10 1 δ. What if he deviates? We might as well look at his best possible deviation. This for sure includes deviating optimally at the first period (of the subgame), since payoffs are discounted (i.e., don t wait some periods and then deviate). The optimal deviation is to pick the low price (here it s the only possible deviation in pure 3

strategies, but in games with, say, a continuum of actions, we must pick the best thing available). After deviates, 1 s strategy s 1 prescribes low prices forever thereafter. The best response to this (remember, optimal deviation!) is to pick low prices as well. Thus, the deviation gives a payoff of u (s 1, s d ) = 15 + δ1 + = 15 + δ 1 δ. Therefore, player will stick to s if u (s 1, s ) u (s 1, s d ) or, 10 1 δ 15 + δ 1 δ, or δ 5 14. Let s do the same thing for player 1 (although player s optimal deviation is more profitable so we are really done already): u 1 (s 1, s ) u 1 (s d 1, s ), or, 10 1 δ 11 + δ 1 δ, or δ 1 10. Thus we need 5/14 to achieve the payoff (10, 10) each period. 3 Note that these strategies yield actions (H, H) in every period if δ 5/14. Although the players will not deviate and the threat of low prices forever will never be realized on the equilibrium path, the fact that the strategy specifies the credible punishment is important as we have seen. d. Follow the reasoning in part c, up to the point when we examine the first category of subgames. Suppose we are considering a subgame starting in period t. If player sticks to s, what is the (expected) continuation payoff? First, he surely earns 10 in the current period t. Next, the stage game in t + 1, if played, yields 10 according to the strategies (s 1, s ). Since the probability of the stage game being played in t + 1 is 1 p, this adds (1 p) 10 to the continuation payoff. Similarly, the expected payoff from the stage game in period 3 If we allow δ to be differ between the players, then δ 1 1/10 and δ 5/14 will be the condition under which they cooperate. 4

t + is (1 p) 10, since the corresponding probability is (1 p). Continuing this, we get the continuation payoff 10 + (1 p)10 + (1 p) 10 + = 10 1 (1 p). Similarly the deviation gives a payoff of 15 + (1 p)1 + (1 p) 1 + = 15 + 1 p 1 (1 p). You might have noticed that we would get these if we replaced every δ in part c by 1 p. This is correct and holds in general; a game with the probability of termination p each period and no discount has exactly the same structure as the corresponding game with discount factor 1 p and zero probability of termination. Indeed, this type of uncertain future is one of the interpretations of the discount factor δ. So the results in part c naturally translates into the current context, when we replace δ by 1 p. By part c, therefore, the strategies sustain cooperation if 1 p 5/14, or p 9/14. Question 3 a. This is a standard Cournot game. Each firm i solves max q i [(50 4(q i + q i ))q i q i ]. The FOC wrt q i is 50 8q i 4q i = 0. Imposing symmetry q i = q i q, we get 50 8q 4q = 0, or q = 4. Therefore q C 1 = q C = 4. b. The joint profit Q = q 1 + q being viewed as a choice variable, it is the monopolist s problem: max[(50 4Q)Q Q]. Q The FOC wrt Q is 50 8Q = 0, or Q = 6. Since the firms produce the same quantity, we have q m 1 = q m = 3. 5

c. Let π C denote the profit that a firm obtains in part a, and π m in part b, so that π C = 64, π m = 7. We will show that the strategies q m if t = 1 s i = q m in period t if (q m, q m ) was played in all periods s < t q C otherwise for i = 1,, constitute a SPE of the game and sustain collusion. As in Question 1, divide the subgames in two categories and let s examine the subgames where monopoly quantities have been observed so far. Fix player i s strategy to s i. What is i s optimal deviation? In the first period of the subgame, player i chooses q m = 3. Therefore the optimal one-period deviation for i in this period is the solution to max[(50 4(3 + q))q q], q which is q d = 4.5. The profit i gets is then π d = 4.5(50 4(4.5 + 3) ) = 81. After this deviation, i s strategy prescribes q C regardless of the history. Therefore player i s BR to that is q C as well. Therefore, (s 1, s ) is a SPE, if 7 64δ 81 + 1 δ 1 δ, or δ 9 17. Since (using the discount rate 5%) δ = 1 1 + 0.05 0.95 9 17, the joint profit-maximization can be achieved in a subgame perfect equilibrium. 6

Question 4 a. We use the same reasoning as above. First solve the monopoly problem max(50 Q/)Q Q to get the monopoly quantity Q = 50. Assuming as above that the firms evenly split the quantity, the quantities that maximize the joint profit are therefore q m 1 = q m = 5. Accordingly, each firm s profit is π m = (50 5) 5 = 65. Now solve the Cournot problem to find the threat payoffs. Each firm i solves ( max 50 q ) i + q i q i, q i whose FOC wrt q i is 50 q i q i / = 0 for i = 1,. Impose symmetry q 1 = q to get q C i = 100 3 33.33, i = 1,, and the profit of each firm π C = 5000 9 555.56. Now we can construct the strategies as above: q m if t = 1 s i = q m in period t if (q m, q m ) was played in all periods s < t q C otherwise for i = 1,. eriving the condition on δ (for (s 1, s ) to be a SPE) follows the same argument as in Question 3: (i) By definition, (s 1, s ) is a SPE if it induces NE in every subgame. (ii) So divide all subgames in two categories: (a) all subgames in which (q m 1, q m ) has been played so far, and (b) all other subgames. 7

(iii) In the subgames of type (b), consider player i s situation. Player i s strategy prescribes q C from now on regardless of the history. Therefore the best response (of i) to this is to play q C at every subsequent period regardless of the history, which is consistent to s i. (iv) In the subgames of type (a), we examine the optimal deviation and compare its payoff to the nondeviation case. Given player i s specified action q m = 5, the optimal one-period deviation solves ( max 50 q ) i + 5 q i q i so that q d i = 75/, and the firm profit is πd = 703.15. Therefore, to sustain the SPE (s 1, s ), it must be π m 1 δ πd + πc δ 1 δ, 65 5000δ/9 703.15 + 1 δ 1 δ, or δ 0.53. b. Consider the same strategies. Cooperation provides payoff π m + δπ m + = πm 1 δ, while the optimal deviation gives payoff 4 π d + δπ d + δ π C + δ 3 π C + δ 4 π C + δ 5 π C + = (π d + δπ d ) + (δ π C + δ 3 π C ) + (δ 4 π C + δ 5 π C ) + = (1 + δ)π d + δ (1 + δ)π C + δ 4 (1 + δ)π C + = (1 + δ)π d + δ (1 + δ)π C [ 1 + δ + δ 4 + ] = (1 + δ)π d + δ (1 + δ)π C 1 δ. Therefore the strategies form a SPE if π m 1 δ (1 + δ)πd + δ (1 + δ)π C 1 δ. 4 Note that the stage payoff in the second period is also π d, because the quantity is already set in the first period and held fixed for two periods. In the third period when they set the quantities, the opponent s punishment kicks in, yielding π C thereafter. 8

ividing both sides by 1 + δ and using (1 + δ)(1 δ) = 1 δ, we have π m 1 δ πd + δ π C 1 δ, or δ 0.73, which is higher than in part a. (Remember we also need to check that the strategies are a NE in subgames where profiles other than (s 1, s ) have been observed.) Intuitively, now it is harder to sustain collusion because payoff from deviation lasts longer and punishment is delayed. 9