IE54500 – Problem Set 4 Dr. David Johnson Fall 2020 1. Pure and Mixed Nash Equilibria Find all of the Nash equilibria in the following 2-player game. In the payoff matrix, (π₯, π¦) indicates a payoff of π₯ for Player A and a payoff of π¦ for Player B. Player B Left Middle Right Top (3,3) (4,8) (5,4) Player A Middle (4,7) (2,6) (2,8) Bottom (6,4) (1,2) (0,1) In this game, the Nash equilibria will be at points where the first value is the maximum among those in the same column (i.e., given Player B’s strategy, it is the maximum payoff among possible actions by Player A), and where the second value is the maximum among those in the same row. For convenience, I will denote the strategies by their payoffs because all of the payoff combinations are unique. The potential pure equilibria in each row satisfying that criterion are therefore (4,8), (2,8), and (6,4). The potential pure equilibria in each column are (6,4), (4,8), and (5,4). Two of these correspond to the same strategies, so the Nash equilibria are (6,4) and (4,8). It is easy to verify that for each of these pairs of strategies, neither player can do better predicated on the other player playing the same action. Checking for mixed strategy equilibria, examine the expected payoff function for Player B, where we denote π1 as the probability Player A plays Top, π2 as the probability Player A plays Middle, π1 as the probability Player B plays Left, and π2 the probability Player B plays Middle: π1 (3π1 + 4π2 + 5(1 − π1 − π2 )) + π2 (4π1 + 2π2 + 2(1 − π1 − π2 )) + (1 − π1 − π2 )(6π1 + π2 ) This has first-order conditions for π1 and π2 , respectively: 3π1 + 4π2 + 5 − 5π1 − 5π2 − 6π1 − π2 = 5 − 8π1 − 2π2 = 0 ⇒ 8π1 + 2π2 = 5 4π1 + 2π2 + 2 − 2π1 − 2π2 − 6π1 − π2 = 2 − 4π1 − π2 = 0 ⇒ 4π1 + π2 = 2 These two equations represent distinct parallel lines, so they do not admit a simultaneous solution for π1 and π2 . This indicates that there is no mixed strategy over all three actions, but we should look for possibilities that mix two of the actions. For example, consider the case where π2 = 0 for Player A. (For the intuition behind exploring this option, observe that Middle is never the best response for Player A to any of Player B’s actions.) In this case, the probability of playing Bottom is 1 − π1 . The expected payoff function for Player A becomes π1 (3π1 + 4π2 + 5(1 − π1 − π2 )) + (1 − π1 )(6π1 + π2 ) which has FOC 3π1 + 4π2 + 5 − 5π1 − 5π2 − 6π1 − π2 = 5 − 8π1 − 2π2 ⇒ 4π1 + π2 = 5/2 Now, we notice that if Player A is playing a strategy where π2 = 0, then Middle dominates Right for Player B. This tells us that Player B would never play Right as part of their best response function; in other words, we know that, given this strategy from Player A, 1 − π1 − π2 = 0. Solving this and the equation above simultaneously, we find that π1 = π2 = 1/2. The payoff function for Player B, on the other hand, becomes π1 (3π1 + 7π2 + 4 − 4π1 − 4π2 ) + (1 − π1 )(8π1 + 2 − 2π1 ) which has FOC 3π1 + 4 − 4π1 − 8π1 − 2 + 2π1 = 2 − 7π1 = 0 ⇒ π1 = 2/7 Therefore, Player A would have a probability of 5/7 of playing Bottom. A mixed-strategy Nash equilibrium is thus (2/7, 0, 5/7) for Player A and (1/2, 1/2, 0) for Player B. Checking other possibilities shows that this is the only mixed-strategy Nash equilibrium of the game. 2. Asymmetric Marginal Costs Consider a Cournot duopoly model where inverse demand is π(π) = π − π but firms have asymmetric marginal costs. Firm 1 has marginal costs π1 and Firm 2 has marginal costs π2 . a) What is the Nash equilibrium if 0 < ππ < π/2 for each firm? b) What is the Nash equilibrium if π1 < π2 < π but 2π2 > π + π1 ? Denote the profit for Firm 1 as π1 and for Firm 2 as π2 . Then for π ≠ π, ππ = ππ (π − ππ − ππ ) − ππ ππ which has FOC π − ππ − ππ πππ = π − ππ − ππ − 2ππ = 0 ⇒ ππ∗ = πππ 2 Subbing in ππ∗ (and switching to subscripts of 1 and 2 for clarity going forward), we can solve π1∗ = π − π1 1 π − π2 − π1 π − 2π1 + π2 + π1∗ π − 2π1 + π2 − ⇒ 2π1∗ = ⇒ 3π1∗ = π − 2π1 + π2 ⇒ π1∗ = 2 2 2 2 3 We also know by symmetry that π2∗ = π−2π2 +π1 . 3 However, all of this does not take the physical system described in the problem into account, which carries implicit constraints that π1 ≥ 0, π2 ≥ 0, and π − π1 − π2 ≥ 0. When 0 < ππ < π/2, this implies that π − π2 − 2π1 > 0, so π1∗ > 0. Similarly, π2∗ > 0. Further, π1 = π1 (π − π1 − π2 − π1 ) = π1 (π − 2π − π1 − π2 π + π2 − 2π1 − π1 ) = π1 ( )>0 3 3 Thus, if 0 < ππ < π/2, all of the implicit constraints clear “as usual” and we end up with both firms producing and receiving positive profits for doing so. In the case where π1 < π2 < π and 2π2 > π + π1 , then we find that π + π1 − 2π2 < 0. However, we found that π2∗ = π − 2π2 + π1 3 so this implies that π2∗ < 0. Effectively, Firm 2’s higher costs push them out of the market because Firm 1 can use their lower costs to undercut them. Therefore, the optimal strategy for Firm 2 is not to produce by setting π2∗ = 0, and this gives Firm 1 a monopoly. At that point, Firm 1 sees π = π − π1 , and π1 = (π − π1 )π1 − π1 π1 . Taking the first-order condition here, we get the usual monopoly result that π1∗ = (π − π1 )/2, and therefore π = (π + π1 )/2. But we are told by assumption that π2 > (π + π1 )/2, so Firm 2’s costs would still be higher than π under the monopoly. Therefore, even at the monopoly pricing, Firm 2 has no incentive to reenter the market and π1∗ = (π − π1 )/2, π2∗ = 0. 3. Subgame Perfect Nash Equilibria In the following dynamic game of complete and perfect information, find all of the Nash equilibria. Which of them are subgame perfect? (Player 1 has action space {πππ, π΅ππ‘π‘ππ}; Player 2 has action space {πΏπππ‘, π ππβπ‘}.) Using backwards induction, we can find a subgame perfect Nash equilibrium as 1: {π΅}, 2: {π → β; π΅ → β}, where π → β means “If player 1 plays π, play β.” We can look for other equilibria that are not subgame perfect by examining other strategies for each player. One is 1: {π΅}, 2: {π → β; π΅ → β}, and another is 1: {π}, 2: {π → β; π΅ → β}. These are not subgame perfect, but it is clear to see that neither player has incentive to deviate. 4. Repeated Bertrand Duopoly Recall the static Bertrand duopoly model: two firms produce homogeneous products and choose their prices simultaneously. Demand for firm π’s product is π − ππ if ππ < ππ , is 0 if ππ > ππ , and is (π − ππ )/2 if ππ = ππ . Marginal costs for both firms are π < π. Now consider an infinitely repeated game based on this model. Show that the firms can use trigger strategies (i.e., a cooperative strategy that switches forever to the single-stage game Nash equilibrium after any deviation) to sustain the monopoly price level in a subgame-perfect Nash equilibrium if and only if each firm has a discount rate πΏ ≤ 1. (Assume that the discount rate is defined such that the present value at time period 0 of profit ππ‘ in time period π‘ is ππ‘ /(1 + πΏ)π‘ .) When cooperating, each firm charges the monopoly price but splits the demand in half. Each firm can initiate a trigger strategy by undercutting the monopoly price ππ by any amount; thus, in the limit they can charge ππ − π for π > 0 arbitrarily small. They would then claim the full demand in that cycle but would be retaliated against by the other firm setting a price of π to ensure they cannot be undercut further. In this model, the payoff in each period is the profit ππ = ππ (ππ − π) = π1π = π−ππ 2 β (ππ − π), so across all periods, (π − ππ ) (π − ππ ) (π − ππ ) 1 (π − ππ ) (ππ − π) + π1 β (ππ − π) + π12 β (ππ − π) + β― = (ππ − π) 2 2 2 1 − π1 2 If the trigger is utilized, then the payoff becomes π1π· = (π − ππ ) β (ππ − π) + π1 β π−π (π − π) = (π − ππ )(ππ − π) 2 where we denote π1 = 1/(1 + πΏ1 ). The strategy of cooperation is sustainable if π1π > π1π· , so 1 π − ππ 1 1 1 (ππ − π) ≥ (π − ππ )(ππ − π) ⇒ ≥ 1 ⇒ π1 ≥ ⇒ πΏ1 ≤ 1 1 − π1 2 1 − π1 2 2 By symmetry, πΏ2 ≤ 1 as well. By generality, this same argument holds no matter what time period the trigger is activated (because up to that point, the profits are identical to cooperation). 5. Static Bayesian Nash Equilibria Find all of the pure-strategy Bayesian Nash equilibria for the following static Bayesian game: i) Nature determines whether the two players’ payoffs are given by Game 1 or Game 2, with each game being equally likely. ii) Player 1 learns whether nature has drawn Game 1 or Game 2, but Player 2 does not. iii) Player 1 chooses either πππ or π΅ππ‘π‘ππ. Player 2 simultaneously chooses either πΏπππ‘ or π ππβπ‘. iv) Payoffs are given by the game drawn by nature: Player 2 Left Right Player 1 Top (1,1) (0,0) Bottom (0,0) (0,0) Game 1 Player 2 Left Right Player 1 Top (0,0) (0,0) Bottom (0,0) (2,2) Game 2 (Note that this is not the only approach for this problem; we could also put the game in its extensive form.) Because we know that each game is equally likely according to Nature, we can write the game in normal form with each cell containing the expected payoffs for each player as follows: Player 2 L T,T 1/2, 1/2 Player 1 T,B 1/2, 1/2 B,T 0, 0 B,B 0, 0 R 0, 0 1, 1 0, 0 1, 1 In this table, (T, B) indicates that Player 1’s strategy would be to play Top if Nature has given Game 1 and Bottom if Game 2, etc. The expected payoffs are the average of each player’s payoff in each game, since each game is known to be equally likely. Now, visual inspection can find the pure-strategy Bayesian Nash equilibria, where each player has no incentive to deviate. These are at [(T,T); L], [(T,B); R] and [(B,B); R]. In all of the equilibria, we must also specify the Players’ beliefs. Player 1 knows what Nature has selected, so they believe with probability 1 that the game is whichever Nature has drawn. Because Player 2 acts simultaneously and there are no further stages in the game, they have no additional information to revise their beliefs; as such, they believe that it is Game 1 with p=0.5 and Game 2 with p=0.5. 6. Pooling Perfect Bayesian Nash Equilibrium Consider the following dynamic signaling game that begins by nature drawing one of three types for Player 1 with equal probability. Player 1 knows their type, but Player 2 does not. Player 1’s action space is {πΏπππ‘, π ππβπ‘}, and Player 2’s action space is {ππ, π·ππ€π}. Specify the conditions under which a pooling perfect Bayesian equilibrium exists where all three Player 1 types play β. (We are not asked this, but we can first observe that L dominates R for type π‘1 , so there can be no pooling equilibrium where all Player 1 types play β.) If Player 1 plays (L, L, L) (in other words, L for each of their three possible types), then Player 2 has gained no new information and therefore believes π(π‘π |β) = 1/3 for all types. If Player 2 plays Up regardless of observing β or β, then Player 1 would have no incentive to deviate since β dominates β for all types if Player 2’s strategy is always to play Up. Player 2 has expected payoff from this strategy of 1 πΌ2 [π’|β] = (1 + 1 + 1) = 1 3 Their expected payoff from a strategy of playing Down is πΌ2 [π’|β] = 0, so playing Up is optimal. (A pure Up strategy is clearly also optimal compared to any mixed strategy, given Player 1’s pooling strategy on β.) However, we require that Player 2’s strategy of playing Up if faced with β be consistent with his beliefs about Player 1’s type if faced with β, so we need for πΌ2 [π’|β] ≥ πΌ2 [π|β]. ∴ 1 β [π(π‘1 |β) + π(π‘2 |β)] ≥ 1 β π(π‘3 |β) and we know 1 − π(π‘3 |β) = π(π‘1 |β) + π(π‘2 |β), so 1 − π(π‘3 |β) ≥ π(π‘3 |β) ⇒ π(π‘3 |β) ≤ 1/2. Therefore, (L,L,L) and (u,u) is an equilibrium with a pooling strategy on L provided that π(π‘3 |β) ≤ 1/2.