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Econ 302 Assignment 4 — Solution These three games are sequential games of perfect information. We solve them by backward induction: start at the nodes/information sets one level above the terminal nodes, and solve for the best responses; after solving those best responses, move up one more level and solve for the best responses there; and so on. 1. In the subgame (in stage 2) in which firm 1 had produced q1 units, firm 2’s best response is to solve: max(a − b(q1 + q2 ))q2 − cq2 . q2 The first-order condition is: a − bq1 − 2bq2 − c = 0, or a − c − bq1 a−c 1 = − q1 . (1) 2b 2b 2 Equation (1) is the equilibrium strategy of firm 2; it specify firm 2’s best-responding action after every contingency (which is a quantity of firm 1, q1 ). Given that player 2 is playing according to the strategy (1), firm 1 in the beginning of the game (stage 1) solves: q2∗ (q1 ) = a−c 1 max a − b q1 + − q1 q1 − 2cq1 . q1 2b 2 The first order condition is: a − bq1 − a−c − 2c = 0, 2 or a − 3c . (2) 2b Equation (2) is the equilibrium strategy of firm 1, and together with the strategy of firm 2 in (1) they form the subgame-perfect equilibrium of this game. In this equilibrium firm 1 produces q1∗ = a−3c units, and firm 2 produces q2∗ (q1∗ ) = a+c 2b 4b a+5c a+c units. The market price in this equilibrium is P = a − b a−3c + = . 2b 4b 4 q1∗ = 1 2. In the subgame (in stage 3) in which firm 1 had produced q1 units and firm 2 had produced q2 units, firm 3’s best response is to solve: max(a − b(q1 + q2 + q3 ))q3 − cq3 . q3 The first-order condition is: a − b(q1 + q2 ) − 2bq3 − c = 0, or a−c 1 a − c − b(q1 + q2 ) = − (q1 + q2 ). (3) 2b 2b 2 Equation (3) is the equilibrium strategy of firm 3; it specify firm 3’s best-responding action after every contingency (which is a quantity of firm 1, q1 , plus a quantity of firm 2, q2 ). In the subgame (in stage 2) in which firm 1 had produced q1 units, and anticipating the strategy (3) of firm 3 in stage 3, firm 2’s best response is to solve: q3∗ (q1 , q2 ) = a−c 1 − (q1 + q2 ) q2 − cq2 . max a − b q1 + q2 + q2 2b 2 The first order condition is b a−c a − q1 − bq2 − − c = 0, 2 2 or a−c 1 − q1 . (4) 2b 2 Equation (4) is the equilibrium strategy of firm 2; it specify firm 2’s best-responding action after every contingency (which is a quantity of firm 1, q1 ). Finally, in the beginning of the game (stage 1), firm 1’s best response, anticipating the strategies (4) and (3), is to solve: q2∗ (q1 ) = a−c 1 a−c 1 a−c 1 max a − b q1 + − q1 + − q1 + − q1 q1 − cq1 q1 2b 2 2b 2 2b 2 3(a − c) 1 + q1 q1 − cq1 = max a − b q1 4b 4 2 The first order condition is a− 3(a − c) b − q1 − c = 0, 4 2 or a−c . (5) 2b Strategies (5), (4) and (3) form a subgame perfect equilibrium. In this equilibrium firm 1 produces q1∗ = a−c units, firm 2 produces q2∗ (q1∗ ) = a−c , and firm 2b 4b 3 produces q3∗ (q1∗ , q2∗ ) = a−c units. Notice that firm 1 and 2 produce the same quantities as if 8b there are only two firms. The market price in this equilibrium is P = a − 78 (a − c) = 81 a + 78 c. q1∗ = 3. Given the huge negative payoff (−1000000000) of death, each player wants to minimize his probability of death. Conditional on being alive, the player wants to kill as many opponents as possible: a payoff of 2 for two dead, 1 for one dead, 0 for zero dead. Here is one subgame perfect equilibrium: In the (last) subgame where Z (assuming that he is alive) is shooting, Z shoots randomly at one of the alive players. That is, if only one other player is alive, Z shoots him; if both X and Y are alive, Z plays the mixed strategy of shooting X with probability 1/2 and shooting Y with probability 1/2. This is a best-response for Z, because he hates X and Y equally. In the subgame where Y (assuming that he is alive) is shooting, Y shoots Z if Z is alive, shoots X if Z is dead. This is a best response for Y, because if Z is alive he will be shot at by Z. In the subgame where X is shooting (the beginning of the game), X shoots shoots into the air. The probability that X dies when he shoots into air is 0.4 × 0.5 = 0.2: Y misses in his shot of Z (probability 0.4), and Z chooses to shoot X (probability 0.5) in the final round. When X shoots Y, the probability that X dies is 0.3 × 1 + 0.7 × 0.2 = 0.44: if X’s bullet hits Y, X dies for sure (shot by Z, who never misses, in the final round); if X’s bullet misses Y, we are back to the scenario of shooting into the air. Likewise, when X shoots Z the probability that X dies is 0.3 × 0.6 + 0.7 × 0.2 = 0.32. Therefore, shooting into the air is the best response. This completes the description of a subgame perfect equilibrium. The game tree in the next page is produced by Haiyun Kevin Chen (notice that D = −1000000000). 3 Air D 1 D D 2 1 X Z chance D D D 1 2 1 Y Air [.6] Hit Z [.3] Hit X Y Z Miss [.4] Air X Z 1 1 D [.6] Hit chance Miss [.7] X Y Z Miss [.4] Air D 1 0 1 D 0 1 1 0 Y Air Y Air D 1 0 1 D 0 1 1 0 X Z Miss [.4] Y Z Air [.6] Hit chance D D D 1 2 1 1 1 D Air Y Air D 1 0 1 D 0 1 1 0 X Z Miss [.4] X Z 1 1 D [.6] Hit chance Miss [.7] X Y Z Miss [.4] Air D 1 0 1 D 0 1 1 0 Y Air Y Z Air Air D 1 0 1 D 0 1 1 0 X Figure 1: Extensive Form Representation of the Three-Way Duel in Assignment 4 D 1 2 1 D D [.6] Hit chance X Y [.3] Hit chance Z Air D D D 1 2 1 Y Z [.6] Hit chance Y Air D 1 0 1 D 0 1 1 0 X Z Miss [.4] X 1 1 D [.6] Hit chance Z X Y Z Miss [.4] Air D 1 0 1 D 0 1 1 0 Y The black letters {X, Y, Z} denote the players who are shooting and the red letters {X, Y, Z, Air} denote the targets at which a player shoots. The blue letters represent chance and its “moves”. Letter D in the payoff vector indicates a player is dead. D 1 0 1 D 0 1 1 0 chance Y X Air Y Air D 1 0 1 D 0 1 1 0 X Z