Answers to Problem Set 4 Economics 703 Spring 016 1. a) The monopolist facing no threat of entry will pick the first cost function. To see this, calculate profits with each one. With the first cost function, the monopolist chooses output, q, to maximize π 1 = (p 3)q = (10 q 3)q = (7 q)q Differentiate with respect to q, set the derivative to zero and solve to get q = 7/. Profits are π 1 = 49/4. With the second cost function, q maximizes π = (p 1)q 10 = (9 q)q 10 Again, differentiate with respect to q and solve to get q = 9/. Profits are π = 81/4 10 = 41/4 < π 1. Thus profits are higher with the first cost function. b) The second firm will only enter the market if he can earn profits of at least 4. First, suppose the monopolist picks the first cost function. Then if the second firm enters, we have a Cournot problem, like the one we analyzed in class earlier. To compute the equilibrium, let qi be the equilibrium output of firm i. Then for each i, qi must solve max(10 q q q j )q c i (q) is firm i s cost function. The first order conditions for firm i are or 10 q j 3 q = 0 q 1 = 7 1 q 1
and q = 7 1 q 1. Solving these two equations together, you get q 1 = q = 7/3. The price will be 10 7/3 7/3 = 16/3. Profits for each firm will be (16/3 3)(7/3) = 49/9. Notice that 49/9 > 4 so that the second firm will enter. Thus if the monopolist picks the first cost function, the second firm will enter and the first firm will earn profits of 49/9. Suppose then that the monopolist chooses the second cost function. Will the second firm enter? Suppose he does. The first firm chooses output, q 1, to maximize (p 1)q 1 10 = (10 q 1 q 1)q 1 10 where q is firm s output. Differentiate with respect to q 1 and set the derivative to zero to get q 1 = 9 q Firm will choose his output to maximize (p 3)q = (7 q 1 q )q Differentiate with respect to q and set the derivative to zero to get q = 7 q 1 Solve these two equations for q 1 and q. Substituting the first equation into the second, or q = 7 (9 q )/ 3q = 5. = 14 9 + q 4 So q = 5/3. Substituting this into the equation above to get q 1, we see that The price would be q 1 = 9 5/3 = 7 5 6 = /6 = 11/3 Firm s profits would be 10 q 1 q = 10 5/3 11/3 = 10 16/3 = 14/3 (14/3 3)q = (5/3)(5/3) = 5/9 which is less than the entry cost of 4. Thus firm would not enter. Therefore, if firm 1 picks the second cost function, there will be no entry and the profits firm 1 earns would
be the same as the profits earned with this cost function as we calculated in (a) or 41/4. Thus we see that the firm 1 earns 49/9 if he chooses the first cost function and 41/4 if he chooses the second one. Since 41/4 > 49/9, he chooses the second one.. a) Since the game has perfect information, backward induction corresponds to sequential rationality at every information set. Hence the unique weak PBE and the unique sequential equilibrium is (ai, c, f). (Note that there is no issue of beliefs because of perfect information.) b) For weak perfect Bayesian equilibria, let s work backwards. Sequential rationality for 1 at his information set in the left hand subgame doesn t pin down his action: depending on his beliefs, he could play either g or h here. However, for any beliefs, the only sequentially rational strategy at his information in the subgame on the right is i. So first suppose 1 plays g in the left hand subgame and i in the right. Then for sequential rationality, must play c in the game on the left and e in the game on the right. Given this, 1 must play b at his initial information set. Since 1 s information set on the left is not reached, we can assign any beliefs we like there, so, for example, we can assign 1 at the node on the left and 0 at the node on the right in that information set. Since 1 s information set on the right is reached, we have to assign 1 to the node on the left there and 0 to the node on the right. With these beliefs, we have sequential rationality for all players. Hence (bgi, ce) is a weak perfect Bayesian equilibrium. Next, suppose 1 plays h in the left hand subgame (and i, as he must, in the right hand subgame). For sequential rationality, must play d and e. Given this, at his initial information set, 1 must play b. Again, we can assign beliefs at the unreached information set to make 1 s strategy sequentially rational. Hence (bhi, de) is a weak perfect Bayesian equilibrium. Which, if either, of these weak PBE s is sequential? In both cases, 1 s information set in the subgame on the right is reached and so weak consistency and consistency say the same thing about the beliefs here. The only question is whether the beliefs at the information set in the left subgame are consistent. In both cases, this is straightforward: any totally mixed strategy for will give beliefs at 1 s information set converging to probability 1 on 1 correctly inferring s action. In both cases, 1 s action is a best response to, so we satisfy sequential rationality and consistency. Hence both weak PBE s are sequential. c) For weak perfect Bayesian equilibria, let s work backwards again. Sequential rationality for 1 does not pin down his strategy at either of those final information sets. So we have four cases to check. First, suppose 1 s strategy is gi at those information sets. Then sequential rationality for implies that his strategy must be ce. Given this, sequential rationality for 1 at his initial information set implies that he plays a. Weak consistency then requires 1 s beliefs at his left hand information set to put probability 1 on the node on the left, a belief which makes g sequentially rational. We can pick any 3
belief for the information set in the right hand subgame and so can pick one making i sequentially rational. Hence (agi, ce) is a weak perfect Bayesian equilibrium. Now suppose 1 plays gj. Now sequential rationality implies that s strategy must be cf and 1 s initial move must be b. Again, the beliefs implied by weak consistency are the ones making 1 s strategy sequentially rational at the information set with positive probability. Again, we can assign anything at the other information set. Hence (bgj, cf) is a weak perfect Bayesian equilibrium. The reasoning for the other two cases is similar: both (bhi, de) and (bhj, df) are weak perfect Bayesian equilibria. The reasoning for sequential for all three cases is similar to part (b) all three equilibria are sequential. d) To compute weak perfect Bayesian equilibria, let s work backward starting with s second information set. Suppose plays h here. Sequential rationality for 1 at his immediately preceding information set would require that he play f. Once we have this, we see that no matter what beliefs has at his first information set, it is optimal for him to play e. Given this, sequential rationality implies that 1 must play c at his first information set. Since this equilibrium reaches every information set, it must be weak perfect Bayesian. So (cf, eh) is weak perfect Bayesian. Recall from the answers to the last problem set that (af, dh) is subgame perfect; we see here that it is not weak perfect Bayesian. So now suppose plays i at that last information set. Again, sequential rationality requires 1 to play f. Using this, s strategy at his preceding information set can be either d or e depending on his beliefs. So we have two subcases here. First, let s suppose plays d here. Then a is the best thing for 1. Since this implies that neither of s information sets are reached, this must be a weak perfect Bayesian equilibrium. So (af, di) is weak perfect Bayesian. (Again, comparing to the answers to the last problem set, we see that this weak PBE is not subgame perfect.) Turning to the second subcase, suppose plays i at his last information set, 1 plays f at his last one, and plays e at his other information set. In this case, 1 is indifferent between b and a. So we have two more subcases: where he plays a and where he plays b. If he plays a, we know we get a weak perfect Bayesian equilibrium. Hence (af, ei) is a weak perfect Bayesian equilibrium. (Again, it is not subgame perfect.) So suppose 1 plays b. Then must put probability 1 on the left hand node in his information set, making e his best strategy there. Again, the lower information set is not reached, so weak perfect Bayesian allows any beliefs there. Hence (bf, ei) is also weak perfect Bayesian. (Again, it is not subgame perfect.) Turning to sequential equilibria, we know that every sequential equilibrium is both weak PBE and subgame perfect. Hence if there is a pure sequential equilibrium, it must 4
be (cf, eh). Note that this equilibrium reaches every information set, so it must be sequential. More specifically, s beliefs at his first information set must put probability 1 on the right hand node to satisfy weak consistency and his beliefs at his second information set must put probability 1 on the left hand node. Taking totally mixed strategies that put probability 1/n on each mistake (i.e., each deviation from these strategies), we see that the probability on s right hand node in the first information set is 1 n 1 1 n which does go to 1 as n. Similarly, s belief on the left hand node in the second information set is [ ] 3 1 1 which also goes to 1 as n. 3. a) The normal form is n [ ] = 1 1 1 1 n, n c d c d a,, 4, 3, 1 a,, 0, 0, 0 b 3, 0, 1 3, 1, 0 b 0, 0, 0 1, 0, 1 e f The pure strategy Nash equilibria are (a, d, e), (b, d, f), and (a, c, f). Since the only subgame is the game itself, all are subgame perfect. Note that (a, d, e) and (b, d, f) reach every information set, so both must be weak perfect Bayesian. (It s worth verifying this for yourself directly!) So consider the equilibrium (a, c, f). This equilibrium is a weak perfect Bayesian equilibrium. To see this, note that s beliefs must put probability 1 on the node where 1 played a and 3 s beliefs are not pinned down by the strategies. So let 3 s beliefs put probability 1 on the right most node in his information set the one where 1 played b and played d. Given these beliefs, 3 s strategy is sequentially rational, as is s. Hence this is a weak perfect Bayesian equilibrium. Turning to sequential equilibrium, we know that every sequential equilibrium is a weak perfect Bayesian equilibrium. Hence we only need to check the three weak perfect Bayesian equilibria: (a, d, e), (b, d, f), and (a, c, f). The first two reach every information set, so they must be sequential. To see this, simply note that any totally mixed strategies that converge to these must generate beliefs converging to the unique weakly consistent beliefs. Hence both satisfy consistency, not just weak consistency. However, (a, c, f) is not sequential. To see this, let µ be 3 s probability on that right most node, where 1 played b and having played d. Clearly, the strategy f is sequentially 5
rational iff µ 1/. Is this consistent? Suppose it is. Let p n be 1 s probability on a and q n be s probability on c in our sequence of totally mixed strategies converging to this equilibrium. We must have p n 1 and q n 1. Let µ n be the probability put on the node in question by Bayes Rule with these strategies. Then µ n = (1 p n )(1 q n ) (1 p n )(1 q n ) + q n (1 p n ) + p n (1 q n ) < (1 p n )(1 q n ) (1 p n )(1 q n ) + p n (1 q n ) = 1 p n. Since p n 1, we must have µ n 0. Hence we cannot converge to these beliefs, so the beliefs are not consistent and the equilibrium is not sequential. b) The normal form is c d c d a 1,, 0 1, 1, 0 a 1,, 0 0, 3, b, 1, 3 0, 3, b, 1, 3 1, 1, 0 e f Note that 1 has a strictly dominant strategy of b. Hence he must play this in every Nash equilibrium. There are two Nash equilibria in pure strategies: (b, d, e) and (b, c, f). Because there is only one subgame, all are subgame perfect. Both are weak perfect Bayesian equilibria. To see this, note that s beliefs must put probability 1 on the right hand node in his information set in either equilibrium. For (b, d, e), have 3 s beliefs put probability 1 on the right hand node in his information set. Given these beliefs, sequential rationality requires 3 to play e. Given this strategy by 3 and s beliefs, sequential rationality requires to play d. Hence this is a weak perfect Bayesian equilibrium. For (b, c, f), let 3 s beliefs put probability 1 on the left hand node in his information set. Given these beliefs, f is sequentially rational. Given this and s beliefs, faces a choice between 1 and 1, so c is sequentially rational for him. Hence, again, this is a weak perfect Bayesian equilibrium. Are the weak PBE s sequential? Since (b, d, e) reaches every information set, it must be sequential. Is (b, c, f)? No there are no totally mixed strategies converging to (b, c, f) which generate beliefs converging to the ones we need to make f sequentially rational. To see this, let p n be the probability 1 makes a mistake (plays a) and let q n be the probability makes a mistake in this equilibrium (plays d). Then by Bayes rule, the probability 3 is at the left hand node of his information set given that the information is reached is p n q n = p nq n = p n. p n q n + (1 p n )q n q n But this must go to zero as n. Hence 3 must put probability zero on the left hand node, meaning that f cannot be sequentially rational given any consistent belief. So this equilibrium is not sequential. 6