14.12 Economic Applications of Game Theory, Midterm 1 (Solutions)

14.12 Game Theory MIDTERM 1 SOLUTIONS Instructions. This is an open book exam; you can use any written material. You may use a calculator. You may not use a computer or any electronic device with wireless communication capacity. You have one hour and 20 minutes. Each question is 25 points, and the breakdown of points within each question is specifed below. Good luck! 1. A normal form Game is depicted below. Player 1 chooses the row (T or B), Player 2 chooses the column (L,M, or R), and Player 3 chooses the matrix (W,X,Y, or Z). (a) (5 points) Write a strategic form game tree for this game, and indicate the payofs on any two terminal nodes of your choice. You don't need to write the payofs at any other terminal nodes. Answer. The strategic form game tree for this game is the following 1 W L M R L M X Y R L B T B T B T B T B T B T 1 1 10 Z M R L M R B T B T B T B T B T B 1 1 -10 (b) (5 points) Find utilities that player 1 can get by playing each of his actions are: UT = 20p + 12(1 - p) = 12 + 8p UB = 16p + 20(1 - p) = 20 - 4p Player 1 will therefore be indiferent between his two actions if and only if UT = UB 12 + 8p = 20 - 4p which implies 2 3 Now, let's assume that player 1 plays T and B with probability a and 1 - a, respectively; then player 2 will get expected utilities given by p= UL = 15a + 20(1 - a) = 20 - 5a UM = 20a + 15(1 - a) = 15 + 5a Therefore, player 2 will be indiferent and willing to randomize if and only if UL = UM 20 - 5a = 15 + 5a which implies 1 2 Thus, if player 3 is forced to play Y , then the only mixed strategy Nash equilibrium a= 2 is given by 1 2 1 p(B) = 2 2 p(L) = 3 1 p(M ) = 3 p(T ) = (c) (10 points) Find all of the rationalizable strategies in the full 3 player game. Show your reasoning. Answer. First of all, remember that, as we saw in problem set 1, in a three player game the set of rationalizable strategies corresponds to the set that survives iter­ ated elimination of strictly dominated strategies only if we allow for the possibility of correlated beliefs. Therefore, if you simply eliminate iteratively all the strictly dominated strategies without specifying that you are assuming correlated beliefs, you lose some points. Alternatevely, you can leave out correlated beliefs and check that all the strategies that survive iterated elimination are indeed best response to some rationalizable strategy. Firt notice that strategy Z is strictly dominated for player 3. Once we eliminate Z, then R for player 2 becomes strictly dominated. There are no other strictly dominated strategies and the algorithm stops here. Thus, if you explicitly allow for correlated beliefs, the solution is (T, B) for player 1, (L, M ) for player 2 and (W, X, Y ) for player 3. However, if you don't mention specifc assumptions about beliefs, then you need to check that each of the remaining strategies is a best response to some rationalizable strategy. In particular, notice that Y is not an optimal response to any rationalizable strategy. To see this, let a and p be the probabilities that T and L are played, respectively. Then player 3 will get utilities Uw = = Ux = = Uy = 10ap + 10a(1 - p) + 10p(1 - a) - 10(1 - a)(1 - p) 20a + 20p - 20pa - 10 -10ap + 10a(1 - p) + 10p(1 - a) + 10(1 - a)(1 - p) 10 - 20ap 2 Now, consider, for example, the mixed strategy p(W ) = p and p(X) = 1 - p. This gives player 3 utility pUw + (1 - p)Ux = 20ap + 20p2 - 20p2 a - 10p + 10 (1 - p) - 20ap(1 - p) = 20ap + 20p2 - 20p2 a - 10p + 10 - 10p - 20ap + 20ap2 = 20p2 - 20p + 10 which is always greater than Uy = 2. Therefore, for each set of beliefs, player 3 can play a mixed strategy that gives him a payof higher than Y . This means that 3 Y is never a best response. Once we eliminate Y , then strategy B for player 1 will be strictly dominated by T . The elimination of B makes strategy M strictly dominated for player 2. Finally, after eliminating M , strategy X will be strictly dominated for player 3. The only rationalizable strategies are therefore T for player 1, L for player 2 and W for player 3. Notice that these strategies form the unique Nash equilibrium found in part b. 2. "Quickies" Part (a) Required. CHOOSE 1 or (b) or (c). (a) (REQUIRED; 15 points) If Bob, Sue and May are rational voters with strict preferences given in the table to the right, with top being better, and all this is common knowledge, what outcome do you expect the binary agenda at left to produce? Answer. By assumption, Bob, Sue and May are rational voters. This means that, when they vote at a particular node, they take into account what will happen in the following nodes. Therefore, we can solve the game by backward induction. Let's start from the penultimate nodes which are circled in the following picture: 4 When asked to vote between Xl and Xo the voters will choose Xo (Sue and May prefer Xo over Xl ). In the same way, Xo will be preferred over X2 and X3 over Xo . Now, consider the node circled in the following graph: where I have eliminated the branches that will not be chosen. If voters are rational then they will understand that, if they choose X2 , the fnal outcome will end up being Xo , while if they opt for X3 , then the fnal outcome will indeed be X3 . In the node circled in the fgure, therefore, the voters will vote for X3 (Bob and May prefer this choice over Xo ). Finally, let's consider the initial node: Rational voters will understand that in the initial node they are not asked to choose between Xl and X2 , but instead between Xo and X3 . Therefore, they will choose X2 and the fnal outcome of the voting agenda will be X3 . (b) (CHOOSE (b) OR (c);10 points) What, if anything, is wrong with the following pattern of choices? (If you don't have a calculator and want to know: 5/6 = 0.83.) • Choice 1: 0.5[$100] + 0.5[$0] = p > q = 0.6[$80] + 0.4[$0]. 5 Choice 2: 1[$80] = r s = (5=6)[$100] + (1=6)[$0]:(10 points) Answer. Notice that lottery p can be rewritten as 0:6[s] + 0:4[$0], while q is equivalent to 0:6[r] + 0:4[$0]. When transformed in this way, it is very easy to see that the two choices violate the independece axiom. The reason is that, if the independence axiom is true, then Choice 1 implies that s r which is the opposite of Choice 2. (c) (If you already answered (b) don’t do this –we won’t grade it!) Consider a Judicial Settlement problem: At each date t = 1; 2; :::n the Plainti¤ makes a settlement o¤er st . The Defendent can either accept or reject each o¤er. (Note that the same player is making o¤ers each period.) If the Defendent accepts at date t, the “game”ends with the Defendent paying st to the Plainti¤, and the Defendent and Plainti¤ paying tcD and tcP to their respective lawyers. If the Defendent rejects at all dates, the case goes to court. The Defendent will lose and have to pay J to the Plainti¤. The Plainti¤ and Defendent will also have to pay lawyer’s fees (n + 1)cP and (n + 1)cD respectively. If it is common knowledge that Plainti¤ and Defendent are sequentially rational, how much will the settlement be, and at what date will it take place? (You don’t have to show the backward induction reasoning explicitly. Just give the answer and 1 or two sentences of intuition.) Answer. Under the assumption of common knowledge of sequential rationality of both players, we can solve the game by backward induction. In the penultimate period, t = n, the Plainti¤ will o¤er a settlement o¤er sn which makes the defendent exactly indi¤erent between accepting the o¤er and rejecting it and 6 paying J + cD in the court. In fact, any ofer smaller than Sn = J + cD will result in a loss for the Plaintf given that the Defendent will accept any ofer smaller or equal to J + cD . In the third to last period, t = n - 1, the Plaintif will again make the Defendent indiferent between accepting and not accepting, that is, he will ofer a settlement of Sn-l = J + 2cD . Proceeding in this way, we fnd that in the initial node of the game, at time t = 1, the Plaintif will make a settlement ofer Sl = J + ncD and the game will end with the Defendent accepting the ofer. 7 3. In this question you are asked to compute the rationalizable strategies in a linear Bertrand­duopoly with discrete prices and fxed "startup" costs. We consider a world where the prices must be an odd multiple of 10 cents, i.e., P = {0.1, 0.3, 0.5, ..., 0.1 + 0.2n, ...} is the set of feasible prices. For each price p, the demand is: Q(p) = max{1 - p, 0}. We have two frms N = {1, 2}, each with 0 marginal cost, but each with a fxed "start­ up" cost k. That is, if the frm produces a positive amount, it must bear the cost k. If it produces 0, it does not have to pay k. Simultaneously, each frm sets a price pi E P . Observing prices pl and p2 , consumers buy from the frm with the lowest price. When prices are equal, they divide the demand equally between the two frms. Each frm i wishes to maximize its proft. pi Q(pi ) - k if pi < pj and Q(pi ) > 0 . 1 i (pl , p2 ) = pi Q(pi )/2 - k pi = pj and Q(pi ) > 0 } { 0 otherwise (a) If k = 0.1 : 1. (5 points) Show that pi = 0.1 is strictly dominated. SOLUTION: Claim: pi = 0.1 is strictly dominated by pi = 0.5. (It is also strictly dominated by every other strategy EXCEPT for pi = 0.9). Let us see this. Suppose pi = 0.1 is played. The payofs will be: 1 i (0.1, pj ) = -0.055 when pj = 0.1 -0.01 when pj > 0.1 The payof, as can be seen, will always be strictly less than zero. Therefore, as long as there is some strategy that ensures a payof of at least 0 in all cases, it will strictly dominate playing 0.01. pi = 0.5 is one of these strategies (but certainly not the only one). If 0.5 is the strictly smaller than the other players strategy, it will yield a payof of 0.15, and if the other player also plays 0.5, the frm will get a payof of 0.025. Finally, if the other player plays strictly less that 0.05, then the frm will get zero. Thus, it strictly dominates 0.1. (3 points credit given for showing that the payof is negative and saying that this is strictly dominated by "not participating". To get full credit, you must have pointed to a strategy (which was a price p) that strictly dominated 0.1 and explained why). 2. (5 points) Show that there are prices greater than the monopoly price (p = 0.5) that are not strictly dominated. SOLUTION: Suppose that the other frm plays 0.1. Then, you do NOT want to "win" the price war, since playing 0.1 will yield a negative payof Playing anything above 0.1 will be a best response as anything strictly above will yield a payof of zero. Thus playing any strategy above the monopoly price will be a best response to the belief that the other frm will be playing 0.1. (They are weakly dominated, but NOT strictly dominated). 8 3. (15 points) Iteratively eliminate all strictly dominated strategies to …nd the set of rationalizable strategies. Explain your reasoning. SOLUTION: First round: Eliminate 0:1 from both players strategies. We know that we can eliminate 0:1 because it is strictly dominated by 0:5 as discussed in part (i). Second round: Eliminate all pi 0:9 from both players strategies. (All are strictly dominated by 0:3) Look at pi = 0:3. Note that: i (0:3; pj ) 0:005 when pj = 0:3 0:11 when pj > 0:3 = So, once 0:1 is eliminated, 0:3 will always produce a payo¤ that is strictly greater than zero. Now consider the payo¤ of playing pi = 0:9 : 8 0 when pj < 0:9 < 0:055 when pj = 0:9 i (0:9; pj ) = : 0:01 when pj > 0:9 Thus, playing 0:9 will never produce a playo¤ greater than zero, so it is strictly dominated by 0:3.. Playing pi > 0:9 will always yield a payo¤ of exactly zero, so will also be strictly dominated by 0:3 which always yields positive payo¤. (once 0:1 has been eliminated).. Third round: Eliminate pi = 0:7 for both players as it is strictly dominated by 0:3. Now we are just left with the strategies 0:3; 0:5; 0:7. Claim: playing 0:3 will strictly dominate playing 0:7. Let us look …rst at playing 0:3: 8 < 0:005 when pj = 0:3 0:11 when pj = 0:5 i (0:3; pj ) = : 0:11 when pj = 0:7 Now look at playing 0:7 : i (0:7; pj ) = 8 < 0 when pj = 0:3 0 when pj = 0:5 : 0:005 when pj = 0:7 Thus, it is clear that playing 0:3 strictly dominates playing 0:7. This leaves 0:3 and 0:5. Fourth round: Eliminate pi = 0:5 for both players as is strictly dominated by 0:3. If the other player plays 0:3; you get 0:005 for playing 0:3 and zero for playing 0:5, and if the other player plays 0:5, you get 0:11 for playing 0:3 and only 0:025 for playing 0:5). Thus 0:3 is the unique rationalizable strategy achieved through the iteration of strictly dominated strategies. 9 (Only partial credit given if the eliminations were made without saying what strategies were actually strictly dominating the strategies that were elimi­ nated. For example, a number of people said that 0.9 could be eliminated in the frst round. This is NOT true, as it is still a best response to 0.1.) 10 4. There are three "dates", t = 1, 2, 3, and two players: Government and Worker. • At t = 1, Worker expends efort to build K E [0, C) units of capital. • At t = 2, Government sets tax rates T K E [0, 1] and T e E [0, 1] on capital­holdings and on labor income. • At t = 3, Worker chooses efort e2 E [0, C) to produce output Ke2 . The payofs of Government and Worker are: Uc = T K K + T e Ke2 and Uw = (1 - T e )Ke2 + (1 - T K )K - K 2 /2 - e22 /2. (a) (20 points) Solve the game by backwards induction. Solution: • At t = 3, worker maximizes Uw taking everything else as given. The frst order conditions give: ek2 = (1 - T e )K. (5 points) • At t = 2, the Government chooses T K and T e to maximize Uc , recognizing that their choice of T e efects ek2 . They thus maximize T K K + T e K(1 - T e )K The frst order condition for T e gives T ke = l2 if K > 0 T ke E [0, 1] if K = 0. (5 points; ­1 for not noting what happens if K = 0). The frst order condition G for T K doesn't have a solution: Uu > 0 if K > 0. Hence, UT K T k = 1 if K > 0 T k E [0, 1] if K = 0. (5 points). • At t = 1, the worker chooses K to maximize Uw , recognizing that his choice will efect the tax rates and his future efort e2 . Plugging in the solutions from t = 2, 3 for K > 0 gives: Uw 1 1 1 = (1 - )K(1 - )K + (1 - 1)K - K 2 /2 - (1 - )K 2 2 2 1 2 1 2 3 2 K - K =- K . = 8 2 8 Hence, the Uw is maximized at K = 0. (5 points.) 11 2 /2 (b) (5 points) Now suppose before the game is played, Government can “delegate” its job to an independent IRS Agent at period t = 0. At t = 0, the Government will o¤er a fraction K 2 [0; 1] of its capital tax revenue and a fraction e 2 [0; 1] of its labor tax revenue to the Agent. The Agent can either Accept or Reject. If the Agent Accepts, she will take the place of the Government in setting tax rates K 2 [0; 1] and e 2 [0; 1] at t = 2: If the Agent Rejects, the game procedes as before. The Agent has payo¤: UA = K KK + " if accept e e Ke2 0 if reject. , where " is a very small but positive “acceptance”cost. The Government’s payo¤ will be: (1 K ) K K + (1 e ) e Ke2 if Agent Accepts UG = : K + Ke K e 2 otherwise Assume that an Agent who accepts will choose the smalest tax rate(s) consistent with sequential rationality. Find an equilbrium of the game using backward induction, and brie‡y comment on it. Solution: As in part (a), at t = 3, e2 = (1 e )K: At t = 2, the same solutions apply if the Agent has Rejected. If the Agent Accepted and k > 0 and e > 0: If the Agent accepted and i = 0 for i = e or k the Agent doesn’t care what i is; by assumption they choose a tax rate of 0. So if the Agent Accepts: e = 12 if K > 0 and e = 0 if K = 0 or = 1 if K > 0 and k 2 0 if K = 0 or >0 =0 e e >0 = 0: k k k At t = 1, K = 0, as in part (a) if the Agent Rejects or if the Agent accepts and both k > 0 and e > 0:. If k > 0 and e = 0, then the worker maximizes (1 0)K(1 0)K + (1 1)K K 2 =2 ((1 > 0 and e 0)K)2 =2 = 0: So any K is equally good. K 2 [0; 1) if If k = 0 and e k =0 > 0, then the worker maximizes 1 1 )K(1 )K + (1 0)K K 2 =2 2 2 3 2 = K K : 8 The …rst order conditions give 4 K = if k = 0 and e > 0: 3 UW = (1 12 (1 1 )K 2 2 =2 • At t = 0, the Agent clearly Rejects if P = 0 and P e = 0. If P = 0 and P e > 0, his payof will be: 4 Pe 4 (1 - 1/2) - c. 2 3 3 So the Agent will Accept if P e > � c, Reject if P e < � c and be willing to do either if P e = � c. If P > 0 and P e = 0, whether or not the Agent will accept depends on what he anticipates K will be (since the worker will be indiferent) Since we're just told to fnd an equilibrium, let's assume K = 0 in thise case. • At t = 0, the Government will get a payof of 0 by ofering P = 0 and P e = 0 (as in (a)) Given the assumption above that the Worker will choose K = 0 if P > 0 and P e = 0, the Government will get a 0 payof by ofering P > 0 and P e = 0 (also as in (a)). If Government ofers P = 0 and P e 2 � c the Agent will Accept, the worker will work, and the Government will get a positive payof, (so long as P e < 1, which is OK to assume, since c is small). It wants to choose the smallest agent share, so it picks P e = � c. 13