Economics 201A - Section 8 (Review)

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1 UC Berkeley Fall 007 Economics 0A - Section 8 (Review) Marina Halac 003 Exam Question : Car Key Gnomes (5 Points) (a) List all of the (pure) strategies for each of the players. Player :. AE. AF 3. BE 4. BF Player :. C. D (b) For each of the players, state which (if any) strategies are equivalent to each other. For Player, BE and BF are equivalent. There are no equivalent strategies for Player. (c) State the backwards-induction strategy for each player. Player s backwards-induction strategy is BF, and Player s is C. (d) Consider the following modification of the game: Decrease Player s payoff on one and only one of the four end nodes of the game, without changing any of Player s other payoffs, and without changing any of Player s payoffs. Choosing which outcome to change and by how much, re-write the game so that Player s backwards-induction payoff in the modified game is higher than Player s backwards-induction payoff in the original game. Rewrite the game changing Player s payoff following F to anything lower than. Now return to the original game, as written above. (e) Write the normal or reduced normal form of the game. (Suggested: reduced normal form.) C D AE,3,8 AF,3 9,0 B 5,0 5,0 (f) Find all the Nash equilibria in the normal or reduced normal form. The unique pure-strategy NE is (B,C). As for the mixed-strategy equilibria, note that if Player plays B, Player is indifferent between C and D. Player may then play C with probability p and D with probability ( p) for any p such that B is still a best response for Player : 5 p + 9( p) p 4/7

2 Note that there are no other mixed-strategy NE. If there were, both players should be mixing. But note that if Player is mixing, Player will never play AE with positive probability (since for any positive probability that Player puts on D, AF strictly dominates AE). And Player will never mix between AF and B, since in that case Player would strictly prefer C to D, and thus Player would not want to mix. Hence, the set of Nash equilibria in the reduced normal form is given by (B; pc, ( p)d) for p [4/7, ] Question : Lewis and Clark Meet Sylvia and Rhonda (5 Points) L C R U,7, 3, M,3 7,8 8,4 D 4,5 5, 9,9 (a) What actions by Player are consistent with Player being rational (making no assumptions about her beliefs about the rationality of Player )? Note that M is a best response to L and D to R, while U is strictly dominated by both M and D. Hence, A () = {M, D} (b) What actions by Player are consistent with Player being rational (making no assumptions about her beliefs about the rationality of Player )? A () = A (0) = {L, C, R} (c) What actions by Player are consistent with Player being rational and believing that Player is rational? Since A () = A (0), A () = A () = {M, D} (d) What actions by Player are consistent with Player being rational and believing that Player is rational? If Player believes that Player is rational and hence will never play U, then L is strictly dominated by both C and R. A () = {C, R} (e) What actions by Player are rationalizable? Note that since A () = A (), we will have that A () = A (3). A () = A (3). Thus, Moreover, note that R = {M, D} (f) What actions by Player are rationalizable?

3 From above, R = {C, R} (g) Find all the Nash equilibria in the game. Of course, we only need to look at the matrix with the rationalizable actions. The set of Nash equilibria is given by { ( 3 (M; C), (D; R), 7 M, 4 7 D; 3 C, )} 3 R Question 3: Lapidary Lapel (5 Points; Parts (a) to (c) worth 0 points in total) A natural interpretation of the game is that Vai-Lam can pay some amount x > 0 to not observe Yuk-Ying s choice of event to go to before he has to make his choice, or can pay y > 0 to observe Yuk-Ying s choice of event. Yuk-Ying knows which choice of payment Vai-Lam has made before she chooses her event. Notice that Yuk-Ying s payoffs are on top. (a) List all the (pure) strategies for each of the players. Yuk-Ying:. BB. BT 3. TB 4. TT Vai-Lam:. xb3b4b5. xb3b4t5 3. xb3t4b5 4. xb3t4t5 5. xt3b4b5. xt3b4t5 7. xt3t4b5 8. xt3t4t5 9. yb3b4b5 0. yb3b4t5. yb3t4b5. yb3t4t5 3. yt3b4b5 4. yt3b4t5 5. yt3t4b5. yt3t4t5 (b) For each of the players, state which (if any) strategies are equivalent to each other. For Yuk-Ying, there are no equivalent strategies. For Vai-Lam, {xb3b4b5, xb3b4t5, xb3t4b5, xb3t4t5 } are equivalent {xt3b4b5, xt3b4t5, xt3t4b5, xt3t4t5 } are equivalent {yb3b4b5, yt3b4b5 } are equivalent {yb3b4t5, yt3b4t5 } are equivalent {yb3t4b5, yt3t4b5 } are equivalent {yb3t4t5, yt3t4t5 } are equivalent 3

4 (c) Besides the game itself, list all other subgames (if any). There are four subgames besides the game itself:. starting at Yuk-Ying s decision node B/T, after x is played. starting at Yuk-Ying s decision node B/T, after y is played 3. starting at Vai-Lam s decision node B4/T4, after B is played 4. starting at Vai-Lam s decision node B5/T5, after T is played (d) Find all the pure-strategy SPNE in the game as a function of x and y. That is, for all possible combinations of x, y > 0, list all the pure-strategy SPNE. Give an intuition for how the set of pure-strategy SPNE depends on x and y, paying special focus on the intuition for the conditions on x and y such that there exists a pure-strategy SPNE where Vai-Lam chooses to pay x and not observe what Yuk-Ying does. Begin at the lowest subgames. Vai-Lam s two subgames at B4/T4 and B5/T5 have unique NE: B4 (because y > y) and T5 (because y > y). Building upwards, at the B/T decision node, Yuk-Ying then chooses B. The left-hand-side subgame beginning at Yuk-Ying s B/T decision node is essentially a Battle of the Sexes with an unknown x added to Vai-Lam s payoffs. There are two purestrategy NE in this subgame (and one mixed-strategy NE, but we are not concerned with that here): (B,B3) and (T,T3). Vai-Lam will then compare ( x) versus ( y), and ( x) versus ( y). Hence, there are four cases:. (BB,xB3B4T5) is a SPNE if ( x) ( y) y x. (BB,yB3B4T5) is a SPNE if ( x) ( y) y x 3. (TB,xT3B4T5) is a SPNE if ( x) ( y) y + x 4. (TB,yT3B4T5) is a SPNE if ( x) ( y) y + x Intuitively, Vai-Lam pays x and does not observe Yuk-Ying s choice when the cost of observing Yuk-Ying s choice is higher (Case ), or when the cost of not observing Yuk-Ying s choice is higher but the additional that Vai-Lam can obtain by not observing Yuk-Ying s choice offsets it. That is, if x is not much more expensive than y, then Vai-Lam chooses to pay x because in the imperfect information subgame Vai-Lam may obtain a higher payoff than in the sequential subgame where he chooses after observing what Yuk-Ying does. Now consider the game Lapidary Lapel, where Yuk-Ying does not observe what payment (x or y) Vai-Lam has chosen. (e) Write out the extensive form of the game. Draw the same game but with a dotted line indicating an informaton set between Yuk- Ying s B/T and B/T decision nodes, and relabel B and T as B and T respectively, because Yuk-Ying now cannot distinguish between the two decision nodes, and hence must be presented with the same choices. (f) Find all the pure-strategy SPNE in the new game as a function of x and y. That is, for all possible combinations of x, y > 0, list all the pure-strategy SPNE. Discuss with subtlety and insight how this set compares to the set of pure-strategy SPNE in the original game. Besides the game itself, there are now only two subgames: those starting at Vai-Lam s B4/T4 and B5/T5 decision nodes. B4 and T5 are still the unique NE in those proper subgames. Hence, to find the SPNE, we can draw the normal form representation taking into account 4

5 the fact that B4 and T5 are always played. Below is the reduced normal form. Hence, there are four cases:. (B,xB3B4T5) is a SPNE if y x. (T,xT3B4T5) is a SPNE if y x 3. (B,yB3B4T5) is a SPNE if x y 4. (B,yT3B4T5) is a SPNE if x y x B3 x T3 y B,-x 0, -x,-y T 0, -x,-x,-y Comparing the pure-strategy SPNE of the two games, we see that the main difference is the absence of the conditions on y +. In the first game, Vai-Lam knew that she would end up meeting Yuk-Ying in any pure-strategy equilibrium. On the right hand side, due to the sequential nature of these subgames, they would always go to her least favorite event, B. On the left hand side, however, the equilibrium in place sometimes called for them going to her favorite event, T. This would give Yuk-Ying one extra payoff, and thus Yuk-Ying would prefer to pay x even if it was up to one unit more expensive than y. That is not the case in the new game, where Yuk-Ying does not observe what choice Vai- Lam made. Yuk-Ying plays the best strategy, B or T (regardless of what payments Vai-Lam makes), and so Vai-Lam best responds by playing the corresponding event and picking the cheapest payment. Hence, Vai-Lam previously stood to gain by choosing a simultaneous over a sequential game. Now, with Yuk-Ying playing the same action in both games, she only cares to minimize the cost of her payments. Question 4: Semi-Sabotage (0 Points) Consider the stage game Work, Laze, Semi-Sabotage: W L W 4,4,5 L 5,, S 0,0 0,0 Consider the repeated game WLSS(T =, δ), where Work, Laze, Semi-Sabotage is played twice, with a discount of δ, 0 < δ, between the first and second period. (a) Either state what the lowest δ such that there exists a Nash equilibrium to the repeated game where both players work in the first period is, or explain why there is no such δ. There is a unique NE in the stage game, (L,L). (Note that for Player, L strictly dominates the other two pure strategies.) Thus, to sustain (W,W) in period, we need to find rewards and punishments, since (W,W) is not a NE of the stage game. The reward must be to play (L,L) in period, since again this is the only NE of the stage game and hence what the players must play on the equilibrium path in the last period. But note that Player does not have the means to punish Player for deviations in period. Player then knows that no matter what he does in period, Player will play L in period. Consequently, Player will always have incentives to deviate from (W,W), and a NE of the repeated game that implements (W,W) in the first period does not exist. 5

6 (b) Either state what the lowest δ such that there exists a SPNE to the repeated game where both players work in the first period is, or explain why there is no such δ. Clearly, if there is no NE where both players work in the first period, there will not be a SPNE where this occurs. (That is, if there is no threat that can make both players work in period, then there is no credible threat that can achieve this either.) Since the stage game has a unique NE, the only SPNE of the finitely repeated game is for the players to unconditionally play that stage-game NE every period. (c) Now consider the infinitely repeated game WLSS(, δ), with a discount factor of δ, 0 < δ <. Give an example of a pair of strategies that, for δ sufficiently close to, constitute a SPNE to WLSS(, δ), and generate the observed ( on-the-equilibrium-path ) behavior of (W, W ), with payoffs (4, 4), each period. Fully specify each player s strategy separately. Suggestion: Employ Nash punishment strategies of the sort we used to prove the weak folk theorem. Consider the following trigger strategy for Player : Play W in period. In period t, play W if (W, W ) was the outcome in periods to t. Otherwise, play L. Similarly, for Player : Play W in period. In period t, play W if (W, W ) was the outcome in periods to t. Otherwise, play L. These strategies constitute a SPNE for values of δ sufficiently close to one. Note that there are two classes of subgames: those following deviation and those following no deviation. For the former, these strategies prescribe that the players will play (L,L) forever. Since this is a NE of the stage game, the strategies constitute a NE in all these subgames. The subgames following no deviation are equal to the whole repeated game. As shown below, these strategies also constitute a NE in these subgames if δ is close to one. Note that if δ were equal to one, then no player would want to deviate, since they would gain in the current period but lose for all subsequent periods. (d) For the strategy pair you proposed in part (c), state for exactly what ranges of values of δ this is a SPNE. (Specify whether any inequalities are weak or strong.) (Recall that t=0 δt M = M/( δ), and t= δt M = δm/( δ).) Comparing the payoff associated with the proposed strategy and the payoff from the most profitable deviation, we have that no player deviates if δ t δ t 4 δ t=0 t= 5 + δ δ δ 3 Hence, for δ 3, the strategies proposed in part (c) constitute a SPNE. Question 5: Stay in Some School (5 Points) Please refer to my Section 7 handout for this question.

7 00 Exam Question (0 points). Question : Syncretic Olla Podrida (0 Points) Consider the game of Syncretic Olla Podrida: L R A B C E D 3 3 G H 7 Syncretic Olla Podrida (a) List all of the (pure) strategies for each of the players. Player :. LG. LH 3. RG 4. RH Player :. AC. AD 3. AE 4. BC 5. BD. BE a) List all of the (pure) strategies for each of the players. Do so in the form: Player :. A strategy. Another strategy... Player :. A strategy,. Another strategy... Use natural shorthand. b) For each of the players, state which (if any) strategies are equivalent to each other according to the definition we have been using. (b) For each of the players, state which (if any) strategies are equivalent to each other. For Player, LG and LH are equivalent. There are no equivalent strategies for Player. c) State the backwards-induction strategy for each player. (c) State the backwards-induction strategy for each player. Player s backwards-induction strategy is LH, and Player s is AC. d) Write either the normal form or the reduced normal form of Syncretic Olla Podrida. (It will be easier to do the next part if you write the reduced normal form, which does not distinguish among equivalent strategies.) (d) Write either the normal form or the reduced normal form of the game. (Suggested: reduced normal form.) e) Find all of the Pure-Strategy Nash equilibria in the normal form of Syncretic Olla Podrida or in the reduced normal form of Syncretic Olla Podrida. AC AD AE BC BD BE L 4,4 4,4 4,4 0,0 0,0 0,0 f) Find all of the NashRG equilibria, in the 3,3normal, form,of Syncretic 3,3, Olla Podrida or in the reduced normal form of Syncretic Olla Podrida such that Player plays L for sure. RH, 3,3 7,, 3,3 7, (e) Find all the pure-strategy Nash equilibria in the normal or reduced normal form. The set of pure-strategy NE in the reduced normal form is {(L, AC), (L, AD), (RG, BC), (RH, BC)} (f) Find all the Nash equilibria in the normal or reduced normal form such that Player plays L for sure. The NE in the reduced normal form such that Player plays L for sure are given by (L; pac, qad, ( p q)ae) for p + 4q 3, p [0, ], q [0, ], (p + q) [0, ] 7

8 Question : Inkhorn Approbation (5 Points) L C R U, 7, 0 3, 4 D 4, 0, 9, 5 (a) What actions by Player (the row player) are consistent with Player being rational (making no assumptions about her beliefs about the rationality of Player )? Note that U is a best response to C and D to L. Hence, A () = {U, D} (b) What actions by Player (the column player) are consistent with Player being rational (making no assumptions about her beliefs about the rationality of Player )? Note that L is a best response to U and C to D. Moreover, if Player believes that Player will play U and D with equal probability, then R is a best response. Hence, A () = {L, C, R} (c) What actions by Player are rationalizable? Since A () = A (0) and A () = A (0), R = {U, D} (d) What actions by Player are rationalizable? From above, R = {L, C, R} (e) Find all the Nash equilibria in the game. Note that there are no pure-strategy NE, and any mixed-strategy NE must have both players mixing. Furthermore, note that Player will not mix between L and C, since if Player is indifferent between L and C, then Player strictly prefers R to L and C. And Player will not mix only between L and R, as in that case Player would not mix. Thus, Player must be mixing between C and R, and we find that the unique Nash equilibrium is ( 5 U, 4 5 D; 7 C, ) 7 R Question 3 (5 points) Question 3: Ossified Unguent (5 Points) Consider Ossified Unguent c x d L R a b y Ossified Unguent 8

9 (a) Write down the normal form or the reduced normal form of this game. xa xb ya yb Lc, -, - -, -, Ld -, -,, -, - R,,,, (b) State all the Nash equilibria in either the normal form or reduced normal form. There are no pure-strategy NE. Consider mixed-strategy NE where Player plays R for sure (with any probability on c and d). Then Player may play xa with probability p and ya with probability ( p) such that the following two conditions hold p ( p) p /3 p + ( p) p /3 Note that there are no other NE in this game. Player will not play Lc and Ld with positive probability, as in that case Player should mix Lc and Ld with equal probability and Player should mix x and y with equal probability in equilibrium (note that the left proper subgame is a Matching Pennies game). But then no matter what probability Player puts on a versus b, Player will always strictly prefer to play R over Lc and Ld. Hence, the set of NE in the reduced normal form is given by (R; pxa, ( p)ya) for p [/3, /3] (c) State (in some natural way) all the subgames of the game. There are three subgames in this game:. starting at Player s decision node x/y, after L is played. starting at Player s decision node a/b, after R is played 3. the game itself (d) List all the SPNE in the game. The unique NE in the Matching Pennies subgame has both players mixing with probability /. The unique NE in the right-hand-side proper subgame has Player choosing a. Building upwards, at the first node, Player will then choose R. Thus, the unique SPNE is ( Rc, Rd; xa, ) ya Question 4: Work-Laze-AsymmetricSabotage (0 Points) Consider Work-Laze-AsymmetricSabotage played twice: Work Laze Sabotage Work 9,9,0,0 Laze 0,,,0 Sabotage,0,0 0,0 Please note that this game is asymmetric. Consider the repeated game W LAS(T =, δ), where WorkLaze-AsymmetricSabotage is played twice, with a discount of δ, 0 < δ, between the first and second period. 9

10 (a) Describe verbally but precisely an example of a subgame of W LAS(T =, δ) that is not the game itself. The second-stage game following history (W,W). (b) Either state what the lowest δ such that there exists a Nash equilibrium to the repeated game where both players work in the first period is, or explain why there is no such δ. Note that the stage game has a unique NE, (L,L). Thus, in the second period, the players must play (L,L) on the equilibrium path. Note further that the worst punishment is to Sabotage. Consider then the following strategies for both players: Work in the first period. In the second period, Laze if the other player played Work in the first period, and Sabotage otherwise. Note that for Player, if Player punishes him in the second period, his payoff is. For these strategies to constitute a NE of the repeated game, we then need 9 + δ 0 + δ δ Hence, since 0 < δ, only for δ = there exists a NE of the repeated game that implements (W,W) in the first period. (c) Either state what the lowest δ such that there exists a SPNE to the repeated game where both players work in the first period is, or explain why there is no such δ. Since the stage game has a unique NE, the only SPNE of the finitely repeated game is for the players to unconditionally play that stage-game NE every period. Hence, there exists no SPNE where both players work in the first period. (d) Now consider the infinitely repeated game WLAS(, δ), with a discount factor δ, 0 < δ <. Give an example of a pair of strategies that, for δ sufficiently close to, constitute a SPNE to WLAS(, δ), and generate the observed ( on-the-equilibrium-path ) behavior of (W, W ), with payoffs (9, 9), each period. Fully specify each player s strategy separately. Consider the following trigger strategy for Player : Play W in period. In period t, play W if (W, W ) was the outcome in periods to t. Otherwise, play L. Similarly, for Player : Play W in period. In period t, play W if (W, W ) was the outcome in periods to t. Otherwise, play L. These strategies constitute a SPNE for values of δ sufficiently close to one. Note that there are two classes of subgames: those following deviation and those following no deviation. For the former, these strategies prescribe that the players will play (L,L) forever. Since this is a NE of the stage game, the strategies constitute a NE in all these subgames. The subgames following no deviation are equal to the whole repeated game, and these strategies also constitute a NE in these subgames if δ is close to one. Note that if δ were equal to one, then no player would want to deviate, since they would gain in the current period but lose 7 for all subsequent periods. Question 5: Stay in School 5 (5 Points) Consider the simple signaling game Stay in School 5. Notice that of potential workers are smart. Note that you are asked to find all equilibria in each part of this problem, not just pure-strategy equilibria. There are no typos in the payoffs (even though they may seem odd, insofar as Dumb types prefer not to be hired). 0

11 Question 5 (5 points) Consider the simple signaling game Stay in School 5 : 9, 0 Hire Hire, 5 No College [0.5] College Smart 3, No Hire R S R No Hire 0, 0 0, 0 Hire Hire 0, 0 No College Dumb [0.5] College 5, 0 No Hire Stay in School 5 No Hire 4, 0 (a) What are all of the (Bayesian) Nash equilibria in Stay in School 5? Give a brief intuition about what Notice actions that might of potential be played workers are bysmart. eachnote typethat ofyou thearesender asked to find Stay all equilibria in School in 5. each part of this problem, not just pure-strategy equilibria. There are no typos in the payoffs I denote the (evenstrategies though they as mayfollows. seem odd, insofar For R, as Dumb (H,NH) typesmeans prefer nothire to beif hired). No College and No Hire if College. For S, (C,NC) means College if Smart, No College if Dumb. a) What are all of the (Bayesian) Nash equilibria in Stay in School 5? Give a brief intuition There is a separating BNE in which the smart type goes to college and is hired and the dumb about what actions might be played by each type of the Sender in Stay in School 5. type does not go to college and is not hired. There is also a continuum of pooling BNE in which both b) What types arechoose all of theno Perfect College, Bayesianand equilibria R chooses in Stay in No School Hire 5? if No College, Hire if College with probability p /. That is, the set of BNE is given by c) What are all of the PBE meeting Signaling Refinement A in Stay in School 5? (Thisisthe weaker of the two signaling refinements, that relies on restricting beliefs based on the sending of some messages by some types being weakly dominated. Give a brief intuition for why the set of PBE meeting Signaling Refinement A does or does not eliminate one or more PBE. {(C, NC; NH, H), (NC, NC; NH, ph + ( p)nh)} for p [0, /] We find that there cannot be an equilibrium where the Dumb type of S goes to College. If d) What are all of the PBE meeting Signaling Refinement B in Stay in School 5? (This there wereisone, the stronger then Rof should the two signaling be choosing refinements, Hireequivalent following hereno to College Cho-Kreps s with Intuitive a higher probability than following Criterion, College. restricting But beliefs thenbased theonsmart sending typeofwould some messages choose byno somecollege, types being and R would not dominated by their equilibrium payoffs. Give a brief intuition for why the set of PBE meeting be best responding. Signaling Refinement Intuitively, B does or since doesthe not eliminate Dumb type one orprefers more PBEnot meeting to be Signaling hired and the Smart type prefers Refinement to be A. hired, either they will separate as in the first BNE we found, or they will pool in No College if the probability that they will be hired if they choose College is low enough. (b) What are all of the Perfect Bayesian equilibria in Stay in School 5? The set of PBE is the same as the set of BNE. There are no out-of-equilibrium messages in the separating BNE, so this equilibrium is for sure perfect Bayesian. For the pooling BNE, note that if R believes that he is facing the Dumb type with probability sufficiently high when he observes College, he will choose No Hire if College with probability at least /. (c) What are all of the PBE meeting Signaling Refinement A in Stay in School 5? (This is the weaker of the two signaling refinements, that relies on restricting beliefs based on the sending of some messages by some types being dominated.) Give a brief intuition for why the set of PBE meeting Signaling Refinement A does or does not eliminate one or more PBE. The set of PBE meeting Signaling Refinement A is the same as the set of PBE (and BNE). This refinement does nothing here since there are no messages dominated for any of the types. The Smart type prefers College to No College if R chooses No Hire if No College, Hire if College, but he prefers No College to College if R chooses Hire if No College, No Hire if College. The Dumb type prefers College to No College if R chooses Hire if No College, No Hire if College, but he prefers No College to College if R chooses No Hire if No College, Hire if College.

12 (d) What are all of the PBE meeting Signaling Refinement B in Stay in School 5? (This is the stronger of the two signaling refinements, equivalent here to Cho-Kreps s Intuitive Criterion, restricting beliefs based on the sending of some messages by some types being dominated by their equilibrium payoffs.) Give a brief intuition for why the set of PBE meeting Signaling Refinement B does or does not eliminate one or more PBE meeting Signaling Refinement A. The separating PBE satisfies this refinement since there are no out-of-equilibrium messages in this equilibrium. The pooling PBE, however, do not survive this refinement. Note that in the pooling equilibria, the Smart type obtains a payoff of 3, while he could obtain a payoff of by choosing College if R chose Hire following College. The Dumb type, on the other hand, obtains a payoff of 5 in the pooling equilibria, and would never have incentives to deviate to College no matter what R believes and chooses following College. Hence, this refinement says that R should believe that he is facing the Smart type if he observes College. But then R should play Hire if College with probability one, and thus the Smart type would deviate. This refinement then rules out all the pooling equilibria. Question : Stay in School and 7 (5 Points) Consider the game Stay in School : It has the exact same payoffs, private-information structure, actions, etc. as Stay in School 5, but: the game is being played after the Who Needs College Anyhow? Act of 0, which has banished the use of any educational information in hiring. Hence, the Receiver chooses Hire/NoHire before observing College/No College by a potential worker. (a) List all of the (pure) strategies for each of the players. For S, I use the same notation as above (i.e., I first write the action corresponding to the Smart type). Sender:. NC,NC. NC,C 3. C,NC 4.C,C Receiver:. H. NH (b) Find all the Bayesian Nash equilibria in Stay in School. The unique BNE of this game is (NC, NC; NH) Now consider the game Stay in School 7: It has the exact same payoffs, private-information structure, actions, etc. as Stay in School 5, but: The game is being played after the Okay We Suppose People Can Say They ve Gone To College If They Want Act of 0, which leaves it illegal for firms to demand educational information in hiring, but allows workers to prove their educational status if they want. Hence, each type of potential worker has four possible messages she can send: College/Reveal, which involves going to college and proving she s gone to college; No College/Reveal which involves not going to college and proving she has not gone to college; College/Hide, which involves going to college and refusing to give evidence of this fact; and No College/Hide which involves not going to college and refusing to give evidence of this fact. The Receiver cannot tell the difference between the last two messages she observes essentially Hide when either is chosen. So the Receiver might observe any of three potential messages before choosing hire or no hire. Assume that neither player s payoffs depend directly on whether college/no college is revealed, but only on the Sender s type, the college vs. no college choice, and the hire vs. NoHire choice. This is not a simple signaling game as we ve defined it (nor is it technically a type of sequential game we ve looked at in the course). But it is a close cousin.

13 (c) Try drawing the game tree (with payoffs and information sets) in the natural way. It won t be so easy (technically, it can be done without crossing any lines, but that is not very easy). Draw the tree with four actions following each of the types nodes. Then connect with a dotted line the arrows corresponding to NCH of Smart, CH of Smart, NCH of Dumb, and CH of Dumb (where H denotes Hide), indicating that R cannot observe the Sender s type nor whether the Sender went to College when the Sender chooses to Hide. Then from the four decisions nodes of R contained in this information set, draw the two lines corresponding to Hire and No Hire. Then draw another information set connecting the arrows corresponding to CR of Smart and CR of Dumb (where R denotes Reveal), and again draw the Hire and No Hire arrows from the two decision nodes of R contained in this information set. Finally, do the same for NCR of Smart and NCR of Dumb. (d) How many (pure) strategies are there in this game for the Sender? You don t need to list them; just give the number. There are two types of Sender and each type has four possible (pure) strategies, so the Sender has a total of (pure) strategies available. (e) How many (pure) strategies are there in this game for the Receiver? Again, just the number. There are three information sets for the Receiver, each with two actions available, so the Receiver has a total of 8 (pure) strategies available. (f) Find all the pure-strategy Nash equilibria that meet the strongest refinement, Signaling Refinement B in Stay in School 7. Again, this is not technically a simple signaling game, but the appropriate generalization here should be clear. Also note that Stay in School 5 was non-generic only in unimportant ways, Stay in School 7 has more important non-genericities. There are two pure-strategy PBE meeting Signaling Refinement B (SRB). First, note that if both types choose Reveal, then the game is as Stay in School 5 on the equilibrium path, and hence the only PBE that meets SRB is the separating one in which Smart chooses College and is hired and Dumb chooses No College and is not hired. By specifying that R chooses No Hire if Hide, this constitutes a PBE meeting SRB (note that only Smart could have incentives to deviate to Hide if R played Hire if Hide, but he would deviate to No College/Hide, in which case No Hire is R s best response.) Next, note that if only Dumb chooses Hide, then again the types separate, and the separating equilibrium we just found would also work. The separating equilibrium would not work, however, if only Smart chooses Hide. In that case, R cannot be choosing Hire if Hide because Smart would deviate to No College/Hide and R would not be best responding. Finally, note that if both types choose Hide, then the game is as Stay in School on the equilibrium path. But there we found that both types would choose No College and R would choose No Hire. But a pooling PBE would not satisfy SRB, as it would have R choosing No Hire (with high probability) if College/Reveal, when SRB will require R to believe that he is facing Smart if he observes College/Reveal. In sum, the set of pure-strategy PBE meeting Signaling Refinement B in Stay in School 7 is {(CR, NCR; NH, H, NH), (CR, NCH; NH, H, NH)} where for S, I use the same notation as above, and for R, I write the action following No College/Reveal first, then following College/Reveal, and finally following Hide. 3