UC BerkeleyFall 2007

Economics 201A - Section 8 (Review)

Marina Halac

1 2003 Exam

Question 1: Car Key Gnomes (25 Points)

(a) List all of the (pure) strategies for each of the players.Player 1: 1. AE 2. AF 3. BE 4. BFPlayer 2: 1. C 2. D

(b) For each of the players, state which (if any) strategies are equivalent to each other.For Player 1, BE and BF are equivalent. There are no equivalent strategies for Player 2.

(c) State the backwards-induction strategy for each player.Player 1’s backwards-induction strategy is BF, and Player 2’s is C.

(d) Consider the following modification of the game: Decrease Player 1’s payoff on one and onlyone of the four end nodes of the game, without changing any of Player 1’s other payoffs, andwithout changing any of Player 2’s payoffs. Choosing which outcome to change and by howmuch, re-write the game so that Player 1’s backwards-induction payoff in the modified gameis higher than Player 1’s backwards-induction payoff in the original game.Rewrite the game changing Player 1’s payoff following F to anything lower than 6.

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 DAE 2,3 6,8AF 2,3 9,0B 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 Player1 plays B, Player 2 is indifferent between C and D. Player 2 may then play C with probabilityp and D with probability (1− p) for any p such that B is still a best response for Player 1:

5 > 2p + 9(1− p) ⇔ p > 4/7

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Note that there are no other mixed-strategy NE. If there were, both players should be mixing.But note that if Player 2 is mixing, Player 1 will never play AE with positive probability (sincefor any positive probability that Player 2 puts on D, AF strictly dominates AE). And Player1 will never mix between AF and B, since in that case Player 2 would strictly prefer C to D,and thus Player 1 would not want to mix.Hence, the set of Nash equilibria in the reduced normal form is given by

(B; pC, (1− p)D) for p ∈ [4/7, 1]

Question 2: Lewis and Clark Meet Sylvia and Rhonda (15 Points)

L C RU 1,7 2,1 3,2M 6,3 7,8 8,4D 4,5 5,6 9,9

(a) What actions by Player 1 are consistent with Player 1 being rational (making no assumptionsabout her beliefs about the rationality of Player 2)?Note that M is a best response to L and D to R, while U is strictly dominated by both Mand D. Hence,

A1(1) = {M,D}

(b) What actions by Player 2 are consistent with Player 2 being rational (making no assumptionsabout her beliefs about the rationality of Player 1)?

A2(1) = A2(0) = {L,C,R}

(c) What actions by Player 1 are consistent with Player 1 being rational and believing that Player2 is rational?Since A2(1) = A2(0),

A1(2) = A1(1) = {M,D}

(d) What actions by Player 2 are consistent with Player 2 being rational and believing that Player1 is rational?If Player 2 believes that Player 1 is rational and hence will never play U, then L is strictlydominated by both C and R.

A2(2) = {C,R}

(e) What actions by Player 1 are rationalizable?Note that since A1(1) = A1(2), we will have that A2(2) = A2(3). Moreover, note thatA1(2) = A1(3). Thus,

R1 = {M,D}

(f) What actions by Player 2 are rationalizable?

2

From above,

R2 = {C,R}

(g) Find all the Nash equilibria in the game.Of course, we only need to look at the 2×2 matrix with the rationalizable actions. The setof Nash equilibria is given by{

(M ;C), (D;R),(

37M,

47D;

13C,

23R

)}Question 3: Lapidary Lapel (25 Points; Parts (a) to (c) worth 10 points in total)

A natural interpretation of the game is that Vai-Lam can pay some amount x > 0 to not observeYuk-Ying’s choice of event to go to before he has to make his choice, or can pay y > 0 to observeYuk-Ying’s choice of event. Yuk-Ying knows which choice of payment Vai-Lam has made beforeshe 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:

1. B1B2 2. B1T2 3. T1B2 4. T1T2Vai-Lam:

1. xB3B4B5 2. xB3B4T5 3. xB3T4B5 4. xB3T4T55. xT3B4B5 6. xT3B4T5 7. xT3T4B5 8. xT3T4T59. yB3B4B5 10. yB3B4T5 11. yB3T4B5 12. yB3T4T5

13. yT3B4B5 14. yT3B4T5 15. yT3T4B5 16. 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

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(c) Besides the game itself, list all other subgames (if any).There are four subgames besides the game itself:

1. starting at Yuk-Ying’s decision node B1/T1, after x is played2. starting at Yuk-Ying’s decision node B2/T2, after y is played3. starting at Vai-Lam’s decision node B4/T4, after B2 is played4. starting at Vai-Lam’s decision node B5/T5, after T2 is played

(d) Find all the pure-strategy SPNE in the game as a function of x and y. That is, for allpossible combinations of x, y > 0, list all the pure-strategy SPNE. Give an intuition for howthe set of pure-strategy SPNE depends on x and y, paying special focus on the intuition forthe conditions on x and y such that there exists a pure-strategy SPNE where Vai-Lam choosesto 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 uniqueNE: B4 (because 1− y > −y) and T5 (because 2− y > −y). Building upwards, at the B2/T2decision node, Yuk-Ying then chooses B2.The left-hand-side subgame beginning at Yuk-Ying’s B1/T1 decision node is essentially aBattle of the Sexes with an unknown −x added to Vai-Lam’s payoffs. There are two pure-strategy NE in this subgame (and one mixed-strategy NE, but we are not concerned withthat here): (B1,B3) and (T1,T3).Vai-Lam will then compare (1 − x) versus (1 − y), and (2 − x) versus (1 − y). Hence, thereare four cases:

1. (B1B2,xB3B4T5) is a SPNE if (1− x) ≥ (1− y) ⇐⇒ y ≥ x

2. (B1B2,yB3B4T5) is a SPNE if (1− x) ≤ (1− y) ⇐⇒ y ≤ x

3. (T1B2,xT3B4T5) is a SPNE if (2− x) ≥ (1− y) ⇐⇒ y + 1 ≥ x

4. (T1B2,yT3B4T5) is a SPNE if (2− x) ≤ (1− y) ⇐⇒ y + 1 ≤ x

Intuitively, Vai-Lam pays x and does not observe Yuk-Ying’s choice when the cost of observingYuk-Ying’s choice is higher (Case 1), or when the cost of not observing Yuk-Ying’s choiceis higher but the additional 1 that Vai-Lam can obtain by not observing Yuk-Ying’s choiceoffsets it. That is, if x is not much more expensive than y, then Vai-Lam chooses to pay xbecause in the imperfect information subgame Vai-Lam may obtain a higher payoff than inthe sequential subgame where he chooses after observing what Yuk-Ying does.

Now consider the game Lapidary Lapel 2, where Yuk-Ying does not observe what payment (xor 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 B1/T1 and B2/T2 decision nodes, and relabel B2 and T2 as B1 and T1 respectively,because Yuk-Ying now cannot distinguish between the two decision nodes, and hence mustbe 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 allpossible combinations of x, y > 0, list all the pure-strategy SPNE. Discuss with subtlety andinsight 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/T4and 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

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the fact that B4 and T5 are always played. Below is the reduced normal form.

x B3 x T3 yB1 2,1-x 0, -x 2,1-yT1 0, -x 1,2-x 1,2-y

Hence, there are four cases:

1. (B1,xB3B4T5) is a SPNE if y ≥ x

2. (T1,xT3B4T5) is a SPNE if y ≥ x

3. (B1,yB3B4T5) is a SPNE if x ≥ y

4. (B1,yT3B4T5) is a SPNE if x ≥ y

Comparing the pure-strategy SPNE of the two games, we see that the main difference is theabsence of the conditions on y + 1. In the first game, Vai-Lam knew that she would endup meeting Yuk-Ying in any pure-strategy equilibrium. On the right hand side, due to thesequential nature of these subgames, they would always go to her least favorite event, B. Onthe left hand side, however, the equilibrium in place sometimes called for them going to herfavorite event, T. This would give Yuk-Ying one extra payoff, and thus Yuk-Ying would preferto 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-Lammakes), and so Vai-Lam best responds by playing the corresponding event and picking thecheapest payment. Hence, Vai-Lam previously stood to gain by choosing a simultaneous overa sequential game. Now, with Yuk-Ying playing the same action in both games, she onlycares to minimize the cost of her payments.

Question 4: Semi-Sabotage (20 Points)

Consider the stage game Work, Laze, Semi-Sabotage:

W LW 4,4 1,5L 5,1 2,2S 0,0 0,0

Consider the repeated game WLSS (T = 2, δ), where Work, Laze, Semi-Sabotage is played twice,with a discount of δ, 0 < δ ≤ 1, between the first and second period.

(a) Either state what the lowest δ such that there exists a Nash equilibrium to the repeated gamewhere 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 1, L strictly dominatesthe other two pure strategies.) Thus, to sustain (W,W) in period 1, we need to find rewardsand punishments, since (W,W) is not a NE of the stage game. The reward must be to play(L,L) in period 2, since again this is the only NE of the stage game and hence what theplayers must play on the equilibrium path in the last period. But note that Player 2 doesnot have the means to punish Player 1 for deviations in period 1. Player 1 then knows thatno matter what he does in period 1, Player 2 will play L in period 2. Consequently, Player1 will always have incentives to deviate from (W,W), and a NE of the repeated game thatimplements (W,W) in the first period does not exist.

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(b) Either state what the lowest δ such that there exists a SPNE to the repeated game where bothplayers 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 SPNEwhere this occurs. (That is, if there is no threat that can make both players work in period1, then there is no credible threat that can achieve this either.) Since the stage game has aunique NE, the only SPNE of the finitely repeated game is for the players to unconditionallyplay that stage-game NE every period.

(c) Now consider the infinitely repeated game WLSS(∞, δ), with a discount factor of δ, 0 < δ < 1.Give an example of a pair of strategies that, for δ sufficiently close to 1, constitute a SPNE toWLSS(∞, δ), and generate the observed (‘on-the-equilibrium-path’) behavior of (W,W ), withpayoffs (4, 4), each period. Fully specify each player’s strategy separately. Suggestion: EmployNash punishment strategies of the sort we used to prove the weak folk theorem.Consider the following trigger strategy for Player 1:

Play W in period 1. In period t, play W if (W,W ) was the outcome in periods 1to t− 1. Otherwise, play L.

Similarly, for Player 2:

Play W in period 1. In period t, play W if (W,W ) was the outcome in periods 1to t− 1. Otherwise, play L.

These strategies constitute a SPNE for values of δ sufficiently close to one. Note that thereare two classes of subgames: those following deviation and those following no deviation. Forthe former, these strategies prescribe that the players will play (L,L) forever. Since this is aNE of the stage game, the strategies constitute a NE in all these subgames. The subgamesfollowing no deviation are equal to the whole repeated game. As shown below, these strategiesalso 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 1 in the current period but lose2 for all subsequent periods.

(d) For the strategy pair you proposed in part (c), state for exactly what ranges of values of δ thisis a SPNE. (Specify whether any inequalities are weak or strong.) (Recall that

∑∞t=0 δt ·M =

M/(1− δ), and∑∞

t=1 δt ·M = δM/(1− δ).)Comparing the payoff associated with the proposed strategy and the payoff from the mostprofitable deviation, we have that no player deviates if

∞∑t=0

δt4 > 5 +∞∑

t=1

δt2 ⇔ 41− δ

> 5 +δ2

1− δ⇔ δ ≥ 1

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Hence, for δ ≥ 13 , the strategies proposed in part (c) constitute a SPNE.

Question 5: Stay in Some School (15 Points)

Please refer to my Section 7 handout for this question.

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2 2006 Exam

Question 1: Syncretic Olla Podrida (20 Points)

Question 1 (20 points).

Consider the game of Syncretic Olla Podrida:

1

2

2

D 1

HG

Syncretic Olla Podrida

L R

A B

C E44

00

16

33

22

71

a) List all of the (pure) strategies for each of the players. Do so in the form: Player 1: 1. Astrategy 2. Another strategy .... Player 2: 1. A strategy, 2. Another strategy ... Use naturalshorthand.

b) For each of the players, state which (if any) strategies are “equivalent” to each otheraccording to the definition we have been using.

c) State the backwards-induction strategy for each player.

d) Write either the normal form or the reduced normal form of Syncretic Olla Podrida. (Itwill be easier to do the next part if you write the reduced normal form, which does notdistinguish among equivalent strategies.)

e) Find all of the Pure-Strategy Nash equilibria in the normal form of Syncretic OllaPodrida or in the reduced normal form of Syncretic Olla Podrida.

f) Find all of the Nash equilibria in the normal form of Syncretic Olla Podrida or in thereduced normal form of Syncretic Olla Podrida such that Player 1 plays L for sure.

2

(a) List all of the (pure) strategies for each of the players.Player 1:1. LG 2. LH 3. RG 4. RH

Player 2:1. AC 2. AD 3. AE4. BC 5. BD 6. BE

(b) For each of the players, state which (if any) strategies are equivalent to each other.For Player 1, LG and LH are equivalent. There are no equivalent strategies for Player 2.

(c) State the backwards-induction strategy for each player.Player 1’s backwards-induction strategy is LH, and Player 2’s is AC.

(d) Write either the normal form or the reduced normal form of the game. (Suggested: reducednormal form.)

AC AD AE BC BD BEL 4,4 4,4 4,4 0,0 0,0 0,0

RG 1,6 3,3 2,2 1,6 3,3 2,2RH 1,6 3,3 7,1 1,6 3,3 7,1

(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 1 playsL for sure.The NE in the reduced normal form such that Player 1 plays L for sure are given by

(L; pAC, qAD, (1− p− q)AE) for 6p + 4q > 3, p ∈ [0, 1], q ∈ [0, 1], (p + q) ∈ [0, 1]

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Question 2: Inkhorn Approbation (15 Points)

L C RU 1, 7 2, 0 3, 4D 4, 0 1, 6 9, 5

(a) What actions by Player 1 (the row player) are consistent with Player 1 being rational (makingno assumptions about her beliefs about the rationality of Player 2)?Note that U is a best response to C and D to L. Hence,

A1(1) = {U,D}

(b) What actions by Player 2 (the column player) are consistent with Player 2 being rational(making no assumptions about her beliefs about the rationality of Player 1)?Note that L is a best response to U and C to D. Moreover, if Player 2 believes that Player 1will play U and D with equal probability, then R is a best response. Hence,

A2(1) = {L,C,R}

(c) What actions by Player 1 are rationalizable?Since A1(1) = A1(0) and A2(1) = A2(0),

R1 = {U,D}

(d) What actions by Player 2 are rationalizable?From above,

R2 = {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 playersmixing. Furthermore, note that Player 2 will not mix between L and C, since if Player 2 isindifferent between L and C, then Player 2 strictly prefers R to L and C. And Player 2 willnot mix only between L and R, as in that case Player 1 would not mix. Thus, Player 2 mustbe mixing between C and R, and we find that the unique Nash equilibrium is(

15U,

45D;

67C,

17R

)

Question 3: Ossified Unguent (15 Points)Question 3 (15 points)

Consider Ossified Unguent

1

L R

2

2 a b

x y

1

c d

Ossified Unguent

6-6

6-6

-66

-66

22

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i) Write down the normal form or the reduced normal form of this game.

ii) State all the Nash equilibria in either the normal form or reduced normal form.

iii) State (in some natural way) all of the subgames in Ossified Unguent.

iv) List all SPNE in Ossified Unguent.

4

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(a) Write down the normal form or the reduced normal form of this game.

xa xb ya ybLc 6, -6 6, -6 -6, 6 -6, 6Ld -6, 6 -6, 6 6, -6 6, -6R 2,2 1,1 2, 2 1,1

(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 1 plays R for sure(with any probability on c and d). Then Player 2 may play xa with probability p and ya withprobability (1− p) such that the following two conditions hold

2 > 6p− 6(1− p) ⇔ p 6 2/32 > −6p + 6(1− p) ⇔ p > 1/3

Note that there are no other NE in this game. Player 1 will not play Lc and Ld with positiveprobability, as in that case Player 1 should mix Lc and Ld with equal probability and Player2 should mix x and y with equal probability in equilibrium (note that the left proper subgameis a Matching Pennies game). But then no matter what probability Player 2 puts on a versusb, Player 1 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, (1− p)ya) for p ∈ [1/3, 2/3]

(c) State (in some natural way) all the subgames of the game.There are three subgames in this game:

1. starting at Player 2’s decision node x/y, after L is played2. starting at Player 2’s decision node a/b, after R is played3. the game itself

(d) List all the SPNE in the game.The unique NE in the Matching Pennies subgame has both players mixing with probability1/2. The unique NE in the right-hand-side proper subgame has Player 2 choosing a. Buildingupwards, at the first node, Player 1 will then choose R. Thus, the unique SPNE is(

12Rc,

12Rd;

12xa,

12ya

)Question 4: Work-Laze-AsymmetricSabotage (20 Points)

Consider Work-Laze-AsymmetricSabotage played twice:

Work Laze SabotageWork 9,9 1,10 1,0Laze 10,1 2,2 1,0

Sabotage 1,0 1,0 0,0

Please note that this game is asymmetric.Consider the repeated game WLAS(T = 2, δ), where WorkLaze-AsymmetricSabotage is played

twice, with a discount of δ, 0 < δ ≤ 1, between the first and second period.

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(a) Describe verbally but precisely an example of a subgame of WLAS(T = 2, δ) that is not thegame 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 gamewhere 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 playersmust play (L,L) on the equilibrium path. Note further that the worst punishment is toSabotage. 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 Sabotageotherwise. Note that for Player 1, if Player 2 punishes him in the second period, his payoff is1. For these strategies to constitute a NE of the repeated game, we then need

9 + δ2 > 10 + δ ⇔ δ ≥ 1

Hence, since 0 < δ ≤ 1, only for δ = 1 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 bothplayers 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 forthe players to unconditionally play that stage-game NE every period. Hence, there exists noSPNE where both players work in the first period.

(d) Now consider the infinitely repeated game WLAS(∞, δ), with a discount factor δ, 0 < δ < 1.Give an example of a pair of strategies that, for δ sufficiently close to 1, constitute a SPNE toWLAS(∞, δ), and generate the observed (‘on-the-equilibrium-path’) behavior of (W,W ), withpayoffs (9, 9), each period. Fully specify each player’s strategy separately.Consider the following trigger strategy for Player 1:

Play W in period 1. In period t, play W if (W,W ) was the outcome in periods 1to t− 1. Otherwise, play L.

Similarly, for Player 2:

Play W in period 1. In period t, play W if (W,W ) was the outcome in periods 1to t− 1. Otherwise, play L.

These strategies constitute a SPNE for values of δ sufficiently close to one. Note that there aretwo classes of subgames: those following deviation and those following no deviation. For theformer, these strategies prescribe that the players will play (L,L) forever. Since this is a NE ofthe stage game, the strategies constitute a NE in all these subgames. The subgames followingno deviation are equal to the whole repeated game, and these strategies also constitute a NEin these subgames if δ is close to one. Note that if δ were equal to one, then no player wouldwant to deviate, since they would gain 1 in the current period but lose 7 for all subsequentperiods.

Question 5: Stay in School 5 (15 Points)

Consider the simple signaling game Stay in School 5.Notice that 1

2 of potential workers are smart. Note that you are asked to find all equilibria ineach part of this problem, not just pure-strategy equilibria. There are no typos in the payoffs (eventhough they may seem odd, insofar as Dumb types prefer not to be hired).

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Question 5 (15 points)

Consider the simple signaling game Stay in School 5 :

Hire Hire

No College College

Smart

No Hire No Hire

Hire Hire

Dumb

No College College

No Hire No Hire

Stay in School 5

3, 1

4, 10

0, 0

0, 0

6, 15

5, 10

9, 0

0, 0

SR R

[0.5]

[0.5]

Notice that 12 of potential workers are smart. Note that you are asked to find all equilibria ineach 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).

a) What are all of the (Bayesian) Nash equilibria in Stay in School 5? Give a brief intuitionabout what actions might be played by each type of the Sender in Stay in School 5.

b) What are all of the Perfect Bayesian equilibria in Stay in School 5?

c) What are all of the PBE meeting Signaling Refinement A in Stay in School 5? (This is theweaker of the two signaling refinements, that relies on restricting beliefs based on the sendingof some messages by some types being weakly dominated. Give a brief intuition for why theset of PBE meeting Signaling Refinement A does or does not eliminate one or more PBE.

d) What are all of the PBE meeting Signaling Refinement B in Stay in School 5? (Thisis the stronger of the two signaling refinements, equivalent here to Cho-Kreps’s “IntuitiveCriterion,” restricting beliefs based on the sending of some messages by some types beingdominated by their equilibrium payoffs. Give a brief intuition for why the set of PBE meetingSignaling Refinement B does or does not eliminate one or more PBE meeting SignalingRefinement A.

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(a) What are all of the (Bayesian) Nash equilibria in Stay in School 5? Give a brief intuitionabout what actions might be played by each type of the Sender in Stay in School 5.I denote the strategies as follows. For R, (H,NH) means Hire if No College and No Hire ifCollege. For S, (C,NC) means College if Smart, No College if Dumb.There is a separating BNE in which the smart type goes to college and is hired and the dumbtype does not go to college and is not hired. There is also a continuum of pooling BNE inwhich both types choose No College, and R chooses No Hire if No College, Hire if Collegewith probability p 6 1/2. That is, the set of BNE is given by

{(C,NC;NH, H), (NC, NC;NH, pH + (1− p)NH)} for p ∈ [0, 1/2]

We find that there cannot be an equilibrium where the Dumb type of S goes to College. Ifthere were one, then R should be choosing Hire following No College with a higher probabilitythan following College. But then the Smart type would choose No College, and R would notbe best responding. Intuitively, since the Dumb type prefers not to be hired and the Smarttype prefers to be hired, either they will separate as in the first BNE we found, or they willpool in No College if the probability that they will be hired if they choose College is lowenough.

(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 inthe 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 whenhe observes College, he will choose No Hire if College with probability at least 1/2.

(c) What are all of the PBE meeting Signaling Refinement A in Stay in School 5? (This is theweaker of the two signaling refinements, that relies on restricting beliefs based on the sendingof some messages by some types being dominated.) Give a brief intuition for why the set ofPBE 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 thetypes. The Smart type prefers College to No College if R chooses No Hire if No College, Hireif College, but he prefers No College to College if R chooses Hire if No College, No Hire ifCollege. The Dumb type prefers College to No College if R chooses Hire if No College, NoHire if College, but he prefers No College to College if R chooses No Hire if No College, Hireif College.

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(d) What are all of the PBE meeting Signaling Refinement B in Stay in School 5? (This isthe stronger of the two signaling refinements, equivalent here to Cho-Kreps’s “Intuitive Crite-rion,” restricting beliefs based on the sending of some messages by some types being dominatedby their equilibrium payoffs.) Give a brief intuition for why the set of PBE meeting SignalingRefinement 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 messagesin this equilibrium. The pooling PBE, however, do not survive this refinement. Note that inthe pooling equilibria, the Smart type obtains a payoff of 3, while he could obtain a payoff of6 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 deviateto College no matter what R believes and chooses following College. Hence, this refinementsays that R should believe that he is facing the Smart type if he observes College. But thenR 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 6: Stay in School 6 and 7 (15 Points)

Consider the game Stay in School 6: 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 CollegeAnyhow? Act of 2012, 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 theSmart type).Sender: 1. NC,NC 2. NC,C 3. C,NC 4.C,CReceiver: 1. H 2. NH

(b) Find all the Bayesian Nash equilibria in Stay in School 6.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-informationstructure, actions, etc. as Stay in School 5, but: The game is being played after the OkayWe Suppose People Can Say They’ve Gone To College If They Want Act of 2022, whichleaves it illegal for firms to demand educational information in hiring, but allows workersto prove their educational status if they want. Hence, each type of potential worker has fourpossible messages she can send: “College/Reveal,” which involves going to college and provingshe’s gone to college; “No College/Reveal” which involves not going to college and provingshe has not gone to college; “College/Hide,” which involves going to college and refusing togive evidence of this fact; and “No College/Hide” which involves not going to college andrefusing to give evidence of this fact. The Receiver cannot tell the difference between thelast two messages — she observes essentially “Hide” when either is chosen. So the Receivermight observe any of three potential “messages” before choosing hire or no hire. Assume thatneither player’s payoffs depend directly on whether college/no college is revealed, but only onthe 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 sequentialgame we’ve looked at in the course). But it is a close cousin.

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(c) Try drawing the game tree (with payoffs and information sets) in the natural way. It won’tbe 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 adotted line the arrows corresponding to NCH of Smart, CH of Smart, NCH of Dumb, andCH of Dumb (where H denotes Hide), indicating that R cannot observe the Sender’s type norwhether the Sender went to College when the Sender chooses to Hide. Then from the fourdecisions nodes of R contained in this information set, draw the two lines corresponding toHire and No Hire. Then draw another information set connecting the arrows correspondingto CR of Smart and CR of Dumb (where R denotes Reveal), and again draw the Hire andNo 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 listthem; just give the number.There are two types of Sender and each type has four possible (pure) strategies, so the Senderhas a total of 16 (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 theReceiver has a total of 8 (pure) strategies available.

(f) Find all the pure-strategy Nash equilibria that meet the strongest refinement, “SignalingRefinement 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 wasnon-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 ifboth 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 Collegeand is hired and Dumb chooses No College and is not hired. By specifying that R chooses NoHire if Hide, this constitutes a PBE meeting SRB (note that only Smart could have incentivesto deviate to Hide if R played Hire if Hide, but he would deviate to No College/Hide, in whichcase No Hire is R’s best response.) Next, note that if only Dumb chooses Hide, then again thetypes separate, and the separating equilibrium we just found would also work. The separatingequilibrium would not work, however, if only Smart chooses Hide. In that case, R cannot bechoosing Hire if Hide because Smart would deviate to No College/Hide and R would not bebest responding. Finally, note that if both types choose Hide, then the game is as Stay inSchool 6 “on the equilibrium path”. But there we found that both types would choose NoCollege and R would choose No Hire. But a pooling PBE would not satisfy SRB, as it wouldhave R choosing No Hire (with high probability) if College/Reveal, when SRB will require Rto 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 NoCollege/Reveal first, then following College/Reveal, and finally following Hide.

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