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Signaling Econ 171

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**Breakfast: Beer or quiche? A Fable ***

*The original Fabulists are game theorists, David Kreps and In-Koo Cho

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**Breakfast and the bully**

A new kid moves to town. Other kids don’t know if he is tough or weak. Class bully likes to beat up weak kids, but doesn’t like to fight tough kids. Bully gets to see what new kid eats for breakfast. New kid can choose either beer or quiche.

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**Preferences Tough kids get utility of 1 from beer and 0 from quiche.**

Weak kids get utility of 1 from quiche and 0 from beer. Bully gets payoff of 1 from fighting a weak kid, -1 from fighting a tough kid, and 0 from not fighting. New kid’s total utility is his utility from breakfast minus w if the bully fights him and he is weak and utility from breakfast plus s if he is strong and bully fights him.

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Nature Tough Weak New Kid New Kid Beer Quiche Beer Bully Quiche Fight Don’t Don’t Fight B Bully 1 1+s -1 Fight -w 1 Don’t Don’t Fight 1 s -1 1-w 1

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**How many possible strategies are there for the bully?**

2 4 6 8

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**What are the possible strategies for bully?**

Fight if quiche, Fight if beer Fight if quiche, Don’t if beer Fight if beer, Don’t if quiche Don’t if beer, Don’t if quiche

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**What are possible strategies for New Kid**

Beer if tough, Beer if weak Beer if tough, Quiche if weak Quiche if tough, Beer if weak Quiche if tough, Quiche if weak

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**Separating equilibrium?**

Is there an equilibrium where Bully uses the strategy Fight if the New Kid has Quiche and Don’t if the new kid has Beer. And the new kid has Quiche if he is weak and Beer if he is strong. For what values of w could this be an equilibrium?

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Best responses? If bully will fight quiche eaters and not beer drinkers: weak kid will get payoff of 0 if he has beer, and 1-w if he has quiche. So weak kid will have quiche if w<1. Tough kid will get payoff of 1 if he has beer and s if he has quiche. So tough kid will have quiche if s>1

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**Suppose w<1 and s<1**

We see that if Bully fights quiche eaters and not beer drinkers, the best responses are for the new kid to have quiche if he is weak and beer if he is strong. If this is the new kid’s strategy, it is a best response for Bully to fight quiche eaters and not beer drinkers. So the outcome where Bully uses strategy “Fight if quiche, Don’t if beer “ and where New Kid uses strategy “Quiche if weak, Beer if tough” is a Nash equilibrium.

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If w>1 Then if Bully uses strategy “Fight if quiche, Don’t if beer”, what will New Kid have for breakfast if he is weak?

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Pooling equilibrium? If $w>1$, is there an equilibrium in which the New Kid chooses to have beer for breakfast, whether or not he is weak. If everybody has beer for breakfast, what will the Bully do? Expected payoff from Fight if beer, Don’t if quiche depends on his belief about the probability that New Kid is tough or weak.

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**Payoff to Bully Let p be probability that new kid is tough.**

If new kid always drinks beer and bully doesn’t fight beer-drinking new kids, his expected payoff is 0. If Bully chooses strategy Fight if Beer, Don’t if Quiche, his expected payoff will be -1/2p+1/2(1-p)=1-2p. Now 1-2p<0 if p>1/2. So if p>1/2, “Don’t fight if Beer, fight if Quiche” is a best response for Bully.

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Pooling equilibrium If p>1/2, there is a pooling equilibrium in which the New Kid has beer even if he is weak and prefers quiche, because that way he can conceal the fact that he is weak from the Bully. If p>1/2, a best response for Bully is to fight the New Kid if he has quiche and not fight him if he has beer.

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**What if p<1/2 and w>1?**

There won’t be a pure strategy equilibrium. There will be a mixed strategy equilibrium in which a weak New Kid plays a mixed strategy that makes the Bully willing to use a mixed strategy when encountering a beer drinker.

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What if s>1? Then tough New Kid would rather fight get in a fight with the Bully than have his favorite breakfast. It would no longer be Nash equilibrium for Bully to fight quiche eaters and not beer drinkers, because best response for tough New Kid would be to eat quiche.

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**Wyatt Earp and the Gun Slinger**

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**A Bayesian gunslinger game**

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**The gunfight game when the stranger is (a) a gunslinger or (b) a cowpoke**

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**What are the strategies?**

Earp Draw Wait Stranger Draw if Gunslinger, Draw if Cowpoke Draw if Gunslinger, Wait if Cowpoke Wait if Gunslinger, Draw if Cowpoke Wait if Gunslinger, Wait if Cowpoke

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**One Bayes Nash equilibrium**

Suppose that Earp waits and the other guy draws if he is a gunslinger, waits if he is a cowpoke. Stranger in either case is doing a best response. If stranger follows this rule, is waiting best for Earp? Earp’s Payoff from waiting is 3/4x1+1/4x8=2.75 Earp’s Payoff from drawing, given these strategies for the other guys is (¾)2+(1/4) 4=2.5 So this is a Bayes Nash equilibrium

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**There is another equilibrium**

Lets see if there is an equilibrium where everybody draws.

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**Alice and Bob Revisited**

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**She loves me, she loves me not? (Bob moves before Alice)**

Nature She loves him She scorns him Bob Bob Go to A Go to B Go to A Go to B Alice Alice Alice Alice Go to B Go to B Go to A Go to A Go to B Go to A Go to A Go to B 1 3 2 2 3 3 2 1 2 1 3

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**What are their strategies?**

For Bob Go to A Go to B Alice has four information sets. I Love him and he’s at A I Love him and he’s at B I Scorn him and he’s at A I Scorn him and he’s at B In each information set, she can go to either A or B. This gives her 2x2x2x2=16 possible strategies.

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**A weakly dominant strategy for Alice**

Go to A if you love he goes to A. Go to B if you love him and he goes B. Go to B if you scorn him and he goes to A. Go to A if you scorn him and he goes to B. (We write this as A/B/B/A) This is weakly dominant but not strictly dominant. Explain. Let’s look for a subgame perfect Nash equilibrium where Alice goes A/B/B/A

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Checking equilibrium Suppose Alice goes where Bob is if she loves him and goes where he is not if she scorns him (A/B/B/A) Payoff to Bob from A is 2p. Payoff from B is 3p+1(1-p)=2p+1. Since 2p+1>2p, for all p>=0, B is his best response to (A/B/B/A). Also A/B/B/A is a best response for Alice to Bob’s B. So we have a Bayes-Nash equilibrium.

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Doubly-blind?

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**Does she or doesn’t she? Simultaneous Play**

Nature She loves him She scorns him Bob Bob Go to A Go to B Go to A Go to B Alice Alice Alice Alice Go to B Go to B Go to A Go to A Go to B Go to A Go to A Go to B 1 3 2 2 3 3 2 1 2 1 3

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**Alice’s (pure) strategies**

Alice doesn’t know what Bob did, so she can’t make her action depend on his choice. She can go to either A or B. She does know whether she loves him or scorns him when she chooses. She has 4 possible strategies A if love, A if scorn A if love, B if scorn B if love, A if scorn B if love, B if scorn

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**Bayes’ Nash equilibrium**

Is there a Bayes’ Nash equilibrium where Bob goes to B and Alice goes to B if she loves Bob, and to A if she scorns him? This is a best response for both Alice types. What about Bob?

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Bob’s Calculations If Bob thinks the probability that Alice loves him is p and Alice will go to B if she loves him and A if she scorns him: His expected payoff from going to B is 3p+1(1-p)=1+2p. His expected payoff from going to A is 2(1-p)+0p=2-2p. Going to B is Bob’s best response to the strategies of the Alice types if 1+2p>=2-2p. Equivalently p>=1/4.

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**Is there a Bayes-Nash equilibrium in pure strategies if p<1/4?**

Yes, Alice goes to B if she loves Bob and A if she scorns him and Bob goes to B. Yes, Alice goes to A if she loves Bob and B if she scorns him and Bob goes to B. Yes there is one, where Alice always goes to A. No there is no Bayes-Nash equilibrium in pure strategies.

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If p<1/4 We showed there is no pure strategy Bayes-Nash equilibrium where Bob goes to B. What if Bob goes to A? The only possible Nash equilibrium would have Alice go to A if she loves him and B if she scorns him. Then payoff to Bob from A would be 2p+0 and payoff from B would be 3p+1(1-p)=2p+1>2p. So if p<1/4, there can’t be a pure strategy Bayes-Nash equilibrium where Bob goes to A.

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**Mixed strategy equilbrium: Bob the stalker**

If Bob thinks it likely that Alice scorns him, then if he uses a pure strategy, he knows Alice would always avoid him. If he uses a mixed strategy, he would catch her sometimes. Let’s look for a mixed strategy for Bob such that Alice, if she scorns Bob would be indifferent between Movies A and B.

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**What about a mixed strategy equilibrium?**

If p<1/4, can we find a mixed strategy for Bob such that Alice is indifferent What if Bob knows Alice scorns him? Consider the Alice type who scorns Bob. If Bob goes to movie A with probability q, When will Alice be indifferent between going to the two movies?

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**The game if Alice hates Bob**

1,2 3,1 2,0 0,3 Alice

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**Making Scornful Alice indifferent**

If Bob goes to Movie A with probability q and Alice Scorns Bob: Alice’s payoff from A is 1q+3(1-q) =3-2q Alice’s payoff from B is 2q+0(1-q)=2q Alice will be indifferent if 3-2q=2q, which implies q=3/4.

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**When will Bob do a mixed strategy?**

Note that if Bob goes to A with probability ¾, and if Alice loves him, her best response is to go to Movie A. If there is an equilibrium where Bob uses a mixed strategy, he must be indifferent between going to A and going to B. Can we find a mixed strategy for Alice to use if she scorns him so that Bob will be indifferent between A and B?

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**Making Bob indifferent**

Let r be the probability that Alice goes to Movie A if she scorns Bob and suppose that Alice always goes to A if she loves Bob. Expected payoffs for Bob are If he goes to A, 2p+(1-p)(2r+0(1-r))=2p-2pr+2r If he goes to B, 1p+(1-p)(1r+3(1-r))=2p+3+2pr-2r He is indifferent between A and B if these are equal. This implies r=(3-4p)/4-4p. Now r is between 0 and 1 if and only if p<=3/4

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Summing up We previously found that if p>=1/4, there is a Bayes-Nash equilibrium in which Bob goes to B and Alice goes to B if she loves him and A if she scorns him. Now we found that whenever p<=3/4, there is a mixed strategy Bayes-Nash equilibrium in which Bob goes to movie A with probability ¾, Alice goes to Movie A if she loves Bob and she goes to movie A with probability r=(3-4p)/(4-4p) if she scorns him. So over the range of p between ¼ and ¾, there are two distinct Bayes-Nash equilibria.

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Midterm next Tuesday A Piece of (If you are prepared…)

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