Lecture 19: Costly Signaling Vs. Cheap Talk – 15.025 Game Theory

Lecture 19: Costly Signaling Vs. Cheap Talk – 15.025 Game Theory

Part III: “Big” Applications Asymmetric Repetition Signaling Information Long-Run Auctions & Credibility & Relationships Market Design Reputation Classes 12-14 Classes 15-18 Classes 19-21 MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 1 Signaling Games How to Make Communication Credible Chapter 1 (Today): Costly Signaling MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 2 Signaling Examples • Entry deterrence: – Incumbent tries to signal its resolve to fight to deter entrants How can strong informed players distinguish themselves? • Credence Goods: – Used car warranties Can weak players signal-jam? Two of your class projects • Social interactions already using this… – Fashion (pitching a lemon, Estonia) MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 3 The beer & quiche model • A monopolist can be either a tough incumbent or a wimp (not tough). • Incumbent earns 4 if the entrant stays out. • Incumbent earns 2 if the entrant enters. • Entrant earns 2 if it enters against wimp. • Entrant earns -1 if it enters against tough. • Entrant gets 0 if it stays out. MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 4 Beer & Quiche • Prior to the entrant’s decision to enter or stay out, the incumbent gets to choose its “breakfast.” • The incumbent can have beer or quiche for breakfast. • Breakfasts are consumed in public. • Eating quiche “costs” 0. • Drinking beer costs differently according to type: – a beer breakfast costs a tough incumbent 1… – but costs a wimp incumbent 3. MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 5 What’s Beer? Toughness Beer Excess Capacity High Output Low Costs Low Prices Beat up Rivals & Deep Pockets Previous Entrants MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 6 Beer & Quiche: first model enter enter 1,-1 Incumbent 2,-1 beer quiche Pr = x Pr = y 3,0 4,0 out tough [½] out Entrant Nature Entrant enter wimp [½] enter -1,2 2,2 Pr = 1-x Pr = 1-y beer quiche Incumbent 4,0 1,0 out out MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 7 Signaling Equilibrium • Can the Incumbent credibly use Beer to signal Toughness? Consistency Checklist 1. Is the Entrant’s strategy optimal given her beliefs? 2. Is Incumbent’s strategy a best response to the Entrant’s strategy? 3. Are Entrant’s beliefs correct given Incumbent’s strategy? MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 8 Beer & Quiche enter enter 1,-1 Incumbent 2,-1 beer quiche Pr = 1 3,0 Pr = 0 4,0 out tough [½] out Entrant Nature Entrant enter wimp [½] enter -1,2 2,2 Pr = 0 Pr = 1 beer quiche Incumbent 4,0 1,0 out out MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 9 Separating Equilibrium • Tough drinks beer. • Wimp eats quiche. • Entrant infers the true type. • Degenerate beliefs (0 or 100%). • Entrant should ignore prior information… … and use strategic information. MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 10 Credible Signals • Why doesn’t Wimp drink beer & deter entry? • It’s too costly • This is the key feature of “credible signaling”! • What if the signal (beer) were a bit less costly? MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 11 Pooling Equilibrium • Suppose the wimp prefers “beer & out” to “quiche & enter” • The “beer signal” can’t work! • If both types drank beer, the entrant would face 50:50 odds, and enter! • Both types of incumbent are then better off w/quiche Cheap beer destroys signaling value • Pooling equilibrium: both eat quiche, entrant enters MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 12 Takeaways 1. Costly signals can be used more credibly: – Warranties are expensive for sellers of bad cars – What is fashion? – Extra capacity must hurt inefficient firms more 2. Delicate balance: – Cheap signals no persuasion – Expensive signals no profit MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 13 Poker call 200 , -200 Got Card raise check 100 , -100 100 , -100 fold good [1/3] No Card Nature call bad [2/3] -200 , 200 -100 , 100 raise check 100 , -100 Got Card fold MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 14 How Did You Play? MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 15 No Separating Equilibrium call 200 , -200 Got Card raise check 100 , -100 Pr = 1 100 , -100 fold good [1/3] No Card Nature call bad [2/3] -200 , 200 Pr = 0 -100 , 100 raise check 100 , -100 Got Card fold MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 16 No Pooling Equilibrium call 200 , -200 Got Card raise check 100 , -100 Pr = 1/3 100 , -100 fold good [1/3] No Card Nature call bad [2/3] -200 , 200 Pr = 2/3 -100 , 100 raise check 100 , -100 Got Card fold MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 17 Bluffing Game (poker) How would you play against a game theory classmate? 1. Can you expect No-Card to always Fold? 2. What about always Call? 3. Can you expect Got-Good-Card to always raise? … That’s a start! 4. Can you expect Got-Bad-Card to always raise? MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 18 Poker: Equilibrium call weakly dominant 200 , -200 Got Card raise check 100 , -100 Pr = x 100 , -100 fold good [1/3] Must randomize No Card Nature call bad [2/3] -200 , 200 Pr = 1-x -100 , 100 raise check 100 , -100 Got Card fold Must randomize MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 19 Poker: Equilibrium • Key property of Nash Equilibrium: “If a player randomizes in equilibrium, she must be indifferent between all the strategies she uses” • Expected payoffs must be equal • Otherwise, the player would choose the better strategy all (not some of) the time… MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 20 Poker: Equilibrium call 200 , -200 Got Card raise check 100 , -100 Pr = 3/4 100 , -100 fold good [1/3] No Card Nature call bad [2/3] -200 , 200 Pr = 1/4 -100 , 100 raise check 100 , -100 Pr = 1/6 Got Card Pr = 5/6 fold MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 21 Calculations 1. If No-Card randomizes it must be that E[u(call)] = E[u(fold)] = -100 200*Pr[bad | raise] – 200*Pr[good | raise] = -100 Pr[good | raise] = ¾ (no-card’s equilibrium belief) 2. We know Pr[raise | good] = 1; in order for Pr[good | raise] = (1/3)/(1/3+ Pr[raise | bad] 2/3) = ¾ it must be Pr[raise | bad] = 1/6. 3. If Bad-Card randomizes, it must be that E[u(raise)] = E[u(check)] = -100 -200*Pr[call]+100*Pr[fold] = -100 Pr[call]=2/3 MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 22 Let’s Compare MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 23 Takeaways 1. Information is valuable, even in zero-sum games 2. Costly signals can be used more credibly Next time • Cheap talk signals are less likely to be effective • What does this have to do with R&D in large Pharma? (aka, CEOs wish scientists were peacocks) MIT Sloan 15.025 S15 Prof. Alessandro Bonatti 24 MIT OpenCourseWare http://ocw.mit.edu 15.025 Game Theory for Strategic Advantage Spring 2015 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms..

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