Static Games of Incomplete Information

1. Static Games of Incomplete Information .

2. Example 1 • Two firms: incumbent (player 1) & entrant (player 2) • Player 1 decides whether to build new plant; player 2 decides whether to enter • Player 1’s cost of building is 1.5 (wp 1-p1) or 3 (wp p1) • How is this game to be played? • It depends on 2’s belief about 1’s probability of building when cost is low (x), and 1’s belief of 2’s probability of entering (y) 2 2 1 1 1’s cost is high 1’s cost is low

3. Example 1 • Harsanyi’s Key Idea (1967-68): i. Player 1’s ‘type’ is determined by a prior move of nature ii. Transforms a game of incomplete info (2 doesn’t know 1’s cost) into one of imperfect info • Equilibrium: 1. Player 2 enters (y=1) if x<1/[2(1-p1], and stays out (y=0) if x>1/[2(1-p1], 2. Low cost player 1 builds (x=1) if y<1/2, and not (x=0) if y>1/2.

4. Bayesian equilibrium • There are i єI players • Players types are drawn from dist p(θ1, θ2,… θI,), where θiєΘi, Θi is finite and p(θ-i| θi) is i’s conditional probability about his rival’s types θ-i • Pure strategies for i are siєSi, and payoff is ui(s1, s2, …,sI, θI, θ2,…,θI,), • A Bayesian equilibrium for above game of incomplete information is a Nash equil of the ‘expanded game’ where each player i’s space of pure strategies is the set of maps from Θi to Si. Given strategy profile s(.), and s/i(.) є , s(.) is a Bayesian equilibrium if

5. Cournot competition with incomplete info • Let firm i’s profit be ui= qi(θi- qi- qj), where θi=α-ci, α being the intercept of the linear demand function • Let θ1=1, and θ2=3/4 w.p. ½ and 5/4 w.p. ½ • Denote q2as q2H if θ2=3/4 , and as q2L if θ2=5/4 • Firm 2’s equil choice must satisfy: q2(θ2)=(θ2- q1)/2 • Firm 1 does not know 2’s type, so its expected payoff is: • This gives q1=(2- q2L - q2H )/4 • Plugging in for q2(θ2) we get (q1=1/3, q2L =11/24, q2H = 5/24) • This is the Bayesian equilibrium

6. War of attrition • Each player i chooses a number siin [0, +∞], • Payoffs are: • i’s type θitakes valuesin[0, +∞], with distribution function P and density p • We look for pure-strategy Bayesian equil (s1(.), s2(.)) • For each θi,si(θi) must satisfy:

7. War of attrition • A. Key step: Look for monotonic strategies that are strictly increasing and continuous in a player’s type 1. To show that: If θi// > θi/implies si //> si /where si //=si (θi//) and si /=si (θi/) Proof: If θi/prefers si (θi/) to si (θi//) then A similar inequality obtains when θi//prefers si (θi//) to si (θi/) Subtracting the two inequalities gives Since, θi// > θi/it must be that si //> si / .

8. War of attrition 2. To show thatstrategies, si (θi) and sj (θj), are strictly increasing Proof: If not, there would be an “atom” for j at s>0, i.e. Pr(sj (θj)=s)>0 Then i assigns probability 0 to [s-є, s] Then any type of j planning to play at s is better-off playing at s-є Thus, no atom at s after all 3. To show that si (θi) is continuous in θi Proof: Similar to above B. Let Φi be inverse function of si : Φ-1i (θi)=si

9. War of Attrition C. Transforming variable of integration from θj to sj, D. The FOC: If si ≡si (θi), then i cannot benefit by playing si +dsi instead of si Cost is dsi if j plays above si +dsi which has probability 1-Pj(Φj(si +dsi )) Expected cost, to first order in dsi , is [1-Pj(Φj(si ))] dsi . Gain is θi= Φi(si) if j plays in [si, si +dsi ], i.e. if θiє [Φj(si ), Φj(si +dsi )] This has probability, pj(Φj(si ))Φ/j(si ) dsi . Equating costs and benefits, the FOC is Φi(si) pj(Φj(si ))Φ/j(si ) = 1-Pj(Φj(si ))

10. War of Attrition • Impose symmetry P1= P2=P • Substitute θ= Φ(s), and use Φ/=1/s/ to get, • Integrating, • If P(.) is exponential, P(θ)=1-exp(-θ), then, S(θ)= θ2/2

11. Double auction • A seller and buyer trade a unit of good • Seller (player 1) has cost c and buyer (player 2) has valuation v . v, cє[0, 1] • Players simultaneously bid b1, b2 є[0, 1] • If b1≤ b2 , they trade at price t= (b1+b2)/2 • With trade 1 gets u1=(b1+b2)/2-c; 2 getsu2=v-(b1+b2)/2. Without trade both get 0 • c distributed as P1 and v distributed as P2 • Let F1(.) and F2(.) be cumulative dist of b1 and b2 • Find the Bayes-Nash equilibrium (s1(.), s2(.)), where si(.):[0, 1]→[0, 1]

12. Double auction • To show that: Bids increase in type. That is, c// > c/implies b1//>b1/where b1 //=s1(c//) and b1/=s1(c /) Proof: Optimization by the seller requires and Combining these inequalities, Since, c// > c/it must be that b1//>b1/ . • The bids are strictly increasing and continuous in types • Similar things hold for the buyer

13. Double auction • Maximization problem for type-c seller • The FOC is: ½ [1-F2(s1(c))]-(s1(c)-c)f2(s1(c))=0 • For the buyer, • And FOC is: (v – s2(v)f1(s2(v))= ½ F1(s2(v))

14. Double auction • Specific case: - P1and P2are uniform dist on [0, 1], and strategies are linear in types s1(c)= α1+β1c and s2(v)= α2+β2v - Then, Fi(b)=Pi(s-1i(b))= s-1i(b)=(b- αi)/βi, so fi(b)=1/ βi, - Plugging into FOCs, we get 2[α1+(β1-1)c]/β2 = [β2 – (α1+β1c)+ α2]/β2 2[(1-β2)v –α2]/β1 = [α2+β2v- α1]/β1 - Solving this system, β1= β2= 2/3;α1=1/4; α2=1/12 - In equilibrium parties trade only if, α2+β2v ≥ α1+β1c - Thus trade occurs only if, v ≥c+1/4 • Too little trading in equilibrium!!

15. First price auction with a continuum of types • Two bidders with a unit of good to trade • Player i’s valuation is θi and belongs to • Players have beliefs P, with density p, about rival’s valuation • Seller imposes reservation price s0> • Player i bids si; gets ui = θi - si if si > sj & ui =0, si < sj • If si = sj both get good w.p. ½ and ui = (θi - si )/2 • Let si (.) be the pure strategy of player i

16. First price auction (continuum of types) • Show that strategies are monotonic, strictly increasing, and continuous in type • To show that: Proof: If then type of player i could slightly lower his bid and still win w.p. 1 • Let Φi be inverse function of si (.):Φ-1i (θi)=son [s0, ], i.e. player i bids s if his valuation is Φi(s) • Type θi maximizes (θi - s)P(Φj(s)) over s • This gives, P(Φj(s)) = [Φi(s)-s]p(Φj(s)) Φj/(s) • There is a similar FOC by switching i and j

17. First price auction (continuum of types) • To show that: There cannot be an asymmetric solution, Φ1(s) ≠ Φ2(s) for all s • Using Φ1 = Φ2= Φ, in FOC, and integrating, we get, • This will give Φ(.), and the inverse function gives s(.)

18. First price auction (with two types) • Each bidder can have types , with < • Corresponding probabilities are and • Seller’s reservation bid is lower than • Key idea: Look for mixed-strategy equilibrium • Type bids , and type randomizes according to a continuous distribution F(s) on • Argue that: • For i of type to play a mixed strategy with support it must be that

19. First price auction (with two types) • Because F( )=0, the constant is • Thus, F(.) is given by • Let G(s)≡ be distribution of bids. Above can be written as • Since F( )=1, implies • Each bidders net utility is 0 when his type is and when his type is

20. Bayesian equil can justify mixed equil • Harsanyi, 1973: A mixed strategy equil of a complete info game can be interpreted as the limit of pure strategy equil of perturbed games of incomplete info • Example: “Grab the Dollar”- complete info version - At times t=0, 1, 2…, two players want to grab a $1 - If only one grabs, he gets 1 and other gets 0 - If both grab at once, dollar destroyed & each gets -1 - If neither grabs, both get 0 - Players have a common discount factor δ - The only symmetric strategy is a mixed strategy, where both grab w.p. p*=1/2 in each period

21. Bayesian equil can justify mixed equil • Example: “Grab the dollar”- complete info version -Consider player 1. By grabbing at time t, he gets δt(1- p*)+ δtp*(-1). By not grabbing, he gets 0. He is indifferent, so δt(1- p*)+ δtp*(-1)=0, and so, p*=1/2 • Consider ‘perturbed’ version of the above game • Example: “Grab the dollar”- incomplete info version -If player i wins, he gets 1+θi, θi is uniform on [-є, є] -Consider symmetric strategy: “si(θi<0)=do not grab; si(θi≥0)=grab” -This is a pure strategy Bayesian equilibrium! Why? -What happens when є→0?

22. Bayesian & mixed equil: 1st price auctions 1. Consider FOC for first-price auctions with continuum of types: P(Φj(s)) = [Φi(s)-s]p(Φj(s))Φj/(s) Let Gj(s)= P(Φj(s)) be dist of bids s. gj(s)=p(Φj(s))Φj/(s) Then FOC becomes: Gj(s)= [Φi(s)-s] gj(s), s> 2. Equivalent condition for two-type case: Differentiating w.r.t s, Consider sequence, Pn(θ), s.t. If Φn(.) is equil strategy for Pn(.), then