Static Games of Incomplete Information

.

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)

Enter Don’t enter

Build 0, -1 2, 0

Don’t build

2, 1 3, 0

Enter Don’t enter

Build 1.5, -1 3.5, 0

Don’t build

2, 1 3, 0

1’s cost is high 1’s cost is low

21

21

Example 1

• Harsanyi’s Key Idea (1967-68): i. Player 1’s ‘type’ is determined by a prior move of

natureii. 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.

Bayesian equilibrium• There are i єI players

• Players types are drawn from dist p(θ1, θ2,… θI,), where θi єΘi, 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 si єSi, 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

Ii 1}{

iiS

iiS

)),(),(),((),(maxarg(.) /

(.)/

i i

iii

iiiiiiiiiSs

i ssups

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 q2 as q2H if θ2=3/4 , and as q2

L 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 - q2

H )/4

• Plugging in for q2(θ2) we get (q1=1/3, q2L =11/24, q2

H = 5/24)

• This is the Bayesian equilibrium

)}1(2

1)1(

2

1{maxarg 2112111

1

LH

q

qqqqqqq

War of attrition

• Each player i chooses a number si in [0, +∞],

• Payoffs are:

• i’s type θi takes values in [0, +∞], with distribution function P and density p

• We look for pure-strategy Bayesian equil (s1(.), s2(.))

• For each θi, si(θi ) must satisfy:

ijji

iji

i sss

sssu

if

if

})({

})())(())(Pr({maxarg)(ijjj

i ss

jjjjjiijjis

ii dpsssss

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

/ .

})({

///////

})({

////

//

/

)()())(Pr())(Pr(

)()())(Pr())(Pr(

ijjj

ijjj

ss

jjjjjijjiijji

ss

jjjjjijjiijji

dpssssss

dpssssss

0)])(Pr())()[Pr(( ////// ijjijjii ssss

War of attrition

2. To show that strategies, si (θi) and sj (θj), are strictly increasingProof: If not, there would be an “atom” for j at s>0, i.e. Pr(sj (θj)=s)>0Then i assigns probability 0 to [s-є, s]Then any type of j planning to play at s is better-off playing at s-єThus, 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

War of AttritionC. 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

θi є [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 ))

i

i

s

jjjjjjjiijjis

ii dssspssPss0

/ })())(()())((1({maxarg)(

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

)(1

)()(/

P

ps

dxxP

xxps

0 )(1

)()(

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, c є[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 gets u2=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]

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

)()2

()()2

( 22/1

2//

122

/12

/1

//1

/1

bdFcbb

bdFcbb

bb

)()2

()()2

( 22//1

2/

122

//12

//1

/1

//1

bdFcbb

bdFcbb

bb

0)]()()[( /12

//12

/// bFbFcc

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))

)()2

(max 22

121

11

bdFcbb

bb

)()2

(max 110

212

2

bdFbb

vb

b

Double auction• Specific case:

- P1 and P2 are 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-1

i(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!!

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

] ,[

First price auction (continuum of types)

A. Show that strategies are monotonic, strictly increasing, and continuous in type

B. To show that:

Proof: If then type of player i could slightly lower his bid and still win w.p. 1

C. Let Φi be inverse function of si (.): Φ-1i (θi)= s on

[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

sss ji )()(

)()( ji ss

s

First price auction (continuum of types)

D. To show that: There cannot be an asymmetric solution, Φ1(s) ≠ Φ2(s) for all s

E. Using Φ1 = Φ2 = Φ, in FOC, and integrating, we get,

F. This will give Φ(.), and the inverse function gives s(.)

s

s xx

dxsP

)())]((ln[

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

p p

] ,[ ss

s

] ,[ ss constant )]()[( ], ,[ sFppssss

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

p)(

psFpps )( )]()[(

s pps

)( p

)(sFpp

psGs )( )()(

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

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*)+ δt p*(-1). By not grabbing, he gets 0. He is indifferent, so δt(1- p*)+ δt p*(-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?

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

psGs )( )()(

)()( )( sgssG

,1

),[ ,

,0

)(lim pPnn

s ,)(lim sn

n