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)
2
Enter
1
Don’t
enter
2
Enter
Don’t
enter
1
Build
0, -1
2, 0
Build
1.5, -1
3.5, 0
Don’t
build
2, 1
3, 0
Don’t
build
2, 1
3, 0
1’s cost is high
1’s cost is low
• 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)
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.
Bayesian equilibrium
• There are i єI players
I
• Players types { }i 1 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

S
set i of maps from Θi to Si. Given strategy profile
s(.), and s/i(.) є Si , s(.) is a Bayesian equilibrium if
i
i
si (.)  arg max  p(i , i )ui ( si/ (i ), si ( i ), (i , i ))
si/ (.)Sii
i
 i
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 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:
1
1
H
q1  arg max { q1 (1  q1  q2 )  q1 (1  q1  q2L )}
2
2
q1
• 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
War of attrition
• Each player i chooses a number si in [0, +∞],
• Payoffs are:
 si
ui  
 i  s j
if s j  si
if s j  si
• 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:
si ( i )  arg max { si Pr( s j ( j )  si ) 
si
 (
i
 s j ( j )) p j ( j )d j }
{ j s j ( j )  si }
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
s
 i/ Pr( s j ( j )  si/ )  si/ Pr( s j ( j )  si/ ) 
j
( j ) p j ( j ) d j 
{ j s j ( j )  si/ }
 i/ Pr( s j ( j )  si// )  si// Pr( s j ( j )  si// ) 
s
j
( j ) p j ( j ) d j
{ j s j ( j )  si// }
A similar inequality obtains when θi// prefers si (θi //) to si (θi /)
Subtracting the two inequalities gives
(i//  i/ )[Pr( s j ( j )  si// )  Pr( s j ( j )  si/ )]  0
Since, θi// > θi/ it must be that si //> si / .
War of attrition
2. To show that strategies, 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
War of Attrition
C. Transforming variable of integration from θj to sj,
si
si (i )  arg max{si (1  Pj ( j ( si ))   (i  s j ) p j ( j ( s j )) /j ( s j )ds j }
si
0
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 ))
War of Attrition
• Impose symmetry P1= P2=P
• Substitute θ= Φ(s), and use Φ/=1/s/ to get,
p ( )
s ( ) 
1  P( )
/
• Integrating,

xp( x)
s( )  
dx
0 1  P( x)
• If P(.) is exponential, P(θ)=1-exp(-θ), then,
S(θ)= θ2/2
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
//
1 b b
b1/  b2 /
/
1
2
(

c
)
dF
(
b
)

(

c
)dF2 (b2 )
2
2
b1/ 2
b1// 2
1
and
/
1 b b
b1//  b2 //
//
1
2
(

c
)
dF
(
b
)

(

c
)dF2 (b2 )
2 2
b1// 2
b1/ 2
1
//
/
//
/
Combining these inequalities, (c  c )[ F2 (b1 )  F2 (b1 )]  0
Since, c// > c/ it must be that b1//>b1/ .
• The bids are strictly increasing and continuous in types
• Similar things hold for the buyer
Double auction
• Maximization problem for type-c seller
b1  b2
max  (
 c)dF2 (b2 )
b1
b1
2
1
• The FOC is: ½ [1-F2(s1(c))]-(s1(c)-c)f2(s1(c))=0
• For the buyer,
b2
max  (v 
b2
0
b1  b2
)dF1 (b1 )
2
• And FOC is: (v – s2(v)f1(s2(v))= ½ F1(s2(v))
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-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!!
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: si ( )  s j ( )  s
Proof: If si ( )  s j ( ) 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, s ], i.e. player i bids s if his valuation is Φi(s)
D. Type θi maximizes (θi - s)P(Φj(s)) over s
E. This gives, P(Φj(s)) = [Φi(s)-s]p(Φj(s)) Φj/(s)
F. There is a similar FOC by switching i and j
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,
s
dx
ln[ P(( s))]   
 ( x)  x
s
F. This will give Φ(.), and the inverse function gives
s(.)
First price auction (with two types)
• Each bidder can have types  ,  with  < 
• Corresponding probabilities are p and p
• 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 [ s, s ]
• Argue that: s  
• For i of type  to play a mixed strategy with support
[ s , s ] it must be that s [ s, s], (  s)[ p  pF ( s)]  constant
First price auction (with two types)
• Because F(  )=0, the constant is (   ) p
• Thus, F(.) is given by (  s)[ p  pF (s)]  (   ) p
• Let G(s)≡ p  pF (s) be distribution of bids. Above can
be written as (  s)G(s)  (   ) p
• Since F( s )=1, implies s  p  p
• Each bidders net utility is 0 when his type is  and
p(   ) when his type is 
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: (  s)G(s)  (   ) p
Differentiating w.r.t s, G ( s)  (  s) g ( s)
0,

n
P ( )   p,
Consider sequence, Pn(θ), s.t. lim
n 

1,
 
  [ ,  )
 
 n ( s)   , s  
If Φn(.) is equil strategy for Pn(.), then lim
n 