Games of Incomplete Information
(Bayesian Games)
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In a game of complete information the players’
payoff functions are common knowledge
In a game of incomplete information, at least one
player is uncertain about another player’s payoff
function
Cournot competition with incomplete
information
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The market price, P is determined by (inverse) market
demand:
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P=a-Q if a>Q, P=0 otherwise.
Each firm decides on the quantity to sell (market share): q1
and q2
Q= q1+q2 total market demand
Both firms seek to maximize profits
The marginal cost of producing each unit of the good:
c1 and c2
c1 is common knowledge, however c2 is known only by firm 2
Firm 1 believes that c2 is “high” cH with probability p and
“low” cL with probability (1-p)
Firm 1’s belief about firm 2’s cost is common knowledge
1
Cournot Competition with Incomplete
Information: Best response of Firm 2
Suppose firm 1 produces q1
! Firm 2’s profits, if it produces q2 are:
– c2q2
π2 = (P-c2)q2 = [a-(q1+ q2)]q2
= (Residual) revenue – Cost
! First order conditions:
d π2/dq2= a - 2q2 – q1 – c2 =
=
RMR
– MC = 0 →
q2=(a-c2- q1)/2
If Firm 2’s type is “high”: qH2=(a-cH- q1)/2
If Firm 2’s type is “low”: qL2=(a-cL- q1)/2
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Cournot Competition with incomplete
information: Best response of Firm 1
Firm 1’s expected profits, if it produces q1 are:
π1 = (P-c1)q1 = p[a-(q1+ qH2)]q1 +
(1-p)[a-(q1+ qL2)]q1 - c1q1
! First order conditions (FOC):
d π1/dq1=p(a - 2q1 - qH2) – (1-p)( a - 2q1 - qL2)- c1
=0→
q1= p(a- c1 -qH2)/2 + (1-p)(a - c1-qL2)/2
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Cournot with complete information: q1=(a - c1– q2)/2
2
Cournot equilibrium under incomplete
information
If Firm 2’s type is “high”: qH2=(a-cH-q1)/2
If Firm 2’s type is “low”: qL2=(a-cL-q1)/2
q1= p(a- c1 -qH2)/2 + (1-p)(a - c1-qL2)/2
q1= (a - 2c1+ p cH+(1-p) cL)/3
qH2= (a-2cH+c1)/3 + (1-p)(cH- cL)/6
qL2= (a-2cL+c1)/3 - p(cH- cL)/6
Comparison with the Cournot equilibrium
under complete information
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Complete information
qC1=(a-2c1+c2)/3
qC2=(a-2c2+c1)/3
q’H= (a-2cH+c1)/3
q’L= (a-2cL+c1)/3
Incomplete information
q1= (a - 2c1+ p cH+(1-p) cL]/3
qH2= (a-2cH+c1)/3 + (1-p)(cH- cL)/6
qL2= (a-2cL+c1)/3 - p(cH- cL)/6
qH2 > q’H
and
qL2 < q’L.
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