Extensive Game with
Imperfect Information III
Topic One:
Costly Signaling Game
Spence’s education game
• Players: worker (1) and firm (2)
• 1 has two types: high ability  H with probability p
H and low ability  L with probability p L .
• The two types of worker choose education level
e H and e L (messages).
• The firm also choose a wage w equal to the
expectation of the ability
• The worker’s payoff is w – e/
Pooling equilibrium
• e H = e L = e*  L pH (H - L)
• w* = pHH + pLL
• Belief: he who chooses a different e is thought
with probability one as a low type
• Then no type will find it beneficial to deviate.
• Hence, a continuum of perfect Bayesian
equilibria
Proof
The "best" deviation is to choose no education.
In this case, the worker gets a wage w '   L .
The low type worker does not have incentive to deviate
 w'  w*
e*
L
  p   p 
L

e*
L
H
H
L
L

e*
L
 p H  H  1  p L   L
 e*   L p H  H   L 
When low type does not have incentive to deviate, so does high type.
QED
Separating equilibrium
•
•
•
•
e L= 0
H (H - L) ≥ e H ≥ L (H - L)
w H = H and w L = L
Belief: he who chooses a different e is thought
with probability one as a low type
• Again, a continuum of perfect Bayesian
equilibria
• Remark: all these (pooling and separating)
perfect Bayesian equilibria are sequential
equilibria as well.
Proof
Low type does not have incentive to deviate to any e  e H
Low type does not have incentive to deviate to e H
 wH 
eH
L
 wL
  L  wH  wL   eH
High type does not have incentive to deviate to any e  0
High type does not have incentive to deviate to e  0
w w 
L
H
eH
H
 eH   H  wH  wL 
QED
The most efficient separating
equilibrium
L type equilibrium payoff
w
H type equilibrium payoff
w
Increase in payoff
w
e
L

H
w 
eH
H
H
 wL
H type payoff
by choosing
e=0
wH
w
wL
eL=0
e
e

H
 wL
e
eH
When does signaling work?
• The signal is costly
• Single crossing condition holds (i.e., signal is
more costly for the low-type than for the
high-type)
Topic Two:
Kreps-Cho Intuitive Criterion
Refinement of sequential
equilibrium
• There are too many sequential
equilibria in the education game. Are
some more appealing than others?
• Cho-Kreps intuitive criterion
– A refinement of sequential equilibrium—
not every sequential equilibrium
satisfies this criterion
•
An example where a sequential
equilibrium is unreasonable
(slided deleted)
Two sequential equilibria with
outcomes: (R,R) and (L,L),
respectively
• (L,L) is supported by belief
that, in case 2’s information set
is reached, with high
probability 1 chose M.
• If 2’s information set is
reached, 2 may think “since M
is strictly dominated by L, it is
not rational for 1 to choose M
and hence 1 must have
chosen R.”
1
L
2,2
R
M
L
1,3
2
2
R
0,0
L
0,0
R
5,1
Beer or Quiche (Slide deleted)
1,1
1,0
N
F
Q
1
B
N
3,1
3,0
2
0.9
strong
Q
1
weak
2
F
0,0
F
N
0.1
c
0,1
B
N
F
1,0
1,1
Why the second equilibrium is
not reasonable? (slide deleted)
• If player 1 is weak she should
realize that the choice for B is
worse for her than following the
equilibrium, whatever the
response of player 2.
• If player 1 is strong and if player 2
correctly concludes from player 1
choosing B that she is strong and
hence chooses N, then player 1 is
indeed better than she is in the
equilibrium.
• Hence player 2’s belief is
unreasonable and the equilibrium
is not appealing under scrutiny.
1,1
1,0
N
F
Q
1
B
3,0
2
0.9 c
strong
F
3,1
0,0
F
N
0.1
weak
2
N
0,1
Q
1
B
N
F
1,0
1,1
Cho-Kreps Intuitive Criterion
• Consider a signaling game. Consider a sequential
equilibrium (β,μ). We call an action that will not reach in
equilibrium as an out-of-equilibrium action (denoted by
a).
• (β,μ) is said to violate the Cho-Kreps Intuitive Criterion if:
– there exists some out-of-equilibrium action a so that one type,
say θ*, can gain by deviating to this action when the receiver
interprets her type correctly, while every other type cannot gain
by deviating to this action even if the receiver interprets her as
type θ*.
• (β,μ) is said to satisfy the Cho-Kreps Intuitive Criterion if
it does not violate it.
Spence’s education game
• Only one separating equilibrium survives the ChoKreps Intuitive criterion, namely: e L = 0 and
e H = L (H - L)
• Any separating equilibrium where e L = 0 and
e H > L (H - L) does not satisfy Cho-Kreps intuitive
criterion.
• A high type worker after choosing an e slightly smaller
will benefit from it if she is correctly construed as a
high type.
• A low type worker cannot benefit from it however.
• Hence, this separating equilibrium does not survive
Cho-Kreps intuitive criterion.
The most efficient separating
equilibrium
L type equilibrium payoff
w
H type equilibrium payoff
wH
wL
eL=0
e
eH
Inefficient separating equilibrium
L type worse off by
deviating to e# if
believed to be High
type
H type better off by
deviating to e# if believed
to be High type
w
L type
equilibrium
payoff
H type equilibrium payoff
wH’
wL
eL=0
e
eH’
e#
eH
Spence’s education game
• All the pooling equilibria are eliminated by the Cho-Kreps
intuitive criterion.
• Let e satisfy w* – e*/ L > H – e/ L and w* – e*/ H > H
– e/ L (such a value of e clearly exists.)
• If a high type work deviates and chooses e and is
correctly viewed as a good type, then she is better off
than under the pooling equilibrium
• If a low type work deviates and successfully convinces
the firm that she is a high type, still she is worse off than
under the pooling equilibrium.
• Hence, according to the intuitive criterion, the firm’s
belief upon such a deviation should be such that the
deviator is a high type rather than a low type.
• The pooling equilibrium break down!
Topic Three:
Cheap Talk Game
Cheap Talk Model
Two players: S (sender) and R (receiver)
S's type  is uniformly distributed in the unit interval [0,1]
In the first stage: S chooses a message m  M  [0,1]
In the second stage: R chooses an action a  A  [0,1]
uS ( , a )  ( a  (  b)) 2 where b  0
uR ( , a )  ( a   ) 2
If S is not allowed to send message, R will choose a  1/ 2.
Does cheap talk matter?
Perfect Information Transmission?
• An equilibrium in which each type will
report honestly does not exist unless b=0.
No information transmission
• There always exists an equilibrium in
which no useful information is transmitted.
• The receiver regards every message from
the sender as useless, uninformative.
• The sender simply utters uninformative
messages.
Some information transmission
There exists a perfect Bayesian equilibrium as follows:
Partition of the unit interval:
[0, x1 ),[ x1 , x2 ),...,[ xk 1 , xk ),...,[ xK 1, xK  1]
Types of S who are in the same segment give the same message
(wlog, m=segment's lower bound)
Types in different segments give different messages
R always choose an action (xk  xk 1 ) / 2 when message xk 1 is received
R holds the belief that: S has a type equally likely in the segment
in which the message is in.
Some information transmission
Consider the incentive of the type   xk 1
If she reports xk 1 , she gets   ( xk  xk 1 ) / 2  b  .
2
If she reports a message in the preceding segment,
she gets   ( xk 2  xk 1 ) / 2  b  .
2
The two must be the same
xk  xk 1
 xk 2  xk 1


 b  
 b
2
2


  xk  xk 1    xk 1  xk 2   4b
each segment exceeds the preceding segment by 4b
Some Information Transmission
d  ( d  4b)  ...  ( d  4( K  1)b)  Kd  2 K ( K  1)b  1
Let K * (b) be the largest integer satisfying 1- 2 K ( K -1)b  0
1- 2 K ( K -1)b
K
and a unique partition [0, d ),[d , 4b),...,[ d  (4 K -1)b,1]
For any K  K * (b), determine a unique number d 
Remark: K * (b) is increasing in b
Final Remark:
• Relationship among different equilibrium
concepts:
• Sequential equilibrium satisfying Chokreps => sequential equilibrium => Perfect
Bayesian equilibrium => subgame perfect
equilibrium => Nash equilibrium