List of Game Theory Toy Models
Zijing Hu
November 9, 2022
*This note is based on ECON 630: Microeconomic Theory II by Dr. Silvana Krasteva, TAMU.
Contents
1. Static Game with Complete Information
1.1 The Linear City Model of Product Differentiation
1.2 Varian’s Model
2. Dynamic Game with Complete Information
2.1 Simultaneous Market Entry
2.2 Rubinstein-Stahl Game
2.3 Repeated Prisoner’s Dilemma
3. Static Game with Incomplete Information
3.1 Cournot with Private Information
3.2 The Battle of Sexes
3.3 The Lover-Hater Battle of the Sexes
3.4 First/Second Price Auctions
4. Dynamic Game with Incomplete Information
4.1 Principle-Agent Model
4.2 Mechanism Design: Monopolistic Seller
5. Dynamic Game with Imperfect Information
5.1 Beer and Quiche Game
5.2 Education Signaling Model
5.3 Spence Signaling Model
1 Static Game with Complete Information
Example 1.1. The Linear City Model of Product Differentiation (MWG 12.C.14). Consider a
city that can be represented as lying on a line segment of length 1. There is a continuum of consumers
whose total number (or, more precisely, measure) is M and who are assumed to be located uniformly
along this line segment. A consumer’s location is indexed by z ∈ [0, 1], the distance from the left end
of the city. At each end of the city is located one supplier of widgets: Firm 1 is at the left end; firm
2, at the right. Widgets are produced at a constant unit cost of c > 0. Every consumer wants at most
1 wedget and derives a gross benefit of v from its consumption. The total cost of buying from firm j
for a consumer located a distance d from firm j is p
j
+ td, where t/2 > 0 can be thought of as the cost
per unit of distance traveled by the consumer in going to and from firm j’s location.
The consumer who is indifferent between buying from the two firms is located at:
z =
t − p
i
+ p
j
2t
1
The utility of this consumer is:
u = v −
t + p
2
+ p
1
2
There are three types of equilibria (p
⋆
i
, p
⋆
j
) in the model:
• The consumer strictly prefers buying to not buying (u > 0). If firm i charges its price slightly
from p
⋆
i
to p
i
, its demand will be determined by the consumer z who is indifferent between
buying from the two firms. Thus, firm i and j chooses (p
⋆
i
, p
⋆
j
) to maximize profits:
(p
i
− c)z = (p
i
− c)
t − p
i
+ p
j
2t
⇒ p
⋆
i
= p
⋆
j
= c + t
Thus, v > c +
3
2
t
• The consumer is strictly prefers not buying to buying (u < 0). If firm i charges its price slightly
from p
⋆
i
to p
i
, its demand will be determined by the consumer who is indifferent between buying
and not buying (v = p
i
+ zt). Thus, firm i chooses p
⋆
i
to maximize profits:
(p
i
− c)z = (p
i
− c)
v − p
i
t
⇒ p
⋆
i
=
v + c
2
Thus, v < c + t
• The consumer is indifferent between buying and not buying (u = 0). Thus,
p
2
+ p
1
= 2v − t
If firm i charges its price slightly higher than from p
⋆
i
, its demand will be determined by the
consumer who is indifferent between buying and not buying. If firm i charges its price slightly
lower than from p
⋆
i
, its demand will be determined by the consumer who is indifferent between
from the two firms. In these two cases, however, neither deviation from p
⋆
i
should not increase
profits. Therefore, we have:
∂
∂p
i
(p
i
− c)
v − p
i
t
≤ 0 ⇒ 2p
i
− v − c ≥ 0
∂
∂p
i
(p
i
− c)
t − p
i
+ p
j
2t
≤ 0 ⇒ p
i
− 2p
j
+ t + c ≥ 0
Taking both firms into consideration and plugging in the utility constraint, we have:
c + t ≤ v ≤ c +
3
2
t
p
1
= p
2
= v +
t
2
Example 1.2. Varian’s Model (simplified version)
• n number of firms
• Constant marginal cost c
• M consumers with unit demand and valuation v-common knowledge
• I-number of consumers that are informed about the price
• U-number of uninformed consumers (standard Bertrand U = 0)
• Consumers visit at most 1 store-informed consumers go to the lowest price firm while uninformed
choose a store at random.
2
The payoff of firm i that charges p ∈ F (·) is given by:
π
∗
(p) =
(1 − F (p))
n−1
∗ I +
U
n
(p − c)
In equilibrium, every p in the support of F (p) should result in the same payoff π
∗
:
F (p) = 1 −
π
∗
I(p − c)
−
U
In
1
n−1
Given that F (¯p) = 1 and F (p) = 0, we have
F (p) = 1 −
U
nI
1
n−1
v − p
p − c
1
n−1
, p ∈
Uv + Inc
In + U
, v
Note that for U = 0, the distribution collapses to F (c) = 1 and p = c Bertrand competition.
2 Dynamic Game with Complete Information
Example 2.1. Simultaneous Market Entry
• Endogenous entry: entry will occur as long as set-up costs K are less than potential profits.
• Consider an inverse demand function p(q) = a − bq.
• Constant MC c ⩾ 0.
• Timing:
• All potential firms simultaneously decide in or out where in is associated with a cost K.
• All entrants J play a Cournot game.
Suppose that J firms have entered the market. Firm i’s optimization problem under Cournot is as
follows:
max
q
t
a − b
J
X
j=1
q
j
− c
q
i
r
i
(q
−i
) =
a − b
P
j=i
q
j
− c
2b
Symmetric NE:
q
∗
i
=
a − c
b(J + 1)
π
∗
i
=
1
b
a − c
J + 1
2
Let J
∗
be the SPNE entrants in the market. No deviation incentives requires
π
∗
(J) ⩾ K
π
∗
(J + 1) < K
From our Cournot example, firm J is exactly indifferent between entry and staying out if
(J + 1)
2
=
(a − c)
2
bK
3
Then J
∗
=
j
(a−c)
√
bK
− 1
k
. We want to maximize the total surplus w.r.t. J :
T S = P S + CS − KJ = J
(a − c)
2
b(J + 1)
2
+ J
2
(a − c)
2
2b(J + 1)
2
− JK
Differentiating w.r.t. J.
(a − c)
2
b(J + 1)
3
− K = 0
(J + 1)
3
=
(a − c)
2
bK
Let
¯
J =
j
(a−c)
2/3
(bK)
1/3
− 1
k
.
¯
J < J
∗
- there is too much entry in SPNE!
Example 2.2. Rubinstein-Stahl Game. There are T + 1 (even) periods numbered {0, 1, ..., T }.
Player 1 and 2 are negotiation over a division of a pie of size 1. Player 1 proposed a division
(x
t
, 1 −x
t
) in even periods {0, 2, ..., T } where x
t
is his share of the pie. If player 2 accepts, game ends.
If player 2 rejects, then in next period he makes a counter offer (x
t+1
, 1 − x
t+1
).
Finite case: they alternate until they reach an agreement or period T + 1. If they fail to reach an
agreement in period T + 1, each one gets 0. Player i’s discount rate is δ
i
. Using backward induction:
• Stage T, 2 proposes (0, 1) and 1 accepts;
• Stage T - 1, player 1 proposes (1 − δ
2
, δ
2
) and 2 accepts;
• Stage T - 2, player 2 proposes (δ
1
(1 − δ
2
), 1 − δ
1
(1 − δ
2
)) and 1 accepts;
• ...
The the Rubinstein-Stahl game has a unique SPNE in which player 1 proposes his share to be:
x
∗
0
= (1 − δ
2
)
1 − (δ
1
δ
2
)
(T +1)/2
1 − δ
1
δ
2
in period 0 and player 2 accepts the offer.
Infinite case:
lim
T →∞
x
∗
0
=
1 − δ
2
1 − δ
1
δ
2
Randomness: Suppose that player i is the proposer in each period with probability α
i
. If at period
0 < t < T , player 1 would propose (x
1
t
, x
2
t
) and player 2 would propose (y
1
t
, y
2
t
). Then, at period t −1,
player 1 and 2 would propose respectively:
(x
1
t−1
, x
2
t−1
) = (1 − δ
2
(α
1
x
2
t
+ α
2
y
2
t
), δ
2
(α
1
x
2
t
+ α
2
y
2
t
))
(y
1
t−1
, y
2
t−1
) = (δ
1
(α
1
x
1
t
+ α
2
y
1
t
), 1 − δ
1
(α
1
x
1
t
+ α
2
y
1
t
))
Example 2.3. Repeated Prisoner’s Dilemma
player 2
C D
player 1
C 5, 5 0, 6
D 6, 0 1, 1
4
Some possible strategies in the repeated Prisoner’s Dilemma game:
• Grim: Play 1 at t = 0; thereafter play C if the players have always played (C, C) in the past,
play D otherwise (i.e., if anyone ever played D in the past).
• Naively Cooperate: Play always C (no matter what happened in the past).
• Tit-for-Tat: Play C at t = 0, and at each t > 0, play whatever the other player played at t − 1.
Strategy profile (Grim, Grim). There are two kinds of histories we need to consider separately
for this strategy profile.
• Cooperation: Histories in which D has never been played by any player. This yields the present
value of
V
C
= 5 + 5δ + 5δ
2
+ ··· = 5/(1 − δ)
The augmented stage game at the given history is
player 2
C D
player 1
C 5 + δV
C
, 5 + δV
C
0 + δV
D
, 6 + δV
D
D 6 + δV
D
, 0 + δV
D
1 + δV
D
, 1 + δV
D
To pass the single-deviation test, (C, C) must be a Nash equilibrium of this game:
5 + δV
C
≥ 6 + δV
D
⇒ δ ≥ 1/5
• Defection: Histories in which D has been played by some one at some date. This yields the
present value of
V
D
= 1 + δ + δ
2
+ ··· = 1/(1 − δ)
The augmented stage game at the given history is
player 2
C D
player 1
C 5 + δV
D
, 5 + δV
D
0 + δV
D
, 6 + δV
D
D 6 + δV
D
, 0 + δV
D
1 + δV
D
, 1 + δV
D
(D, D) is the only Nash equilibrium of this game and thus passes the single-deviation test.
Strategy profile (Tit-for-tat, Tit-for-tat). If (C, D) is played at t, then according to (Tit-for-tat,
Tit-for-tat) the sequence of plays starting at t + 1 will be
(D, C)(C, D)(D, C)(C, D) ···
with t + 1-present value of
6
1 − δ
2
,
6δ
1 − δ
2
= (6, 0) + δ(0, 6) + δ
2
(6, 0) + +δ
3
(0, 6) + δ
4
(6, 0) ···
The reduced game at any t for any previous history is
player 2
C D
player 1
C 5 +
5δ
1−δ
, 5 +
5δ
1−δ
0 +
6δ
1−δ
2
, 6 +
6δ
2
1−δ
2
D 6 +
6δ
2
1−δ
2
, 0 +
6δ
1−δ
2
1 +
δ
1−δ
, 1 +
δ
1−δ
Consider (C, C) or (C, D) is played at t − 1, then (C, C) or (D, C) should be the Nash equilibrium
respectively at t. This requires δ = 1/5.
5
3 Static Game with Incomplete Information
Example 3.1. Cournot with Private Information. There are 2 firms in the market. The inverse
demand is given by p(q) = a − q with q = q
1
+ q
2
. Firm 1’s marginal cost c is common knowledge.
Firm 2’s marginal cost c = {c
H
, c
L
} is private information to firm 2: c
H
with probability p and c
L
with
probability 1 − p. The two firms play Cournot by simultaneously choosing their quantities. Baysian
Nash equilibrium specifies a strategy for each player and type (q
∗
1
, q
∗
2
(c
H
) , q
∗
2
(c
L
)).
Given the conjecture q
∗
1
, the type c
i
with i = {L, H} maximizes his profit:
max
q
2
(a − q
∗
1
− q
2
− c
i
) q
2
q
2
(q
∗
1
| c
i
) =
a − q
∗
1
− c
i
2
Firm 1 maximizes its expected surplus given (q
∗
2
(c
H
) , q
∗
2
(c
L
)) :
max
q
1
p (a − q
1
− q
∗
2
(c
H
) − c) q
1
+ (1 − p) (a − q
1
− q
∗
2
(c
L
) − c) q
1
q
∗
1
= p
a − q
∗
2
(c
H
) − c
2
+ (1 − p)
a − q
∗
2
(c
L
) − c
2
The two equations should be simultaneously satisfied in equilibrium.
q
∗
1
=
a − 2c + pc
H
+ (1 − p)c
L
3
q
∗
2
(c
H
) =
a − 2c
H
+ c
3
+
1 − p
6
(c
H
− c
L
)
q
∗
2
(c
L
) =
a − 2c
L
+ c
3
−
p
6
(c
H
− c
L
)
Example 3.2. The Battle of Sexes. Assume that both θ
1
and θ
2
are independent draws from a
uniform distribution [0, x]. We shall look for a Bayesian equilibrium in which player 1 goes to the fight
if θ
1
exceeds some critical value x
1
and goes to the ballet otherwise, and player 2 goes to the ballet if
θ
2
exceeds some critical value x
2
and goes to the fight otherwise.
F B
F 2 + θ
1
, 1 0, 0
B 0, 0 1, 2 + θ
2
The probability that player i goes to the ballet is:
σ
i
(θ
i
) = Pr [θ
i
> x
i
] = 1 − Pr [θ
i
≤ x
i
] = 1 −
x
i
x
player i’s expected payoffs from going to the fight and going to the ballet are:
E [u
i
(F | θ
i
, θ
j
)] = (2 + θ
i
) (1 − σ
j
(θ
j
)) + (0)σ
j
(θ
j
) =
x
j
x
(2 + θ
i
)
E [u
i
(B | θ
i
, θ
j
)] = (0) (1 − σ
j
(θ
j
)) + (1)σ
j
(θ
j
) = 1 −
x
j
x
To make both players indifferent between going to the fight and going to the ballet:
x
j
x
(2 + x
i
) = 1 −
x
j
x
⇒ x
1
= x
2
=
−3 +
√
9 + 4x
2
6
If the uncertainty disappears, the probabilities of player 1 playing F and player 2 playing B both
converge to 2/3. These are exactly the probabilities of the mixed strategy Nash equilibrium of the
complete information case.
lim
x→0
h
1 −
x
i
x
i
= 1 − lim
x→0
"
d
dx
(−3 +
√
9 + 4x)
d
dx
(2x)
#
= 1 − lim
x→0
2(9 + 4x)
−1/2
2
=
2
3
Example 3.3. The Lover-Hater Battle of the Sexes. Suppose that player 2 has perfect informa-
tion and two types, l and h. Type l loves going out with player 1 whereas type h hates it. Player 1
has only one type and is uncertain about player 2’s type and believes the two types are equally likely.
F B
F 2, 1 0, 0
B 0, 0 1, 2
type l
F B
F 2, 0 0, 2
B 0, 1 1, 0
type h
• player 2 of type l. The expected utilities from playing F and B are
E [u
2
(σ
1
, F | l)] = σ
1
(F )
E [u
2
(σ
1
, B | l)] = 2 (1 − σ
1
(F ))
and type l’s best response is
BR
2
(σ
1
| l) =
1 if σ
1
(F ) > 2/3
[0, 1] if σ
1
(F ) = 2/3
0 if σ
1
(F ) < 2/3
That is, type l will go to the fight whenever σ
1
(F ) > 2/3.
• player 2 of type h. The expected utilities from playing F and B are
E [u
2
(σ
1
, F | h)] = 1 − σ
1
(F )
E [u
2
(σ
1
, B | h)] = 2σ
1
(F )
and so type h ’s best response is
BR
2
(σ
1
| h) =
1 if σ
1
(F ) < 1/3
[0, 1] if σ
1
(F ) = 1/3
0 if σ
1
(F ) > 1/3
That is, type h will go to the fight whenever σ
1
(F ) < 1/3.
• player 1. Given player 2’s strategy σ
2
≡ (σ
2
(· | l), σ
2
(· | h) , player 1’s expected payoffs to playing
F and B are
E [u
1
(F, σ
2
)] =
1
2
σ
2
(F | l)(2) +
1
2
σ
2
(F | h)(2) = σ
2
(F | l) + σ
2
(F | h)
E [u
1
(B, σ
2
)] =
1
2
(1 − σ
2
(F | l)) +
1
2
(1 − σ
2
(F | h)) = 1 −
σ
2
(F | l) + σ
2
(F | h)
2
and so player 1’s best response is
BR
1
(σ
2
) =
1 if σ
2
(F | l) + σ
2
(F | h) > 2/3
[0, 1] if σ
2
(F | l) + σ
2
(F | h) = 2/3
0 if σ
2
(F | l) + σ
2
(F | h) < 2/3
That is, player 1 will go to the fight whenever σ
2
(F | l) + σ
2
(F | h) > 2/3.
7
Pooling Equilibrium
• Is there a pure strategy equilibrium where both h and l choose F , that is, σ
2
(F | l) = σ
2
(F |
h) = 1 ? player 1’s best response to these strategies is to play F with probability 1 , but in
that case type h ’s optimal response is to play B with probability 1. Therefore, there is no such
equilibrium.
• Is there a pure strategy pooling equilibrium where σ
2
(F | l) = σ
2
(F | h) = 0 ? In this case
player 1’s best response is to play B with probability 1 , to which type h ’s optimal response is
to play F with probability 1 . Therefore, there is no such equilibrium.
Separating Equilibrium
• Is there a pure strategy equilibrium where type l chooses F and type h chooses B, that is,
σ
2
(F | l) = 1, and σ
2
(F | h) = 0 ? player 1’s best response is to choose F with probability 1 ,
to which the specified strategies for both types of player 2 are best responses. Therefore,
(σ
1
(F ), σ
2
(F | l), σ
2
(F | h)) = (1, 1, 0)
is a pure-strategy separating Bayesian equilibrium.
• Is there a pure strategy separating equilibrium where σ
2
(F | l) = 0 and σ
2
(F | h) = 1 ? player
l’s best response is to play F with probability 1 , to which type h ’s strategy is not a best
response. Therefore, there is no such equilibrium.
Semi-separating Equilibrium
• Is there a semi-separating equilibrium, in which both types of player 2 mix? No, because this
would require σ
1
(F ) = 2/3 and σ
1
(F ) = 1/3 at the same time. Therefore, there is no such
equilibrium.
• Is there an equilibrium where only type l mixes? Suppose that type l mixes in equilibrium. This
implies that player 1 mixes with σ
1
(F ) = 2/3, which in turn implies σ
2
(F | h) = 0. Since we
also require that σ
2
(F | l) + σ
2
(F | h) = 2/3 (or else player 1 would not be mixing), it follows
that σ
2
(F | l) = 2/3. Therefore,
(σ
1
(F ), σ
2
(F | l), σ
2
(F | h)) = (2/3, 2/3, 0)
is a semi-separating Bayesian equilibrium.
• Is there an equilibrium where only type h mixes? Suppose that type h mixes in equilibrium.
This implies that player 1 must also be mixing with σ
1
(F ) = 1/3. This in turn means that type l
’s optimal action is to play B, or σ
2
(F | l) = 0. Since we also require σ
2
(F | l) + σ
2
(F | h) = 2/3
for player 1 to mix, we conclude that σ
2
(F | h) = 2/3. Therefore,
(σ
1
(F ), σ
2
(F | l), σ
2
(F | h)) = (1/3, 0, 2/3)
is a semi-separating Bayesian equilibrium.
Example 3.4. First/Second Price Auctions. There is a single object for sale, and N potential
buyers are bidding for the object. Bidder i assigns a value of X
i
to the object—the maximum amount
a bidder is willing to pay for the object. Each X
i
is independently and identically distributed on some
interval [0, ω] according to the increasing distribution function F . Bidder i knows the realization x
i
of X
i
and only that other bidders’ values are independently distributed according to F . Bidders are
risk neutral; they seek to maximize their expected profits. All components of the model other than the
realized values are assumed to be commonly known to all bidders. In particular, the distribution F is
common knowledge, as is the number of bidders. A strategy for a bidder is a function β
i
: [0, ω] → R
+
,
which determines his or her bid for any value.
8
Second-price sealed-bid auction. it is a weakly dominant strategy to bid according to β
II
(x) = x.
Suppose that p
1
= max
j=1
b
j
and consider an alternative bid z
1
. If p
1
> x
1
> z
1
, p
1
> z
1
> x
1
,
x
1
> z
1
> p
1
, or x
1
> p
1
> z
1
, the outcomes won’t change. If z
1
> p
1
> x
1
, the bidder gets negative
payoffs. If x
1
> p
1
> z
1
, the bidder loses and his payoffs decrease. Let Y
1
= max
j=1
X
j
. Then, the
expected payment by a bidder with value x can be written as
m
II
(x) = F (x)
N−1
E [Y
1
|Y
1
< x] = G(x)E [Y
1
|Y
1
< x]
First-price sealed-bid auction. Suppose that bidders j = 1 follow the symmetric, increasing, and
differentiable equilibrium strategy β. Suppose bidder 1 receives a signal, X
1
= x, and bids b. We wish
to determine the optimal b. Note that 0 = β(0) < b < β(ω). His expected payoff is
G
β
−1
(b)
(x − b)
Maximizing this with respect to b yields the first-order condition:
G(x)β
′
(x) + g(x)β(x) = xg(x) ⇒
d
dx
(G(x)β(x)) = xg(x)
Thus, we have
β(x) =
1
G(x)
Z
x
0
yg(y)dy
= E [Y
1
| Y
1
< x]
Now we prove that it is indeed optimal for a bidder with value x to bid β(x). Suppose that the bidder
pretends to have another value z and bids β(z):
Π(b, x) = G(z)[x − β(z)]
= G(z)x − G(z)E [Y
1
| Y
1
< z]
= G(z)x −
Z
z
0
yg(y)dy
= G(z)x − G(z)z +
Z
z
0
G(y)dy
= G(z)(x − z) +
Z
z
0
G(y)dy
Compare it with the original strategy:
Π(β(x), x) − Π(β(z), x) = G(z)(z − x) −
Z
z
x
G(y)dy ≥ 0
The equilibrium bid can be rewritten as
β(x) =
1
G(x)
Z
x
0
yg(y)dy = x −
Z
x
0
G(y)
G(x)
dy = x −
Z
x
0
F (y)
F (x)
N−1
dy
As N goes to infinity, the equilibrium bid β
I
(x) approaches x.
The expected payment by a bidder with value x is
m
I
(x) = G(x)E [Y
1
|Y
1
< x]
Revenue Comparison. The ex ante expected payment of a particular bidder in either auction is
9
E
m
A
(X)
=
Z
ω
0
m
A
(x)f(x)dx
=
Z
ω
0
Z
x
0
yg(y)dy
f(x)dx
=
Z
ω
0
Z
ω
y
f(x)dx
yg(y)dy
=
Z
ω
0
y(1 −F (y))g(y)dy
Notice that the density of Y
(N)
2
, the second highest of N values, f
(N)
2
(y) = N(1 − F (y))f
(N−1)
1
(y).
The expected revenue accruing to the seller is
E
R
A
= N × E
m
A
(X)
= N
Z
ω
0
y(1 −F (y))g(y)dy
=
Z
ω
0
yf
(N)
2
(y)dy
= E
h
Y
(N)
2
i
Reserve Price in Second-Price Auctions
m
II
(x, r) = rG(r) +
Z
x
r
yg(y)dy
Reserve Price in First-Price Auctions
β
I
(x) = E [max {Y
1
, r} | Y
1
< x]
= r
G(r)
G(x)
+
1
G(x)
Z
x
r
yg(y)dy
m
I
(x, r) = G(x)β
I
(x) = rG(r) +
Z
x
r
yg(y)dy
Revenue Comparison.
E
m
A
(X, r)
=
Z
ω
r
m
A
(x, r)f(x)dx
= r(1 − F (r))G(r) +
Z
ω
r
y(1 −F (y))g(y)dy
Suppose that the seller attaches a value x
0
∈ [0, ω). This means that if the object is left unsold, the
seller would derive a value x
0
from its use. The seller would not set a reserve price r that is lower
than x
0
. Then the overall expected payoff of the seller from setting a reserve price r ≥ x
0
is
Π
0
= N × E
m
A
(X, r)
+ F (r)
N
x
0
Differentiating this with respect to r, we obtain
dΠ
0
dr
= N[1 − F (r) − rf(r)]G(r) + N G(r)f(r)x
0
10
Now recall that the hazard rate function associated with the distribution F is defined as λ(x) =
f(x)/(1 −F (x)). Thus, we can write
dΠ
0
dr
= N [1 − (r − x
0
) λ(r)] (1 − F (r))G(r)
The relevant first-order condition implies that the optimal reserve price r
∗
must satisfy
(r
∗
− x
0
) λ (r
∗
) = 1
4 Dynamic Game with Incomplete Information
Example 4.1. Principle-Agent Model. One principle P and one agent A that is hired for a one
time project. The project’s observable profit π ∈ [π, ¯π]. π is a draw from a conditional distribution
F (π|e) where F
e
(π|e) < 0. The effort level e is private information of A. The principal offers a
contract to the agent. The agent chooses whether the accept or reject. If he accepts, he chooses e.
Given e, nature moves and draws π from F (π|e). The agent obtains w(π).
Full information: first-best contract
• Suppose that e is observable by the principal.
• The agent has a separable utility: U
A
(w, e) = v(w) − e, v
′
(w) > 0
• The principal is risk neutral: U
P
(π, w) = π − w
The contract can specify an effort e and a compensation w(π). Then the principal solves:
max
e,w
e
(π)
Z
(π − w
e
(π)) dF (π | e) ⇒ min
w
e
(π)
Z
w
e
(π)dF (π | e)
subject to
Z
v (w
e
(π)) dF (π | e) − e ≥ U
0
A
The Lagrangian is given by
L =
Z
(w
e
(π) −λv (w
e
(π))) dF (π | e) + λ(e + U
0
A
)
The optimality conditions yield the optimal wage
w
⋆
e
= v
−1
(U
0
A
+ e)
and the principals expected payoff
U
P
(π, w
⋆
e
) = E [π|e] − v
−1
(U
0
A
+ e)
Unobservable effort and risk-neutral agent: first-best contract
• The agent is risk-neutral v(w) = w
• Suppose that the principal offers a compensation schedule w(π) = π − p
Then, the agent will maximize:
max
e
E [π|e] − p − e
Note that the solution also solves the principals expected payoff in the full information case. Therefore,
anticipating this effort, the principal offers a contract:
11
p
∗
= E [π | e
∗
] − e
∗
− U
0
A
⇒ w (p
∗
) = π −
E (π | e
∗
) − e
∗
− U
0
A
Unobservable effort and risk-averse agent: second-best contract
• Let e = {e
L
, e
H
} with e
L
< e
H
. F(π|e
H
) < F (π|e
L
)
Suppose that the principal wants to implement effort e. Then, his optimization problem is
max
e,w
e
(π)
Z
(π − w
e
(π)) dF (π | e) ⇒ min
w
e
(π)
Z
w
e
(π)dF (π | e)
subject to
Z
v (w
e
(π)) dF (π | e) − e ≥ U
0
A
Z
v (w
e
(π)) dF (π | e) − e ≥
Z
v (w
e
(π)) dF (π | e
′
) − e
′
Let λ ≥ 0 and µ ≥ 0 be the Lagrange multipliers of the IR and IC constraints, respectively.
L =
Z
w
e
H
(π)dF (π | e
H
)
− λ
Z
v (w
e
H
(π)) dF (π | e
H
) − e
H
− U
0
A
− µ
Z
v (w
e
H
(π)) dF (π | e
H
) − e
H
−
Z
v (w
e
H
(π)) dF (π | e
L
) + e
L
The F.O.C. with respect to w
e
H
(π) gives:
1
v
′
(w
e
H
(π))
= λ + µ
1 −
f(π|e
L
)
f(π|e
H
)
Both constraints need to be binding
• µ = 0 implies fixed wage, which would implement e
L
• λ = 0 implies that there exists π such that v
′
(w
e
H
(π)) < 0
Example 4.2. Mechanism Design: Monopolistic Seller. One seller and one buyer. The seller
chooses the price t(x) for each quality x that the buyer purchases.
• The seller’s payoff is given by U
s
(x, t) = t(x) − c(x)
• The buyer’s payoff is given by U
b
(x, t, θ) = v(x, θ) −t(x), where θ is private information.
• Single-crossing property: v
x
(x, θ) is increasing in θ
• The seller makes a take-it-or-leave-it offer t(x) to the buyer.
• The agent can have one of two possible types: Θ = {θ
L
, θ
H
}, with θ
H
> θ
L
Suppose the principal’s prior is given by Pr {θ = θ
L
} = π ∈ (0, 1). The principal solves:
max
⟨(x
L
,t
L
),(x
H
,t
H
)⟩
π [t
L
− c (x
L
)] + (1 − π) [t
H
− c (x
H
)]
s.t. v (x
L
, θ
L
) − t
L
≥ v (0, θ
L
) (IR
L
)
v (x
H
, θ
H
) − t
H
≥ v (0, θ
H
) (IR
H
)
v (x
L
, θ
L
) − t
L
≥ v (x
H
, θ
L
) − t
H
(IC
L
)
v (x
H
, θ
H
) − t
H
≥ v (x
L
, θ
H
) − t
L
(IC
H
)
12
• IR
H
is redundant. We show that [IR
L
and IC
H
] ⇒ IR
H
:
v (x
H
, θ
H
) − t
H
− v (0, θ
H
)
(IC
H
)
≥ v (x
L
, θ
H
) − t
L
− v (0, θ
H
)
SCP
≥ v (x
L
, θ
L
) − t
L
− v (0, θ
L
)
(IR
L
)
≥ 0
Thus, IR
H
can be discarded.
• IR
L
binds. Otherwise increasing both t
L
and t
H
by a small ε > 0 would preserve IR
L
, not affect
IC
H
and IC
L
, and raise profits.
• IC
H
binds. Otherwise, increasing t
H
by a small ε > 0 would preserve IC
H
, not affect IR
L
, relax
IC
L
, and raise profits.
• IC
L
is redundant. Assume not, and let {(x
′
H
, t
′
H
) , (x
′
L
, t
′
L
)} be a solution to the reduced problem
subject to only IR
L
and IC
H
. If IC
L
is not redundant then this solution violates IC
L
: type θ
L
strictly prefers (x
′
H
, t
′
H
) to (x
′
L
, t
′
L
). If t
′
H
−c (x
′
H
) ≥ t
′
L
−c (x
′
L
), the principal can raise profits by
giving both types (x
H
, t
H
+ ε ) (recall that IR
H
can be discarded). If t
′
H
−c (x
′
H
) < t
′
L
−c (x
′
L
),
she can do better by giving both types (x
L
, t
L
). Note that IR
L
is satisfied in both cases because
(x
L
, t
L
) satisfies it and (x
H
, t
H
) is strictly preferred by type θ
L
. IC
H
is trivially satisfied in both
cases. Thus, we have a contradiction, and IC
L
can be discarded.
Plugging in binding IR
L
and IC
H
, we have:
max
x
H
,x
L
∈X
total expected surplus
z }| {
π [v (x
L
, θ
L
) − c (x
L
)] + (1 − π) [v (x
H
, θ
H
) − c (x
H
)]
− (1 − π) [v (x
L
, θ
H
) − v (x
L
, θ
L
)]
| {z }
information rent of high type
We see that the objective function is additively separable in x
L
and x
H
, hence the program can be
broken into two:
max
x
H
∈X
(1 − π) [v (x
H
, θ
H
) − c (x
H
)]
max
x
L
∈X
π [v (x
L
, θ
L
) − c (x
L
)] − (1 − π) [v (x
L
, θ
H
) − v (x
L
, θ
L
)]
Conclusions
• No rent for the low type
• High type has a positive rent when x
L
> 0
• Efficiency at the top x
H
= x
⋆
H
• Downward distortion at the bottom x
L
< x
⋆
L
5 Dynamic Game with Imperfect Information
Example 5.1. Beer and Quiche Game. Player 1 has two types: strong (t
s
) and weak (t
w
). The
strong type likes beer for breakfast, while the weak type likes quiche. Player 1 is ordering his breakfast,
while player 2, who is a bully, is watching and contemplating whether to pick a fight with player 1.
Player 2 would like to pick a fight if player 1 is weak but not fight if he is strong. His payoffs are such
that if he assign probability more than 1/2 to weak, he prefers a fight, and if he assigns probability
more than 1/2 to strong, then he prefers not to fight. Player 1 would like to avoid a fight: he gets 1
utile from the preferred breakfast and 2 utiles from avoiding the fight. Before observing the breakfast
player 2 finds it more likely that player 1 is strong. If player 2 sees Beer, he assigns probability 0.9 to
strong and does not fight; if he sees Quiche, he assigns probability 1 on weak and fights. Let us check
that this is indeed a sequential equilibrium.
13
Pooling Equilibrium
t
s
t
w
.9
beer
quiche
.1
beer
quiche
.9
3, 1
don
′
t
1, 0
duel
0
2, 1
don
′
t
0, 0
duel
1
3, 0
don
′
t
1, 1
duel
.1
2, 0
don
′
t
0, 1
duel
• Sequential Rationality. Player 2 would choose don’t fight if observing beer and fight if
observing quiche. If player 1’s type is strong, beer is the dominant strategy. If player 1’s type
is week, he would also choose beer given player 2’s belief.
• Consistency. For the information set after bear:
Pr (t
s
| beer, s
∗
) =
Pr (t
s
) Pr (beer | t
s
, s
∗
)
Pr (t
s
) Pr (beer | t
s
, s
∗
) + Pr (t
w
) Pr (beer | t
w
, s
∗
)
=
(.9)(1)
(.9)(1) + (.1)(1)
= .9
= b
∗
(t
s
| beer)
For the information set after quiche, suppose that weak type trembles with probability ε while
the strong type trembles with probability zero:
Pr (t
w
| quiche, ε) =
(.1)ε
(.1)ε + (.9)(0)
= 1
Separating Equilibrium
t
s
t
w
.9
beer
quiche
.1
beer
quiche
1
3, 1
don
′
t
1, 0
duel
0
2, 1
don
′
t
0, 0
duel
1
3, 0
don
′
t
1, 1
duel
0
0, 0
don
′
t
−2, 1
duel
• Sequential Rationality. Player 2 would choose don’t fight if observing beer and fight if
observing quiche. If player 1’s type is strong, beer is the dominant strategy. If player 1’s type
is week, he would choose quiche given player 2’s belief.
14
• Consistency. For the information set after bear:
Pr (t
s
| beer, s
∗
) =
Pr (t
s
) Pr (beer | t
s
, s
∗
)
Pr (t
s
) Pr (beer | t
s
, s
∗
) + Pr (t
w
) Pr (beer | t
w
, s
∗
)
=
(1)(1)
(1)(1) + (0)(1)
= 1
= b
∗
(t
s
| beer)
For the information set after quiche:
Pr (t
w
| quiche, s
∗
) =
Pr (t
w
) Pr (quiche | t
w
, s
∗
)
Pr (t
w
) Pr (quiche | t
w
, s
∗
) + Pr (t
s
) Pr (quiche | t
s
, s
∗
)
=
(1)(1)
(1)(1) + (0)(1)
= 1
= b
∗
(t
w
| quiche)
Semi-separating Equilibrium
t
s
t
w
.2
beer
quiche
.8
beer
quiche
.5
3, 1
don
′
t
1, 0
duel
0
2, 1
don
′
t
0, 0
duel
1
3, 0
don
′
t
1, 1
duel
.5
2, 0
don
′
t
0, 1
duel
• Sequential Rationality. If both types play beer, player 2 must assign probability 0.8 to weak
type after observing beer, and he must fight by sequential rationality. In that case, t
w
must play
quiche as a best reply. One can also check that there is no separating equilibrium. For example,
if strong type has beer and the weak type has quiche, then player 2 would learn player’s type
after the choice of breakfast and would fight only after quiche. In that case, weak type would
like to deviate. Therefore in a sequential equilibrium, at least one of the types must be playing
a mixed strategy. Write p
B
and p
Q
for the probabilities of “don’t” (i.e. “don’t duel”) after beer
and quiche respectively. Write U
B
(t) and U
Q
(t) for the expected payoffs from beer and quiche
for type t, respectively. Then,
U
B
(t
s
) − U
Q
(t
s
) = 2 + U
B
(t
w
) − U
Q
(t
w
) > 0
Sequential rationality requires that t
s
must play beer with probability 1. Therefore, in equilib-
rium, t
s
plays beer and t
w
mixes.
• Consistency. For the information set after quiche:
Pr (t
w
| quiche) =
Pr (t
w
) Pr (quiche | t
w
)
Pr (t
w
) Pr (quiche | t
w
) + Pr (t
s
) Pr (quiche | t
s
)
=
Pr (quiche | t
w
) (.8)
Pr (quiche | t
w
) (.8) + (0)(.2)
= 1
Given that player 2 must fight after quiche and t
w
mixes, we have
p
Q
= 1, −1 + 2(p
B
− p
Q
) = 0 ⇒ p
B
= 1/2
15
This shows that player 2 is indifferent between two actions, indicating that
0.5 = Pr (t
w
| beer) =
Pr (t
w
) Pr (beer | t
w
)
Pr (t
w
) Pr (beer | t
w
) + Pr (t
s
) Pr (beer | t
s
)
=
Pr (beer | t
w
) (.8)
Pr (beer | t
w
) (.8) + (1)(.2)
This shows that Pr (beer | t
w
) = 1/4
Example 5.2. Education Signaling Model A firm is considering hiring a worker, who can be one
of two types: high or low (i.e. θ = {H, L} ). The worker knows his own type and the firm only knows
that the probability of the worker being of high type is
1
3
. The high ability worker generates revenue of
π(H) = 2 for the firm and the low ability worker- π(L) = 0. If hired, the firm will pay the worker a
fixed wage w = 1. The high-ability worker has the possibility of an alternative occupation earning him
a payoff of
1
2
while the low-ability worker’s outside option is 0.
Suppose now that before the firm makes its hiring decision, the worker can choose to invest in educa-
tion. Let e = {1, 0} denote the education strategy of the worker where e = 1 stands for the worker’s
choice to acquire education. The cost for the high type of acquiring education is c(1 | H) =
1
6
and
the cost for the low type of acquiring education is c(1 | L) =
2
3
. While education does not affect the
workers’ productivity inside or outside the firm, it is observable to the firm. Let µ(θ | e) denote the
firm’s belief that the worker is of type θ after observing the education choice.
2/3 1/3
L
e = 1
e = 0
H
e = 1
e = 0
−2/3, 0
h = 0
1/3, −1
h = 1
0, 0
h = 0
1, −1
h = 1
1/2, 0
h = 0
1, 1
h = 1
1/3, 0
h = 0
5/6, 1
h = 1
Separating Equilibrium
Consider first a fully separating equilibrium with e
∗
(H) = 1 and e
∗
(L) = 0. Then, µ(H | 1) = 1 and
µ(L | 0) = 1. Given this belief, it is optimal for the firm to set h
∗
(1) = 1 and h
∗
(0) = 0. However,
given this belief structure and the firm’s optimal response, the low type will have incentives to deviate
to e = 1 since it will result in a payoff of
1
3
compared to 0 . Therefore, e
∗
(H) = 1 and e
∗
(L) = 0
cannot be part of a PBE. Similarly, one can show that e
∗
(H) = 0 and e
∗
(L) = 1 cannot be part of a
PBE. Thus, a fully separating equilibrium does not exist.
Pooling Equilibrium
In any pooling equilibrium on e
∗
, by Bayes’ rule, µ (H | e
∗
) =
1
3
. Therefore, the firm’s optimal strategy
is h
∗
(e
∗
) = 0. This, in turn implies that the only pooling equilibrium possible is e
∗
(L) = e
∗
(H) = 0.
Suppose the contrary, i.e., e
∗
(L) = e
∗
(H) = 1. Then, given h
∗
(e
∗
) = 0, both the high and the low
type will have strict incentives to deviate to e = 0 because it would result in strictly higher payoff in-
dependent of the firm’s strategy h(0). Consider now e
∗
(L) = e
∗
(H) = 0. PBE involves µ (H | e
∗
) =
1
3
and h
∗
(e
∗
) = 0. To assign µ(H | 1), we need to make sure that none of the types has incentives to
16
deviate to e = 1. One possible off-equilibrium belief is µ(H | 1) =
1
3
resulting in h
∗
(1) = 0.
Semi-separating Equilibrium
PBE(α(e | θ),
ˆ
h(e), µ(θ | e)), where α(e | θ) stands for the prob bility of type θ choosing e and
ˆ
h(e)
stands for the probability of the firm making a job offer to the worke fter observing e, or explain
why such equilibrium does not exists. Solution: For the employer to mix after observing e, his belief
µ(H | e) needs to satisfy:
µ(H | e) − (1 − µ(H | e)) = 0 =⇒ µ(H | e) =
1
2
By Bayes’ rule, this implies that:
µ(H | e) =
1
3
α(e | H)
1
3
α(e | H) +
2
3
α(e | L)
=
1
2
Therefore, from the above equation α(e | H) = 2α(e | L). Let us specify a PBE, in which α(1 | H) =
2α(1 | L). By Bayes’ rule,
µ(H | 0) =
1
3
α(0 | H)
1
3
α(0 | H) +
2
3
α(0 | L)
=
1
3
(1 − α(1 | H))
1
3
(1 − α(1 | H)) +
2
3
(1 − α(0 | L))
where that last equality follows from the fact that α(1 | θ) + α(0 | θ) = 1. Taking into account
α(1 | H) = 2α(1 | L) and substituting in the above equation:
µ(H | 0) =
1
3
−
2
3
α(1 | L)
1 −
2
3
α(1 | L)
<
1
2
Therefore, the firm’s optimal strategy upon observing e = 0 is
ˆ
h(0) = 0. Finally, L is indifferent
between e = 0 and e = 1 if:
1
3
ˆ
h(1) −
2
3
(1 −
ˆ
h(1)) = 0 =⇒
ˆ
h(1) =
2
3
It is straightforward to verify that for
ˆ
h(1) =
2
3
, α(1 | H) = 1. Therefore, α(1 | L) =
1
2
. To sum up,
the following strategy set and belief structure constitutes a PBE:
α(1 | H) = 1, α(1 | L) =
1
2
,
ˆ
h(0) = 0,
ˆ
h(1) =
2
3
, µ(H | 1) =
1
2
, µ(L | 0) = 1
Example 5.3. Spence Signaling Model. Suppose that a worker can choose to acquire e units of
education, where e is any nonnegative number. The worker will have to study hard to obtain her
education, and this creates disutility (studying for exams, doing homework, etc.). Assume that a
worker of true ability θ expends e/θ in disutility. The game proceeds as follows:
• Nature moves and chooses a worker type, H or L. The type is revealed to the worker but not to
the employer.
• The worker then chooses e units of education. This is perfectly observed by the employer.
• The employer observes e and forms an estimate of θ. He then pays the worker a salary equal to
this estimate, which is just the conditional expectation of θ given e, written E [θ | e].
• The H-worker’s payoff is E [θ | e] − (e/H), and the L-worker’s payoff is E [θ | e] − (e/L).
The employer’s expected payoff is zero. Assume that the worker’s choice of education is visible to the
world at large so that perfect competition must push her wage to E [θ | e], the conditional expectation
of θ given e.
17
If e is a possible choice of the high worker and e
′
a possible choice of the low worker, then it must be
that e ≥ e
′
.Two constraints:
E [θ | e] −
e
H
≥ E [θ | e
′
] −
e
′
H
E [θ | e
′
] −
e
′
L
≥ E [θ | e] −
e
L
Separating Equilibrium In this case, L must choose e = 0, for there is nothing to be gained in
making a positive effort choice. Thus,
H −
e
H
≥ L, L ≥ H −
e
L
⇒ e
⋆
∈ [(H − L)L, (H − L)H]
Any outcome in which the low type chooses 0 and the high type chooses some e
⋆
is supportable as a
separating equilibrium. One of beliefs could be: the employer believes that any e < e
⋆
comes from
the low type, while any e > e
⋆
comes from the high type.
Pooling Equilibrium. If the firm sees that signal e
⋆
it simply pays out the expected value calculated
using the prior beliefs: pH + (1 − p)L. For both types not to deviate, it must be that
pH + (1 − p)L −
e
⋆
θ
≥ L
when binding we have θ = L. Any e
⋆
between 0 and the upper bound specified above could implement
the pooling equilibrium. One of beliefs could be: the employer believes that any action e = e
⋆
is taken
by the low type. The consistency is checked by
Pr(L | e
′
, s
⋆
) =
Pr(L)ε
Pr(L)ε + Pr(H) · 0
= 1
Semi-separating Equilibrium. An example: the low type chooses 0 while the high type randomizes
between 0 with probability q and some e with probability 1 − q. If the employer sees e he knows the
type is high. If he sees 0 the posterior probability of the high type there is equal to
Pr(H | 0) =
Pr(H) Pr(0 | H)
Pr(H) Pr(0 | H) + Pr(L) Pr(0 | L)
=
pq
pq + (1 − p)
So the employer pays
pq
pq + (1 − p)
H +
1 − p
pq + (1 − p)
L
But the high type must be indifferent between the announcement of 0 and that of e, because he
willingly randomizes. It follows that
pq
pq + (1 − p)
H +
1 − p
pq + (1 − p)
L = H −
e
H
Firm could believe that all other e-choices come from low types.
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