GAME THEORY
Spring 2015
Lecture 7
Lecturer: Pingzhong Tang
1
Scribe: Weiran Shen
Increasing difference
We assume that each player i’s type tiQis drawn from some probability distribution Pi , which is independent
of all other players’ types. So P (t) = i Pi (ti ).
Definition 1
• A function f : R → R is increasing if and only if for all x̂ > x, f (x̂) > f (x);
• A function f : R → R is strictly increasing if and only if for all x̂ > x, f (x̂) > f (x).
Now consider a two-player Bayesian game where the set of actions for player 1 is A1 = {T, B}. Player 1
has many possible types and each type is denoted by t1 . Player 2 may have many actions c2 and also many
types t2 . Assume t1 is independent of t2 .
The difference in player 1’s payoff in switching from B to T is u1 (T, c2 , t1 , t2 ) − u1 (B, c2 , t1 , t2 ), which is a
function of c2 , t1 and t2 .
Definition 2 u1 satisfies (weakly/strictly) increasing difference if this difference is a (weakly/strictly) increasing function of t1 , for any given c2 and t2 .
Interpretation: Player 1 finds T relatively more attractive than B for higher types.
Example: In the public goods game from the previous lecture, we have:
(
c1
If player 2 calls
u1 (c1 , Don’t) − u1 (c1 , Call) =
c1 − 1 If player 2 doesn’t call.
is increasing in c1 .
2
Cutoff strategies
Definition 3 (Cutoff strategy) Player 1 uses a cutoff strategy if there exists some type θ such that for
each type t1 of player 1:
• if t1 > θ, then player 1 chooses T for sure.
• if t1 6 θ, then player 1 chooses B for sure.
1
2
Lecture 7
• if t1 = θ, then player 1 is indifferent between the two choices.
Remark 1 If θ increases, then the probability of player 1 choosing T decreases.
Theorem 4 If u1 satisfies increasing differences, then mo matter what strategy player 2 may use, player 1
always want to use a cutoff strategy.
A direct conclusion from the above theorem is that when looking for a Bayesian Nash equilibrium, it is
without loss of generality to look for those where player 1 uses a cutoff strategy.
More generally, player 1’s action can be (or be mapped to) any number in some range (e.g. auctions),
we say that u1 satisfies increasing differences if for every pair c1 > d1 of actions in A1 , u1 (c1 , c2 , t1 , t2 ) −
2
u2 (d1 , c2 , t1 , t2 ) is an increasing function of t1 , for all c2 , t2 . If u1 is differentiable, then ∂c∂ 1u∂t11 > 0.
Theorem 5 If u1 satisfies increasing differences, then no matter what player 2 uses, player 1 has some best
response s1 (t1 ) that is weakly increasing in t1 . Also, when u1 satisfies strictly increasing differences, then
any of player 1’s best response strategies must be weakly increasing.
Proof: We only prove the second part of the theorem.
Suppose r1 and t1 are two possible types of player 1 and that r1 > t1 . Let δ2 be some action of player 2 and
c1 = s∗1 (t1 ) and d1 = s∗1 (r1 ) be the best responses when player 1’s type is t1 and r1 , respectively. Our goal
is to prove that d1 > c1 .
Since c1 is the best response to δ2 when player 1’s type is t1 , we have:
IEu1 (c1 , δ1 , t1 , t2 ) − IEu1 (d1 , δ1 , t1 , t2 ) > 0
Similarly,
IEu1 (c1 , δ1 , r1 , t2 ) − IEu1 (d1 , δ1 , r1 , t2 ) 6 0
We cannot have c1 > d1 , otherwise it contradicts to the definition of increasing differences.
Example: The set of all possible types of player 1 is Φ1 = {0, 0.1, 0.2, 0.3}, each with probability 14 . Player
2 has only 1 type so he has no private information. The action set for the players are A1 = {T, B} and
A2 = {L, R}. For any type t1 of player 1, the payoff matrix is: u1 (T, δ2 , t1 ) − u1 (B, δ2 , t1 ) increases in t1 for
T
B
L
t1 , 0
1, 0
R
t1 , -1
-1, 3
both δ2 = L and δ2 = R.
Recall that if u1 satisfies increasing differences, it is without loss of generality to assume that player 1 always
uses a cutoff strategy. Suppose in the above example player 1 plays T if t1 > θ and plays B if t1 < θ.
However, θ may vary for different actions of player 2. For a fixed θ, player 2 thinks of the action of player
1 as a mixed strategy. For example, if θ = 0.15, then player 2 thinks that player 1 will play T and B both
with probability 0.5, since the probabilities for player 1 to have a type greater or less than 0.15 are both 0.5.
Now we assume player 2’s strategy is q · L + (1 − q) · R. Then player 1 is indifferent between T and B if his
type is θ.
qθ + (1 − q)θ = q · 1 + (1 − q) · (−1)
θ+1
q=
2
Lecture 7
3
We use support enumeration for player 2 to find the Bayesian Nash equilibria for the game.
It can be shown that no equilibrium exists if the support for player 2 is {L} or {R}. For {L, R}, assume
player 1’s mixed strategy (from player 2’s point of view) is p · T + (1 − p) · B.
u2 (L) = u2 (R)
0 = −1 · p + 3 · (1 − p)
3
p=
4
We can conclude that θ ∈ [0, 0.1], i.e. player 1 will choose B for sure if his type is 0, and choose T otherwise.
1 11
Combining q = θ+1
2 , we have q ∈ [ 2 , 20 ].
Example: Player 1’s type is drawn from a uniform distribution over [0, 1]. ε is a number satisfying 0 < ε < 1.
The payoff matrix is as follows:
T
B
L
εt1 , 0
1, 0
R
εt1 , -1
-1, 3
The utility function for player 1 u1 satisfies increasing differences. So we can assume that there exists
0 6 θ1 6 1 and player 1 will play T if t1 > θ1 and play B if t1 < θ1 .Since t1 ∼ U[0, 1], player 2 thinks
of player 1’s strategy as (1 − θ1 ) · T + θ1 · B. Again, we use support enumeration to find the equilibria of
the game. There is no equilibrium if the support of player 2’s strategy is {L} of {R}. When the support is
{L, R}, we have:
0 = −1 · (1 − θ1 ) + 3 · θ1
1
θ1 =
4
So player 1 will play T if t1 < 14 and play B if t1 > 43 . Let player 2’s strategy be q · L + (1 − q) · R. If θ1 = 41 ,
player 1 is indifferent between T and B. So we have:
u1 (T ) = εθ1 = q + (1 − q) · (−1) = u1 (B)
1 + 1/4 · ε
q=
2
When ε → 0, player 2’s strategy becomes 21 L + 12 R, and player 1’s strategy becomes 14 T + 43 B.
From these examples we know that a pure Bayesian Nash corresponds to a mixed strategy Nash equilibrium
(aka. purification of mixed strategies by Bayesian games).
3
Labor market1
Suppose there are workers seeking jobs in a labor market. Each worker has a type (productivity or ability).
The workers may choose different education levels. The labor market then observes their educations and
set wages for the workers. We assume that education has nothing to do with productivity and low-ability
workers can get a high education. However, the cost of obtaining an education c(e, θ) is related to the
worker’s type, where e is the worker’s education and θ is the worker’s type. Also, we assume c(e, θ) has the
following property:
1 For
detailed discussion, please refer to Dynamic Games with Incomplete Information.
4
Lecture 7
•
∂c
∂e
•
∂2c
∂e∂θ
> 0, meaning higher education costs more.
< 0, meaning education is less costly for high type workers.
A competitive market believes that a worker with education e is of high type with probability µ(e). Also a
competitive wage is set to be w(e) = θH · µ(e) + θL · (1 − µ(e)). The utility for the worker then, is naturally
u(θ, e) = w(e) − c(e, θ).
The key observation is as follows:
∂2c
∂2u
=−
>0
∂e∂θ
∂e∂θ
Therefore the worker will use an increasing strategy, i.e. workers of higher types will indeed choose get a
higher education.
There are two extreme cases for this game:
1. Separating equilibrium. This holds if e(θH ) 6= e(θL ). In this case µ(e(θH )) = 1 and µ(e(θL )) = 0. So
w(e(θ)) = θ as if the labor market knows the private type. In addition, lower type wokers will choose
e(θL ) = 0 and get the same wage with lower education costs.
2. Pooling equilibrium. This is the case when each worker chooses the same education level. In this case,
w(e) = λθH + (1 − λ)θL .