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# EC941 Game Theory - Lecture 2

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Lecture 2
1
Structure of the Lecture

Bayesian Games and Nash Equilibrium

Bayesian Games and Information

Cournot Duopoly with Private Information

Public Good Provision with Private Information

Juries and Information Aggregation
2
What is a Bayesian Game?
• Some details of the game are not common
knowledge.
• Some players have more information than others.
=&gt;solution concepts and techniques must
adjust!
3
For example:
• Firm may not know rival firm’s cost in the
Cournot game.
• A seller may not know a buyer’s valuation.
• A buyer may not know the quality of the
seller’s product.
• A bidder may not know other bidders’
valuations.
4
Example:
• You want to buy a used car. If the value to the
3
seller is 𝑣, it is worth 𝑣 to you.
2
• You know 𝑣~𝑈 0,100 . What should you
offer?
(the seller will only accept if 𝑏 ≥ 𝑣!)
=&gt; Our beloved lemons problem! (Akerlof 1970)
5
You need to take into account what the other
player would do conditional on his private
information
=&gt;best responses take into account how each
type of the other player would best respond and
how that affects your payoff weighing by the
probabilities of that type arising.
6
Bayesian equilibrium is a collection of strategies
such that each type of each player best
responds given her beliefs about other players’
types and their strategies.
7
• Incomplete information can be about:
Payoff functions
Actions available to other players
Beliefs of others about others’ payoffs, others’
beliefs, others’ strategies….
8
• Harsanyi (1968): Games of incomplete
information can be thought of as games of complete
but imperfect information where
• Nature makes the first move but not everyone
observes nature’s move.
• Nature (randomly) assigns types to players.
• Later won the Nobel prize for his work.
9
A Bayesian Game: Battle of the Sexes
with Private Information
• Player 1 prefers Bach and player 2 Stravinsky.
• Player 1 prefers to be with player 2.
• Player 2 knows player 1’s preferences.
• Player 1 thinks that with probability &frac12; player 2 wants
to go out with her, and with probability &frac12; player 2
wants to avoid her.
10
Battle of the Sexes
•
•
•
•
Players: The pair of people.
States: The set of states is W ={meet, avoid}.
Strategies: The strategies of each player are Si={B, S}.
Types: Player 1 has a single type t1= W.
Player 2 has two types {meet} and {avoid}.
• Beliefs: Player 1 assigns probability 1/2 to each state.
Player 2 knows the state.
11
Information Structure of the Battle of
the Sexes
2 B
B
2, 1
S
0, 0
2 B
S
B
0, 0
1, 2
S
S
2, 0
0, 2
0, 1
1, 0
1
There are two possible states, one in which the players’
payoffs are as in the left table and one in which these
payoffs are as in the right table. Nature chooses the state
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in the beginning of the game.
Payoffs: The payoffs vi (s, meet) of each player i for all
possible strategy pairs s are given in the left panel, and
the payoffs vi (s, avoid) are given in the right panel.
2
B
S
B
S
B
2
2, 1
0, 0
0, 0
1, 2
B
S
S
2, 0
0, 2
0, 1
1, 0
1
13
2 B
B
2, 1
S
0, 0
2
S
B
B
0, 0
1, 2
S
S
2, 0
0, 2
0, 1
1, 0
1
• Player 2 knows the state -whether she wishes to
meet or avoid player 1- whereas player 1 does not;
player 1 assigns probability 1/2 to each state.
• From player 1’s point of view, player 2 has two
possible types, one whose preferences are given in
the left matrix, type meet, and one with preferences
given in the right matrix, type avoid.
14
2
B
S
B
2
S
2, 1
0, 0
0, 0
1, 2
B
B
S
S
2, 0
0, 2
0, 1
1, 0
1
•
•
Player 1 does not know player 2’s type. To choose her
strategy, she forms a belief on each types’ strategy.
Given these beliefs and her belief about the
probability of each type, she calculates the payoff of
her strategies.
15
Bayesian Nash Equilibrium of the
Battle of the Sexes
A Bayesian Nash equilibrium (in this setup) is a triple of
strategies, one for player 1 and one for each type of player 2:
1. player 1’s strategy is optimal, given the strategies of the two
types of player 2 and player 1’s beliefs on player 2’s type.
2. the strategy of each type of player 2 is optimal, given player
1’s strategy.
16
B
S
B,B
B,S
S,B
2
1
1
0
1
2
1
0
1
2
S,S
Player 1’s
expected
payoff for
the four
possible pure
strategies of
player 2
• We can then characterise best response of
player 1 to pure strategies of player 2.
• Next, ask this: for this best response of player
1, does the player 2 want to play these
strategies?
• So, check : (B,(B,B)),(B,(BS)), (B,(S,B)) and
(S,(S,S))
17
2
B
S
B
2
S
2, 1
0, 0
0, 0
1, 2
B
B
S
S
2, 0
0, 2
0, 1
1, 0
1
So, check (B,(B,B)),(B,(B,S)), (B,(S,B)) and (S,(S,S)). The first
component is the strategy of player 1 and the second is the
pair of strategy of the two types of player 2.
18
B
B
S
S
2, 1
0, 0
B
B
0, 0
1, 2
S
2, 0
0, 2
0, 1
1, 0
2 2
S
1
(B,(B,S)) is a Bayesian Nash equilibrium:
1. Given the strategies (B, S) of player 2, player 1’s strategy
B is optimal given his belief.
2. Given that player 1 chooses B, B is optimal for player 2
type meet and S is optimal for player 2 type avoid.
19
B
B
S
S
2, 1
0, 0
B
B
0, 0
1, 2
S
2, 0
0, 2
0, 1
1, 0
2 2
S
1
There are no other (pure-strategy) Nash equilibria…
1. If player 1 plays S, then the two types of player 2
respond (S, B). The type who wants to meet 1 chooses
S and the type who wants to avoid 1 chooses B.
2. But given that the two types of player 2 respond (S, B),
player 1’s best response is B.
20
B
B
S
S
2, 1
0, 0
B
B
0, 0
1, 2
S
2, 0
0, 2
0, 1
1, 0
2 2
S
1
• What would happen if the probability of type meet is
𝜌 instead of &frac12;? (in pure strategies.)
21
• Solve the equilibrium for any 𝜌 ∈ [0,1]?
22
•
•
•
2
For 𝜌 ∈ [ ,1] both (B,(B,S)) and (S,(S,B))
3
1 2
For 𝜌 ∈ [ , ] , (B,(B,S))
3 3
1
For 𝜌 &lt; no pure strategy NE! check this!
3
23
Mixed strategy equilibria:
• The mixed strategy equilibria of the battle of the
sexes are calculated as follows. Call m type meet and
v type avoid.
• Player 1 is indifferent between S and B if and only if
2sm(B) + 2sv(B) = 1- sm(B) + 1- sv(B).
• Player 2 of type m is indifferent between S and B if
and only if s1(B) = 2(1- s1(B)), i.e. s1(B) = 2/3.
• Player 2 of type v is indifferent between S and B if
and only if 2s1(B) = 1- s1(B), i.e. s1(B) = 1/3.
24
• If type m is indifferent between S and B, then
s1(B) = 2/3, and thus type v chooses S.
• If type v is indifferent between S and B, then
s1(B) = 1/3, and thus type m chooses S.
• Hence there are two mixed strategy Bayesian Nash
Equilibria:
1. s1(B) = 1/3, sm(B) = 0, sv(B) = 2/3.
2. s1(B) = 2/3, sm(B) = 2/3, sv(B) = 0.
25
Formal definition:
Definition: A Bayesian game consists of
• a set of players I
• a set of states W
and for each player i:
• a set of strategies Si
• a set of types for each player Ti.
• for each type 𝑡𝑖, a belief 𝑝(𝑡−𝑖 |𝑡𝑖), a probability
distribution over other players’ types
• a payoff function vi over pairs (s, w), where s is a
strategy profile and w is a state.
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• Each state is a complete description of the players’
relevant characteristics, including their preferences and
their information.
• At the start of the game, nature chooses a state w.
The players do not observe it.
• Each player 𝑖 has a type ti . The types of a player 𝑖 are
disjoint sets, and cover the set of states W. The set of
types 𝑇𝑖 is a partition of W.
27
•
The set Ti reflects the quality of player i’s information.
If Ti is composed of sets of single states, then player i
always knows perfectly which state has occurred.
•
If Ti doesn’t contain any subsets, then player i never
has any information about the state.
•
If Ti contains more than one set, but not all of them
are single sets, then player i has partial information.
•
28
Bayesian Nash Equilibrium
• A strategy profile 𝑠 ∗ = {𝑠1∗ , … . 𝑠𝑛∗ } is a
Bayesian Nash equilibrium if and only if for each
type 𝑡𝑖 of each player 𝑖:
𝑠𝑖∗ 𝑡𝑖 ∈
𝑎𝑟𝑔𝑚𝑎𝑥𝑠𝑖
p(𝑡−𝑖 | ti) vi(si(𝑡𝑖 )s−i (𝑡−𝑖 )).
𝑡−𝑖 ∈𝑇−𝑖
29
Example:
2 L
U 1, -1
D
-1, 1
2
R
L
U
-1, 1
1, -1
D
R
1, 1
-1, -1
-1, 1
1, -1
1
The probability of left is &frac12;.
Find the Bayesian Nash Equilibria.
30
2 L
U 1, -1
D
-1, 1
2
R
L
U
-1, 1
1, -1
D
R
1, 1
-1, -1
-1, 1
1, -1
1
31
• (U, (R,L)) is the unique pure strategy Bayesian
NE. Realize that L is a dominant strategy of
player 2 type right.
• What about mixed?
Any mixed strategy (𝜎1 𝑈 , (R,L)) with 𝜎1 𝑈 ≥
1
1
. When 𝜎1 𝑈 &lt; , player 2 plays (L,L) but then
2
2
player 1 should have played U. So this is not
possible to arise.
32
Cournot’s Duopoly Game with
Imperfect Information
• Two firms compete a-la Cournot.
• Their costs are linear: Ci (qi) = ciqi, i=1,2.
• The demand is linear:
P (Q) = a – Q if a &gt; Q, P(Q) = 0 if a &lt; Q.
• Both firms know that firm 1’s unit cost is c.
• Firm 2 knows its own unit cost.
• Firm 1 believes that firm 2’s cost is cL with probability 𝜇
and cH with probability 1 − 𝜇; with 0 &lt; 𝜇 &lt; 1, cL &lt; cH.
33
• Players: Firm 1 and firm 2.
• States: 𝜔 ∈{L, H}.
• Strategies: Each firm’s strategies are all non-negative
numbers.
• Types: Firm 1 has one type: {L,H}. Firm 2 has two types:
{H} and {L}.
• Beliefs: Firm 1 assigns probability 𝜇 to state L and
probability 1 − 𝜇 to state H. Firm 2 knows the state.
• Payoff functions: The firms’ payoffs are their profits:
u1(q1, q2;w) = qi [a – (q1 + q2,{w})] - cqi
u2,{w}(q1, q2;w) = q2,{w}[a –(q1+ q2,{w})] − c w q2,{w}.
34
Bayesian Nash Equilibrium of the
Cournot’s Duopoly Game
• To find the best response functions of player 2’s
types w ∈ {L, H}, differentiate p2,{w} with respect to
q2,{w}, to find the first order condition:
dp2,{w}(q1, q2,{w}) /dq2,{w}
= a – q2,{w} – (q1+ q2,{w}) – cw = 0.
• Best Response functions:
b2,{w} (q1) = [a – q1 – cw]/2.
35
• For player 1, the expected profit is
p1(q1, q2) = q1 [a – (q1 + 𝜇 q2L +(1- 𝜇)q2H )] – cq1
• The best response function is found as follows:
dp1(q1, q2)/dq1 = a – 2q1–(𝜇 q2L +(1- 𝜇)q2H ) – c =0.
b1(q2) = [a – (𝜇 q2L+(1- 𝜇)q2H ) – c]/2.
• The Bayesian Nash Equilibrium is found by solving the
system defined by the best response functions:
q*1= [a – 2c + 𝜇 cL +(1- 𝜇)cH]/3,
q*2L= [2a + 2c - 3cL - 𝜇 cL - (1- 𝜇)cH]/6,
q*2H= [2a + 2c - 3cH - 𝜇 cL - (1- 𝜇)cH]/6.
36
Public Good Provision
• A public good is provided to a group of n people if at
least one person pays the cost of the good, c.
• Each person’s valuation of the good is vi. If the good
is not provided, then each individual’s payoff is 0.
• Each person i knows her own valuation vi, but
does know anyone else’s valuation.
• It is common knowledge that valuations are
independently and identically distributed, with
cumulative distribution function F.
37
• All n individuals simultaneously submit contributions
of either c or 0 (no intermediate contributions
allowed).
• If all individuals submit 0, then the good is not
provided and each individual’s payoff is 0.
• If at least one individual submits c then the good is
provided, each individual i who submits c obtains the
payoff vi - c, and each i who submits 0 obtains vi.
38
Bayesian Game Representation
• Players: The set of n individuals.
• States: The set of all profiles (v1, . . . , vn) of valuations,
where
0 &lt; vi &lt; 1 for all i.
• Strategies: Each player’s set of strategies is {0, c}.
• Types: Each player i’s has a type which consists of her
valuation vi and of the set of all profiles v-i of opponents’
valuations.
39
• Beliefs: Each type of player i assigns probability
F(v1)F(v2) &middot; &middot; &middot; F(vi −1)F(vi+1) &middot; &middot; &middot; F(vn) to the event that
the valuation of every other player j is less than vj.
• Payoff functions: Player i’s payoff in state (v1, . . . , vn)
is ui(v1, . . . , vn) = 0 if no one contributes, ui(v1, . . . , vn) =
vi if i does not contribute but some other player does,
ui(v1, . . . , vn) = vi − c if i contributes.
40
Bayesian Nash Equilibria
The game has a symmetric Nash equilibrium in which every
player i contributes if and only if vi ≥ v∗.

Consider player i. Suppose that every other player j
contributes if and only if vj ≥ v∗.

The probability that at least one of the other players
contributes is 1− (F(v∗))n−1.
41
• Player i’s expected payoff is [1 − (F(v∗))n−1] vi if she does
not contribute and vi − c if she contributes.
• The conditions for player i to not contribute when vi &lt; v∗
and contribute when vi ≥ v∗ are:
(1−(F(v∗))n−1)vi &gt; vi−c if vi&lt; v∗,
(1−(F(v∗))n−1)vi ≤ vi−c if vi ≥ v∗:
I.e. vi(F(v∗))n−1 &lt; c if vi &lt; v∗, vi(F(v∗))n−1 ≥ c if vi ≥ v∗.
• Hence, in equilibrium, v∗(F(v∗))n−1 = c.
42
• As the number of individuals increases, is the good
more or less likely to be provided in this equilibrium?
• The probability that the good is provided is the
probability that at least for one i, vi ≥ v∗, which is equal
to 1 − (F(v∗))n.
• In equilibrium, this probability is equal to 1 − cF(v∗)/v∗.
• The value of v∗ increases as n increases.
• As n increases the change in the probability that the
good is provided increases if F(v∗)/v∗ decreases in v∗,
whereas it decreases if F(v∗)/v∗ increases in v∗.
43
Juries
• In a trial, jurors are presented with evidence on the
guilt or innocence of a defendant.
• They may interpret the evidence differently.
• Each juror votes either to convict or acquit the
defendant.
• A unanimous verdict is required for conviction: the
defendant is convicted if and only if every juror votes to
convict her.
44
• In deciding how to vote, each juror considers the
costs of convicting an innocent person and of acquitting
a guilty person, on the basis of her information.
• When voting, she conditions her choice on being
pivotal, i.e. changing the outcome of the trial.
• The event of being pivotal is informative: It gives the
juror information about the private information held by
the other jurors, as this determines their votes.
45
A Model of Jury Vote
• Each juror comes to the trial with the belief that the
defendant is guilty with probability π.
• Given the defendant’s status (guilty or innocent),
each juror receives a signal.
• The probability that the signal is “guilty” when the
defendant is guilty is p, and the probability that the
signal is “innocent” if the defendant is innocent is q.
• Jurors are more likely to get a correct signal: p &gt; 1/2
and q &gt; 1/2, and hence p &gt; 1 − q.
46
• Each juror wishes to convict a guilty defendant and
acquit an innocent one.
• Each juror’s payoffs are:
−z if innocent defendant convicted
−(1 − z) if guilty defendant acquitted.
0 if an innocent defendant is acquitted, or if a guilty
defendant is convicted.
47
• Let r be the updated probability of the defendant’s
guilt, given a juror’s information.
• Her expected payoff if the defendant is acquitted is
−r(1 − z) + (1 − r) &middot; 0 = −r(1 − z)
and her expected payoff if the defendant is convicted is
r &middot; 0 − (1 − r)z = −(1 −r)z.
• She prefers the defendant to be acquitted if r &lt; z, and
convicted if r &gt; z.
48
Bayesian Game of Jury Vote
• Players The set of n jurors.
• States The states are the set of all profiles (X, s1, . . .
, sn) where X ∈ {G, I} and sj ∈ {g, b} for every juror j.
X = G if the defendant is guilty,
X = I if she is innocent.
si = g if player i receives the signal “guilty,”
si = b if player i receives the signal “innocent.”
49
• Strategies The set of strategies of each player is
v = {C, Q}, where C is voting to convict,
and Q is voting to acquit.
• Types Each player i’s type consists in a signal si
together with the set of all profiles (X, s-i), where s-i
denotes i’s opponents’ signals.
50
Beliefs:
• Type g of a player i believes that the state is
(G, s1, . . . , sn) with probability Pr(G|g)pk−1(1− p)n-k
and (I, s1, . . . , sn) with probability Pr(I|g)(1-q)k−1qn-k,
where k is the number of players j (including i) with
sj=g.
• Type b believes that the state is (G, s1, . . . , sn) with
probability Pr(G|b)pk(1− p)n-k-1 and (I, s1, . . . , sn) with
probability Pr(I|b)qn-k-1(1− q)k, where k is the number of
players j for whom sj = g.
51
• Payoff functions The payoff function of each player i is:
ui(v, X) = 0
if v ≠ (C,...,C) and X=I
or if v = (C,...,C) and X=G,
ui(v, X) = −z
if v = (C,...,C) and X = I
ui(v, X) =−(1 − z) if v ≠ (C,...,C) and X = G,
where X is the first component of the state, giving the
defendant’s true status.
52
Bayes’ rule:
Pr(A|B)=
Pr(A∩B)
Pr(B)
𝑃(𝐵|𝐴)(𝐴)
=
𝑃(𝐵)
53
Bayesian Nash Equilibrium
• One juror problem: Suppose there is a single juror with
signal b. To see if she prefers conviction or acquittal, we
find the probability Pr(G|b) that the defendant’s is guilty.
By the Bayes’ Rule:
Pr(b|G) Pr(G)
Pr(G|b) =
=
Pr(b|G)Pr(G)+Pr(b|I)Pr(I)
(1 − p)π
(1 − p)π + q(1 − π)
• The juror votes Acquittal if and only if:
(1 − p)π
z≥
(1 − p)π + q(1 − π)
54
• The juror votes Convict given g if and only if:
=&gt;
P(G|g)≥ z
pπ
z≤
pπ + (1−q)(1 − π)
Then, truthful if and only if:
(1 − p)π
pπ
≤𝑧≤
(1 − p)π + q(1 − π)
pπ + (1−q)(1 − π)
55
• Suppose there are n jurors. We show that truthful voting
is not a Bayesian Equilibrium.
• Suppose by contradiction that every juror other than 1
votes truthfully (acquit if her signal is b, convict if it is g).
• Consider type b of juror 1. Her vote has no effect on the
outcome unless every other juror’s signal is g.
• Hence, she votes Acquittal if the probability
Pr(G|b,g,...,g) that the defendant is guilty, given juror 1’s
signal is b and every other juror’s signal is g, is at most z.
56
• Specifically,
Pr(G|b,g,...,g) =
Pr(b,g,...,g|G)Pr(G)
[Pr(b,g,...,g|G)Pr(G)+Pr(b,g,...,g|I)Pr(I)]
(1 − p)𝐩𝑛−1 π
=[(1 − p) 𝑝𝑛−1π + q(1 − q) n−1 (1−π)]
• Hence, type b of juror 1 optimally votes for acquittal iff:
(1 − p)𝐩𝑛−1 π
z ≥
[(1 − p) 𝑝𝑛−1 π + q(1 − q) n−1 (1−π)]
=
1
1+
𝑞
1−𝑝
1−𝑞 𝑛−1 1−𝜋
𝑝
𝜋
• Comparing this to the single jury problem, we see that unless z is large
or n is small, voting truthfully is not a Bayesian Nash Equilibrium.
57
Exercise: two jurors
• Now suppose there are 2 jurors. The
defendant is convicted if and only if both vote
to convict.
• Find the value range of z for which there exists
a Nash equilibrium in which each juror votes
according to her signal.
58
59
Exercise: n jurors
• Find the conditions under which for n number
of jurors, there exists a NE in which every juror
votes for acquittal regardless of her signal?
• In addition, find conditions under which every
juror votes for conviction regardless of her
signal?
60
61
Mixed strategy equilibrium:
• Under some conditions there is a symmetric mixed
strategy equilibrium in which each type g juror votes for
conviction, and each type b juror randomizes.
• Denote by β the mixed strategy of each juror of type b.
• Each type b juror is indifferent between voting
conviction and acquittal.
• Hence the mixed strategy β is such that:
z = Pr(G | signal b, n−1 votes for C)
62
Pr(b|G)(Pr(vote C| G))n−1Pr(G)
z=
Pr(b|G)(Pr(vote C|G))n−1Pr(G)+Pr(b|I)(Pr(vote C|I))n−1Pr(I)
(1−p)(p+(1−p)β) n−1π
(1−p)(p+(1−p)β)n−1π +q(1−q+qβ) n−1(1 − π)
• This simplifies to:
(1−p)(p+(1−p)β)n−1π(1 − z) = q(1 − q + qβ)n−1(1 − π)z.
63
hence:
[pX − (1 − q)]
β=
[ q − (1 − p)X]
1/(n-1)
where
π(1 − p)(1 − z)
X=
[(1 − π)qz]
64
• When n is large, X is close to 1, and hence β nears 1:
a juror who interprets the evidence as pointing to
innocence very likely nonetheless votes for conviction.
• An interesting property of this equilibrium is that the
probability that an innocent defendant is convicted
increases as n increases: the larger the jury, the more
likely an innocent defendant is to be convicted.
i.e., Pr(C,n|I) = {1 – (1-β)(1-q)}n increases in n.
65
Summary of the Lecture

Bayesian Games and Nash Equilibrium

Bayesian Games and Information

Cournot Duopoly with Private Information

Public Good Provision with Private Information

Juries and Information Aggregation
66
Preview of the Next Lecture

Jonathan Cave will talk about:
Auctions and Mechanism Design
67
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