stata代写-ECON 2112
时间:2022-02-14
Mixed strategies, Dominant and Dominated Strategies,
Applications
Week 2, ECON 2112
Arghya Ghosh
Mixed Strategies
Mixed Strategies
Recall our definition of a game:
Definition 1 (Normal Form game)
An n-player normal form game G = (S1, S2, . . . ,Sn; u1, u2, . . . , un) consists of
ˆ for each player i = 1, 2, . . . , n, a set of strategies Si ; and
ˆ for each player i = 1, 2, . . . , n a utility function ui that, for each strategy profile
(s1, s2, . . . , sn) ∈ S1 × S2 · · · × Sn specifies a real number ui (s1, s2, . . . , sn) ∈ R.
ˆ (informally) A pure strategy is a “definite” choice.
ˆ A mixed strategy is a probability distribution over the possible “definite” choices.
1
Mixed Strategies
Recall our definition of a game:
Definition 1 (Normal Form game)
An n-player normal form game G = (S1, S2, . . . ,Sn; u1, u2, . . . , un) consists of
ˆ for each player i = 1, 2, . . . , n, a set of pure strategies Si ; and
ˆ for each player i = 1, 2, . . . , n a utility function ui that, for each pure strategy
profile (s1, s2, . . . , sn) ∈ S1 × S2 · · · × Sn specifies a real number
ui (s1, s2, . . . , sn) ∈ R.
ˆ (informally) A pure strategy is a “definite” choice.
ˆ A mixed strategy is a probability distribution over the possible “definite” choices.
1
Mixed strategies
Example 2
If player 1 has two strategies, e.g., S1 = {T ,B}, examples of mixed strategies are:
1
2
T +
1
2
B,
1
3
T +
2
3
B,
αT + (1− α)B for any 0 ≤ α ≤ 1
Definition 3 (Mixed Strategy)
A mixed strategy for player i is a probability distribution pi over his set of pure
strategies Si .
2
Mixed strategies
Example 4
If player 1 has two strategies, e.g., S1 = {T ,B}, we could also write the mixed
strategy
p1 = αT + (1− α)B
by just writing out the coefficients
p1 = (α, 1− α),
so long as we have imposed some order on the strategies. This has the advantage of
instantly resembling a coordinate in the Cartesian plane.
This notation tends to thus generalise more naturally when we have more strategies,
but for the purposes of this course we will stick to the former notation. Be aware of
this subtle difference if you consult other textbooks.
3
Nash equilibrium with mixed strategies
Definition 5 (Nash Equilibrium)
The mixed strategy profile p∗ = (p∗1 , p

2 , . . . , p

n) is a Nash equilibrium of G if for each
i = 1, . . . , n
ui (p

1 , . . . , p

i−1, p

i , p

i+1, . . . , p

n) ≥ ui (p∗1 , . . . , p∗i−1, pi , p∗i+1, . . . , p∗n)
for every other mixed strategy pi available to player i .
Note:
Each of the p∗i ’s in the mixed strategy profile p
∗ are themselves distributions over
player i ’s strategies (i = 1, . . . , n). So p∗i might look something like
1
2T +
1
2B (for
example).
4
Mixed Strategies
Consider an n-player normal form game G = (S1, . . . ,Sn; u1, . . . , un) where n is a finite
number and Si is finite for every i = 1, . . . , n.
Theorem 1 (Nash, 1950)
The game G has at least one Nash equilibrium, possibly, in mixed strategies.
Proof.
Unfortunately well beyond the scope of this course... take ECON6101!
Note:
Every pure strategy si can be expressed as a mixed strategy that gives probability one
to si and probability zero to every other strategy.
5
Mixed Strategies and Nash equilibrium
Suppose that p∗ = (p∗1 , . . . , p

n) is a Nash Equilibrium of some game G .
ˆ No player can unilaterally deviate from p∗ and obtain a higher payoff. Moreover,
it is enough to check the following:
p∗ = (p∗1 , . . . , p

n) is a Nash equilibrium if for every player i = 1, . . . , n:
ui (p

1 , . . . , p

i−1, p

i , p

i+1, . . . , p

n) ≥ ui (p∗1 , . . . , p∗i−1, si , p∗i+1, . . . , p∗n)
for every si ∈ Si (intuition?).
ˆ If players other than i play according to (p∗1 , . . . , p

i−1, p

i+1, . . . , p

n) then player i
is indifferent between any two pure strategies used with strictly positive
probability by p∗i .
6
Mixed Strategies and Nash equilibrium
Suppose that p∗ = (p∗1 , . . . , p

n) is a Nash Equilibrium of some game G .
ˆ No player can unilaterally deviate from p∗ and obtain a higher payoff. Moreover,
it is enough to check the following:
p∗ = (p∗1 , . . . , p

n) is a Nash equilibrium if for every player i = 1, . . . , n:
ui (p

1 , . . . , p

i−1, p

i , p

i+1, . . . , p

n) ≥ ui (p∗1 , . . . , p∗i−1, si , p∗i+1, . . . , p∗n)
for every si ∈ Si (intuition?).
ˆ If players other than i play according to (p∗1 , . . . , p

i−1, p

i+1, . . . , p

n) then player i
is indifferent between any two pure strategies used with strictly positive
probability by p∗i .
6
Mixed Strategies
Example 6
H T
H 1,−1 −1, 1
T −1, 1 1,−1
Under the Nash equilibrium (12H +
1
2T ,
1
2H +
1
2T ) player 1 is indifferent between H
and T and player 2 is indifferent between H and T .
7
Nash equilibrium: Equivalent definition
Consider an n-player normal form game G = {S1,S2, . . . ,Sn; u1, . . . , un} and a mixed
strategy profile (p1, p2, . . . , pn).
Definition 7
The set of player i ’s best responses against (p1, . . . , pi−1, pi+1, . . . , pn) is the set of
player i ’s strategies that solve:
max
p′i
ui (p1, . . . , pi−1, p′i , pi+1, . . . , pn).
Furthermore, we denote as Ri (p1, p2, . . . , pn) the set of player i ’s best responses
against (p1, p2, . . . , pn).
8
Nash equilibrium: Equivalent definition
Definition 8
The strategy profile (p∗1 , p

2 , . . . , p

n) is a Nash equilibrium if for every i = 1, . . . , n we
have p∗i ∈ Ri (p∗1 , p∗2 , . . . , p∗n).
9
Mixed Strategies
Example 9 (Battle of the sexes)
L R
T 2, 1 0, 0
B 0, 0 1, 2
ˆ The game has two Nash equilibria in pure strategies (T , L) and (B,R).
ˆ What about Nash equilibria in mixed strategies?
10
Mixed Strategies
Example 10 (Battle of the sexes)
L R
T 2, 1 0, 0
B 0, 0 1, 2
ˆ Let x be the probability that player 1 plays T .
ˆ Let y be the probability that player 2 plays L.
R1(y)

x = 1 if y > 13
0 ≤ x ≤ 1 if y = 13
x = 0 if y < 13 .
R2(x)

y = 1 if x > 23
0 ≤ y ≤ 1 if x = 23
y = 0 if x < 23 .
ˆ The NE are: (T , L), (B,R) and (23T +
1
3B,
1
3L +
2
3R)
11
Mixed Strategies
Example 10 (Battle of the sexes)
L R
T 2, 1 0, 0
B 0, 0 1, 2
ˆ Let x be the probability that player 1 plays T .
ˆ Let y be the probability that player 2 plays L.
R1(y)

x = 1 if y > 13
0 ≤ x ≤ 1 if y = 13
x = 0 if y < 13 .
R2(x)

y = 1 if x > 23
0 ≤ y ≤ 1 if x = 23
y = 0 if x < 23 .
ˆ The NE are: (T , L), (B,R) and (23T +
1
3B,
1
3L +
2
3R)
11
Mixed Strategies
Example 10 (Battle of the sexes)
L R
T 2, 1 0, 0
B 0, 0 1, 2
ˆ Let x be the probability that player 1 plays T .
ˆ Let y be the probability that player 2 plays L.
R1(y)

x = 1 if y > 13
0 ≤ x ≤ 1 if y = 13
x = 0 if y < 13 .
R2(x)

y = 1 if x > 23
0 ≤ y ≤ 1 if x = 23
y = 0 if x < 23 .
ˆ The NE are: (T , L), (B,R) and (23T +
1
3B,
1
3L +
2
3R)
11
Mixed Strategies
Example 10 (Battle of the sexes)
L R
T 2, 1 0, 0
B 0, 0 1, 2
ˆ Let x be the probability that player 1 plays T .
ˆ Let y be the probability that player 2 plays L.
R1(y)

x = 1 if y > 13
0 ≤ x ≤ 1 if y = 13
x = 0 if y < 13 .
R2(x)

y = 1 if x > 23
0 ≤ y ≤ 1 if x = 23
y = 0 if x < 23 .
ˆ The NE are: (T , L), (B,R) and (23T +
1
3B,
1
3L +
2
3R) 11
Dominant and Dominated
Strategies
Motivation
ˆ Under what conditions can we say that a strategy is unequivocally better than
another?
ˆ Certainly this should be the case if a strategy always gives a larger payoff than a
second strategy no matter what the other players do.
12
Dominant Strategies
Example 11 (Prisoners’ Dilemma)
Not confess Confess
Not confess −1,−1 −9, 0
Confess 0,−9 −6,−6
R1(Not Confess) =
{Confess}
R1(Confess) =
{Confess}
R2(Not Confess) =
{Confess}
R2(Confess) =
{Confess}
ˆ For any player, whatever the other player does, Confess pays more than Not
Confess.
ˆ We say that the strategy Not Confess is strictly dominated by Confess, or that
the strategy Confess strictly dominates the strategy Not Confess. 13
Dominant Strategies
Example 11 (Prisoners’ Dilemma)
Not confess Confess
Not confess −1,−1 −9, 0
Confess 0,−9 −6,−6
R1(Not Confess) = {Confess}
R1(Confess) =
{Confess}
R2(Not Confess) =
{Confess}
R2(Confess) =
{Confess}
ˆ For any player, whatever the other player does, Confess pays more than Not
Confess.
ˆ We say that the strategy Not Confess is strictly dominated by Confess, or that
the strategy Confess strictly dominates the strategy Not Confess. 13
Dominant Strategies
Example 11 (Prisoners’ Dilemma)
Not confess Confess
Not confess −1,−1 −9, 0
Confess 0,−9 −6,−6
R1(Not Confess) = {Confess}
R1(Confess) = {Confess}
R2(Not Confess) =
{Confess}
R2(Confess) =
{Confess}
ˆ For any player, whatever the other player does, Confess pays more than Not
Confess.
ˆ We say that the strategy Not Confess is strictly dominated by Confess, or that
the strategy Confess strictly dominates the strategy Not Confess. 13
Dominant Strategies
Example 11 (Prisoners’ Dilemma)
Not confess Confess
Not confess −1,−1 −9, 0
Confess 0,−9 −6,−6
R1(Not Confess) = {Confess}
R1(Confess) = {Confess}
R2(Not Confess) = {Confess}
R2(Confess) =
{Confess}
ˆ For any player, whatever the other player does, Confess pays more than Not
Confess.
ˆ We say that the strategy Not Confess is strictly dominated by Confess, or that
the strategy Confess strictly dominates the strategy Not Confess. 13
Dominant Strategies
Example 11 (Prisoners’ Dilemma)
Not confess Confess
Not confess −1,−1 −9, 0
Confess 0,−9 −6,−6
R1(Not Confess) = {Confess}
R1(Confess) = {Confess}
R2(Not Confess) = {Confess}
R2(Confess) = {Confess}
ˆ For any player, whatever the other player does, Confess pays more than Not
Confess.
ˆ We say that the strategy Not Confess is strictly dominated by Confess, or that
the strategy Confess strictly dominates the strategy Not Confess. 13
Normal Form Games
Definition 12 (Strictly Dominated Strategy)
Given a game G , the (mixed) strategy pi is strictly dominated by the (mixed)
strategy p′i 6= pi if for every pure strategy profile of the opponents
(s1, . . . , si−1, si+1, . . . , sn) we have:
ui (s1, . . . , si−1, pi , si+1, . . . , sn) < ui (s1, . . . , si−1, p′i , si+1, . . . , sn).
ˆ If a player has a strictly dominated strategy, it will never be used in a Nash
equilibrium.
14
Equilibrium in Dominant Strategies: Example
Example 13 (3x3 game)
L C R
T 1, 1 0, 2 5, 1
M 2, 0 1, 1 6, 0
B 0, 3 0, 6 5, 5
ˆ T and B are strictly dominated by M.
ˆ L and R are strictly dominated by C .
ˆ (M,C ) is the unique Nash equilibrium.
15
Strictly Dominated Strategies: Properties
ˆ If a strategy is strictly dominated, it will never be used in a Nash equilibrium and,
moreover:
Theorem 2
Let G be an n-player normal form game such that player i has a strictly dominated
strategy in G . Let G ′ be the n-player normal form game that is obtained from G by
removing such a strictly dominated pure strategy. Then G and G ′ have the same set
of Nash equilibria.
16
Strictly Dominated Strategies: Properties
Example 14
3 player game
L C R
T 1, 1, 1 0, 0, 1 1, 2, 1
B 0, 1, 1 2, 0, 1 0, 0, 1
X
L C R
T 1, 1, 0 1, 3, 0 2, 1, 0
B 1, 1, 0 2, 0, 0 2, 3, 0
Y
(T ,R,X ) is the unique Nash equilibrium of the original game.
The process is called iterated deletion of strictly dominated strategies.
17
Strictly Dominated Strategies: Properties
Example 14
3 player game
L C R
T 1, 1, 1 0, 0, 1 1, 2, 1
B 0, 1, 1 2, 0, 1 0, 0, 1
X
L C R
T 1, 1, 0 1, 3, 0 2, 1, 0
B 1, 1, 0 2, 0, 0 2, 3, 0
Y
(T ,R,X ) is the unique Nash equilibrium of the original game.
The process is called iterated deletion of strictly dominated strategies.
17
Strictly Dominated Strategies: Properties
Example 14
3 player game
L
C
R
T 1, 1, 1
0, 0, 1
1, 2, 1
B 0, 1, 1
2, 0, 1
0, 0, 1
X
L C R
T 1, 1, 0 1, 3, 0 2, 1, 0
B 1, 1, 0 2, 0, 0 2, 3, 0
Y
(T ,R,X ) is the unique Nash equilibrium of the original game.
The process is called iterated deletion of strictly dominated strategies.
17
Strictly Dominated Strategies: Properties
Example 14
3 player game
L
C
R
T 1, 1, 1
0, 0, 1
1, 2, 1
B 0, 1, 1 2, 0, 1 0, 0, 1
X
L C R
T 1, 1, 0 1, 3, 0 2, 1, 0
B 1, 1, 0 2, 0, 0 2, 3, 0
Y
(T ,R,X ) is the unique Nash equilibrium of the original game.
The process is called iterated deletion of strictly dominated strategies.
17
Strictly Dominated Strategies: Properties
Example 14
3 player game
L C
R
T
1, 1, 1 0, 0, 1
1, 2, 1
B 0, 1, 1 2, 0, 1 0, 0, 1
X
L C R
T 1, 1, 0 1, 3, 0 2, 1, 0
B 1, 1, 0 2, 0, 0 2, 3, 0
Y
(T ,R,X ) is the unique Nash equilibrium of the original game.
The process is called iterated deletion of strictly dominated strategies.
17
Strictly Dominated Strategies: Properties
Example 14
3 player game
L C
R
T
1, 1, 1 0, 0, 1
1, 2, 1
B 0, 1, 1 2, 0, 1 0, 0, 1
X
L C R
T 1, 1, 0 1, 3, 0 2, 1, 0
B 1, 1, 0 2, 0, 0 2, 3, 0
Y
(T ,R,X ) is the unique Nash equilibrium of the original game.
The process is called iterated deletion of strictly dominated strategies.
17
Strictly Dominated Strategies: Properties
Example 15 (3 player game)
L C R
T 2, 1, 1 0, 0, 1 0, 0, 1
B 0, 0, 1 1,−1, 1 1, 2, 1
X
L C R
T 1, 1, 0 1, 3, 0 2, 1, 0
B 1, 1, 0 2, 0, 0 2, 3, 0
Y
We have reduced the original game to the Battle of the Sexes between players 1 and 2.
18
Strictly Dominated Strategies: Properties
Example 15 (3 player game)
L C R
T 2, 1, 1 0, 0, 1 0, 0, 1
B 0, 0, 1 1,−1, 1 1, 2, 1
X
L C R
T 1, 1, 0 1, 3, 0 2, 1, 0
B 1, 1, 0 2, 0, 0 2, 3, 0
Y
We have reduced the original game to the Battle of the Sexes between players 1 and 2.
18
Strictly Dominated Strategies: Properties
Example 15 (3 player game)
L
C
R
T 2, 1, 1
0, 0, 1
0, 0, 1
B 0, 0, 1
1,−1, 1
1, 2, 1
X
L C R
T 1, 1, 0 1, 3, 0 2, 1, 0
B 1, 1, 0 2, 0, 0 2, 3, 0
Y
We have reduced the original game to the Battle of the Sexes between players 1 and 2.
18
Strictly Dominated Strategies: Properties
Example 15 (3 player game)
L
C
R
T 2, 1, 1
0, 0, 1
0, 0, 1
B 0, 0, 1
1,−1, 1
1, 2, 1
X
L C R
T 1, 1, 0 1, 3, 0 2, 1, 0
B 1, 1, 0 2, 0, 0 2, 3, 0
Y
We have reduced the original game to the Battle of the Sexes between players 1 and 2.
18
Weakly Dominated Strategies: Motivation
Example 16
Rational players will never play strictly dominated strategies
L R
T 1, 1 1, 0
B 0, 1 0, 0
ˆ If player 2 plays L player 1 is better off playing T .
ˆ If player 2 plays R player 1 is better off playing T .
ˆ Therefore, player 1 will never play B. (And player 2 will never play R.)
19
Weakly Dominated Strategies: Motivation
Example 16
Rational players will never play strictly dominated strategies
L R
T 1, 1 1, 0
B 0, 1 0, 0
ˆ If player 2 plays L player 1 is better off playing T .
ˆ If player 2 plays R player 1 is better off playing T .
ˆ Therefore, player 1 will never play B. (And player 2 will never play R.)
19
Weakly Dominated Strategies: Motivation
Example 16
Rational players will never play strictly dominated strategies
L R
T 1, 1 1, 0
B 0, 1 0, 0
ˆ If player 2 plays L player 1 is better off playing T .
ˆ If player 2 plays R player 1 is better off playing T .
ˆ Therefore, player 1 will never play B. (And player 2 will never play R.)
19
Weakly Dominated Strategies: Motivation
Example 16
Rational players will never play strictly dominated strategies
L R
T 1, 1 1, 0
B 0, 1 0, 0
ˆ If player 2 plays L player 1 is better off playing T .
ˆ If player 2 plays R player 1 is better off playing T .
ˆ Therefore, player 1 will never play B. (And player 2 will never play R.)
19
Weakly Dominated strategies.
What happens in the following situation?
L R
T 1, 1 0, 0
B 0, 0 0, 0
ˆ If player 2 plays L player 1 is better off playing T .
ˆ If player 2 plays R player 1 is indifferent between T and B.
ˆ That is, for player 1, strategy T always does at least as good as strategy B and
sometimes does strictly better.
20
Weakly Dominated strategies.
Definition 17 (Weakly Dominated Strategy (informal))
The strategy pi is weakly dominated by p

i if, whatever the other players play, p

i is
always at least as good as pi and sometimes strictly better.
Definition 18 (Weakly Dominated Strategies)
Given a game G we say that strategy pi is weakly dominated by strategy p

i if for
every pure strategy profile of the opponents s−i = (s1, . . . , si−1, si+1, . . . , sn)
ui (s−i , pi ) ≤ ui (s−i , p′i )
and there exists at least one pure strategy profile s ′−i such that
ui (s

−i , pi ) < ui (s

−i , p

i ).
21
Weakly and Strictly Dominated Strategies: Comparison
Definition 12 (Strictly Dominated Strategy)
Given a game G we say that the (mixed) strategy pi is strictly dominated by the
(mixed) strategy p′i 6= pi if for every s−i
ui (s−i , pi ) < ui (s−i , p′i ).
Definition 18 (Weakly Dominated Strategies)
Given a game G we say that strategy pi is weakly dominated by p

i 6= si if for every
s−i
ui (s−i , pi ) ≤ ui (s−i , p′i ),
and there exists at least one s ′−i such that
ui (s

−i , pi ) < ui (s

−i , p

i ).
22
Weakly Dominated strategies.
What happens in the following situation?
L R
T 1, 1 0, 0
B 0, 0 0, 0
ˆ Strategy T is always at least as good as strategy B, and sometimes (when player
2 plays L) is strictly better.
ˆ We may be interested in equilibria where neither strictly nor weakly dominated
strategies are played. A Nash equilibrium satisfying this property is called
admissible.
ˆ Nonetheless, note that (B,R) is a Nash equilibrium.
ˆ However, (T , L) is the unique admissible (or undominated) Nash equilibrium.
23
Dominated Strategies: Summary
To Summarise
ˆ Players will never use strictly dominated strategies in a Nash equilibrium.
ˆ Therefore, if the objective is to compute Nash equilibria, it is safe to apply
iterated deletion of strictly dominated strategies.
ˆ As we have seen, there are Nash equilibria where players play weakly dominated
strategies.
ˆ Therefore, if the objective is to compute Nash equilibria, it is not safe to apply
iterated deletion of weakly dominated strategies.
24
Dominated Strategies: Summary
To Summarise
ˆ Players will never use strictly dominated strategies in a Nash equilibrium.
ˆ Therefore, if the objective is to compute Nash equilibria, it is safe to apply
iterated deletion of strictly dominated strategies.
ˆ As we have seen, there are Nash equilibria where players play weakly dominated
strategies.
ˆ Therefore, if the objective is to compute Nash equilibria, it is not safe to apply
iterated deletion of weakly dominated strategies.
24
Dominated Strategies: Summary
To Summarise
ˆ Players will never use strictly dominated strategies in a Nash equilibrium.
ˆ Therefore, if the objective is to compute Nash equilibria, it is safe to apply
iterated deletion of strictly dominated strategies.
ˆ As we have seen, there are Nash equilibria where players play weakly dominated
strategies.
ˆ Therefore, if the objective is to compute Nash equilibria, it is not safe to apply
iterated deletion of weakly dominated strategies.
24
Dominated Strategies: Summary
To Summarise
ˆ Players will never use strictly dominated strategies in a Nash equilibrium.
ˆ Therefore, if the objective is to compute Nash equilibria, it is safe to apply
iterated deletion of strictly dominated strategies.
ˆ As we have seen, there are Nash equilibria where players play weakly dominated
strategies.
ˆ Therefore, if the objective is to compute Nash equilibria, it is not safe to apply
iterated deletion of weakly dominated strategies.
24
Applications
Normal form games
Definition 1 (Finite Normal Form Game)
An n-player finite normal form game G = (S1, S2, . . . ,Sn; u1, u2, . . . , un) consists of
ˆ for each player i = 1, . . . , n a set of pure strategies Si , and
ˆ for each player i = 1, . . . , n a utility function ui defined on the set of pure strategy
profiles.
ˆ So far the strategy set Si has always been finite.
ˆ However, we can also work with Si infinite.
ˆ The definitions of Nash equilibrium, strictly and weakly dominated strategies do
not change.
25
Normal form games
Definition 1 (Finite Normal Form Game)
An n-player finite normal form game G = (S1, S2, . . . ,Sn; u1, u2, . . . , un) consists of
ˆ for each player i = 1, . . . , n a set of pure strategies Si , and
ˆ for each player i = 1, . . . , n a utility function ui defined on the set of pure strategy
profiles.
ˆ So far the strategy set Si has always been finite.
ˆ However, we can also work with Si infinite.
ˆ The definitions of Nash equilibrium, strictly and weakly dominated strategies do
not change.
25
Normal form games
Definition 1 (Finite Normal Form Game)
An n-player finite normal form game G = (S1, S2, . . . ,Sn; u1, u2, . . . , un) consists of
ˆ for each player i = 1, . . . , n a set of pure strategies Si , and
ˆ for each player i = 1, . . . , n a utility function ui defined on the set of pure strategy
profiles.
ˆ So far the strategy set Si has always been finite.
ˆ However, we can also work with Si infinite.
ˆ The definitions of Nash equilibrium, strictly and weakly dominated strategies do
not change.
25
Applications
Cournot model of duopoly.
Cournot model of duopoly: Assumptions.
ˆ There are two firms, 1 and 2 competing in a duopolistic market.
ˆ These firms produce an identical good.
ˆ Firms choose the quantities that they are going to produce.
ˆ The market clearing price when Q = q1 + q2 is the total amount in the market is:
p =
a− Q if Q < a (where a > 0)0 if Q ≥ a.
ˆ Producing q1 units of the good costs cq1 dollars to firm 1.
ˆ Producing q2 units of the good costs cq2 dollars to firm 2. We assume c < a.
26
Cournot model of duopoly: Assumptions.
ˆ There are two firms, 1 and 2 competing in a duopolistic market.
ˆ These firms produce an identical good.
ˆ Firms choose the quantities that they are going to produce.
ˆ The market clearing price when Q = q1 + q2 is the total amount in the market is:
p =
a− Q if Q < a (where a > 0)0 if Q ≥ a.
ˆ Producing q1 units of the good costs cq1 dollars to firm 1.
ˆ Producing q2 units of the good costs cq2 dollars to firm 2. We assume c < a.
26
Cournot model of duopoly: Assumptions.
ˆ There are two firms, 1 and 2 competing in a duopolistic market.
ˆ These firms produce an identical good.
ˆ Firms choose the quantities that they are going to produce.
ˆ The market clearing price when Q = q1 + q2 is the total amount in the market is:
p =
a− Q if Q < a (where a > 0)0 if Q ≥ a.
ˆ Producing q1 units of the good costs cq1 dollars to firm 1.
ˆ Producing q2 units of the good costs cq2 dollars to firm 2. We assume c < a.
26
Cournot model of duopoly: Assumptions.
ˆ There are two firms, 1 and 2 competing in a duopolistic market.
ˆ These firms produce an identical good.
ˆ Firms choose the quantities that they are going to produce.
ˆ The market clearing price when Q = q1 + q2 is the total amount in the market is:
p =
a− Q if Q < a (where a > 0)0 if Q ≥ a.
ˆ Producing q1 units of the good costs cq1 dollars to firm 1.
ˆ Producing q2 units of the good costs cq2 dollars to firm 2. We assume c < a.
26
Cournot model of duopoly: Assumptions.
ˆ There are two firms, 1 and 2 competing in a duopolistic market.
ˆ These firms produce an identical good.
ˆ Firms choose the quantities that they are going to produce.
ˆ The market clearing price when Q = q1 + q2 is the total amount in the market is:
p =
a− Q if Q < a (where a > 0)0 if Q ≥ a.
ˆ Producing q1 units of the good costs cq1 dollars to firm 1.
ˆ Producing q2 units of the good costs cq2 dollars to firm 2. We assume c < a.
26
Cournot model of duopoly: Assumptions.
ˆ There are two firms, 1 and 2 competing in a duopolistic market.
ˆ These firms produce an identical good.
ˆ Firms choose the quantities that they are going to produce.
ˆ The market clearing price when Q = q1 + q2 is the total amount in the market is:
p =
a− Q if Q < a (where a > 0)0 if Q ≥ a.
ˆ Producing q1 units of the good costs cq1 dollars to firm 1.
ˆ Producing q2 units of the good costs cq2 dollars to firm 2. We assume c < a.
26
Cournot model of duopoly: The game.
ˆ Players: Firms 1 and 2.
ˆ Strategy set of firm 1: any number larger or equal to zero. That is, S1 = [0,∞).
ˆ Strategy set of firm 2: any number larger or equal to zero. That is, S2 = [0,∞).
ˆ For each strategy profile (q1, q2) we need to specify payoffs to player (firm) 1 and
player (firm) 2.
ˆ pi1(q1, q2) = pq1 − cq1 =
{
(a− q1 − q2)q1 − cq1 if q1 + q2 < a
−cq1 if q1 + q2 ≥ a.
ˆ pi2(q1, q2) = pq2 − cq2 =
{
(a− q1 − q2)q2 − cq2 if q1 + q2 < a
−cq2 if q1 + q2 ≥ a.
27
Cournot model of duopoly: The game.
ˆ Players: Firms 1 and 2.
ˆ Strategy set of firm 1: any number larger or equal to zero. That is, S1 = [0,∞).
ˆ Strategy set of firm 2: any number larger or equal to zero. That is, S2 = [0,∞).
ˆ For each strategy profile (q1, q2) we need to specify payoffs to player (firm) 1 and
player (firm) 2.
ˆ pi1(q1, q2) = pq1 − cq1 =
{
(a− q1 − q2)q1 − cq1 if q1 + q2 < a
−cq1 if q1 + q2 ≥ a.
ˆ pi2(q1, q2) = pq2 − cq2 =
{
(a− q1 − q2)q2 − cq2 if q1 + q2 < a
−cq2 if q1 + q2 ≥ a.
27
Cournot model of duopoly: The game.
ˆ Players: Firms 1 and 2.
ˆ Strategy set of firm 1: any number larger or equal to zero. That is, S1 = [0,∞).
ˆ Strategy set of firm 2: any number larger or equal to zero. That is, S2 = [0,∞).
ˆ For each strategy profile (q1, q2) we need to specify payoffs to player (firm) 1 and
player (firm) 2.
ˆ pi1(q1, q2) = pq1 − cq1 =
{
(a− q1 − q2)q1 − cq1 if q1 + q2 < a
−cq1 if q1 + q2 ≥ a.
ˆ pi2(q1, q2) = pq2 − cq2 =
{
(a− q1 − q2)q2 − cq2 if q1 + q2 < a
−cq2 if q1 + q2 ≥ a.
27
Cournot model of duopoly: The game.
ˆ Players: Firms 1 and 2.
ˆ Strategy set of firm 1: any number larger or equal to zero. That is, S1 = [0,∞).
ˆ Strategy set of firm 2: any number larger or equal to zero. That is, S2 = [0,∞).
ˆ For each strategy profile (q1, q2) we need to specify payoffs to player (firm) 1 and
player (firm) 2.
ˆ pi1(q1, q2) = pq1 − cq1 =
{
(a− q1 − q2)q1 − cq1 if q1 + q2 < a
−cq1 if q1 + q2 ≥ a.
ˆ pi2(q1, q2) = pq2 − cq2 =
{
(a− q1 − q2)q2 − cq2 if q1 + q2 < a
−cq2 if q1 + q2 ≥ a.
27
Cournot model of duopoly: The game.
ˆ Players: Firms 1 and 2.
ˆ Strategy set of firm 1: any number larger or equal to zero. That is, S1 = [0,∞).
ˆ Strategy set of firm 2: any number larger or equal to zero. That is, S2 = [0,∞).
ˆ For each strategy profile (q1, q2) we need to specify payoffs to player (firm) 1 and
player (firm) 2.
ˆ pi1(q1, q2) = pq1 − cq1 =
{
(a− q1 − q2)q1 − cq1 if q1 + q2 < a
−cq1 if q1 + q2 ≥ a.
ˆ pi2(q1, q2) = pq2 − cq2 =
{
(a− q1 − q2)q2 − cq2 if q1 + q2 < a
−cq2 if q1 + q2 ≥ a.
27
Cournot model of duopoly: The game.
ˆ Players: Firms 1 and 2.
ˆ Strategy set of firm 1: any number larger or equal to zero. That is, S1 = [0,∞).
ˆ Strategy set of firm 2: any number larger or equal to zero. That is, S2 = [0,∞).
ˆ For each strategy profile (q1, q2) we need to specify payoffs to player (firm) 1 and
player (firm) 2.
ˆ pi1(q1, q2) = pq1 − cq1 =
{
(a− q1 − q2)q1 − cq1 if q1 + q2 < a
−cq1 if q1 + q2 ≥ a.
ˆ pi2(q1, q2) = pq2 − cq2 =
{
(a− q1 − q2)q2 − cq2 if q1 + q2 < a
−cq2 if q1 + q2 ≥ a.
27
Doupoly and Cournot competition: Solution.
Definition 19 (Nash equilibrium for two firms in competition)
A strategy profile (q∗1 , q

2) is a Nash equilibrium if
pi1(q

1 , q

2) ≥pi1(q1, q∗2) for every q1 ∈ [0,∞)
pi2(q

1 , q

2) ≥pi2(q∗1 , q2) for every q2 ∈ [0,∞)
ˆ Fix the production of firm 2 to q2. What is the best strategy that firm 1 can take?
ˆ Choose q1 that satisfies:
max
q1
pi1(q1, q2) = max
q1
(a− q1 − q2)q1 − cq1 = max
q1
(a− c − q2)q1 − q21 .
28
Doupoly and Cournot competition: Solution.
Definition 19 (Nash equilibrium for two firms in competition)
A strategy profile (q∗1 , q

2) is a Nash equilibrium if
pi1(q

1 , q

2) ≥pi1(q1, q∗2) for every q1 ∈ [0,∞)
pi2(q

1 , q

2) ≥pi2(q∗1 , q2) for every q2 ∈ [0,∞)
ˆ Fix the production of firm 2 to q2. What is the best strategy that firm 1 can take?
ˆ Choose q1 that satisfies:
max
q1
pi1(q1, q2) = max
q1
(a− q1 − q2)q1 − cq1 = max
q1
(a− c − q2)q1 − q21 .
28
Duopoly and Cournot competition: Solution.
ˆ Assuming q2 < a− c , the optimal q1 is:
R1(q2) =
1
2
(a− c − q2).
ˆ Similarly for firm 2:
R2(q1) =
1
2
(a− c − q1).
ˆ Thus the profile (q∗1 , q

2) is a Nash equilibrium if it solves
2q∗1 =(a− c − q∗2),
2q∗2 =(a− c − q∗1).
29
Duopoly and Cournot competition: Solution.
ˆ Assuming q2 < a− c , the optimal q1 is:
R1(q2) =
1
2
(a− c − q2).
ˆ Similarly for firm 2:
R2(q1) =
1
2
(a− c − q1).
ˆ Thus the profile (q∗1 , q

2) is a Nash equilibrium if it solves
2q∗1 =(a− c − q∗2),
2q∗2 =(a− c − q∗1).
29
Duopoly and Cournot competition: Solution.
ˆ Assuming q2 < a− c , the optimal q1 is:
R1(q2) =
1
2
(a− c − q2).
ˆ Similarly for firm 2:
R2(q1) =
1
2
(a− c − q1).
ˆ Thus the profile (q∗1 , q

2) is a Nash equilibrium if it solves
2q∗1 =(a− c − q∗2),
2q∗2 =(a− c − q∗1).
29
Duopoly and Cournot competition: Solution.
ˆ Solving simultaneously, we find that
q∗1 = q

2 =
a− c
3
.
ˆ Recall that we assumed qi < a− c (i = 1, 2) in order to derive this equilibrium.
We must check that this is the case. Remember that c < a (one of our initial
assumptions in the model) which implies that a− c > 0 and hence
a− c > 0 =⇒ a− c
3
> 0 =⇒ q∗i > 0,
as assumed.
30
Duopoly and Cournot competition: Solution.
ˆ Solving simultaneously, we find that
q∗1 = q

2 =
a− c
3
.
ˆ Recall that we assumed qi < a− c (i = 1, 2) in order to derive this equilibrium.
We must check that this is the case. Remember that c < a (one of our initial
assumptions in the model) which implies that a− c > 0 and hence
a− c > 0 =⇒ a− c
3
> 0 =⇒ q∗i > 0,
as assumed.
30
Applications
Application 2: Bertrand model of duopoly
(to be covered in Q & A session)
Bertrand model of duopoly: Assumptions.
ˆ Two firms 1 and 2 that produce exactly the same good.
ˆ The demand is going to absorb Q units of the good, no matter at what price.
ˆ Each firm i has to choose a price out of the interval [0,∞).
ˆ Buyers will always buy from the firm with the lowest price.
ˆ If both firms set the same price, each firm will sell Q2 units.
ˆ For each firm, producing 1 unit of the good costs c dollars.
31
Bertrand model of duopoly: Assumptions.
ˆ Two firms 1 and 2 that produce exactly the same good.
ˆ The demand is going to absorb Q units of the good, no matter at what price.
ˆ Each firm i has to choose a price out of the interval [0,∞).
ˆ Buyers will always buy from the firm with the lowest price.
ˆ If both firms set the same price, each firm will sell Q2 units.
ˆ For each firm, producing 1 unit of the good costs c dollars.
31
Bertrand model of duopoly: Assumptions.
ˆ Two firms 1 and 2 that produce exactly the same good.
ˆ The demand is going to absorb Q units of the good, no matter at what price.
ˆ Each firm i has to choose a price out of the interval [0,∞).
ˆ Buyers will always buy from the firm with the lowest price.
ˆ If both firms set the same price, each firm will sell Q2 units.
ˆ For each firm, producing 1 unit of the good costs c dollars.
31
Bertrand model of duopoly: Assumptions.
ˆ Two firms 1 and 2 that produce exactly the same good.
ˆ The demand is going to absorb Q units of the good, no matter at what price.
ˆ Each firm i has to choose a price out of the interval [0,∞).
ˆ Buyers will always buy from the firm with the lowest price.
ˆ If both firms set the same price, each firm will sell Q2 units.
ˆ For each firm, producing 1 unit of the good costs c dollars.
31
Bertrand model of duopoly: Assumptions.
ˆ Two firms 1 and 2 that produce exactly the same good.
ˆ The demand is going to absorb Q units of the good, no matter at what price.
ˆ Each firm i has to choose a price out of the interval [0,∞).
ˆ Buyers will always buy from the firm with the lowest price.
ˆ If both firms set the same price, each firm will sell Q2 units.
ˆ For each firm, producing 1 unit of the good costs c dollars.
31
Bertrand model of duopoly: Assumptions.
ˆ Two firms 1 and 2 that produce exactly the same good.
ˆ The demand is going to absorb Q units of the good, no matter at what price.
ˆ Each firm i has to choose a price out of the interval [0,∞).
ˆ Buyers will always buy from the firm with the lowest price.
ˆ If both firms set the same price, each firm will sell Q2 units.
ˆ For each firm, producing 1 unit of the good costs c dollars.
31
Bertrand model of duopoly: The game.
ˆ Two players.
ˆ Strategy sets S1 = S2 = [0,∞).
ˆ Payoffs for firm 1 (for firm 2 are analogous):
pi1(p1, p2) =

(p1 − c)Q if p1 < p2
1
2(p1 − c)Q if p1 = p2
0 if p1 > p2
32
Bertrand model of duopoly: The game.
ˆ Two players.
ˆ Strategy sets S1 = S2 = [0,∞).
ˆ Payoffs for firm 1 (for firm 2 are analogous):
pi1(p1, p2) =

(p1 − c)Q if p1 < p2
1
2(p1 − c)Q if p1 = p2
0 if p1 > p2
32
Bertrand model of duopoly: The game.
ˆ Two players.
ˆ Strategy sets S1 = S2 = [0,∞).
ˆ Payoffs for firm 1 (for firm 2 are analogous):
pi1(p1, p2) =

(p1 − c)Q if p1 < p2
1
2(p1 − c)Q if p1 = p2
0 if p1 > p2
32
Bertrand model of duopoly: The solution.
ˆ Nash equilibrium: Prices (p∗1 , p

2) such that
pi1(p

1 , p

2) ≥pi1(p1, p∗2) for every p1 ∈ [0,∞)
pi2(p

1 , p

2) ≥pi2(p∗1 , p2) for every p2 ∈ [0,∞]
ˆ The unique Nash equilibrium is p∗1 = p

2 = c.
33
Applications
Application 3: The problem of the
commons (self study)
The Problem of the commons: Assumptions.
ˆ There are n farmers in a village.
ˆ Each farmer has to decide how many goats gi (i = 1, . . . , n) to own.
ˆ Goats will have to share the same limited space, hence, the benefit of having a
goat will decrease on the total number of goats.
ˆ We will assume that the benefit of having a goat when the total number of goats
is G is equal to A− G 2 (where A is a positive constant and G = ∑ni=1 gi ).
ˆ The cost of buying and caring for a goat is c .
ˆ We assume A > c.
ˆ For simplicity, we will assume that goats are perfectly divisible.
34
The Problem of the commons: Assumptions.
ˆ There are n farmers in a village.
ˆ Each farmer has to decide how many goats gi (i = 1, . . . , n) to own.
ˆ Goats will have to share the same limited space, hence, the benefit of having a
goat will decrease on the total number of goats.
ˆ We will assume that the benefit of having a goat when the total number of goats
is G is equal to A− G 2 (where A is a positive constant and G = ∑ni=1 gi ).
ˆ The cost of buying and caring for a goat is c .
ˆ We assume A > c.
ˆ For simplicity, we will assume that goats are perfectly divisible.
34
The Problem of the commons: Assumptions.
ˆ There are n farmers in a village.
ˆ Each farmer has to decide how many goats gi (i = 1, . . . , n) to own.
ˆ Goats will have to share the same limited space, hence, the benefit of having a
goat will decrease on the total number of goats.
ˆ We will assume that the benefit of having a goat when the total number of goats
is G is equal to A− G 2 (where A is a positive constant and G = ∑ni=1 gi ).
ˆ The cost of buying and caring for a goat is c .
ˆ We assume A > c.
ˆ For simplicity, we will assume that goats are perfectly divisible.
34
The Problem of the commons: Assumptions.
ˆ There are n farmers in a village.
ˆ Each farmer has to decide how many goats gi (i = 1, . . . , n) to own.
ˆ Goats will have to share the same limited space, hence, the benefit of having a
goat will decrease on the total number of goats.
ˆ We will assume that the benefit of having a goat when the total number of goats
is G is equal to A− G 2 (where A is a positive constant and G = ∑ni=1 gi ).
ˆ The cost of buying and caring for a goat is c .
ˆ We assume A > c.
ˆ For simplicity, we will assume that goats are perfectly divisible.
34
The Problem of the commons: Assumptions.
ˆ There are n farmers in a village.
ˆ Each farmer has to decide how many goats gi (i = 1, . . . , n) to own.
ˆ Goats will have to share the same limited space, hence, the benefit of having a
goat will decrease on the total number of goats.
ˆ We will assume that the benefit of having a goat when the total number of goats
is G is equal to A− G 2 (where A is a positive constant and G = ∑ni=1 gi ).
ˆ The cost of buying and caring for a goat is c .
ˆ We assume A > c.
ˆ For simplicity, we will assume that goats are perfectly divisible.
34
The Problem of the commons: Assumptions.
ˆ There are n farmers in a village.
ˆ Each farmer has to decide how many goats gi (i = 1, . . . , n) to own.
ˆ Goats will have to share the same limited space, hence, the benefit of having a
goat will decrease on the total number of goats.
ˆ We will assume that the benefit of having a goat when the total number of goats
is G is equal to A− G 2 (where A is a positive constant and G = ∑ni=1 gi ).
ˆ The cost of buying and caring for a goat is c .
ˆ We assume A > c.
ˆ For simplicity, we will assume that goats are perfectly divisible.
34
The Problem of the commons: Assumptions.
ˆ There are n farmers in a village.
ˆ Each farmer has to decide how many goats gi (i = 1, . . . , n) to own.
ˆ Goats will have to share the same limited space, hence, the benefit of having a
goat will decrease on the total number of goats.
ˆ We will assume that the benefit of having a goat when the total number of goats
is G is equal to A− G 2 (where A is a positive constant and G = ∑ni=1 gi ).
ˆ The cost of buying and caring for a goat is c .
ˆ We assume A > c.
ˆ For simplicity, we will assume that goats are perfectly divisible. 34
The Problem of the commons: The game.
ˆ Number of players: n.
ˆ Strategy sets: For each farmer (player) the interval [0,∞).
ˆ Given a strategy profile (g1, . . . , gn) the payoff to farmer i is:
pii (g1, . . . , gn) = gi (A− (g1 + · · ·+ gn)2)− cgi .
35
The Problem of the commons: The game.
ˆ Number of players: n.
ˆ Strategy sets: For each farmer (player) the interval [0,∞).
ˆ Given a strategy profile (g1, . . . , gn) the payoff to farmer i is:
pii (g1, . . . , gn) = gi (A− (g1 + · · ·+ gn)2)− cgi .
35
The Problem of the commons: The game.
ˆ Number of players: n.
ˆ Strategy sets: For each farmer (player) the interval [0,∞).
ˆ Given a strategy profile (g1, . . . , gn) the payoff to farmer i is:
pii (g1, . . . , gn) = gi (A− (g1 + · · ·+ gn)2)− cgi .
35
The Problem of the commons: Solution.
ˆ Take the first order condition of gi (A− (g1 + · · ·+ gn)2)− cgi .
ˆ A− (g1 + · · ·+ gn)2 − 2(g1 + · · ·+ gn)gi − c = 0.
ˆ Aggregate over every farmer:
ˆ nA− n(g1 + · · ·+ gn)2 − 2(g1 + · · ·+ gn)2 − nc = 0, that is:
ˆ n(A− c) = (2 + n)(g1 + · · ·+ gn)2.
ˆ From where we can obtain a value for the total number of goats in a Nash
equilibrium:
G ∗ = g1 + · · ·+ gn =
[
n
2 + n
(A− c)
] 1
2
36
The Problem of the commons: Solution.
ˆ Take the first order condition of gi (A− (g1 + · · ·+ gn)2)− cgi .
ˆ A− (g1 + · · ·+ gn)2 − 2(g1 + · · ·+ gn)gi − c = 0.
ˆ Aggregate over every farmer:
ˆ nA− n(g1 + · · ·+ gn)2 − 2(g1 + · · ·+ gn)2 − nc = 0, that is:
ˆ n(A− c) = (2 + n)(g1 + · · ·+ gn)2.
ˆ From where we can obtain a value for the total number of goats in a Nash
equilibrium:
G ∗ = g1 + · · ·+ gn =
[
n
2 + n
(A− c)
] 1
2
36
The Problem of the commons: Solution.
ˆ Take the first order condition of gi (A− (g1 + · · ·+ gn)2)− cgi .
ˆ A− (g1 + · · ·+ gn)2 − 2(g1 + · · ·+ gn)gi − c = 0.
ˆ Aggregate over every farmer:
ˆ nA− n(g1 + · · ·+ gn)2 − 2(g1 + · · ·+ gn)2 − nc = 0, that is:
ˆ n(A− c) = (2 + n)(g1 + · · ·+ gn)2.
ˆ From where we can obtain a value for the total number of goats in a Nash
equilibrium:
G ∗ = g1 + · · ·+ gn =
[
n
2 + n
(A− c)
] 1
2
36
The Problem of the commons: Solution.
ˆ Take the first order condition of gi (A− (g1 + · · ·+ gn)2)− cgi .
ˆ A− (g1 + · · ·+ gn)2 − 2(g1 + · · ·+ gn)gi − c = 0.
ˆ Aggregate over every farmer:
ˆ nA− n(g1 + · · ·+ gn)2 − 2(g1 + · · ·+ gn)2 − nc = 0, that is:
ˆ n(A− c) = (2 + n)(g1 + · · ·+ gn)2.
ˆ From where we can obtain a value for the total number of goats in a Nash
equilibrium:
G ∗ = g1 + · · ·+ gn =
[
n
2 + n
(A− c)
] 1
2
36
The Problem of the commons: Solution.
ˆ Take the first order condition of gi (A− (g1 + · · ·+ gn)2)− cgi .
ˆ A− (g1 + · · ·+ gn)2 − 2(g1 + · · ·+ gn)gi − c = 0.
ˆ Aggregate over every farmer:
ˆ nA− n(g1 + · · ·+ gn)2 − 2(g1 + · · ·+ gn)2 − nc = 0, that is:
ˆ n(A− c) = (2 + n)(g1 + · · ·+ gn)2.
ˆ From where we can obtain a value for the total number of goats in a Nash
equilibrium:
G ∗ = g1 + · · ·+ gn =
[
n
2 + n
(A− c)
] 1
2
36
The Problem of the commons: Solution.
ˆ Take the first order condition of gi (A− (g1 + · · ·+ gn)2)− cgi .
ˆ A− (g1 + · · ·+ gn)2 − 2(g1 + · · ·+ gn)gi − c = 0.
ˆ Aggregate over every farmer:
ˆ nA− n(g1 + · · ·+ gn)2 − 2(g1 + · · ·+ gn)2 − nc = 0, that is:
ˆ n(A− c) = (2 + n)(g1 + · · ·+ gn)2.
ˆ From where we can obtain a value for the total number of goats in a Nash
equilibrium:
G ∗ = g1 + · · ·+ gn =
[
n
2 + n
(A− c)
] 1
2
36
The Problem of the commons: Social Optimum.
ˆ What is the socially optimal number of goats?
ˆ From a utilitarian standpoint: whatever number G maximizes:
G (A− G 2)− cG
ˆ Taking the first order condition:
A− G 2 − 2G 2 − c = 0
ˆ From where we can obtain a value for the socially optimal number of goats:
G ∗∗ =
[
1
3
(A− c)
] 1
2
ˆ Exercise: Check that G ∗∗ < G ∗ whenever n > 1. Why is this?
37
The Problem of the commons: Social Optimum.
ˆ What is the socially optimal number of goats?
ˆ From a utilitarian standpoint: whatever number G maximizes:
G (A− G 2)− cG
ˆ Taking the first order condition:
A− G 2 − 2G 2 − c = 0
ˆ From where we can obtain a value for the socially optimal number of goats:
G ∗∗ =
[
1
3
(A− c)
] 1
2
ˆ Exercise: Check that G ∗∗ < G ∗ whenever n > 1. Why is this?
37
The Problem of the commons: Social Optimum.
ˆ What is the socially optimal number of goats?
ˆ From a utilitarian standpoint: whatever number G maximizes:
G (A− G 2)− cG
ˆ Taking the first order condition:
A− G 2 − 2G 2 − c = 0
ˆ From where we can obtain a value for the socially optimal number of goats:
G ∗∗ =
[
1
3
(A− c)
] 1
2
ˆ Exercise: Check that G ∗∗ < G ∗ whenever n > 1. Why is this?
37
The Problem of the commons: Social Optimum.
ˆ What is the socially optimal number of goats?
ˆ From a utilitarian standpoint: whatever number G maximizes:
G (A− G 2)− cG
ˆ Taking the first order condition:
A− G 2 − 2G 2 − c = 0
ˆ From where we can obtain a value for the socially optimal number of goats:
G ∗∗ =
[
1
3
(A− c)
] 1
2
ˆ Exercise: Check that G ∗∗ < G ∗ whenever n > 1. Why is this?
37
The Problem of the commons: Social Optimum.
ˆ What is the socially optimal number of goats?
ˆ From a utilitarian standpoint: whatever number G maximizes:
G (A− G 2)− cG
ˆ Taking the first order condition:
A− G 2 − 2G 2 − c = 0
ˆ From where we can obtain a value for the socially optimal number of goats:
G ∗∗ =
[
1
3
(A− c)
] 1
2
ˆ Exercise: Check that G ∗∗ < G ∗ whenever n > 1. Why is this? 37
Applications
Application 4: Hotelling model of political
competition
Political competition (Hotelling): Assumptions
ˆ Consider an election with two parties, 1 and 2.
ˆ Each party has to announce its political ideology.
ˆ Based on these announcements the voters will cast their vote.
ˆ The party with most votes wins the election. In case of a tie, a fair coin is tossed
to determine the winner.
ˆ Each voter will cast her vote for the political party that is closest to her political
ideology.
38
Political competition (Hotelling): Assumptions
ˆ Consider an election with two parties, 1 and 2.
ˆ Each party has to announce its political ideology.
ˆ Based on these announcements the voters will cast their vote.
ˆ The party with most votes wins the election. In case of a tie, a fair coin is tossed
to determine the winner.
ˆ Each voter will cast her vote for the political party that is closest to her political
ideology.
38
Political competition (Hotelling): Assumptions
ˆ Consider an election with two parties, 1 and 2.
ˆ Each party has to announce its political ideology.
ˆ Based on these announcements the voters will cast their vote.
ˆ The party with most votes wins the election. In case of a tie, a fair coin is tossed
to determine the winner.
ˆ Each voter will cast her vote for the political party that is closest to her political
ideology.
38
Political competition (Hotelling): Assumptions
ˆ Consider an election with two parties, 1 and 2.
ˆ Each party has to announce its political ideology.
ˆ Based on these announcements the voters will cast their vote.
ˆ The party with most votes wins the election. In case of a tie, a fair coin is tossed
to determine the winner.
ˆ Each voter will cast her vote for the political party that is closest to her political
ideology.
38
Political competition (Hotelling): Assumptions
ˆ Consider an election with two parties, 1 and 2.
ˆ Each party has to announce its political ideology.
ˆ Based on these announcements the voters will cast their vote.
ˆ The party with most votes wins the election. In case of a tie, a fair coin is tossed
to determine the winner.
ˆ Each voter will cast her vote for the political party that is closest to her political
ideology.
38
Political competition (Hotelling): Assumptions
ˆ To be concrete, suppose that the space of political ideologies is the interval
[0, 1] ⊂ R.
ˆ Parties hence have to choose ideologies I1, I2 in [0, 1]
ˆ Suppose that each “point” in [0, 1] is inhabited by a voter who has that ideology.
E.g. there is a voter with ideology pi/4.
Socialist Alternative
0
UNSW Conservatives
1
0.5 pi/4
39
Political competition (Hotelling): Assumptions
ˆ To be concrete, suppose that the space of political ideologies is the interval
[0, 1] ⊂ R.
ˆ Parties hence have to choose ideologies I1, I2 in [0, 1]
ˆ Suppose that each “point” in [0, 1] is inhabited by a voter who has that ideology.
E.g. there is a voter with ideology pi/4.
Socialist Alternative
0
UNSW Conservatives
1
0.5 pi/4
39
Political competition (Hotelling): Assumptions
ˆ To be concrete, suppose that the space of political ideologies is the interval
[0, 1] ⊂ R.
ˆ Parties hence have to choose ideologies I1, I2 in [0, 1]
ˆ Suppose that each “point” in [0, 1] is inhabited by a voter who has that ideology.
E.g. there is a voter with ideology pi/4.
Socialist Alternative
0
UNSW Conservatives
1
0.5 pi/4
39
Hotelling model of Political competition: The game.
ˆ Players: parties, 1 and 2.
ˆ Strategies: S1 = S2 = [0, 1].
ˆ Payoffs:
ˆ Let I1 < I2. Party 1 gets a share of the votes equal to
1
2 (I1 + I2).
ˆ If 1
2
(I1 + I2) >
1
2
then Party 1 gets a payoff equal to 1 and Party 2 gets a payoff equal
to 0.
ˆ If 1
2
(I1 + I2) =
1
2
then both parties get a payoff equal to 1
2
.
ˆ If I1 = I2 then both parties will get a payoff equal to
1
2 .
40
Hotelling model of Political competition: The game.
ˆ Players: parties, 1 and 2.
ˆ Strategies: S1 = S2 = [0, 1].
ˆ Payoffs:
ˆ Let I1 < I2. Party 1 gets a share of the votes equal to
1
2 (I1 + I2).
ˆ If 1
2
(I1 + I2) >
1
2
then Party 1 gets a payoff equal to 1 and Party 2 gets a payoff equal
to 0.
ˆ If 1
2
(I1 + I2) =
1
2
then both parties get a payoff equal to 1
2
.
ˆ If I1 = I2 then both parties will get a payoff equal to
1
2 .
40
Hotelling model of Political competition: The game.
ˆ Players: parties, 1 and 2.
ˆ Strategies: S1 = S2 = [0, 1].
ˆ Payoffs:
ˆ Let I1 < I2. Party 1 gets a share of the votes equal to
1
2 (I1 + I2).
ˆ If 1
2
(I1 + I2) >
1
2
then Party 1 gets a payoff equal to 1 and Party 2 gets a payoff equal
to 0.
ˆ If 1
2
(I1 + I2) =
1
2
then both parties get a payoff equal to 1
2
.
ˆ If I1 = I2 then both parties will get a payoff equal to
1
2 .
40
Hotelling model of Political competition: The game.
ˆ Players: parties, 1 and 2.
ˆ Strategies: S1 = S2 = [0, 1].
ˆ Payoffs:
ˆ Let I1 < I2. Party 1 gets a share of the votes equal to
1
2 (I1 + I2).
ˆ If 1
2
(I1 + I2) >
1
2
then Party 1 gets a payoff equal to 1 and Party 2 gets a payoff equal
to 0.
ˆ If 1
2
(I1 + I2) =
1
2
then both parties get a payoff equal to 1
2
.
ˆ If I1 = I2 then both parties will get a payoff equal to
1
2 .
40
Hotelling model of Political competition: The game.
ˆ Players: parties, 1 and 2.
ˆ Strategies: S1 = S2 = [0, 1].
ˆ Payoffs:
ˆ Let I1 < I2. Party 1 gets a share of the votes equal to
1
2 (I1 + I2).
ˆ If 1
2
(I1 + I2) >
1
2
then Party 1 gets a payoff equal to 1 and Party 2 gets a payoff equal
to 0.
ˆ If 1
2
(I1 + I2) =
1
2
then both parties get a payoff equal to 1
2
.
ˆ If I1 = I2 then both parties will get a payoff equal to
1
2 .
40
Hotelling model of Political competition: The game.
ˆ Players: parties, 1 and 2.
ˆ Strategies: S1 = S2 = [0, 1].
ˆ Payoffs:
ˆ Let I1 < I2. Party 1 gets a share of the votes equal to
1
2 (I1 + I2).
ˆ If 1
2
(I1 + I2) >
1
2
then Party 1 gets a payoff equal to 1 and Party 2 gets a payoff equal
to 0.
ˆ If 1
2
(I1 + I2) =
1
2
then both parties get a payoff equal to 1
2
.
ˆ If I1 = I2 then both parties will get a payoff equal to
1
2 .
40
Hotelling model of Political competition: The game.
ˆ Players: parties, 1 and 2.
ˆ Strategies: S1 = S2 = [0, 1].
ˆ Payoffs:
ˆ Let I1 < I2. Party 1 gets a share of the votes equal to
1
2 (I1 + I2).
ˆ If 1
2
(I1 + I2) >
1
2
then Party 1 gets a payoff equal to 1 and Party 2 gets a payoff equal
to 0.
ˆ If 1
2
(I1 + I2) =
1
2
then both parties get a payoff equal to 1
2
.
ˆ If I1 = I2 then both parties will get a payoff equal to
1
2 .
40
Hotelling model of Political competition: The solution.
0 1
0.50.3 0.8
I1 I2
41
Hotelling model of Political competition: The solution.
0 1
0.50.3 0.8
I1 I2
1
2
(I1 + I2)
41
Hotelling model of Political competition: The solution.
0 1
0.50.3 0.8
I1 I2
1
2
(I1 + I2)
41
Hotelling model of Political competition: The solution.
0 1
0.50.3 0.8
I1 I2
1
2
(I1 + I2)
ˆ In the above example, party 1 wins (though party 2 is not playing a best
response!).
ˆ Nash equilibrium: Ideologies I ∗1 and I

2 such that no party can unilaterally increase
their chances of winning the election.
ˆ The unique Nash equilibrium is I1 = I2 =
1
2 .
41
Applications
Application 5: Bertrand model of duopoly
(version 2) (to be covered in QA session)
Bertrand model of duopoly (v. 2): Assumptions.
ˆ Consider a duopoly with differentiated products. Producing one unit of either
good costs c to the firm.
ˆ Firms choose prices p1 and p2. There is a “sustitutability” parameter b < 1, s.t.
ˆ If firms choose prices p1 and p2, consumers demand
q1(p1, p2) = a− p1 + bp2
from firm 1.
ˆ If firms choose prices p1 and p2, consumers demand
q2(p1, p2) = a− p2 + bp1
from firm 2.
42
Bertrand model of duopoly (v. 2): Assumptions.
ˆ Consider a duopoly with differentiated products. Producing one unit of either
good costs c to the firm.
ˆ Firms choose prices p1 and p2. There is a “sustitutability” parameter b < 1, s.t.
ˆ If firms choose prices p1 and p2, consumers demand
q1(p1, p2) = a− p1 + bp2
from firm 1.
ˆ If firms choose prices p1 and p2, consumers demand
q2(p1, p2) = a− p2 + bp1
from firm 2.
42
Bertrand model of duopoly (v. 2): Assumptions.
ˆ Consider a duopoly with differentiated products. Producing one unit of either
good costs c to the firm.
ˆ Firms choose prices p1 and p2. There is a “sustitutability” parameter b < 1, s.t.
ˆ If firms choose prices p1 and p2, consumers demand
q1(p1, p2) = a− p1 + bp2
from firm 1.
ˆ If firms choose prices p1 and p2, consumers demand
q2(p1, p2) = a− p2 + bp1
from firm 2.
42
Bertrand model of duopoly (v. 2): Assumptions.
ˆ Consider a duopoly with differentiated products. Producing one unit of either
good costs c to the firm.
ˆ Firms choose prices p1 and p2. There is a “sustitutability” parameter b < 1, s.t.
ˆ If firms choose prices p1 and p2, consumers demand
q1(p1, p2) = a− p1 + bp2
from firm 1.
ˆ If firms choose prices p1 and p2, consumers demand
q2(p1, p2) = a− p2 + bp1
from firm 2.
42
Bertrand model of duopoly (v 2): The game.
Therefore, the game is given by:
ˆ Set of players, firms 1 and 2,
ˆ Each firm has to choose a price in the interval [0,∞),
ˆ Firm i tries to maximize profits given by pii (pi , p−i ) = (pi − c)qi ,
43
Bertrand model of duopoly (v 2): Solution.
ˆ Take as given p2 and compute R1(p2).
ˆ Given p2 firm one chooses a price p1 so that
max
p1
pi1(p1, p2) = max
p1
(p1 − c)q1 = max
p1
(p1 − c)(a− p1 + bp2)
ˆ R1(p2) =
1
2(a + bp2 + c).
ˆ And since everything is symmetric, R2(p1) =
1
2(a + bp1 + c).
ˆ The profile (p∗1 , p

2) is a Nash equilibrium if it solves:
p∗1 =
1
2
(a + bp∗2 + c)
p∗2 =
1
2
(a + bp∗1 + c).
(p∗1, p

2) =
a−c
2−b
44
Bertrand model of duopoly (v 2): Solution.
ˆ Take as given p2 and compute R1(p2).
ˆ Given p2 firm one chooses a price p1 so that
max
p1
pi1(p1, p2) = max
p1
(p1 − c)q1 = max
p1
(p1 − c)(a− p1 + bp2)
ˆ R1(p2) =
1
2(a + bp2 + c).
ˆ And since everything is symmetric, R2(p1) =
1
2(a + bp1 + c).
ˆ The profile (p∗1 , p

2) is a Nash equilibrium if it solves:
p∗1 =
1
2
(a + bp∗2 + c)
p∗2 =
1
2
(a + bp∗1 + c).
(p∗1, p

2) =
a−c
2−b
44
Bertrand model of duopoly (v 2): Solution.
ˆ Take as given p2 and compute R1(p2).
ˆ Given p2 firm one chooses a price p1 so that
max
p1
pi1(p1, p2) = max
p1
(p1 − c)q1 = max
p1
(p1 − c)(a− p1 + bp2)
ˆ R1(p2) =
1
2(a + bp2 + c).
ˆ And since everything is symmetric, R2(p1) =
1
2(a + bp1 + c).
ˆ The profile (p∗1 , p

2) is a Nash equilibrium if it solves:
p∗1 =
1
2
(a + bp∗2 + c)
p∗2 =
1
2
(a + bp∗1 + c).
(p∗1, p

2) =
a−c
2−b
44
Bertrand model of duopoly (v 2): Solution.
ˆ Take as given p2 and compute R1(p2).
ˆ Given p2 firm one chooses a price p1 so that
max
p1
pi1(p1, p2) = max
p1
(p1 − c)q1 = max
p1
(p1 − c)(a− p1 + bp2)
ˆ R1(p2) =
1
2(a + bp2 + c).
ˆ And since everything is symmetric, R2(p1) =
1
2(a + bp1 + c).
ˆ The profile (p∗1 , p

2) is a Nash equilibrium if it solves:
p∗1 =
1
2
(a + bp∗2 + c)
p∗2 =
1
2
(a + bp∗1 + c).
(p∗1, p

2) =
a−c
2−b
44
Bertrand model of duopoly (v 2): Solution.
ˆ Take as given p2 and compute R1(p2).
ˆ Given p2 firm one chooses a price p1 so that
max
p1
pi1(p1, p2) = max
p1
(p1 − c)q1 = max
p1
(p1 − c)(a− p1 + bp2)
ˆ R1(p2) =
1
2(a + bp2 + c).
ˆ And since everything is symmetric, R2(p1) =
1
2(a + bp1 + c).
ˆ The profile (p∗1 , p

2) is a Nash equilibrium if it solves:
p∗1 =
1
2
(a + bp∗2 + c)
p∗2 =
1
2
(a + bp∗1 + c).
(p∗1, p

2) =
a−c
2−b
44
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