1 Economics & Evolution. 2 Cournot Game 2 players Each chooses quantity q i ≥ 0 Player i’s...
-
Upload
arthur-gibson -
Category
Documents
-
view
219 -
download
3
Transcript of 1 Economics & Evolution. 2 Cournot Game 2 players Each chooses quantity q i ≥ 0 Player i’s...
1
Economics & Evolution
2
Cournot Game
• 2 players
• Each chooses quantity qi ≥ 0
• Player i’s payoff is: qi(1- qi –qj )
Inverse demand(price)
No cost
• Player 1’s best response (given q2 ):
11 1 2 q 1 2q 1 - q - q ' = 1 - 2q - q = 0
21
1 - qq =
2similarly, 1
2
1 - qq =
2BR1 BR2
3
q1
q2
21
1 - qq =
2BR1
12
1 - qq =
2BR2
Nash (Cournot) Equilibrium ( ⅓ , ⅓ )
4
tt+1 21
1 - qq =
2
tt+1 12
1 - qq =
2
A Dynamic ProcessA Dynamic Process
t = 1, 2, 3, ..... di.. sc.. rete
t+1 t1 0 11 1= -
1 1 02 2q q
t+1 t1 1t+1 t2 2
1 0 1q q1 1= -
1 1 02 2q q
5
q1
q2
t+11q
tq
t+12q
t+1q
6
t+1 t1 0 11 1= -
1 1 02 2q q
A steady state:
1 0 11 1= -
1 1 02 2q q
0 1 11 1=
1 0 12 2I + q
2 1 1=
1 2 1q
-12 1 2 -11
=1 2 -1 23
2 1 2 -1
1 2 -1 2
4 - 1 = 3
2 - 2 = 0
7
2 1 1=
1 2 1q
-12 1 2 -11
=1 2 -1 23
2 -1 1 11 1 1= =
-1 2 1 13 3 3q e
Define:
t tt t 1
3
1= -
3d q e q d e
t+1 t0 11 1 1 1
= -1 03 2 2 3
d e e d e
The difference equation:
becomes
t+1 t0 11 1= -
1 02 2q e q
t+1 t0 11= -
1 02d dor
8
t+1 t0 11= -
1 02d d
t+1 t+11 2t+1 t+12 1
d d1= -
2d d
,t+1 t t+1 t1 2 2 1
1 1d = d d = d
2 2
tid 0
t t 1 1
3 3q d e e Convergence to
Nash Equilibrium
9
q1
q2
A
Show that if then for all 0 t:q A t q A
B
C
Show that if then t t+1q B q C
10
Change the model:Players are very rarely allowed to revise their strategy.At period t, player i is allowed to choose his best
response with probability p, where p ~ 0.
The players alternate:
For odd 's:
For even 's:
t t -1 t t -11 1 2 2 2
t t -1 t t -11 1 2 2 1
q = BR q , q = q
q = q , q = BR q
t
t
11
q1
q2
0q1q
2q4q
For odd 's:
For even 's:
t t -1 t t -11 1 2 2 2
t t -1 t t -11 1 2 2 1
q = BR q , q = q
q = q , q = BR q
t
t
.
:For even odd
t -21 2
t t -1 t -21 1 1 2
t t -12 2 1 2
t - 1
=
q
q = q BR q ,
q = BR q BR BR
t
.
t -22
t -22t
2
1 - q1 - 1+ q2 =
2 4q =1
3
12
1 1 1 2 1t + Δt t + Δt × t tq = q BR q q
Continuous Time
Let the time interval between periods approach 0.tq tqA change of notation:
1q 1 2BR q
In time Δt the individual advances only part of the path
13
1 1 1 2 1t + Δt t + Δt × t tq = q BR q q
Continuous Time
Let the time interval between periods approach 0.
1 11 2 1
t + Δt - t= t t
Δt
q qBR q q t 0
•
1 1 2 1q = BR q t q t
tq tq
similarly • 1
2 2
1-q tq = q t
2
A change of notation:
21
1-q tq t
2
21
1-q tq t
2
14
• 2
1 1 t1-q t
q = q2
• 1
2 2
1-q tq = q t
2
Continuous Time
• •
1 2 1 21
- -2
t tq q = q q
denote: 1 2t t d t q q
1
2t d(t)= - d
1
2t
d(t) = -d
1- t2td = Ce
•
1
21 1
1 1 t2
t1+ -q t
qq2t
=
1- t2•
11
3 t2
1+q = q2Ce
1- t2•
11
+ t32
1+q q = Ce2
ln
1= -
2d t
15
1- t2•
11
+ t32
1+q q = Ce2
16
1- t2•
11
+ t32
1+q q = Ce2
1- t2• 3 3t t
2 21
1+
3 t2
1+ Ceq q e = e2
1- t2•
11
+3 t2
1+ Ceq q =2 X
1- t2e
2
3 3t t t2 21 t 1 Cq e = e e
2
2
3 3t t t2 21 t 1 Cq e = e e + K
3
2
1 3- t t2 2
1 t 1 Cq = e + Ke3
13
17
t tt+1 2 31
1 - q - qq =
2
1 1 1 2 3π q 1 - q - q - q=
Cournot model, Three Firms
discrete time
11 2 3
1
π1 - 2q - q - q
q=
t tt+1 1 32
1 - q - qq =
2
t tt+1 1 23
1 - q - qq =
2
t+1 t+1 t+1 t t t1 2 3 1 2 3
3q + q + q = q + q + q
2+
t+1 t3Q = Q
2
3Q = Q ,
2
3Q =
4
t t t+1 t t+1 t3 3 3 3d = Q - d + = - - d d = -d
4 4 2 4
18
Cournot model, Three Firms
discrete time
t+1 td = -d 3
4 ttttd = -1 C, Q -1 C
does not converge !!!!!
Continuous time
t
2 31 1 1
1 - q t - qq t + Δt = q t + Δt - q t
2
2 31 1=
1-q t -q tq -q t
2
Similarly for i = 2, 3. Add the three equations:
=3Q - 2Q t2
19
Cournot model, Three Firms
Continuous time
=3Q - 2Q t2
+ =3Q 2Q2 X 2te
•
2t 2t=3Qe e2
2t 2t= +3Qe e K4
-2t= +3Q Ke4
2 31 1=
1-q t -q tq -q t
2
1 31
21 1-=
1- -q t -q tq
q t 1 q t+ q2 2
t
1 1= -1-Q t 1q q t
2 2
-2t
1 1
-= -
31- Ke 14q q t2 2
20
Cournot model, Three Firms
Continuous time
-2t
1 1
-= -
31- Ke 14q q t2 2
21
Cournot model, Three Firms
Continuous time
-2t
1 1
-= -
31- Ke 14q q t2 2
-2t
1 1
-= -
1 Ke 14q q t2 2
-2t1 1+ =
1 1q q t -Ce2 8 X
1t2e
1t
21
1 32 2
- tq e =
t1 e -Ce8
1 32 2
1t - t21 = +
t1q e e Ce4
-2t1 = +
1q Ce4
14
22
Fictitious PlayFictitious Play
• Two players repeatedly play a normal form game
• Each player observes the frequencies of the strategies played by the other player in the past (fictitious mixed strategy)(fictitious mixed strategy)
• Each player chooses a best response to the fictitious mixed strategy of his opponent.
23
Updating the frequenciesUpdating the frequencies:
t t t1 2 n(f , f , ........, f )+ (0,0, ......,0,1,0, ....,0)
t + 1t
oror:
t t t1 2 n(f , f , ........, f )+ (0,0, ......,0,1,0, ....,0)Δ
t + Δt
24
A B
A 0 , 2 3 , 0
B 2 , 1 1 , 3
Let p(t),q(t) be the frequencies of the
second second strategy played by played 1,2
Analysis of the stage-game:
q
0
1
(A)
(B)
(A) p 1(B)
An ExampleAn Example
BRBR11
BRBR22
Nash Equilibrium p = q = 1/2
25
A B
A 0 , 2 3 , 0B 2 , 1 1 , 3 q
0
1
(A)
(B)
(A) p 1(B)
An ExampleAn Example
As long as ( p(t) , q(t) ) is in the first quadrant,
the best responses are: ( B , A ).
26
A B
A 0 , 2 3 , 0B 2 , 1 1 , 3 q
0
1
(A)
(B)
(A) p 1(B)
An ExampleAn Example
the best responses are: ( B , A )
p(t) increases, q(t) decreases (with t)
27
A B
A 0 , 2 3 , 0B 2 , 1 1 , 3 q
0
1
(A)
(B)
(A) p 1(B)
An ExampleAn Example
λtp t + λτp t + τ =
λ t + τ
τ 1 - p ttp t + τ
p t + τ - = - =t +
p tτ
p tt + τ
=t p 1 - p t
tp t +λ λτp t + τ =
λ t + τ
=ap t 1 -t
•
=tp 1
tp t + τp t + τ =
t + τ
tp = t - a
28
A B
A 0 , 2 3 , 0B 2 , 1 1 , 3 q
0
1
(A)
(B)
(A) p 1(B)
An ExampleAn Example
=ap t 1 -t
tq t + 0 τq tq t + τ - = - = -
tq t
+ τq t
t + τ
=t q -q t
λtq t + 0q t + τ =
λ t + τ
tq = 0
=bq tt
29
A B
A 0 , 2 3 , 0B 2 , 1 1 , 3 q
0
1
(A)
(B)
(A) p 1(B)
An ExampleAn Example
=ap t 1 -t
=bq tt
30
A B
A 0 , 2 3 , 0B 2 , 1 1 , 3
q(t)
0
1
1 - p(t) 1
An ExampleAn Example
=ap t 1 -t
=bq tt
=1- p t a/t = const.
b/tq t
31
A B
A 0 , 2 3 , 0B 2 , 1 1 , 3
q(t)
0
1
1 - p(t) 1
An ExampleAn Example
=ap t 1 -t
=bq tt
=1- p t a/t = const.
b/tq t
Best responses in this quadrant are (B , B )
p(t) , q(t) increase (with t )
32
A B
A 0 , 2 3 , 0B 2 , 1 1 , 3
0
1
1
An ExampleAn Example
Best responses in the quadrant are:
(B , A )(B , B )
(A , B )
(A , A )
33
A B
A 0 , 2 3 , 0B 2 , 1 1 , 3
0
1
1
An ExampleAn Example
Best responses in the quadrant are:
(B , A )(B , B )
(A , B )
(A , A )
34
A B
A 0 , 2 3 , 0B 2 , 1 1 , 3
0
1
1
An ExampleAn Example
Does it converge? (B , A )(B , B )
(A , B )
(A , A )
at
at+1
t t
t+1
a a + 1/2=
a 1/2
t+1 t
1 1= 1+
2a 2adefine: t
t
1 s =
2a
t+1 ts = 1+ st 0s = t + s
t0
1a =
2 t + s0
converges !!!
Does convergence mean that Does convergence mean that they play the equilibrium?they play the equilibrium?
35
A B
A 0 , 2 3 , 0B 2 , 1 1 , 3
0
1
1
An ExampleAn Example
(B , A )(B , B )
(A , B )
(A , A )
What does an outside observer see?What does an outside observer see?
(B , A )
(B , B )
(A , B )
(A , A )
How much time is spent in each quadrant ???How much time is spent in each quadrant ???
36
A B
A 0 , 2 3 , 0B 2 , 1 1 , 3
0
1
1
An ExampleAn Example
(B , A )(B , B )
(A , B )
(A , A )
time is spent in each quadranttime is spent in each quadrant
0 0
1t , p ,
2
1 1
1t , ,q
2,2 2
1t , p
2
t+1 t
1 1= 1+
2a 2a 1 0
1 1= 1+
2 1/2 - q 2 1/2 - p
2 1
1 1= 1+
2 p - 1/2 2 1/2 - q
(to be used later)
37
A B
A 0 , 2 3 , 0B 2 , 1 1 , 3
0
1
1
An ExampleAn Example
(B , A )(B , B )
(A , B )
(A , A )
time is spent in each quadranttime is spent in each quadrant
0 0
1t , p ,
2
1 1
1t , ,q
2,2 2
1t , p
2 =
ap t 1 -t =
bq tt
00
=ap t 1 -t
00a = t 1 - p
0
0 =bq tt
0tb =2
0 0=
t 1 - pp t 1 -
t 0=
tq t
2t
38
A B
A 0 , 2 3 , 0B 2 , 1 1 , 3
0
1
1
An ExampleAn Example
(B , A )(B , B )
(A , B )
(A , A )
time is spent in each quadranttime is spent in each quadrant
0 0
1t , p ,
2
1 1
1t , ,q
2,2 2
1t , p
2 0 0=
t 1 - pp t 1 -
t 0=
tq t
2t
0
11
0= =
t 1 - p1p t 1 -2 t
1 0 0t = 2t 1 - p
1 0 0 0-t t = 2 1 - p - 1 t
39
A B
A 0 , 2 3 , 0B 2 , 1 1 , 3
0
1
1
An ExampleAn Example
(B , A )(B , B )
(A , B )
(A , A )
0 0
1t , p ,
2
1 1
1t , ,q
2,2 2
1t , p
2
1 0 0 0-t t = 2 1 - p - 1 t
time spent in the time spent in the firstfirst quadrantquadrant
0 0=
t 1 - pp t 1 -
t 0=
tq t
2t
1 1=
t 1 - pp t 1 -
t 1 1t 1 - q
q t = 1 -t
analogously: