economics & evolution
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Economics & Evolution. Number 3. The replicator dynamics (in general). The Replicator Dynamics in the Generalized Rock Scissors, Paper. A =. ? ? ?. The Replicator Dynamics in the Generalized Rock Scissors, Paper. A =. The Replicator Dynamics in the Generalized Rock Scissors, Paper. A =. - PowerPoint PPT PresentationTRANSCRIPT
1
Economics & EvolutionNumber 3
2
The replicator dynamics (in general)
The fitness of strategy ii i : f = π e , x
A general normal form game with strategies strategies: i , i = 1, ..,n n n e
.The population of players after reproduction: ii
ii+ e τx x π e , x
The proportion of players after reproduction:
i i
i
i
j jj
i
j
+ τt
τ
e
+ τ
+
x x π e , xx
x x π e , x
i i
i+ τ
1+ τπ x, x
x x π e , x
•
i ii= x - π x, xx π e , x
3
The Replicator Dynamics in the Generalized Rock Scissors, Paper
a > -1, 2 + a > 1
.The only equilibrium is: 1/3, 1/3, 1/3
The replicator dynamics :
t
1 1 1 2= x x + 2 + a x - x Ax x
1 2+a 0
0 1 2+a
2+a 0 1
A =
? ? ?
•
i ii= x - π x, xx π e , x
t2 2 2 3
t3 3 3 1
= x x + 2 + a x - x Ax
= x x + 2 + a x - x Ax
x
x
For , is an ESS.a > 0 1/3, 1/3, 1/3
4
The Replicator Dynamics in the Generalized Rock Scissors, Paper
a > -1, 2 + a > 1
1 2+a 0
0 1 2+a
2+a 0 1
A =
t1 1 1 2
t2 2 2 3
t3 3 3 1
= x x + 2 + a x - x Ax
= x x + 2 + a x - x Ax
= x x + 2 + a x - x Ax
x
x
x
ln .define : 1 2 3h x = x x x
ln.
3
i1 2 3
i=1 i
•x x x
x =t x
xh t
1 2 3 1 2= x + x + x + 2 + a x + x + x - 3x Ax
t= 3 + a - 3x Ax
22
1 2 3 1 2 2 3 1 3
2t1 2 2 3 1 3
1 = x + x + x = x + 2 x x + x x + x x
ax Ax = 1+ a x x + x x + x x = 1+ 1 - x
2
2a= 3 x - 1
2
5
The Replicator Dynamics in the Generalized Rock Scissors, Paper
a > -1, 2 + a > 1
1 2+a 0
0 1 2+a
2+a 0 1
A =
t1 1 1 2
t2 2 2 3
t3 3 3 1
= x x + 2 + a x - x Ax
= x x + 2 + a x - x Ax
= x x + 2 + a x - x Ax
x
x
x
ln .define : 1 2 3h x = x x x
ln.
3
i1 2 3
i=1 i
•x x x
x =t x
xh t
1 2 3 1 2= x + x + x + 2 + a x + x + x - 3x Ax
t= 3 + a - 3x Ax 2a= 3 x - 1
2
min argmin
=1
2i
2 2
e
1 1 1 1x = x = , ,
3 3 3 3
23 x - 1 0
6
The Replicator Dynamics in the Generalized Rock Scissors, Paper
a > -1, 2 + a > 1
1 2+a 0
0 1 2+a
2+a 0 1
A =
t1 1 1 2
t2 2 2 3
t3 3 3 1
= x x + 2 + a x - x Ax
= x x + 2 + a x - x Ax
= x x + 2 + a x - x Ax
x
x
x
ln .define : 1 2 3h x = x x x
x =h 2a3 x - 1
2
1 2 3
•x = 0 x x x = Consth →→
S2 S1
S3
1 2 3
1 2 3
1 x x x
9 x x x 0
Intersection of a hyperbola with the triangle
7
The Replicator Dynamics in the Generalized Rock Scissors, Paper
a > -1, 2 + a > 1
1 2+a 0
0 1 2+a
2+a 0 1
A =
t1 1 1 2
t2 2 2 3
t3 3 3 1
= x x + 2 + a x - x Ax
= x x + 2 + a x - x Ax
= x x + 2 + a x - x Ax
x
x
x
ln .define : 1 2 3h x = x x x
x =h 2a3 x - 1
2
S2 S1
S3
For 1 2 3
•: x = 0, x x x = C ta = 0 onsh
Does not converge to equlibrium
?
8
The Replicator Dynamics in the Generalized Rock Scissors, Paper
a > -1, 2 + a > 1
1 2+a 0
0 1 2+a
2+a 0 1
A =
t1 1 1 2
t2 2 2 3
t3 3 3 1
= x x + 2 + a x - x Ax
= x x + 2 + a x - x Ax
= x x + 2 + a x - x Ax
x
x
x
2ax = 3 x - 1
2h
S2 S1
S3 For 1 2 3
•: x < 0, a < 0 x x x h
Moves away from equlibrium
9
The Replicator Dynamics in the Generalized Rock Scissors, Paper
a > -1, 2 + a > 1
1 2+a 0
0 1 2+a
2+a 0 1
A =
t1 1 1 2
t2 2 2 3
t3 3 3 1
= x x + 2 + a x - x Ax
= x x + 2 + a x - x Ax
= x x + 2 + a x - x Ax
x
x
x
2ax = 3 x - 1
2h
S2 S1
S3
For 1 2 3
•: x > 0, a > 0 x x x h
Moves towards equlibrium
10
Let ξ(t, x0) denote the replicator dynamics of a given game, beginning at x0.
,If is a strictly dominated strategy and
then
i 0
0i
e x IntΔ
t, x 0.ξ
Lemma :
Let dominate and let ,
i i
x Δy e ε = π y - e , x > 0.minProof :
ln lnDefine n
i i i ij=1
V x = x - y x
i
ni j
i ij=1 jx
x xV = - y
x jik
ij=1
= π e - x, x - π e - x, xy
(here it is used that x0 is in the interior)
i i= π e - x, x π y - x, x = -π y - e , x- -ε
i -V i 0x
Replicator Dynamics and Strict Dominance
11
If is weakly dominated by , and then
for if then
(if is always present then must disappea r)
ji i
0 0 0j i
j j
e y π y - e ,e > 0
x IntΔ, t, x 0 t, x 0.
e e
ξ ξ
Lemma :
Define as before:
ii iV x e - y, xV x = π Proof :
by intergrating:
i jV x <εx 0,
j0
i
t
0dtV x -V x - x
,If never vanishes then
hence,
and
i
t
j j0
i
- x x V
x 0.
Replicator Dynamics and Weak Dominance
.
If then
hj
h
ji
i ih
π y - e ,e = ε > 0,
π y - e , x = x π y - e ,e εx
12
., weakly dominates
1 20 1 e
0 0eExample :
Replicator Dynamics and Weak Dominance
, x 11 2 2
•2 21
2 1 2 2 1
• 2
1
1
1 1
x , e = xπ x, x = x
x= x
x = 1 - x
- x x = x = 1 - x
x
x
> 0
.brings its own destruction, by giving advantage to 1 2ee
e1 vanishes !!!
13
, when is present. >
1 2 3A = 1 1 11 1 00 0 0
e eeExample :
Replicator Dynamics and Weak Dominance
t1 2 1 2
1t 2t1 2
Ax = x + x x + x
Ax = 1, e Ax = x + x
x
e
, • •
1 21 2
1 2
x x= 1 - x, x = x + x - π x, x
x x
1 • •
2 1 21 2 3
1 2
xd ln
x x x= - = 1 - x - x = x > 0
dt x x
x1/x2 increases as long as x3 > 0.
14
, when is present. >
1 2 3A = 1 1 11 1 00 0 0
e eeExample :
Replicator Dynamics and Weak Dominance
1 • •
2 1 21 2 3
1 2
xd ln
x x x= - = 1 - x - x = x > 0
dt x x
x1/x2 increases as long as x3 > 0.
S3 S1
S2
x1/x2 = Constant
15
The set of stationary points (in the Rep. Dyn. ) :
0 iΔ = x π e ,x = π x,x i supp x
Replicator Dynamics and Nash Equilibria
max
If is N.E. then
iz Δ
x
π e ,x = π z,x i supp x
So is stationary.x
.
If then
for all pure strategies
so, is BR to itself.
0
ii
IntΔ Δ
x
x
π e ,x = π x,x e
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Define : 000Δ = Δ IntΔ
Replicator Dynamics and Nash Equilibria
.We have shown that : 000 NEΔ Δ Δ
. 000 NEΔ IntΔ Δ IntΔ Δ IntΔ
0000 NE 00 NE Δ Δ IntΔ Δ Δ Δ IntΔ
is a c onve se x t00 0Δ Δ IntΔ
:Moreover, if: an d then as long as
.
00
0
α + β = 1x, y Δ x + y
x + y Δ
Therefore, if then the whole line connecting
consists of N.E. including the last point on the fronti
.
er
NEx, y Δ IntΔx, y
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: If and then is a N.E. 0 0x IntΔ, ξ t, x x x Lemma
Replicator Dynamics and Nash Equilibria
: If is not N.E., then there is s.t. i ix e π e - x,x = ε > 0.Proof
hence for sufficiently large i 0 0 εt : e - ξ t, x ,ξ t, x > .2
.
or:
and Contradiction.
i
•ε ti 2
i
i
x ε> . x e Tx 2x
x
Q.E.D
: If and then is a N.E. 0 0x IntΔ, ξ t, x x x Lemma
: If is not N.E., then there is s.t. i ix e π e - x,x = ε > 0.Proof
hence for sufficiently large i 0 0 εt : e - ξ t, x ,ξ t, x > .2
.
or:
and Contradiction.
i
•ε ti 2
i
i
x ε> . x e Tx 2x
x
limProve that if:
then is a N.E.
T 0
T 0
1 ξ t, x dt = xT
x
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A point is stable if there exists a neighborhood of ,
s.t. for all x U x
y U : lim ξ t, y = x.
Definition :
Replicator Dynamics and Stability
A point is Lyapunov stable if for all neighborhoods
of , there exists a neighborhood of ,
s.t. for all
xU x V x
y V : ξ t, y U.
Definition :
Lyapunov: If the process starts close, it remains close.
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Replicator Dynamics and Stability
A point is Lyapunov stable if for all neighborhoods
of , there exists a neighborhood of ,
s.t. for all
xU x V x
y V : ξ t, y U.
Definition :
Lemma: If x0 is Lyapunov stable then it is a N.E.
If there exists s.t. 0 0i ie π e - x , x > 0 :Proof :
By continuity there exists a neighborhood of and a s.t.
0
i
V δ > 0
y V : e - y, y > δ
x
π
But then : and starting from for a small :
,
But and this contradicts Lyapunov stability.
•
0 ii
i
0 i 0 δti i
0i
1 - α
1 - α +
1 - α +
x> δ x + αe α > 0
x
ξ t, 1 - α x + αe > x α e
x α > 0
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Replicator Dynamics and Stability
Against : does better than .
When is present then dominates it.
Against strategy does bett r, e .
3 3 1
3 2
2 1
e e e
e e
e e .
1 2 3
1 2 3
1 2 3
, e , e )
, e , e )
, e , e )
(e
(e
(e
Example: Example: A stable point need not be Lyapunov stable.
is , a N. . E
1A = 0 1 00 0 2 e0 0 1
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Replicator Dynamics and StabilityExample: Example: A stable point need not be Lyapunov stable.
is , a N. . E
1A = 0 1 00 0 2 e0 0 1
S2 S1
S3On the edges:
There are therefore close trajectories:
22
Stability ConceptsStability Concepts
A population plays the strategy p,
A small group of mutant enters, playing the strategy q
The population is now (1-ε)p+ εq
The fitness of p is:
Π p, 1 - ε p + εq = 1 - ε Π p, p + εΠ p,q
The fitness of q is:
Π q, 1 - ε p + εq = 1 - ε Π q, p + εΠ q,q
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:
The strategy is an Evolutionarily Stable Strate y if
g
i.
p
q
e.
1
ε 0 < ε < ε
- ε Π p, p
Π p, 1 - ε p + εq >
+ εΠ p,q > 1 - ε Π
Π q, 1 -
q, p + ε
ε p + εq
Π q,q
Definition:
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.
The strategy is an Evolutionarily Stable Strategy iff
(i)
and
(ii) if then
p
q p :
Π p, p Π q, p
Π p, p = Π q, p Π p,q Π q,q>
Lemma:
ESS Nash Equilibrium
( If p is an ESS then p is the best response to p )( If p is a strict equilibrium then it is an ESS.)
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: If is an E.S.S. then for all
1 - ε Π p, p + εΠ p,q > 1 - ε Π q, p + ε
p ε < ε
Π q,q
Proof:
.: By taking
Π p, p
ε 0
Π q, p
.If Π p, p = Π q, p
. for all > 0 : Π p,q > Π q,qε
Q.E.D.
26
.
If
(i)
and
(ii) if then
q p :
Π p, p Π q, p
Π p, p = Π q, p Π p,q Π q,q>
Proof:
. Choose a ifq p : Π p, p > Π q, p
. Then for sufficiently small ε :
1 - ε Π p, p + εΠ p,q > 1 - ε Π q, p + εΠ q,q
.if then [by (ii)]Π p, p = Π q, p Π p,q Π q,q>
.and for all
1 - ε Π p, p 1 - ε Π q
ε > 0 :
+ εΠ p,q > +p εΠ q,q,
Q.E.D.
27
ESS is Nash Equilibrium,But not all Nash Equilibria are ESS
s t
s 0 ,0 1 , 0
t 0 , 1 2 , 2
(s,s) is not an ESS, t can invade and does better !!
t is like s against s, but earns more against itself.
(t,t) is an ESS, t is the unique best response to itself.
(t,t) is a strict Nash Equilibrium
28
ESS does not always exist
R S P
R 0 , 0 1 , -1 -1 , 1
S -1 , 1 0 , 0 1 , -1
P 1 , -1 -1 , 1 0 , 0
Rock, Scissors, Paper
The only equilibrium is α = (⅓, ⅓, ⅓)
But α can be invaded by R
π α,α = π R,α = 0
π α,R = π R,R = 0
There is no distinction between α, R
There is no ESS(the only candidate is not an ESS)
29
Exercise:
Given a matrix MM of player 1’s payoffs in a symmetric game GGMM.Obtain a matrix NN by adding to each column of MM a constant.
(NNij=MMij+cj)
Show that the two games: GGM M ,G,GN N have the same eqilibria,same ESS, and the same Replicator Dynamics
30
ESS of 2x2 games
a11 , a11 a12 , a21
a21 , a12 a22 , a22
a11 a12
a21 a22
a11 + c1 a12 + c2
a21 + c1 a22 + c2
b1 0
0 b2
Given a symmetric game,
c1 = -a21
c2 = -a12
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ESS of 2x2 games
b1 0
0 b2
If: b1 > 0, b2 < 0
The first strategy is the unique equilibrium of this game, and it is a strict one.
Hence it is the unique ESS.
P.D
32
ESS of 2x2 games
b1 0
0 b2
If: b1 > 0, b2 > 0
Both pure equilibria are strict.Hence they are ESS.
Coordination
The mixed strategy equilibrium: 2
1 2
1 2 b
b + bx = λe + 1- λ e , λ =
is not an ESS.
1 1 11 21
1 2
=b b
bx,e = e ,b + b
e <
, and is not an ESS. 1 1 1 x,e e ,e x <
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ESS of 2x2 games
b1 0
0 b2
If: b1 < 0, b2 < 0
The only symmetric equilibrium is the mixed one.
Chicken
2
1 2
1 2 b
b + bx = λe + 1- λ e , λ = is ESS.
. x, y y, y >
All strategies get the same payoff against x
To show that x is an ESS we should show that for
all strategies y :
34
ESS of 2x2 games
b1 0
0 b2
If: b1 < 0, b2 < 0
Chicken
. x, y y, y >
2 1 1 21 1 2 2
1 2 1 2 1 2
=b b b b
x, y b y + b y =b + b b + b b + b
2
1 2
1 2 b
b + bx = λe + 1- λ e , λ =
1 22 211
1 2 1 222
= +yb 0
y, y y , y b y b yy0 b
1 2 221
221 1 2 2 1 2+b y b 1 - y = b + b y - 2b y + b
(-)
= x,x
35
ESS of 2x2 games
b1 0
0 b2
If: b1 < 0, b2 < 0
Chicken
. x, y y, y >
1 2
1 2
= =b b
x, y x,xb + b
2
1 2
1 2 b
b + bx = λe + 1- λ e , λ =
211 2 2 1 2y, y = b + b y - 2b y + b
The maximum of
is at: =
211 2 2 1 2
21
1 2
y, y = b + b y - 2b y + b
by
b + b 1= x
.
hence for all
but
=
y x : x, x > π y, y
x,x π x, y > π y, y
36
How many ESS can there be?
If is an ESS, then is an ESSnot i i
i ii H i H
.x = x e y = y e Lemma :
,
There is a finite no. of un-nested Supports.
Moreover, if is an interior ESS
then it is the ESS
n1 2 i
1.
2. x ,x ,....., x x > 0
.unique
For all hence jj H : e , x = π x,x y,x = π x,x .Proof :
By the second condition of ESS: . x, y π y, y>
is not a BR to itself, it is not a N.E. and therefore no ESS.yQ.E.D.
37
It can be shown that there is a uniform invasion barrier.
.
ESS can defined as:
is an Evolutionarily Stable Strategy i f
p> 0 q p
Π p, 1 - ε p+ εq > Π q, 1 - ε p+ εq
q p > 0 ε 0 < ε < ε
38
ESS is not stable against two mutants !!!
-1 0
0 -2Chicken
The unique ESS is the mixed strategy
2 1x = , .3 3
If and invaded with equal groups .
The population is:
The first coordinate is , the second .
.
1 2
1 2
e e ε/2
e + e 2 ε 1 εω = 1- ε x + ε = 1- ε + , 1 - ε +2 3 2 3 2
2 1< >3 3
The unique BR to is . 1ω e
is not an ESS, for against the population
the invader does better than 1
x ω,
e x.
39
Strategy is a Neutrally Stable Strategy (NSS) if
pε y
i) π x,x π y,x
ii) π x,x = π y,x π x, y π y, y
Definition :
ESS NSS NE
,
Rock, Scissors, Paper
The only NE is it is not ESS, but it
is NSS.
1 2 0
0 1 2
2 0 1
1 1 1, ,
3 3 3
40
NSS may not exist.
,The NE are:
1 2 3.
1 1 0
0 1 1
1 0 1
1 1 1, , e ,e ,e .
3 3 3x =
.There is no NSS.
is no NSS for: but
1 1 1 3 1 3 3 1 3e π e ,e = π e ,e π e ,e > π e ,e
is no NSS, for: but 1 1 1 1 2x π e ,x = π x,x 1 = π e ,e > π x,e = .3