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

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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

?

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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

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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

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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

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., 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 !!!

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, 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.

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, 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

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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:

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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:

24

.

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.

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.

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.

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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

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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

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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

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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