6 Evolutionary Game Theory & Replicator Dynamics Flashcards

Question

What is the relationship between the notion of stability in nonlinear dynamics and the concepts of evolutionary stability and Nash equilibrium?

Answer 1

notions are close but do not perfectly overlap, which gives rise to what is referred to as “Folk theorems”. given ODEs we can find equilibria, we saw that all equilibria correspond to a pure strategy of each species taking over we also saw that there can be 0 or 1 coexistence equilirbium, by demanding [*] to vanish, relates to BCT x˙ᵢ = xᵢ[Πᵢ(x) − Π¯(x)] = xᵢ[(Ax)ᵢ − xT Ax] = xᵢ[*]

Answer 2

(a) NE of the underlying stage game= equilibria of the REs , while strict NE of the underlying stage game are attractors (asymptotically equilibria) of the REs. (b) A stable equilibrium of the REs is a NE of the underlying stage game. (c) If a solution of the REs converges to some x, then x is a NE of the stage game. (d) The evolutionary stable strategy ESSes of the stage game is an attractor of the REs Moreover, interior ESSes (see Theorem 5.2) of the underlying stage games are global attractors of the REs (6.4), i.e. their basin of attraction is the whole simplex S. Importantly, the converse of statements (a)-(d) are not true. Proofs, along with examples and counter-examples, can, e.g., be found in Ref.(4) (see Sec. 6.6). (e) For two-player symmetric matrix games with two pure strategies (2×2 symmetric games), things are fortunately very simple: in this case, x is an ESS if and only if it is an attractor of the REs (6.4).

Answer 3

- A **Nash equilibrium** (NE) of a stage game is an **equilibrium** of the associated replicator equations (REs) - **Strict Nash equilibria are attractors**(asympotic stable equilibria) of the associated REs - A **stable equilibrium of the REs** is **NE** of the underlying stage game - **Evolutionary stable strategies** (ESSes) of a stage game are **attractors** of the underlying REs Note: converse statements are not true! (e.g. attractors of the REs are not ESSes of the stage game)

Answer 4

In this special, but important case: ESSes <=> ATTRACTORS OF THE REs

Answer 5

symmetric two-player games with two pure strategies (2 × 2 games) REs are scalar ODEs (:D) infinitely large and unstructured population where individuals (players) have two pure strategies at their disposal, and are randomly paired to play the underlying symmetric 2 × 2 stage game.

Answer 6

“cooperation” (C) e𝒸 = (1, 0)^T “defection” (D) e𝒹 = (0, 1)^T **A**= vs |C |D C a b D c d Frequency of species/strategy C is x, frequency of species/strategy D is 1-x (Fractions)

Answer 7

x(t) = (x(t), 1 − x(t))^T = x(t)e𝒸 + (1 − x(t))e𝒹

Answer 8

Expected payoff of C = fitness of C in state x(t) : Π𝒸 = e𝒸^T**A**x(t) = ax(t) + b(1 − x(t)) =(1 0) **A** ( x, 1-x)^T playing c against x (note x is x(t)) Expected payoff of D = fitness of D in state x(t) : Π𝒹 = e𝒹^T**A**x(t) = cx(t) + d(1 − x(t)) =(1 0) **A** ( x, 1-x)^T avrg payoff/fitness: Π¯(t) = x(t)^T **A**x(t) = x(t)Π𝒸(t) + (1 − x(t))Π𝒹(t) =(x, 1-x) **A** (x, 1-x) =(x 1 − x) (Π𝒸) (Π𝒹) = xΠ𝒸 + (1 − x)Π𝒹 = [ax + b(1 − x)] [cx + d(1 − x)] = (Π𝒸) (Π𝒹) x is the fraction of cooperation in pop x Π𝒸 the fitness of these + 1-x pop of defectors x Π𝒹

Answer 9

uses avrg payoff/fitness x˙ = x[Π𝒸 (x)−Π¯(x)] = x(1−x)[Π𝒸 (x)−Π𝒹(x)] = x(1−x)[b−d+x(a−b−c+d)] SINGLE scalar ODE k=2 case solved by integration, separation solution gives the frequency x(t) of C according to the replicator dynamics, frequency for D is 1-x(t)

Answer 10

EQ: to the x=0 (all D individuals) and x=1 (all C individuals) and when b>d and c>a, or b

Answer 11

(important replicator eq property: we can get the same replicator equations by adding/subtracting a constant in each column of payoff matrix) so payoff matrix **A~** = [a − c 0] [0 d − b] gives same RE as **A** rescaling the payoff matrix

Answer 12

t(x) = C + ∫₀ˣ (dy)/(y(1 − y) {b − d + (a + d − b − c)y}) which would need to be inverted to get x(t) [See Q3(d) of Sheet 3]

Answer 13

uses payoff matrix **A~** = [a − c 0] [0 d − b] quadrants: c-a - b-d **When c> a and d> b**: defection dominance (class of PD game) =>D is attractor and sNE **When cC is attractor and sNE **When c>a and d is ESS and attractor => stable coexistence of C and D **When c b:** coordination games (like stag-hunt) => C and D are ESSes and attractors, is a NE (not an ESS) and an unstable equilibrium of the RE. Coordination game exhibits bistability. One of C and D has a larger basin of attraction => strategy that is risk dominant Game with a=c and b=d is neutral and its RE is x˙ = 0 ⇒ x(t) = constant if x∗ closer to C than D (left bottom quadrant) then basin of attraction to c larger, risk dominant and the other would be pareto dominant diagram

Answer 14

quadrants: c-a DOMINANCE |ANTI-COORDINATION --------------------------------b-d COORDINATION| DOMINANCE games: |Hawk Gove snowdrift -------------------------------------------------- stag-hunt| prisoner's dilemma Diagrams: ₓ₌₀ ₓ₌₁ |ₓ₌₀ ₓ∗ ₓ₌₁ D → C | D→◯←C ------------------------------ D←◯→C | D←C ₓ₌₀ ₓ∗ ₓ₌₁ |ₓ₌₀ ₓ₌₁ explanation fitness: Π𝒸>Π𝒹 so C dominance |cooperation x∗ = (x∗, 1 − x∗) stable ----------- ESSes and atractors 0,1 coxistence unstable|Π𝒸<Π𝒹 so D dominance cooperation or defection attractors: dominance | Π𝒸<Π𝒹 so D dominance

Answer 15

The replicator dynamics of symmetric games with more than 2 pure strategies is much richer than the case of 2x2 games seen so far. Here, we look at the rock-paper-scissors (RPS) games with the cyclic dominance of three pure strategies and find oscillatory behaviour

Answer 16

Rock-paper-scissors game: reference model for cyclic dominance. 3 pure strategies in cyclic competition: Rock (R) which wins against Scissors (S) and loses vs. Paper (P), which in turn loses to Scissors (S) pure strategies eᵣ= (1, 0, 0)^T for R, eₚ = (0, 1, 0)^T for P, eₛ = (0, 0, 1)^T for S Here k=3 pure strategies/species => (k-1)=2 coupled REs P dominates over R R dominates over S S dominates over P cyclic no species dominates over all others

Answer 17

Population games: pure strategies interpreted as species and represented by unit vectors:

Answer 18

Aᴿᴾˢ = [ R P S ] [R 0 −1 1] [P 1 0 −1] [S −1 1 0] => win: +1, draw: 0, loss: -1 => in this formulation of zero-sum RPS: one player gains what the other loses zero sum: why? because +1 and -1

Answer 19

x1 : frequency of R x2 : frequency of P x3 : frequency of S **x** = (x₁, x₂, x₃)^T = x₁**eᵣ** + x₂**eₚ** + x₃**eₛ**

Answer 20

to R-player: Πᵣ= Π₁ = eᵣ^T Aᴿᴾˢx = (**Aᴿᴾˢx**)₁ = x₃ − x₂ = 1 − 2x₂ − x₁ to P-player: Πₚ = Π₂ = eₚ^T Aᴿᴾˢx = (**Aᴿᴾˢx**)₂ = x₁ − x₃ = 2x₁ − x₂ − 1 to S-player: Πₛ = Π₃ = eₛ^TAᴿᴾˢx = (**Aᴿᴾˢx**)₃ = x₂ − x₁ (bc they sum to 1)

Answer 21

(payoff matrix here is antisymmetric so easier) Π¯ = x^T Aᴿᴾˢx = 0 zero sum game average

Answer 22

x˙ᵢ = xᵢ(Aᴿᴾˢx)ᵢ− x^T Aᴿᴾˢx = xᵢ(Aᴿᴾˢx)ᵢ (for i = 1, 2, 3) quadratic in xᵢ => explicitly: x˙₁ = x₁(x₃ − x₂), x˙₂ = x₂(x₁ − x₃), x˙₃ = x₃(x₂ − x₁) 2 linear coupled ODEs (if I know 1,2 i know 3) x₃= 1- x₂ − x₁

Answer 23

x₃= 1- x₂ − x₁ x˙₁ = x₁(1 − x₁ − 2x₂),=f₁ x˙₂ = x₂(2x₁ + x₂-1)=f₂ non linear system of planar ODES can't solve: look for EQ

Answer 24

x˙₁ = x˙₂ =0 are EQ: only R only P only S x = (1, 0, 0),(0, 1, 0),(0, 0, 1) corresponding to 1 species taking over COEXISTENCE x∗ = (x₁∗, x₂∗, 1 − x₁∗ − x₂∗) such that 1 − x₁∗ − 2x₂∗ = 2x₁∗ + x₂∗ − 1 = 0 thus x₁∗ = x₂∗ = x₃∗ = 1/3 x∗ = (1/3, 1/3, 1/3)^T Each species /pure strategy coexist with the same frequency

Answer 25

J(x₁, x₂) = [∂f₁/∂x₁ ∂f₁/∂x₂] [∂f₂/∂x₁ ∂f₂/∂x₂] = [1 − 2(x₁ + x₂) −2x₁] [ 2x₂ 2(x₁ + x₂) − 1] evaluated jacobian at each equilibrium and find the eigenvalues x = (1, 0, 0),(0, 1 , 0),(0, 0, 1) saddles one +ve one-ve eigenvalue (+1,-1) x∗ : J(x∗) = J∗ [−1/3 −2/3] [2/3 1/3] eigenvalues of J* are ±i/√3 PURELY IMAGINARY x∗ non hyperbolic so no conclusion about stability of from linear stability analysis

Answer 26

this is conserved by the RES if i look for time derivative of K, this is 0 (d/dt)K= K˙= x˙₁x₂x₃ + x₁x˙₂x₃ + x₁x₂x˙₃ = x₁x₂x₃(x₃ − x₂) + x₁x₂x₃(x₁ − x₃) + x₁x₂x₃(x₂ − x₁) = 0 Due to the time derivative being 0, if given IC for **x** then we can find K(t)=K(0), which define neutrally stable closed orbits surrounding x∗

Answer 27

Coexistence eq not stable but said to be neutrally/imaginary stable with neutrally closed orbits surrounding x∗ SKETCH t against x values, x∗ =1/3 oscillations of x₁(t),x₂(t),x₃(t) oscillate about 1/3

Answer 28

x∗ =1/3,..., surrounding closed orbits of x₁(t),x₂(t),x₃(t) Closed orbits around set by K(0)=constant triangle with RPS vertices

Answer 29

Generalized RPS game (Q4 of Example Sheet 3): Aᵍᴿᴾˢ = [ R P S] [R 0 −ϵ 1] [P 1 0 −ϵ] [S −ϵ 1 0] with ϵ ≥ 0 When the gRPS game is a non-zero-sum game. When we recover the zero-sum RPS game Here: R vs. P = P vs. S = S vs. R and R vs. S= P vs. R = S vs. P => symmetry (all species on same footing) In gRPS game, win: +1, draw: 0, loss −ϵ NO LONGER ZERO SUM if not = 1. no longer antisymmetric

Answer 30

Like for the Lotka-Volterra model, replicator dynamics of zero-sum RPS game lead to oscillatory behaviour set by initial condition => is marginally stable centre and oscillations about it are not robust (depend on initial condition)

Answer 31

Expected payoffs are Πᵢ = (Aᵍᴿᴾˢx)ᵢ ⇒ Πᵣ(x) = Π₁(x) = x₃ − ϵx₂, Πₚ (x) = Π₂(x) = x₁ − ϵx₃, Πₛ(x) = Π₃(x) = x₂ − ϵx₁ Π¯ ̸= 0 when ϵ ̸= 1 Average payoff is Π¯(x) = x^T Aᵍᴿᴾˢx = x₁Π₁ + x₂Π₂ + x₃Π₃ = (1 − ϵ)(x₁x₂ + x₁x₃ + x₂x₃) REs:

Answer 32

x˙ᵢ = xᵢ(Aᵍᴿᴾˢx)ᵢ − x^T Aᵍᴿᴾˢx ⇒ x˙₁= x₁ [x₃ − ϵx₂ − (1 − ϵ)(x₁x₂ + x₁x₃ + x₂x₃)] x˙₂= x₂[x₁ − ϵx₃ − (1 − ϵ)(x₁x₂ + x₁x₃ + x₂x₃)] x˙₃ = x₃ [x₂ − ϵx₁ − (1 − ϵ)(x₁x₂ + x₁x₃ + x₂x₃)]

Answer 33

If it exists, it is unique by BCT: x₃∗ − ϵx₂∗ = x₁∗ − ϵx₃∗ = x∗₂ − ϵx₁∗ = (1 − ϵ)(x₁∗x₂∗ + x∗₁x∗₃ + x∗₂x∗₃) invariant under 1 -> 2 ->3 -> 1 and under 1-> 3 -> 2-> 1 Symmetry here suggests x∗1 = x∗2 = x∗3 = 1/3, physical We can indeed verify that Π₁(x∗) = Π₂(x∗) = Π₃(x∗) = Π¯(x∗) = 1 − ϵ^3 when x∗ = (1/3, 1/3, 1/3)^T By Bishop-Cannings Theorem, x∗ = 1/3(1, 1, 1) is a coexistence equilibrium of the REs and NE of the gRPS game

Answer 34

When ϵ > 1: replicator dynamics connection of the saddle equilibria (1,0,0), (0,1,0), (0,0,1) forming what is called a heteroclinic cycle. The frequency of each species approaches 0 or 1 in turn (heteroclinic oscillate not periodically) coexistence x∗ is unstable

Answer 35

x∗ is the ESS and global attractor of the gRPS game=> replicator dynamics leads trajectories spiralling towards in the phase plane. spiralling inwards This corresponds to damped oscillations of xᵢ → xᵢ∗ = 1/3

Answer 36

a) ϵ > 1 b) ϵ = 1 c) ϵ < 1

Answer 37

didnt matter before player 1 player 2 same options same payoff So far we've focused on symmetric games, but there are asymmetric games in which players have different roles and used different payoff matrix. 2 types of players payoff matrices **A and B** => bimatrix games different “types” or “roles” (e.g., adults vs. juveniles, males vs. females, etc

Answer 38

asymmetric games: each type of players has its own payoff matrix and knows of which type it is. Asymmetric games are thus played between players of different types (in different roles) choosing among the same set of pure strategies, say cooperation (C) and defection (D), using payoff matrices **A**= (Aᵢⱼ) and **B**=(Bᵢⱼ) ≠ A

Answer 39

**A**= (Aᵢⱼ) and **B**=(Bᵢⱼ) ≠ A pure strategies **e**C=(1,0)^T **e**D=(0,1)^T A P1-individual uses payoff matrix against a P2-player, and a P2 individual uses payoff matrix against a P1-player **A**= P1 ↓ || P2 → | C D C| A₁₁ A₁₂ D|A₂₁ A₂₂ **B**= P2 ↓ || P1 → | C D C| B₁₁ B₁₂ D|B₂₁ B₂₂

Answer 40

P1 uses strategy **x**=Σxᵢ**e**ᵢ each pure strategy i with frequency 0 ≤ xᵢ≤ 1 , Σᵢ xᵢ = 1 P2 uses strategy **y**=Σyᵢ**e**ᵢ each pure strategy i with frequency 0 ≤ yᵢ≤ 1 , Σᵢ yᵢ = 1 giving STRATEGY PAIR (P1 player, P2 player) (x,y) Expected payoff player to P1 is Π⁽¹⁾(x, y) = x^T Ay Expected payoff player to P2 is Π⁽²⁾(x, y) = x^T Ay

Answer 41

playing with fractions: **x**= 0.5 **e**𝒸 +0.5**e**𝒹 =(0.5,0.5)^T **y** = (1/3)**e**𝒸 +(2/3)**e**𝒹 STRATEGY PROFILE (**x,y**)= ( (1/2,1/2)^T, (1/3, 2/3)^T) Meaning: P1 population consists of half C's and half D's, P2 population consists of a fraction 1/3 of C and 2/3 of D. This description, does not say if the population of P2 is larger than that of P1 (we tacitely assume that both groups are "very large")

Answer 42

Notions of Nash equilibrium (NE) and strict Nash equilibrium (sNE) are direct generalizations:

Answer 43

(x∗, y∗) is a NE if x∗^T **A**y∗≥ x^T **A**y∗ and y∗^T Bx∗ ≥ y^T Bx∗ for all (x, y) ̸= (x∗, y∗) **(x∗, y∗) is NE if x∗ is best reply to y∗ AND y∗ is best reply to x∗** (a NE is the best reply to itself)

Answer 44

(x∗, y∗) is an sNE if x∗T Ay∗ > x^T Ay∗ and y∗T Bx∗ > y^T Bx∗ for all (x, y) ̸= (x∗, y∗) in bimatrix games (x∗, y∗) is an evolutionary stable strategy (ESS) pair IFF it is a sNE

Answer 45

(Ay∗)₁ = · · · = (Ay∗)ₖ = x∗^T Ay∗ and (Bx∗)₁ = · · · = (Bx∗)ₖ = y∗^T Bx∗ In bimatrix games, all sNE/ESSes are necessarily pure strategies=> a mixed-strategy NE is never an ESS in a bimatrix game

Answer 46

In this game Husband (Player 1) and Wife (Player 2) have to decide whether to watch a football game or go to the theatre (events are mutually exclusive). The husband prefers the football and the wife the theatre, and both are happier (higher payoff) if they pick a common activity. This leads to a bimatrix game in which each player has two pure strategies, “watch football” (F) or “go to theatre” (T), and the two “roles” are husband (P1) and wife (P2). payoff matrices: A = vs. F T F 2 0 T 0 1 and B = vs. F T F 1 0 T 0 2

Answer 47

By inspection of A and B, it is clear that the pair of pure strategies (F,F), corresponding to the Husband and Wife watching football, and (T,T) are strict NE and therefore ESSes. The strategy pair (F,F) yields a payoff 2 to the husband and 1 to the wife, and we thus say that the payoff is (2, 1) to (husband, wife). Similarly, the strategy pair (T,T) yields a payoff (1, 2). Is there also a mixed NE (x∗, y∗)? To answer this question, we look at the mixed strategy x ∗ = (x ∗ , 1 − x ∗ ) T for the husband to play F with frequency 0 < x∗ < 1 and T with frequency 1 − x ∗ , and y ∗ = (y ∗ , 1 − y ∗ ) T for the wife to play F and T with respective frequencies 0 < y∗ < 1 and 1 − y ∗

Answer 48

In population bimatrix games, there are two very large interacting populations: P1-individuals using the payoff matrix B =/ A versus P2-players and P2-individuals using the payoff matrix versus P1-players. Interactions/games are only between P1 and P2 individuals. Expected payoff player to P1 playing pure strategy i: Πᵢ⁽¹⁾ ≡ Π⁽¹⁾(eᵢ, y) = eᵢ^T Ay = (Ay)ᵢ Expected payoff player to P2 playing pure strategy i: Πᵢ⁽²⁾ ≡ Π⁽²⁾(eᵢ, y) = eᵢ^T By = (By)ᵢ Average payoff to P1 and P2 in population stat (x, y) Π¯⁽¹⁾ = x^T Ay and Π¯ ⁽²⁾= y^T Bx respectively giving Replicator equations (REs):

Answer 49

x˙ᵢ = xᵢ[Πᵢ⁽¹⁾ − Π¯⁽¹⁾] = xᵢ [(Ay)ᵢ − x^T Ay] y˙ᵢ = yᵢ[Πᵢ⁽²⁾− Π¯⁽²⁾] = yᵢ [(Bx)ᵢ − y^T Bx] i=1,..,k set of 2(k-1) coupled nonlinear ODEs 0≤ xᵢ, yᵢ ≤ 1 Σᵢ xᵢ=1 Σᵢ yᵢ=1 (links to growth rate? difference between expected payoff and average payoff)

Answer 50

xᵢ=0 or 1 yᵢ=0 or 1 when k=2 (x₁, y₁) = (1 − x₂, 1 − y₂) = {(0, 0),(0, 1),(1, 0),(1, 1)} => when k=2, are 4 equilibria All P1 and P2 Ds all P1 are D's, P2 are Cs P1 are Cs P2 are Ds all P1 are C's and all P2 are Ds There is a unique interior equilibrium that is not asymptotically stable (not ESS) if (Ay∗)ᵢ = x∗T Ay∗ and (Bx∗)ᵢ = y∗T Bx∗ for all i = 1, . . . , k have a physical solution with 0 < xᵢ∗,yᵢ∗ < 1 (by demanding what is in [*] to vanish) In two-player bimatrix games, the interior equilibrium point is either unstable or surrounded by closed orbits.

Answer 51

P1 and P2 individuals, conceal in their palms a penny either with its face up or face down. Both coins are revealed simultaneously. If both coins show the same (both are heads or both are tails), P1 wins and P2 loses, whereas P1 loses and P2 wins if one shows head and the other tail.

Answer 52

Here: 2 players with 2 pure strategies, "show head" (H) represented by **e**ₕ=(1,0)^T or "show tail" (T) represented by **e**ₜ= (0,1)^T H and T are played with respective frequencies by P1, and frequencies x₁ ≡ x and x₂ ≡ 1 − x by the P2 player y₁ ≡ y and y₂ ≡ 1 − y

Answer 53

**A**= vs H T H 1 -1 T -1 1 **B**= vs H T H -1 1 T 1 -1 win +1 loss -1 Πₕ⁽¹⁾= (**Ay**)₁ = 2y-1=Πₜ⁽¹⁾ (first component of **Ay**) Πₕ⁽²⁾= (**Bx**)₁= 1 -2x = =Πₜ⁽²⁾ Π~⁽¹⁾=(2x-1)(2y-1) Π~⁽²⁾=-(2x-1)(2y-1) second average payoff uses B instead

Answer 54

x˙ = x[Π⁽¹⁾ −Π¯⁽¹⁾] = −2x(1−x)(1−2y), y˙ = y[Π⁽²⁾ −Π¯⁽²⁾] = 2y(1−y)(1−2x)

Answer 55

Equilibrium points: (x, y) = {(0, 0),(0, 1),(1, 0),(1, 1)} (0,0) P1 and P2 always playing T, (0,1) P1 always playing T and P2 always playing H, (1,0) P1 always playing H and P2 always playing T, (1,1) playing H **coexistence equilibrium** (x*,y*)=(1/2,1/2) for which Πₕ⁽¹⁾ = Πₜ⁽¹⁾ = Π¯⁽¹⁾ = 0 = Πₕ⁽²⁾ = Πₜ⁽²⁾=Π¯⁽²⁾ half-half is a coexistence equilibrium

Answer 56

Jacobian of REs is J(x, y) = 2* [2(x + y) − 4xy − 1 2x(1 − x)] [−2y(1 − y) −[2(x + y) − 4xy − 1] Evaluated at each equilibrium=> (0, 0),(0, 1),(1, 0),(1, 1) saddle points and thus unstable (one positive and one negative eigenvalue of J). **coexistence** J at (x*,y*) has purely imaginary eigenvalues => (x*,y*) is non-hyperbolic and we cannot conclude from linear stability.

Answer 57

In fact, REs conserve the quantity H = x(1 − x)y(1 − y) that is H˙ = 0 ⇒ H(t) = H(0) (initial value, constant) Hence is a marginally stable centre: it is surrounded by closed orbits set by the initial condition we could also do phase plan analysis to see this

6 Evolutionary Game Theory & Replicator Dynamics Flashcards

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