Exercise 11. Solution. EVOLUTION AND THE THEORY OF GAMES (Spring 2009) EXERCISES Show that every 2 2 matrix game with payoff matrix

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1 EOLUTION AND THE THEORY OF GAMES (Spring 009) EXERCISES Exercise 11 Show that every matrix game with payoff matrix a,a c,b b,c d,d with a c or b d has at least one ESS. One should distinguish the following cases: If a > c then x 1 is an ESS. If d > b then x is an ESS. Suppose now a c and d b but with at least one strict inequality. In order to occur, by Bishop-Cannings, a mixed ESS requires x 1 p = a p + b (1 p) x p = c p + d (1 p) to be equal The hipotheses made on the utility guarantee x 1 p = x p p = b d b d + c a (0.1) 0 p 1 It remain to verify whether the ESS conditions are satisfied. The first is certainly not. Let J = (p, 1 p) the strategy satisfying Bishop-Cannings. Then for any other strategy I = (q, 1 q) the equality I J = q x 1 p +(1 q) x p = x 1 p = J J Hence the first ESS condition is not satified. The second ESS condition requires This means J I > I I J I = p x 1 q +(1 p) x q > q x 1 q +(1 q) x q = I I for any (q, 1 q). Unfolding the products and using the expression of (0.1) one gets into J I I I = (c a + b d)(p q) > 0 (0.) which proves that the second ESS condition is verified. 1

2 Exercise 1 Extend the Hawk-Dove game with a third strategy called Retaliator (R) who plays Dove against Dove but Hawk against Hawk. How does Retaliator play against itself? (There are several possibilities here.) Give the payoff matrix of the Hawk-Dove-Retaliator game and calculate all ESSs. We suppose that the retaliator plays Dove versus herself. The resulting payoff matrix is Hawk Dove Retaliator Hawk C C Dove 0 Retaliator C Note that as the game is symmetric it is sufficient to analyse the strategies of the row player. In order to find ESS conditions we need to look for symmetric Nash equilibria. Pure ESS s Suppose > C. In such a case Consider the Hawk vs Hawk game: U(H H) = C the first ESS condition is degenerate. The second requires = U(R H) (0.3) U(H R) = C > U(R R) = which is never satisfied (independently of the sign of C). The utility of the Dove vs Dove game satisfies (0.4) U(D D) = < = U(H D) (0.5) which invalidates both the first and the second ESS conditions (independently of the sign of C).

3 The of the Retaliator vs Retaliator game satisfies U(R R) = = = U(D R) U(R R) = > C = U(D R) (0.6) Hence we need to check the second ESS condition requiring U(R D) = > U(D D) = (0.7) which is false. No ESS independently of the sign of C. Suppose now < C: also in such a case no ESS is possible. The above analysis rules out pure ESS solutions. It remains to check the existence of mixed ESS solutions. Mixed ESS s Invoking Bishop-Cannings lemma we can construct a candidate ESS as solution of U p (H) = U(H H) p 1 + U(H D) p + U(H R) (1 p 1 p ) = w U p (D) = U(D H) p 1 + U(D D) p + U(D R) (1 p 1 p ) = w U p (R) = U(R H) p 1 + U(R D) p + U(R R) (1 p 1 p ) = w (0.8) The solution is p 1 = 0, p = C C +, w = (0.9) The second ESS requires then that for every q = (q 1, q, 1 q 1 q ) U q (H)p 1 + U q (D)p + U q (R)(1 p 1 p ) > U q (H) q 1 + U q (D) q + U q (R)(1 q 1 q ) (0.10) Inserting (0.9) one gets into q 1 C [( C) (q 1 + q ) C] (C + ) > 0 (0.11) For C > is semi-positive defined but vanishes for every q 1 = 0. The conclusion is that the second ESS condition does not hold also for mixed equilibria. Exercise 13 Extend the Hawk-Dove game with a third strategy called Bully (B) who plays Hawk against Dove but Dove against Hawk. How does Bully play against itself? (Again, there are multiple possibilities.) Give the payoff matrix of the Hawk-Dove-Bully game and calculate all ESSs. We analyse the game assuming that the Bully plays Dove against herself. 3

4 Hawk Dove Bully Hawk C Dove 0 0 Bully 0 Let us first inquire pure strategies. Pure ESS s Suppose > C. Consider the Hawk vs Hawk game. It satisfies the first ESS U(H H) = C > 0 = U(D H) = U(B H) (0.1) The Dove vs Dove cannot be an ESS U(D D) = < U(D H) (0.13) independently of the sign of C. The Bully vs Bully cannot either be an ESS as U(B B) = < = U(H B) (0.14) independently of the sign of C. If = C then Hawk vs Hawk fails to satisfy the first ESS condition. The second requires U(H D) = > = U(D D) & U(H R) = > = U(H R) (0.15) the second ESS is satisfied. if < C then Hawk vs Hawk is not an ESS. Mixed ESS s Let us now consider mixed ESS. 4

5 As the C condition guarantess that Hawk is an ESS, there cannot be a mixed strategy with worthwhile support on the Hawk strategy profile. However, if we eliminate the Hawk profile from the game the corresponding mixed strategy must satisfy U p (D) = p = w which is inconsistent.thus there cannot be an ESS. If < C then we can look for mixed strategies with full support The solution is U p (H) = C U p (B) = p + (1 p ) = w (0.16) U p (D) = p = w p 1 + p + (1 p 1 p ) = w U p (B) = p + (1 p 1 p ) = w (0.17) p 1 = 1, p = 1 C, w = C (0.18) which is not a consistent probability measure on the full support. On the other hand if we consider mixed ESS with non-full support. Using Bishop-Cannings we can look for a solution having support only on the Hawk-Bully strategy profiles The solution is U p (H) = C p 1 + (1 p 1 ) = w U p (B) = (1 p 1) = w (0.19) p = C Inserting in the first ESS condition we then get into & w = (C ) C (0.0) [U p (H)p 1 + U p (D)p + U p (R)(1 p 1 p )] p1 = C,p =0 [U p (H)q 1 + U p (D)q + U p (R)(1 q 1 q )] p1 = C,p =0 = (C ) q C which shows that degeneration occurs for (0.1) q = 0 (0.) The second ESS condition yields [U q (H)p 1 + U q (D)p + U q (R)(1 p 1 p )] p1 = C,p =0 [U q (H)q 1 + U q (D)q + U q (R)(1 q 1 q )] p1 = C,p =0 = ( C q 1) (0.3) 5

6 i.e. fails only for this means that q 1 = C p = [ C, 0, 1 ] C is an ESS. From the Hawk-Dove game we also know that a second mixed strategy profile exists for (0.4) (0.5) p = [ C, 1 C, 0] (0.6) Proceeding as indicated above it is straightforward to check that this is indeed the case. Exercise 14 Analyze the Retaliator-Bully game. We assume that once confronted with thenselves both Retaliator and Bully play Dove. In such a case we can assume that with equal probability the Bully Retaliator is an Hawk vs Hawk or a Dove vs Dove game. This choice is however arbitrary. Later in the course we will see that the natural way to model the Bully-Retaliator game is in the form of a multi-stage game where given an initial conditions Bully an Retaliator define deterministic algorithm to respond to the opponent strategy. Retaliator Retaliator Bully C Bully C In such a case we have > C 0 > C (0.7) which ensures that both the Retaliator and the Bully are ESS s when C >. The game is completely degenerate if = C. Finally if > C we can look for mixed ESS. with solution U p (R) = p 1 + C (1 p 1 ) = w U p (B) = C p 1 + (1 p 1) = w (0.8) p 1 = 1 & 6 w = 3 C 4 (0.9)

7 The second ESS condition for such a candidate mixed ESS yields U q (R) + U q (R) q U q (R) (1 q) U q (R) = ( C)(1 q) 4 which insures that the second ESS condition is ensured for the mixed ESS for > C. > 0 (0.30) Exercise 15 Animals can invest time and resources into growing weapons such as antlers etc. to improve their chances in a pairwise contest for some resource of value. Suppose that the individual with the biggest investment always wins. Give the payoff function and calculate all ESSs. What is the fundamental difference with the War of Attrition? Think about how a population of individuals with the same fixed level of investment would evolve if most of the time new mutants have about the same level of investment but sometimes a mutant appears with a largely different level of investment. u(t 1 t ) = α(t 1 ) if t 1 > t α(t 1) if t 1 = t α(t 1 ) if t 1 < t (0.31) with α a strictly monotonic function of time. To check the existence of ESS s we observe that Case of pure ESS. The inequality gives u(t t) > u(t s) t > s (0.3) > (0.33) which cannot be satisfied for > 0. In consequence we can rule out the existence of pure ESS unless we fix In such a case t 1 > t is not possible and yields the condition t 1 = 0 (0.34) u(0 0) > u(t s) s (0.35) > 0 (0.36) which is satisfied. So the only possible pure ESS is not to pay the cost of developing weapons. However this lead to certain defeat against armed opponents. 7

8 Mixed ESS. The Bishop-Cannings lemma gives for the candidate mixed ESS the equation i.e. U(t, [τ]) = t which rules out the existence of a mixed ESS. 0 ds [ α(t)] p τ (s) t ds α(t) p τ (s) (0.37) du dt = p τ (t) = 0 (0.38) 8