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The Structure of Networks with emphasis on information and social networks T-214-SINE Summer 2011 Chapter 7 Ýmir Vigfússon

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Game theory Regular game theory ◦ Individual players make decisions ◦ Payoffs depend on decisions made by all ◦ The reasoning about what other players might do happens simultaneously Evolutionary game theory ◦ Game theory continues to apply even if no individual is overtly reasoning or making explicit decisions ◦ Decisions may thus not be conscious ◦ What behavior will persist in a population?

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Background Evolutionary biology ◦ The idea that an organism‘s genes largely determine its observable characteristics (fitness) in a given environment More fit organisms will produce more offspring This causes genes that provide greater fitness to increase their representation in the population ◦ Natural selection

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Evolutionary game theory Key insight ◦ Many behaviors involve the interaction of multiple organisms in a population ◦ The success of an organism depends on how its behavior interacts with that of others Can‘t measure fitness of an individual organism ◦ So fitness must be evaluated in the context of the full population in which it lives Analogous to game theory! ◦ Organisms‘s genetically determined characteristics and behavior = Strategy ◦ Fitness = Payoff ◦ Payoff depends on strategies of organisms with which it interacts = Game matrix

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Motivating example Let‘s look at a species of a beetle ◦ Each beetle‘s fitness depends on finding and processing food effectively ◦ Mutation introduced Beetles with mutation have larger body size Large beetles need more food What would we expect to happen? ◦ Large beetles need more food ◦ This makes them less fit for the environment ◦ The mutation will thus die out over time But there is more to the story...

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Motivating example Beetles compete with each other for food ◦ Large beetles more effective at claiming above-average share of the food Assume food competition is among pairs ◦ Same sized beetles get equal shares of food ◦ A large beetle gets the majority of food from a smaller beetle ◦ Large beetles always experience less fitness benefit from given quantity of food Need to maintain their expensive metabolism

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Motivating example The body-size game between two beetles Something funny about this ◦ No beetle is asking itself: “Do I want to be small or large?“ Need to think about strategy changes that operate over longer time scales ◦ Taking place as shifts in population under evolutionary forces SmallLarge Small5, 51, 8 Large8, 13, 33, 3

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Evolutionary stable strategies The concept of a Nash equilibrium doesn‘t work in this setting ◦ Nobody is changing their personal strategy Instead, we want an evolutionary stable strategy ◦ A genetically determined strategy that tends to persist once it is prevalent in a population Need to make this precise...

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Evolutionarily stable strategies Suppose each beatle is repeatedly paired off with other beetles at random ◦ Population large enough so that there are no repeated interactions between two beetles A beetle‘s fitness = average fitness from food interactions = reproductive success ◦ More food thus means more offspring to carry genes (strategy) to the next generation Def: ◦ A strategy is evolutionarily stable if everyone uses it, and any small group of invaders with a different strategy will die off over multiple generations

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Evolutionarily stable strategies Def: More formally ◦ Fitness of an organism in a population = expected payoff from interaction with another member of population ◦ Strategy T invades a strategy S at level x (for small x) if: x fraction of population uses T 1-x fractionof population uses S ◦ Strategy S is evolutionarily stable if there is some number y such that: When any other strategy T invades S at any level x < y, the fitness of an organism playing S is strictly greater than the fitness of an organism playing T

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Is Small an evolutionarily stable strategy? Let‘s use the definition ◦ Suppose for some small number x, a 1-x fraction of population use Small and x use Large ◦ In other words, a small invader population of Large beetles What is the expected payoff to a Small beetle in a random interaction? ◦ With prob. 1-x, meet another Small beetle for a payoff of 5 ◦ With prob. x, meet Large beetle for a payoff of 1 ◦ Expected payoff: 5(1-x) + 1x = 5-4x Motivating example

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Is Small an evolutionarily stable strategy? Let‘s use the definition ◦ Suppose for some small number x, a 1-x fraction of population use Small and x use Large ◦ In other words, a small invader population of Large beetles What is the expected payoff to a Large beetle in a random interaction? ◦ With prob. 1-x, meet a Small beetle for payoff of 8 ◦ With prob. x, meet another Large beetle for a payoff of 3 ◦ Expected payoff: 8(1-x) + 3x = 8-5x Motivating example

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Expected fitness of Large beetles is 8-5x Expected fitness of Small beetles is 5-4x ◦ For small enough x (and even big x), the fitness of Large beetles exceeds the fitness for Small ◦ Thus Small is not evolutionarily stable What about the Large strategy? ◦ Assume x fraction are Small, rest Large. ◦ Expected payoff to Large: 3(1-x) + 8x = 3+5x ◦ Expected payoff to Small: 1(1-x) + 5x = 1+4x ◦ Large is evolutionarily stable

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Motivating example Summary ◦ A few large beetles introduced into a population consisting of small beetles ◦ Large beetles will do really well: They rarely meet each other They get most of the food in most competitions ◦ Population of small beetles cannot drive out the large ones So Small is not evolutionarily stable

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Motivating example Summary ◦ Conversely, a few small beetles will do very badly They will lose almost every competition for food ◦ A population of large beetles resists the invasion of small beetles ◦ Large is thus evolutionarily stable The structure is like prisoner‘s dilemma ◦ Competition for food = arms race ◦ Beetles can‘t change body sizes, but evolutionarily forces over multiple generations are achieving analogous effect

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Motivating example Even more striking feature! ◦ Start from a population of small beetles ◦ Evolution by natural selection is causing the fitness of the organisms to decrease over time Does this contradict Darwin‘s theory? ◦ Natural selection increases fitness in a fixed environment ◦ Each beetle‘s environment includes all other beetles ◦ The environment is thus changing It is becoming increasingly more hostile for everyone This naturally decreases the fitness of the population

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Evolutionary arms races Lots of examples ◦ Height of trees follows prisoner‘s dilemma Only applies to a particular height range More sunlight offset by fitness downside of height ◦ Roots of soybean plants to claim resources Conserve vs. Explore Hard to truly determine payoffs in real- world settings

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Evolutionary arms races One recent example with known payoffs ◦ Virus populations can play an evolutionary version of prisoner‘s dilemma ◦ Virus A Infects bacteria Manifactures products required for replication ◦ Virus B Mutated version of A Can replicate inside bacteria, but less efficiently Benefits from presence of A ◦ Is B evolutionarily stable?

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Virus game Look at interactions between two viruses ◦ Viruses in a pure A population do better than viruses in pure B population ◦ But regardless of what other viruses do, higher payoff to be B Thus B is evolutionarily stable ◦ Even though A would have been better ◦ Similar to the exam-presentation game AB A1.00, 1.000.65, 1.99 B1.99, 0.650.83, 0.83

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What happens in general? Under what conditions is a strategy evolutionarily stable? ◦ Need to figure out the right form of the payoff matrix ◦ How do we write the condition of evolutionary stability in terms of these 4 variables, a,b,c,d? ST Sa, ab, c Tc, bd, d Organism 2 Organism 1

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What happens in general? Look at the definition again ◦ Suppose again that for some small number x: A 1-x fraction of the population uses S An x fraction of the population uses T What is the payoff for playing S in a random interaction in the population? ◦ Meet another S with prob. 1-x. Payoff = a ◦ Meet T with prob. x. Payoff = b ◦ Expected payoff = a(1-x)+bx Analogous for playing T ◦ Expected payoff = c(1-x)+dx

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What happens in general? Therefore, S is evolutionarily stable if for all small values of x: ◦ a(1-x)+bx > c(1-x)+dx ◦ When x is really small (goes to 0), this is a > c ◦ When a=c, the left hand side is larger when b > d In other words ◦ In a two-player, two-strategy symmetric game, S is evolutionarily stable pricely when either a > c, or a = c, and b > d

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What happens in general? Intuition ◦ In order for S to be evolutionarily stable, then: Using S against S must be at least as good as using T against S Otherwise, an invader using T would have higher fitness than the rest of the population ◦ If S and T are equally good responses to S S can only be evolutionarily stable if those who play S do better against T than what those who play T do with each another Otherwise, T players would do as well against the S part of the population as the S players

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Relationship with Nash equilibria Let‘s look at Nash in the symmetric game ◦ When is (S,S) a Nash equilibrium? ◦ S is a best response to S: a ≥ c Compare with evolutionarily stable strategies: ◦ (i) a > c or (ii) a = c and b > d Very similar! ST Sa, ab, c Tc, bd, d

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Relationship with Nash equilibria We get the following conclusion ◦ Thm: If strategy S is evolutionary stable, then (S,S) is a Nash equilibrium Does the other direction hold? ◦ What if a = c, and b < d? ◦ Can we construct such an example?

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From yesterday Stag Hunt ◦ If hunters work together, they can catch a stag ◦ On their own they can each catch a hare ◦ If one hunter tries for a stag, he gets nothing Two equilibria, but “riskier“ to hunt stag ◦ What if other player hunts hare? Get nothing ◦ Similar to prisoner‘s dilemma Must trust other person to get best outcome! Hunt StagHunt Hare Hunt Stag 4, 40, 3 Hunt Hare 3, 03, 3

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Counterexample Modify the game a bit Want: a = c, and b < d Hunt StagHunt Hare Hunt Stag 4, 40, 3 Hunt Hare 3, 03, 3 ST Sa, ab, c Tc, bd, d

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Counterexample Modify the game a bit Want: a = c, and b < d ◦ We‘re done! Hunt StagHunt Hare Hunt Stag 4, 40, 4 Hunt Hare 4, 03, 3 ST Sa, ab, c Tc, bd, d

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Relationship with Nash equilibria We get the following conclusion: ◦ Thm: If strategy S is evolutionarily stable, then (S,S) is a Nash equilibrium Does the other direction hold? ◦ What if a = c, and b < d? ◦ Can we construct such an example? Yes! However! Look at a strict Nash equilibrium ◦ The condition gives a > c ◦ Thm: If (S,S) is a strict Nash equilibrium, then strategy S is evolutionarily stable The equilibrium concepts refine one another

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Summary Nash equilibrium ◦ Rational players choosing mutual best responses to each other‘s strategy ◦ Great demands on the ability to choose optimally and coordinate on strategies that are best responses to each other Evolutionarily stable strategies ◦ No intelligence or coordination ◦ Strategies hard-wired into players (genes) ◦ Successful strategies produce more offspring Yet somehow they are almost the same!

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Mixed strategies It may be the case that no strategy is evolutionarily stable ◦ The Hawk-Dove game is an example Hawk does well in all-Dove population Dove does well in population of all Hawks ◦ The game even has two Nash equilibria! Yesterday we introduced mixed strategies to study this ◦ How should we define this in our setting?

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Mixed strategies Suppose: ◦ Organism 1 plays S with probability p Plays T with probability 1-p ◦ Organism 2 plays S with probability q Plays T with probability 1-q Expected payoff for organism 1 ◦ Probability pq of (S,S) pairing, giving a ◦ Probability p(1-q) of (S,T) pairing, giving b ◦ Probability (1-p)q of (T,S) pairing, giving c ◦ Probability p(1-q) of (T,T) pairing, giving d In total: ◦ V(p,q) = pqa+p(1-q)b+(1-p)qc+(1-p)(1-q)d

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Mixed strategies Fitness of an organism = expected payoff in a random interaction More precisely: ◦ Def: p is an evolutionary stable mixed strategy if there is a small positive number y s.t. when any other mixed strategy q invades p at any level x

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Mixed strategies Let‘s dig into this condition p is an evolutionarily stable mixed strategy if: ◦ For some y and any x < y, the following holds for all mixed strategies q ≠ p: (1-x)V(p,p) + xV(p,q) > (1-x)V(q,p) + xV(q,q) This parallels what we saw earlier for mixed Nash equilibria ◦ If p is an evolutionarily stable mixed strategy then V(p,p) ≥ V(q,p), ◦ Thus p is a best response to q ◦ So (p,p) is a mixed Nash equilibrium

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Example: Hawk-Dove See book

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Interpretation Can interpret this in two ways ◦ All participants in the population are mixing over two possible pure strategies with given probability Members genetically the same ◦ Population level: 1/3 of animals hard-wired to play D and 2/3 are hard-wired to always play H

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