Yale 11 and 12 Evolutionary Stability: cooperation, mutation, and equilibrium.

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Presentation transcript:

Yale 11 and 12 Evolutionary Stability: cooperation, mutation, and equilibrium

Evolutionary Process Simplified model Symmetric two player games Random Matching – keep track of average payoffs Relative successful strategies grow. Assume no gene redistribution Asexual reproduction Large population. entire population is all playing the same (possibly mixed) strategy, S, (hard wired). suppose there is a mutation, S’. Will it thrive or will it die out? If all possible mutations die out, the original population is evolutionary stable. CD C2,20,3 D3,01,1

Is cooperation evolutionary stable? Randomly matching yields an expected utility of 2. As we deviate from the “C” strategy, what happens? Play C is not evolutionary stable. Invader D doesn’t die out. Is any other pure strategy evolutionary stable? (any mutation from D dies out, D is ES) If (s,s) is not a NE, then s is NOT evolutionary stable. Evolutionary solutions CAN be bad. If a strategy is strictly dominated, then it is not ES

Is “c” evolutionary stable? abc a2,20,0 b 1,1 c0,01,10,0 What can invade? What percentage of population can pick b and be happy? In pure strategies, must be on diagonal to be evolutionary stable

Can we find a NE which is not evolutionary stable? b,b is NE, but not ES Defn: : s* is Evolutionary stable if for all small mutants s’ (1-e)U 1 (s*,s*) + e U 1 (s*,s’) > (1-e)U 1 (s’,s*) + e U 1 (s’,s’) what s* expects > what invader expects ab a1,10,0 b

Defn 2: In a symmetric two player game, s* is ES in pure strategies if (s*,s*) is a symmetric NE, ie U 2 (s*,s*) > U 2 (s*,s’) for all s’ (strict ES) (s*,s*) is a symmetric NE if U 2 (s*,s*) = U 2 (s*,s’) then U 2 (s*,s’) > U 2 (s’, s’) (non-strict) ES of the most important ideas in sociology and one of the most important ideas in economics.

Is (a,a) ES? Are both a,a and b,b ES? Are strict, but both are not efficient. ab a1,1 b 0,0 ab a2,20,0 b 1,1

Chicken variant straightswerve Straight0,02,1 swerve1,20,0 No symmetric NE. Look at mixed NE. (2/3, 1/3) and (2/3, 1/3) are NE Monomorphic – only one type out there. Polymorphic – two types out there. Can have a stable polymorphic population.

2/31/3 Chickenstraightswerve 2/3Straight0,02,1 1/3swerve1,20,0 Allowing that original strategy might be mixed and a mutation might be mixed. 2/3, 1/3) and (2/3, 1/3) are NE Does (2/3, 1/3) satisfy definition? How to deviations do against it? Well, we picked the ratio so we would be indifferent to change – so it does as well. Generally, a mixed NE will never be strict. Check rule (2).

Does (2/3, 1/3) satisfy definition of ES? (p*,p*) is a symmetric NE if U(p*,p*) = U(p*,p’) then NEED u(p*,p’) > u(p’, p’). Have to check this for all possible mutations. Here is a heuristic argument. Suppose there is a mutation that is more aggressive: It will die out. It does very badly against itself – so u(p*,p’) > u(p’, p’). Suppose there is a mutation that is more passive? It will die out. It does very badly against itself – so u(p*,p’) > u(p’, p’). So, we have a mixed evolutionary stable state.

1 < v < 2 Biologist version of rock/paper/scissors Only hope is (1/3,1/3,1/3) but it is not ESS. There is NO ESS – so lots of cycling around. abc a1,1v,00,v b 1,1v,0 c 0,v1,1

1 < v < 2 Check (1/3,1/3,1/3) Mixed NE will always be a weak NE so check u(p*,p’) > u(p’,p’) (1+v)/3 < 1 So no ESS abc a1,1v,00,v b 1,1v,0 c 0,v1,1