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Tabela de conteúdos
During the first four billion years or life on Earth, the primary means of information transfer was genetic.
Game Theory and Animal Behaviour
S. E. Riechert
- Games spiders play: behavioural variability in territorial disputes. Behavioural Ecology and Sociobiology 3:135-162. 1979.
- Games spiders play II: Resource assessment strategies. Behavioural Ecology and Sociobiology 6:121-128. 1979.
- The consequences of being territorial: spiders, a case of study. American Naturalist 117:871-892. 1981.
- Spider interaction and strategies: communication vs. coercion. 281-315. In Spider communication, Mechanisms and ecological significance. Princeton University Press. 1982.
Adaptive dynamics in space
| Lion, Sebastien |
Over the last twenty years, the role of spatial self-structuring as a template for evolution has drawn much attention. Spatial structure can be an important component of the eco-evolutionary feedback loop: the evolution of a trait (e.g. migration) can shape the local structure of the population, which in turn creates new selective pressures on the evolving trait. As a consequence, the evolution of spatially structured populations often displays very different features from the evolution of well-mixed populations.
ESS for Chicken and Nash Equilibrium
I is stable iff:
E(I,I) > E(J,I)
or
E(I,I) = E(J,I) and E(I,J) > E(J,J)
| S0 | S1 | |
|---|---|---|
| S0 | 0 | -1 |
| S1 | 1 | -10 |
S0 is not ESS against S1:
E(I,I) > E(J,I) E(S0,S0) > E(S1,S0) 0 > 1 (FALSE)
E(I,I) = E(J,I) and E(I,J) > E(J,J) E(S0,S0) = E(S1,S0) and E(S0,S1) > E(S1,S1) 0 = 1 (FALSE)
S1 is not ESS against S0:
E(I,I) > E(J,I) E(S1,S1) > E(S0,S1) -10 > -1 (FALSE)
E(I,I) = E(J,I) and E(I,J) > E(J,J) E(S1,S1) = E(S0,S1) and E(S1,S0) > E(S0,S0) -10 = -1 (FALSE)
Adding Neq (S0.1) to the game, we have:
| S0 | S0.1 | S1 | |
|---|---|---|---|
| S0 | 0 | -0.1 | -1 |
| S0.1 | 0.1 | -0.1 | -1.9 |
| S1 | 1 | -0.1 | -10 |
S0.1 is ESS against S0:
E(I,I) > E(J,I) E(S0.1,S0.1) > E(S0,S0.1) -0.1 > -0.1 (FALSE)
E(I,I) = E(J,I) and E(I,J) > E(J,J) E(S0.1,S0.1) = E(S0,S0.1) and E(S0.1,S0) > E(S0,S0) -0.1 = -0.1 and +0.1 > 0 (TRUE)
S0.1 is ESS against S1:
E(I,I) > E(J,I) E(S0.1,S0.1) > E(S1,S0.1) -0.1 > -0.1 (FALSE)
E(I,I) = E(J,I) and E(I,J) > E(J,J) E(S0.1,S0.1) = E(S1,S0.1) and E(S0.1,S1) > E(S1,S1) -0.1 = -0.1 and -1.9 > -10 (TRUE)