Lecture of April 5, 2019: Evolutionary Stable Strategies
Chiara Mocenni
Course on Game Theory
Evolutionary Stable Strategies (ESS)
A strategy robust with respect to evolutive pressure from other strategies is said Evolutionary Stable Strategy (ESS) Let’s consider a population of individuals that have been preprogrammed to play a given strategy (predominant strategy in the population), pure or mixed, in the framework of repeated games between random couples of players Suppose to include in the population a little number of individuals that are playing another pure or mixed strategy The predominant strategy is said evolutionary stable if for all mutant strategies there exist an invasion barrier such that the number of individuals that are playing the mutant strategy is below such barrier, in the sense that the predominant strategy is stil providing a higher payoff
This approach considers symmetric interactions between all possible couples of individuals of a large population
The concept of evolutionary stable strategy is based on the relationship that exists between the payoffs of a game and the diffusion of a given strategy in a population: game payoffs represent the gain in terms of biological fitness (reproducing capacity) resulting from all the interactions between the members of the population
ESS strategies generalize biological evolution in a darwinian sense, i.e. the idea that individuals choosing fittest strategies have a higher probability to survive. Moreover, the fitness itself depends on the strategies chosen by the other members of the population
Similary to the Nash equilibrium, evolutionary stability does not explain how the population has reached such a result The idea of ESS can be used to explain the robustness of human behavior in several situations, such as in economic and social contexts
Indeed, if a small number of individuals attempts to use a strategy different from the prevalent, and earns a lower payoff, then these individuals will not be induced to change their previous behavior
From this point of view, the ESS strategy is said norm
Let ˆx be a predominant strategy and x a diverging strategy that is played with probability
Then, the mixed strategy will be x + (1 − )ˆx
The strategy ˆx is said evolutionary stable if for each diverging strategy x we have that the inequality
xTA(x + (1 − )ˆx) < ˆxTA(x + (1 − )ˆx) is verified for all > 0 smaller than a value ¯(x) > 0, said invasion barrier
Rewrite the inequality as
(1 − )(ˆxTAˆx − xTAˆx) + (ˆxTAx − xTAx) > 0.
If tends to 0, we have that the strategy ˆx is evolutionary stable if the following conditions are satisfied:
a Equilibrium
ˆxTAˆx ≥ xTAˆx, for all x ∈ S ; b Stability
if x 6= ˆx e ˆxTAˆx = xTAˆx, then ˆxTAx > xTAx.
The strategy ˆx is a best response to itself, but this fact alone is not enough to avoid invasion by other strategies, indeed when the equality (and not the inequality) holds there exist the alternative best response x
Then we have to verify that ˆx provides a higher payoff if played against x than playing x itself.
If a strategy is a strict Nash equilibrium, then it is ESS If a strategy is ESS, then it is a Nash equilibrium
Theorem
A strategy ˆx ∈ S is ESS iff
ˆxTAy > yTAy for all y 6= ˆx in some interval of S .
It follows that if a strategy ESS exists in the simplex, then it is unique (see the following slide).
We define support C (x ) of one strategy x the set of pure strategies to which x assigns non null probability.
Let ˆx be a ESS strategy internal to the simplex. Suppose now that it exists another ESS strategy y 6= ˆx such that C (y ) ⊆ C (ˆx ). Then yTAˆx = ˆxTAˆx .
From the condition on stability of ESS strategies it follows that yTAy < ˆxTAy , then y can’t be a Nash equilibrium and then can’t be ESS.
Thus, if multiple ESS strategies exist, then they must belong to the borders of the simplex. Moreover, if a ESS strategy exists internal to the simplex, then it is unique.
Exercise. Verify that in the Hawk-Doves game, described by the payoff matrix
A =
G −C
2 G
0 G2
(1) the only ESS strategy is ˆx = (G /C ,C −GC ), with C > G .
To verify this, we calculate ˆxTAy , ˆxTAˆx , yTAˆx and yTAy with y = (y , 1 − y ) generic strategy.
ˆ
xTAy = G2(1 +GC − 2y ) ˆ
xTAˆx = G2(1 −GC) yTAˆx = G2(1 −GC) yTAy =G2(1 −GCy2)
ˆ
x is Nash iff ˆxTAˆx ≥ yTAˆx , i.e. iff G2(1 −GC) ≥ G2(1 −GC).
This is true in the nonstrict case.
Furthermore, ˆx is ESS iff ˆxTAy > yTAy , ∀y 6= ˆx iff
G
2(1 +GC − 2y ) −G2(1 −GCy2) > 0, ∀y 6= ˆx iff
G
Cy2− 2y +GC > 0∀y 6= ˆx , which is verified for all y 6= G2. Then, the mixed strategy ˆx is a Nash equilibrium evolutionary stable (ESS) for the Hawk-Dove game. Notice that ˆx is the unique ESS strategy and the unique Nash equilibrium of the game.
Exercise. Verify that in the game described by the payoff matrix
A =
0 6 −4
−3 0 5
−1 3 0
(2)
the mixed strategy x = (1/3, 1/3, 1/3) is not ESS. Sol. The pure strategy e1 is ESS.
Exercise. Verify the presence of pure strategies and Nash equilibria in the game described by the payoff matrix:
A =
1 0 0 0 1 0 0 0 1
. (3)
Sol. In the game there are 3 ESS (e1, e2, e3) and 1 mixed NE.
Exercise. Verify the presence of pure strategies and Nash equilibria in the game described by the payoff matrix:
A =
0 1 −1
−1 0 1
1 −1 0
. (4)
Sol. In the game there are no ESS strategies and 1 mixed NE (1/3, 1/3, 1/3).
NE and ESS pure in games with 2 strategies
In a game with 2 strategies, described by the payoff matrix A =
a b c d
, the following properties hold:
Strategy s1 is a strict NE if a > c;
Strategy s1 is a NE if a ≥ c;
Strategy s2 is a strict NE if d > b;
Strategy s2 is a NE if d ≥ b.
The pure strategy s1 is said evolutionary stable if a > c, or a = c and b > d .
Under these conditions the evolutive selection favours s2 on s1. In the general case of n pure strategies , let π(ei, ej) be the payoff of strategy ei with respect to ej.
i A strategy ek is a strict NE if π(ek, ek) > π(ei, ek), ∀i 6= k;
ii A strategy ek is a NE if π(ek, ek) ≥ π(ei, ek), ∀i 6= k.
iii A strategy ek is ESS if π(ek, ek) > π(ei, ek), ∀i 6= k, or π(ek, ek) = π(ei, ek) e π(ek, ei) > π(ei, ei) ∀i 6= k.
Mixed ESS strategies
Mixed NE are never strict. Then, for this kind of strategies it will be always necessary to verify the condition of stability.
Classification of ESS in symmetric games with 2 strategies
In categories I and IV, where the coefficients a1 and a2 have opposite signs (e.g. the prisoner’s dilemma), one of the 2 strategies dominates the other, then a unique strict NE exists, then it is also ESS
In category II, where a1 > 0 and a2> 0, the 2 pure strategies are strict NE and then they are also ESS, then the internal mixed NE can’t be ESS
In category III (e.g. Hawk-Dove game), there is only one internal mixed NE. Then, in this case we have to verify the condition of stability to know if it is ESS. See previous exercise.
