dove-hawk game

Game

Suppose two types of animals, hawks (aggressive) and doves (peaceful).

Hawk aggressively repels invader of territory, dove yields. If both doves, one leaves with probability $1/2$ . If both are hawks, fight ensures, both injured, at most one will remain in the territory and produce offspring.

		Invader
		Dove ( $y$ )	Hawk ( $1-y$ )
Defender	Dove	$4, 4$	$2, 8$
	Hawk	$8, 2$	$1, 1$

"Single species" population, as both players have same set of strategies, payoff functions satisfy $u_1(s_1,s_2) = u_2(s_1,s_2)$ , game is symmetric.

Describe a mutation, an individual in the population characterized by a particular behavior: type hawk or type dove. Type $x$ where $0 \leq x \leq 1$ (dove with probability $x$ ), hawk with probability $1-x$ .

Mutation population game:

		Population
		Dove ( $y$ )	Hawk ( $1-y$ )
Mutation	Dove	$4, 4$	$2, 8$
	Hawk	$8, 2$	$1, 1$

The column player is the population, the row player as the mutation chooses its type.

Population is implementing a fixed mixed strategy [ $y$ (Dove), $(1-y)$ (Hawk)], behaving with probability $y$ as a dove, $1-y$ as a hawk.

Expected payoff of a mutation from a random encounter is

$4y + 2(1-y)$ if it is a dove,
$8y + (1-y)$ if it is a hawk,
$x(4y + 2(1-y)) + (1-x)(8y + (1-y))$ if of type $x$ .

Equal payoffs

		Population
		Dove ( $y$ )	Hawk ( $1-y$ )
Mutation	Dove	$4, 4$	$2, 2$
	Hawk	$2, 2$	$2, 2$

In this case the dove strategy ( $x^* = 1$ ) is an evolutionarily stable strategy:

(1-\epsilon) u_1(x, x^*) + \epsilon u_1(x,x) = y \cdot 2 + (1-y) \cdot 2 < y \cdot 4 + (1-y) \cdot 2 = (1-\epsilon) u_1(x^*, x^*) + \epsilon u_1(x^*, x)

| note that $\varepsilon$ represents the proportion of mutations |

Notes

example of an evolutionary game

References

M. Maschler, E. Solan, and Shmuel Zamir, Game Theory, Cambridge University Press, 2013, p. 187-189.