# **Chapter 7.Evolutionary Game Theory**
[ToC]

Evolutionary game theory shows that the basic ideas of game theory can be applied to situations in which no individual is overtly reasoning or even making explicit decisions.

- strategy: organism's characterics/behaviors
- payoff: fitness
## **7.1 Fitness as a Result of Interaction**
- example
- Small beatles and large beatles
- payoff: the fitness that each beetle gets from a given food-related interaction
- 
- Difference from game in Chapter 6: They can't choose their strategies.

## **7.2 Evolutionarily Stable Strategies**
- For evolutionary settings, the notion similar to equilibrium is ***evolutionarily stable strategy*** – a genetically determined strategy that tends to persist once it is prevalent in a population.
- A strategy is *evolutionarily stable* if, when the whole population is using this strategy, any small group of invaders using a different strategy will eventually die off over multiple generations.
- Formal def.
- The ***fitness*** of an organism is the expected payoff it receives from an interaction with a random member of the population.
- A strategy $T$ invades a strategy $S$ at level $x$, for some small positive number $x$, if an $x$ fraction of the underlying population uses $T$ and a $1 − x$ fraction of the underlying population uses $S$.
- A strategy $S$ is *evolutionarily stable* if there is a (small) positive 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.
### Evolutionarily Stable Strategies in the First Example
- Whether **Small** is evolutionarily stable?
- Suppose that, for some small $x\gt 0$, a $1 − x$ fraction of the population uses **Small** and an $x$ fraction of the population uses **Large**
- 
- expected payoff to a **small** beetle = $5(1 − x) + 1x = 5 − 4x$
- expected payoff to a **large** beetle = $8(1 − x) + 3x = 8 − 5x$
- We get: for small enough values of x , the expected fitness of large beetles in this population exceeds the expected fitness of small beetles. Therefore, Small is not *evolutionarily stable*.
- Whether **Large** is evolutionarily stable?
- Suppose that, for some very small positive number $x$, a $1 − x$ fraction of the population uses **Large** and an $x$ fraction of the population uses **Small**.
- 
- expected payoff to a **large** beetle = $3(1 − x) + 8x = 3 + 5x$
- expected payoff to a **small** beetle = $(1 − x) + 5x = 1 + 4x$
- the expected fitness of large beetles in this population exceeds the expected fitness of small beetles, and so Large is evolutionarily stable.
- The game between small and large beetles has precisely the structure of a Prisoner’s Dilemma game
- other example in natural: heights of trees, root systems of plants(Conserve/Explore)

## **7.3 A General Description of Evolutionarily Stable Strategies**
- For a very small positive number $x$
- use $S$ with part $1-x$
- use $T$ with part $x$
- Expected payoff
- $E[$ payoff to an organism playing $S$ $]$ = $a(1 − x) + bx$
- $E[$ payoff to an organism playing $T$ $]$ = $c(1 − x) + dx$

- $S$ is ***evolutionarily stable*** if $a(1 − x) + bx > c(1 − x) + dx$ for sufficiently small $x$ > 0.
$\Rightarrow$ $S$ is ***evolutionarily stable*** when either (i) $a > c$ or (ii) $a = c$ and $b > d$.

## **7.4 Relationship between Evolutionary and Nash Equilibria**
- $(S, S)$ is a *Nash equilibrium* if $a \ge c$
- $S$ is *evolutionarily stable* when either (i) $a > c$ or (ii) $a = c$ and $b > d$.

Then we get:
- If strategy $S$ is *evolutionarily stable*, then $(S, S)$ is a *Nash equilibrium*.
- It is possible to have a game where $(S, S)$ is a *Nash equilibrium*, but $S$ is not *evolutionarily stable*.
- Example: modified Stag Hunt game
- 
- (Hunt Stag, Hunt Stag) is still a *Nash equilibrium*, but Hunt Stag is not an *evolutionarily stable* strategy.
- Relationship between *evolutionarily stable* strategies and *strict Nash equilibrium*
- the condition for $(S, S)$ to be a *strict Nash equilibrium* is $a > c$
- if $(S, S)$ is a *strict Nash equilibrium*, then $S$ is *evolutionarily stable*.
- Despite the similarities between the conclusions of *evolutionary stability* and *Nash equilibrium*, they are built on very different underlying stories.
## **7.5 Evolutionarily Stable Mixed Strategies**

### Defining Mixed Strategies in Evolutionary Game Theory

:::info
Each individual plays a particular mixed strategy - that is,
they are genetically configured to choose randomly from certain options with certain probabilities.
:::

### Evolutionarily Stable Mixed Strategies in the General Symmetric Game


:::info
**Notation**
Mixed strategy $p$ indicating that the organism plays S with probability $p$ and plays T with probability $(1-p)$.
:::
When Organism 1 uses the mixed strategy $p$ and Organism 2 uses the mixed strategy $q$

The expected payoff for Organism 1 is
$V(p,q) = pqa + p(1-q)b + (1-p)qc + (1-p)(1-q)d$

:::info
**Definition.**
In the General Symmetric Game, $p$ is an evolutionarily stable mixed strategy if
there is a (small) positive number $y$ such that when any other mixed strategy $q$
invades $p$ at any level $x < y$, the fitness of an organism playing $p$ is strictly
greater than the fitness(expected payoff) of an organism playing $q$
:::

In other words.
:::success
$p$ is a evolutionarily stable mixed strategy, if

$\exists y$ such that for any $x < y$,

$$(1-x)V(p,p) + xV(p,q) > (1-x)V(q,p) + xV(q,q)$$

holds $\forall q \neq p$, where $q$ is the mixed strategy invade $p$
:::

* If $p$ is a evolutionarily stable mixed strategy
then $V(p,p) \geq V(q,p)$, implies strategy $(p,p)$ is a mixed Nash equilibrium