Evolutionary Stable Strategy

Unlike previous optimization examples, in this case we are considering not just an individuals choice, but also the effect of the rest of the members in the population. This model is from Hamilton and May (1973).

We have an annual plant (dies each year after producing seeds) that lives in small sites that are limited, and only one plant can occupy one site to full growth - one seed will survive and outcompete the others. We will consider this factor to be random (no seeds have an advantage in germination rates). Also each site is the same as far as conditions.

Dispersal is a risky process - many dispersed seeds will not land on appropriate sites.

Assume each plant produces a large number of seeds $F$ .

$s$ is the probability that one seed survives dispersal and lands in one of the $N$ suitable sites. We assume $N$ is also very large. Since there is a limit to the number of sites, each site is occupied each year.

$d$ is the fraction of the seeds that are sent for dispersal, instead of just dropping them on the same site. A population with low $d$ is following a "resident strategy"

Imagine a new mutant with a high $d$ appears in a resident strategy population with mostly low $d$ . (new mutation higher dispersal rate = $d_{m u t}$ )

Optimal reply

What should $d_{m u t}$ be to have the highest number of surviving seeds in response to the population rate $d$ ?

The plant has $(1 - d_{m u t}) F$ seeds that do not disperse but stay where the mother plant is.
Other plants have also sent seeds that have landed in this mutant's site. The total population has:
- $N$ plants
- $N F$ seeds
- $N F d$ dispersed seeds
- $N F d s$ dispersed seeds arriving safely at all sites
- $\frac{N F d s}{N} = F d s$ seeds arrive a specific site
So, at the mutant's site, a total of $(1 - d_{m u t}) F + F d s$ seeds arrive before competition starts. Since we are assuming competition is random, the probability of one of the mutant's own seeds maturing out of the total seeds landing at the site is:

\frac{(1 - d_{m u t}) F}{(1 - d_{m u t}) F + F d s} = \frac{1 - d_{m u t}}{1 - d_{m u t} + s d}

This tells us how many of the mutants seeds survive at its own site. But it also has seeds that it has sent to other sites. The probability of one of its seeds landing on another suitable site, outcompeting the others and surviving to adulthood:

\frac{1}{(1 - d) F + F d s + 1}

Since $F$ is very large we can ignore the +1 in the denominator (the one extra seed added by the mutant won't affect the probability much, its going to be quite low).

The mutant will create $F d_{m u t} s$ successful seeds. So, just like before, the number of sites that it will successfully colonize is:$$sd_{mut}\over 1-d+sd$$ When we combine this with our equation earlier from 3, we can get its total Fitness Function:

W (d_{m u t}, d) = \frac{1 - d_{m u t}}{1 - d_{m u t} + s d} + \frac{s d_{m u t}}{1 - d + s d}

Since this depends on both the mutant and natural population's dispersal method we write it as a function with two parameters.

If the normal dispersal rate $d$ is known, we can take the partial derivative to find the optimal value of $d_{m u t}$ (treating $d$ as a constant). I'm skipping the calculations, but if we do this we can get the first derivative:

\frac{\partial W (d_{m u t}, d)}{\partial d_{m u t}} = \frac{- s d}{(1 - d_{m u t} + s d)^{2}} + \frac{s}{1 - d + s d}

At the optimal value, this derivative is equal to zero. This is solvable but its really ugly. This tells us the best dispersal value for a mutant in a population with the strategy $d$ -the best "reply" to this value.

What if the optimal reply is the same as the existing value?

In this case, we call this an "evolutionary stable strategy" - the population should reach this value after a period of selection. To find this value $d_{E S S}$ , we use our partial derivative from earlier, except we consider the case when $d_{m u t} = d$ . Again, we evaluate for when the derivative is zero to find the optimal value:

0 = \frac{- s d}{(1 - d + s d)^{2}} + \frac{s}{1 - d + s d}

0 = - d + 1 - d + s d

d_{E S S} = \frac{1}{2 - s}

So simple!

Conclusions

If $s = 1$ then $d_{E S S} = 1$ . The evolutionary stable strategy is to send 100% of your seeds away. It doesn't make sense to evolve to keep any seeds at home, since this is only one site and no seeds will die during dispersal. Someone else can have your current site.
When $s \to 0$ , about half the seeds will be distributed ( $d_{E S S} \to 0.5$ ). These seeds will surely die, but it's better to take the risk since only one will survive in your current site - this is not enough to pass on your genes.