Price 3.0

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A model with frequency dependence

We go back to the asexual model, but now we will take a model with frequency dependence. This will be a bit more complicated, but we can still use a basic, well known model. Suppose there are $N$ individuals. They are paired randomly, and play a game in those pairs. The game gives them payoffs according to the following payoff matrix


\begin{array}{ccc}
& A & B \\ 
A & 1 & 0 \\ 
B & 0 & 1
\end{array}

If qi = 1, then that means that individual $i$ plays strategy A, and qi = 0 means that individual i plays strategy B. Depending on the match, each individual gets a payoff πi. The next generation is drawn, as before, one individual at a time. But now the probability that i is drawn as a parent depends on the payoff, and not just on the own genotype.


\mathbb{P}\left( i\text{ is chosen}\right) =\frac{\alpha \pi _{i}+\beta }{\sum_{j=1}^{N}\left( \alpha \pi _{j}+\beta \right) }

We do the same procedure as before, but with a few different starting points.

An A mutant in a population of B's

Suppose q1 = 1 and qi = 0 for i = 2,...,N. The probability of any individual i = 2,...,N to be matched with individual 1 is 1 / (N − 1).

The probability with which an individual i with qi = 1, who is matched to an individual j with qj = 0, is chosen is:



p_{1,0,1}=\frac{\beta }{\left( N-2\right) \alpha +N\beta }

The probability with which an individual i with qi = 0, who is matched to an individual j with qj = 0, is chosen is:


p_{0,0,1}=\frac{\alpha +\beta }{\left( N-2\right) \alpha +N\beta }

The probability with which an individual i with qi = 0, who is matched to an individual j with qj = 1, is chosen is:


p_{0,1,1}=\frac{\beta }{\left( N-2\right) \alpha +N\beta }

We can also compute the covariance of these two actual random variables. Let us start by computing expectations...


\mathbb{E}\left[ X\right] =\frac{1}{N}

[show details]

For symmetry reasons...

\mathbb{E}\left[ Y\right] =1


\mathbb{E}\left[ XY\right] =\frac{\beta }{\left( N-2\right) \alpha +N\beta }

[show details]

The Covariance is thus...


Cov\left( X,Y\right) =\mathbb{E}\left[ XY\right] -\mathbb{E}\left[ X\right]  \mathbb{E}\left[ Y\right] =\frac{\beta }{\left( N-2\right) \alpha +N\beta }-\frac{1}{N}


A B mutant in a population of A's

Suppose q1 = 0 and qi = 1 for i = 2,...,N. The probability of any individual i = 2,...,N to be matched with individual 1 is 1 / (N − 1).

The probability with which an individual i with qi = 0, who is matched to an individual j with qj = 1...

p_{0,1,N-1}=\frac{\beta }{\left( N-2\right) \alpha +N\beta }

The probability with which an individual i with qi = 1, who is matched to an individual j with qj = 1...


p_{1,1,N-1}=\frac{\alpha +\beta }{\left( N-2\right) \alpha +N\beta }

The probability with which an individual i with qi = 1, who is matched to an individual j with qj = 0...


p_{1,0,N-1}=\frac{\beta }{\left( N-2\right) \alpha +N\beta }

So...

\mathbb{P}\left( Y=k|X=0\right) =\binom{N}{k}\left( p_{0,1,N-1}\right)^{k}\left( 1-p_{0,1,N-1}\right) ^{N-k}=

\mathbb{P}\left( Y=k|X=1\right) =\frac{1}{N-1}\binom{N}{k}\left(p_{1,0,N-1}\right) ^{k}\left( 1-p_{1,0,N-1}\right) ^{N-k}+\frac{N-2}{N-1}\binom{N}{k}\left( p_{1,1,N-1}\right) ^{k}\left( 1-p_{1,1,N-1}\right) ^{N-k}


This is pretty well defined, so we can also compute the covariance of these two actual random variables. Let us start with the expectations... \mathbb{E}\left[ X\right] =\frac{N-1}{N}

[show details]


For symmetry reasons...

\mathbb{E}\left[ Y\right] =1


\mathbb{E}\left[ XY\right] = \frac{\left( N-2\right) \alpha +\left( N-1\right) \beta }{\left(N-2\right) \alpha +N\beta }

[show details]

Thus....

Cov\left( X,Y\right) =\mathbb{E}\left[ XY\right] -\mathbb{E}\left[ X\right] \mathbb{E}\left[ Y\right] =\frac{\left( N-2\right) \alpha +\left( N-1\right)\beta }{\left( N-2\right) \alpha +N\beta }-\frac{N-1}{N}

Comparing the two covariances

If we compare these two covariances, for instance for β = 0 and α = 1, then in the first case Cov(X,Y) = − 1 / N, while in the second case Cov(X,Y) = + 1 / N. This should not feel strange; it is a coordination game, and if everyone plays B, then playing A is a bad idea, but if everyone else plays A too, then playing A is a good idea. The model covariance is therefore not constant, and changes with the frequency of strategy A.

The Price equation for this very simple model

Here we have a nested fruit machine; you can choose a starting point, keep it fixed, and draw a few transitions, then change your starting point, and draw a few transitions from there. You will notice that the Price equation varies with every transition, along with the likelyhood of the different transitions that depend on the initial population.