Price 3.0

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== A ''B'' mutant in a population of ''A'''s  ==
== A ''B'' mutant in a population of ''A'''s  ==
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Suppose <math>q_{1}=0</math> and <math>q_{i}=1</math> for <math>i=2,...,N</math>. The probability of any individual <math>i=2,...,N</math> to be matched with individual <math>1</math> is <math>1/(N-1)</math>.
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The probability with which an individual <math>i</math> with <math>q_{i}=0</math>, who is matched to an individual <math>j</math> with <math>q_{j}=1</math>...
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<math>p_{0,1,N-1}=\frac{\beta }{\left( N-2\right) \alpha +N\beta } </math>
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The probability with which an individual <math>i</math> with <math>q_{i}=1</math>, who is matched to an individual <math>j</math> with <math>q_{j}=1</math>...
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<math>
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p_{1,1,N-1}=\frac{\alpha +\beta }{\left( N-2\right) \alpha +N\beta }
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</math>
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The probability with which an individual <math>i</math> with <math>q_{i}=1</math>, who is matched to an individual <math>j</math> with <math>q_{j}=0</math>...
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<math>
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p_{1,0,N-1}=\frac{\beta }{\left( N-2\right) \alpha +N\beta }
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</math>

Revision as of 21:41, 18 September 2010


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 }

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