Price 3.0
From Evolution and Games
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| - | == | + | [[Category:Price equation]] |
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| + | == A model with frequency dependence == | ||
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| + | 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 | ||
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| + | <math> | ||
| + | \begin{array}{ccc} | ||
| + | & A & B \\ | ||
| + | A & 1 & 0 \\ | ||
| + | B & 0 & 1 | ||
| + | \end{array} | ||
| + | </math> | ||
| + | |||
| + | If <math>q_{i}=1</math>, then that means that individual $i$ plays strategy <math>A</math>, and <math>q_{i}=0</math> means that individual <math>i</math> plays strategy <math>B</math>. Depending on the match, each individual gets a payoff <math>\pi _{i}</math>. The next generation | ||
| + | is drawn, as before, one individual at a time. But now the probability that <math>i</math> is drawn as a parent depends on the payoff, and not just on the own genotype. | ||
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| + | <math> | ||
| + | \mathbb{P}\left( i\text{ is chosen}\right) =\frac{\alpha \pi _{i}+\beta }{\sum_{j=1}^{N}\left( \alpha \pi _{j}+\beta \right) } | ||
| + | </math> | ||
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| + | We do the same procedure as before, but with a few different starting points. | ||
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| + | == An ''A'' mutant in a population of ''B'''s == | ||
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| + | Suppose <math>q_{1}=1</math> and <math>q_{i}=0</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}=1</math>, who is matched to an individual <math>j</math> with <math>q_{j}=0</math>, is chosen is: | ||
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| + | <math> | ||
| + | p_{1,0,1}=\frac{\beta }{\left( N-2\right) \alpha +N\beta } | ||
| + | </math> | ||
| + | |||
| + | 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}=0</math>, is chosen is: | ||
| + | |||
| + | <math> | ||
| + | p_{0,0,1}=\frac{\alpha +\beta }{\left( N-2\right) \alpha +N\beta } | ||
| + | </math> | ||
| + | |||
| + | 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>, is chosen is: | ||
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| + | <math> | ||
| + | p_{0,1,1}=\frac{\beta }{\left( N-2\right) \alpha +N\beta } | ||
| + | </math> | ||
| + | |||
| + | We can also compute the covariance of these two actual random variables. Let us start by computing expectations... | ||
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| + | <math> | ||
| + | \mathbb{E}\left[ XY\right] =\frac{\beta }{\left( N-2\right) \alpha +N\beta } | ||
| + | </math> | ||
| + | <toggledisplay> | ||
| + | <math>\mathbb{E}\left[ XY\right] =\sum_{x=0,1}\sum_{y=0}^{N}xy\mathbb{P}\left(X=x,Y=y\right) \text{ (because }xy=0\text{ if either }x=0\text{ or }y=0\text{)} </math> | ||
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| + | <math>=\sum_{y=0}^{N}y\mathbb{P}\left( X=1,Y=y\right) = </math> | ||
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| + | <math>=\sum_{k=0}^{N}y\mathbb{P}\left( X=1\right) \mathbb{P}\left(Y=k|X=1\right) = </math> | ||
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| + | <math>=\frac{1}{N}\sum_{k=0}^{N}k\binom{N}{k}\left( p_{1,0,1}\right) ^{k}\left(1-p_{1,0,1}\right) ^{N-k} </math> | ||
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| + | <math>=p_{1,0,1}= </math> | ||
| + | |||
| + | <math>=\frac{\beta }{\left( N-2\right) \alpha +N\beta }</math> | ||
| + | |||
| + | </toggledisplay> | ||
Revision as of 21:06, 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
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.
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:
The probability with which an individual i with qi = 0, who is matched to an individual j with qj = 0, is chosen is:
The probability with which an individual i with qi = 0, who is matched to an individual j with qj = 1, is chosen is:
We can also compute the covariance of these two actual random variables. Let us start by computing expectations...
![\mathbb{E}\left[ XY\right] =\frac{\beta }{\left( N-2\right) \alpha +N\beta }](/wiki/images/math/b/c/1/bc1ba1eb3290d07e011aad2aaf8c6036.png)
![\mathbb{E}\left[ XY\right] =\sum_{x=0,1}\sum_{y=0}^{N}xy\mathbb{P}\left(X=x,Y=y\right) \text{ (because }xy=0\text{ if either }x=0\text{ or }y=0\text{)}](/wiki/images/math/f/1/4/f146e34f77cefe890894be133473bf2a.png)
