Price 3.0

From Evolution and Games

(Difference between revisions)
Jump to: navigation, search
(→A B mutant in a population of A's)
(→A B mutant in a population of A's)
Line 115: Line 115:
This is pretty well defined, so we can also compute the covariance of these two actual random variables. Let us start with the expectations...
This is pretty well defined, so we can also compute the covariance of these two actual random variables. Let us start with the expectations...
 +
<math>\mathbb{E}\left[ X\right] =\frac{N-1}{N}</math>
 +
 +
<toggledisplay>
 +
<math>\mathbb{E}\left[ X\right] =\sum_{x=0,1}x\mathbb{P}\left( X=x\right) = </math>
 +
 +
<math>=\mathbb{P}\left( X=1\right) = \frac{N-1}{N}</math>
 +
</toggledisplay>
 +
 +
 +
For symmetry reasons...
 +
 +
<math>\mathbb{E}\left[ Y\right] =1 </math>
 +
 +
 +
<math>\mathbb{E}\left[ XY\right] = \frac{\left( N-2\right) \alpha +\left( N-1\right) \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>
 +
 +
<math>=\sum_{y=0}^{N}y\mathbb{P}\left( X=1,Y=y\right) = </math>
 +
 +
<math>=\sum_{k=0}^{N}y\mathbb{P}\left( X=1\right) \mathbb{P}\left(Y=k|X=1\right) = </math>
 +
 +
<math>=\frac{N-1}{N}\sum_{k=0}^{N}\left[ \frac{1}{N-1}k\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}k\binom{N}{k}\left( p_{1,1,N-1}\right) ^{k}\left( 1-p_{1,1,N-1}\right) ^{N-k}\right] </math>
 +
 +
<math>=p_{1,0,N-1}+\left( N-2\right) p_{1,1,N-1} </math>
 +
 +
<math>=\frac{\beta }{\left( N-2\right) \alpha +N\beta }+\left( N-2\right) \frac{\alpha +\beta }{\left( N-2\right) \alpha +N\beta }= </math>
 +
 +
<math>=\frac{\left( N-2\right) \alpha +\left( N-1\right) \beta }{\left(N-2\right) \alpha +N\beta }</math>
 +
</toggledisplay>
 +
 +
Thus....
 +
 +
<math>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} </math>

Revision as of 22:11, 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 }

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}

Retrieved from "http://evolutionandgames.nl/wiki/index.php?title=Price_3.0"