Price 3.0 - Revision history

Matthijs.van.veelen: /* What would a proper statistician do? */

2010-09-20T11:08:32Z

What would a proper statistician do?

← Older revision		Revision as of 11:08, 20 September 2010
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	Try to figure out the transition probabilities for the suggested model as a function of the current state.		Try to figure out the transition probabilities for the suggested model as a function of the current state.

-	== What would a ~~proper~~ statistician do? ==	+	== What would a statistician do? ==

	Sample often for different starting points. Fit data to the model, estimate <math>\alpha</math> and <math>\beta </math>. Test if the payoff matrix is as assumed.		Sample often for different starting points. Fit data to the model, estimate <math>\alpha</math> and <math>\beta </math>. Test if the payoff matrix is as assumed.

Matthijs.van.veelen: /* What would a probability theorist do? */

2010-09-20T11:08:23Z

What would a probability theorist do?

← Older revision		Revision as of 11:08, 20 September 2010
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	== What would a probability theorist do? ==		== What would a probability theorist do? ==

-	Try to figure out the transition probabilities for the suggested model as a function of the current state~~. It seems that Nowak, Liebermann, Ohtsuki and Tarnita are a proper probability theorists, because that is what they did already~~.	+	Try to figure out the transition probabilities for the suggested model as a function of the current state.

	== What would a proper statistician do? ==		== What would a proper statistician do? ==

Matthijs.van.veelen: /* What would a proper probability theorist do? */

2010-09-20T11:08:04Z

What would a proper probability theorist do?

← Older revision		Revision as of 11:08, 20 September 2010
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	<java_applet code="tutorial3.Tutorial3.class" height="480" width="600" archive="Price.jar" />		<java_applet code="tutorial3.Tutorial3.class" height="480" width="600" archive="Price.jar" />

-	== What would a ~~proper~~ probability theorist do? ==	+	== What would a probability theorist do? ==

	Try to figure out the transition probabilities for the suggested model as a function of the current state. It seems that Nowak, Liebermann, Ohtsuki and Tarnita are a proper probability theorists, because that is what they did already.		Try to figure out the transition probabilities for the suggested model as a function of the current state. It seems that Nowak, Liebermann, Ohtsuki and Tarnita are a proper probability theorists, because that is what they did already.

Julian.garcia at 22:39, 18 September 2010

2010-09-18T22:39:52Z

← Older revision		Revision as of 22:39, 18 September 2010
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	[[Category:Price equation]]		[[Category:Price equation]]
		+	[[Price \| Go to the front-page of Price equation tutorials ]]
		+

	== A model with frequency dependence ==		== A model with frequency dependence ==

Julian.garcia: /* The Price equation for this very simple model */

2010-09-18T22:32:57Z

The Price equation for this very simple model

← Older revision		Revision as of 22:32, 18 September 2010
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	== The Price equation for this very simple model ==		== 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.	+	Here we have a nested sampling procedure; you can choose a starting point (by clicking on each individual in the parent population), 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.

		+	<java_applet code="tutorial3.Tutorial3.class" height="480" width="600" archive="Price.jar" />

	== What would a proper probability theorist do? ==		== What would a proper probability theorist do? ==

Julian.garcia: /* So how does the Price equation help here? */

2010-09-18T22:30:35Z

So how does the Price equation help here?

← Older revision		Revision as of 22:30, 18 September 2010
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	Not. Can't think of anything remotely useful about the Price equation. The extra problem here is the frequency dependence. Sometimes this is referred		Not. Can't think of anything remotely useful about the Price equation. The extra problem here is the frequency dependence. Sometimes this is referred
-	to as the problem that the Price Equation is not dynamically sufficient. That is a step in the right direction, but still inaccurate, because it only describes one of the symptoms, and not the cause. It is the ''model'' we have here that gives us, after a bit of computing, the transition probabilities for all frequencies of <math>A</math>, and thereby all properties of the dynamical system. The ''models'' in versions 1.0 and 2.0 just concerned transitions leaving from one given starting point. A ~~\textit{~~model} can therefore be described as dynamically sufficient or insufficient. Because the Price equation can only deal with one step at at time, it tends to inspire people to think of implicit assumptions that pertain to dynamically insufficient models. That, however, does not mean that the Price Equation cannot be used in combination with a dynamically sufficient model. It can; as shown in the applet above. But the question is: used for what? And the answer is: for nothing.	+	to as the problem that the Price Equation is not dynamically sufficient. That is a step in the right direction, but still inaccurate, because it only describes one of the symptoms, and not the cause. It is the ''model'' we have here that gives us, after a bit of computing, the transition probabilities for all frequencies of <math>A</math>, and thereby all properties of the dynamical system. The ''models'' in versions 1.0 and 2.0 just concerned transitions leaving from one given starting point. A ''model'' can therefore be described as dynamically sufficient or insufficient. Because the Price equation can only deal with one step at at time, it tends to inspire people to think of implicit assumptions that pertain to dynamically insufficient models. That, however, does not mean that the Price Equation cannot be used in combination with a dynamically sufficient model. It can; as shown in the applet above. But the question is: used for what? And the answer is: for nothing.

Julian.garcia: /* The Price equation for this very simple model */

2010-09-18T22:30:18Z

The Price equation for this very simple model

← Older revision		Revision as of 22:30, 18 September 2010
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	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.		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.
		+
		+
		+	== What would a proper probability theorist do? ==
		+
		+	Try to figure out the transition probabilities for the suggested model as a function of the current state. It seems that Nowak, Liebermann, Ohtsuki and Tarnita are a proper probability theorists, because that is what they did already.
		+
		+	== What would a proper statistician do? ==
		+
		+	Sample often for different starting points. Fit data to the model, estimate <math>\alpha</math> and <math>\beta </math>. Test if the payoff matrix is as assumed.
		+
		+	== So how does the Price equation help here? ==
		+
		+	Not. Can't think of anything remotely useful about the Price equation. The extra problem here is the frequency dependence. Sometimes this is referred
		+	to as the problem that the Price Equation is not dynamically sufficient. That is a step in the right direction, but still inaccurate, because it only describes one of the symptoms, and not the cause. It is the ''model'' we have here that gives us, after a bit of computing, the transition probabilities for all frequencies of <math>A</math>, and thereby all properties of the dynamical system. The ''models'' in versions 1.0 and 2.0 just concerned transitions leaving from one given starting point. A \textit{model} can therefore be described as dynamically sufficient or insufficient. Because the Price equation can only deal with one step at at time, it tends to inspire people to think of implicit assumptions that pertain to dynamically insufficient models. That, however, does not mean that the Price Equation cannot be used in combination with a dynamically sufficient model. It can; as shown in the applet above. But the question is: used for what? And the answer is: for nothing.

Julian.garcia: /* Comparing the two covariances */

2010-09-18T22:27:22Z

Comparing the two covariances

← Older revision		Revision as of 22:27, 18 September 2010
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	If we compare these two covariances, for instance for <math>\beta =0</math> and <math>\alpha =1</math>, then in the first case <math>Cov(X,Y) =-1/N</math>, while in the second case <math>Cov( X,Y) =+1/N</math>. This should not feel strange; it is a coordination game, and if everyone plays <math>B</math>, then playing <math>A</math> is a bad idea, but if everyone else plays <math>A</math> too, then playing <math>A</math> is a good idea. The model covariance is therefore not constant, and changes with the frequency of strategy <math>A</math>.		If we compare these two covariances, for instance for <math>\beta =0</math> and <math>\alpha =1</math>, then in the first case <math>Cov(X,Y) =-1/N</math>, while in the second case <math>Cov( X,Y) =+1/N</math>. This should not feel strange; it is a coordination game, and if everyone plays <math>B</math>, then playing <math>A</math> is a bad idea, but if everyone else plays <math>A</math> too, then playing <math>A</math> is a good idea. The model covariance is therefore not constant, and changes with the frequency of strategy <math>A</math>.
		+
		+	== 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.

Julian.garcia: /* A B mutant in a population of A's */

2010-09-18T22:25:32Z

A B mutant in a population of A's

← Older revision		Revision as of 22:25, 18 September 2010
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	<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>		<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>
		+
		+	== Comparing the two covariances ==
		+
		+	If we compare these two covariances, for instance for <math>\beta =0</math> and <math>\alpha =1</math>, then in the first case <math>Cov(X,Y) =-1/N</math>, while in the second case <math>Cov( X,Y) =+1/N</math>. This should not feel strange; it is a coordination game, and if everyone plays <math>B</math>, then playing <math>A</math> is a bad idea, but if everyone else plays <math>A</math> too, then playing <math>A</math> is a good idea. The model covariance is therefore not constant, and changes with the frequency of strategy <math>A</math>.

Julian.garcia: /* A B mutant in a population of A's */

2010-09-18T22:11:38Z

A B mutant in a population of A's

← Older revision		Revision as of 22:11, 18 September 2010
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	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>