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Consider a population of two altruist with coefficient of relatedness $r$. The average inclusive fitness of this population will be $w_{0} + br -c$. Like in this example, assignment of inclusive fitness I've seen in books and articles follow the rule that you should not count changes in inclusive fitness that result from the action of your neighbor. For instance, in our example, the calculation of average inclusive fitness include the increase in indirect fitness to the actor (i.e., $br$), the costs (i.e., $c$), but do not include the effects of the altruist behavior of the neighbor (i.e., $b$). That is, the average inclusive fitness is not $(w_{0} + br -c) + b$.

My question is: why is it that, in inclusive fitness models, you do not count changes in inclusive fitness that result from the action of your neighbor? Wouldn't it make the model more accurate to take into account the effects of your neighbor's behavior? In other words, in our example, why isn't $(w_{0} + br -c) + b$ a better value for the average inclusive fitness than simply $(w_{0} + br -c)$?


Note:

As @Corvus pointed out, if we include changes in inclusive fitness that result from the action of your neighbor, we will double-count benefits. Consider the example mentioned above once again of a population of two altruists with a coefficient of relatedness $r$:

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If we write $w_1 = w_2= (w_{0} + br -c) + b$, the help of 1 on 2 appears as $rb$ in $w_1$ and as $b$ in $w_2$. If we write the fitness in the right way, $w_1 = w_2= (w_{0} + br -c)$, the help of 1 on 2 is only counted once (i.e., $br$ in $w_1$). However, I cannot help but ask "Why not count the help of 1 on 2 twice?" We are counting the help of 1 on 2 twice, but each time we are assigning it to different individuals, 1 and 2.

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    $\begingroup$ This is a very good question and in short the answer is that inclusive fitness is only an approximation or what natural selection maximizes. That said, the quantity you propose is not a better explanation; it ends up double-counting benefits. I'll try to write up a more detailed description soon. $\endgroup$ – Corvus Apr 22 '15 at 21:34
  • $\begingroup$ It is good to have questions about social evolution. It is a superficially easy field that very quickly gets quite complicated when you dig into it. +1 $\endgroup$ – Remi.b Apr 24 '15 at 15:38
  • $\begingroup$ @Remi.b Initially I thought that it wouldn't take me long to have a good grasp of the basic models of social evolution. I was sooo wrong. $\endgroup$ – falsum Apr 24 '15 at 15:45
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Generalities

There are plenty of ways at looking at phenotypes that affect the fitness of the carrier and of other individuals. One of them is group selection and another one is kin selection. Those two concepts are just two different ways of looking at the same processes. Now let's consider only the kin selection way of looking at these processes.

You can either consider the impact of the phenotype of a focus individual to neighbour individuals or the impact of neighbour individuals to the focus individual. Looking at it both ways, would yield to count the effect of your phenotype on twice. It doesn't make much sense to talk about the mean inclusive fitness of a population. You can only talk about the mean fitness of a population.

The confusion

The reason for all these confusion boils down to what $B$ and $C$ really mean. One of the issue is that you confuse $B$ and $b$, $C$ and $c$. The formulation $rB>C$ is a oversimplification of the reality. Hamilton did not use this formulation at first and interpreting $b$ as the benefit for the carrier might be misleading. It is important to understand Hamilton's rule in its original formulation and it is important to understand the evolutionary game theory that underlies the evolution of social traits.

What is Hamilton's rule

One cannot study the evolution of social traits under the kin selection framework if (s)he doesn't understand the underlying game theory. $rB>C$ assumes that the game we're playing is the Prisoner's dilemna. You can learn more about game theory. You can learn more about evolutionary game theory on wiki or in this book. Here is a Khan academy video on Prisoner's dilemna

Let's assume we are playing prisoner's dilemna. The fitness of an individual which cooperate is by definition $w_o + b - c$ (note the letters are not capitalized). If you cooperate and the other don't, your fitness is $w_o - c$. If you don't cooperate and the other does cooperate, you fitness is $w_o + b$. If nobody cooperates, your fitness if $w_o$. And by definition , $b>c$. Knowing the frequency of people that cooperate in the population $y$ and knowing and your probability of cooperating is $x$. Then, the level of altruism (frequency of cooperations) increases in the population if and only if

$$R\cdot\frac{dw(x,y)}{dx}>\frac{dw(x,y)}{dy}$$, where $R$ is the coefficient of relatedness which can itself be expressed as a correlation between the variables $x$ and $y$. $w(x,y)$ is the fitness of the the individual being altruistic with probability $x$ in a population where individuals cooperate with probability $y$ and $\frac{dw(x,y)}{dx}$ is the partial derivative of the fitness function with respect to $x$. By definition, $\frac{dw(x,y)}{dx} = B$ and $\frac{dw(x,y)}{dy}=C$ (capital letters).

The mean fitness of the population therefore depends on the frequency of cooperation $y$. Let's assume for simplicity that $y=1$ (equilibrium), then the mean fitness of the population is $w_o + b - c$ (note that the letters, $b$ and $c$ are not capital letter), and the variance in fitness is null. All individuals has a fitness of $w_o + b -c$ as all individuals are altruistic and perform action that has a negative impact on the fitness $-c$ and a positive impact on their fitness $b$.

In short

$b$ and $c$ in your question, correspond to $b$ and $c$ in my answer and not to $B$ and $C$. If you're playing prisoner's dilemna, then by definition, if everybody cooperate (no variance in the population), then everybody has a fitness of $w_o + b - c$ and the mean fitness is $w_0 + b - c$ (as there is no variance). Now the question of whether everybody ends up cooperating depends on $B$, $C$ and $R$.

In your edit, you write $rb$ and $c$ but you get confused about the meaning of $b$ and $c$. If you are playing prisoner's dilemna, when you cooperate with someone that is also cooperating, you inclusive fitness is $b-c + r(b-c)$ and if you cooperate with someone that is not cooperating, your inclusive fitness is $b-c + r(b)$.

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  • $\begingroup$ What is the quantity $w(x,y)$? $\endgroup$ – falsum Apr 24 '15 at 15:30
  • $\begingroup$ Sorry, it was unclear. See edit. $w(x,y)$ is the fitness of an individual cooperating with probability $x$ in a population that cooperate with probability $y$ $\endgroup$ – Remi.b Apr 24 '15 at 15:33
  • $\begingroup$ You correct me if I am wrong, but your point seems to be that if we formulate Hamilton's rule using PD, my question doesn't arise. The reason seems to be that the derivation of Hamilton's rule via PD only deals with direct fitness. That is, instead of measuring the degree of genetic relatedness, $R$ measures positive assortment (i.e., how likely an altruist will interact with another altruist). $\endgroup$ – falsum Apr 24 '15 at 17:06
  • $\begingroup$ See the "in short" paragraph that I added. I am not sure I am answering your question. You will let me know. $\endgroup$ – Remi.b Apr 24 '15 at 22:05
  • $\begingroup$ I think I get what you're saying. This is an interesting way of deriving Hamliton's rule. If I understood it correctly, it looks for the conditions in which $dw/dx > 0$. Once we use the chain rule, we get the version of Hamilton's rule you wrote where $R = \beta_{yx} = dy/dx$. $\endgroup$ – falsum Apr 27 '15 at 20:01

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