Hamilton's inclusive fitness approach

Question

The underlying intuition of Hamilton's model of inclusive fitness is that we should study social behaviors from the point of view of actors -- rather than the recipients. To build his model, Hamilton expresses the genotype of the actor $j$ in terms of the genotype of the recipient of the behavior, $i$. The genotype of $j$ is decomposed in two parts, ``genes which are copies by direct replication of genes in $i$; the other part consists of non-replica genes'' (Hamilton 1970, p. 1219). Hamilton (1970) further defines $q_{i}$ as the gene frequency of the replica part, $b_{ij}$ represents the replica fraction, and $q$ is the average gene frequency in the population. From these definitions Hamilton (1970) jumps to the equality: \begin{equation} E (q_{j}) = \frac{1}{1-b_{i}}\left\{ (b_{ij} - b_{i})q_{i} + (1-b_{ij})q\right\} \end{equation} where \begin{align*} b_{i} = \frac{1}{n}\sum_{j}b_{ij} \end{align*}

How did Hamilton derive the above equation?

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Here is what I think Hamilton is doing. My impression is that the above equation expresses $E(q_j | q_i)$ as a linear regression on $q_i$. In other words, I think the above equation is equivalent to:

$E(q_j | q_i) = E(q_j) + \beta (q_i - E(q_i))$

$E(q_j | q_i) = q + \beta (q_i - q)$

In fact, this equation is equivalent to Hamilton's equation if the regression coefficient is:

$\beta = (b_{ij} - b_i) / (1 - b_i)$

However, I haven't been able to derive this regression coefficient. Given that $\beta = Cov (q_j, q_i)/Var (q_i)$, I suspect that the way to go is to rewrite $q_j$ and $q_i$ in terms of $b_{ij}$ and $b_i$ and calculate the regression coefficient.

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Reference:

Hamilton 1970 "Selfish and Spiteful Behaviour in an Evolutionary Model" http://www.nature.com/nature/journal/v228/n5277/abs/2281218a0.html

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Answer 1

It is not a regression (not at this stage of the paper, a regression will be done latter)

The only complicated thing to understand is $b_i$, which is the 'base relatedness', ie how $i$ is related to a random individual (to be compared to how related it is to individuals with whom it interacts).

To simplify let's first consider the situation where $b_i = 0$:

$E(qj) = b_{i,j} q_i + (1-b_{i,j}) q$ is just the translation of 'the gene frequency of the replica part is q_i' and 'the gene frequency of the non replica part is $q$'; because $b_{i,j}$ is the fraction of the replica part, ie the chances that our locus of interest belongs to the replica part of individual $i$ in individual $j$.

Now let's re-introduce $b_i$. The idea is to compare relatedness of the two individuals $i$ and $j$ to the average relatedness of $i$ with a randomly picked individual in the population (this random relatedness is exactly $bi$). This is important because $q$ already accounts for this 'random relatedness'.

So instead of giving probability $b_{i,j}$ to $q_i$, we give it probability $b_{i,j}-b_i$, which is the chance that the allele of interest is present because of the replica fraction being higher than random. And because now the quantity varies between 0 and $1-b_i$ we normalize it by $1-b_i$

The underlying intuition of Hamilton's model of inclusive fitness is that we should study social behaviors from the point of view of actors -- rather than the recipients

Not exactly, it is saying that we should study social behaviors from the point of view of the alleles causing it, that may be shared between the actors and the recipients. But this paper is not the paper that introduces inclusive fitness, quite the opposite it is the paper that tries to reconciliate kin selection with Price equation.

Answer 2

From the limited information, I can provide the following but I am not sure if this is what you are looking for. Also, I still don't see the statement where the author concludes we get the Linear regression model $E(q_i) = Aq_i + C$ which is odd notation since it says the expected value is a linear regression. In fact, if it is a linear regression, it should say $q_j = Aq_i + C$. $$ A = \frac{\text{cov}(q_i,q_j)}{\text{var}(q_i)}\quad\text{and}\quad C = E[q_j] - \frac{\text{cov}(q_i,q_j)}{\text{var}(q_i)}E[q_i]. $$ Now, we can write the variance and covariance as \begin{align} \text{var}(q_i) &= E[q_i^2]-E^2[q_i]\\ \text{cov}(q_i,q_j) &= E[q_iq_j]-E[q_i]E[q_j] \end{align} where the expected value of a discrete random variable $X$ is calculated as $$ E[X] = \sum_{i=1}^Nx_ip_X[x_i] $$ where $p_X[x_i]$ is probability of $x_i$ and $N$ can be countable infinite.

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The mean of conditional PDFs comes up in optimal prediction where the minimum mean square error is $E_{Y\mid X}[Y\mid x]$. This optimal prediction covers linear and nonlinear. For the standard Gaussian PDF, the optimal prediction will be linear since $E_{Y\mid X}[Y\mid x]=\rho x$ where $\rho$ is the correlation coefficient. I am going to short hand $E_{Y\mid X}[Y\mid x]$ to $E[Y\mid x]$ $$ E[Y\mid x] = \mu_Y + \frac{\rho\sigma_Y}{\sigma_X}(x-\mu_X) $$ where $\mu_i$ $i=X,Y$ is the mean and $\sigma_i$ is the standard deviation. If $X$ and $Y$ are not Gaussian, then the model can be nonlinear. Are there examples of non Gaussian be linear probably.