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?

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.


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

Excerpt from Hamilton's paper

  • 2
    $\begingroup$ Hamilton's papers are always sooo complicated! $\endgroup$
    – Remi.b
    Jan 8, 2015 at 17:01
  • $\begingroup$ Agreed! I think his 1964 papers are even worse. I suppose the above equation is a linear regression of $q_j$ on $q_i$. But I still don't see how Hamilton arrived at the equation above. $\endgroup$
    – falsum
    Jan 8, 2015 at 17:53
  • $\begingroup$ Sad to see that research-level question often don't receive as much attention as basic questions. I hope you'll get some more upvotes and will get someone less lazy than me that will be ready to dive into Hamilton's article. $\endgroup$
    – Remi.b
    Jan 20, 2015 at 1:08
  • $\begingroup$ I can't access the paper so can you provide more detail about the model. Also, $E(q_j|q_i)$ conditional expected value? $\endgroup$
    – dustin
    Jan 20, 2015 at 6:39
  • $\begingroup$ @dustin: Hamilton does not use conditional probabilities. I'm the one that wrote that way when I was explaining my question. I changed this now. $\endgroup$
    – falsum
    Jan 20, 2015 at 14:29

2 Answers 2


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.

  • $\begingroup$ you say that $(b_{i,j}-b_i) \in [0,1-b_i]$. My impression is that $(b_{i,j}-b_i) \in [-b_i,1-b_i]$. $(b_{i,j}-b_i)$ will be negative when the individual $j$ is less related to individual $i$ than average. $\endgroup$
    – falsum
    Apr 17, 2015 at 14:34
  • $\begingroup$ I suppose the lower bound of $\frac{(b_{ij} - b_{i})}{(1-b_{i})}$ is $- \frac{b_{i}}{(1-b_{i})}$ (since $0\leq b_{ij} \leq 1$). However, this doesn't seem right because, otherwise, $\frac{(b_{ij} - b_{i})}{(1-b_{i})}$ could assume values in the interval $[0, -\infty)$. $\endgroup$
    – falsum
    Apr 17, 2015 at 22:03

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.

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.

  • $\begingroup$ You're right. Hamilton does not say that this equation is a linear regression. This is something that I assumed (I changed my question to reflect that). My guess is that Hamilton's above equation expresses $E(q_j\mid q_i) = E(q_j) + \beta (q_i - E(q_i))$, where $\beta$ is the regression coefficient. $\endgroup$
    – falsum
    Jan 20, 2015 at 16:35
  • $\begingroup$ @falsum what do you mean by $E[q_j\mid q_i]$. I know it has a conditional expected value. Is that what you mean? $\endgroup$
    – dustin
    Jan 20, 2015 at 16:42
  • $\begingroup$ Yes. This is what I mean. My guess is that Hamilton is providing an equation that is trying to predict the gene frequency of a random actor (i.e., $q_j$) from a given $q_i$. $\endgroup$
    – falsum
    Jan 20, 2015 at 16:46
  • $\begingroup$ @falsum If we use that definition, there is a chance it will be a nonlinear predictor not linear. $\endgroup$
    – dustin
    Jan 20, 2015 at 16:56

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