Even more gory details¹¹

Let $x_{mn,i}$ be an indicator variable such that $x_{mn,i} = 1$ if allele $m$ from individual $n$ is of type $i$ and is 0 otherwise. Clearly, the sample frequency $\hat p_i = \frac{1}{2N}\sum_{m=1}^2\sum_{n=1}^Nx_{mn,i}$, and $E(\hat p_i) = p_i$, $i=1\dots A$. Assuming that alleles are sampled independently from the population

$$\begin{eqnarray*} E(x^2_{mn,i}) &=& p_i \\ E(x_{mn,i}x_{mn',i}) = E(x_{mn,i}x_{... ...2 + \sigma_{x_{mn,i}x_{m'n',i}} \\ &=& p_i^2 + p_i(1-p_i)\theta \end{eqnarray*}$$

where $\sigma_{x_{mn,i}x_{m'n',i}}$ is the intraclass covariance for the indicator variables and

$$\theta = \frac{\sigma^2_{p_i}}{p_i(1-p_i)}$$ (1)

is the scaled among population variance in allele frequency in the populations from which this population was sampled. Using (1) we find after some algebra

$$\sigma^2_{\hat p_i} = p_i(1-p_i)\theta + \frac{p_i(1-p_i)(1-\theta)}{2N} \quad .$$

A natural estimate for $\theta$ emerges using the method of moments when an analysis of variance is applied to indicator variables derived from samples representing more than one population.