The gory details

You knew that I wouldn't be satisfied with a numerical example, didn't you? You knew there had to be some algebra coming, right? Well, here it is. Let

$$\begin{eqnarray*} D_i &=& x_{11,i} - p_{1i}p_{2i} \\ D_t &=& \bar x_{11} - \bar p_1\bar p_2 \quad , \end{eqnarray*}$$

where $\bar x_{11} = \frac{1}{K} \sum_{k=1}^K x_{11,k}$, $\bar p_1 = \frac{1}{K} \sum_{k=1}^K p_{1k}$, and $\bar p_2 = \frac{1}{K} \sum_{k=1}^K p_{2k}$. Given these definitions, we can now caclculate $D_t$.

$$\begin{eqnarray*} D_t &=& \bar x_{11} - \bar p_1\bar p_2 \\ &=& \frac{1}{K} ... ...r p_2) + \bar D \\ &=& \mbox{Cov}(p_1, p_2) + \bar D \quad , \end{eqnarray*}$$

where $\mbox{Cov}(p_1, p_2)$ is the covariance in allele frequencies across populations and $\bar D$ is the mean within-population gametic disequilibrium. Suppose $D_i = 0$ for all subpopulations. Then $\bar D = 0$, too (obviously). But that means that

$$\begin{eqnarray*} D_t &=& \hbox{Cov}(p_1, p_2) \quad . \end{eqnarray*}$$

So if allele frequencies covary across populations, i.e., $\mbox{Cov}(p_1, p_2) \ne 0$, then there will be non-random association of alleles into gametes in the sample, i.e., $D_t \ne 0$, even if there is random association alleles into gametes within each population.⁶

Returning to the example in Table 1

$$\begin{eqnarray*} \mbox{Cov}(p_1, p_2) &=& 0.5(0.6-0.65)(0.4-0.3) + 0.5(0.7-0.6... ...\ \bar x_{22} &=& (0.35)(0.70) - 0.005 \\ &=& 0.24 \quad . \end{eqnarray*}$$