I was reading Tajima's 1989 paper on his test for neutrality.
Tajima, Fumio. "Statistical method for testing the neutral mutation hypothesis by DNA polymorphism." Genetics 123.3 (1989): 585-595.
Here is the question: Suppose we have three sequences labelled $C,D,E$, and their genealogy follows $\{\{CD\}E\}$ (i.e. $C$ and $D$ coalesce first before they coalesce with $E$). Let $B$ be the most recent common ancestor of $C$ and $D$.
Now, let the random variable $k_{ij}$ be the number of nucleotide differences between sequence $i$ and sequence $j$, then Tajima shows that $k_{BC}$ and $k_{BD}$ have a non zero covariance.
But aren't mutations in the branch $BC$ independent from mutations in the branch $BD$? I was thinking that the total numbers of mutations in the brand $BC$ and $BD$ are two independent identically distributed variables, so $k_{BC}$ and $k_{BD}$ are independent, then why they have a non-zero covariance?
============Update=============
Now I have some basic ideas, but haven't worked out the full answer.
Tajima's definition of $k_{ij}$ is neither independent of the sample size, nor for a fixed coalescent time. (See his 1983 paper: Tajima, Fumio. "Evolutionary relationship of DNA sequences in finite populations." Genetics 105.2 (1983): 437-460.)
For example, in a sample size of 3, if you pick 2 individuals, and condition on that the two coalesce first, their coalescent time will follow: \begin{align*} \mathbb{P}(t=T)=p(T)=\frac{3}{2N}e^{-\frac{3}{2N}T} \end{align*} Now conditioning on a fixed coalescent time $t$, the number of mutations under the infinite sites model in each branch either from $B$ to $C$ or from $B$ to $D$ will follow a poisson distribution with parameter $\mu t$, where $\mu$ is the mutation rate per sequence per generation. Let this poisson random variable be $\xi_t$ in branch $BC$ and $\eta_t$ in branch $BD$. Then \begin{align*} k_{BC}=\sum_{t=0}^{\infty}\xi_tp(t)\\ k_{BD}=\sum_{t=0}^{\infty}\eta_tp(t) \end{align*} If we only consider the partial sum of the above series, \begin{align*} k_{BC}^{(n)}=\sum_{t=0}^{n}\xi_tp(t)\\ k_{BD}^{(m)}=\sum_{t=0}^{n}\eta_tp(t) \end{align*} then $k_{BC}^{(n)}$ and $k_{BD}^{(n)}$ clearly have a zero covariance, because $\xi_t$ and $\eta_t$ are independent poisson variables, hence \begin{align*} \mathbb{E}(k_{BC}^{(n)}k_{BD}^{(n)})&=\mathbb{E}(\sum_{t=0}^{n}\xi_tp(t)\sum_{t=0}^{n}\eta_tp(t))=\sum_{i=0}^{n}\sum_{j=0}^np(i)p(j)\mathbb{E}(\xi_i\eta_j)=\sum_{i=0}^{n}\sum_{j=0}^np(i)p(j)\mathbb{E}\xi_i\mathbb{E}\eta_j\\ &=\mathbb{E}k_{BC}^{(n)}\mathbb{E}k_{BD}^{(n)}, \end{align*} so their covariance is zero.
But as $n\to+\infty$, how $k_{BC}^{(n)}k_{BD}^{(n)}$ converges to $k_{BC}k_{BD}$ is questionable. It will not uniformly converge to $k_{BC}k_{BD}$ because otherwise we can first computing the expectation then take the limit, which gives us a zero covariance. Tajima didn't explicitly show us how he computed the covariance by summing together three infinite series (Line 7, Pg. 448, 1983's paper). I tried directly working on that series but failed in the last sum. His result is correct, though, I hope someone can give some hint on why there is inherent correlation between these seemingly independent random variables.
=======Update: a simple explanation has been posted==============
