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?


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==============

  • $\begingroup$ Could you please highlight where in the paper (which equation) is it claimed that $cov(k_{BC}, K_{BD}) ≠ 0$? $\endgroup$
    – Remi.b
    Jan 27, 2018 at 22:51
  • $\begingroup$ If the ancestor $B$ is not considered as a single sequence but as a population of sequences, then if B has many polymorphic sites $S$ and a high average heterozygosity $\pi$, then drift will have an important say in the values $k_{BC}$ and $k_{BD}$ and both values will be large. If the ancestor $B$, has $S=0$, then the only contributors to $k_{BC}$ and $k_{BD}$ are new mutations and hence both values will be small. $\endgroup$
    – Remi.b
    Jan 27, 2018 at 22:55
  • 1
    $\begingroup$ Hi Remi, the equation is at Pg. 586 at the middle right panel, roughly 5 lines above Equation (15). It is originally derived from His 1983 paper: Evolutionary relationships between DNA sequences in finite populations $\endgroup$ Jan 28, 2018 at 0:11
  • $\begingroup$ But the coalescent process only considers the relationship among sampled individuals, so I think each label should be a unique sequence. $\endgroup$ Jan 28, 2018 at 0:12
  • $\begingroup$ Hmmm.... Yes, I now agree with you. The graph makes it pretty clear that 'B','C', 'D' refers to individuals sequences. Would be worth going into this 1983 paper. +1 in the meantime. Good question! $\endgroup$
    – Remi.b
    Jan 28, 2018 at 0:15

1 Answer 1


Here is a simple answer to my question. The reason two numbers of total mutations accumulated in two divergent branches are not independent of each other is because they experience the same amount of coalescent time.

While the stationary Poisson mutation processes are independent from each other as long as they happen in different branches of a genealogy, they are likely to produce a similar number of mutations if two such processes happen together for a same amount of time. Thus, the non-zero part of the covariance between $k_{BC}$ and $k_{BD}$ is not from mutation processes themselves, but from the shared coalescent time $T$.


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