Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 39 / 1 / i / Solution

During the epoch with $j$ ancestral lineages, Kingman's coalescent has waiting time $T_j\sim \mathrm{Exp}((\genfrac{}{}{0ex}{}{j}{2}))$. In the stated time units, each lineage accumulates mutations at rate $Nu=\theta /2$, so the mutation count on a neutral coalescent tree decomposes asThe law of total expectation and the mean of an exponential distribution give $\mathbb{E}{Y}_j=(j\theta /2)/\left(\genfrac{}{}{0ex}{}{j}{2}\right)=\theta /(j-1)$. ThereforeThis calculation counts mutation events. Equating the count with observed segregating sites additionally uses the infinite sites mutation model.