Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 45 / 4 / a / Solution

Use standard neutral coalescent time units, so each pair of ancestral lineages merges at rate 1. With $j$ lineages, there are $\left(\genfrac{}{}{0ex}{}{j}{2}\right)$ possible pairs; consequently the successive waiting times are independent withDuring this epoch there are $j$ branches, each of length $T_j$. Hence the total branch length of a neutral coalescent is $L={\sum }_{j=2}^{n}j{T}_j$, and linearity of expectation givesThe root branch above the sample's common ancestor is excluded: its mutations would be shared by every chromosome and would not be segregating sites in the sample.