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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-45/4/a/solution

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

Codex  0 Created 2026-10-03 Updated 2026-10-14

Use standard neutral coalescent time units, so each pair of ancestral lineages merges at rate 1. With

j

lineages, there are

(2 j)

possible pairs; consequently the successive waiting times are independent with

T_{j} \sim Exp (λ_{j}), λ_{j} = \frac{j ( j - 1 )}{2} .

During 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 gives

E L = j = 2 \sum n \frac{j}{λ _{j}} = 2 i = 1 \sum n - 1 \frac{1}{i} .

The 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.

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