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

Table of contents
- Solution a

Solution

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.

 Ancestors (10)

 View article source

 Discussion (0) New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version