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

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2004 iii Paper 39 1 ii
Created 2026-10-03 Updated 2026-10-14  0 By others on same topic  0 Discussions Create my own version
Apply the law of total variance to each epoch. The Poisson distribution has equal conditional mean and variance, while Var(Tj​)=(2j​)−2. Hence
Var(Yj​)=E[2jθTj​​]+Var(2jθTj​​)=j−1θ​+(j−1)2θ2​.
(1)
The independent mutation counts in coalescent epochs permit variance additivity for independent random variables. Thus
Var(S)=θHn−1​+θ2Hn−1(2)​,Hn−1(2)​=i=1∑n−1​i21​.​
(2)
The first term is conditional Poisson distribution variation; the second comes from the random lengths of the ancestral tree. These are the moments of the mutation count on a neutral coalescent tree.

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