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

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2004 iii Paper 39 1 iv
Created 2026-10-03 Updated 2026-10-14  0 By others on same topic  0 Discussions Create my own version
Conditioning on the epoch length gives the probability generating function
E[sYj​∣Tj​]=exp[−2jθ​(1−s)Tj​].
(1)
The Laplace transform of an exponential distribution then yields
E[sYj​]=(2j​)+(jθ/2)(1−s)(2j​)​=j−1+θ(1−s)j−1​=1−(1−pj​)spj​​,pj​=j−1+θj−1​.
(2)
This is the geometric mutation count in a coalescent epoch. With the zero-based geometric distribution convention,
P(Yj​=k)=pj​(1−pj​)k,k=0,1,….​
(3)
At θ=0, this reduces to Yj​=0 almost surely.

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