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

Table of contents
- Solution ii

Solution

Apply the law of total variance to each epoch. The Poisson distribution has equal conditional mean and variance, while

Var (T_{j}) = (2 j)^{- 2}

. Hence

Var (Y_{j}) = E [\frac{j θ T _{j}}{2}] + Var (\frac{j θ T _{j}}{2}) = \frac{θ}{j - 1} + \frac{θ ^{2}}{( j - 1 ) ^{2}} .

The independent mutation counts in coalescent epochs permit variance additivity for independent random variables. Thus

Var (S) = θ H_{n - 1} + θ^{2} H_{n - 1}^{(2)}, H_{n - 1}^{(2)} = i = 1 \sum n - 1 \frac{1}{i ^{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.

 Ancestors (10)

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