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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-45/3/iv/solution

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Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 45 / 3 / iv / Solution by

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

g (θ, t) = π (θ) q (t)

as the proposal for rejection sampling. Since the Poisson probability

a_{k} (θ, t)

lies in

[0, 1]

, the following algorithm is valid:

Draw $θ$ from the proper prior distribution $π$ .
Independently draw $T_{j}$ from the exponential distributions with rates $(2 j)$ , for $j = 2, \dots, n$ , and calculate $L = \sum j T_{j}$ .
Draw $U$ uniformly on $[0, 1]$ , independently of all previous draws. Retain $(θ, T)$ if $U \leq e^{- θ L /2} (θ L /2)^{k} / k!$ ; otherwise restart.

The joint density of a proposed pair together with its acceptance event is

g (θ, t) a_{k} (θ, t)

. Its integral is

m_{k}

, so conditioning on acceptance gives precisely

g a_{k} / m_{k}

, the desired posterior density. Repeating independent proposal trials until each acceptance produces independent posterior observations. This is Bayesian rejection sampling for a segregating-site count.

An equivalent implementation simulates

S^{*} \sim Poisson (θ L /2)

and accepts exactly when

S^{*} = k

. Its acceptance probability is the same Poisson mass. Propose from the joint prior and accept according to the exact observed-count likelihood. No approximation to the posterior distribution is involved in the accepted draws.

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