Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 35 / 1 / h / Solution

A uniform prior on $(0,T)$ makes the posterior density proportional to the Poisson change-point posterior likelihood above. The zeros trick introduces an observed zero with Poisson distribution mean $K-\log L(\theta )$, making its likelihood function equal to $e^{-K}L(\theta )$. Choose $K=T+n\log 2+1$; since $\log L(\theta )\le n\log 2$, this mean is strictly positive. Rough BUGS code, with zero=0 supplied as data, isFor $n=0$, omit the array and set j <- 0. Monitor the sampled theta to obtain its posterior mean, credible interval and interval probabilities.

There is also an exact sampling method. On $(t_j,t_{j+1})$ the posterior density is proportional to $2^{-j}{e}^{\theta }$, so choose the interval with weightsThen draw $U$ from a uniform distribution on $(0,1)$ and set $\theta =\log (e^{t_j}+U(e^{t_{j+1}}-e^{t_j}))$. These weighted interval draws sample the posterior directly, without asking a local Markov chain Monte Carlo update to cross its discontinuities.