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

The likelihood function for the ordered arrival times of a Poisson process, including the absence of further arrivals before $T$, isThe arrival-time constraint does not depend on $\lambda$. Equivalently, the Poisson distribution of the total count gives $L(\lambda )=e^{-\lambda T}(\lambda T{)}^n/n!$, which differs by a parameter-independent factor. Conditional on the count, the Poisson process conditional arrival times have density $n!/T^n$, independent of $\lambda$. Thus the total count contains all the information about the rate when the exposure is known; it is a sufficient statistic.

A conjugate prior is a family of prior distributions whose members remain in that family after updating by the likelihood function. Here multiplying a shape-rate gamma distribution density by the Poisson process likelihood changes its power of $\lambda$ and its exponential rate, leaving a gamma distribution. The parameters change with the observations; conjugate prior does not mean that the Bayesian posterior equals the prior distribution.

By Poisson-gamma conjugacy, the posterior density is proportional toThus the shape-rate posterior isIts posterior mean is $A/B$ and its variance is $A/B^2$. The parameter $B$ is a rate, rather than a scale.

Independent increments of the Poisson process give $M\mid \lambda \sim \mathrm{Poisson}(\lambda )$ for a one-hour interval. The Poisson distribution therefore givesThis conditional probability still depends on the unknown rate.

The posterior predictive probability averages the conditional probability over the Bayesian posterior. Using its gamma distribution density,For fixed $a,b$ and $n/T\to \hat{\lambda }<\infty$,Consequently $p_0/e^{-n/T}\to 1$, and both tend to $e^{-\hat{\lambda }}$. This is reasonable because the posterior mean tends to the observed rate and the posterior variance tends to zero. Averaging $e^{-\lambda }$ then approaches evaluating it at the estimated rate: with abundant observations, predictive uncertainty about the rate vanishes.

There is a notation problem here. The posterior predictive probability $p_0$ calculated above is already a fixed number conditional on $a,b,n,T$. Literally,No simulation is needed for that interpretation. The known exposure $T$ is also necessary, despite its omission from this part's list of inputs.

The natural uncertain quantity is instead $q(\lambda )=\mathbb{P}(M=0\mid \lambda )=e^{-\lambda }$. For its posterior probability of being below one half, draw the rate from its gamma distribution Bayesian posterior and average an indicator. Rough BUGS code isThe monitored average of belowHalf estimates $\mathbb{P}(\lambda >\log 2\mid \mathcal{D})$, equivalently $1-F_{\mathrm{Gamma}(A,B)}(\log 2)$. The equality boundary has zero probability. This code samples the already updated Bayesian posterior; adding the count likelihood function again would count the data twice.

For an Inhomogeneous Poisson process, the likelihood function is the product of the intensities at arrivals times the exponential of minus the integrated intensity. The integrated intensity is $\theta +2(T-\theta )=2T-\theta$. There are $j(\theta )$ arrivals at rate one and $n-j(\theta )$ at rate two, soSet $t_0=0$ and $t_{n+1}=T$ to include changes before the first or after the last arrival. An arrival exactly at the change has probability zero, so the convention for $j$ at that point does not affect the Bayesian posterior.

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.