Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 216 / 3 / b

Use the printed parameterization $\lambda ={\sigma }^2$, $v={\sigma }_0^2$, and $a=\sqrt{\rho }$. Let $T=365$, $X$ be the design matrix with rows $x_i^T$, and $r=\theta -X\beta$. Conditional on $\theta$ and the parameters, $Y$ contributes no extra dependence on $Z$ by conditional independence.

The stationary autoregressive normal distribution has prior density proportional toIts Gaussian autoregressive conditional precision is the tridiagonal matrix $J$ withCombining this prior with $r\mid Z\sim N(\lambda Z,vI)$ and completing the square givesThe precision matrix $Q$ is positive definite. Compute its lower Cholesky decomposition $Q=L{L}^T$, solve $Qm=h$, draw $\xi \sim N(0,I_T)$, and solve $L^T\eta =\xi$. Then return $Z=m+\eta$. This Gaussian sampling from a precision Cholesky factor is exact because $\mathrm{Cov}(\eta )=L^{-T}{L}^{-1}=Q^{-1}$. Tridiagonal structure gives linear cost in $T$ and preserves the conditional correlation between days. Notice that the solve uses $L^T$, not $L$.

These conditionals are normalized for every fixed positive ${\sigma }^2,{\sigma }_0^2$, despite the joint posterior impropriety explained in part (a). If the intended model uses $\sigma Z_i$ instead, replace the observation coefficient $\lambda$ by $\sigma$ throughout; the information and precision terms then use $\sigma /v$ and ${\sigma }^2/v$.

Given the parameters and $Z$, the coordinates are conditionally independent. Write $m_i=x_i^T\beta +{\sigma }^2{Z}_i$, $v={\sigma }_0^2$, and let $y_i$ denote the observed count. Their full conditional distributions have scalar densitiesEach is proper: the binomial distribution likelihood factor is bounded, and the remaining factor is a proper normal distribution density.

One completely specified exact algorithm is rejection sampling from $N(m_i,v)$. DefineFor $0<y_i<n_i$,for $y_i=0$ or $y_i=n_i$, take $M_i=1$, and for $n_i=0$ take $H_i=M_i=1$. Propose $t\sim N(m_i,v)$ and accept with probability $H_i(t)/M_i$, repeating until acceptance. The accepted density is proportional to the proposal times $H_i$, exactly $q_i$. Its acceptance probability is positive, so it terminates almost surely, although it can be inefficient.

An efficient exact alternative is adaptive rejection sampling. The log density hasThus it is a log-concave probability density. Choose one tangent point with positive derivative in the left tail and one with negative derivative in the right tail; these exist because the Gaussian term dominates there. Their tangent upper hull gives an integrable piecewise exponential rejection envelope. Sample that envelope, accept using the target-to-envelope ratio, and add evaluated points to improve it. Concavity guarantees the upper bound, so accepted samples remain exact. Sample each coordinate independently by either method; a finite number of Metropolis steps would not provide the requested exact draws.