Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 40 / 5

The linear predictor is unchanged by ${\alpha }_j\mapsto {\alpha }_j+c$, ${\beta }_t\mapsto {\beta }_t+d$, $\mu \mapsto \mu -c-d$. Without constraints, different parameter triples therefore give the same rates and likelihood. Identifiability requires choosing one representative of this redundancy.

Taking the first levels as reference yields ${\alpha }_1={\beta }_1=0$. ThenThese identify all three free parameters, while ${\lambda }_{22}={\lambda }_{21}{\lambda }_{12}/{\lambda }_{11}$ is the model's no-interaction constraint. The zero reference effects fix a parametrization; they do not remove the baseline rate or impose equal rates across groups.

Use the standard interpretation that the counts are conditionally independent given the parameters, with mutually independent stated priors. Let $a_2={\alpha }_2$, $b_2={\beta }_2$, and defineUnder the reference constraints, the Poisson distribution likelihood, up to data-only factors, isMultiplying by the shape-rate gamma prior gives the Poisson-gamma conjugacy with unequal exposuresHere $b$ is a rate, matching the probability density function convention on the PDF's first page. The factor $I$ represents the number of replicates per cell, not the total count.

An exact Gibbs sampler update is available despite the non-Gaussian intercept: draw ${\theta }^{\prime }$ from the conditional gamma distribution law in part (ii), then setTogether with conditional updates of the two remaining effects, this targets the required joint posterior in the log-intercept coordinates. The log-gamma prior for a Poisson log-intercept includes the change-of-variable factor: with $d\theta /d\mu =e^{\mu }$,Using $(a+T-1)\mu$ instead would omit the Jacobian and give the wrong conditional law for $\mu$.

The full conditional log probability density functions of the effects are, up to constants,These are Gaussian-prior Poisson log-effect conditional probability density functions. The exponential likelihood term prevents a standard normal distribution conjugate update. A Metropolis-within-Gibbs algorithm conveniently updates one effect at a time without computing its conditional normalizing constant.

For example, propose $a_2^{\prime }=a_2+s_{\alpha }Z$ with $Z\sim N(0,1)$, and accept withUse an analogous normal distribution random-walk proposal for $b_2$. The proposal is symmetric and lives on the appropriate unrestricted real parameter space, so no proposal ratio is needed. It makes local moves compatible with a unimodal conditional probability density function. Tune the positive scale in a pilot or freeze it after warm-up; a scale far too large causes rejection and a scale far too small moves slowly.

A useful scale can be inferred from the conditional curvature: ${\ell }_{\alpha }^{\prime \prime }(a_2)=-I\theta (1+e^{b_2})e^{a_2}-{\sigma }_1^{-2}<0$, with the analogous formula for $b_2$. A normal proposal near the conditional mode with variance approximately $-1/{\ell }^{\prime \prime }$ is another sensible choice, but an asymmetric or state-dependent proposal must include its full reverse-to-forward proposal-density ratio. Metropolis–Hastings is convenient, not logically mandatory; these concave conditionals also admit other specialized samplers.

Introduce a model indicator $M\in \{0,1\}$. In $M=0$, set $\mu =0$, hence $\theta =1$, and retain the two group effects. In $M=1$, include $\mu$ with the prior induced by the gamma baseline. Choose positive prior model probabilities $p_0,p_1$ summing to one; the supplied within-model parameter priors alone do not specify these probabilities.

Let $r=(a_2,b_2)$, use the same proper prior $p(r)$ in both models, and denote their likelihoods by $L_0(r)$ and $L_1(\mu ,r)$. The unnormalized joint model targets areFor a birth move, draw an auxiliary $u$ from a positive probability density function $q(u)$ and map $(r,u)$ to $({\mu }^{\prime }=u,r^{\prime }=r)$. The reverse death move simply deletes $\mu$. This is a dimension-matching bijection with absolute Jacobian one. If the birth and death selection probabilities are $b_0$ and $d_1$, respectively, the birth and death moves for Bayesian variable selection rule givesThe reverse acceptance is the reciprocal ratio, capped at one, evaluated at the current $\mu$. Choosing $q=p_{\mu }$ cancels the added-parameter prior; a data-informed proposal may mix better but must retain its proposal factor. Within each model, also update its continuous parameters to ensure exploration.

For an explicit likelihood ratio, with $E$ and $T$ from part (ii),This supplies a complete reversible-jump Markov chain Monte Carlo move. Its acceptance probabilities enforce detailed balance between the two dimensions. The lower-dimensional state is a separate point-mass model, not a zero-probability equality test inside the continuous full-model prior.

Under the reference convention ${\alpha }_1=0$, the hypothesis becomes absence of the free group contrast ${\alpha }_2$. Introduce an inclusion indicator $J$: set ${\alpha }_2=0$ when $J=0$, and use its stated proper normal prior when $J=1$. Assign positive prior probabilities $q_0,q_1$ to those models, and retain $\mu ,{\beta }_2$ in both.

A birth draws $u$ from a proposal probability density function $g(u)$ and appends it as ${\alpha }_2^{\prime }=u$; the reverse death deletes it. The Jacobian is one. At fixed $\theta ,{\beta }_2$, the likelihood ratio isThus the birth acceptance ratio before capping iswhere $p_{\alpha }$ is the normalized $N(0,{\sigma }_1^2)$ prior. Again, taking $g=p_{\alpha }$ cancels that factor. Use the reciprocal death rule and ordinary within-model parameter updates.

After warm-up, the Markov chain Monte Carlo occupation fraction estimates the desired probability:This requires the chain to explore both models and satisfy its usual ergodic convergence conditions. Under the original continuous normal prior alone, the equality event has posterior probability zero; the explicit model indicator supplies the necessary prior atom. If intercept selection from part (v) is also performed, use two indicators and count all sampled states with $J=0$, regardless of the intercept indicator.