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Suppose a Markov kernel satisfies $PV(x)\le M<\infty$ for a measurable $V\ge 1$. For $0<\rho <1$ and $R>M/\rho$, the sublevel set $C=\{V\le R\}$ gives $PV\le \rho V+M{1}_C$. If $C$ is a small set, this is a geometric drift condition. The hypotheses of an irreducible Markov chain and an aperiodic Markov chain are also required for the usual geometric ergodicity theorem.

Use the shape-rate convention for every gamma distribution, with $a,c>0$. Let $A_i=a+y_i$ and $s={\sum }_i{\theta }_i$. The Poisson distribution likelihood and the Gamma–Poisson hierarchical model giveAll factors involving the coordinate being updated must be retained, including $b^{na}$ from the conditional gamma distributions. The full conditional distributions areFor a blocked Gibbs sampler, draw $b$ from the second full conditional distribution, then redraw all the ${\theta }_i$ independently from the first ones. Repeating these two steps preserves the joint posterior distribution; observing $\theta$ after each complete sweep gives the Markov chain used in the next part. Standard gamma distribution sampling works for noninteger as well as integer shapes. The positive observation-period convention $t_i>0$ is used for the subsequent geometric drift condition.

Take observation periods to have positive lengths $t_i>0$, and let $P$ be the Markov kernel of a sweep which first draws $b$ given the current $\theta$ and then draws the new vector ${\Theta }^{\prime }$. Write $A_i=a+y_i$. Conditional on $b$, independence and the gamma distribution mean and variance giveAveraging over the full conditional distribution of $b$ preserves this bound, uniformly in the current state. Therefore $PV(\theta )\le M:=1+D<\infty$. This is the mechanism in bounded conditional moments imply a geometric drift.

Choose $0<\rho <1$, $R>M/\rho$, and $C=\{\theta :V(\theta )\le R\}$. Outside $C$, $PV\le M<\rho V$; inside $C$, $PV\le M\le \rho V+M$. HenceFor completeness, $C$ is a small set, even though it approaches the boundary of the positive orthant. Choose $\delta >0$ small enough that $B=[\delta ,2\delta {]}^n\subset C$. On $C$, $s={\sum }_i{\theta }_i\le \sqrt{R-1}$. The conditional gamma distribution of $b$ has fixed shape $na+c$ and rate $1+s$ in a compact positive interval. Its probability density function therefore has a positive common lower bound on $b\in [1,2]$. For those $b$, the product probability density function of ${\Theta }^{\prime }$ similarly has a positive common lower bound on $B$, since every shape is positive and every rate $b+t_i$ lies in a compact positive interval. Integrating over $b\in [1,2]$ gives $P(\theta ,\cdot )\ge \epsilon \eta (\cdot )$ on $C$, where $\eta$ is the uniform distribution on $B$ and $\epsilon >0$.

The transition probability density function is positive throughout the positive orthant, giving an irreducible Markov chain. Since $B\subset C$, the box minorization includes starts in the same box, which gives an aperiodic Markov chain. The proper prior distributions and bounded Poisson distribution likelihood have positive finite evidence, so the invariant posterior distribution is proper. Together with the geometric drift condition, these hypotheses imply geometric ergodicity. In particular, the argument proves the required drift rather than assuming that every Gibbs sampler is geometrically ergodic.

The positive-period convention matters. If zero lengths are allowed, the printed assertion can fail: take $n=1$, $a=c=1$, $t_1=y_1=0$. Then $b\mid \theta \sim \mathrm{Gamma}(2,1+\theta )$ and ${\Theta }^{\prime }\mid b\sim \mathrm{Gamma}(1,b)$, soThis example has a proper posterior distribution but cannot satisfy a finite quadratic geometric drift condition. Thus the proof establishes the intended claim for positive observation periods, not an unrestricted extension to zero exposure.

The geometric drift condition requires a measurable $V:\mathcal{X}\to [1,\infty )$, a small set $C$, constants $\lambda \in (0,1)$ and $b<\infty$, andSmallness means that for some $m\ge 1$, $\epsilon >0$, and probability measure $\nu$, $P^m(x,\cdot )\ge \epsilon \nu (\cdot )$ for every $x\in C$. With irreducibility and aperiodicity, this drift-minorisation condition implies geometric ergodicity. The function $V$ controls the chain away from the small set; inside that set the additive constant allows unrestricted bounded drift.

Let $A=\mathrm{Leb}(\mathcal{C})>0$ and let $\mu$ be the uniform distribution on the body. The hit-and-run kernel density lower bound impliesFor the last range, $\mathcal{C}$ lies in a radius-$D$ disc about any of its points, so $A\le \pi D^2$ and $\epsilon \le 1/2$. This global Doeblin condition proves irreducibility with respect to $\mu$ and aperiodicity. The entire state space is a small set; $V=1$, $\lambda =1/2$, and $b=1/2$ satisfy the geometric drift condition.

In fact the conclusion is stronger. Write $P=\epsilon \Pi +(1-\epsilon )R$, where $\Pi (x,\cdot )=\mu$ and $R$ is the residual Markov kernel. Since $\mu P=\mu$, also $\mu R=\mu$. Each transition has a probability $\epsilon$ of resetting to stationarity, and after such a reset every subsequent distribution remains stationary. Equivalently,Thus the algorithm has uniform geometric ergodicity, using the full-dimensional and boundary conventions stated in part (a).