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First require a proper target: $0<Z={\int }_{{\mathbb{R}}^d}\pi (x)dx<\infty$ and $\bar{\pi }=\pi /Z$. The Metropolis–Hastings algorithm uses the Metropolis–Hastings acceptance probabilitySufficient general conditions are phi-irreducibility, aperiodicity, and a drift-minorisation condition establishing geometric ergodicity of the resulting chain, together with a stationary $(2+\delta )$th moment for the observable being averaged. These imply the central limit theorem for a geometrically ergodic Markov chain. The observable's moment condition must be included: conditions on the sampler alone cannot give the theorem for every arbitrary function.

A concrete stronger condition, directly in terms of the proposal, isTogether with a bounded observable, this is an especially simple sufficient answer. The accepted proposal density is $\min \{q(y\mid x),\bar{\pi }(y)q(x\mid y)/\bar{\pi }(x)\}$, so it is at least $ϵ\bar{\pi }(y)$. The kernel satisfies a global minorization condition with the target, hence uniform geometric ergodicity, positive Harris recurrence, and aperiodicity. An independent proposal whose importance weight $\bar{\pi }(y)/q(y)$ is uniformly bounded is one example. No claim of these properties follows merely from writing down a positive proposal.

Let $P$ be the Markov kernel and $\mu$ its invariant probability measure. Using the conventiongeometric ergodicity means that there are $r\in (0,1)$ and a finite measurable state-dependent prefactor $M(x)$ such thatHere total variation distance is used, and the rate $r$ does not depend on $x$. The prefactor need not be uniformly bounded; that stronger property is uniform geometric ergodicity. For a deterministic starting state, this estimate must hold at that state. If the definition is made only $\mu$-almost everywhere, the subsequent assertions concern those starting states where the prefactor is finite. The reversible Markov chain assumption means $\mu (dx)P(x,dy)=\mu (dy)P(y,dx)$.

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).