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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.

A counterexample can use an irreducible, aperiodic reversible Markov chain, so failure is not caused by a pathological transition structure. Take two binary coordinates, enumerate the four states as $00,01,10,11$, and let $\pi$ be their uniform distribution. Use the Markov kernelIts rows sum to one and it is symmetric, proving detailed balance with $\pi$. Every entry is positive. Let the univariate parametric family be $\tilde{\mathcal{Q}}=\{\mathrm{Bernoulli}(r):0\le r\le 2/5\}$.

For a product initial probability distribution with parameters $r_1,r_2$, additivity of the Kullback-Leibler divergence givesEach summand has derivative $\log (r_i/(1-r_i))<0$ on $(0,2/5]$, so the unique optimum at $t=0$ has $r_1=r_2=2/5$. ThusOn the other hand the allowed initial probability distribution ${\delta }_{00}$, obtained with $r_1=r_2=0$, is sent by one step to $\pi$. Its Kullback-Leibler divergence is zero, the smallest possible value. ConsequentlyThe transition reduces Kullback-Leibler divergence for each input, but may reduce it by different amounts for different inputs. It therefore need not preserve the ordering of competing initial probability distributions, which is exactly why Markov chain variational inference must optimize the initial parameters for its chosen number of steps.

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 $\mu (dx)P(x,dy)=\mu (dy)P(y,dx)$.

Put $B=\Vert f{\Vert }_{\infty }$, $v={\mathrm{Var}}_{\mu }(f)$, and $N_n=n-m_n$. We first obtain a uniform-in-$f$ stationary bound, then compare the burned-in trajectory with a stationary one.

For a reversible Markov chain, geometric ergodicity gives an absolute L2 spectral gap of a reversible Markov chain. Thus, for some $r<1$, the Markov kernel viewed as an operator satisfiesOne can use the rate in the total-variation estimate. To see why, reversibility makes $P$ a self-adjoint operator. For bounded mean-zero $h$ supported on a set where $M$ is bounded,The spectral theorem for normal operators writes the left side as $\int t^{2k}d{\nu }_h(t)$. Any positive spectral mass where $|t|>r$ would contradict the inequality as $k\to \infty$. These test functions are dense in $L_0^2(\mu )$: truncate a function and restrict it to $\{M\le j\}$, then correct its mean using a bounded function supported on a fixed positive-measure set where $M$ is bounded. Hence the spectral restriction extends to the entire mean-zero space.

Let $(W_i)$ be the stationary chain. For $h=f-\mu f$, the stationary covariance bound for reversible Markov chains givesConsequently

The initial law of the burned-in trajectory is $P^{m_n}(x,\cdot )$. Its total variation distance from $\mu$ is at most ${\epsilon }_n=M(x)r^{m_n}$. Attaching the same conditional future evolution cannot increase that distance, so the entire length-$N_n$ path laws differ by at most ${\epsilon }_n$. For either path the sum $S$ has $|S|\le B{N}_n$. The dual bound for total variation distance gives a difference of at most $2B^2{N}_n^2{\epsilon }_n$ in the second moments, and at most $4B^2{N}_n^2{\epsilon }_n$ in the squares of the means. This burn-in total variation comparison yieldsSince $m_n=\lfloor n^{1/3}\rfloor$, $N_n{r}^{m_n}\to 0$. Dividing by $N_n$ givesThe constant depends on the chain, not on $f$.

For the planar uniform distribution with a Lebesgue measure density, assume that the compact convex set $\mathcal{C}$ has nonempty interior, equivalently positive area. This assumption is missing from the wording: a line segment or singleton would not have such a planar uniform density.

The hit-and-run sampler at a current point $x$ chooses $u=(\cos \Theta ,\sin \Theta )$ with $\Theta$ uniform on $[0,2\pi )$, finds the intervaland samples $T$ uniformly on that interval, setting $x^{\prime }=x+Tu$. Convexity makes the intersection an interval and compactness makes it bounded.

If $x$ is in the interior, every direction has positive chord length, and the next point is in the interior almost surely. For a definition valid at every boundary point as well, redraw the direction whenever the chord has zero length. We use this convention in the following parts. Positive-length directions have positive angular probability at every $x$, since an interior ball is visible from $x$, so the redraw terminates almost surely.

The Markov kernel has the uniform distribution on $\mathcal{C}$ as its invariant law, as follows from the density symmetry proved next. Repeating the step therefore produces approximate uniform samples after convergence. The boundary convention matters: if singleton chords were instead treated as holding moves, a square's corner would have a positive atom at its starting point, contradicting part (b)'s claim of an everywhere absolutely continuous kernel.

Write $\ell (x,u)=t_{+}(x,u)-t_{-}(x,u)$ and letThis is the accepted fraction of directions when zero-length chords are redrawn; it equals one in the interior. The Markov kernel integrates a test function $h$ asThe signed polar representation has Jacobian determinant $|t|$. Each point $y\ne x$ is represented twice, once with positive $t$ and once with negative $t$ and the opposite direction. Since the chord is the same in either representation, the hit-and-run kernel density isfor almost every $y\in \mathcal{C}$, and it is zero outside the body. Thus the transition is absolutely continuous with respect to planar Lebesgue measure. Redrawing removes the only potential zero-chord atoms.

Let $D=\mathrm{diam}(\mathcal{C})>0$. For distinct $x,y\in \mathcal{C}$, the chord contains their segment, so it has positive length, andHence $p(x,y)\ge 1/(\pi D^2)$. Values on the diagonal, and on other null sets where a density is only specified almost everywhere, can be assigned this same lower bound. This gives a version satisfyingFor interior $x,y$, $a(x)=a(y)=1$ and the chord length is unchanged on interchanging its endpoints, so $p(x,y)=p(y,x)$. A convex body's boundary has area zero; for example contractions about an interior point fill the interior and their areas tend to the body's area. Thus this symmetry holds almost everywhere for two uniform points, proving detailed balance and stationarity of the uniform distribution. Without positive area or the boundary convention in part (a), the literal assertion need not hold.

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).

Let $K(t,\cdot )$ be the Markov kernel of the Metropolis–Hastings algorithm. Away from the possibility of remaining at $t$, its transition density isMultiplication by the target density givesThis is detailed balance for every pair of distinct states.

Writefor the rejection probability, soFor every measurable $B\subseteq \Theta$, detailed balance and the Tonelli theorem giveHence $\mu$ is an invariant distribution of the chain.

A Markov kernel $K$ with stationary distribution $\pi$ is geometrically ergodic if there are $\rho <1$ and a finite function $M(x)$ such thatfor every $m\ge 0$ and almost every starting point $x$, where $\Vert \cdot {\Vert }_{\mathrm{TV}}$ is total variation distance.

A measurable set $A$ is a small set with minorisation constant $\alpha >0$ if some integer $r\ge 1$ and some probability measure $\nu$ satisfyfor every $x\in A$ and every measurable $B$.

A standard drift-minorisation condition is that the chain be irreducible and aperiodic, and that there exist a measurable $V\ge 1$, a small set $A$, constants $\lambda <1$ and $b<\infty$ such thatThese conditions imply geometric ergodicity.

Suppose $X\to Y\to Z$ is a Markov chain, so $I(X;Z\mid Y)=0$. The chain rule for mutual information givesand alsobecause conditional mutual information is nonnegative. Therefore $I(X;Z)\le I(X;Y)$. Similarly,while $I(X,Y;Z)=I(X;Z)+I(Y;Z\mid X)\ge I(X;Z)$, so $I(X;Z)\le I(Y;Z)$. These are the two data processing inequalities. In particular, applying any deterministic function or Markov kernel to either argument cannot increase mutual information.

Let $K(y\mid x)$ be a Markov kernel, and let $P_Y=PK$, $Q_Y=QK$ be the output probability distributions. Applying the log-sum inequality for each $y$ to $a_x=P(x)K(y\mid x)$ and $b_x=Q(x)K(y\mid x)$ givesSumming over $y$ and using ${\sum }_{y}K(y\mid x)=1$ yields the data processing inequality for relative entropy

Suppose each coordinate has stationary distribution $\pi$ for the same Markov kernel. Updating one coordinate using independent new randomness while leaving the others fixed preserves the product distribution ${\pi }^{\otimes m}$. For a two-state update with $P(1\to 1)=a$ and $P(0\to 1)=b$, its stationary occupied probability is $p=b/(1-a+b)$. Independent occupied-state indicators then have total mean $mp$ and variance $mp(1-p)$. The conclusion requires the update selection and transition randomness not to introduce dependence on the other coordinates' states.

A regular conditional distribution of a random variable $X$ given $Y$ is a Markov kernel $K$ such that, for all measurable sets $A,B$,For each $y$, $K(y,\cdot )$ is a probability measure; for each $B$, $K(\cdot ,B)$ is measurable. Thus it assigns an actual probability distribution to a specified observation even when $P(Y=y)=0$. Such kernels exist for standard Borel state spaces, including real-valued variables, and their uniqueness is only up to $P_Y$-null sets. With a joint density and positive marginal density, the ratio of the two densities defines the usual conditional density kernel. Values outside the marginal support or on a null set are not uniquely determined by the joint law.

A row-stochastic matrix has nonnegative entries and each row sums to one. For a time-homogeneous Discrete-time Markov chain, its transition matrix has entriesIf a state has zero probability at time $n$, this row is specified by the model's Markov kernel, rather than recovered by conditioning on that zero-probability event. Rows index departure states and columns arrival states. Writing the current probability distribution as a row vector ${\mu }_n$ givesThe row-sum condition ensures that matrix multiplication preserves total probability. Under a column convention the evolution uses the transpose; the convention must accompany the formula.

For the sunny/rainy transition matrix, direct matrix multiplication givesIn particular, $(P^2{)}_{SS}=0.8\times 0.8+0.2\times 0.4=0.72$. The two terms sum the disjoint paths $S\to S\to S$ and $S\to R\to S$. Requiring both subsequent days to be sunny has probability $0.64$; requiring only the second day to be sunny permits the extra path through rain.

Generally $(P^n{)}_{ij}$ is the $n$-step transition probability. The identity $P^{m+n}=P^m{P}^n$ is the discrete-time Chapman-Kolmogorov equation: sum over the intermediate state. This makes matrix powers useful for forecasting many steps without enumerating every path individually.

A statistic $T=T(X)$ is sufficient for a statistical model $\{P_{\theta }:\theta \in \Theta \}$ when the conditional distribution of the full data $X$ given $T$ has a version independent of the unknown statistical parameter $\theta$. Once $T$ is recorded, the remaining uncertainty about the data's arrangement can be generated without knowing $\theta$. ^[ref]

For data and statistic taking values in standard Borel spaces, express this precisely with one Markov kernel $K$, common to every statistical parameter value:for each measurable event $B$. Thus $K(t,\cdot )$ is a regular conditional distribution of $X$ given $T=t$. Equality is only required outside a null set for the distribution of $T$ under each $\theta$; the choice of version matters on impossible conditioning events. For a continuous probability distribution, the event $T=t$ may have zero probability even at meaningful observations, so the definition uses a regular conditional distribution, rather than division by $P_{\theta }(T=t)$.

A sufficient statistic is a property of a specified statistical model, rather than of a formula for summarizing data in isolation. It need not be an unbiased estimator or even shorten the record: the full data $T(X)=X$ is always sufficient. An estimator instead targets a quantity such as a mean. An ancillary statistic has a parameter-independent marginal probability distribution, whereas a sufficient statistic has a parameter-independent conditional full-data law. A minimal sufficient statistic permits the greatest reduction among sufficient statistics. A complete statistic satisfies a separate condition about zero expectations; a complete sufficient statistic must satisfy both conditions.

Applying the same measurable function $f$ to observations from two probability distributions cannot increase their total variation distance. Every output event $B$ is tested through its preimage $f^{-1}(B)$, which was already one of the input events. ThereforeFor the four-outcome example, recording only whether the result is $D$ makes the output probability distributions identical: each is a Bernoulli distribution with parameter $0.3$. The original total variation distance was $0.3$, but this coarse measurement leaves zero discrepancy. Discarding information can make two sources less distinguishable.

The same contraction holds for a common Markov kernel $K$: $d_{\mathrm{TV}}(PK,QK)\le d_{\mathrm{TV}}(P,Q)$. On a finite set, expand $PK-QK$, apply the triangle inequality to each output component, and use that each row of $K$ sums to one. A common randomized postprocessing step cannot increase distinguishability either.