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The symmetric group has $n!$ elements, not the printed order $n$. For $n\ge 2$ the identity and the $n-1$ adjacent transpositions have probabilities $1/n$, so the specified transition matrix is normalized. Its stationary distribution is uniform and it satisfies detailed balance, since each transposition is its own inverse. Adjacent transpositions generate the symmetric group, making this an irreducible Markov chain; the positive holding probability makes it an aperiodic Markov chain.

For a fixed label $i$, right multiplication swaps positions, so $f(\sigma )={\sigma }^{-1}(i)$ is the label's position. Under the uniform distribution on a finite set it is uniform on $\{1,\dots ,n\}$. ThereforeFor each generator $s_j=(j,j+1)$, the squared change $[f(\sigma s_j)-f(\sigma ){]}^2$ is one if the label occupies either of the two positions, and zero otherwise. Its stationary expected value is $2/n$, givingThus the tagged-label Rayleigh quotient for adjacent transpositions gives the corrected bound

The printed factor six cannot be obtained: the hint's variance of a uniform distribution is wrong. Directly, $\mathrm{Var}(U)={\int }_0^1{u}^{2}du-({\int }_0^{1}udu{)}^2=1/12$. Bounded convergence in distribution of $U_n$ does imply convergence of its first two moments, hence convergence of its variance, but the limit is $1/12$.

More strongly, the claimed asymptotic bound itself is false for this kernel. The Aldous spectral gap theorem states that the interchange process on a finite connected weighted graph has the same spectral gap as the single-label random walk with identical edge rates; see Theorem 1.1 and its proof. Here $P-I$ is that interchange process generator on a path graph with edge rates $1/n$. For the tagged-label generator, direct substitution of $g_{\ell }(k)=\cos (\ell \pi (k-1/2)/n)$ gives eigenvalues $-2(1-\cos (\ell \pi /n))/n$, $0\le \ell <n$, including the reflecting endpoint equations. These $n$ distinct eigenvalues exhaust the tagged-label function space. Consequently the theorem yieldswhich exceeds $6/n^3$ asymptotically. This use of a substantial external theorem is only to certify the printed claim's failure; the corrected factor-twelve upper bound above follows directly from the requested test function and the variational formula.

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.

Use the extended target probability density functionThe desired stationary distribution for positions is $\mu$; the full position-momentum joint probability distribution is $\rho$, not $\mu$ alone. Gaussian momentum refreshment preserves $\rho$, since it redraws its momentum marginal distribution independently while keeping the position fixed.

Let $R(x,p)=(x,-p)$ and $T=T_{\epsilon ,L}$ be the surrogate leapfrog integration map. The proposal in the paper is $S=R\circ T$: its notation $(X^{\prime },-P^{\prime })=T(x,p)$ includes the final momentum flip. Reversibility gives $RTR=T^{-1}$, so $S^2=I$. Both $R$ and $T$ preserve volume. Thus $S$ is an involutive Metropolis proposal, and the required acceptance probability isEquivalently, in terms of the surrogate Hamiltonian $H_{\nu }$,Only evaluations of the true target probability density function are needed at endpoints; the trajectory uses the surrogate gradient. Target and surrogate normalizing constants cancel.

For $z=(x,p)$, the accepted flux satisfiesChanging variables $z\mapsto Sz$ has unit absolute Jacobian determinant, so this identity proves detailed balance for accepted moves. The rejection mass stays at the same point and is also reversible. The accept/reject step therefore preserves $\rho$. Since momentum refreshment also preserves $\rho$, their composition and either phase of the alternating process preserve $\rho$. Projecting onto positions proves the position stationary distribution is exactly the original target. Stationarity alone does not establish uniqueness or convergence from every start; those require additional irreducible Markov chain and aperiodic Markov chain hypotheses.

There is a genuine defect in the printed smoothing claim. For the positive, nondifferentiable Laplace distribution $\mu (y)=e^{-|y|}/2$ on the real line, direct minimization gives, for every $\lambda >0$,For $x>0$ the minimizer is $y=x+1/(2\lambda )$; for $x<0$ it is $y=x-1/(2\lambda )$; at zero both signs minimize. Hence normalization leaves $\nu =\mu$, still nondifferentiable at zero. The displayed construction does not in general produce the asserted smooth surrogate. For example, the also nondifferentiable target $\mu (y)\propto e^{-y^4-|y|}$ makes that infimum $-\infty$ for every $x$, since the negative quartic term dominates the quadratic penalty. The invariance proof above is valid conditional on actually having a usable smooth surrogate, as the subsequent algorithm assumes.

A corrected sufficient construction is Moreau smoothing of a negative log-density. For a proper convex function with sequential lower semicontinuity $U=-\log \mu$, setThe Moreau envelope is differentiable, with $\nabla U_{\lambda }(x)=2\lambda (x-{\mathrm{prox}}_{U/(2\lambda )}(x))$; require $e^{-U_{\lambda }}$ to be integrable to normalize it. The signs differ from the printed formula. For the Laplace distribution, this givesa genuinely differentiable potential with integrable exponential tails. Using that surrogate with the boxed acceptance probability still targets the original Laplace distribution.

Conditional on $X_0=x$, the random variable $T$ is the first return time $T_x^{+}$. The Kac's lemma mean-recurrence formula for a finite irreducible Markov chain givesAveraging over $X_0\sim \pi$ therefore yields

Write the target posterior density as $\pi (\theta )$ and the proposal distribution density as $q({\theta }^{\prime }\mid \theta )$. The Metropolis–Hastings algorithm accepts a proposed move withFor distinct states,which is symmetric in $\theta$ and ${\theta }^{\prime }$. The rejection probability supplies the diagonal part, so the entire transition kernel satisfies detailed balance. Integrating the detailed-balance identity over the starting state proves $\int \pi (\theta )P(\theta ,d{\theta }^{\prime })=\pi ({\theta }^{\prime })d{\theta }^{\prime }$. Hence the posterior is a stationary distribution; an irreducible Markov chain that is also an aperiodic Markov chain converges uniquely to it.

A continuous-time Markov chain is irreducible if every state can reach every other state with positive probability. A state $i$ is recurrent if, after leaving $i$, the chain returns to $i$ with probability one; an irreducible chain is recurrent if one, and hence every, state is recurrent. A recurrent state is positive recurrent if its expected continuous first return time is finite, and an irreducible chain is positive recurrent when all its states are.

Put $K=N+s$, the total capacity. The customer count is an M-M-s-K queue, hence a birth-death process on $\{0,1,\dots ,K\}$. Its nonzero transition rates areThe first rate is zero at $K$ because an arrival finding the shop full is rejected. The second rate is the number of busy servers times the individual service rate.

Write $\rho =\lambda /\mu$. The detailed balance for a birth-death process recursion isThereforewhereThis is the stationary distribution of an M-M-s-K queue.

The state space is finite, and every state communicates with every other through successive births and deaths because $\lambda ,\mu >0$. The chain is therefore an irreducible Markov chain, so its invariant distribution is unique. Finally the adjacent-state identitiesare detailed balance, while all nonadjacent rates vanish in both directions. Part a now shows that the queue is reversible in equilibrium.

Write $q_i=-g_{ii}$ for the total rate of leaving $i$. The jump chain of a continuous-time Markov chain $X$ is the discrete-time chain $Y=(Y_n)$ recording the states occupied after successive jumps. Its transition probabilities arewhenever $q_i>0$.

The chain $X$ is irreducible if every state can be reached from every other state with positive probability. A state $i$ is recurrent if, after leaving $i$, the process returns to $i$ with probability one; an irreducible chain is recurrent when one, and hence every, state is recurrent.

The continuous-time Markov chain and its jump chain have exactly the same successive states. If $Y$ is transient, it visits each state only finitely often, so $X$ is transient. If $Y$ is recurrent, it visits a fixed state $i$ infinitely often. At those visits, $X$ has independent exponential holding times of rate $q_i<\infty$. Their sum diverges almost surely, so explosion cannot occur before all those returns. Hence $X$ returns to $i$ infinitely often and is recurrent. Thus, for an irreducible chain,

True. If the Markov chain with transition matrix $P$ is irreducible, then for every pair of states $i,j$ there is a pathwith $P_{i_{r-1}{i}_r}>0$ at every step. The support assumption gives $Q_{i_{r-1}{i}_r}>0$ for every edge of the same path. Thus every state can reach every other state under $Q$, so the second chain is also an irreducible Markov 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.

Center the board at the origin and map each white square $(i,j)$ toThe white squares become the integer points of the diamondand a bishop move changes exactly one coordinate. From $(u,v)$ the chain can move first to $(u,0)$ and then to $(0,0)$, since both points lie in $D_n$. Reversing such paths connects any two states, so the bishop random walk is an irreducible Markov chain.

For finite irreducible Markov chains that are aperiodic Markov chains and reversible Markov chains, the ratio definition of cutoff and $T_n=t_{\mathrm{mix}}^{(n)}(1/4)\to \infty$ imply the Peres product condition, with ${\gamma }_n=1-{\lambda }_{2,n}$. If ${\gamma }_n{T}_n\le B$ on a subsequence, then ${\gamma }_{\ast ,n}\le {\gamma }_n$ and the eigenvalue lower bound for total variation mixing giveFor sufficiently small fixed $\epsilon$, this contradicts the limit one forced by cutoff for Markov chains. The same argument proves the stronger conclusion ${\gamma }_{\ast ,n}{T}_n\to \infty$.

For a finite irreducible Markov chain with at least two states satisfying detailed balance, its Dirichlet form of a Markov chain obeysExpand $f-\pi (f)$ in an orthonormal eigenbasis. The denominator is ${\sum }_{j\ge 2}{a}_j^2$, and the numerator is ${\sum }_{j\ge 2}(1-{\lambda }_j)a_j^2$. Their ratio is at least $1-{\lambda }_2$ and equality holds for the second eigenfunction. Thus the optimal Poincare inequality for a reversible Markov chain constant is $1/\gamma$, for the ordinary rather than absolute spectral gap.

An irreducible Markov chain is positive recurrent when its mean return times are finite. For a nonexplosive Markov chain on a countable state space, this is equivalent to the existence of a stationary distribution on its irreducible class. For continuous time, a return cycle includes departure and return, rather than counting immediate residence as a return.

For a finite irreducible Markov chain satisfying detailed balance with $\pi$, its transition matrix is a self-adjoint operator for $⟨f,g{⟩}_{\pi }={\sum }_x\pi (x)f(x)g(x)$. For a real orthonormal eigenbasis $(f_j)$ with eigenvalues $({\lambda }_j)$,Indeed, apply the spectral decomposition of $P^t$ to the function $1_{\{y\}}/\pi (y)$, whose coefficient against $f_j$ is $f_j(y)$. For $t=0$, every eigenvalue multiplier is interpreted as one.