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On a compact metric space $X$, a conservative Feller semigroup is a strongly continuous semigroup of positive linear maps on $C(X)$ satisfying $P_{t}1=1$. Positivity means $f\ge 0⟹P_{t}f\ge 0$. These conditions imply sup-norm contraction. Conversely, for a contraction semigroup with $1\in D(L)$ and $L1=0$, differentiation on the generator domain gives $\frac{d}{dt}{P}_{t}1=P_{t}L1=0$, so $P_{t}1=1$. On the real space $C(X)$, if $0\le f\le 1$, then $\Vert 1-P_{t}f{\Vert }_{\infty }\le \Vert 1-f{\Vert }_{\infty }\le 1$, implying $P_{t}f\ge 0$. Scaling proves positivity for every nonnegative $f$. This is the unital contraction positivity criterion. For complex $C(X)$ the same conclusion follows because each functional $f\mapsto P_{t}f(x)$ has norm one and value one on $1$, hence is a positive linear functional.

For each $t,x$, the Riesz-Markov-Kakutani representation theorem represents that functional by a unique Borel probability measure $p_t(x,dy)$:Continuity of $P_{t}f$ gives weak continuity of $x\mapsto p_t(x,\cdot )$. Approximating indicators of open sets increasingly by continuous functions proves Borel measurability in $x$; a monotone-class argument then gives a Markov kernel. The semigroup property and uniqueness of representing measures yield $p_0(x,\cdot )={\delta }_x$ and the Chapman-Kolmogorov equationThese are the required transition probabilities.

An invariant probability measure for a semigroup satisfies ${\int }_X{P}_{t}fd\mu ={\int }_{X}fd\mu$ for every $t\ge 0$ and $f\in C(X)$. Equivalently its distribution is preserved by the Markov kernel. For $1\le p<\infty$, Jensen inequality gives $|P_{t}f{|}^p\le P_t(|f{|}^p)$ pointwise. Integrating and using invariance provesAlso $\Vert P_{t}f-f{\Vert }_{L^p(\mu )}\le \Vert P_{t}f-f{\Vert }_{\infty }\to 0$. Continuous functions are dense in these $L^p$ spaces for a finite Borel measure on a compact metric space, so the contraction extends to a strongly continuous semigroup on $L^p(\mu )$.

For real $f$ with $f,f^2\in D(L)$, the kernel form of Jensen inequality gives $P_t(f^2)-(P_{t}f{)}^2\ge 0$, with equality at $t=0$. Take its right derivative to obtain the generator square inequalityThis is nonnegativity of the associated carré du champ.

Each time-average functional ${\varphi }_n$ is positive with ${\varphi }_n(1)=1$, hence represents a Borel probability measure ${\mu }_n$. The compactness of probability measures on a compact metric space gives a subsequence converging weakly to a probability measure $\mu$. Explicitly, $C(X)$ is a separable Banach space, so Banach-Alaoglu theorem and a countable dense family give a subsequence converging on every continuous function, rather than merely a net. For fixed $t\ge 0$, the semigroup property gives the Krylov-Bogolyubov time-average argument:Its absolute value is at most $2t\Vert f{\Vert }_{\infty }/n$. Pass to the subsequential limit, observing that $P_{t}f\in C(X)$, to get ${\int }_X{P}_{t}fd\mu ={\int }_{X}fd\mu$. Thus the limiting measure is invariant.

Let $S_k={\sum }_{\ell =1}^k{Y}_{\ell }$, $S_0=0$, be the epochs of the renewal process. The definition uses the next epoch at or after $n$, so $X_n=0$ at an epoch. If $X_n=i\ge 1$, no renewal occurs before $n+i$ and necessarily $X_{n+1}=i-1$.

If $X_n=0$, time $n$ is a renewal epoch, and the next unused interarrival time is independent of the entire observed past with the original distribution. To justify the random index, condition on each possible number $k$ of renewals up to $n$: the event identifying $k$ depends only on $Y_1,\dots ,Y_k$, whereas $Y_{k+1}$ is independent of those variables. Summing over $k$ preserves that distribution and independence. Therefore the residual lifetime Markov chain has transition probabilitiesIts reachable state space is $E=\{j\ge 0:\mathbb{P}(Y_1>j)>0\}$, either an initial finite segment or all nonnegative integers. The conditional next-step law depends only on $X_n$, proving the Markov property. Initially $X_0=0$. The reset is $Y-1$, rather than $Y$, because one unit of time has elapsed by the next step.

The first departure from the no-symptoms state has an exponential distribution with rate $\lambda$. Spending the entire interval there means no departure at all, soThis is the holding time survival probability. The transition probability $p_{11}(t)$ only says that the patient is symptom-free at the endpoint, allowing an onset and recovery in between, and is therefore not the answer. The formula also covers $\lambda =0$, when state 1 is absorbing.

Visits can be missed or rescheduled because of illness, clinic availability or patient choice; follow-up may end through dropout or right censoring. The correct panel-observed multi-state likelihood uses each observed interval length ${\Delta }_{ik}$:Thus the four aggregated one-year counts are generally no longer sufficient; records must retain actual durations. Numerical maximum likelihood estimation uses the corresponding transition probabilities, or a matrix exponential of the generator when a larger model is fitted. One must not treat a two-year interval as a single one-year transition or as two known transitions through an unobserved intermediate state.

This conditional calculation is valid when visit scheduling and loss to follow-up are ignorable given the modeled information. If symptomatic patients are systematically examined earlier or leave the study differently, observation itself may be informative; the visit or dropout process may require a joint statistical model. Simply replacing one year by the actual interval does not correct informative observation.

Condition on each patient's initial observed state and assume independent patient trajectories and observation times that do not carry additional information about the unobserved path. The Markov property factors each trajectory's likelihood function into its successive one-year transition probabilities. ThusMore explicitly, with $q=\lambda +\mu >0$ and $E=e^{-q}$,Successive transitions within one patient need not be independent unconditionally: this is a product of conditional factors justified by the Markov property. If initial-state probabilities are modeled rather than conditioned on, their likelihood contributions must also be included. Zero-count terms contribute zero; a positive count attached to a zero model probability makes the log-likelihood equal to $-\infty$.

Let $R$ interchange the two complete graphs, sending $v_i$ to $v_i^{\prime }$ and fixing $w$. This involution preserves the transition matrix of the simple random walk. A mirror coupling under an involution starts $X$ at $v_i$, keeps $Y_t=R(X_t)$ until the hitting time $\tau$ of $w$, and then makes the two walks move identically. Each coordinate has the correct transition probabilities, and they coalesce precisely at $\tau$. The coupling inequality for total variation gives

There is a genuine error in the printed expected hitting time bound, as well as an unspecified initial law. For $n\ge 2$, write $a={\mathbb{E}}_{v_1}\tau$ and $b={\mathbb{E}}_{v_i}\tau$ for $i\ne 1$. From every clique vertex there is a positive chance of reaching $w$ within two steps, uniformly over the finite state space for each fixed $n$, so $\tau$ has finite expected value. First-step analysis, using the degree of a vertex $n$ at $v_1$ and $n-1$ at the other clique vertices, givesSolving these linear equations yieldsIn particular even the attachment vertex exceeds the printed bound by one. The valid uniform replacement is ${\max }_x{\mathbb{E}}_x\tau \le n^2$, the hitting time of the bridge vertex between two cliques formula.

We must also handle arbitrary starting states, including $w$, before applying the mirror argument. Identify related vertices and consider the lumped Markov chain $Z$ on $\{1,\dots ,n,w\}$. Its transition matrix $Q$ has $Q(1,j)=1/n$ for $j\ne 1$, $Q(1,w)=1/n$, $Q(i,j)=1/(n-1)$ for $i\ne 1$ and $j\ne i$, and $Q(w,1)=1$, with all other entries zero. Here the row for $i\ne 1$ only ranges over $j\in \{1,\dots ,n\}$.

For $n\ge 3$ and $j\in \{2,\dots ,n\}$, direct two-step calculation givesAll these entries are at least $1/(2n)$. Thus $Q^2(z,\cdot )\ge \alpha \nu (\cdot )$ for every $z$, where $\nu$ is the uniform distribution on a finite set $\{2,\dots ,n\}$ and $\alpha =(n-1)/(2n)\ge 1/3$. This is a Doeblin condition. At each two-step block couple the two quotient endpoints to the same sample from $\nu$ with probability $\alpha$, using the residual probability distributions otherwise. The first successful block has expected value at most $1/\alpha$, so the alignment time $A$ has $\mathbb{E}A\le 6$. Lift each endpoint coupling of probability distributions to the original walks by their conditional two-step path laws; this preserves every marginal path law. The success test uses fresh block randomness and so permits restarting the walks with their original transition matrix at the aligned endpoints.

At alignment the original walks are equal or related. Use identical transitions in the first case and the mirror coupling under an involution in the second. The resulting coalescing coupling has coalescence time $T$ satisfying $\mathbb{E}T\le 6+n^2$ for every pair of initial states. Therefore Markov inequality and the pairwise mixing diameter implyFor the hint's particular starting states, the quotient one-step laws from two different clique indices have a common component of mass at least $1-2/n$: for two ordinary indices it is $(n-2)/(n-1)$; for one attachment index it is $(n-2)/n$. A maximal coupling therefore aligns those indices with probability $1-O(1/n)$, as suggested. The two-step argument above additionally covers $w$.

The nonlazy walk is an aperiodic Markov chain for $n\ge 3$, because the connected graph contains a triangle. For $n=1,2$ it is a periodic Markov chain on a bipartite graph; each bipartition class has stationary mass $1/2$, and the walk stays on a single class at each time. Hence $d(t)\ge 1/2$ and the $1/4$ mixing time is infinite. The asymptotic conclusion is valid for $n\ge 3$.

The lazy Markov chain has holding probability $1/2$ and transition probability $1/(2n)$ to each neighbour in the hypercube graph. Its stationary distribution is $\pi (x)=2^{-n}$ by symmetry.

For $(x,y)$ choose the path changing the differing coordinates in increasing order. Its length is at most $n$. Fix a directed edge $(u,u^{(k)})$ that flips coordinate $k$. A path uses this edge exactly when $x_k=u_k$, $y_k=-u_k$, $y_j=u_j$ for $j<k$, and $x_j=u_j$ for $j>k$. The coordinates $x_j$ for $j<k$ and $y_j$ for $j>k$ are free. Thus exactly $2^{n-1}$ ordered pairs use it, whileThe unweighted load divided by capacity equals $n$; multiplying by the maximum length yields $\rho \le n^2$. The canonical paths Poincare bound therefore proves the slightly stronger ${\mathrm{Var}}_{\pi }(f)\le n^2\mathcal{E}(f,f)$, and in particular the requestedOne can keep the actual lengths: among these $2^{n-1}$ pairs, coordinate $k$ always differs, and each of the other $n-1$ coordinates differs for half the pairs. Their mean path length is $(n+1)/2$, giving exact congestion $\rho =n(n+1)/2$. This refinement still has order $n^2$.

Define interaction-picture states by $|{\psi }_I(t)⟩=e^{i{H}_{0}t/\hslash }|{\psi }_S(t)⟩$ and $V_I(t)=e^{i{H}_{0}t/\hslash }V(t)e^{-i{H}_{0}t/\hslash }$. Differentiation cancels the free Hamiltonian operator and yieldsAssume the interaction is switched off sufficiently fast in the remote past for the following integral to converge. Starting in $|n⟩$, the first iteration of the integral equation givesfor $m\ne n$. The picture change only multiplies the measured amplitude by a phase. Squaring therefore givesFor the harmonic oscillator, $|1⟩=a^{†}|0⟩$ is normalized and $⟨1|(a+a^{†})|0⟩=1$. The energy difference is $\hslash \omega$, andHence the lowest nonzero transition probability isThe stronger remainder here follows from parity: every insertion of $a+a^{†}$ reverses oscillator parity, so the amplitude between $|0⟩$ and $|1⟩$ has only odd powers of $\lambda$. The general displayed $O({\lambda }^3)$ estimate remains valid.

Conditional on $Y_m=i$, the holding time $V_m=U_m/g(i)$ has an exponential distribution of rate $g(i)$, and after that holding time the next state is $j$ with transition probability $p_{ij}$. The memoryless property of exponential random variables and the Markov property of $(Y_m)$ therefore make $(X_t)$ a continuous-time Markov chain.

Moreover $g(i)<1/\epsilon$, so $V_m\ge \epsilon U_m$. The strong law of large numbers gives ${\sum }_m{U}_m=\infty$ almost surely, hence $J_m\to \infty$ and the process is nonexplosive. Its Q-matrix isEquivalently, if $D_g$ is the diagonal matrix with entries $g(i)$, then

A random time $\tau$ is a stopping time for the natural filtration $({\mathcal{F}}_n)$ of a Markov chain when $\{\tau \le n\}\in {\mathcal{F}}_n$ for every $n$. The Strong Markov property says that, conditionally on $\tau <\infty$ and $X_{\tau }=i$, the process after $\tau$ is a fresh copy of the chain started from $i$, independent of the pre-$\tau$ history.

Before capture define the half-separationFor $D_n=k\ge 2$, independence of the two moves gives the transition probabilitiesIgnore the holding steps. The resulting embedded chain is a biased random walk that moves left with probability $q$ and right with probability $1-q$. Its expected number of moves needed to descend one level is $1/(2q-1)$; this follows either from first-step analysis or from its drift $1-2q$. A non-holding move occurs with probability $1/2$ at each time, so its mean waiting time is two. Hence, for every $m\ge 2$, the expected time to go from separation $2m$ to $2(m-1)$ is

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,

Use the Poisson distribution with probability mass function ${\pi }_n=e^{-\lambda }{\lambda }^n/n!$ on ${\mathbb{N}}_0$. A symmetric nearest-neighbour proposal with a reflecting boundary isAll other entries vanish. In particular, the boundary proposal does not force a move to one: retaining the probability $q(0,0)=1/2$ makes $q(0,1)=q(1,0)$.

The adjacent ratios of the Poisson distribution are ${\pi }_{n+1}/{\pi }_n=\lambda /(n+1)$ and ${\pi }_{n-1}/{\pi }_n=n/\lambda$. The Metropolis–Hastings acceptance probability consequently giveswhere the symbol $u_n$ denotes the upward transition probability. Thus the complete transition matrix iswith no downward entry at zero and all other entries zero. In matrix form,Every row is nonnegative and sums to one. An executable Python step of this Poisson Metropolis chain is:

def poisson_step(n, lam, rng):
    if lam <= 0 or n < 0 or int(n) != n:
        raise ValueError("require lam > 0 and an integer n >= 0")
    y = n + 1 if rng.random() < 0.5 else max(0, n - 1)
    ratio = lam / (n + 1) if y > n else n / lam if y < n else 1.0
    return y if rng.random() < min(1.0, ratio) else n

Supply a pseudorandom number generator such as random.Random(2021) and retain each returned state, including repeats. No normalizing constant or evaluation of a factorial is needed by the algorithm.

For each adjacent pair, the detailed balance equations follows directly:For nonadjacent distinct states both fluxes vanish, and diagonal fluxes agree trivially. Since ${\sum }_n{\pi }_n=e^{-\lambda }{\sum }_n{\lambda }^n/n!=1$, this is a genuine stationary distribution, rather than merely an invariant measure of infinite mass.

Every upward and downward transition probability between neighbouring states is positive for $\lambda >0$. Any two states can therefore be joined by a finite path of positive probability: the Markov chain is irreducible. Moreover,State zero has period one; Markov chain irreducibility makes every state aperiodic. Finally, the standard countable-state recurrence theorem states that an irreducible Markov chain admitting a stationary distribution of total mass one is positive recurrent, with that stationary distribution unique. The countable-state Markov chain convergence theorem then states that Markov chain irreducibility, positive recurrence and aperiodicity imply $P^k(i,j)\to {\pi }_j$ for all states $i,j$. The hypotheses just verified hold on the infinite state space ${\mathbb{N}}_0$, not merely on a numerical truncation.^[ref]

Consequently, the unique stationary distribution and the limiting distribution are both $\mathrm{Poisson}(\lambda )$, from every initial state. The countable-state Markov chain ergodic theorem also gives $N^{-1}{\sum }_{k=1}^{N}f(X_k)\to {\sum }_n{\pi }_{n}f(n)$ almost surely whenever ${\sum }_n{\pi }_n|f(n)|<\infty$.^[ref]

The history $\sigma (S_0,\dots ,S_n)$ is contained in $\sigma (Y_1,\dots ,Y_n)$. The increment $Y_{n+1}$ is independent of this history by the given independence of random variables. ThereforeThis depends only on the current state and not on time. It is precisely the time-homogeneous Markov property, so $S_n$ is a Markov chain with transition probabilitiesThe conditional identity is an identity that holds almost surely; histories of zero probability require no division by their probabilities.

Let $d=\gcd (a,b)>1$ and put $a^{\prime }=a/d$, $b^{\prime }=b/d$. Then $a^{\prime },b^{\prime }$ are coprime integers, and $p=b^{\prime }/(a^{\prime }+b^{\prime })$. The six requested revisions are as follows.

(b)(i): communicating classes. Every increment is divisible by $d$, so the residue class modulo $d$ is preserved. Conversely, if $j-i$ is divisible by $d$, Bézout's identity applied to $a^{\prime },b^{\prime }$ gives a path just as in (b)(i). Thus the original Markov chain is reducible on $\mathbb{Z}$, with exactly the $d$ closed communicating classesThe specified walk starting at zero stays in $d\mathbb{Z}$.

(b)(ii): period of a state in a Markov chain. The return equation $au=bv$ reduces to $a^{\prime }u=b^{\prime }v$. Hence $u=b^{\prime }k$, $v=a^{\prime }k$. For every state, the period of a state in a Markov chain is thereforeThis is still at least two, so no state is aperiodic.

(b)(iii): stationary distribution. The conclusion is unchanged: there is no stationary distribution on $\mathbb{Z}$. The maximum-mass proof in (b)(iii) applies verbatim, without using coprimality. Equivalently, each closed communicating class, after translating and dividing its state labels by $d$, is the same centered random walk with step sizes $a^{\prime },-b^{\prime }$ and has no stationary distribution.

(b)(iv): occupation of zero. On $d\mathbb{Z}$, the rescaled walk $S_n/d$ has exactly the transition probabilities of the coprime walk with parameters $a^{\prime },b^{\prime }$. The return-probability argument in (b)(iv) therefore makes this class null recurrent. The renewal process proof of the occupation limit applies unchanged:Starting in another residue class would make every visit to zero impossible, also giving occupation proportion zero.

(c)(i): modified communicating classes. These are unchanged: $C=\{-c-b,\dots ,c+a\}$ is one finite closed communicating class, and all other states are transient singleton classes. The original proof used consecutive reset destinations and never used coprimality. Resetting to the $b$ consecutive integers $c-b+1,\dots ,c$ connects different residue classes, so it would be incorrect to split $C$ according to residue classes modulo $d$.

(c)(ii): limiting distribution. The column-sum calculation, the irreducible Markov chain property on a finite class, positive self-transition probabilities, and geometric entrance argument also required no coprimality. Thus for every initial state the same convergence in distribution holds:In particular, the denominator $2c+a+b+1$ is unchanged; it is not divided by $d$.

Under the specified evolution rule, conditional on the entire history, the next event's probability distribution depends only on the current population size. This is the time-homogeneous Markov property. Birth and immigration both increase the size by one, while death decreases it by one. Writing $q=1-p$, the transition probabilities on ${\mathbb{Z}}_{\ge 0}$ areand all other entries are zero. In particular $P(0,1)=1$ and $d_0=0$. The row sum is $[kp+\theta +kq]/(k+\theta )=1$. Every neighboring pair communicates, so this is an irreducible Markov chain, specifically a birth-death-immigration jump chain.

Differentiate ${\dot{c}}_2=-i\gamma e^{2i\Omega t}{c}_1$ and substitute the equation for ${\dot{c}}_1$. The result isPutting $c_2=e^{i\Omega t}u$ removes the first derivative and gives $\ddot{u}+{\kappa }^{2}u=0$, where $\kappa =\sqrt{{\Omega }^2+{\gamma }^2}$. For $\kappa >0$ the general solution isThe initial spinor fixes $c_1(0)=1,c_2(0)=0$, and the first-order coupled equation fixes the second initial datum ${\dot{c}}_2(0)=-i\gamma$. Thus $A=0$, $B=-i\gamma /\kappa$, andTheir squared magnitudes sum to one. If both $\Omega$ and $\gamma$ vanish, the initial amplitudes simply remain $(1,0)$; the formulas have that continuous limit.

Source convention correction. For positive $\gamma$, the printed expression for $c_2$ lacks the factor $-i$ required by the stated basis and initial vector. A concrete test is its derivative at $t=0$: the printed expression gives $+\gamma$, whereas the Schrödinger equation gives $-i\gamma$. Multiplying only $c_2$ by a phase changes the relative quantum phase. One may change the overall phase of the entire spinor, but must then transform $c_1$ and its initial amplitude as well. The printed modulus nevertheless gives the correct transition probability.

For nonzero Rabi detuning,The sketch starts at zero and oscillates between zero and ${\gamma }^2/({\gamma }^2+{\Omega }^2)<1$, with period $\pi /\sqrt{{\gamma }^2+{\Omega }^2}$ when the coupling is nonzero. Relative to resonance, the oscillations have a smaller maximum and a shorter period at fixed coupling. The sign of the Rabi detuning changes the amplitudes' phases but leaves this transition probability unchanged. For $|\Omega |\gg |\gamma |$, its maximum is approximately ${\gamma }^2/{\Omega }^2$, so population transfer is strongly suppressed. The plotted detuned example has $|\Omega |=|\gamma |$, giving maximum $1/2$.

Multiply each time-sliced path weight by $\exp [-{\sum }_{j}V(x_j)\Delta t]$. A nonnegative finite $V$ is a local killing rate: the Feynman-Kac weight attenuates paths in proportion to the time spent in regions where $V>0$. Let that rate tend to infinity outside the allowed interval. Paths that exit are removed, leaving Brownian motion killed at its first boundary hit.

Consequently the required kernel is the Gaussian path measure restricted to surviving paths, equivalentlywith the infinite-barrier interpretation stated above. It solves ${\partial }_{t}K=D{\partial }_x^{2}K$ in the interior with absorbing Dirichlet boundary conditions at $\pm L$. A reflecting boundary condition for a diffusion would describe a different process.

Strictly, this is a transition probability density in the final position, jointly with survival; a probability for an interval of final positions includes an integral over $x_f$. Its integral over the whole allowed interval is at most one. It is not the normalized endpoint density conditional on survival, which would require division by the survival probability. In these diffusion units $V$ and the eigenvalues of $H$ have units of inverse time.

A Poisson Metropolis chain samples a Poisson distribution by proposing an upward or downward unit step with equal probability and reflecting a downward proposal at zero into a self-proposal. Its upward and downward transition probabilities are the displayed $u_n,d_n$, with $d_0=0$, and the diagonal entry is $1-u_n-d_n$. For ${\pi }_n=e^{-\lambda }{\lambda }^n/n!$, adjacent stationary fluxes equal $\frac{1}{2}\min ({\pi }_n,{\pi }_{n+1})$. Hence detailed balance holds. Positive adjacent probabilities imply an irreducible Markov chain; $P(0,0)\ge 1/2$ implies aperiodicity. The normalized invariant Poisson distribution implies positive recurrence, so the countable-state Markov chain convergence theorem applies. The boundary proposal probability is essential: forcing every proposal at zero to go upward would require a different acceptance ratio.