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For a continuous-time Markov chain whose jump times tend to infinity, the chain and its jump chain have the same directed paths of positive-probability jumps. They are therefore irreducible simultaneously, and a state is recurrent for one exactly when it is recurrent for the other: the event of returning after departure depends only on the sequence of visited states.

Add independent rate-$\mu$ resets to zero to a linear birth process with immigration. The resulting irreducible continuous-time Markov chain is nonexplosive by domination of births and finiteness of Poisson reset counts. When $\mu =\lambda$, an excursion from one reaches $n$ with probability $2/(n+1)$, so its jump chain is recurrent with infinite expected jump count. The elapsed return-cycle time instead has meanThus the continuous-time chain is positive recurrent while its jump chain is null recurrent. Its stationary probabilities are ${\pi }_0=1/2$ and ${\pi }_n=1/[(n+1)(n+2)]$ for $n\ge 1$.

Let $(A_t,B_t)$ be a Markov jump process on integer populations, with rateswhen $a,b>0$, where $\kappa >0$. Make both axes absorbing, so the fight ends when one population reaches zero. Start at $(N,b_N)$, with $b_N=\lfloor \alpha N\rfloor$; when $\alpha N$ is integral this is the printed initial condition exactly. The harmless rounding convention makes the model defined for every integer $N$.

Use the embedded jump chain $(A_j,B_j)$ rather than physical time. Its probabilities of the two losses are respectively $b/(a+b)$ and $a/(a+b)$, independently of $\kappa$. It reaches an axis after at most $N+b_N\le 2N$ steps; hold it constant thereafter. PutBefore absorption its two increments are $-2a+1$ and $2b-1$, soAfter absorption set ${\mu }_j=0$. Consequentlyis a discrete-time martingale. All populations are at most $N$, so each martingale difference is bounded in absolute value by $2N+2\le 4N$. The total drift up to step $2N$ is at most $2N$. This is the squared-population near-invariant for stochastic attrition.

The two-sided Azuma-Hoeffding inequality states that a martingale with $m$ differences bounded by $C$ satisfies $\mathbb{P}(|M_m|\ge z)\le 2\exp (-z^2/(2m{C}^2))$. Its exponential estimate follows by applying the conditional Hoeffding lemma to each centered difference, iterating, and optimizing the exponential Markov bound. With $m=2N$ and $C=4N$ it gives

Let $c=\sqrt{1-{\alpha }^2}>0$, and let $X^N=A_{2N}/N$ be the final proportion. If the first population survives, then the second is zero and $D_{2N}/N^2=(X^N{)}^2$. Hence $|X^N-c|>\delta$ impliesIf the first population is extinguished, $X^N=0$ and $D_{2N}\le 0$. This can satisfy the indicated deviation event only when $\delta <c$, and then $|D_{2N}-c^2{N}^2|\ge c^2{N}^2>c\delta N^2$. Thus the same implication holds in both cases.

The initial rounding error obeys $|D_0-c^2{N}^2|=|{\alpha }^2{N}^2-b_N^2|\le 2N$. Combining it with the drift bound shows that the deviation event forcesfor all sufficiently large $N$. Set $\eta =c\delta /2$ in the preceding concentration inequality. We concludefor sufficiently large $N$, and thereforeThe symmetric stochastic attrition process thus has exponential concentration of terminal stochastic attrition survivors. The deterministic equations conserve $a^2-b^2$ and predict the same surviving fraction, but the martingale argument supplies the stronger exponential probability estimate required here.

The jump chain records the successive states visited at actual jumps of the continuous-time Markov chain. Put $q_i=-q_{ii}$. For an irreducible chain with at least two states, all $q_i>0$, and its transition probabilities are ${\hat{P}}_{ij}=q_{ij}/q_i$ for $j\ne i$, with ${\hat{P}}_{ii}=0$. The holding time in state $i$ is exponential with mean $1/q_i$.

If $\pi Q=0$, set $C={\sum }_i{\pi }_i{q}_i$ and ${\hat{\pi }}_i={\pi }_i{q}_i/C$. ThenConversely, $\hat{\pi }\hat{P}=\hat{\pi }$ implies that ${\pi }_i=c{\hat{\pi }}_i/q_i$, normalized to sum one, satisfies $\pi Q=0$. ThereforeJump observations overrepresent states with short holding times. A one-state chain has no actual jumps; its trivial equilibrium can be handled separately, without using this division by $q_i=0$.

Write $q_{ij}=c{m}_{ij}$ for $i\ne j$, with a common positive constant $c$, and put $d_i={\sum }_{j\ne i}{m}_{ij}$. The infinitesimal generator matrix is symmetric because the graph is undirected. Thus the uniform distribution ${\pi }_i=1/|I|$ satisfies detailed balance, and the continuous-time random walk is reversible when started uniformly.

Its jump chain chooses a link uniformly from those incident to its current vertex:The flux ${\hat{\pi }}_i{\hat{P}}_{ij}=m_{ij}/{\sum }_j{d}_j$ is symmetric, so this random walk on a multigraph is also reversible. The continuous-time equilibrium is uniform because higher-degree vertices have shorter holding times; at jump times the equilibrium is degree-weighted. If the intended proportionality constant depends on the starting vertex, $q_{ij}=c_i{m}_{ij}$, reversibility still holds with ${\pi }_i\propto 1/c_i$, and the same jump chain results.

Use the equilibrium correspondence from the preceding part. For $i\ne j$,Consequently the detailed balance equations for the jump chain hold exactly when those for the continuous-time Markov chain hold. The reversibility characterizations already given prove the equivalence. Both chains here are started in their respective stationary distributions, which generally differ.

At a positive-rate state the jump chain moves to either other state with probability $1/2$, independently of the holding rate. At a zero-rate state there is no subsequent jump; extend the chain as permanently absorbed there, giving an identity row. This convention is needed to speak of all powers of its transition matrix in the absorbing cases.

When all rates are positive,Its unique equilibrium is uniform and its powers tend to the matrix with all rows $(1/3,1/3,1/3)$. This differs from the holding-time-weighted equilibrium of the continuous-time chain unless the rates coincide.

With one zero-rate state the unique jump-chain equilibrium is its point mass and the limit has all rows that point mass. With two zero-rate states the equilibria are all mixtures on them, and the limiting rows are exactly the absorbing rows and half-half transient row found in part (d).

A state is recurrent if, after departure, its return probability is one. For an irreducible nonexplosive continuous-time Markov chain, this is equivalent to recurrence of its jump chain and is shared by all states.

A state of an irreducible continuous-time Markov chain is transient if, after departure, its return probability is less than one. Equivalently its jump chain visits that state only finitely often almost surely. This is a class property.

A recurrent state is null recurrent if its expected elapsed return-cycle time is infinite. Recurrence itself agrees with that of the jump chain, but positive recurrence need not agree, because large states can have very short holding times.

For this linear birth process with catastrophes, the off-diagonal entries of the Q-matrix arewith all other off-diagonal entries zero. The diagonal entries are $q_{00}=-\lambda$ and $q_{nn}=-[\lambda (n+1)+\mu ]$ for $n\ge 1$. At zero there is no separate catastrophe transition, because it would not change the state.

The chain is irreducible: every positive state can jump to zero, and from zero any finite sequence of births has positive probability. To prove nonexplosion, couple it with the linear birth process with immigration having birth rate $\lambda (n+1)$ and no catastrophes. Its mean population is finite at each finite time, obtained from $m^{\prime }=\lambda (m+1)$, or its exponential holding-time sum diverges since ${\sum }_{n}1/[\lambda (n+1)]=\infty$. Couple births monotonically and implement catastrophes with an independent rate-$\mu$ Poisson clock, resetting the smaller population. On a finite time interval the dominating pure-birth process has finitely many births, and the catastrophe clock finitely many rings. Thus the original chain is nonexplosive.

For its jump chain, the transitions are

When $\mu =\lambda$, an excursion after the jump from zero to one visits $n$ before its reset with probabilityThis tends to zero, so a catastrophe eventually occurs with probability one. Both the original chain and its jump chain are recurrent.

The expected number of jumps in a return cycle isThus the jump chain is null recurrent. But its holding time at $n\ge 1$ has mean $1/[\lambda (n+2)]$, and the initial holding time at zero has mean $1/\lambda$. The expected elapsed cycle time isThe continuous-time Markov chain is therefore positive recurrent. Its equilibrium distribution, computed as mean time in a state per mean cycle time, isThe entries sum to one by telescoping. The contrast is explained by progressively shorter holding times at the large states, not by a change in return probabilities.

A time-homogeneous continuous-time Markov chain on a countable state space is a stochastic process with the Markov property: conditional on the current state, the law of its future depends on neither its past nor the current time. In the usual conservative, nonexplosive setting its sample paths are right-continuous step functions. Its Q-matrix $Q=(q_{ij})$ has $q_{ij}\ge 0$ for $j\ne i$, $q_i=-q_{ii}={\sum }_{j\ne i}{q}_{ij}<\infty$, and infinitesimal transition probabilities ${\mathbb{P}}_i(X(h)=j)=q_{ij}h+o(h)$ for $j\ne i$.

On entering $i$, the holding time has distribution $\mathrm{Exp}(q_i)$; independently, the next state is $j$ with probability $q_{ij}/q_i$. The successive states form the jump chain, whose transition probabilities areIf $q_i=0$, the state is absorbing, its holding time is infinite, and the jump-chain convention is $p_{ii}=1$. Nonexplosion ensures this construction defines the process for all finite times rather than accumulating infinitely many jumps in finite time.

Because $J_m\to \infty$ almost surely, no infinite portion of the jump chain is hidden inside a finite time interval. After either chain leaves a state $i$, the event that it ever returns is therefore the same event for both chains. Hence irreducibility and recurrence pass between a CTMC and its jump chain gives

The jump chain of $(X_t)$ is exactly $(Y_m)$: from $i$ it visits $j$ next with probability $p_{ij}$. Consequently the two chains have exactly the same directed paths of positive-probability transitions. Thus

Let $\nu$ be an invariant distribution for the continuous-time Markov chain. Using $Q=D_g(P-I)$, its stationarity equation $\nu Q=0$ becomesThus the measure with weights ${\nu }_{i}g(i)$ is invariant for the irreducible positive-recurrent jump chain. Its invariant distribution $\pi$ is unique, so ${\nu }_{i}g(i)=c{\pi }_i$. Normalization gives the invariant-measure transfer between a jump chain and a CTMC formulaThe denominator is finite and nonzero because $\epsilon <g(j)<1/\epsilon$. Thus $\nu$ is a probability distribution and $(X_t)$ is positive recurrent.

The total jump rate is ${\lambda }_0$ at zero and ${\lambda }_i(1+{\rho }_i)$ at $i>0$. Hence the jump chain has transition probabilitiesand, for $i>0$,All other entries vanish.

Here the total holding rate isand therefore the jump chain has state-independent transitionsIt is an upward biased random walk on $\mathbb{Z}$. By the strong law of large numbers, $Y_n/n⟶1/3$ almost surely, so $Y_n\to +\infty$. The jump chain, and hence $X$, is transient.

Counting measure is invariant for this jump chain: for every $j$, its incoming mass is $1/3+2/3=1$. The invariant-measure transfer between a jump chain and a CTMC therefore gives an invariant measure proportional to $q_i^{-1}=3^{-|i|-1}$. Sincethe normalized invariant distribution of a continuous-time Markov chain isThis does not contradict transience because an invariant distribution of an explosive chain need not imply positive recurrence.

It remains to verify explosion. Starting from zero, the probability that the biased walk ever reaches $-k$ is $2^{-k}$. Once it reaches any state, its probability of returning after departure isso its expected total number of visits to that state, conditional on reaching it, is $1/(1-2/3)=3$. If $V_i$ is the number of visits to $i$, thenConsequently the expected sum of all holding times isThe nonnegative random variable $T_{\infty }$ is therefore finite almost surely. By explosion by summable holding times, $X$ makes infinitely many jumps by time $T_{\infty }$, so it is explosive.

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,

Put $q_x=-q_{x,x}={\sum }_{y\ne x}{q}_{x,y}$. The jump chain records the successive states visited by $X$ and has transition probabilitiesBy the exponential holding-time construction, conditional on the current state $x$, the destination of the next jump has this distribution independently of the earlier path. Hence the jump chain has the Markov property.