The Cemetery State After Explosion

Why Add a Cemetery State?

A continuous-time chain on a countable state space can make infinitely many jumps before a finite time. In that case the usual jump-by-jump construction specifies the path only up to the accumulation time: there is no final ordinary state to occupy after the jumps accumulate. Adjoining a cemetery state gives a precise convention for the rest of the path. It also makes explicit the difference between a chain that explodes and one that simply stays forever in an ordinary absorbing state.

Let \(S\) be a finite or countable state space and let \(Q=(q_{ij})\) be a rate matrix: for \(i\ne j\), \(q_{ij}\ge0\), each exit rate \(q_i=\sum_{j\ne i}q_{ij}\) is finite, and \(q_{ii}=-q_i\). These conditions make each row sum to zero. The minimal jump construction uses, when the chain is at \(i\) with \(q_i>0\), an exponential holding time with rate \(q_i\), followed by a jump to \(j\ne i\) with probability \(q_{ij}/q_i\). When \(q_i=0\), the chain stays at \(i\) forever. In what follows, all random variables are defined on a probability space \((\Omega,\mathcal F,\mathbb P)\).

The interval convention includes the jump time \(S_n\) in the new state \(Y_n\), so paths are right-continuous. At explosion there is no last \(Y_n\) to assign: \(X_\zeta\) is defined to be \(\Delta\). The extended process takes values in \(S\cup\{\Delta\}\). If \(\zeta=\infty\), it never visits \(\Delta\). This is the cemetery convention for the minimal process; it does not assert that every chain explodes or that explosion is an ordinary jump from some particular state.

Explosion Time and the Markov Property

Let \((\mathcal F_t)_{t\ge0}\) be the natural filtration of the constructed path, containing the states and jumps observed up to time \(t\). A random time \(\tau\) is a stopping time for this filtration if \(\{\tau\le t\}\in\mathcal F_t\) for every \(t\ge0\). The explosion time has this property: each \(S_n\) is the time of the \(n\)th jump, hence a stopping time, and \(\{\zeta\le t\}=\bigcap_{n\ge1}\{S_n\le t\}\). This identifies explosion using observations available by time \(t\).

Proof. The \(n\)th jump time \(S_n\) is determined by the path up to that jump; thus \(\{S_n\le t\}\in\mathcal F_t\). Since \(S_n\) is nondecreasing and \(\zeta=\lim_n S_n\), the event \(\{\zeta\le t\}\) equals \(\bigcap_{n\ge1}\{S_n\le t\}\), which belongs to \(\mathcal F_t\). This proves the stopping-time assertion.

For the Markov assertion, first consider the event \(\{X_t=\Delta\}\). On this event \(\zeta\le t\), and the process stays at \(\Delta\), so its future is determined by \(X_t\). On \(\{X_t\in S\}\), there is a finite index \(n\) such that \(S_n\le t<S_{n+1}\) and \(X_t=Y_n\). Condition on the history through \(S_n\), including \(Y_n=i\), and on the elapsed holding time \(a=t-S_n\) and the fact that \(H_n>a\). The holding time at \(i\) is exponential with rate \(q_i\). Its memoryless property gives, for \(u,v\ge0\), \(\mathbb P(H_n-a>u+v\mid H_n>a+u)=e^{-q_i v}=\mathbb P(H_n>v)\). Thus the residual holding time has the original exponential law, independent of the elapsed time. The next destination has probabilities \(q_{ij}/q_i\), and the unused holding times and jump choices follow the same construction from that destination. Hence, conditional on the observed history, the future has exactly the law of a fresh construction started at \(i=X_t\). If \(q_i=0\), the process stays at \(i\), which also depends only on its current state. Combining the live and cemetery cases gives the stated conditional-expectation identity. \(\square\)

For \(i,j\in S\), define the minimal transition probabilities by \(p_t^{\min}(i,j)=\mathbb P_i(X_t=j,\ t<\zeta)\). Their row sums need not equal one: \(\sum_{j\in S}p_t^{\min}(i,j)=\mathbb P_i(\zeta>t)\). The extended transition kernel restores total probability by assigning the missing mass to \(\Delta\): \[ \overline p_t(i,j)=p_t^{\min}(i,j),\qquad \overline p_t(i,\Delta)=\mathbb P_i(\zeta\le t),\qquad \overline p_t(\Delta,\Delta)=1. \] Thus every row sums to one. The Markov property just proved implies the Chapman–Kolmogorov relation for this extended kernel: \(\overline p_{t+s}(i,k)=\sum_{\ell\in S\cup\{\Delta\}}\overline p_t(i,\ell)\overline p_s(\ell,k)\). The cemetery state records probability that would otherwise be missing from the minimal transition probabilities.

Worked Constructions

What the Cemetery Convention Does—and Does Not—Say

After extension, the process is an ordinary probability-valued Markov process on \(S\cup\{\Delta\}\); before extension, the minimal transition probabilities on \(S\) may be substochastic because some probability has exploded by time \(t\). This is useful when stating transition probabilities or conditioning on the state at a later time: the cemetery state makes the outcomes exhaustive without pretending that the chain has a final ordinary state.

The convention does not add a direct jump rate from each ordinary state to \(\Delta\). At state \(i\), the prescribed finite exit rate \(q_i\) governs the next ordinary jump, and its destination probabilities are \(q_{ij}/q_i\) when \(q_i>0\). Explosion can result from infinitely many such jumps and their shrinking holding times, not from a single ordinary jump to a state “at infinity.” The cemetery state is an assignment at the accumulation time, not a new jump destination in the rate table.

Common mistake. In the explosive pure-birth example, one might try to add \(q_{n,\Delta}=2^n\) while retaining \(q_{n,n+1}=2^n\) and \(q_{nn}=-2^n\). The resulting row total would be \(-2^n+2^n+2^n=2^n\ne0\), so it would not be a rate-matrix row; it would also give a direct cemetery jump at every visit to \(n\), which is not the construction. The correction is to keep \(q_{n,n+1}=2^n\), \(q_{nn}=-2^n\), and \(q_{n,\Delta}=0\) as ordinary rate data, then set \(X_t=\Delta\) when \(t\ge\zeta\). This preserves the specified rates and records explosion at the time it actually occurs.