Jumps, Holding Times, and Lifetime
A continuous-time birth-death chain can make many jumps in a short interval if its jump rates become large at the states it visits. In an extreme case, infinitely many jumps occur before a finite time. This is called explosion. It is different from reaching an absorbing state, where jumps stop, and from growing to arbitrarily large states only as time tends to infinity. The distinction matters because the usual description of the chain by its states at each finite time must be handled carefully if its jumps accumulate at a finite time.
Let \((X_t)_{t\ge0}\) be a continuous-time birth-death chain on \(S=\{0,1,2,\ldots\}\), defined on a probability space \((\Omega,\mathcal F,\mathbb P)\). At state \(i\), let the birth rate be \(\lambda_i\ge0\) and the death rate be \(\mu_i\ge0\), with \(\mu_0=0\). Assume these rates are finite at each state. Write \(q_i=\lambda_i+\mu_i\) for the total rate of leaving \(i\). The generator has entries \(q_{i,i+1}=\lambda_i\), \(q_{i,i-1}=\mu_i\) when \(i\ge1\), \(q_{ii}=-q_i\), and zero elsewhere; each row sums to zero. If \(q_i=0\), the chain stays at \(i\) forever.
When \(q_i>0\), the standard waiting-time construction gives an exponential holding time with rate \(q_i\). At the end of that holding time, the chain jumps up with probability \(\lambda_i/q_i\) or down with probability \(\mu_i/q_i\). The successive states visited at jump times form the embedded jump chain. Conditional on that sequence of states, the holding times are independent exponentials with the corresponding rates. This construction is the starting point for defining the lifetime; no criterion for deciding explosion from the rates alone is needed here.
For a path with infinitely many jumps, let \(H_n=S_{n+1}-S_n\) be its \(n\)th holding time. Then \(S_m=\sum_{n=0}^{m-1}H_n\), and explosion is exactly the event that this infinite sum is finite. Each individual holding time is positive almost surely when its rate is finite and positive. Explosion, if it occurs, is therefore caused by the accumulation of infinitely many positive holding times, not by a single jump taking zero time.
Why a Finite Set Cannot Contain an Explosion
Proof. Fix a finite set \(K\subseteq S\). If \(q_i=0\) for every \(i\in K\), no jump can occur from a visit to \(K\), so the assertion holds. Otherwise let \(M=\max_{i\in K}q_i>0\). At any visit to \(i\in K\) with \(q_i>0\), the conditional probability that the holding time is at least \(1/M\) is \(\mathbb P(H\ge 1/M\mid X=i)=e^{-q_i/M}\ge e^{-1}\). At a state with \(q_i=0\), the holding time is infinite, which also satisfies the lower bound. Consider the successive visits to \(K\). Given the path up to any such visit, the probability that its holding time is less than \(1/M\) is at most \(1-e^{-1}\). By conditioning successively, the probability that the holding times at the first \(r\) visits are all less than \(1/M\) is at most \((1-e^{-1})^r\). This bound tends to zero as \(r\to\infty\). The same argument applies after any fixed visit number, so the probability that only finitely many visits have holding time at least \(1/M\) is zero: it is a countable union over possible last such visits of events with probability zero. Consequently, infinitely many visits to \(K\) entail infinitely many holding times of at least \(1/M\), and their sum is infinite. Such a path cannot make infinitely many jumps in finite time. \(\square\)
In particular, explosion cannot consist of endlessly jumping back and forth within a fixed finite range. For a birth-death chain, if explosion occurs, the path must eventually leave every fixed finite set and not return to it. This proposition does not say whether explosion occurs: rates that increase at large states may permit very short holding times, but the path must actually reach and continue through those states.
Three Different Lifetime Behaviors
Worked Example: A Pure-Birth Chain That Explodes
Start at \(X_0=0\), set \(\lambda_i=2^i\) and \(\mu_i=0\) for every \(i\ge0\). The generator has \(q_{i,i+1}=2^i\), \(q_{ii}=-2^i\), and all other entries zero, so each row sums to zero. The embedded chain visits \(0,1,2,\ldots\) in order. Its holding times \(H_i\) are independent exponentials with rates \(2^i\). Therefore
The expectation calculation uses monotone convergence on the partial sums of nonnegative holding times. Since \(\mathbb E[\zeta]=2<\infty\), \(\zeta\) is finite almost surely: if \(\mathbb P(\zeta=\infty)>0\), its expectation would be infinite. Thus this chain makes infinitely many births in finite time with probability one. At each jump it moves to the next larger state, so its population tends to infinity along the jump sequence. The finite value \(2\) is the mean lifetime, not a deterministic explosion time. As a check, the \(m\)-jump partial lifetime has mean \(\sum_{i=0}^{m-1}2^{-i}=2(1-2^{-m})\), which increases to \(2\).
Worked Example: Unbounded Growth Without Explosion
Now start at \(X_0=0\) and set \(\lambda_i=1\), \(\mu_i=0\) for every \(i\ge0\). The generator has \(q_{i,i+1}=1\), \(q_{ii}=-1\), and zero elsewhere, again giving row sum \(0\) at every state. The chain visits the states in order and has independent holding times \(H_i\) with rate \(1\). In particular, \(\mathbb P(H_i\ge1)=e^{-1}\) for every \(i\).
For any fixed \(N\), independence gives \(\mathbb P(H_i<1\text{ for }N\le i\le N+r)=(1-e^{-1})^{r+1}\), which tends to zero as \(r\to\infty\). Thus the probability that \(H_i<1\) for every \(i\ge N\) is zero. Taking the countable union over \(N\), with probability one infinitely many holding times are at least \(1\). Their sum is infinite, so \(\zeta=\infty\) almost surely. Nevertheless, the chain reaches every state: after \(n\) jumps it is at state \(n\), and these jump times tend to infinity. The population grows without bound, but not in finite time. This directly contrasts with the preceding example, where the rates increase with the state.
Check: The probability of no holding time of at least \(1\) among \(H_N,\ldots,H_{N+r}\) is exactly \((1-e^{-1})^{r+1}\); it tends to zero, not to a positive probability of an explosion.
Worked Example: Absorption Is Not Explosion
Let \(X_0=3\), set \(\lambda_i=0\) for all \(i\), and take \(\mu_1=1\), \(\mu_2=2\), \(\mu_3=3\), and \(\mu_0=0\). For states \(1,2,3\), the generator has \(q_{i,i-1}=\mu_i\), \(q_{ii}=-\mu_i\), and all other entries zero; at state \(0\), its row is all zero. These rows each sum to zero. The chain makes exactly three jumps before reaching the absorbing state \(0\). Its holding times before those jumps have rates \(3,2,1\), respectively. Hence the absorption time \(T_{\{0\}}\) is finite almost surely and has mean
The lifetime is still \(\zeta=\infty\), because the chain stays at \(0\) after absorption and makes no further jumps. The sum \(11/6\) is the mean time to absorption, not the lifetime before explosion. A path with finitely many jumps followed by permanent residence in a state is not explosive.
Check: There are three holding times, with means \(1/3\), \(1/2\), and \(1\). Their sum is \((2+3+6)/6=11/6\), as stated; after these three jumps the rate at state \(0\) is \(q_0=0\).
Reading a Lifetime Calculation Correctly
The holding-time representation separates two questions. First, what sequence of states does the embedded jump chain visit? Second, how much calendar time does the chain spend at those states? For a pure-birth chain started at \(i_0\), the state sequence is deterministic, so the lifetime is the sum of independent exponential holding times with rates \(\lambda_{i_0},\lambda_{i_0+1},\ldots\), provided the chain makes infinitely many jumps. For a general birth-death chain, the visited states are random, and the holding-time rates depend on that path. A calculation based only on how high the chain can go does not settle whether its jump times accumulate.
Absorption changes the number of jumps, not the definition of explosion. Once the chain reaches a state with \(q_i=0\), it stays there and its lifetime is infinite under the convention used here. By contrast, at an explosive lifetime the chain has made infinitely many jumps by time \(\zeta<\infty\); the state in \(S\) after that time is not supplied by the jump construction, so an extension to a cemetery state is a convention for recording what happens beyond the lifetime.
Common mistake. A tempting but incorrect argument for the constant-rate pure-birth example is: “The chain makes infinitely many jumps, so it must explode.” Infinitely many jumps alone do not imply a finite sum of holding times. Here each holding time has mean \(1\), and the calculation above shows that infinitely many are at least \(1\) almost surely; therefore their sum is infinite. The correction is to examine the accumulated holding times, not just the number of jumps or the fact that the population becomes arbitrarily large.
The next question is when the holding-time sum is finite for a pure-birth process with general rates. That requires a criterion involving the whole sequence of rates; the examples here illustrate lifetime behavior without asserting such a general test.
Check Your Understanding
For each question, distinguish carefully between a finite absorption time and a finite explosion time.
- State the definition of explosion in terms of jump times, and explain what \(\zeta<\infty\) means.
- In the pure-birth example with rates \(\lambda_i=2^i\), write the holding-time series and compute its expected value.
- Why does reaching arbitrarily large states not by itself imply explosion? Use the constant-rate example.
- In the pure-death example, how many jumps occur before absorption, and why is the lifetime nevertheless infinite?
- Explain why a path that visits one fixed finite set infinitely often cannot make infinitely many jumps in finite time when all rates at states are finite.