When Infinitely Many Jumps Fit into Finite Time
A continuous-time Markov chain can move through infinitely many states without making infinitely many jumps in a finite interval. The distinction is the time spent between jumps. If the holding times shrink quickly enough along a path, their infinite sum can be finite: the jump times then accumulate at a finite time. This phenomenon is called explosion.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with a countable state space \(S\). Consider a jump path starting from \(J_0\), where \(J_n\) is the state occupied after \(n\) jumps. At a state \(i\) with finite positive exit rate \(q_i\), the holding time has the exponential distribution with rate \(q_i\), so its mean is \(1/q_i\). Given the jump chain, write \(H_n\) for the holding time at \(J_n\). The successive jump times are the partial sums \(\sigma_0=0\) and \(\sigma_n=\sum_{k=0}^{n-1}H_k\) for \(n\ge1\).
For a path that continues to jump, \(H_n>0\) almost surely, so the jump times increase. The question is whether their limit is finite. In the rate-table description, a valid row of a generator \(Q=(q_{ij})\) has \(q_{ij}\ge0\) for \(j\ne i\), \(q_{ii}=-q_i\), and \(\sum_jq_{ij}=0\); when \(q_i>0\), the holding time at \(i\) is exponential with rate \(q_i\). The embedded jump chain chooses its next state according to the off-diagonal rates. Thus the rates affect explosion by setting the sizes of the holding times along the realized jump path.
Two Useful Pathwise Results
Proof. By definition, the \(n\)th jump time is \(\sigma_n=\sum_{k=0}^{n-1}H_k\). These partial sums are increasing because each \(H_k>0\). Their limit is the infinite series \(\sum_{k=0}^{\infty}H_k\), with infinity allowed. Infinitely many jumps occur by a finite time exactly when this limit is finite. Therefore the path is explosive exactly when the series is finite. \(\square\)
This equivalence identifies the issue but does not, by itself, decide whether the sum is finite. The next criterion gives a direct sufficient test. It uses the usual path construction from holding times: conditional on the jump chain, \(H_n\) has an exponential distribution with rate \(q_{J_n}\).
Proof. Conditional on the jump chain, the mean of \(H_n\) is \(1/q_{J_n}\), since \(H_n\) is exponential with that rate. Because the terms are nonnegative, the expectation of their infinite sum is the sum of their expectations (the monotone convergence theorem). Consequently, \(\mathbb{E}\zeta=\sum_{n=0}^{\infty}\mathbb{E}[1/q_{J_n}]\), interpreted as an extended expectation. If the right-hand side is finite, then \(\zeta<\infty\) almost surely: a nonnegative extended random variable that equals infinity with positive probability has infinite expectation. The path has infinitely many jumps by assumption, so finite \(\zeta\) means it is explosive. If \(\mathbb{E}\zeta<\infty\), the same argument directly gives \(\zeta<\infty\) almost surely. \(\square\)
A complementary sufficient condition rules out explosion. It applies when rates are bounded above along every possible path, not just on average.
Proof. Use independent unit-rate exponential variables \(E_0,E_1,\ldots\) to represent the holding times along the path: \(H_n=E_n/q_{J_n}\). Since \(q_{J_n}\le C\), we have \(H_n\ge E_n/C\). Each \(E_n\) satisfies \(\mathbb{P}(E_n\ge1)=e^{-1}>0\). For any fixed \(m\), the probability that \(E_n<1\) for every \(n\ge m\) is \(\lim_{r\to\infty}(1-e^{-1})^{r+1}=0\), by independence. Taking the countable union over \(m\) shows that \(E_n\ge1\) occurs infinitely often almost surely. Each such term contributes at least \(1/C\) to the sum of holding times. Therefore the sum diverges almost surely, and the path is not explosive. \(\square\)
Worked Examples
Worked Example: Rates That Increase Geometrically
Consider the countable state space \(S=\{1,2,\ldots\}\), with every jump from \(n\) going to \(n+1\). Let the exit rate at \(n\) be \(q_n=2^n\). The generator row at state \(n\) has \(q_{nn}=-2^n\), \(q_{n,n+1}=2^n\), and all other entries zero. Its row sum is \(-2^n+2^n=0\); its off-diagonal entries are nonnegative.
Starting at state \(1\), the holding time \(H_n\) at state \(n\) has mean \(1/2^n\). Since the jump chain moves upward at every jump, the expected accumulation time is
The geometric sum equals \(1\), so the expected accumulation time is finite. The finite-expected-sum theorem implies \(\zeta<\infty\) almost surely: this path makes infinitely many jumps in finite time. Check: the rate at state \(n\) is \(2^n\), so the corresponding mean holding time is \(1/q_n=1/2^n\), as used in the sum.
Worked Example: A Countable Path with Bounded Rates
On \(S=\{1,2,\ldots\}\), again let every jump go from \(n\) to \(n+1\), but give every state exit rate \(3\). The row at each state has diagonal entry \(-3\) and next-state entry \(3\), so each generator row sums to \(-3+3=0\).
The holding times can be represented as \(H_n=E_n/3\), where the \(E_n\) are independent unit-rate exponential variables. If \(E_n\ge1\), then \(H_n\ge1/3\). The events \(E_n\ge1\) occur infinitely often almost surely, as shown in the bounded-rate proposition. Therefore the sum of holding times is infinite almost surely, even though the chain visits the successive states \(1,2,3,\ldots\).
Check: there are infinitely many terms at least \(1/3\) almost surely, so their partial sums are unbounded. Thus the jump times do not accumulate at any finite time. Visiting infinitely many states is not, on its own, evidence of explosion.
Worked Example: Infinite Expected Time but Positive Explosion Probability
This example shows why an infinite expected accumulation time does not prove that explosion has probability zero. Take states \(0\), \(F_n\), and \(L_n\) for \(n\ge1\). From \(0\), jump to \(F_1\) or \(L_1\) with probability \(1/2\) each, at total rate \(1\). Along the \(F\)-states, jump from \(F_n\) to \(F_{n+1}\) at rate \(2^n\); along the \(L\)-states, jump from \(L_n\) to \(L_{n+1}\) at rate \(1/n\). The row at \(0\) sums to \(-1+1/2+1/2=0\); the \(F_n\) row sums to \(-2^n+2^n=0\); and the \(L_n\) row sums to \(-1/n+1/n=0\). All off-diagonal rates are nonnegative.
The initial holding time has mean \(1\). Conditional on choosing the \(F\)-branch, the expected total holding time after that initial jump is \(\sum_{n=1}^{\infty}1/2^n=1\). That branch is therefore explosive almost surely by the finite-expected-sum theorem. Conditional on choosing the \(L\)-branch, write the holding time at \(L_n\) as \(nE_n\), where \(E_n\) is unit-rate exponential. Since \(nE_n\ge E_n\) for every \(n\ge1\), and \(E_n\ge1\) infinitely often almost surely, the holding-time sum on the \(L\)-branch is infinite almost surely. The explosion probability from state \(0\) is consequently \(\tfrac12\).
Nevertheless, \(\mathbb{E}\zeta=\infty\). To see this, the \(L\)-branch is chosen with probability \(1/2\), and the expected sum of its holding times is \(\sum_{n=1}^{\infty}n=\infty\). Its contribution to the unconditional expected accumulation time is therefore infinite. Check: the probability of explosion is strictly positive, \(1/2\), even though the expected accumulation time is infinite. An infinite expectation alone cannot distinguish explosion almost surely from a mixture of explosive and nonexplosive paths.
Common Pitfall: Inferring Probability from an Expectation
A tempting but incorrect calculation is: “\(\mathbb{E}\zeta=\infty\), so \(\zeta=\infty\) almost surely and there is no explosion.” The branching example disproves that inference: its expected accumulation time is infinite, but it explodes with probability \(1/2\). An infinite expectation can result from rare large values, or from a mixture of paths with finite and infinite accumulation times. By contrast, a finite expectation of a nonnegative accumulation time does imply that it is finite almost surely. The direction of the implication matters.
To test a proposed path for explosion, first determine whether it makes infinitely many jumps. Then add the holding times along that path. If their sum has finite expectation, the path explodes almost surely; if its rates are uniformly bounded, the path does not explode. When neither criterion applies, neither conclusion follows from these tests alone: the holding-time sum must be analyzed more directly.
Check Your Understanding
Use the definitions and criteria above to answer the following questions.
- A path makes infinitely many jumps, and \(\sum_{n=0}^{\infty}H_n=\tfrac72\). Is it explosive? State the defining condition.
- For deterministic jumps \(n\to n+1\) with rates \(q_n=4^n\), compute the expected accumulation time starting at \(1\), and state what the finite-expectation criterion implies.
- Why does visiting infinitely many distinct states not, by itself, imply explosion?
- In the branching example, which branch is explosive, and what is the explosion probability from state \(0\)?
- Give the error in the claim that an infinite expected accumulation time proves almost-sure nonexplosion.