The Explosion Criterion For Pure-Birth Processes

Holding Times and Explosion

A pure-birth process moves through the states in increasing order, making exactly one upward jump at each transition. Explosion means that it makes infinitely many such jumps in a finite amount of time. The key question is therefore whether the successive holding times have a finite total. Since the holding time in state $n$ has mean $1/\lambda_n$, it is tempting to compare explosion directly with the sum of these means. A proof is needed: an infinite expected total alone does not establish that the total is infinite almost surely.

Work on a probability space $(\Omega,\mathcal F,\mathbb P)$ supporting independent random variables $(H_n)_{n\ge0}$, where $H_n$ has the exponential distribution with rate $\lambda_n$, and $0<\lambda_n<\infty$. Start the process in state $0$. Its state space before explosion is $S=\{0,1,2,\ldots\}$, and its time index is continuous, $t\ge0$. Define the successive jump times by $S_0=0$ and $S_{k+1}=\sum_{n=0}^{k}H_n$. The process is in state $k$ for $S_k\le t<S_{k+1}$. Its lifetime, or explosion time, is $T_\infty=\lim_{k\to\infty}S_k=\sum_{n=0}^\infty H_n$, with the sum allowed to be infinite. The natural filtration is $\mathcal F_t=\sigma(X_s:0\le s\le t)$ before the lifetime.

The corresponding pure-birth rates are $q_{n,n+1}=\lambda_n$, $q_{n,n}=-\lambda_n$, and $q_{n,j}=0$ for other $j$. For each row $n$, the generator entries sum to $-\lambda_n+\lambda_n=0$. The holding-time construction specifies the minimal process up to its lifetime; if $T_\infty<\infty$, it has made infinitely many jumps by that time.

Definition: The pure-birth process is nonexplosive if $\mathbb P(T_\infty=\infty)=1$. It is explosive if $\mathbb P(T_\infty<\infty)=1$. In the positive-rate pure-birth setting considered here, the criterion below shows that one of these alternatives holds.

A Criterion for Sums of Exponential Holding Times

Proposition: Let $(H_n)_{n\ge0}$ be independent exponential random variables with rates $0<\lambda_n<\infty$. Then $$ \sum_{n=0}^\infty H_n=\infty\ \text{almost surely} \quad\Longleftrightarrow\quad \sum_{n=0}^\infty \frac{1}{\lambda_n}=\infty. $$ If the deterministic sum on the right is finite, the random sum on the left is finite almost surely.

Proof. Write $R_N=\sum_{n=0}^N H_n$ and $R=\lim_{N\to\infty}R_N$, allowing $R=\infty$. First suppose that $\sum_n1/\lambda_n<\infty$. Since each exponential variable has mean $\mathbb E[H_n]=1/\lambda_n$, monotone convergence gives $$ \mathbb E[R]=\lim_{N\to\infty}\mathbb E[R_N] =\lim_{N\to\infty}\sum_{n=0}^N\frac{1}{\lambda_n} =\sum_{n=0}^\infty\frac{1}{\lambda_n}<\infty. $$ A nonnegative extended random variable that is infinite with positive probability has infinite expectation: if $\mathbb P(R=\infty)>0$, then $\mathbb E[R]\ge K\mathbb P(R=\infty)$ for every $K>0$, which contradicts finiteness as $K\to\infty$. Thus $R<\infty$ almost surely.

Now suppose $\sum_n1/\lambda_n=\infty$. For an exponential variable of rate $\lambda_n$, direct integration gives $\mathbb E[e^{-H_n}]=\int_0^\infty e^{-x}\lambda_ne^{-\lambda_nx}\,dx=\lambda_n/(\lambda_n+1)$. Independence therefore gives $$ \mathbb E[e^{-R_N}] =\prod_{n=0}^N\frac{\lambda_n}{\lambda_n+1} =\exp\left(-\sum_{n=0}^N\log\left(1+\frac{1}{\lambda_n}\right)\right). $$ For $x>0$, $\log(1+x)=\int_0^x(1+t)^{-1}\,dt\ge x/(1+x)$. Hence $\log(1+x)\ge x/2$ when $0<x\le1$; and when $x\ge1$, $\log(1+x)\ge\log2\ge1/2$, since $\log2=\int_0^1(1+t)^{-1}\,dt\ge1/2$. Thus $\log(1+x)\ge\frac12\min(x,1)$. If $\sum_n1/\lambda_n$ diverges, then $\sum_n\min(1/\lambda_n,1)$ diverges: either infinitely many terms $1/\lambda_n\ge1$, or eventually every term is less than $1$, in which case the two tails have the same terms. It follows that $\sum_n\log(1+1/\lambda_n)=\infty$ and therefore $\mathbb E[e^{-R_N}]\to0$. Because $R_N$ increases to $R$, $e^{-R_N}$ decreases to $e^{-R}$, with the convention $e^{-\infty}=0$. Dominated convergence now gives $\mathbb E[e^{-R}]=0$. Since $e^{-R}\ge0$, this implies $e^{-R}=0$ almost surely, or $R=\infty$ almost surely. Both implications are proved. $\square$

The Pure-Birth Explosion Criterion

Theorem: For a pure-birth process started in state $0$, with independent exponential holding times of rates $\lambda_n\in(0,\infty)$, $$ \mathbb P(T_\infty=\infty)=1 \quad\Longleftrightarrow\quad \sum_{n=0}^\infty\frac{1}{\lambda_n}=\infty. $$ If $\sum_{n=0}^\infty1/\lambda_n<\infty$, then $\mathbb P(T_\infty<\infty)=1$.

Proof. In the holding-time construction, $T_\infty=\sum_{n=0}^\infty H_n$. The holding times are independent exponentials with the rates specified by the process. Applying the proposition to this sequence gives the stated alternatives. In particular, divergence of the reciprocal-rate sum rules out explosion almost surely, while convergence makes the lifetime finite almost surely. $\square$

The criterion is unchanged if the process starts in any fixed state $m\in\{0,1,2,\ldots\}$. In that case $T_\infty=\sum_{n=m}^\infty H_n$. Removing the finitely many terms with indices $0,\ldots,m-1$ cannot change whether a series of positive terms converges or diverges. Thus the test is determined by the tail of the rate sequence, not by the particular finite starting state.

Worked Applications

Worked Example: Constant Birth Rate

Let $\lambda_n=1$ for every $n\ge0$, and start in state $0$. For each state $n$, the generator has entries $q_{n,n}=-1$ and $q_{n,n+1}=1$, with all other entries zero; the row sum is $-1+1=0$. The reciprocal-rate partial sum through state $N$ is $$ \sum_{n=0}^N\frac{1}{\lambda_n} =\sum_{n=0}^N1=N+1. $$ As $N\to\infty$, this diverges. The criterion gives $T_\infty=\infty$ almost surely, so this process is nonexplosive.

Check: The holding-time means are all $1$, and their partial sums of means equal $N+1$, as obtained directly from the rate formula. The conclusion is almost-sure divergence by the proposition, not merely a statement about the expectation of the lifetime.

Worked Example: Exponentially Increasing Rates

Let $\lambda_n=2^{n+1}$. In state $n$, the only nonzero generator entries are $q_{n,n}=-2^{n+1}$ and $q_{n,n+1}=2^{n+1}$, whose sum is zero. The reciprocal-rate partial sum is a finite geometric sum: $$ \sum_{n=0}^N\frac{1}{2^{n+1}} =1-\frac{1}{2^{N+1}}. $$ Indeed, multiplying the sum by $2$ and subtracting the original sum leaves $1-2^{-(N+1)}$. As $N\to\infty$, the sum tends to $1$. The criterion implies that the process explodes almost surely.

The expected lifetime can also be checked directly. For each $N$, $\mathbb E[\sum_{n=0}^N H_n]=1-2^{-(N+1)}$; by monotone convergence, $\mathbb E[T_\infty]=1$. This finite expectation is consistent with almost-sure explosion. It does not say that explosion occurs before every fixed positive time: it says the random lifetime is finite with probability one.

Check: At state $0$, the rate is $2^{0+1}=2$, so the generator entries are $-2$ and $2$, and the holding-time mean is $1/2$, matching the first term of the reciprocal-rate sum.

Worked Example: Quadratically Increasing Rates

Let $\lambda_n=(n+1)(n+2)$. At state $n$, the generator entries are $q_{n,n}=-(n+1)(n+2)$ and $q_{n,n+1}=(n+1)(n+2)$, so every row sums to zero. The reciprocal rate has the partial-fraction form $$ \frac{1}{(n+1)(n+2)} =\frac{1}{n+1}-\frac{1}{n+2}. $$ Consequently, $$ \sum_{n=0}^N\frac{1}{(n+1)(n+2)} =\sum_{n=0}^N\left(\frac{1}{n+1}-\frac{1}{n+2}\right) =1-\frac{1}{N+2}. $$ The limit is $1$, so the process explodes almost surely, and monotone convergence gives $\mathbb E[T_\infty]=1$.

Check: The first two rates are $\lambda_0=(1)(2)=2$ and $\lambda_1=(2)(3)=6$. Their reciprocal contributions are $1/2$ and $1/6$, and the displayed partial-sum formula at $N=1$ gives $1-1/3=2/3=1/2+1/6$.

Interpreting the Test Correctly

The criterion compares the rates through their reciprocals because $1/\lambda_n$ is the mean holding time in state $n$. Large rates make individual holding times short, but explosion depends on the accumulated time over the entire sequence of states. A sequence of rapidly shrinking mean holding times can still have an infinite total; the constant-rate example illustrates the other side, where each holding time has the same positive mean and the total diverges.

Common mistake. A tempting but incorrect test is to add the rates and declare explosion when $\sum_n\lambda_n=\infty$. For the constant-rate example, $\sum_{n=0}^N\lambda_n=\sum_{n=0}^N1=N+1$ diverges, so that incorrect rule would predict explosion. But $\sum_{n=0}^N1/\lambda_n=N+1$ also diverges, and the proved criterion says the process is nonexplosive almost surely. The rates themselves are not the holding-time means: those means are their reciprocals.

A second logical pitfall is to use only $\mathbb E[T_\infty]=\sum_n1/\lambda_n$ and infer almost-sure behavior from the expectation. A finite expectation does imply $T_\infty<\infty$ almost surely for a nonnegative random variable, as used in the proof. An infinite expectation, however, does not by itself imply $T_\infty=\infty$ almost surely. The divergent-sum direction required the independence and Laplace-transform argument above. That distinction is why the reciprocal-rate criterion is stronger than a comparison of expected holding times alone.

Check Your Understanding

For each rate sequence, first evaluate the reciprocal-rate series, then state what the criterion implies.

If $\lambda_n=3$ for all $n\ge0$, is the process explosive or nonexplosive?
If $\lambda_n=3^{n+1}$, compute the infinite reciprocal-rate sum and determine the almost-sure behavior of $T_\infty$.
For $\lambda_n=(n+1)(n+3)$, write $1/\lambda_n$ as a difference of two fractions and decide whether the process explodes.
Why does $\mathbb E[T_\infty]<\infty$ imply $T_\infty<\infty$ almost surely?
Explain why knowing only that $\mathbb E[T_\infty]=\infty$ is not the proof of the divergent-sum direction of the criterion.