Nonexplosion of Exponential Arrival Sequences

Why Arrival Times Cannot Accumulate

A sequence of positive waiting times can have an infinite sum, so that its arrival times continue without bound. But in principle, positive waiting times can also have a finite sum: then infinitely many arrivals occur before some finite time. This phenomenon is called explosion. For independent exponential waiting times, explosion has probability zero. The key is to find waiting times that are long enough to force the partial sums upward, and to show that such waiting times occur infinitely often.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((\tau_k)_{k\ge1}\) be independent and identically distributed exponential random variables with rate \(\lambda>0\), so \(\mathbb{P}(\tau_k\ge x)=e^{-\lambda x}\) for \(x\ge0\). Define the arrival times by \(S_0=0\) and \(S_n=\sum_{k=1}^n\tau_k\) for \(n\ge1\). The count by time \(t\ge0\) is \(N(t)=\#\{n\ge1:S_n\le t\}\), initially allowing this count to be infinite.

The Long-Gap Argument

Fix a threshold \(c>0\). Each event \(A_k=\{\tau_k\ge c\}\) represents a waiting time that adds at least \(c\) to the cumulative arrival time. The events are independent, and each has probability \(e^{-\lambda c}>0\). We first prove that these events occur infinitely often, without invoking a limit theorem.

Lemma. If \((A_k)_{k\ge1}\) are independent events with the same probability \(p>0\), then with probability one infinitely many \(A_k\) occur.

Proof. Fix an integer \(m\ge1\). For each \(r\ge0\), independence gives \(\mathbb{P}(\bigcap_{k=m}^{m+r}A_k^c)=(1-p)^{r+1}\). These events decrease as \(r\) increases, and their intersection is the event that no \(A_k\) occurs for any \(k\ge m\). By continuity of probability for decreasing events, the probability of that intersection is \(\lim_{r\to\infty}(1-p)^{r+1}=0\). If only finitely many \(A_k\) occur, then there is some \(m\) after which none occur. The event of finitely many occurrences is therefore contained in the countable union, over \(m\ge1\), of events of probability zero. Its probability is zero. \(\square\)

Proof. Use the lemma with \(c=1/\lambda\). The exponential tail formula gives \(\mathbb{P}(\tau_k\ge1/\lambda)=e^{-\lambda/\lambda}=e^{-1}>0\), and the events \(A_k=\{\tau_k\ge1/\lambda\}\) are independent. Thus, almost surely, infinitely many indicators \(\mathbf{1}_{A_k}\) equal \(1\). For each \(n\),

The sum on the right counts how many of the first \(n\) waiting times are at least \(1/\lambda\). Since infinitely many such events occur almost surely, that count tends to infinity as \(n\to\infty\). Because \(1/\lambda>0\), the lower bound tends to infinity, so \(S_n\to\infty\) almost surely. On any outcome with this divergence, and for any finite \(t\), there is an index \(n_0\) such that \(S_n>t\) for all \(n\ge n_0\). Hence only finitely many indices satisfy \(S_n\le t\). The same pathwise argument applies to every finite \(t\), so all bounded intervals contain only finitely many arrivals on that outcome. \(\square\)

The choice \(c=1/\lambda\) makes the probability of a long gap the same, \(e^{-1}\), for every positive rate. It is convenient but not essential: any fixed \(c>0\) gives probability \(e^{-\lambda c}>0\), and the same proof shows that gaps of at least \(c\) occur infinitely often almost surely. In particular, the argument does not require a separate estimate for each observation time \(t\).

Worked Rate Calculations

What the Pathwise Statement Means

For any fixed outcome with \(S_n\to\infty\), the sequence of arrival times eventually exceeds each finite time \(t\). Thus \(N(t)=\#\{n\ge1:S_n\le t\}\) is finite. This is a statement about the entire arrival sequence on an outcome, not just about the count at one selected time. No countable-union argument over possible real values of \(t\) is needed: divergence itself implies finiteness for every finite \(t\) on the same outcome.

The positivity of the rate matters to the argument. For \(\lambda>0\), every chosen threshold \(c>0\) has a strictly positive chance \(e^{-\lambda c}\) of being exceeded. A rate-zero Poisson process is instead the process with no arrivals; it has no positive-rate exponential interarrival sequence to which this proof applies, and it is nonexplosive by definition.

Common mistake. One might compute \(\mathbb{E}[S_n]=n/\lambda\), observe that this tends to infinity, and conclude that \(S_n\to\infty\) almost surely. Diverging expectations alone do not justify almost-sure divergence. For example, let \(Z_n=n^2\) with probability \(1/n\) and \(Z_n=0\) otherwise, independently for \(n\ge1\). Then \(\mathbb{E}[Z_n]=n\), which tends to infinity. Yet \(Z_n=0\) occurs infinitely often almost surely: for each fixed \(m\), the probability that \(Z_n\ne0\) for every \(n\ge m\) is \(\lim_{r\to\infty}\prod_{n=m}^{m+r}(1/n)=0\), and a countable union over \(m\) still has probability zero. Thus \(Z_n\) does not tend to infinity almost surely. The exponential-arrival proof avoids this error by establishing a pathwise lower bound from infinitely many independent long gaps.

The argument also explains the role of independence. It is not enough to know only the marginal probability of a long gap; the proof that long gaps occur infinitely often used the factorization of probabilities for disjoint waiting times. For iid exponential interarrivals, both the positive tail probability and independence are available, so the conclusion follows directly.