Arithmetic and Nonarithmetic Interarrival Distributions

Why the Interarrival Law’s Lattice Structure Matters

Renewal limits can depend on whether interarrival times lie on a common lattice. If every interarrival is a whole-number multiple of a fixed span, renewal epochs are confined to that same lattice. Otherwise, the epochs are not restricted to any such grid. This distinction is not the same as “discrete versus continuous”: a law can have atoms and still be nonarithmetic.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((X_k)_{k\ge1}\) be independent, identically distributed, strictly positive real-valued interarrivals with common distribution \(F\). Write \(S_n=\sum_{k=1}^n X_k\) for the \(n\)th renewal epoch and \(N(t)=\max\{n\ge0:S_n\le t\}\) for the renewal count, where \(S_0=0\). Arithmeticity is a property of the law of one interarrival \(X\), and does not require a finite mean. When discussing long-run renewal rates, the usual finite-mean condition is \(0<\mu=\mathbb{E}[X]<\infty\); the classification below does not itself depend on that condition.

The maximal span is the coarsest lattice that contains all possible interarrivals, up to events of probability zero. A law may have many spans: if \(h\) is a span, then \(h/r\) is also a span for every positive integer \(r\). The maximal span records the largest one, rather than an arbitrary choice of grid.

A Characteristic-Function Test

For a real-valued random variable \(X\), its characteristic function is \(\phi(\theta)=\mathbb{E}[e^{i\theta X}]\), for real \(\theta\). The following criterion makes the lattice condition testable through this function.

Proof. Suppose first that \(\phi(\theta)=1\) for some \(\theta>0\). Since \(|e^{i\theta X}|=1\),

The random variable inside the expectation is nonnegative, so it equals zero almost surely. Thus \(e^{i\theta X}=1\) almost surely, which means \(\theta X\in2\pi\mathbb{Z}\) almost surely. Therefore \(X\in(2\pi/\theta)\mathbb{Z}\) almost surely and the law is arithmetic. Conversely, if \(X\in h\mathbb{Z}\) almost surely, then \(e^{i(2\pi/h)X}=1\) almost surely, so \(\phi(2\pi/h)=1\).

For the statement about the maximal span, choose any span \(h\), and write \(X=hY\), where \(Y\) is a positive integer-valued random variable. The greatest common divisor \(g\) of the possible integer values of \(Y\) exists: the gcds of finite subsets are positive integers, and their smallest value is attained. The integer gcd property implies that \(Y/g\) is integer-valued almost surely, so \(d=hg\) is a span. It is maximal: if \(h'\) is any other span, then the values \(hg\) are integer combinations of finitely many possible values of \(X\) (the finite-subset gcd property). Each such value is a multiple of \(h'\), so \(hg/h'\) is an integer and \(h'\le hg=d\). Finally, writing \(X=dZ\), the integer values of \(Z\) have gcd \(1\). If \(\phi(\theta)=1\), then \(\theta dZ\in2\pi\mathbb{Z}\) almost surely. The gcd-\(1\) property gives \(\theta d/(2\pi)\in\mathbb{Z}\); hence \(\theta=2\pi r/d\) for a positive integer \(r\). Each such \(\theta\) does give \(\phi(\theta)=1\), since \(X\in d\mathbb{Z}\). \(\square\)

This criterion uses \(\phi(\theta)=1\), not merely \(|\phi(\theta)|=1\). The latter condition only says that \(e^{i\theta X}\) is constant almost surely; that constant need not be \(1\), and it does not by itself show that \(X\) lies in a lattice through zero.

Finding the Maximal Span

For a finite set of positive integer support points, the maximal span is their greatest common divisor. More generally, if the possible values are integer multiples of some base \(c>0\), divide those integer multipliers by their gcd and multiply the gcd by \(c\). This is often the quickest way to find the maximal span.

Continuous and Mixed Laws

Mixed laws need not be nonarithmetic. For example, if \(X=2\) with probability \(\tfrac14\) and \(X=4\) with probability \(\tfrac34\), then the law is arithmetic with maximal span \(2\): both support points are multiples of \(2\), and a larger span would have to divide both \(2\) and \(4\) as integer multiples, which is impossible. The presence of multiple possible values, or a mixture in the everyday sense, does not determine arithmeticity; the question is whether the whole law is concentrated on one lattice.

Why the Distinction Matters for Renewal Behavior

If \(X_k\in d\mathbb{Z}\) almost surely, then every renewal epoch \(S_n=X_1+\cdots+X_n\) lies in \(d\mathbb{Z}\) almost surely. Thus the renewal process can only renew at lattice times. In contrast, a nonarithmetic interarrival law has no such common grid. This difference can create periodic behavior in local renewal quantities and is why renewal-limit results distinguish arithmetic and nonarithmetic assumptions. The particular limit statements and their hypotheses belong to the corresponding renewal theorems; the classification here identifies which case a model occupies.

For a direct illustration, take deterministic interarrivals \(X_k=d\). Then \(S_n=nd\), and for \(t\ge0\), \(N(t)=\lfloor t/d\rfloor\). The count changes only at multiples of \(d\); shifting an observation window relative to that grid can change whether it contains a renewal. This law is arithmetic with maximal span \(d\), since its only possible value \(d\) is a multiple of \(h\) precisely when \(d/h\) is an integer, and the largest such \(h\) is \(d\).

Common pitfall. The smallest possible interarrival is not necessarily the maximal span. For the law supported on \(\{6,10\}\), one might incorrectly take \(6\) as the span because it is the smaller support point. But \(10/6\) is not an integer, so \(6\mathbb{Z}\) does not contain the support. The correct calculation uses both support points: their common integer divisor is \(2\), and the maximal span is \(2\). Another mistake is to assume that any law with an atom is arithmetic. The mixed law above has an atom of mass \(\tfrac12\), but no lattice can contain the continuous component with positive probability, so it is nonarithmetic.