Inverting the Epoch Strong Law to Count Renewals

From Long-Run Epochs to Long-Run Counts

Renewal theory describes two ways to measure the same sequence of events. The renewal epochs record the times at which events occur; the renewal count records how many have occurred by a given time. If the average time between events settles to a positive value, it is natural to expect the number of events per unit time to settle to its reciprocal. The key step is to invert the epoch sequence carefully, preserving the inequalities at the event-time boundary.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \(X_1,X_2,\ldots\) be independent, identically distributed random variables with \(\mathbb{P}(X_k>0)=1\) and mean \(\mu=\mathbb{E}[X_1]\) satisfying \(0<\mu<\infty\). Define \(S_0=0\) and \(S_n=\sum_{k=1}^n X_k\) for \(n\ge1\). The renewal count is \(N(t)=\max\{n\ge0:S_n\le t\}\), for \(t\ge0\). Thus an event occurring exactly at time \(t\) is included in \(N(t)\). Since the interarrivals are positive and the epoch strong law gives \(S_n/n\to\mu>0\) almost surely, \(S_n\to\infty\) almost surely, so \(N(t)\) is finite for each finite \(t\), almost surely.

The proof separates into a deterministic fact and a probabilistic application. The deterministic fact needs only an increasing sequence of epochs whose average spacing has a positive finite limit. Independence and identical distribution enter when we use the strong law to establish that limit for the random epochs.

Proof. Since \(s_n/n\to\mu>0\), the sequence \(s_n\) tends to infinity. Consequently \(n(t)\to\infty\) as \(t\to\infty\): for any fixed integer \(r\), if \(t\ge s_r\), then \(n(t)\ge r\). For all sufficiently large \(t\), write \(n=n(t)\ge1\). By the definition of the maximum, \(s_n\le t<s_{n+1}\). The first inequality, with positive denominators, gives \(n/t\le n/s_n\). The second gives \((n+1)/t>(n+1)/s_{n+1}\), and hence \(n/t>(n+1)/s_{n+1}-1/t\). As \(t\to\infty\), we have \(n\to\infty\), so \(n/s_n\to1/\mu\) and \((n+1)/s_{n+1}\to1/\mu\). Also \(1/t\to0\). These upper and lower bounds therefore imply \(n(t)/t\to1/\mu\). \(\square\)

Proof of the theorem. By the strong law of large numbers for iid random variables with finite mean, \(S_n/n\to\mathbb{E}[X_1]=\mu\) almost surely. This is the strong law for renewal epochs. On every sample point in the probability-one event where this convergence holds and every \(X_k>0\), the realized epochs satisfy the deterministic lemma’s hypotheses. Applying that lemma to those epochs gives \(N(t)/t\to1/\mu\) on that event. Thus the convergence is almost sure. \(\square\)

This is an inversion argument, not a claim that the count is itself a sum of independent increments. It uses only the two epochs that surround \(t\): the last one at or before \(t\), and the first one after \(t\). In the notation of the theorem, the exact bounds are \(S_{N(t)}\le t<S_{N(t)+1}\). The strict inequality on the right follows because \(N(t)\) is the largest index whose epoch is at most \(t\).

What the Inversion Does—and Does Not—Say

The result concerns a pathwise long-run ratio: outside an event of probability zero, the realized count divided by elapsed time approaches \(1/\mu\). It is not a statement about the expected count \(\mathbb{E}[N(t)]\), nor does it say that \(N(t)/t\) equals \(1/\mu\) at each finite time. Convergence of expectations is a different question and needs its own justification; the almost-sure limit alone does not permit expectations to be passed through a limit without additional conditions.

The hypotheses ensure a finite, positive average spacing. If the positive finite mean assumption is dropped altogether, the conclusion need not even have a finite rate. For example, take the deterministic, non-identically distributed sequence \(X_k=1/k^2\). Its partial sums satisfy \(S_n\le\sum_{k=1}^\infty 1/k^2<\infty\), since the series converges. For every \(t\) greater than or equal to this finite sum, infinitely many epochs satisfy \(S_n\le t\); thus \(N(t)=\infty\). The count-per-time conclusion fails. This example is not within the theorem’s iid, positive-finite-mean assumptions; it illustrates why an epoch sequence that tends to infinity with a positive finite average spacing is essential to the inversion.

Common mistake: reversing a bound. From \(S_{N(t)}\le t\), one might incorrectly write \(N(t)/t\ge N(t)/S_{N(t)}\). For the two-point example at \(t=7\), that would say \(4/7\ge4/6\), which is false: \(4/7=24/42\), while \(4/6=28/42\). Because \(S_{N(t)}\le t\) and the numerators are positive, the correct direction is \(N(t)/t\le N(t)/S_{N(t)}\). The lower bound comes instead from the next epoch: \(t<S_{N(t)+1}\), which yields \(N(t)/t>(N(t)+1)/S_{N(t)+1}-1/t\). Keeping these two sources of the bounds distinct avoids the error.