The Renewal Measure and Its Atoms

From Renewal Epochs to a Measure

A renewal process can be described not only by the number of renewals before a time, but also by the distribution of all its renewal epochs. The measure obtained by adding the distributions of those epochs records how much renewal mass lies in any chosen time set. In particular, its mass at a single point tells us how often, in expectation, renewal epochs occur exactly there. This makes atoms important: a deterministic or lattice-valued interarrival distribution may produce positive mass at specific times, while a continuous interarrival distribution may instead produce a density.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((X_k)_{k\ge1}\) be independent and identically distributed random variables taking values in \((0,\infty)\) almost surely, with common distribution \(F\). The \(n\)th renewal epoch is \(S_n=X_1+\cdots+X_n\), for \(n\ge1\). Thus the epoch index \(n\) is discrete, while the epochs take values in the time space \([0,\infty)\). Write \(F^{*n}\) for the distribution of \(S_n\), equivalently the \(n\)-fold convolution of \(F\). Only positive renewal epochs are included in the measure below; there is no epoch at time zero in this convention.

This is a measure because each \(F^{*n}\) is a measure and a countable sum of nonnegative measures is countably additive. More explicitly, if \((B_r)_{r\ge1}\) are disjoint Borel sets, then countable additivity of each \(F^{*n}\), followed by rearrangement of nonnegative sums, gives \(U(\bigcup_r B_r)=\sum_r U(B_r)\). The value is finite on bounded time intervals under the assumption \(X_k>0\) almost surely. Here is a justification, which also makes clear that the measure is not merely a formal sum.

Fix \(T<\infty\). Since \(X_1>0\) almost surely, there is some \(\varepsilon>0\) for which \(q=\mathbb{P}(X_1\ge\varepsilon)>0\). If \(S_n\le T\), at most \(m=\lfloor T/\varepsilon\rfloor\) of the first \(n\) interarrivals can be at least \(\varepsilon\). The number of such interarrivals has a binomial distribution with parameters \(n\) and \(q\). If \(q=1\), then \(S_n\ge n\varepsilon\), so only finitely many \(n\) can satisfy \(S_n\le T\). If \(0<q<1\), then \(\mathbb{P}(S_n\le T)\le\sum_{r=0}^{m}\binom{n}{r}q^r(1-q)^{n-r}\). For each fixed \(r\), this term is bounded by a constant times \(n^r(1-q)^n\), whose sum over \(n\) is finite because a geometric factor with ratio less than one dominates a fixed power of \(n\). Therefore \(U([0,T])=\sum_{n\ge1}\mathbb{P}(S_n\le T)<\infty\).

Atoms: Add the Contributions from Every Epoch

Proof. Apply the definition of \(U\) to the Borel set \(\{t\}\). This gives the displayed sum, with equality at a point (not an interval) required in every term. Since each \(X_k>0\) almost surely, \(S_n>0\) almost surely for every \(n\ge1\); hence every term in the sum at \(t=0\) is zero. If \(F\) has no atoms, then for any fixed \(n\ge2\), condition on \(X_1,\ldots,X_{n-1}\). For every possible value \(s\) of their sum, the conditional probability that \(S_n=t\) is \(\mathbb{P}(X_n=t-s)=0\), because \(F\) assigns zero probability to every singleton. Averaging these zero conditional probabilities gives \(\mathbb{P}(S_n=t)=0\). The same conclusion for \(n=1\) follows directly from \(F(\{t\})=0\). Thus \(U(\{t\})=0\) for every \(t\). Conversely, if \(F\) has an atom at some \(t\), then \(t>0\) and \(U(\{t\})\ge F^{*1}(\{t\})=F(\{t\})>0\). So \(U\) has an atom. This proves both directions. \(\square\)

The atom formula matters because an atom of \(U\) need not come only from the first renewal epoch. Several different epoch numbers can contribute to the same time. The sum is over all \(n\), and each term is the probability of the exact equality \(S_n=t\). It is not enough to inspect \(F(\{t\})\) alone. Conversely, a continuous distribution has no atoms in its renewal measure, even though the renewal measure can still have substantial mass in every interval.

Densities: When the Renewal Measure Is Spread Continuously

For a continuous interarrival distribution with density \(f\) on \((0,\infty)\), the distributions of the later epochs also have densities. Write \(f^{*1}=f\), and, for \(n\ge1\), define the convolution recursively by \(f^{*(n+1)}(t)=\int_0^t f^{*n}(s)f(t-s)\,ds\) for \(t>0\). This is the density of \(S_{n+1}\). The renewal measure then has a density formed by adding the densities for all positive epoch numbers.

Proof. For each \(n\), independence and the convolution formula give \(F^{*n}(B)=\int_B f^{*n}(t)\,dt\) for every Borel set \(B\subseteq[0,\infty)\). Therefore, using the definition of \(U\) and interchanging nonnegative sums and integrals,

The interchange is valid for nonnegative functions, including when an integral is initially allowed to be infinite. The sum in parentheses is consequently a density of \(U\), understood up to equality almost everywhere. An absolutely continuous measure assigns zero mass to every singleton because a singleton has Lebesgue measure zero. At \(t=0\), positive interarrivals also give \(U(\{0\})=0\). \(\square\)

Reading the Measure Correctly

The renewal measure records expected epoch mass by location in time; it is not itself a probability distribution. Its total mass on all of \([0,\infty)\) is generally infinite, even though it is finite on every bounded interval. For a set \(B\), \(U(B)\) adds the probabilities that each positive epoch falls in \(B\). The later tutorial on the renewal function considers the expected number of renewals in a time interval; the measure gives the underlying set-by-set quantity from which such counts can be expressed.

A common mistake is to calculate an atom by looking only at the interarrival law. In the two-point example, this would give the incorrect calculation \(U(\{3\})=F(\{3\})=0\). The correction is to include every epoch number: \(F^{*2}(\{3\})=\tfrac12\) and \(F^{*3}(\{3\})=\tfrac18\), so \(U(\{3\})=\tfrac58\). Another convention to watch is whether a source includes an epoch at time zero. The definition here sums over \(n\ge1\); adding an \(n=0\) term would add a unit atom at zero and would describe a different measure.