Renewals: Event Times and Event Counts
A renewal process describes successive occurrences of events whose gaps in time follow a common probabilistic rule. Examples include successive component failures after replacement, repeated arrivals, or the completion times of successive tasks. Two objects are central, and they should not be confused: the renewal epochs are the times at which events occur, while the renewal count records how many events have occurred by a given time. The distinction matters especially at the exact time of an event and at time zero.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \((X_k)_{k\ge1}\) be independent, identically distributed random variables with \(X_k>0\) almost surely. Each \(X_k\) is an interarrival time: it is the gap between consecutive renewal epochs. Define the ordinary renewal epochs by \(S_0=0\) and \(S_n=\sum_{k=1}^{n}X_k\) for \(n\ge1\). The process is indexed by continuous time \(t\ge0\), and its count \(N(t)\) takes values in \(\mathbb{N}_0=\{0,1,2,\ldots\}\).
“Ordinary” specifies the initial convention: the process starts at time zero with its first event after the random waiting time \(X_1\). The gaps are strictly positive, so successive epochs are strictly increasing. We use the non-strict inequality \(S_n\le t\) in the count: an event occurring exactly at time \(t\) has already occurred by time \(t\). Accordingly, an alternative count using \(S_n<t\) would describe events strictly before \(t\), not events by \(t\).
Why the Count Is Well Defined
The maximum in the definition of \(N(t)\) must exist and be finite for every finite \(t\). The assumptions \(X_k>0\) almost surely and independence are enough; a finite mean is not required. First, the epoch sequence is strictly increasing on every outcome where all gaps are positive. More importantly, the epochs tend to infinity almost surely, so only finitely many can lie below any fixed finite time.
Proof. Since \(S_{n+1}=S_n+X_{n+1}\) and \(X_{n+1}>0\), we have \(S_{n+1}>S_n\) for every \(n\). Now fix \(t\ge0\) and suppose \(S_n\to\infty\). There is an index \(m\) such that \(S_m>t\). Strict increase then gives \(S_n>t\) for every \(n\ge m\). Hence only finitely many indices \(n\ge1\) satisfy \(S_n\le t\). Because the epochs increase, those indices are exactly \(1,\ldots,N(t)\), with none if \(N(t)=0\). The sum of their indicators is therefore \(N(t)\), as claimed. \(\square\)
Proof. Since \(X_1>0\) almost surely, the events \(\{X_1\ge1/m\}\), for positive integers \(m\), have union of probability one. At least one such event has positive probability: otherwise their countable union would have probability zero. Choose \(\varepsilon>0\) with \(p=\mathbb{P}(X_1\ge\varepsilon)>0\). By identical distribution and independence, the events \(A_k=\{X_k\ge\varepsilon\}\) are independent and each has probability \(p\). For any fixed \(r\ge1\), the probability that none of \(A_r,\ldots,A_{r+\ell}\) occurs is \((1-p)^{\ell+1}\), which tends to zero as \(\ell\to\infty\). Thus the probability that no \(A_k\) occurs at or after \(r\) is zero. The event that only finitely many \(A_k\) occur is the countable union, over \(r\ge1\), of the events that none occurs at or after \(r\), so it also has probability zero. Almost surely, infinitely many gaps are at least \(\varepsilon\). Their partial sums therefore grow without bound, giving \(S_n\to\infty\). The proposition now ensures that \(N(t)\) is finite for every finite \(t\). \(\square\)
Strict positivity is an important part of this statement. If zero gaps are allowed, the epochs need not be strictly increasing, and infinitely many events can occur at the same finite time. For instance, if \(X_k=0\) for every \(k\), then \(S_n=0\) for every \(n\); the set in the definition of \(N(0)\) has no largest element. The count is not a finite integer-valued quantity. In contrast, a gap distribution may have substantial probability near zero and still satisfy the definition, provided every gap is strictly positive almost surely.
Worked Examples: Using the Definitions
Worked Example: Counting at an Exact Renewal Epoch
Suppose a realized sequence of positive gaps is \(X_1=\tfrac23\), \(X_2=1\), and \(X_3=\tfrac53\). Compute the first three positive renewal epochs, then find \(N(1)\), \(N(\tfrac53)\), and \(N(\tfrac{10}3)\).
Starting with \(S_0=0\), add the gaps in order: \(S_1=0+\tfrac23=\tfrac23\), \(S_2=\tfrac23+1=\tfrac53\), and \(S_3=\tfrac53+\tfrac53=\tfrac{10}3\). Since the epochs are increasing, \(N(1)=0\) if no epoch is at most 1, but \(S_1=\tfrac23\le1\), so \(N(1)=1\). At \(t=\tfrac53\), both \(S_1=\tfrac23\le\tfrac53\) and \(S_2=\tfrac53\le\tfrac53\), while \(S_3=\tfrac{10}3>\tfrac53\); hence \(N(\tfrac53)=2\). At \(t=\tfrac{10}3\), the first three epochs are all at most \(t\), so \(N(\tfrac{10}3)=3\).
Check: The indicator representation gives \(N(\tfrac53)=\mathbf{1}_{\{\frac23\le\frac53\}}+\mathbf{1}_{\{\frac53\le\frac53\}}+\mathbf{1}_{\{\frac{10}3\le\frac53\}}=1+1+0=2\). The equality at the second epoch is included, as required by the \(\le\) convention.
Worked Example: A Positive Interarrival Law with Infinite Mean
Let the iid interarrival times have density \(f(x)=1/(1+x)^2\) for \(x>0\), and \(f(x)=0\) otherwise. Check that this is a positive interarrival distribution, determine its mean, and calculate the probability of no renewal by time \(t=2\).
The density integrates to one: \(\int_0^\infty (1+x)^{-2}\,dx=[-(1+x)^{-1}]_0^\infty=1\). It assigns probability one to positive values, so the strict-positivity assumption holds. For \(R>0\), the truncated first moment is \(\int_0^R x(1+x)^{-2}\,dx=\log(1+R)+(1+R)^{-1}-1\). As \(R\to\infty\), this tends to infinity, so \(\mathbb{E}[X_1]=\infty\).
Nevertheless, the renewal process is well defined. There is no renewal by time \(t\) exactly when the first gap exceeds \(t\), because \(S_1=X_1\) and all later epochs occur after \(S_1\). Thus \(\mathbb{P}(N(t)=0)=\mathbb{P}(X_1>t)=\int_t^\infty(1+x)^{-2}\,dx=1/(1+t)\). At \(t=2\), this is \(1/(1+2)=\tfrac13\).
Check: The survival probability at \(t=2\) is also \(1-F(2)\), where \(F(2)=\int_0^2(1+x)^{-2}\,dx=1-\tfrac13=\tfrac23\). Therefore \(1-F(2)=\tfrac13\), in agreement with the direct calculation. Infinite mean does not prevent the epochs from tending to infinity almost surely; it is not needed for the definition.
Worked Example: A Delayed Initial Event
To compare initial conventions, suppose the first event occurs after a fixed delay \(D=\tfrac12\), and subsequent gaps along the realized path are \(1,\tfrac32,2\). Find the event times and count events by \(t=2\) and \(t=3\).
The event times are \(D=\tfrac12\), \(D+1=\tfrac32\), \(D+1+\tfrac32=3\), and then \(3+2=5\). By time \(2\), the first two events have occurred and the event at time \(3\) has not, so the delayed count is \(2\). By time \(3\), the event at time \(3\) is included, so the count is \(3\).
Check: The ordered event times are \(\tfrac12\le2\), \(\tfrac32\le2\), \(3>2\), giving two events by \(2\); and \(\tfrac12\le3\), \(\tfrac32\le3\), \(3\le3\), giving three by \(3\). A delayed process has an initial waiting time \(D\) before its first event, unlike the ordinary convention in which the first gap \(X_1\) itself is the wait from time zero. If \(D\) is random, it is commonly assumed nonnegative; any further independence assumptions should be stated as part of the model.
Initial Epochs, Assumptions, and a Common Pitfall
For the ordinary process, the epoch \(S_0=0\) is a starting reference, not an event counted by \(N(t)\). Some applications instead choose to count the initial point, or specify a delay before the first event. These choices change the count convention and must be stated; they are not interchangeable notational details. A delayed process can be defined by event times \(D,D+X_1,D+X_1+X_2,\ldots\), with \(D\ge0\). Its count by time \(t\) counts those event times that are at most \(t\). If \(D=0\) and the initial point is treated as an event, that count includes an event at zero, unlike the ordinary count defined here.
The basic ordinary-process assumptions are precise: the interarrivals are defined on one probability space, are iid, and are strictly positive almost surely. No density, continuity, or finite-mean assumption is needed to define the epochs and count. In particular, the gap distribution may be discrete, continuous, or a mixture, and it may assign positive probability to particular values. Additional assumptions become relevant for specific questions, such as calculating distributions or long-run rates, but should not be silently added to the definition.
Common pitfall: using a strict endpoint inequality. A tempting but incorrect formula is \(N(t)=\sum_{n\ge1}\mathbf{1}_{\{S_n<t\}}\). In the first example, at \(t=\tfrac53=S_2\), this formula gives \(\mathbf{1}_{\{\frac23<\frac53\}}+\mathbf{1}_{\{\frac53<\frac53\}}+\mathbf{1}_{\{\frac{10}3<\frac53\}}=1+0+0=1\). That counts events strictly before \(t\), not events by \(t\). With the ordinary renewal-count convention, replace \(<\) by \(\le\): the second indicator is then one and \(N(\tfrac53)=2\).
Keeping epochs and counts separate makes later renewal calculations easier to interpret. The epochs \(S_n\) answer “when does the \(n\)th event occur?”; the count \(N(t)\) answers “how many events have occurred by time \(t\)?” Their relation is given by the defining inequality \(S_n\le t\), including its endpoint. The choice \(S_0=0\) and the exclusion of that reference point from the count complete the ordinary-process convention.
Check Your Understanding
Use the ordinary convention unless a question specifies a delayed start. In each count, include an event occurring exactly at the stated time.
- If \(S_1=\tfrac14\), \(S_2=1\), and \(S_3=\tfrac74\), what are \(N(1)\) and \(N(\tfrac32)\)?
- Why does \(N(0)=0\) for the ordinary process even though \(S_0=0\)?
- Which assumption ensures that the epochs are strictly increasing? Give a sequence of gaps for which that property fails if zero gaps are allowed.
- Does the definition of an ordinary renewal process require \(\mathbb{E}[X_1]<\infty\)? Explain using the infinite-mean example.
- If a delayed process has its first event at time \(D=\tfrac34\), what is its count at \(t=\tfrac34\), and how does that compare with the ordinary process’s initial-epoch convention?