The Renewal Count and Its Convention
In a renewal process, successive events occur after successive interarrival times. The renewal function records the expected number of events by a given time. Its equation is a useful example of how a count over an entire sequence of events can be expressed using the first interarrival and a fresh copy of the process. The convention about whether the initial point at time zero is counted must be fixed before writing the equation.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \(X_1,X_2,\ldots\) be independent and identically distributed random variables with distribution function \(F(t)=\mathbb{P}(X_1\le t)\), and assume \(0<X_1<\infty\) almost surely. Define the renewal epochs \(S_0=0\) and \(S_n=X_1+\cdots+X_n\) for \(n\ge1\). The renewal count \(N(t)\), for continuous time \(t\ge0\), is the number of epochs strictly after zero and no later than \(t\): \(N(t)=\#\{n\ge1:S_n\le t\}\). Thus \(N(t)\) takes values in the state space \(\mathbb{N}_0\), and \(N(0)=0\).
These expectations are finite for bounded \(t\) under the assumption \(X_1>0\) almost surely. To see why, fix \(T<\infty\). There is some \(\varepsilon>0\) for which \(p=\mathbb{P}(X_1\ge\varepsilon)>0\), because the events \(\{X_1\ge1/m\}\), \(m\ge1\), have union \(\{X_1>0\}\). If \(S_n\le T\), at most \(r=\lfloor T/\varepsilon\rfloor\) of the first \(n\) interarrivals can be at least \(\varepsilon\). Consequently, for \(0<p<1\), \(\mathbb{P}(S_n\le T)\le\sum_{j=0}^{r}\binom{n}{j}p^j(1-p)^{n-j}\). This bound is summable over \(n\): it is a finite sum of terms bounded by a polynomial in \(n\) times a geometric sequence. If \(p=1\), then \(S_n\ge n\varepsilon\), so the probabilities are zero for \(n>T/\varepsilon\). In either case, summing \(\mathbb{P}(S_n\le T)\) gives a finite value. In particular, \(U\) is bounded on every bounded time interval.
The probability space and the interarrival law specify a time-homogeneous renewal process: after the first event, the remaining interarrivals are independent of the first and have the same law. This fresh-start property is what produces a renewal equation. The integral below is with respect to the probability distribution \(F(dx)\), and includes atoms at both endpoints \(0\) and \(t\) when they have mass. Here \(F(t)=\mathbb{P}(X_1\le t)\), so an event exactly at time \(t\) is counted.
Proof. Fix \(t\ge0\), and let \(N'(s)\) be the number of renewals by time \(s\) formed from the sequence \(X_2,X_3,\ldots\), with \(N'(s)=0\) for \(s<0\). If \(X_1>t\), then \(N(t)=0\). If \(X_1\le t\), the first renewal contributes one event and the subsequent count is \(N'(t-X_1)\). Therefore $$ N(t)=\mathbf{1}_{\{X_1\le t\}}\bigl(1+N'(t-X_1)\bigr). $$ Condition on the sigma-field \(\sigma(X_1)\). The sequence \(X_2,X_3,\ldots\) is independent of \(X_1\) and has the same law as the original sequence, so the conditional expectation of \(N'(t-X_1)\), when \(X_1\le t\), is \(U(t-X_1)\). Taking expectations gives $$ U(t)=\mathbb{P}(X_1\le t)+\mathbb{E}\!\left[\mathbf{1}_{\{X_1\le t\}}U(t-X_1)\right] =F(t)+\int_{[0,t]}U(t-x)\,F(dx). $$ This proves the equation, including at \(t=0\): since \(X_1>0\) almost surely, \(F(0)=0\) and \(U(0)=0\).
To prove uniqueness, let \(V\) be another locally bounded measurable solution, and set \(h(t)=U(t)-V(t)\). Subtracting their equations gives \(h(t)=\int_{[0,t]}h(t-x)F(dx)\). Iterating this identity \(n\) times yields $$ h(t)=\mathbb{E}\!\left[h(t-S_n)\mathbf{1}_{\{S_n\le t\}}\right]. $$ For fixed \(t\), local boundedness gives a finite constant \(C_t\) such that \(|h(s)|\le C_t\) for \(0\le s\le t\). Hence \(|h(t)|\le C_t\mathbb{P}(S_n\le t)\). The summability bound above implies \(\mathbb{P}(S_n\le t)\to0\) as \(n\to\infty\), so \(h(t)=0\). Since \(t\) was arbitrary, \(U=V\), proving uniqueness. \(\square\)
Counting or Excluding the Initial Epoch
Some authors include the epoch \(S_0=0\) in the count. Write \(\widetilde{N}(t)=1+N(t)\) and \(\widetilde{U}(t)=\mathbb{E}[\widetilde{N}(t)]=1+U(t)\) for \(t\ge0\). Then \(\widetilde{U}(0)=1\), while \(U(0)=0\). Substituting \(U=\widetilde{U}-1\) into the renewal equation gives the corresponding equation for the inclusive convention: $$ \widetilde{U}(t)=1+\int_{[0,t]}\widetilde{U}(t-x)\,F(dx). $$ The extra \(1\) counts the initial epoch. It is not an additional renewal after time zero.
The equation uses \(F(t)=\mathbb{P}(X_1\le t)\), not \(\mathbb{P}(X_1<t)\). This matters at atoms: if the next renewal occurs exactly at the observation time, it is included in \(N(t)\). In the integral, the endpoint \(x=t\) contributes \(U(0)=0\) under the time-zero-excluded convention; the first renewal at that endpoint is already counted by \(F(t)\).
Worked Example: Deterministic Interarrivals
Suppose every interarrival equals \(2\), so \(X_k=2\) almost surely for each \(k\). Then \(F(t)=0\) for \(t<2\) and \(F(t)=1\) for \(t\ge2\), and \(F(dx)\) places all its mass at \(x=2\). Find \(U(t)\).
For \(0\le t<2\), no renewal has occurred, so \(U(t)=0\). For \(t\ge2\), the equation becomes \(U(t)=1+U(t-2)\). Each use of this recurrence reduces the time argument by \(2\), stopping once it is below \(2\). Thus $$ U(t)=\left\lfloor\frac{t}{2}\right\rfloor,\qquad t\ge0. $$ At an exact multiple of \(2\), the renewal at that time is included. For example, at \(t=4\), the equation gives \(U(4)=1+U(2)=1+(1+U(0))=2\), matching the two renewals at times \(2\) and \(4\). At \(t=5\), it gives \(U(5)=1+U(3)=2\), matching the renewals at times \(2\) and \(4\).
Check: At \(t=5\), the formula gives \(\lfloor5/2\rfloor=2\), and the equation gives \(F(5)+U(3)=1+1=2\). At \(t=1\), the formula gives \(0\), and both the distribution term and the integral are zero.
Two Further Interarrival Laws
For continuous interarrival distributions, the integral equation can often be checked by direct integration. When the distribution has atoms, it becomes a finite sum at each fixed time if there are only finitely many possible interarrival values. In either case, the definition through the renewal count provides an independent way to check a computed answer.
Worked Example: Exponential Interarrivals
Let the interarrivals be exponential with rate \(2\). Thus \(F(t)=1-e^{-2t}\) for \(t\ge0\), with density \(2e^{-2x}\). We verify that \(U(t)=2t\).
Substitute this candidate into the right-hand side of the renewal equation: $$ F(t)+\int_0^t U(t-x)\,2e^{-2x}\,dx =1-e^{-2t}+4\int_0^t(t-x)e^{-2x}\,dx. $$ Integration gives $$ \int_0^t(t-x)e^{-2x}\,dx =\frac{t}{2}-\frac14+\frac14e^{-2t}, $$ so the integral term is \(2t-1+e^{-2t}\). Adding \(F(t)\) gives \(1-e^{-2t}+2t-1+e^{-2t}=2t\), as required.
Check at \(t=\tfrac12\): The candidate gives \(U(\tfrac12)=1\). The distribution term is \(1-e^{-1}\), and the integral term, from the expression just computed, is \(2(\tfrac12)-1+e^{-2(\frac12)}=e^{-1}\). Their sum is \(1-e^{-1}+e^{-1}=1\).
There is also a check from the renewal-count definition. The density of \(S_n\), the sum of \(n\) exponential interarrivals of rate \(2\), is \(2^n x^{n-1}e^{-2x}/(n-1)!\) for \(x>0\). Summing these densities over \(n\ge1\) gives \(2e^{-2x}\sum_{k\ge0}(2x)^k/k!=2\). Integrating this summed density from \(0\) to \(t\) gives \(U(t)=2t\), in agreement with the equation.
Worked Example: Interarrivals of One or Two Time Units
Suppose \(X_k=1\) or \(2\), each with probability \(\tfrac12\). Compute \(U(1)\), \(U(2)\), and \(U(3)\). The distribution has mass \(\tfrac12\) at each of \(1\) and \(2\); in particular, both endpoint renewals and all events at the observation time are included.
Since \(U(0)=0\), the renewal equation at \(t=1\) gives \(U(1)=F(1)+\tfrac12U(0)=\tfrac12\). At \(t=2\), \(F(2)=1\), so \(U(2)=1+\tfrac12U(1)+\tfrac12U(0)=1+\tfrac14=\tfrac54\). At \(t=3\), \(F(3)=1\), and both atoms of the interarrival law contribute: $$ U(3)=1+\tfrac12U(2)+\tfrac12U(1) =1+\tfrac12\cdot\frac54+\tfrac12\cdot\frac12 =1+\frac58+\frac14=\frac{15}{8}. $$
Check from the renewal count: The first renewal is always by time \(3\), contributing \(1\). For two interarrivals, the four equally likely ordered pairs have sums \(2,3,3,4\), so \(\mathbb{P}(S_2\le3)=3/4\). For three interarrivals, only the ordered triple \((1,1,1)\) has sum at most \(3\), so \(\mathbb{P}(S_3\le3)=1/8\). For four or more interarrivals the sum is at least \(4\), so the probability is zero. Thus \(U(3)=1+3/4+1/8=15/8\), confirming the equation calculation.
Common Convention Errors
A frequent error is to use the equation for the count excluding time zero while reporting the count that includes time zero. For deterministic interarrivals of length \(2\), at \(t=1\) the count excluding zero is \(N(1)=0\), hence \(U(1)=0\). The count including zero is \(\widetilde{N}(1)=1\), hence \(\widetilde{U}(1)=1\). Applying the equation \(U(t)=F(t)+\int_{[0,t]}U(t-x)F(dx)\) to the inclusive quantity would give \(0\) at \(t=1\), not \(1\). The correction is to use \(\widetilde{U}(t)=1+\int_{[0,t]}\widetilde{U}(t-x)F(dx)\) for the inclusive convention.
A second error is to omit the endpoint \(x=t\) or to use a strict inequality in the event count. With deterministic interarrivals of length \(2\), \(N(2)=1\), not \(0\): the renewal at time \(2\) is counted because the definition uses \(S_n\le t\). Correspondingly, \(F(2)=1\). The distribution term records the first renewal, while the integral term records the expected renewals after that first one. Keeping those roles and the time-zero convention separate makes the equation’s terms interpretable and its endpoint behavior consistent.
Check Your Understanding
Use the convention that \(U(t)\) excludes the epoch at zero and counts renewals satisfying \(S_n\le t\).
- If every interarrival equals \(3\), find \(U(0)\), \(U(3)\), and \(U(8)\).
- For exponential interarrivals of rate \(4\), verify the renewal equation using the candidate \(U(t)=4t\).
- For interarrivals equal to \(1\) or \(2\) with equal probabilities, compute \(U(2)\) from the renewal equation and check it by summing \(\mathbb{P}(S_n\le2)\).
- Write the renewal equation for the expected count when the initial epoch at zero is included.
- Explain why a renewal exactly at time \(t\) is counted under the convention used here.