Deterministic Interarrival Times

Equally Spaced Renewal Epochs

A renewal process records events separated by successive interarrival times. When every interarrival time is the same positive constant, the randomness in the event times disappears: the epochs are equally spaced. The main care needed is at the endpoints. Counting events at or before time $t$ is not the same as counting events strictly before $t$ when $t$ itself is a renewal epoch, and time zero requires separate attention because the first renewal occurs after one interarrival time, not at time zero.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $(X_k)_{k\ge1}$ be the interarrival sequence, with each $X_k$ taking values in $(0,\infty)$, and define the $n$th renewal epoch by $S_n=\sum_{k=1}^n X_k$ for $n\ge1$. The renewal count $N(t)$, for continuous time $t\ge0$, counts epochs at or before $t$. It is a process with state space $\{0,1,2,\ldots\}$. In this tutorial, assume there is a constant $c>0$ such that $\mathbb{P}(X_k=c)=1$ for every $k$. No independence assumption is needed.

Definition: The inclusive renewal count is $N(t)=\sum_{n\ge1}\mathbf{1}_{\{S_n\le t\}}$. The strict renewal count is $N_{<}(t)=\sum_{n\ge1}\mathbf{1}_{\{S_n<t\}}$. Both counts begin at zero: no renewal has occurred at time $0$.

Although each equality $X_k=c$ holds almost surely, the countably many equalities hold simultaneously on an event of probability one: the intersection of countably many probability-one events has probability one. On that event, $S_n=nc$ for every $n\ge1$. Thus the pathwise calculation reduces to counting positive integers $n$ that satisfy an endpoint inequality.

Counting at and before an Endpoint

Theorem: If $X_k=c$ almost surely for every $k$, where $c>0$, then for every $t\ge0$, $$ S_n=nc,\qquad N(t)=\left\lfloor\frac{t}{c}\right\rfloor,\qquad N_{<}(t)=\max\left\{0,\left\lceil\frac{t}{c}\right\rceil-1\right\} $$ almost surely. For $t=0$, both counts are zero. For $t>0$ with $t/c$ a positive integer, the inclusive count exceeds the strict count by one; at all other positive times they are equal.

Proof. Intersect the probability-one events $\{X_k=c\}$ over all $k\ge1$. On this intersection, summing the first $n$ interarrivals gives $S_n=nc$. Fix any $t\ge0$ and write $x=t/c$, so $x\ge0$. The inclusive count is the number of positive integers $n$ satisfying $n\le x$. If $x=0$, there are no such integers and $\lfloor x\rfloor=0$. If $x>0$, the positive integers not exceeding $x$ are $1,\ldots,\lfloor x\rfloor$, so their number is $\lfloor x\rfloor$, including zero when $0<x<1$. Hence $N(t)=\lfloor t/c\rfloor$.

For the strict count, we must count positive integers $n<x$. At $x=0$, that set is empty, so its size is zero. If $0<x<1$, it is also empty; in this case $\lceil x\rceil-1=0$. If $x=m$ is a positive integer, the eligible integers are $1,\ldots,m-1$, giving $m-1=\lceil x\rceil-1$. If $x>0$ is not an integer, write $m=\lfloor x\rfloor$; then $m<x<m+1$, so the eligible integers are $1,\ldots,m$, with none when $m=0$. Their number is $m=\lceil x\rceil-1$. These cases give $N_{<}(t)=\max\{0,\lceil t/c\rceil-1\}$ for all $t\ge0$. Finally, at $t=0$ both counts are zero. At a positive integer $x=m$, the inclusive count is $m$ and the strict count is $m-1$. At positive noninteger $x$, both counts equal $\lfloor x\rfloor$. This proves the endpoint comparison. $\square$

The floor and ceiling functions encode different inequalities: $\lfloor x\rfloor$ counts positive integers satisfying $n\le x$, while $\lceil x\rceil-1$ counts those satisfying $n<x$ when $x>0$. The maximum with zero in the strict formula makes the result valid at $t=0$ as well. It also keeps the count nonnegative when $0\le t<c$.

Worked Example: An Endpoint That Is a Renewal Epoch

Let every interarrival time equal $c=2$, and count renewals at and before $t=6$, then strictly before $t=6$. The first epochs are $S_1=2$, $S_2=4$, $S_3=6$, and $S_4=8$. Since $t/c=6/2=3$, the inclusive formula gives $N(6)=\lfloor3\rfloor=3$. The strict formula gives $N_{<}(6)=\max\{0,\lceil3\rceil-1\}=\max\{0,3-1\}=2$.

Check: The epochs at or before $6$ are $2,4,6$, which gives three; the epochs strictly before $6$ are $2,4$, which gives two. The difference is exactly the renewal at the positive boundary time $6$.

Worked Example: A Time Between Renewal Epochs

Let $c=\tfrac54$ and $t=\tfrac{17}{8}$. The ratio is $$ \frac{t}{c}=\frac{17}{8}\cdot\frac45=\frac{68}{40}=\frac{17}{10}. $$ Since $\tfrac{17}{10}$ lies strictly between $1$ and $2$, the inclusive count is $N(t)=\lfloor\tfrac{17}{10}\rfloor=1$. The strict count is $N_{<}(t)=\max\{0,\lceil\tfrac{17}{10}\rceil-1\}=\max\{0,2-1\}=1$.

Check: The first epoch is $S_1=\tfrac54=\tfrac{10}{8}<\tfrac{17}{8}$. The second is $S_2=\tfrac52=\tfrac{20}{8}>\tfrac{17}{8}$. Thus exactly one epoch satisfies either endpoint convention; no renewal occurs at $t$.

The Renewal Function

The renewal function is the expected number of renewals by time $t$. Under the inclusive convention used here, it is $m(t)=\mathbb{E}[N(t)]$ for $t\ge0$. In general, an expected count can vary because the renewal epochs are random. With deterministic interarrivals, the count is itself nonrandom, so its expectation equals its pathwise value.

Corollary: For deterministic interarrival time $c>0$, the renewal function is $$ m(t)=\mathbb{E}[N(t)]=\left\lfloor\frac{t}{c}\right\rfloor,\qquad t\ge0. $$

Proof. The theorem gives $N(t)=\lfloor t/c\rfloor$ almost surely. Since this is a finite constant for each fixed $t\ge0$, its expectation is that same constant: $\mathbb{E}[N(t)]=\lfloor t/c\rfloor$. At $t=0$, the formula gives zero, in agreement with the fact that $S_n=nc>0$ for every $n\ge1$. $\square$

This renewal function is a step function. It stays constant between consecutive epochs and increases by one at each positive epoch $c,2c,3c,\ldots$. Its value at each jump includes the renewal at that time; immediately before the jump, the strict count is one smaller. This is why the endpoint convention must be stated whenever a count or expected count is reported.

Worked Example: Computing the Renewal Function

Suppose $c=\tfrac32$, and find the renewal function at $t=\tfrac{15}{4}$. First compute the ratio: $$ \frac{t}{c}=\frac{15}{4}\cdot\frac23=\frac{30}{12}=\frac52. $$ Therefore $m(\tfrac{15}{4})=\lfloor\tfrac52\rfloor=2$. In this deterministic process, this expectation is also the actual count on every path in the probability-one event where all interarrivals equal $\tfrac32$.

Check: The first three epochs are $S_1=\tfrac32=\tfrac64$, $S_2=3=\tfrac{12}{4}$, and $S_3=\tfrac92=\tfrac{18}{4}$. The first two are at or before $\tfrac{15}{4}$, while the third is after it, so the count is two. At the boundary $t=3$, the ratio is $3/(\tfrac32)=2$, so $m(3)=\lfloor2\rfloor=2$; strictly before $3$, only the epoch $\tfrac32$ is counted.

Worked Example: Time Zero and the First Renewal

Let $c=4$. At time $t=0$, the inclusive count is $N(0)=\lfloor0/4\rfloor=0$, and the strict count is $N_{<}(0)=\max\{0,\lceil0/4\rceil-1\}=\max\{0,-1\}=0$. At time $t=4$, the inclusive count is $N(4)=\lfloor4/4\rfloor=1$, while the strict count is $N_{<}(4)=\max\{0,\lceil4/4\rceil-1\}=\max\{0,1-1\}=0$.

Check: The first renewal epoch is $S_1=4$. There are no epochs before time zero, and there is no renewal at time zero. At time four, the inclusive convention counts $S_1=4$, whereas the strict convention does not. The computed counts, zero and one inclusively at the two respective times, match the epoch definition.

Endpoint Pitfalls and Interpretation

Common mistake: using the strict formula without its zero-time case. A tempting formula is $N_{<}(t)=\lceil t/c\rceil-1$. At $t=0$, it gives $\lceil0\rceil-1=0-1=-1$, which cannot be a renewal count and disagrees with the definition: no epoch $S_n$, $n\ge1$, is strictly before zero. The correction is $$ N_{<}(t)=\max\left\{0,\left\lceil\frac{t}{c}\right\rceil-1\right\},\qquad t\ge0. $$ Equivalently, one may use $\lceil t/c\rceil-1$ only for $t>0$, while stating separately that $N_{<}(0)=0$.

A related overstatement is that the inclusive and strict counts differ whenever $t/c$ is an integer. The integer zero is an exception: when $t=0$, both counts are zero because renewals are indexed from $n=1$, not from $n=0$. They differ by one precisely when $t/c$ is a positive integer. A further common error is to insert an initial renewal at time zero; doing so would change the count by one and would no longer match the definition $S_n=\sum_{k=1}^nX_k$.

Deterministic interarrivals provide a direct reference case for renewal calculations: the epochs, counts, and expected count can all be checked from the same arithmetic sequence. The floor formula answers an inclusive counting question; the strict formula answers a different endpoint question. Neither formula should be used until the intended convention and the value $t=0$ have been checked.

Check Your Understanding

For each question, state whether the count includes the endpoint and verify your answer by listing the relevant epochs.

If $c=3$ and $t=9$, calculate $N(t)$ and $N_{<}(t)$. Explain why the counts differ.
If $c=2$ and $t=5$, calculate both counts and identify the epochs that are counted.
For an arbitrary $c>0$, calculate $N(0)$ and $N_{<}(0)$ from their definitions, then check both formulas.
For deterministic interarrival time $c=\tfrac23$, find the renewal function at $t=\tfrac53$.
Why is $\lceil t/c\rceil-1$ not a valid strict-count formula for every $t\ge0$ unless it is modified?