The Renewal Function as an Expected Count

Counting Renewals on Average

A renewal count answers how many renewals have occurred by a specified time. The renewal function answers a different, averaged question: what is the expected number of renewals by that time? Keeping the count, its expectation, and the measure that records expected renewals in sets of times distinct is essential. In particular, an expected count can exceed $1$, even though the probability of at least one renewal cannot.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, and let $X_1,X_2,\ldots$ be independent, identically distributed, strictly positive interarrival times with distribution $F$. Define the renewal epochs by $S_0=0$ and $S_n=X_1+\cdots+X_n$ for $n\ge1$. Assume $S_n\to\infty$ almost surely, so that only finitely many renewals occur by any finite time. The renewal count at time $t\ge0$ is $N(t)=\max\{n\ge0:S_n\le t\}$. Thus an epoch exactly at $t$ is included. The state space of the count is $\mathbb{N}_0=\{0,1,2,\ldots\}$, and its time index is continuous, $t\ge0$.

Definition: The renewal function is $U(t)=\mathbb{E}[N(t)]$, for $t\ge0$. It is the expected number of renewals in $(0,t]$, not the probability that a renewal occurs by time $t$. Since $N(t)$ is nonnegative, its expectation is well-defined as a value in $[0,\infty]$.

The count starts at zero: $N(0)=0$, because every interarrival time is strictly positive. Also, $U(t)$ is nondecreasing: for $s\le t$, every epoch counted by $N(s)$ is also counted by $N(t)$, so $N(s)\le N(t)$ on every sample path and hence $U(s)\le U(t)$. These facts are useful boundary and consistency checks.

The Renewal Function and the Renewal Measure

The renewal measure records expected numbers of epochs in sets of times, rather than just up to a single time. For a Borel set $B\subseteq(0,\infty)$, define $\mathcal{R}(B)=\mathbb{E}[\sum_{n\ge1}\mathbf{1}_{\{S_n\in B\}}]$, where $\mathbf{1}_{\{S_n\in B\}}$ is $1$ when the $n$th epoch lies in $B$ and $0$ otherwise. This is a measure on $(0,\infty)$; its mass on a set is the expected number of renewal epochs in that set. It need not have total mass $1$, and it need not have a density.

Theorem: With the inclusive endpoint convention $N(t)=\max\{n:S_n\le t\}$, the renewal function is the cumulative mass of the renewal measure: $$ U(t)=\mathcal{R}((0,t])\qquad(t\ge0). $$

Proof. Fix $t\ge0$. On each sample path, the epochs $S_1,S_2,\ldots$ are strictly increasing and positive. Therefore the epochs in $(0,t]$ are exactly $S_1,\ldots,S_{N(t)}$, with none if $N(t)=0$. Their number is $N(t)$, so $\sum_{n\ge1}\mathbf{1}_{\{S_n\in(0,t]\}}=N(t)$ on every sample path. Taking expectations gives $\mathbb{E}[\sum_{n\ge1}\mathbf{1}_{\{S_n\in(0,t]\}}]=\mathbb{E}[N(t)]$, including if the expectation is infinite. By the definition of $\mathcal{R}$, the left side is $\mathcal{R}((0,t])$, and the right side is $U(t)$. $\square$

The endpoint matters. An epoch at $t$ contributes to both $N(t)$ and $\mathcal{R}((0,t])$. If one instead counts epochs strictly before $t$, the corresponding interval is $(0,t)$, not $(0,t]$. When there is no atom at $t$, these two choices give the same value; when there is an atom, they can differ.

Using Epoch Probabilities in Finite-Horizon Models

For calculations, it is often convenient to ask whether the $k$th epoch has occurred by time $t$. The event $\{S_k\le t\}$ is exactly the event that at least $k$ renewals have occurred by $t$. If there is a deterministic bound $M$ such that $N(t)\le M$ almost surely, the count is a finite sum of these event indicators. This yields a direct route from epoch probabilities to the expected count.

Theorem (Finite-horizon epoch-probability identity): If $N(t)\le M$ almost surely for some finite integer $M$, then $$ U(t)=\sum_{k=1}^{M}\mathbb{P}(S_k\le t). $$ Terms for epochs that cannot occur by $t$ are zero.

Proof. On any sample path, $S_k\le t$ holds exactly for $k=1,\ldots,N(t)$, because the epochs increase strictly. Since $N(t)\le M$, this gives the finite pointwise identity $N(t)=\sum_{k=1}^{M}\mathbf{1}_{\{S_k\le t\}}$. Taking expectations and using linearity of expectation for this finite sum gives $\mathbb{E}[N(t)]=\sum_{k=1}^{M}\mathbb{E}[\mathbf{1}_{\{S_k\le t\}}]=\sum_{k=1}^{M}\mathbb{P}(S_k\le t)$. The left side is $U(t)$, proving the formula. $\square$

The deterministic bound is a convenient sufficient condition for this finite calculation, not part of the definition of $U$. For example, if $X_k\ge c>0$ almost surely, then $S_n\ge nc$, so $S_n\le t$ implies $n\le t/c$. In the examples below, we use a positive minimum interarrival time or a finite discrete time horizon to know exactly how many epoch probabilities can be nonzero.

Worked Example: Deterministic Interarrival Times

Suppose every interarrival time equals $2$, so $X_k=2$ almost surely. Find $U(5)$, using the inclusive convention that a renewal at time $5$ would be counted.

The first epochs are $S_1=2$, $S_2=2+2=4$, and $S_3=2+2+2=6$. Since $6>5$, no later epoch can occur by time $5$, so $N(5)\le2$. The epoch probabilities are $\mathbb{P}(S_1\le5)=1$, $\mathbb{P}(S_2\le5)=1$, and $\mathbb{P}(S_3\le5)=0$. Thus the finite-horizon identity gives $U(5)=1+1=2$.

Check: Directly, the renewals by time $5$ occur at times $2$ and $4$, so $N(5)=2$ on every sample path and $\mathbb{E}[N(5)]=2$, as calculated. At time $4$, the second renewal is included because the endpoint is inclusive; the count is also $2$.

Worked Example: Two-Point Interarrivals

Let each interarrival time be $1$ or $2$, each with probability $\tfrac12$, independently. Find $U(3)$. Since each interarrival time is at least $1$, at most $3$ renewals can occur by time $3$.

The first epoch always occurs by time $3$, so $\mathbb{P}(S_1\le3)=1$. For $S_2\le3$, the only excluded pair of interarrival times is $(2,2)$: its sum is $2+2=4>3$. The other three pairs, $(1,1),(1,2),(2,1)$, each have probability $\tfrac14$. Therefore $\mathbb{P}(S_2\le3)=\tfrac34$. For $S_3\le3$, all three interarrival times must equal $1$, giving $\mathbb{P}(S_3\le3)=\tfrac12\cdot\tfrac12\cdot\tfrac12=\tfrac18$. The finite-horizon identity now gives $$ U(3)=1+\frac34+\frac18=\frac{8}{8}+\frac{6}{8}+\frac{1}{8}=\frac{15}{8}. $$

Check by the count distribution: Three renewals occur only for $(1,1,1)$, with probability $\tfrac18$. Exactly two occur when the first two interarrivals sum to at most $3$, but the first three do not; this has probability $\tfrac34-\tfrac18=\tfrac58$. Exactly one occurs when the first two interarrivals are $(2,2)$, with probability $\tfrac14$. The probabilities add to $\tfrac18+\tfrac58+\tfrac14=\tfrac18+\tfrac58+\tfrac28=1$. The expectation from this distribution is $3\cdot\tfrac18+2\cdot\tfrac58+1\cdot\tfrac14=\tfrac38+\tfrac{10}{8}+\tfrac28=\tfrac{15}{8}$, agreeing with the epoch calculation.

Worked Example: Bernoulli Renewals on a Discrete Grid

At each integer time, suppose a renewal occurs with probability $\tfrac13$, independently of the other integer times. Equivalently, the interarrival times are independent geometric random variables with success probability $\tfrac13$, taking values $1,2,\ldots$. Find the expected number of renewals by time $3$, including a renewal at time $3$.

At most $3$ renewals can occur by time $3$. For the $k$th renewal to occur by time $3$, there must be at least $k$ renewals in the first three time slots. The probability of no renewal in any one slot is $1-\tfrac13=\tfrac23$, so $$ \mathbb{P}(S_1\le3)=1-\left(\frac23\right)^3 =1-\frac{8}{27}=\frac{19}{27}. $$

For two or more renewals, there must be exactly two or exactly three successes among the three slots. The probability of exactly two is $\binom{3}{2}(\tfrac13)^2(\tfrac23)=3\cdot\tfrac19\cdot\tfrac23=\tfrac{6}{27}$; the probability of exactly three is $(\tfrac13)^3=\tfrac{1}{27}$. Hence $\mathbb{P}(S_2\le3)=\tfrac{6}{27}+\tfrac{1}{27}=\tfrac{7}{27}$. Three renewals by time $3$ requires success in every slot, so $\mathbb{P}(S_3\le3)=\tfrac{1}{27}$. Adding the epoch probabilities gives $U(3)=\tfrac{19}{27}+\tfrac{7}{27}+\tfrac{1}{27}=\tfrac{27}{27}=1$.

Check by direct expectation: The count is the sum of the three slot indicators. Each indicator has expectation $\tfrac13$, so linearity of expectation gives $\mathbb{E}[N(3)]=\tfrac13+\tfrac13+\tfrac13=1$. This agrees with the epoch-probability calculation.

What the Expected Count Does—and Does Not—Measure

The renewal function compresses the distribution of $N(t)$ into one number: its mean. It does not determine the probability of a renewal, and it does not say how renewals are distributed across subintervals. The renewal measure contains more set-by-set information: $\mathcal{R}(B)$ is the expected number of epochs in $B$, while $U(t)$ is just its cumulative value on $(0,t]$. A measure can assign mass greater than $1$, because several renewal epochs may fall in the same set.

Common mistake: replacing an expected count by a probability. In the two-point example, $U(3)=\tfrac{15}{8}$, but $\mathbb{P}(N(3)\ge1)=1$, since the first interarrival time is at most $2$. The incorrect calculation $U(3)=\mathbb{P}(N(3)\ge1)=1$ counts only whether at least one renewal occurred. The correct expectation counts every renewal: the events $\{N(3)\ge1\}$, $\{N(3)\ge2\}$, and $\{N(3)\ge3\}$ contribute $1$, $\tfrac34$, and $\tfrac18$, respectively, for a total of $\tfrac{15}{8}$. Thus $U(t)$ may exceed $1$, whereas a probability cannot.

Endpoint conventions should be checked before using any formula. With the convention here, an epoch exactly at $t$ is included, so all epoch events in the finite-horizon calculation use $S_k\le t$, not $S_k<t$. At a time where the renewal-epoch distribution has an atom, changing this inequality can change the answer. Finally, when an infinite collection of possible epochs must be handled directly, the finite-horizon calculation above is not itself a proof of an infinite series expansion; the finite argument applies when the number of possible renewals by the chosen time is bounded.

Check Your Understanding

For each question, use the inclusive endpoint convention and distinguish an expected count from an event probability.

If $X_k=3$ almost surely, what are $N(6)$ and $U(6)$? Which epochs are included at time $6$?
For interarrival times equal to $1$ or $2$, each with probability $\tfrac12$, calculate $\mathbb{P}(S_2\le2)$. Explain which ordered pairs contribute.
Why can a renewal measure assign mass greater than $1$ to $(0,t]$, while the probability of at least one renewal by time $t$ cannot exceed $1$?
In the Bernoulli-grid example, what event corresponds to $S_2\le3$, and why does its probability include outcomes with exactly two as well as exactly three renewals?
If a process counts epochs strictly before $t$, which interval should replace $(0,t]$ in the renewal-measure expression?

The Renewal Function as an Expected Count

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