Asymptotic Renewal Rates in Worked Models

From Interarrival Means to Long-Run Rates

A renewal process records events that recur after successive waiting times. Once the mean time between events is known, the elementary renewal theorem gives the long-run expected number of events per unit time. This tutorial applies that result to three models: fixed waiting times, exponential waiting times, and waiting times with two possible values. The calculations are simple, but the assumptions and the meaning of “rate” matter: a long-run expected rate is not generally the expected count over a short time interval.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $(X_k)_{k\ge1}$ be independent and identically distributed, strictly positive interarrival times, with common distribution $F$. Define the $n$th renewal epoch by $S_0=0$ and $S_n=\sum_{k=1}^n X_k$ for $n\ge1$. The renewal count and renewal function are

N(t)=\max\{n\ge0:S_n\le t\},\qquad m(t)=\mathbb{E}[N(t)].

Thus an event occurring exactly at time $t$ is included in $N(t)$. The maximum is finite almost surely for strictly positive iid interarrivals: for some $\varepsilon>0$, $\mathbb{P}(X_1\ge\varepsilon)>0$; independence implies that such interarrivals occur infinitely often almost surely, so $S_n$ tends to infinity. The interarrival mean is denoted by $\mu=\mathbb{E}[X_1]$.

Theorem (Elementary renewal theorem): If the interarrival times are iid, strictly positive almost surely, and $0<\mu=\mathbb{E}[X_1]<\infty$, then $$ \lim_{t\to\infty}\frac{m(t)}{t}=\frac{1}{\mu}. $$ The theorem concerns the expected count $m(t)$. The corresponding almost-sure frequency limit $N(t)/t\to1/\mu$ follows from the strong law for renewal epochs and its inversion, established in the earlier material on renewal processes. The proof of the elementary renewal theorem is given in the earlier material on that theorem; here we use its stated hypotheses and conclusion.

The quantity $1/\mu$ has units of events per unit time. It is the reciprocal of the mean waiting time, not generally the mean of the reciprocal waiting time. Each example below checks the interarrival assumptions, computes $\mu$, and then applies the theorem. In the exponential model we also derive the exact finite-time expectation, which makes the distinction between a limit and an exact formula explicit.

Fixed Interarrival Times

Worked Example: A Renewal Every Three Time Units

Suppose every interarrival time equals $3$ with probability $1$. The interarrivals are iid and strictly positive, and their mean is $\mu=3$. The elementary renewal theorem therefore predicts a long-run expected rate of $1/3$ renewal per time unit. In this deterministic case, the count can also be calculated exactly.

Here $S_n=3n$. The condition $S_n\le t$ is equivalent to $n\le t/3$, so, including an event at the endpoint, $N(t)=\lfloor t/3\rfloor$. As the count is deterministic, $m(t)=\lfloor t/3\rfloor$. Write $t=3q+r$, where $q$ is a nonnegative integer and $0\le r<3$. Then

\frac{m(t)}{t}=\frac{q}{3q+r}.

As $t$ tends to infinity, $q$ tends to infinity, and $0\le r<3$. Hence $q/(3q+r)\to1/3$, as predicted by the theorem. At the special endpoint $t=3q$ with $q\ge1$, the count is exactly $q$, not $q-1$; this checks that the convention $S_n\le t$ is being used correctly. More generally, for a fixed interarrival time $a>0$, the same argument gives $N(t)=\lfloor t/a\rfloor$ and limiting rate $1/a$.

Check: The computed mean is $3$, and its reciprocal is $1/3$. The exact ratio differs from $1/3$ at finite $t$, but the decomposition $t=3q+r$ shows that its limit is $1/3$.

This example also shows why a long-run rate does not describe every finite-time count. A deterministic process has no renewal at all between consecutive multiples of $3$, even though its average count per unit time approaches $1/3$. The theorem averages over the full interval from $0$ to $t$; it does not assert that every short time window contains approximately its length divided by $\mu$ renewals.

Exponential Interarrivals

Worked Example: Exponential Waiting Times with Rate Two

Suppose $X_k$ are iid exponential random variables with rate $\lambda=2$, so their density is $f(x)=2e^{-2x}$ for $x>0$. Its integral over $(0,\infty)$ is $1$, and the mean is $\mu=1/2$. The interarrivals are strictly positive almost surely with finite positive mean, so the elementary renewal theorem gives the long-run rate $1/\mu=2$.

In fact, we can calculate the renewal function exactly. For $n\ge1$, the sum $S_n$ has density \[ f_{S_n}(s)=\frac{2^n s^{n-1}e^{-2s}}{(n-1)!},\qquad s>0. \] This formula follows by induction from convolution: it is the exponential density for $n=1$; if it holds for $n$, convolution with $2e^{-2x}$ gives \[ \int_0^s \frac{2^n u^{n-1}e^{-2u}}{(n-1)!}\,2e^{-2(s-u)}\,du =\frac{2^{n+1}s^n e^{-2s}}{n!}, \] which is the claimed formula with $n+1$ in place of $n$.

For $n\ge1$, the event $N(t)=n$ means $S_n\le t$ and $X_{n+1}>t-S_n$. By independence of $X_{n+1}$ from $S_n$, and by the exponential tail probability $\mathbb{P}(X_{n+1}>u)=e^{-2u}$ for $u\ge0$,

\begin{aligned} \mathbb{P}(N(t)=n) &=\int_0^t \frac{2^n s^{n-1}e^{-2s}}{(n-1)!}\,e^{-2(t-s)}\,ds\\ &=e^{-2t}\frac{2^n}{(n-1)!}\int_0^t s^{n-1}\,ds\\ &=e^{-2t}\frac{(2t)^n}{n!}. \end{aligned}

For $n=0$, $N(t)=0$ exactly when $X_1>t$, which has probability $e^{-2t}$; this is also $e^{-2t}(2t)^0/0!$. Thus the displayed probability formula holds for every integer $n\ge0$, including $t=0$. Using the expectation of a nonnegative integer-valued variable and the exponential series,

\begin{aligned} m(t) &=\sum_{n=0}^{\infty}n\,\mathbb{P}(N(t)=n)\\ &=e^{-2t}\sum_{n=1}^{\infty}n\frac{(2t)^n}{n!}\\ &=e^{-2t}(2t)\sum_{r=0}^{\infty}\frac{(2t)^r}{r!} =2t. \end{aligned}

Check: Substituting $t=3$ into the exact expression gives $m(3)=2\cdot3=6$. Dividing by $3$ gives an expected count per time unit of $2$, exactly the reciprocal of the mean $1/2$. In contrast to the deterministic example, the equality $m(t)=2t$ holds for every $t\ge0$, not only asymptotically.

The exponential example has a special finite-time property: its counts have the Poisson distribution, as the calculation above establishes, so the expected count is exactly linear in time. The elementary renewal theorem does not claim this for general interarrival laws. For a general renewal process, it gives only the limiting ratio $m(t)/t$; finite-time counts can depend on the full distribution, not just its mean.

Two Possible Interarrival Times

Worked Example: Short or Long Waiting Times

Let each interarrival be $1$ with probability $1/3$ and $3$ with probability $2/3$, independently across renewals. The probabilities are nonnegative and sum to $1/3+2/3=1$, and both possible waiting times are strictly positive. The mean is

\mu=1\cdot\frac13+3\cdot\frac23=\frac13+2=\frac73.

This mean is finite and positive, so the elementary renewal theorem applies. The long-run expected renewal rate is the reciprocal:

\lim_{t\to\infty}\frac{m(t)}{t}=\frac{1}{7/3}=\frac37.

Check: The mean calculation gives $\mu=7/3$, and $(7/3)(3/7)=1$. Thus $3/7$ is the reciprocal mean, in renewals per unit time. This check verifies the rate calculation without implying that $m(t)=3t/7$ at finite $t$; the theorem asserts a limit, not that exact identity.

A tempting but incorrect calculation is to average the reciprocal waiting time. In this example it gives \[ \mathbb{E}\!\left[\frac{1}{X_1}\right] =\frac13\cdot1+\frac23\cdot\frac13 =\frac13+\frac29=\frac59, \] which is not the long-run rate $3/7$. The reason is that the long-run number of events per unit time is determined by the total elapsed time divided by the number of interarrivals. The relevant average elapsed time per event is $\mathbb{E}[X_1]$, so the limiting rate is its reciprocal. Averaging $1/X_1$ instead weights each possible waiting time as if it generated a separate long-run rate, which is not the renewal count calculation.

Why the Finite-Mean Assumption Matters

The positivity and finite-mean assumptions are not decorative. Here is a specific iid law that violates the finite-mean condition and does not have a positive limiting rate. Let \[ \mathbb{P}(X_1=2^k)=2^{-(k+1)},\qquad k=0,1,2,\ldots. \] The probabilities sum to $1$, since $\sum_{k=0}^{\infty}2^{-(k+1)}=1$, and $X_1\ge1$. But its mean is infinite: \[ \mathbb{E}[X_1]=\sum_{k=0}^{\infty}2^k\,2^{-(k+1)} =\sum_{k=0}^{\infty}\frac12=\infty. \]

For this law, the renewal rate is zero rather than a positive reciprocal mean. To see this, fix a positive integer $M$ and set $Y_k=\min(X_k,M)$. The variables $Y_k$ are bounded iid random variables. The strong law of large numbers gives $n^{-1}\sum_{k=1}^nY_k\to\mathbb{E}[Y_1]$ almost surely. Since $X_k\ge Y_k$, it follows that $\liminf_{n\to\infty}S_n/n\ge\mathbb{E}[Y_1]$ almost surely. As $M$ increases, $\mathbb{E}[Y_1]$ increases without bound, because $\mathbb{E}[X_1]=\infty$. Hence $S_n/n\to\infty$ almost surely.

On a sample path where $S_n/n\to\infty$, the count satisfies $N(t)/t\to0$: for any $\varepsilon>0$, eventually $S_n\ge n/\varepsilon$, so $N(t)\le\varepsilon t$ for all sufficiently large $t$. Also $N(t)\le t$, since every interarrival is at least $1$. Therefore $0\le N(t)/t\le1$ for $t\ge1$, and dominated convergence yields $m(t)/t\to0$. This example shows why the finite positive mean in the elementary renewal theorem cannot be dropped while retaining its positive-rate conclusion.

Common mistake. The incorrect calculation $\mathbb{E}[1/X_1]$ in the two-point example substitutes an average of individual reciprocal waiting times for the reciprocal of the average waiting time. The correction is to calculate $\mu=\mathbb{E}[X_1]$ first and then use $1/\mu$. A second frequent error is to state the theorem as an exact finite-time formula: only the exponential example above has been shown to satisfy $m(t)=t/\mu$ for every $t$; for the deterministic and two-point examples, the theorem gives the asymptotic ratio.

Check Your Understanding

For each question, use the renewal count convention $S_n\le t$, so a renewal at the endpoint is included.

If every interarrival equals $5$, calculate $N(17)$, $m(17)$, and the long-run rate.
For exponential interarrivals with rate $\lambda=4$, find the mean interarrival time and the exact expected count at $t=2$.
If an interarrival equals $2$ with probability $1/4$ and $6$ with probability $3/4$, calculate its mean and the elementary renewal rate.
For the two-point law in the previous question, calculate $\mathbb{E}[1/X_1]$. Explain why it is not the renewal rate.
Which assumption of the elementary renewal theorem fails for the law $\mathbb{P}(X_1=2^k)=2^{-(k+1)}$, and what is the limiting value of $m(t)/t$ for that law?