Finite-Mean and Infinite-Mean Interarrivals

Interarrival Means and Renewal Rates

A renewal process records repeated events separated by random waiting times. If the mean waiting time is finite, the long-run expected number of events per unit time is governed by its reciprocal. If the mean is infinite, the expected count grows more slowly than linearly. The distinction depends on the assumptions about the interarrivals, and the infinite-mean conclusion can be proved by comparing the original waiting times with bounded truncations.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $X_1,X_2,\ldots$ be independent and identically distributed random variables with $X_k>0$ almost surely. They represent the interarrival times. Define the renewal epochs by $S_0=0$ and $S_n=X_1+\cdots+X_n$ for $n\ge1$. At continuous time $t\ge0$, define the renewal count and its expectation by

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

Thus a renewal exactly at time $t$ is included, while the initial time $S_0=0$ is not counted as a renewal. Strict positivity of the interarrivals makes the epochs strictly increasing. They also diverge almost surely: there is an $\varepsilon>0$ with $\mathbb{P}(X_1\ge\varepsilon)>0$, and independence implies that $X_k\ge\varepsilon$ occurs infinitely often almost surely. Consequently $N(t)$ is finite almost surely for each fixed $t$.

Definition: The interarrival mean is $\mu=\mathbb{E}[X_1]$, interpreted as an extended nonnegative expectation. Under $X_1>0$ almost surely, $\mu>0$, but it may be finite or infinite. A linear expected renewal rate $r$ means that $m(t)/t\to r$ as $t\to\infty$.

The renewal-rate statements below concern the expectation $m(t)$, not an almost-sure limit of the random ratio $N(t)/t$. For finite mean, the standard elementary renewal theorem gives the rate $1/\mu$; its proof belongs to the later renewal-limit material. Here we prove a useful upper bound that applies in both mean regimes, and use it to establish the infinite-mean conclusion directly.

A Truncation Bound

For a fixed $b>0$, truncate each interarrival at $b$ by setting $Y_k=\min(X_k,b)$. The variables $Y_k$ remain independent and identically distributed, are strictly positive, and are bounded above by $b$. Write $\nu_b=\mathbb{E}[Y_1]$, and let $M_b(t)$ be the renewal count formed from the partial sums of the $Y_k$, with the same endpoint convention as $N(t)$. Since $Y_k\le X_k$, the truncated epochs occur no later than the original epochs, so $N(t)\le M_b(t)$.

Proposition: If $X_k>0$ almost surely and are i.i.d., then for every $b>0$ and $t\ge0$, $$ m(t)\le \frac{t+b}{\nu_b}-1,\qquad \nu_b=\mathbb{E}[\min(X_1,b)]. $$ In particular, $m(t)$ is finite for each fixed $t$.

Proof. Let $M=M_b(t)$, and let $R_k=Y_1+\cdots+Y_k$, with $R_0=0$. The next truncated renewal occurs after $t$, so $R_{M+1}>t$. Also $R_M\le t$ and $Y_{M+1}\le b$, giving $R_{M+1}\le t+b$. For each $k\ge1$, the event $\{M+1\ge k\}$ is exactly $\{R_{k-1}\le t\}$; it depends only on $Y_1,\ldots,Y_{k-1}$ and is independent of $Y_k$. Summing nonnegative terms and using this independence gives $$ \mathbb{E}[R_{M+1}] =\sum_{k\ge1}\mathbb{E}\!\left[Y_k\mathbf{1}_{\{M+1\ge k\}}\right] =\nu_b\sum_{k\ge1}\mathbb{P}(M+1\ge k) =\nu_b\,\mathbb{E}[M+1]. $$ The last equality is the tail-sum identity for a positive integer-valued random variable; it holds also for extended expectations. Since $R_{M+1}\le t+b$, this identity implies $\mathbb{E}[M+1]\le(t+b)/\nu_b$, so $M$ has finite expectation. Finally, $N(t)\le M$, and therefore $m(t)\le\mathbb{E}[M]\le(t+b)/\nu_b-1$. $\square$

The bound compares the original process with one whose interarrivals cannot exceed $b$. The truncated mean $\nu_b$ is positive, and it increases to $\mu$ as $b$ increases, with the limit allowed to be infinite. This lets the same calculation say something in either case.

Finite and Infinite Mean

Theorem (Expected-rate comparison): Suppose the interarrivals are i.i.d. and strictly positive almost surely.

If $0<\mu=\mathbb{E}[X_1]<\infty$, then $\limsup_{t\to\infty}m(t)/t\le1/\mu$. The standard elementary renewal theorem strengthens this to $m(t)/t\to1/\mu$; its proof comes later in the course.
If $\mathbb{E}[X_1]=\infty$, then $m(t)/t\to0$.

Proof. Fix $b>0$. The proposition implies $$ \frac{m(t)}{t}\le\frac{1+b/t}{\nu_b}-\frac{1}{t}\qquad (t>0). $$ Taking the limit superior as $t\to\infty$ gives $\limsup_{t\to\infty}m(t)/t\le1/\nu_b$. If $\mu<\infty$, then $\nu_b=\mathbb{E}[\min(X_1,b)]$ increases to $\mu$ by monotone convergence. Letting $b\to\infty$ gives the claimed upper bound $1/\mu$. If $\mu=\infty$, monotone convergence gives $\nu_b\to\infty$. The limit superior is therefore at most $1/\nu_b$ for every $b$, hence at most zero. Since $m(t)/t\ge0$, its limit is zero. $\square$

The finite-mean upper bound alone does not prove that the count reaches that rate. The equality $m(t)/t\to1/\mu$ is a separate standard renewal limit, under the stated i.i.d., positive, finite-mean assumptions. In contrast, the infinite-mean zero-rate conclusion follows from the truncation bound itself. Neither conclusion says that the process stops: zero linear rate means the expected count divided by elapsed time tends to zero.

Worked Example: Deterministic Interarrivals

Suppose $X_k=2$ almost surely for every $k$. These interarrivals are strictly positive and i.i.d., with $\mu=\mathbb{E}[X_1]=2$. The epochs are $S_n=2n$, so the count is $N(t)=\lfloor t/2\rfloor$, where the floor includes an event at an exact endpoint.

At $t=10$, the epochs at or before $10$ are $2,4,6,8,10$, so $N(10)=5$ and $m(10)/10=5/10=1/2$. For general $t$, $N(t)=\lfloor t/2\rfloor$, and $t/2-1<\lfloor t/2\rfloor\le t/2$. Dividing by $t>0$ and taking $t\to\infty$ gives $m(t)/t\to1/2=1/\mu$. Check: at $t=10$, $1/\mu=1/2=5/10$, agreeing with the exact count.

Worked Example: Exponential Interarrivals

Let $X_k$ have the exponential distribution with rate $2$, so its density is $2e^{-2x}$ for $x\ge0$. The interarrivals are positive almost surely and have finite mean $$ \mu=\int_0^\infty x\,2e^{-2x}\,dx=\frac12. $$ This is the Poisson special case of a renewal process: $N(t)$ has a Poisson distribution with parameter $2t$.

Using that distribution, for $t\ge0$, $$ m(t)=\sum_{k=0}^{\infty}k e^{-2t}\frac{(2t)^k}{k!} =2t\,e^{-2t}\sum_{j=0}^{\infty}\frac{(2t)^j}{j!} =2t. $$ The re-indexing uses $k/k!=1/(k-1)!$ for $k\ge1$, and the final series is $e^{2t}$. Thus $m(t)/t=2=1/\mu$ for $t>0$. Check: at $t=3$, the parameter is $2\cdot3=6$, the expected count is $6$, and $m(3)/3=6/3=2=1/(1/2)$.

Worked Example: Infinite-Mean Interarrivals

Suppose $X_k$ has survival function $\mathbb{P}(X_k>x)=1/(1+x)$ for $x\ge0$. Its density on $(0,\infty)$ is $1/(1+x)^2$, so the interarrivals are strictly positive. The tail-integral formula gives $$ \mathbb{E}[X_k]=\int_0^\infty\mathbb{P}(X_k>x)\,dx =\int_0^\infty\frac{1}{1+x}\,dx=\infty. $$ For a truncation level $b>0$, $$ \nu_b=\mathbb{E}[\min(X_k,b)] =\int_0^b\mathbb{P}(X_k>x)\,dx =\int_0^b\frac{1}{1+x}\,dx=\log(1+b). $$ The proposition gives $m(t)\le(t+b)/\log(1+b)-1$. Keeping $b$ fixed and letting $t\to\infty$ yields $\limsup m(t)/t\le1/\log(1+b)$. As $b\to\infty$, $\log(1+b)\to\infty$, so $m(t)/t\to0$, as the theorem states. Check: at truncation level $b=3$, the mean of the truncated interarrival is $\log(4)$, and the resulting limit-superior bound is $1/\log(4)$; increasing $b$ makes this bound tend to zero.

What the Rate Does—and Does Not—Say

The reciprocal $1/\mu$ is a long-run expected count rate in the finite-mean setting, not a formula for $m(t)$ at every time. The deterministic example happens to have an exact floor formula; a general interarrival distribution need not. The infinite-mean result is also a rate statement, not a claim that $N(t)$ is zero or bounded. For the distribution in the last example, $\mathbb{P}(X_1\le4)=1-1/5=4/5$, so a renewal by time $4$ has positive probability even though $m(t)/t\to0$.

Common mistake. A tempting but invalid calculation is to take the finite-mean expression $1/\mu$, substitute $\mu=\infty$, and conclude that there are no renewals. For the last example this would incorrectly imply $N(4)=0$, despite $\mathbb{P}(X_1\le4)=4/5$. The correct conclusion is only that the normalized expected count $m(t)/t$ tends to zero as $t$ grows. Individual renewals still occur, and the expected count at a fixed time can be positive.

Keep the assumptions attached to the conclusions. Strictly positive interarrivals rule out zero waiting times and ensure the epochs increase. Independence and identical distribution allow the same truncated mean to be used at every step. Finite mean is required for the reciprocal-rate benchmark; infinite mean instead yields the zero expected linear rate proved above. A claim about almost-sure sample-path rates is a different limit statement and is not proved by this expected-count argument.

Check Your Understanding

For each question, identify the assumptions used and distinguish a fixed-time count from a long-run rate.

If $X_k=3$ almost surely, write $N(t)$ using the chosen endpoint convention and find $\lim_{t\to\infty}m(t)/t$.
Why does truncating at $b$ make the renewal epochs no later than those of the original process?
For an interarrival law with $\mathbb{E}[X_1]=\infty$, what does $m(t)/t\to0$ assert, and what does it not assert?
In the truncation proof, why is $\{M_b(t)+1\ge k\}$ independent of $Y_k$?
Which part of the finite-mean conclusion follows from the proved truncation bound, and which stronger conclusion is the standard elementary renewal theorem?

Finite-Mean and Infinite-Mean Interarrivals

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