Blackwell Limits for Delayed Renewals

Why an Initial Delay Can Change the Start but Not the Rate

A renewal process begins with an event at time zero and then has independent, identically distributed gaps. In applications, observation may begin before the first renewal but the first event may occur only after a waiting period. This initial delay changes which renewals appear in a finite observation window. If the delay is independent and finite almost surely, however, it does not keep shifting renewals farther and farther into the future. The limiting local rate is therefore inherited from the ordinary renewal process, subject to the usual lattice qualification.

Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space. Let $X_1,X_2,\ldots$ be independent, identically distributed, strictly positive random variables with distribution $F$ and mean $\mu=\mathbb E[X_1]$, where $0<\mu<\infty$. Define $S_0=0$ and $S_n=X_1+\cdots+X_n$ for $n\ge1$. Let $D$ be a nonnegative random variable, independent of the entire sequence $(X_n)_{n\ge1}$, with $\mathbb P(D<\infty)=1$. The delayed renewal epochs are $D+S_n$, $n\ge0$.

Definition: The delayed renewal count and delayed renewal measure are $$ N_D(t)=\sum_{n\ge0}\mathbf 1_{\{D+S_n\le t\}},\qquad U_D(B)=\sum_{n\ge0}\mathbb P(D+S_n\in B), $$ for $t\ge0$ and Borel sets $B\subseteq[0,\infty)$. Thus $U_D((t,t+h])=\mathbb E[N_D(t+h)-N_D(t)]$ for $t\ge0$ and $h>0$. The ordinary renewal measure is $U(B)=\sum_{n\ge0}\mathbb P(S_n\in B)$. The term $n=0$ includes the initial epoch, at $D$ in the delayed process and at $0$ in the ordinary process.

The interarrival law is nonarithmetic if its support is not contained in $d\mathbb Z$ for any $d>0$. It is arithmetic with span $d>0$ if $d$ is the largest positive number for which $X_1\in d\mathbb Z$ almost surely. Blackwell’s renewal theorem for the ordinary process is a standard result established in the earlier material on Blackwell limits: for nonarithmetic interarrivals, $U((t,t+h])\to h/\mu$ for every fixed $h>0$; for arithmetic interarrivals of span $d$, the lattice version gives $U((t,t+kd])\to kd/\mu$ for each positive integer $k$. The lattice statement uses intervals of length a whole number of spans; arbitrary lengths can retain phase effects.

A Transfer Result for Delayed Renewals

Theorem (Blackwell limits with a finite independent delay): Under the assumptions above, if $X_1$ is nonarithmetic, then for every fixed $h>0$, $$ U_D((t,t+h])\longrightarrow \frac{h}{\mu}\qquad (t\to\infty). $$ If $X_1$ is arithmetic with span $d$, then for every positive integer $k$, $$ U_D((t,t+kd])\longrightarrow \frac{kd}{\mu}\qquad (t\to\infty). $$ The delay need only be finite almost surely; it need not have finite mean or be bounded.

The proof needs a uniform bound on ordinary renewal mass in a fixed-length interval. This also ensures that averaging over a potentially unbounded delay is justified.

Lemma: If $X_1>0$ almost surely and $\mathbb E[X_1]<\infty$, then $U([0,h])<\infty$ for each finite $h$, and $$ \sup_{y\in\mathbb R}U\bigl((y,y+h]\bigr)\le U([0,h]). $$

Proof. Since $X_1>0$ almost surely, there is an $\varepsilon>0$ for which $p=\mathbb P(X_1\ge\varepsilon)>0$. Indeed, the events $\{X_1\ge 1/m\}$, $m\ge1$, have union $\{X_1>0\}$, so at least one has positive probability. Set $m=\lfloor h/\varepsilon\rfloor$. If $S_n\le h$, at most $m$ of the first $n$ interarrivals can be at least $\varepsilon$. If $p=1$, then $S_n\ge n\varepsilon$ almost surely, so $\mathbb P(S_n\le h)=0$ for $n>m$; hence $U([0,h])\le m+1$. If $0<p<1$, independence gives $$ \mathbb P(S_n\le h)\le \sum_{r=0}^{m}\binom nr p^r(1-p)^{n-r}, $$ where terms with $r>n$ are taken to be zero. For fixed $r\le m$, $\binom nr\le n^r$, and $(1-p)^{n-r}=(1-p)^n(1-p)^{-r}$. Thus the right side is bounded by a fixed polynomial in $n$ times $(1-p)^n$. The sum over $n$ is finite because a geometric sequence with ratio $1-p<1$, multiplied by any fixed-degree polynomial, is summable. Since $U([0,h])=\sum_{n\ge0}\mathbb P(S_n\le h)$, this proves its finiteness in both cases.

For the interval bound, it suffices to consider $y\ge0$, since when $y<0$ the part of $(y,y+h]$ in the renewal process is contained in $[0,h]$. If there is a renewal in $(y,y+h]$, let $S_r$ be the first one there. The number of renewals in that interval, including $S_r$, is at most the number of renewals starting at $0$ in the next $h$ units after $S_r$. The interarrivals following $S_r$ are independent of the interarrivals up to $S_r$ and have the original law. Consequently, conditional on the first renewal in the interval, the expected number counted is at most $U([0,h])$. If no renewal falls there, the count is zero. Taking expectations gives the claimed bound. $\square$

Proof. Fix $h>0$ and write $g_t(x)=U((t-x,t+h-x])$ for $x\ge0$. Independence and the definition of $U_D$ give $$ U_D((t,t+h])=\mathbb E\!\left[U((t-D,t+h-D])\right]=\mathbb E[g_t(D)]. $$ For each fixed $x<\infty$, $t-x\to\infty$ as $t\to\infty$. In the nonarithmetic case, ordinary Blackwell convergence therefore gives $g_t(x)\to h/\mu$. In the arithmetic case with $h=kd$, it gives $g_t(x)\to kd/\mu$, including when $t-x$ is not a lattice point, as covered by the interval form of the lattice theorem. The lemma bounds $g_t(x)$ by $U([0,h])$, uniformly in $t$ and $x$. This bound is integrable with respect to the probability distribution of $D$. Since $D<\infty$ almost surely, the dominated convergence theorem applies to $\mathbb E[g_t(D)]$, yielding the asserted limits. $\square$

The proof identifies the two essential ingredients: each fixed delay leaves the ordinary process’s distant local limit unchanged, and the uniform interval bound permits averaging over the random delay. Independence is what allows the conditional future interarrivals to retain their original renewal law.

Worked Examples

Worked Example: A Deterministic Lattice with a Shifted Start

Let every interarrival equal $2$, so $\mu=2$ and the arithmetic span is $d=2$. Let the independent delay be the constant $D=1$. The delayed epochs are $1,3,5,7,\ldots$. For an interval of length $4=2d$, compute its renewal count and compare it with the limiting rate.

For any real $t\ge0$, the interval $(t,t+4]$ contains exactly two points of the lattice $1+2\mathbb Z_{\ge0}$: the left endpoint is excluded and the right endpoint is included, and the interval spans two lattice gaps. Thus $U_D((t,t+4])=2$ for every $t\ge0$. The theorem predicts $4/\mu=4/2=2$, agreeing exactly with the count. At $t=0$, the interval $(0,4]$ contains the epochs $1$ and $3$, also giving $2$.

Check: The $n$th delayed epoch is $1+2n$. The two epochs in $(t,t+4]$ are the first two lattice points strictly after $t$, and each is at most $t+4$; hence the count is exactly $2$, not merely asymptotically $2$.

Worked Example: A Random Delay and a Lattice-Multiple Window

Keep the deterministic interarrival $X_n=2$, but now let $D=0$ or $D=1$, each with probability $1/2$, independently of the interarrivals. Compute the expected number of delayed renewals in $(0,2]$, and verify the long-run rate for any interval of length $2$.

If $D=0$, the epochs are $0,2,4,\ldots$, so $(0,2]$ contains just the epoch $2$. If $D=1$, the epochs are $1,3,5,\ldots$, so $(0,2]$ contains just the epoch $1$. Therefore $$ U_D((0,2])=\tfrac12(1)+\tfrac12(1)=1. $$ For any real $t\ge0$, each of the two possible delayed processes has spacing $2$, so its interval $(t,t+2]$ contains exactly one epoch. Averaging the two conditional counts gives $U_D((t,t+2])=\tfrac12(1)+\tfrac12(1)=1$. The Blackwell rate is $2/\mu=2/2=1$, as required.

Check: The interval has length one lattice span, $d=2$, and the theorem gives $d/\mu=2/2=1$. The initial delay changes the location of the epoch in the interval, but not the number of epochs in this lattice-multiple window.

Worked Example: An Exponential Nonlattice Renewal Process

Let the interarrivals be exponential with rate $1$, so $\mu=1$, and let $D=2$ deterministically. The interarrival law is nonarithmetic. Compute the expected count in $(0,1]$, and then in $(4,6]$.

The interval $(0,1]$ lies before the initial epoch at time $2$, so it contains no delayed renewal and $U_D((0,1])=0$. For $(4,6]$, the initial epoch at $2$ is not in the interval. The epochs after $2$ form a rate-$1$ Poisson process shifted by $2$, because sums of independent exponential interarrivals are Poisson arrival times. The interval $(4,6]$, shifted back by $2$, has length $2$, so its expected number of Poisson arrivals is $1\cdot2=2$. Thus $U_D((4,6])=2$. The theorem predicts $h/\mu=2/1=2$.

Check: The first calculation uses the interval endpoint convention: the epoch at $2$ is outside $(0,1]$. In the second, the initial epoch is again outside the interval, while the Poisson increment has mean rate times duration, $1\cdot(6-4)=2$.

Lattice Qualifications and Finite-Time Effects

The phrase “the same limiting rate” must be read with the interval convention and the arithmetic structure in view. In the nonarithmetic case, every fixed interval length $h>0$ has the limit $h/\mu$. In the arithmetic case, the theorem guarantees this form for $h=kd$, a positive integer multiple of the span. For other lengths, the local mass can depend on the position of the interval relative to the lattice, and a finite delay does not generally remove that dependence.

For example, with $X_n=2$ and $D=1$, the interval $(2m,2m+3]$ contains the two delayed epochs $2m+1$ and $2m+3$, whereas $(2m+1,2m+4]$ contains only $2m+3$, for every integer $m\ge0$. The two sequences of interval counts are respectively $2$ and $1$, even though the interval length is always $3$. The unqualified value $h/\mu=3/2$ is not a limit along either sequence. This is why the lattice conclusion is stated for lengths that are integer multiples of $d$.

Finite-time counts can differ even when the limiting local rate agrees. In the exponential example, the delayed process has no renewal in $(0,1]$, while the limiting rate for intervals of length $1$ is $1$. The delay can also change the expected cumulative count: the delayed process with $D=2$ has $N_D(0)=0$, while the ordinary process includes its epoch $S_0=0$ in $N_0(0)$. Such early-time differences are compatible with the transfer theorem, which concerns intervals translated to infinity.

Common mistake. A tempting but incorrect argument is to say that $D$ is finite almost surely, so it can simply be ignored for all sufficiently large $t$. There need not be a deterministic time after which every possible delay has passed: $D$ may be unbounded. The valid argument conditions on each fixed value $D=x$, applies the ordinary Blackwell limit as $t-x\to\infty$, and then uses the uniform interval bound and dominated convergence to average over $D$. A second mistake is to use the nonarithmetic formula $h/\mu$ for every lattice interval length; the explicit length-$3$ example shows why the lattice restriction matters.

Takeaway: An independent initial delay that is finite almost surely changes the finite-time placement of renewals, but not the Blackwell local rate. Nonlattice renewals have limit $h/\mu$ for every fixed $h>0$; lattice renewals have the corresponding limit $kd/\mu$ for windows of length $kd$, with arbitrary window lengths potentially retaining phase effects.

Check Your Understanding

For each question, specify the interval endpoint convention and the relevant arithmetic or nonarithmetic assumption.

State the assumptions on $D$ and on the interarrivals that allow the delayed Blackwell limit to be proved.
For arithmetic interarrivals of span $d$, what limit does the theorem give for $U_D((t,t+3d])$?
Why is a uniform bound on ordinary renewal mass in intervals of fixed length needed when $D$ is unbounded?
If $X_n=2$ and $D=1$, count the epochs in $(2m+1,2m+5]$ for an integer $m\ge0$, and compare with the Blackwell value.
Does the theorem require $\mathbb E[D]<\infty$? Explain which assumption on $D$ is used in the proof instead.

Blackwell Limits for Delayed Renewals

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