From Local Renewal Rates to Window Counts
Suppose events occur at renewal epochs: after each event, a new interarrival time is sampled, independently of the previous ones. A fixed-window question asks for the expected number of events in the interval immediately after a large time \(t\) and up to \(t+h\), where the window length \(h>0\) is held fixed. The long-run event rate suggests an answer of about \(h/\mu\), where \(\mu\) is the mean interarrival time. Blackwell’s theorem makes this precise in the nonlattice case. In the lattice case, the location of the window relative to the lattice can matter as much as its length.
Let \(X_1,X_2,\ldots\) be independent, identically distributed, strictly positive interarrival times on a probability space \((\Omega,\mathcal F,\mathbb P)\). Write \(S_0=0\), \(S_n=X_1+\cdots+X_n\) for \(n\ge1\), and \(N(t)=\sum_{n\ge1}\mathbf 1_{\{S_n\le t\}}\) for \(t\ge0\). Assume \(0<\mu=\mathbb E[X_1]<\infty\). The renewal measure is \(U(B)=\sum_{n\ge0}\mathbb P(S_n\in B)\) for measurable sets \(B\subseteq[0,\infty)\). It includes the initial epoch \(S_0=0\), but for windows whose left endpoint tends to infinity that epoch contributes nothing.
The last identity follows by writing the count as \(\sum_{n\ge1}\mathbf 1_{\{t<S_n\le t+h\}}\) and taking expectations term by term. The terms are nonnegative, so the expectation of the sum equals the sum of the expectations. Since \(0\notin(t,t+h]\), adding the \(n=0\) term in the definition of \(U\) does not change the value.
Blackwell’s Theorem and the Fixed-Window Consequences
Thus, in a nonarithmetic model, the expected count in \((t,t+h]\) tends to \(h/\mu\) for every fixed \(h>0\). In an arithmetic model, count the lattice points in the window and use the local limit \(d/\mu\) for each one. The two conclusions coincide when the window length is an integer multiple of the span. For other lengths, shifting the window can change how many lattice points it contains.
Proof of the fixed-window consequences. In the nonarithmetic case, the identity \(\mathbb E[N(t+h)-N(t)]=U((t,t+h])\) and Blackwell’s nonarithmetic limit give the claimed limit \(h/\mu\) directly. For a closed or open endpoint variant, the only possible change is the renewal mass at one or both endpoints. Fix \(\varepsilon>0\). Since \(\{t\}\subset(t-\varepsilon,t]\), for \(t>\varepsilon\) we have \(U(\{t\})\le U((t-\varepsilon,t])\). Blackwell’s theorem applied to this interval of length \(\varepsilon\) gives \(\limsup_{t\to\infty}U(\{t\})\le\varepsilon/\mu\). This holds for every \(\varepsilon>0\), so \(U(\{t\})\to0\). Applying the same argument with \(t+h\) in place of \(t\) shows \(U(\{t+h\})\to0\) as well. Therefore including or excluding either endpoint does not alter the nonarithmetic limit.
In the arithmetic case, write \(h/d=m+r\), where \(m\) is a nonnegative integer and \(0\le r<1\). For any real \(x=t/d\), the number of lattice points \(kd\) in \((t,t+h]\) is \(\lfloor x+h/d\rfloor-\lfloor x\rfloor\). If \(r=0\), this number is always \(m\). All the \(m\) indices \(k\) in the window tend to infinity as \(t\to\infty\); hence each of their renewal masses tends to \(d/\mu\), and the window expectation tends to \(md/\mu=h/\mu\). If \(r>0\), choose \(t_n=nd\), for which the count is \(m\), and \(t'_n=(n+1-r/2)d\), for which it is \(m+1\). Along these two sequences the expectations tend, respectively, to \(m d/\mu\) and \((m+1)d/\mu\). These limits differ because \(d/\mu>0\), so the window expectation has no limit over all real \(t\). When \(m=0\), the first limit is zero; the argument still applies. \(\square\)
Worked Example: A Nonlattice Model with an Exact Window Mean
Let the interarrivals be exponential with rate \(1/2\), so their mean is \(\mu=2\). Find the limiting expected count in \((t,t+\tfrac32]\), and check the result directly from the renewal measure.
The exponential law is continuous, so it is nonarithmetic. Blackwell’s theorem predicts a limit of \(h/\mu=(\tfrac32)/2=\tfrac34\). For a direct check, the density of \(S_n\) for \(n\ge1\) is the Erlang density \[ f_{S_n}(s)=\frac{(1/2)^n s^{n-1}e^{-s/2}}{(n-1)!},\qquad s>0. \] Summing these densities gives the density of the positive-time part of \(U\): \[ \sum_{n\ge1}f_{S_n}(s) =\frac12 e^{-s/2}\sum_{j\ge0}\frac{(s/2)^j}{j!} =\frac12 e^{-s/2}e^{s/2} =\frac12. \] The atom at \(S_0=0\) is outside the window for \(t\ge0\). Therefore \[ \mathbb E[N(t+\tfrac32)-N(t)] =U((t,t+\tfrac32]) =\int_t^{t+3/2}\frac12\,ds =\frac12\cdot\frac32 =\frac34. \] This holds for every \(t\ge0\), not just in the limit. As a check, the exact value agrees with \(h/\mu=(\tfrac32)/2=\tfrac34\). There is no endpoint correction: the positive renewal epochs have a density and therefore place zero mass at any fixed endpoint.
Worked Example: A Lattice Model and an Endpoint Change
Let each interarrival be \(2\) or \(4\), each with probability \(\tfrac12\). Find the limiting expected count in \((4n,4n+2]\), and compare it with the count in the closed interval \([4n,4n+2]\).
The row of probabilities for the two possible interarrivals sums to \(\tfrac12+\tfrac12=1\), and the mean is \(\mu=\tfrac12(2)+\tfrac12(4)=3\). The span is \(d=2\), so the limiting renewal mass at each distant lattice point is \(d/\mu=2/3\).
For an explicit check, let \(u_k=U(\{2k\})\). Then \(u_0=1\), \(u_1=\tfrac12\), and for \(k\ge2\), \(u_k=\tfrac12u_{k-1}+\tfrac12u_{k-2}\), by conditioning on the last interarrival. The formula \(u_k=\tfrac23+\tfrac13(-\tfrac12)^k\) matches \(u_0=1\) and \(u_1=\tfrac12\). For \(k\ge2\), substituting the formula into the recurrence gives \[ \frac12\left(\frac23+\frac13(-\tfrac12)^{k-1}\right) +\frac12\left(\frac23+\frac13(-\tfrac12)^{k-2}\right) =\frac23+\frac13(-\tfrac12)^k, \] so the formula holds for every \(k\). In particular, \(u_k\to\tfrac23\).
The half-open interval \((4n,4n+2]\) contains exactly the lattice point \(4n+2\). Its expected count is \(u_{2n+1}\), which tends to \(2/3\). The closed interval \([4n,4n+2]\) also includes \(4n\), so its expected count is \(u_{2n}+u_{2n+1}\), which tends to \(\tfrac23+\tfrac23=\tfrac43\). The left endpoint contributes a nonvanishing limiting mass. This is why the endpoint convention must be stated in a lattice problem.
Check: at \(n=1\), \(u_2=\tfrac12u_1+\tfrac12u_0=\tfrac34\), and \(u_3=\tfrac12u_2+\tfrac12u_1=\tfrac58\). The formula gives \(u_2=\tfrac23+\tfrac1{12}=\tfrac34\) and \(u_3=\tfrac23-\tfrac1{24}=\tfrac58\), confirming both the recurrence and the computed terms.
Worked Example: A Fixed Window That Has No Lattice Limit
Let every interarrival equal \(2\), and take a window of length \(3\). Compute the expected count for two sequences of window locations tending to infinity.
Here \(\mu=2\), the span is \(d=2\), and the renewal epochs are exactly \(2,4,6,\ldots\). For \(t=4n\), the interval \((4n,4n+3]\) contains only \(4n+2\), so its count is \(1\). For \(t=4n+1\), the interval \((4n+1,4n+4]\) contains \(4n+2\) and \(4n+4\), with the latter included because the right endpoint is closed. Its count is \(2\). Both sequences of left endpoints tend to infinity, but their counts have different limits. Thus no overall limit exists.
The theorem’s lattice prediction explains the discrepancy: \(h/d=3/2\) is not an integer, so the number of lattice points in the window can vary. Along these two sequences the limiting counts are \(1\cdot(d/\mu)=1\) and \(2\cdot(d/\mu)=2\), since \(d/\mu=2/2=1\). If the window instead had length \(4=2d\), every translated half-open window would contain exactly two lattice points and the expected count would be \(2\).
Common Mistakes and a Practical Rule
A common mistake is to apply \(h/\mu\) to every fixed window, regardless of arithmetic structure. In the deterministic example, this would give \(3/2\). But the count is exactly \(1\) along \(t=4n\) and exactly \(2\) along \(t=4n+1\), so \(3/2\) is not a limit. The correction is to check whether the interarrival law is arithmetic before applying a fixed-window limit. In the lattice case, count the lattice points in the specified interval and multiply each distant point’s limiting mass by that count.
A second mistake is to treat \((t,t+h]\) and \([t,t+h]\) as interchangeable. In a nonarithmetic model, endpoint masses vanish at large times, so the limiting expected counts agree. In a lattice model, endpoint masses can approach \(d/\mu\), as the two endpoints in the \(\{2,4\}\) example demonstrate. Keep the event being counted explicit: \(N(t+h)-N(t)\) counts \((t,t+h]\), not \([t,t+h]\).
Check that interarrivals are strictly positive, i.i.d., and have finite positive mean \(\mu\); then determine whether the law is arithmetic.
For \(N(t+h)-N(t)\), use \((t,t+h]\). Do not silently add or remove an endpoint.
For a nonarithmetic law, the limit is \(h/\mu\). For span \(d\), sum the limiting mass \(d/\mu\) over the lattice points included.
If \(h/d\) is an integer, every translated half-open window contains that many lattice points. Otherwise, different phases can give different counts and different subsequential limits.
Check Your Understanding
In each question, keep the interval convention explicit and justify the limiting calculation.
- A nonarithmetic renewal process has mean interarrival time \(5\). What is the limiting expected count in \((t,t+2]\)?
- A lattice renewal process has span \(3\) and mean \(6\). What is the limiting expected count in a half-open window of length \(6\)?
- For span \(d=2\) and window length \(h=5\), how many lattice points can a translated interval \((t,t+h]\) contain?
- Why can replacing \((t,t+h]\) by \([t,t+h]\) change a lattice limit but not a nonarithmetic one?
- For interarrivals equal to \(2\) or \(4\) with equal probabilities, what is the limiting expected count in \((4n,4n+4]\)?