From a Smoothed Limit to an Interval Limit
Blackwell’s renewal theorem describes a local feature of a renewal process: far out in time, the expected number of renewals in an interval of fixed width approaches a constant that depends only on the interval width and the mean interarrival time. In the nonarithmetic case, the limit does not oscillate with the position of the interval. A useful proof strategy is to first obtain limits for averages against continuous functions, then squeeze the indicator of an interval between continuous approximations.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space carrying independent, identically distributed random variables \(X_1,X_2,\ldots\) with common distribution \(F\). Assume throughout the theorem that \(X_k>0\) almost surely and \(0<\mu=\mathbb{E}[X_1]<\infty\). Define \(S_0=0\) and \(S_n=X_1+\cdots+X_n\) for \(n\ge1\). The renewal count is \(N(t)=\max\{n\ge0:S_n\le t\}\) for \(t\ge0\). The renewal measure \(U\) assigns to a Borel set \(C\subseteq[0,\infty)\) the expected number of renewal epochs in that set, including \(S_0\): \(U(C)=\sum_{n\ge0}\mathbb{P}(S_n\in C)\). In particular, \(U((t,t+h])=\mathbb{E}[N(t+h)-N(t)]\) for \(t\ge0\) and \(h>0\).
The key input in the smoothing approach is a limit for continuous test functions. Here a test function \(g:\mathbb{R}\to\mathbb{R}\) is continuous and zero outside a bounded interval. The smoothed renewal limit is the standard result that, under the theorem’s hypotheses, for every such \(g\), $$ \lim_{t\to\infty}\int_{[0,\infty)}g(x-t)\,U(dx) =\frac{1}{\mu}\int_{\mathbb{R}}g(y)\,dy. $$ This is the renewal-measure smoothing lemma. Its proof uses the renewal transform and a technical analysis of its behavior near frequency zero and away from zero; that Fourier-analytic proof is beyond this tutorial. The lemma is stated with its hypotheses explicitly, rather than treating the smoothing step as an elementary estimate. The argument below proves how this smoothed limit yields the interval theorem.
Continuous Bounds for an Interval
Fix \(h>0\), and choose \(0<\varepsilon<h/2\). Define continuous functions \(g_-\) and \(g_+\) as follows. The upper function \(g_+\) is zero outside \([-\varepsilon,h+\varepsilon]\), rises linearly from zero to one on \([-\varepsilon,0]\), equals one on \([0,h]\), and falls linearly to zero on \([h,h+\varepsilon]\). The lower function \(g_-\) is zero outside \([0,h]\), rises linearly from zero to one on \([0,\varepsilon]\), equals one on \([\varepsilon,h-\varepsilon]\), and falls linearly to zero on \([h-\varepsilon,h]\). Then $$ g_-(y)\le \mathbf{1}_{(0,h]}(y)\le g_+(y) $$ for every real \(y\). In particular, the inequalities hold at both endpoints: \(g_-(0)=0\), while \(g_+(0)=g_+(h)=1\).
Proof. The integrals of the two triangular bounds are their rectangular plateau areas plus the areas of their ramps. For \(g_-\), the plateau has width \(h-2\varepsilon\), and the two ramps each have area \(\varepsilon/2\). Thus \(\int_{\mathbb{R}}g_-(y)\,dy=h-\varepsilon\). For \(g_+\), the plateau has width \(h\), and its two ramps each have area \(\varepsilon/2\), so \(\int_{\mathbb{R}}g_+(y)\,dy=h+\varepsilon\). The lower bound \(h-2\varepsilon\) in the proposition is weaker than the lower integral \(h-\varepsilon\), but will suffice. Integrating the pointwise inequalities against the translated measure \(U\) gives $$ \int g_-(x-t)\,U(dx)\le U((t,t+h])\le \int g_+(x-t)\,U(dx). $$ The smoothing lemma sends the left and right sides to \((h-\varepsilon)/\mu\) and \((h+\varepsilon)/\mu\), respectively. Therefore the liminf is at least \((h-\varepsilon)/\mu\), and hence also at least the weaker bound \((h-2\varepsilon)/\mu\); the limsup is at most \((h+\varepsilon)/\mu\). These bounds hold for every allowed \(\varepsilon\). Letting \(\varepsilon\downarrow0\) forces the liminf and limsup both to equal \(h/\mu\), proving the claimed limit. \(\square\)
This proof separates two tasks. The smoothing lemma supplies convergence for continuous averages. The approximation argument then transfers that convergence to the discontinuous interval indicator. The strict-left, closed-right interval convention matters: the pointwise bounds above explicitly cover both endpoints, so atoms at \(t\) or \(t+h\) do not invalidate the squeeze.
Worked Example: An Exponential Interarrival Law
Suppose \(X_k\) has exponential distribution with rate \(\lambda>0\), so its density is \(f(x)=\lambda e^{-\lambda x}\) for \(x>0\). This law is nonarithmetic because it has a density and is not concentrated on any countable lattice. Its mean is \(\mu=1/\lambda\). We can verify the interval limit directly, rather than relying only on the theorem.
For \(n\ge1\), the sum \(S_n\) has the gamma density $$ f_n(x)=\lambda e^{-\lambda x}\frac{(\lambda x)^{n-1}}{(n-1)!},\qquad x>0. $$ For \(x>0\), the density of the part of \(U\) away from its atom at zero is the sum of these densities: $$ \sum_{n=1}^{\infty}f_n(x) =\lambda e^{-\lambda x}\sum_{n=1}^{\infty}\frac{(\lambda x)^{n-1}}{(n-1)!} =\lambda e^{-\lambda x}e^{\lambda x} =\lambda. $$ Thus \(U\) has constant density \(\lambda\) on \((0,\infty)\), in addition to its atom at zero. For \(t\ge0\) and \(h>0\), the interval \((t,t+h]\) excludes zero and has length \(h\), so $$ U((t,t+h])=\lambda h=\frac{h}{1/\lambda}=\frac{h}{\mu}. $$ This equality holds for every \(t\ge0\), which checks the limiting value as well as the mean used in it.
Worked Example: Smoothing Bounds for a Nonarithmetic Two-Point Law
Let \(X_k=1\) or \(X_k=\sqrt{2}\), each with probability \(1/2\). These variables are strictly positive and have mean \(\mu=(1+\sqrt{2})/2\). The law is nonarithmetic: if both possible values belonged to \(d\mathbb{Z}\), their ratio would be rational, whereas \(1/\sqrt{2}\) is irrational.
Take \(h=2\) and \(\varepsilon=1/4\). The lower and upper continuous bounds from the proof have integrals \(h-\varepsilon=2-1/4=7/4\) and \(h+\varepsilon=2+1/4=9/4\). Since \(1/\mu=2/(1+\sqrt{2})\), the smoothing lemma and the pointwise bounds give $$ \frac{7}{2(1+\sqrt{2})} \le \liminf_{t\to\infty}U((t,t+2]) \le \limsup_{t\to\infty}U((t,t+2]) \le \frac{9}{2(1+\sqrt{2})}. $$ The wider lower bound stated in the proposition also holds, since \((h-2\varepsilon)/\mu=(3/2)\,2/(1+\sqrt{2})=3/(1+\sqrt{2})\), which is smaller than \(7/[2(1+\sqrt{2})]\). Taking \(\varepsilon\) arbitrarily close to zero makes the lower and upper integral bounds converge to \(2/\mu=4/(1+\sqrt{2})\). The theorem therefore gives $$ \lim_{t\to\infty}U((t,t+2])=\frac{4}{1+\sqrt{2}}. $$ This example illustrates the approximation step; it does not require computing the renewal measure at each finite \(t\).
Where the Assumptions Enter
Nonarithmeticity is used in the smoothed renewal limit: it rules out nonzero frequency resonances that would preserve a periodic oscillation. The finite, positive mean supplies the normalization \(1/\mu\) in that limit. Strict positivity of the interarrivals gives well-ordered renewal epochs, while the finite-mean assumption rules out an infinite value for the normalization. The smoothing proof does not replace these hypotheses with conditions on the interval approximation; it relies on them before the squeeze begins.
Worked Example: Why the Lattice Case Oscillates
Let \(X_k=1\) almost surely. Then \(\mu=1\), and the renewal epochs are \(S_n=n\) for integers \(n\ge0\). This is arithmetic with span \(1\). Set \(h=1/2\). For integer \(n\ge1\), the strict-left, closed-right interval \((n,n+1/2]\) contains no integer renewal epoch, so \(U((n,n+1/2])=0\). For \(t=n+1/2\), the interval \((n+1/2,n+1]\) contains exactly the epoch \(n+1\), including its right endpoint, so \(U((n+1/2,n+1])=1\). The two sequences of values do not converge to the nonarithmetic prediction \(h/\mu=1/2\).
Common mistake. Applying the nonarithmetic theorem to this deterministic interarrival law would give \(U((t,t+1/2])\to1/2\). That calculation uses the correct mean, \(1\), but ignores the arithmetic hypothesis. The direct counts above alternate between \(0\) and \(1\) along two sequences of positions, so there is no such limit.
The main practical lesson is to distinguish local smoothing from a pointwise interval count. The continuous-function limit is the difficult renewal-theoretic input; the interval approximation is a separate, explicit squeeze. Nonarithmeticity removes persistent lattice oscillations, and letting the ramp width shrink removes the approximation error. The resulting limit concerns expected counts in a fixed window, not the distribution of the count or a claim that individual sample paths have a fixed number of renewals in every distant window.
Check Your Understanding
Use the definitions and bounds in the tutorial to answer the following questions.
- Why does the renewal measure include an atom at zero, and why does that atom not affect \(U((t,t+h])\) when \(t\ge0\)?
- For \(h=3\) and \(0<\varepsilon<3/2\), compute the integrals of the lower and upper continuous approximations.
- Explain which step of the smoothing argument uses nonarithmeticity and which quantity supplies the limiting normalization.
- For deterministic interarrivals \(X_k=1\), compute \(U((n,n+1/2])\) and \(U((n+1/2,n+1])\) for integer \(n\ge1\), respecting the interval endpoints.
- In the exponential example, verify the renewal-measure density by summing the gamma densities and state the resulting limit for a window of width \(h\).