Boundary and Lattice Pitfalls in Renewal Limits

Four Checks Before Using a Renewal Limit

A renewal limit can be applied correctly and still give the wrong answer if its hypotheses or conventions do not match the model. The most common mismatches concern the lattice span of the interarrival distribution, whether interval endpoints are included, whether the mean interarrival time is finite, and whether the process starts with an ordinary or delayed first renewal. These distinctions matter especially for local limits, which count renewal mass in a fixed-width interval.

Let \(X_1,X_2,\ldots\) be independent, identically distributed, strictly positive interarrival times on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Write \(F\) for their common distribution, \(\mu=\mathbb{E}X_1\), \(S_0=0\), and \(S_n=X_1+\cdots+X_n\) for \(n\ge1\). The ordinary renewal measure is \(U(A)=\sum_{n=0}^{\infty}\mathbb{P}(S_n\in A)\) for measurable \(A\subseteq[0,\infty)\). It includes the initial renewal at \(S_0=0\). For \(t\ge0\), the renewal count \(N(t)=\max\{n\ge0:S_n\le t\}\) counts arrivals after time zero, so \(N(t)=\sum_{n\ge1}\mathbf{1}_{\{S_n\le t\}}\).

Definition: The interarrival distribution is arithmetic if there is a largest \(d>0\) such that \(\mathbb{P}(X_1\in\{d,2d,3d,\ldots\})=1\). This \(d\) is its lattice span. If there is no such \(d\), the distribution is nonarithmetic. A local renewal limit concerns \(U((t,t+h])\) for a fixed width \(h>0\) as \(t\) increases; the endpoint convention is part of the expression.

For nonarithmetic interarrivals with \(0<\mu<\infty\), Blackwell's renewal theorem gives \(U((t,t+h])\to h/\mu\) for each fixed \(h>0\). For arithmetic interarrivals with span \(d\) and \(0<\mu<\infty\), the lattice version instead accounts for the lattice sites in the interval: along \(t=md+r\), with fixed \(r\in[0,d)\), its limit is \(d/\mu\) times the number of integers \(k\) satisfying \(r<kd\le r+h\). These are standard Blackwell theorems established in the earlier material on renewal limits. The strict inequality at the left endpoint and the weak inequality at the right are not interchangeable when the renewal measure has atoms.

Worked Examples: Locating the Pitfalls

Worked Example: The Lattice Span Changes the Local Limit

Let each interarrival time have the geometric distribution on the positive integers, with \(\mathbb{P}(X_i=k)=2^{-k}\) for \(k\ge1\). Its probabilities sum to \(\sum_{k\ge1}2^{-k}=1\), its lattice span is \(d=1\), and its mean is \(\mu=2\). Find the renewal mass at each integer and compare the interval \((m,m+\tfrac12]\) with the nonarithmetic expression \(h/\mu\).

For \(n\ge1\), the event \(S_k=n\) requires \(k\) positive integer interarrivals summing to \(n\). There are \(\binom{n-1}{k-1}\) such ordered compositions, and each has probability \(2^{-n}\). Therefore

U(\{n\})=\sum_{k=1}^{n}\mathbb{P}(S_k=n) =\sum_{k=1}^{n}\binom{n-1}{k-1}2^{-n} =2^{-n}2^{n-1} =\tfrac12 .

At zero, \(U(\{0\})=1\), since \(S_0=0\) and all later \(S_k\) are positive. For every integer \(m\ge1\), the interval \((m,m+\tfrac12]\) contains no integer, so \(U((m,m+\tfrac12])=0\). The tempting nonarithmetic answer would be \(h/\mu=(\tfrac12)/2=\tfrac14\), which is wrong here. For the interval \((m,m+1]\), there is exactly one renewal site, \(m+1\), and its mass is \(\tfrac12\). This agrees with the lattice value \(d/\mu=1/2\).

Check: The row of probabilities defining \(X_i\) is a distribution because \(\sum_{k\ge1}2^{-k}=1\). The calculated mass in \((m,m+1]\) is \(U(\{m+1\})=\tfrac12\), exactly \(1/2\), while the half-width interval has mass zero. The discrepancy is the lattice effect, not an arithmetic error.

The lattice formula is sensitive to the position of the interval relative to the lattice, not just its length. When the interval width is a multiple of \(d\), the number of included sites is the same for every phase away from the initial boundary; otherwise the count can vary with the phase. Replacing the lattice theorem by \(h/\mu\) discards this information.

Worked Example: Changing an Endpoint Changes the Answer

Take deterministic interarrivals \(X_i=1\). Then \(\mu=1\), \(S_n=n\), and \(U\) assigns mass one to each nonnegative integer. Compare the intervals \((m,m+1]\) and \([m,m+1]\) for integer \(m\ge1\).

The first interval contains just \(m+1\), so \(U((m,m+1])=1\). The closed interval contains both \(m\) and \(m+1\), so \(U([m,m+1])=2\). Thus the endpoints contribute an additional unit of renewal mass, even though the interval width is one in both expressions. More generally, for any measure \(U\),

\begin{aligned} U([t,t+h])&=U((t,t+h])+U(\{t\}),\\ U([t,t+h))&=U((t,t+h])+U(\{t\})-U(\{t+h\}),\\ U((t,t+h))&=U((t,t+h])-U(\{t+h\}). \end{aligned}

Each identity follows by splitting the interval into disjoint pieces. For example, \([t,t+h]\) is the disjoint union of \(\{t\}\) and \((t,t+h]\); additivity of \(U\) gives the first identity. The other two follow by adding the left endpoint and, where needed, removing the right endpoint.

Check: With \(t=m\), \(h=1\), and unit mass at each integer, the formulas give \(U([m,m+1])=1+1=2\), \(U([m,m+1))=1+1-1=1\), and \(U((m,m+1))=1-1=0\). The answers differ because the endpoint atoms are real renewal mass.

This example also warns against transferring an interval limit between conventions without checking endpoint atoms. In a nonarithmetic setting, endpoint atoms may become negligible under suitable hypotheses; in a lattice setting, they can persist at every large time. The exact interval named in the theorem must therefore be preserved.

Worked Example: Infinite Mean Does Not Give the Finite-Mean Rate

Let \(X_i\) take positive integer values with \(\mathbb{P}(X_i=k)=c/k^2\), where \(c=(\sum_{k\ge1}k^{-2})^{-1}\). The normalizing constant is finite and positive, so these probabilities sum to one. The mean is infinite because \(\mathbb{E}X_i=c\sum_{k\ge1}1/k=\infty\).

For this model the finite-mean renewal-rate formula \(N(t)/t\to1/\mu\) cannot be used: its hypothesis \(0<\mu<\infty\) fails. In fact, \(N(t)/t\to0\) almost surely. To see why, fix any positive integer \(K\). The variables \(X_i\wedge K\) are bounded independent, identically distributed random variables. The strong law of large numbers gives

\liminf_{n\to\infty}\frac{S_n}{n} \ge \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n(X_i\wedge K) =\mathbb{E}(X_1\wedge K).

As \(K\) increases, \(\mathbb{E}(X_1\wedge K)\) increases without bound, because \(\mathbb{E}X_1=\infty\). Hence \(S_n/n\to\infty\) almost surely. If \(N(t)=n\) is sufficiently large, then \(S_n\le t\), and for any fixed \(\varepsilon>0\), eventually \(S_n\ge n/\varepsilon\). Consequently \(n\le\varepsilon S_n\le\varepsilon t\). This proves \(N(t)/t\to0\) almost surely.

Check: The mean diverges by the harmonic series, so \(1/\mu\) is not a positive finite renewal rate. The derived count rate is zero, not the positive finite-mean rate one might obtain by incorrectly inserting a finite number for \(\mu\). This count-rate conclusion does not, by itself, determine a fixed-window local limit; that requires its own hypotheses and analysis.

The strong law used here is the standard law for bounded independent, identically distributed variables; it applies separately for each fixed \(K\). The argument only uses those bounded variables and then lets \(K\) increase. It does not assert a finite-mean Blackwell limit when the mean is infinite.

Worked Example: An Initial Delay Shifts the Lattice

Suppose the interarrivals are deterministic, \(X_i=1\), but the first renewal occurs after a deterministic delay \(Y=\tfrac12\). The delayed epochs are \(Y+S_n=\tfrac12+n\) for \(n\ge0\). Compare this process with the ordinary process, whose epochs are \(S_n=n\), on the interval \((m,m+\tfrac12]\) for integer \(m\ge1\).

For the ordinary process, the interval contains no integer and has renewal mass zero. For the delayed process, it contains exactly the epoch \(m+\tfrac12\), with mass one. The interarrival law after the first renewal and its mean are unchanged; the difference comes entirely from the starting phase. In particular, treating a delayed process as ordinary can change a lattice-window count even at arbitrarily large times.

Check: The ordinary epochs in the interval would have to be integers strictly greater than \(m\) and at most \(m+\tfrac12\), and there are none. The delayed epochs satisfy \(m<m+\tfrac12\le m+\tfrac12\), so exactly one is included. Both endpoint tests use the stated interval \((m,m+\tfrac12]\).

What a Delayed Renewal Measure Actually Counts

Let \(Y\ge0\) be a first delay, independent of the interarrivals, with distribution \(G\). Define the delayed renewal measure \(V(A)=\sum_{n=0}^{\infty}\mathbb{P}(Y+S_n\in A)\); unlike the ordinary measure, it need not have mass at zero. For measures \(G\) and \(U\), their convolution is \((G*U)(A)=\int U(A-y)\,G(dy)\), where \(A-y=\{x\ge0:x+y\in A\}\).

Proposition: Under the independence assumption above, the delayed renewal measure satisfies \(V=G*U\).

Proof. For any measurable \(A\subseteq[0,\infty)\), condition on \(Y=y\). Independence gives \(\mathbb{P}(Y+S_n\in A\mid Y=y)=\mathbb{P}(S_n\in A-y)\). Sum these nonnegative probabilities over \(n\ge0\), then integrate over \(G(dy)\). Interchanging the sum and integral is valid for nonnegative terms. The result is \(V(A)=\int\sum_{n\ge0}\mathbb{P}(S_n\in A-y)\,G(dy)=\int U(A-y)\,G(dy)=(G*U)(A)\). \(\square\)

This identity makes the initial delay explicit: the delayed measure is an average of shifted ordinary renewal measures. It does not make the two measures identical, nor does it justify using an ordinary lattice phase for a delayed process. In applications, first identify the law of \(Y\), then use the convolution or an appropriate delayed-renewal result.

A Practical Diagnostic

Before applying a renewal limit, check the model in this order:

Check the mean. The finite-mean Blackwell and renewal-rate formulas require \(0<\mu<\infty\). A proper interarrival distribution can still have infinite mean.
Check the span. Determine whether the interarrivals lie on a lattice and, if so, find the largest span \(d\). Do not replace the lattice limit by \(h/\mu\) without accounting for lattice sites.
Write the interval exactly. Record whether its left and right endpoints are included. If \(U\) has atoms there, changing either endpoint changes its mass.
Check the start. An ordinary process has epochs \(S_n\) starting with \(S_0=0\); a delayed process has epochs \(Y+S_n\). A delay can shift a lattice and alter local counts.

Common mistake. For unit deterministic interarrivals, one might write \(U([m,m+1])\to1\) by applying the interval-length expression \(h/\mu\). This calculation silently substitutes a nonarithmetic formula and also overlooks the closed left endpoint. Directly counting the atoms gives \(U([m,m+1])=U(\{m\})+U(\{m+1\})=1+1=2\). For the half-open interval \((m,m+1]\), the correct count is one. Always match the theorem's span and endpoints to the interval being counted.

Check Your Understanding

For each question, state the relevant convention or hypothesis before calculating.

For the geometric interarrivals with probabilities \(\mathbb{P}(X_i=k)=2^{-k}\), what is \(U((m,m+2])\) for integer \(m\ge1\)? Compare it with the lattice limit formula.
With deterministic unit interarrivals, compute \(U((m,m+1))\) and explain which endpoint atom is excluded.
For the law \(\mathbb{P}(X_i=k)=c/k^2\), identify the failed hypothesis in the finite-mean renewal-rate formula.
With unit interarrivals and delay \(Y=\tfrac12\), how many delayed renewal epochs lie in \((m+\tfrac12,m+\tfrac32]\) for integer \(m\ge0\)?
State the measure identity relating a delayed renewal measure to the ordinary renewal measure and the initial-delay distribution.