Renewal Convolutions After an Initial Delay
A renewal response often begins only after an initial waiting time: a component is installed after a delivery, a study begins after enrollment, or a system starts operating after setup. The response at time \(t\) can be written as a sum of contributions from each renewal epoch up to \(t\). The key renewal theorem describes the long-run limit of this sum. With an appropriate finite initial delay, the same limit holds: the delay affects the transient response, but not its asymptotic constant.
Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Let \(X_1,X_2,\ldots\) be independent, identically distributed, strictly positive interarrival times with distribution \(F\) and finite mean \(\mu=\mathbb{E}X_1\), where \(0<\mu<\infty\). Set \(S_0=0\) and \(S_n=X_1+\cdots+X_n\) for \(n\ge1\). Let \(D\) be a nonnegative, almost surely finite random variable independent of all the \(X_n\). The delayed renewal epochs are \(D+S_n\), \(n\ge0\). Their counting process \(N_D(t)=\sum_{n\ge0}\mathbf{1}_{\{D+S_n\le t\}}\) has continuous time \(t\ge0\) and state space the nonnegative integers.
Thus \(H_D(t)\) is the expected sum of the forcing contributions from all delayed renewal epochs that have occurred by time \(t\). Independence of \(D\) and the interarrivals gives the useful identity \(H_D(t)=\mathbb{E}[\mathbf{1}_{\{D\le t\}}H(t-D)]\). No assumption that \(D\) has a density is needed; it may, for example, be constant or have point masses.
The Limit and Its Conditions
A function \(g:[0,\infty)\to[0,\infty)\) is directly Riemann integrable (DRI) if its upper and lower Riemann sums on successively finer equal-width partitions are finite and converge to the same finite integral. In particular, its upper sums on unit intervals are summable. This condition rules out forcing functions with too much mass in narrow, repeated spikes; it is stronger than merely requiring \(\int_0^\infty g(u)\,du<\infty\).
The ordinary key renewal theorem used here is a standard result: under the nonarithmetic, finite-mean and DRI assumptions, \(H(t)\to \mu^{-1}\int_0^\infty g(u)\,du\). Its proof is part of the earlier material on the key renewal theorem. We prove completely that this ordinary limit extends to an independent, almost surely finite delay.
Proof. Write \(c=\mu^{-1}\int_0^\infty g(u)\,du\). For each fixed finite \(d\ge0\), the ordinary key renewal theorem gives \(H(t-d)\to c\) as \(t\to\infty\), since \(t-d\to\infty\). To average these limits over \(D\), we need a bound that does not depend on \(t\).
By Blackwell’s renewal theorem for nonarithmetic renewals, \(U((x,x+1])\to1/\mu\) as \(x\to\infty\), so these masses are bounded for large \(x\). The renewal measure is finite on bounded intervals: because \(X_1>0\) almost surely, there is an \(\varepsilon>0\) with \(\mathbb{P}(X_1\ge\varepsilon)>0\); on any bounded time interval, only finitely many such increments can occur, and the expected number of trials to obtain any fixed finite number of them is finite. Consequently, the masses of intervals of length one are bounded over all locations. Let \(C<\infty\) be a uniform bound, enlarged if necessary to cover closed or half-open endpoints.
Partition the values of \(t-s\) into unit intervals. If \(a_k=\sup_{u\in[k,k+1)}g(u)\), the DRI condition gives \(\sum_{k\ge0}a_k<\infty\). Each corresponding interval of \(s\)-values has \(U\)-mass at most \(C\), so \(0\le H(x)\le C\sum_{k\ge0}a_k\) for every \(x\ge0\). This is a finite bound independent of \(x\); also \(H(x)=0\) for \(x<0\).
For each outcome with \(D=d<\infty\), the quantity \(\mathbf{1}_{\{D\le t\}}H(t-D)\) converges to \(c\): eventually \(t\ge d\), and then the ordinary limit applies. The uniform bound just established dominates these random quantities by an integrable constant. The dominated convergence theorem therefore gives $$ H_D(t)=\mathbb{E}[\mathbf{1}_{\{D\le t\}}H(t-D)]\longrightarrow \mathbb{E}[c]=c. $$ This proves the delayed limit. \(\square\)
The theorem has a simple interpretation. The integral of \(g\) measures the total contribution associated with one unit of renewal mass, while the long-run renewal mass per unit time is \(1/\mu\). A finite initial delay shifts the ordinary response in time. Since that shift is finite almost surely, it cannot change the eventual constant. The uniform bound is what makes this statement valid even when \(D\) has a heavy tail or no finite mean.
Worked Calculations
Worked Example: A Fixed Setup Delay
Let the interarrivals be exponential with mean \(\mu=2\), let the fixed delay be \(D=3\), and take \(g(u)=e^{-u}\) for \(u\ge0\). Find the delayed convolution and its limit.
The interarrival rate is \(1/2\). For exponential interarrivals, the renewal epochs after time zero form a Poisson process of rate \(1/2\). Equivalently, the density of \(S_n\), for \(n\ge1\), is \(e^{-s/2}(s/2)^{n-1}/(2(n-1)!)\); summing these densities over \(n\ge1\) gives \(1/2\). Thus \(U(ds)=\delta_0(ds)+\tfrac12\,ds\), where \(\delta_0\) is unit mass at zero. This is a renewal measure, not a transition matrix, so there are no transition rows to check.
For \(x\ge0\), the ordinary convolution is $$ H(x)=g(x)+\frac12\int_0^x g(x-s)\,ds =e^{-x}+\frac12\int_0^x e^{-u}\,du =e^{-x}+\frac12(1-e^{-x}) =\frac12+\frac12e^{-x}. $$
The delayed response is zero for \(t<3\), and for \(t\ge3\) it is $$ H_D(t)=H(t-3)=\frac12+\frac12e^{-(t-3)}. $$
Check: The theorem predicts \(\mu^{-1}\int_0^\infty e^{-u}\,du=\tfrac12\cdot1=\tfrac12\). The explicit expression tends to \(\tfrac12\) as \(t\to\infty\), as required. At \(t=3\), it gives \(H_D(3)=1\), which is also \(H(0)=g(0)=1\): only the renewal epoch at the delayed start contributes.
Worked Example: A Two-Point Random Delay
Keep the exponential interarrivals of mean \(2\) and the forcing function \(g(u)=e^{-u}\), but let \(\mathbb{P}(D=0)=\tfrac13\) and \(\mathbb{P}(D=2)=\tfrac23\). Compute the response for \(t\ge2\).
The ordinary convolution, computed from \(U(ds)=\delta_0(ds)+\tfrac12 ds\), is \(H(x)=\tfrac12+\tfrac12e^{-x}\) for \(x\ge0\). Since the two possible delays are both no greater than \(t\), conditioning on \(D\) gives $$ \begin{aligned} H_D(t) &=\frac13H(t)+\frac23H(t-2)\\ &=\frac13\left(\frac12+\frac12e^{-t}\right) +\frac23\left(\frac12+\frac12e^{-(t-2)}\right)\\ &=\frac12+\frac16e^{-t}+\frac13e^{-(t-2)}. \end{aligned} $$
Check: At \(t=2\), direct conditioning gives \(\tfrac13H(2)+\tfrac23H(0)=\tfrac13(\tfrac12+\tfrac12e^{-2})+\tfrac23=\tfrac56+\tfrac16e^{-2}\). Substitution into the displayed formula gives \(\tfrac12+\tfrac16e^{-2}+\tfrac13=\tfrac56+\tfrac16e^{-2}\), the same value. As \(t\to\infty\), both exponential terms vanish, leaving \(\tfrac12\), independent of the two delay probabilities.
Worked Example: A Compactly Supported Forcing Function
Use exponential interarrivals of mean \(2\), a fixed delay \(D=1\), and \(g(u)=1-u/4\) for \(0\le u\le4\), with \(g(u)=0\) for \(u>4\). Find the limiting delayed response and verify it directly for sufficiently large \(t\).
The forcing function is nonnegative and continuous on \([0,4]\), vanishes after \(4\), and is DRI. Its integral is $$ \int_0^\infty g(u)\,du =\int_0^4\left(1-\frac{u}{4}\right)du =\left[u-\frac{u^2}{8}\right]_0^4 =4-2=2. $$ The key renewal constant is therefore \(\tfrac12\cdot2=1\).
For \(x\ge4\), \(g(x)=0\) and the full support of \(g\) lies inside the integral in the renewal convolution. Hence $$ H(x)=g(x)+\frac12\int_0^x g(u)\,du =0+\frac12\int_0^4\left(1-\frac{u}{4}\right)du =1. $$ For the delayed response, \(H_D(t)=H(t-1)\). Thus \(H_D(t)=1\) for \(t\ge5\), not merely in the limit.
Check: Substituting \(t=5\) gives \(H_D(5)=H(4)=g(4)+\tfrac12\int_0^4g(u)\,du=0+\tfrac12(2)=1\). This agrees with the theorem’s constant.
What Can Go Wrong
Common mistake: adding the initial delay to the mean interarrival time. In the first example, this incorrect calculation would use \(1/(\mu+D)=1/(2+3)=1/5\) as the asymptotic constant. The delay \(D=3\) occurs once, before the renewal sequence begins; it is not repeated between renewals. The correct calculation uses the interarrival mean \(\mu=2\) and the forcing integral \(\int_0^\infty e^{-u}\,du=1\), giving \(1/2\). The exact expression \(H_D(t)=\tfrac12+\tfrac12e^{-(t-3)}\) verifies the correction.
Nonarithmeticity is also essential for a continuous-time limit of this form. For example, let every interarrival equal \(1\), so \(S_n=n\), and take \(g(u)=\mathbf{1}_{[0,1/2]}(u)\). This function is DRI and has integral \(1/2\), but it produces no limit: if \(t=m\) is an integer, exactly one renewal epoch contributes and \(H(m)=1\); if \(t=m+\tfrac34\), no renewal epoch contributes and \(H(t)=0\). The values alternate between \(1\) and \(0\) along these two sequences, so a continuous-time limit does not exist. This example shows why the nonarithmetic assumption cannot be dropped from the stated theorem.
For applying the result, check the assumptions in order: interarrivals are strictly positive with finite, nonzero mean; their distribution is nonarithmetic; the forcing function is DRI; and the delay is nonnegative, finite almost surely, and independent of the interarrivals. The theorem gives a limit, not an exact formula at finite \(t\); exact transient values require evaluating the convolution itself.
Check Your Understanding
Use the theorem’s assumptions explicitly when answering.
- If \(\mu=4\) and \(\int_0^\infty g(u)\,du=6\), what is the delayed renewal limit for any independent, almost surely finite delay?
- For the exponential interarrival example with \(\mu=2\) and \(g(u)=e^{-u}\), let \(D=4\) deterministically. Write \(H_D(t)\) for \(t\ge4\) and check its limit.
- Does the theorem require \(\mathbb{E}D<\infty\)? State the condition on \(D\) that is required.
- In the deterministic unit-interarrival example with \(g=\mathbf{1}_{[0,1/2]}\), compare \(H(m)\) and \(H(m+\tfrac34)\) for integers \(m\ge0\). What hypothesis fails?
- Explain why the proposed constant \(1/(\mu+\mathbb{E}D)\) is not the key renewal constant when \(\mathbb{E}D\) exists.