From Local Renewal Rates to Forced Equations
Blackwell's renewal theorem describes the asymptotic renewal mass in a fixed-width window. A forced renewal equation asks a related question: what happens when each renewal contributes an amount that depends on how much time remains? The key renewal theorem answers this by combining the long-run renewal rate with the total accumulated forcing. It is useful in waiting-time calculations, reliability models, and other problems where a quantity is repeatedly renewed and then weighted by elapsed time.
Let \(X_1,X_2,\ldots\) be independent and identically distributed, strictly positive random variables on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), with common distribution \(F\) and finite mean \(\mu=\mathbb{E}[X_1]\). Define the renewal times \(S_0=0\) and \(S_n=X_1+\cdots+X_n\) for \(n\ge1\). Time is continuous, \(t\ge0\). The renewal measure is \(U(B)=\sum_{n\ge0}\mathbb{P}(S_n\in B)\) for measurable \(B\subseteq[0,\infty)\); it includes the unit mass at \(S_0=0\).
Here the upper sum at width \(h\) uses \(h\sum_{k\ge0}\sup_{kh\le s<(k+1)h}g(s)\), and the lower sum uses the corresponding infima. For a nonnegative function, DRI requires these sums to be finite and their difference to vanish as the mesh becomes fine. A continuous function with compact support is DRI; so is \(g(t)=e^{-bt}\) for \(b>0\). The integral in the theorem below is the ordinary Lebesgue integral \(\int_0^\infty g(s)\,ds\).
The factor \(1/\mu\) is the long-run renewal rate. The integral measures the total forcing contributed by one renewal across possible remaining times. The theorem says that, in the limit, the accumulated contributions equal the total forcing multiplied by the long-run rate. The hypotheses are substantive: finite positive mean supplies a finite, nonzero rate, and nonarithmeticity prevents persistent oscillations tied to a common time grid. The key renewal theorem is a standard result; its general proof requires renewal-measure estimates beyond the arguments developed here, so it is stated rather than proved.
Why the Convolution Solves the Equation
Before applying the theorem, it helps to establish how the renewal convolution relates to the forced equation. The result below also makes precise when the equation has a unique locally bounded solution. For \(n\ge1\), write \(F^{*n}\) for the distribution of \(S_n\), and let \(F^{*0}\) be unit mass at zero.
Proof. Fix \(T<\infty\), and choose \(\theta>0\). Since \(X_1>0\) almost surely, \(r=\mathbb{E}[e^{-\theta X_1}]<1\). For every \(n\ge1\), the event \(S_n\le T\) implies \(e^{-\theta S_n}\ge e^{-\theta T}\), so $$ \mathbb{P}(S_n\le T)\le e^{\theta T}\mathbb{E}[e^{-\theta S_n}] =e^{\theta T}r^n. $$ Let \(M_T=\sup_{0\le s\le T}|g(s)|<\infty\). The absolute value of the \(n\)th term in the proposed series, for \(0\le t\le T\), is at most \(M_T\mathbb{P}(S_n\le T)\). The geometric bound makes the series converge uniformly on \([0,T]\), so it defines a locally bounded function. Separating its \(n=0\) term gives \(g(t)\). In each remaining term, write \(S_n=X_1+(X_2+\cdots+X_n)\); independence makes the sum of those terms equal to \(\int_{[0,t]}z(t-x)\,F(dx)\). Thus the series solves the equation. If \(z_1,z_2\) are two locally bounded solutions, their difference \(d\) satisfies \(d=F*d\), hence, by iterating, \(d=F^{*n}*d\). On \([0,T]\), \(|d(t)|\le \sup_{[0,T]}|d|\,\mathbb{P}(S_n\le T)\), which tends to zero by the same bound. Therefore \(d(t)=0\) for \(0\le t\le T\). Since \(T\) was arbitrary, the solution is unique. \(\square\)
Worked Applications
Worked Example: Exponential Interarrivals and Exponential Forcing
Suppose the interarrival times have exponential distribution with rate \(2\), so \(F(dx)=2e^{-2x}\,dx\) for \(x\ge0\), and take \(g(t)=e^{-t}\) for \(t\ge0\). Find the solution and its limiting value.
The mean is \(\mu=1/2\), and the exponential distribution is nonarithmetic. Also, \(g\) is DRI and $$ \int_0^\infty e^{-s}\,ds=1. $$ The theorem therefore predicts the limit \(1/(1/2)=2\).
For this interarrival law, \(F^{*n}\) has density \(2^n x^{n-1}e^{-2x}/(n-1)!\) for \(n\ge1\). Summing these densities over \(n\ge1\) gives $$ \sum_{n\ge1}\frac{2^n x^{n-1}e^{-2x}}{(n-1)!} =2e^{-2x}\sum_{m\ge0}\frac{(2x)^m}{m!}=2. $$ Thus \(U\) is unit mass at zero plus density \(2\) on \([0,\infty)\), and the convolution solution is $$ z(t)=e^{-t}+2\int_0^t e^{-s}\,ds =e^{-t}+2(1-e^{-t})=2-e^{-t}. $$ As \(t\to\infty\), \(e^{-t}\to0\), so \(z(t)\to2\), as predicted.
Check by substitution. The forced equation requires \(z(t)=e^{-t}+\int_0^t2e^{-2x}z(t-x)\,dx\). Substituting \(z(t-x)=2-e^{-(t-x)}\), its integral term is $$ 2\int_0^t e^{-2x}(2-e^{-(t-x)})\,dx =2(1-e^{-t}). $$ Adding \(g(t)=e^{-t}\) gives \(2-e^{-t}=z(t)\). This verifies the equation as well as the limiting constant.
Worked Example: Compactly Supported Forcing
Keep exponential interarrivals, now with rate \(3\), and use \(g(t)=1-t\) for \(0\le t\le1\), with \(g(t)=0\) for \(t>1\). Find the forced-equation solution and its limit.
The interarrival mean is \(\mu=1/3\), the law is nonarithmetic, and \(g\) is DRI because it is bounded, piecewise continuous, and zero beyond \(1\). Its integral is $$ \int_0^\infty g(s)\,ds=\int_0^1(1-s)\,ds =\left[s-\frac{s^2}{2}\right]_0^1=\frac12. $$ The key renewal theorem gives the limiting value \(\frac{1}{1/3}\cdot\frac12=\frac32\).
For rate \(3\), the renewal measure is unit mass at zero plus density \(3\). Therefore $$ z(t)=g(t)+3\int_0^t g(s)\,ds. $$ For \(0\le t\le1\), the integral is \(t-t^2/2\), so \(z(t)=1-t+3t-\frac32t^2=1+2t-\frac32t^2\). For \(t>1\), the integral is \(1/2\) and \(g(t)=0\), so \(z(t)=3/2\). At \(t=1\), the first formula also gives \(1+2-\frac32=3/2\). Hence the solution is exactly \(3/2\) for \(t\ge1\), agreeing with the theorem's limit.
Check by substitution. For \(t>1\), the renewal-convolution formula gives \(g(t)+3\int_0^t g(s)\,ds=0+3(1/2)=3/2\). At \(t=1\), it gives \(g(1)+3\int_0^1g(s)\,ds=0+3(1/2)=3/2\). The computed tail value is consistent at the endpoint and thereafter.
Worked Example: Arithmetic Interarrivals Can Prevent a Limit
Let every interarrival time equal \(1\), so \(F\) is unit mass at \(1\) and \(\mu=1\). Take \(g(t)=1\) for \(0\le t<1/2\) and \(g(t)=0\) otherwise. This is DRI, and its integral is \(1/2\). What happens to the convolution?
Here \(S_n=n\), so $$ (U*g)(t)=\sum_{n\ge0}g(t-n). $$ Write \(t=m+r\), where \(m\) is a nonnegative integer and \(0\le r<1\). If \(n<m\), then \(t-n\ge1\), so that term is zero. If \(n>m\), then \(t-n<0\), so it is zero by the extension of \(g\). Only \(n=m\) can contribute. Consequently, \(z(t)=1\) when \(0\le r<1/2\), and \(z(t)=0\) when \(1/2\le r<1\).
For every integer \(m\ge0\), \(z(m+1/4)=1\) and \(z(m+3/4)=0\). Both sequences of times tend to infinity, but the corresponding values differ, so \(z(t)\) has no limit. Although the expression \(\mu^{-1}\int_0^\infty g(s)\,ds=1/2\) is defined, it is not the limit here: the interarrival law is arithmetic, violating a key hypothesis.
Interpreting the Conditions and Avoiding Pitfalls
The first two examples show how the theorem turns a convolution problem into an integral. Once the hypotheses hold, the detailed shape of the interarrival distribution affects the transient solution but not the limiting constant beyond its mean. The third example shows why that simplification cannot be used without checking nonarithmeticity. A common grid of possible renewal times can preserve periodic variation indefinitely.
Common mistake: multiplying by the mean instead of dividing by it. With exponential interarrivals of rate \(2\) and \(g(t)=e^{-t}\), the mean is \(1/2\) and the forcing integral is \(1\). The incorrect calculation \(\mu\int g=(1/2)\cdot1=1/2\) reverses the renewal rate. The theorem gives \(\mu^{-1}\int g=2\), and the exact solution \(z(t)=2-e^{-t}\) confirms that its limit is \(2\). The reciprocal mean, not the mean, is the long-run number of renewals per unit time.
When using the theorem, check the assumptions before computing: interarrivals must be strictly positive, their mean must be finite and nonzero, their distribution must be nonarithmetic, and the forcing must be DRI. The integral is over \([0,\infty)\), with the forcing extended by zero to negative arguments in the convolution. The conclusion is a limit as continuous time \(t\) tends to infinity; it is not a statement about convergence in probability or an assertion that every finite-time convolution equals its limit.
Check Your Understanding
For each question, identify the hypotheses you use and show the calculation that supports your answer.
- If \(\mu=4\) and \(\int_0^\infty g(s)\,ds=3\), what limiting value does the key renewal theorem give, provided its other hypotheses hold?
- For exponential interarrivals with rate \(5\) and \(g(t)=e^{-2t}\), compute the predicted limiting value.
- In the arithmetic example, why do the values at \(m+1/4\) and \(m+3/4\) rule out a limit as \(t\to\infty\)?
- State the conditions on the interarrival law and forcing function needed to apply the key renewal theorem.
- For exponential interarrivals with rate \(3\) and \(g(t)=1-t\) on \([0,1]\), compute \(z(1)\) from the renewal-convolution formula.