From a Limiting Distribution to a Mean
At a late observation time in a renewal process, the forward recurrence time is the wait from that observation until the next renewal. Its limiting distribution describes the residual wait seen at a typical time, rather than the length of a newly begun interarrival interval. Integrating this limiting distribution gives a simple expression for its mean, but the expression depends on the second moment of the interarrival time.
Let \((X_k)_{k\ge1}\) be independent, identically distributed, strictly positive random variables on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). They are the interarrival times, with common distribution \(F\), survival function \(\overline F(x)=\mathbb{P}(X_1>x)\), and mean \(\mu=\mathbb{E}[X_1]\). Assume first that \(0<\mu<\infty\). Define \(S_0=0\), \(S_n=\sum_{k=1}^n X_k\), and \(N(t)=\max\{n\ge0:S_n\le t\}\). The forward recurrence time at \(t\) is \(R(t)=S_{N(t)+1}-t\). In particular, it is the time to the next renewal, so a renewal exactly at \(t\) is not counted as the next one.
Here, nonarithmetic means that the interarrival distribution is not concentrated on the integer multiples of any fixed positive number. The formula defines a probability distribution because \(\int_0^\infty \overline F(u)\,du=\mu\). Its tail form will be especially useful: a residual wait exceeds \(x\) when the observation falls sufficiently far from the end of the interval containing it. The tail formula weights long intervals more heavily than a fresh interarrival law does.
The mean of a nonnegative random variable can be obtained by integrating its tail. Applying that identity to \(R_\infty\) gives the mean residual life.
Proof. For any nonnegative random variable \(Y\), the tail-integral identity is \(\mathbb{E}[Y]=\int_0^\infty\mathbb{P}(Y>x)\,dx\), with either side allowed to be infinite. Using the stated limiting tail and Tonelli’s theorem for the nonnegative integrand, $$ \mathbb{E}[R_\infty] =\frac{1}{\mu}\int_0^\infty\int_x^\infty \overline F(u)\,du\,dx =\frac{1}{\mu}\int_0^\infty u\,\overline F(u)\,du. $$ The second equality follows by changing the order of integration: for a fixed \(u\ge0\), the variable \(x\) ranges from \(0\) to \(u\), an interval of length \(u\). Also, since \(X_1\ge0\), the same tail-integral identity applied to \(X_1^2\) gives $$ \mathbb{E}[X_1^2]=\int_0^\infty \mathbb{P}(X_1^2>y)\,dy =2\int_0^\infty u\,\mathbb{P}(X_1>u)\,du. $$ The last equality is the substitution \(y=u^2\); the strict tail event gives the same integral at any atoms. Substitution into the expression for \(\mathbb{E}[R_\infty]\) proves the formula, including the infinite case. Since \(0<\mu<\infty\), the ratio is finite exactly when the second moment is finite. \(\square\)
Comparing Residual Life with an Interarrival Time
The factor of \(1/2\) matters. The quantity \(\mathbb{E}[X_1^2]/\mu\) is not the mean residual life: it is twice that value. If the variance is finite, the comparison with the mean interval \(\mu\) can be made directly: $$ \mathbb{E}[R_\infty] =\frac{\mu^2+\operatorname{Var}(X_1)}{2\mu} =\frac{\mu}{2}\left(1+\frac{\operatorname{Var}(X_1)}{\mu^2}\right). $$ The ratio \(\operatorname{Var}(X_1)/\mu^2\) is the squared coefficient of variation. It follows that the mean residual life is greater than \(\mu\) when this ratio exceeds \(1\), equal to \(\mu\) when it equals \(1\), and less than \(\mu\) when it is below \(1\). The inspection effect does not imply that the residual wait always exceeds the mean interval. When the second moment is infinite, the limiting residual mean is infinite and therefore exceeds the finite interarrival mean.
Worked Example: Exponential Interarrivals
Suppose the interarrival times are exponential with rate \(2\), so \(\overline F(u)=e^{-2u}\) for \(u\ge0\). This is a nonarithmetic law. Compute and check the limiting residual-life mean.
The mean and second moment of an exponential random variable with rate \(2\) are $$ \mu=\int_0^\infty e^{-2u}\,du=\frac12,\qquad \mathbb{E}[X_1^2]=2\int_0^\infty u e^{-2u}\,du=\frac12. $$ Thus the theorem gives \(\mathbb{E}[R_\infty]=(\frac12)/(2\cdot\frac12)=\frac12\).
We can check this from the entire limiting tail. For \(x\ge0\), $$ \mathbb{P}(R_\infty>x) =\frac{1}{1/2}\int_x^\infty e^{-2u}\,du =2\cdot\frac12 e^{-2x} =e^{-2x}. $$ At \(x=0\) the tail is \(1\), and its integral over \(x\ge0\) is \(\int_0^\infty e^{-2x}\,dx=\frac12\), matching the moment calculation. Here the mean residual life equals the mean interarrival time.
Worked Example: Erlang Interarrivals with a Smaller Residual Mean
Let the interarrival density be \(f(x)=x e^{-x}\) for \(x\ge0\), the gamma distribution with shape \(2\) and rate \(1\). It is nonarithmetic. Find its limiting residual mean and compare it with \(\mu\).
Using \(\int_0^\infty x^k e^{-x}\,dx=k!\), the moments are $$ \mu=\int_0^\infty x^2e^{-x}\,dx=2,\qquad \mathbb{E}[X_1^2]=\int_0^\infty x^3e^{-x}\,dx=6. $$ Consequently, \(\mathbb{E}[R_\infty]=6/(2\cdot2)=3/2\), which is less than the interarrival mean \(2\).
For an independent check, integration of the density gives \(\overline F(u)=e^{-u}(1+u)\). Hence $$ \mathbb{P}(R_\infty>x) =\frac12\int_x^\infty e^{-u}(1+u)\,du =\frac12 e^{-x}(x+2). $$ At \(x=0\) this tail is \(\frac12\cdot2=1\). Its integral is $$ \frac12\int_0^\infty e^{-x}(x+2)\,dx =\frac12(1+2)=\frac32, $$ which verifies the answer and its comparison with \(\mu=2\).
Worked Example: A Mixture with a Larger Residual Mean
Suppose each interarrival time is drawn from an equal mixture of an exponential distribution with rate \(1\) and an exponential distribution with rate \(1/3\). The mixture is nonarithmetic. Compute its mean residual life and verify it from the limiting tail.
The component means are \(1\) and \(3\), and their second moments are \(2\) and \(18\), respectively. Averaging with the two mixture probabilities gives $$ \mu=\frac12(1)+\frac12(3)=2,\qquad \mathbb{E}[X_1^2]=\frac12(2)+\frac12(18)=10. $$ Therefore \(\mathbb{E}[R_\infty]=10/(2\cdot2)=5/2\), greater than the interarrival mean \(2\).
The mixture survival function is \(\overline F(u)=\frac12e^{-u}+\frac12e^{-u/3}\). Its limiting residual tail is $$ \mathbb{P}(R_\infty>x) =\frac12\int_x^\infty\left(\frac12e^{-u}+\frac12e^{-u/3}\right)du =\frac14e^{-x}+\frac34e^{-x/3}. $$ At \(x=0\), this is \(\frac14+\frac34=1\). Integrating the tail gives \(\frac14(1)+\frac34(3)=\frac52\), confirming the moment result. The same formula thus allows the residual mean to be below or above the mean interval, depending on the variability of the interarrival law.
Conditions, Limits, and a Common Mistake
The finite-mean condition is needed to normalize the limiting distribution: its proposed total mass is \(\mu^{-1}\int_0^\infty\overline F(u)\,du=1\). For example, an interarrival law with Pareto survival \(\overline F(u)=1/u\) for \(u\ge1\) has infinite mean, and this integral is infinite, so the stated normalized limit formula does not define a probability distribution. Even when the mean is finite, the nonarithmetic condition cannot simply be omitted for convergence at every deterministic time. If every interarrival is exactly \(1\), then \(R(n)=1\) at integer \(n\), whereas \(R(n+\frac12)=\frac12\). These two sequences of observation times give different residual values, so there is no single limiting distribution as \(t\to\infty\).
There is a second limit issue: convergence in distribution of \(R(t)\) to \(R_\infty\) does not, by itself, establish that \(\mathbb{E}[R(t)]\) converges to \(\mathbb{E}[R_\infty]\). Such a conclusion generally needs an additional condition, such as uniform integrability of the family of residual times. The theorem proved here computes the mean of the limiting distribution; it does not assert convergence of the finite-time means.
Common mistake. In the Erlang example, using \(\mathbb{E}[X_1^2]/\mu=6/2=3\) as the mean residual life misses the factor \(1/2\). The tail calculation checks the correction directly: its integral is \(\frac12\int_0^\infty e^{-x}(x+2)\,dx=\frac32\), not \(3\). The residual is a wait to the end of an interval, not the entire length of the interval containing the observation.
The key comparison can be summarized when the variance is finite:
| Interarrival variability | Mean residual life | Comparison with \(\mu\) |
|---|---|---|
| \(\operatorname{Var}(X_1)<\mu^2\) | \(\frac{\mu}{2}(1+\operatorname{Var}(X_1)/\mu^2)\) | Less than \(\mu\) |
| \(\operatorname{Var}(X_1)=\mu^2\) | \(\mu\) | Equal to \(\mu\) |
| \(\operatorname{Var}(X_1)>\mu^2\) | \(\frac{\mu}{2}(1+\operatorname{Var}(X_1)/\mu^2)\) | Greater than \(\mu\) |
| \(\mathbb{E}[X_1^2]=\infty\), with \(0<\mu<\infty\) | \(\infty\) | Greater than the finite \(\mu\) |
Check Your Understanding
Use the limiting tail formula and the moment identity to check each conclusion.
- For nonarithmetic interarrivals with finite positive mean, write the limiting tail \(\mathbb{P}(R_\infty>x)\) in terms of \(\overline F\).
- If \(\mu=3\) and \(\mathbb{E}[X_1^2]=18\), compute the mean residual life and compare it with \(\mu\).
- For finite variance, what condition makes the mean residual life equal to the mean interarrival time?
- Why does an infinite second moment, together with a finite positive mean, imply an infinite limiting residual mean?
- Give the deterministic interarrival example showing why the nonarithmetic condition matters for a limit at every time \(t\).