Equilibrium Age and Residual Life Distributions

From Renewal Cycles to Equilibrium Marginals

At a fixed time, the age of a renewal process is the time elapsed since its most recent renewal, while the residual life is the time from observation until the next renewal. Both depend on where the observation falls within its current interarrival interval. Long intervals are more likely to contain a randomly chosen observation time, so these quantities generally do not have the original interarrival distribution. Under appropriate assumptions, each has a limiting marginal distribution as time becomes large.

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, \(\overline F(u)=\mathbb P(X_1>u)\) for \(u\ge0\), and \(\mu=\mathbb E[X_1]\). Assume \(0<\mu<\infty\). Define renewal epochs \(S_0=0\), \(S_n=X_1+\cdots+X_n\) for \(n\ge1\), and \(N(t)=\max\{n:S_n\le t\}\). The age and residual life at time \(t\) are \(A(t)=t-S_{N(t)}\) and \(R(t)=S_{N(t)+1}-t\). In particular, if \(t\) is itself a renewal epoch, then \(A(t)=0\).

The density in this definition is normalized because, by integrating the indicator \(\mathbf 1_{\{X_1>u\}}\), \(\int_0^\infty \overline F(u)\,du=\mathbb E[X_1]=\mu\). This equilibrium law has no atoms, even when the interarrival distribution itself has atoms. The key renewal theorem gives the pointwise-in-time convergence that leads to this law. We use its standard form here: if the interarrival distribution is nonarithmetic with finite positive mean and \(z\) is directly Riemann integrable on \([0,\infty)\), then \(\int_{[0,t]}z(t-s)\,U(ds)\to \mu^{-1}\int_0^\infty z(u)\,du\), where \(U=\sum_{n\ge0}F^{*n}\) is the renewal measure. This is the standard key renewal theorem from renewal limit theory; its proof is not repeated here.

The Limiting Age and Residual-Life Laws

Proof. Fix \(t\ge0\). Let \(U\) be the renewal measure, so \(U(B)=\sum_{n\ge0}\mathbb P(S_n\in B)\) for measurable \(B\subseteq[0,\infty)\). The renewal at \(S_n=s\) is followed by an independent interval distributed as \(X_1\). To have \(A(t)>x\), the last renewal \(s\) must satisfy \(t-s>x\), and its following interval must exceed \(t-s\). The events indexed by the last renewal are disjoint, so $$ \mathbb P(A(t)>x)=\int_{[0,t]} \mathbf 1_{\{t-s>x\}}\overline F(t-s)\,U(ds). $$ The strict inequality is required: if \(t-s=x\), the age is not greater than \(x\). Set \(z_A(u)=\mathbf 1_{\{u>x\}}\overline F(u)\). This function is directly Riemann integrable: it is bounded, has only a possible jump at \(x\), and is dominated in integral by the integrable nonincreasing function \(\overline F\). The key renewal theorem therefore gives the limit \(\mu^{-1}\int_x^\infty\overline F(u)\,du\); changing the value at the single point \(x\) does not change this integral.

For residual life, conditional on a renewal at \(s\le t\), the next renewal is more than \(x\) time units after \(t\) exactly when the following interarrival time exceeds \(t+x-s\). Again, partitioning by the last renewal gives $$ \mathbb P(R(t)>x)=\int_{[0,t]}\overline F(t+x-s)\,U(ds). $$ Here \(z_R(u)=\overline F(u+x)\) is nonincreasing and integrable, hence directly Riemann integrable. The key renewal theorem gives the limit \(\mu^{-1}\int_0^\infty\overline F(u+x)\,du=\mu^{-1}\int_x^\infty\overline F(v)\,dv\), using \(v=u+x\).

Finally, \(\int_0^\infty\overline F(u)\,du=\mu\), so the proposed limiting tail equals \(1\) at \(x=0\), and it decreases to \(0\) as \(x\to\infty\). Its complementary distribution function is \(\mu^{-1}\int_0^x\overline F(u)\,du\), which is continuous in \(x\). The tail limits thus give convergence of the age and residual-life distribution functions at every point, hence convergence in distribution to this same law. \(\square\)

The theorem establishes equality of the two marginal limits. It does not assert that age and residual life are independent, or specify their joint distribution. Those are separate questions: knowing two marginal laws alone does not determine a joint law.

Tails and Equilibrium Expectations

The tail expression is often more convenient than the density. It gives the probability that equilibrium age, or equilibrium residual life, exceeds a specified threshold. In particular, for any \(x\ge0\), that probability is the integrated interarrival survival beyond \(x\), divided by the mean interarrival time. It is not generally just \(\overline F(x)\): the integration accounts for how intervals of different lengths are represented at observation times.

Proof. The tail formula is the definition of the equilibrium distribution function in complementary form. For a nonnegative random variable, the tail integral identity gives \(\mathbb E[Y]=\int_0^\infty\mathbb P(Y>x)\,dx\), allowing an infinite expectation. Substituting the tail and using Tonelli’s theorem for the nonnegative integrand, $$ \mathbb E[Y] =\frac{1}{\mu}\int_0^\infty\int_x^\infty\overline F(u)\,du\,dx =\frac{1}{\mu}\int_0^\infty u\,\overline F(u)\,du. $$ The last equality follows by reversing the order of integration over \(0\le x\le u<\infty\): for each \(u\), the range of \(x\) has length \(u\). Also, \(X_1^2=\int_0^\infty 2u\,\mathbf 1_{\{X_1>u\}}\,du\) almost surely, up to an endpoint of zero Lebesgue measure. Taking expectations and applying Tonelli gives \(\mathbb E[X_1^2]=2\int_0^\infty u\overline F(u)\,du\), including when this is infinite. Combining these identities proves the stated formula. \(\square\)

In particular, the equilibrium mean is finite exactly when the interarrival time has finite second moment. This is a statement about the mean of the limiting equilibrium law; convergence in distribution by itself does not guarantee convergence of the finite-time expectations \(\mathbb E[A(t)]\) or \(\mathbb E[R(t)]\).

Why the Nonarithmetic Condition Matters

Finite mean alone is not enough for the pointwise-in-time conclusion. Consider deterministic interarrivals \(X_k=1\) for every \(k\). The mean is \(1\), but the distribution is arithmetic. If \(t=n+r\) with integer \(n\ge0\) and \(0\le r<1\), then \(A(t)=r\); for \(r<1\), \(R(t)=1-r\). At integer times \(t=n\), \(A(t)=0\) and \(R(t)=1\). In particular, along \(t=n\), the age is always \(0\); along \(t=n+1/2\), it is always \(1/2\). There is no single limiting age distribution as \(t\to\infty\). The equilibrium formula would give a uniform distribution on \([0,1]\), but that describes a time-averaged phase, not the pointwise limit for this process.

Common mistake. For the uniform \((0,2)\) interarrival example, using \(\overline F(1)=1/2\) as the equilibrium tail at \(1\) gives the wrong answer. The tail formula instead gives \(\int_1^2(1-u/2)\,du=1/4\), since \(\mu=1\). The survival probability of one interarrival interval answers a different question; the equilibrium tail integrates the available interval lengths above the threshold and normalizes by the mean. Always check the nonarithmetic and finite-mean assumptions, and distinguish the interarrival tail from the equilibrium tail.