The Joint Equilibrium Law of Age and Residual Life

Why the Joint Law Matters

At a late observation time in a renewal process, the age is the time since the most recent renewal and the residual life is the time until the next one. Their separate equilibrium distributions do not describe how the two quantities fit together. In particular, both are determined by the length of the same renewal interval. The joint law makes that connection explicit and lets us calculate probabilities involving both quantities at once.

Let $X_1,X_2,\ldots$ be independent and identically distributed, strictly positive interarrival times on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Assume their common distribution has a density $f$, is nonarithmetic, and has finite positive mean $\mu=\mathbb{E}[X_1]$. Set $S_0=0$ and $S_n=X_1+\cdots+X_n$ for $n\ge1$. At time $t\ge0$, let $N(t)=\max\{n:S_n\le t\}$, and define the age $A(t)=t-S_{N(t)}$ and residual life $R(t)=S_{N(t)+1}-t$. Thus $A(t)\ge0$, $R(t)>0$ except at renewal instants, and $A(t)+R(t)$ is the length of the interval containing $t$.

Theorem (Joint equilibrium law): Under these assumptions, $(A(t),R(t))$ converges in distribution as $t\to\infty$ to a pair $(A,R)$ with density $$ g(a,r)=\frac{f(a+r)}{\mu},\qquad a\ge0,\ r\ge0. $$ Consequently, for a measurable region $D$ in the nonnegative quadrant, its limiting probability is the integral of this density over $D$, whenever the boundary of $D$ has probability zero under the limiting law.

The density is concentrated according to the sum $a+r$: the larger the interarrival density at that total interval length, the greater the joint density there. Its normalization follows directly by changing variables from $(a,r)$ to $(a,x=a+r)$: the integral of $f(a+r)$ over the quadrant is $\int_0^\infty x f(x)\,dx=\mu$.

Deriving the Joint Limit

The derivation uses the key renewal theorem, a standard renewal-theory result established in the earlier material on renewal limits. In the form needed here, if the interarrival law is nonarithmetic with finite mean $\mu$, and $h$ is directly Riemann integrable, then $$ \int_{[0,t]}h(t-s)\,U(ds)\ \longrightarrow\ \frac{1}{\mu}\int_0^\infty h(v)\,dv, $$ where $U(ds)=\sum_{n\ge0}\mathbb{P}(S_n\in ds)$ is the renewal measure. We use this standard theorem rather than reproving it here.

Proof. First take a continuous function $g_0$ with compact support on the nonnegative quadrant. If the current renewal interval began at time $s$, then its age at time $t$ is $v=t-s$; if its length is $x>v$, its residual life is $x-v$. Summing over all possible renewal epochs and then over the next interval gives $$ \mathbb{E}\!\left[g_0(A(t),R(t))\right] =\int_{[0,t]}h(t-s)\,U(ds),\qquad h(v)=\int_{(v,\infty)}g_0(v,x-v)\,F(dx), $$ where $F$ is the distribution of an interarrival time. This identity counts each possible interval containing $t$; because interarrivals are strictly positive, the renewal epochs are ordered, and the containing interval is unique with probability one at a fixed non-atomic observation time.

The function $h$ is zero for sufficiently large $v$, since $g_0$ has compact support. It is directly Riemann integrable under the stated density and continuity assumptions, so the key renewal theorem applies. Its integral is $$ \int_0^\infty h(v)\,dv =\int_0^\infty f(x)\left(\int_0^x g_0(v,x-v)\,dv\right)dx. $$ Therefore the limit of the expectation is $$ \frac{1}{\mu}\int_0^\infty\int_0^x g_0(v,x-v)\,dv\,f(x)\,dx =\int_0^\infty\int_0^\infty g_0(a,r)\frac{f(a+r)}{\mu}\,dr\,da. $$ The proposed density integrates to one, since $\int_0^\infty\int_0^\infty f(a+r)\,dr\,da=\int_0^\infty x f(x)\,dx=\mu$. Convergence of expectations for continuous compactly supported functions to a probability law implies convergence in distribution. This proves the claimed joint limit. $\square$

The strict inequality $x>v$ in the integral simply excludes an interval whose endpoint is exactly the observation time. With a continuous interarrival distribution, this boundary has probability zero, so it does not change the limiting density or region probabilities.

Marginals, Total Interval Length, and Dependence

Integrating the joint density over one coordinate gives the marginal survival probabilities. For $a,r\ge0$, write $\overline F(u)=\mathbb{P}(X_1>u)$. Then $$ \mathbb{P}(A>a)=\frac{1}{\mu}\int_a^\infty \overline F(u)\,du,\qquad \mathbb{P}(R>r)=\frac{1}{\mu}\int_r^\infty \overline F(u)\,du. $$ For example, the first identity follows by integrating $f(u+v)/\mu$ over $u>a,\ v\ge0$, then integrating first over $u+v$. The corresponding identity for $R$ follows by interchanging the coordinates. Also, the sum $A+R$, the length of the interval containing the observation time, has density $x f(x)/\mu$ for $x>0$: integrating the joint density along $a+r=x$ contributes a segment of length $x$.

Theorem (Independence characterization): Under the joint equilibrium law above, $A$ and $R$ are independent if and only if $X_1$ has an exponential distribution.

Proof. For $x,y\ge0$, integrating the joint density over $a>x,\ r>y$ gives $$ \mathbb{P}(A>x,R>y) =\frac{1}{\mu}\int_{x+y}^\infty (z-x-y)f(z)\,dz =\frac{1}{\mu}\int_{x+y}^\infty \overline F(u)\,du. $$ Define $H(s)=\mu^{-1}\int_s^\infty\overline F(u)\,du$. This is the marginal tail of either $A$ or $R$, and $H(0)=1$. Independence is equivalent to $H(x+y)=H(x)H(y)$ for all $x,y\ge0$. The function $H$ is continuous and nonincreasing. It is positive everywhere: if $H(c)=0$ for some $c>0$, the multiplicative identity gives $H(c/2)=0$, and iteration contradicts continuity at $0$ and $H(0)=1$. Hence $\log H$ is continuous and additive on the nonnegative reals, so $\log H(s)=-\kappa s$ for some $\kappa\ge0$. Since $H(s)\to0$ as $s\to\infty$, $\kappa>0$. Differentiating $H(s)=e^{-\kappa s}$ at points where the survival function is continuous gives $\overline F(s)=\mu\kappa e^{-\kappa s}$; using $\overline F(0)=1$ gives $\mu\kappa=1$. Thus the interarrival law is exponential with mean $\mu$. Conversely, for an exponential density $f(x)=\lambda e^{-\lambda x}$, $\mu=1/\lambda$, and the joint density factors as $\lambda e^{-\lambda a}\lambda e^{-\lambda r}$, which proves independence. $\square$

Worked Probability Calculations

Worked Example: Exponential Interarrivals

Suppose interarrivals are exponential with rate $2$, so $f(x)=2e^{-2x}$ for $x\ge0$. Find the limiting probability that the age exceeds $1/2$ and the residual life exceeds $1$.

The mean is $\mu=1/2$. Substitution into the joint density gives $g(a,r)=4e^{-2(a+r)}$. Therefore $$ \begin{aligned} \mathbb{P}(A>1/2,R>1) &=\int_{1/2}^\infty\int_1^\infty 4e^{-2(a+r)}\,dr\,da\\ &=\left(\int_{1/2}^\infty 2e^{-2a}\,da\right) \left(\int_1^\infty 2e^{-2r}\,dr\right)\\ &=e^{-1}e^{-2}=e^{-3}. \end{aligned} $$ Check: the two marginal tails are $e^{-1}$ and $e^{-2}$, and their product is $e^{-3}$, as required by the factorization of the density.

Worked Example: A Joint Tail for Uniform Interarrivals

Let $X_1$ be uniform on $(0,2)$, so $f(x)=1/2$ on that interval and $\mu=1$. Find $\mathbb{P}(A>1/2,R>1/2)$ in equilibrium.

The joint density is $1/2$ where $a\ge0,\ r\ge0,\ a+r<2$, and zero elsewhere. In the requested region, put $u=a-1/2$ and $v=r-1/2$. Then $u>0,\ v>0,\ u+v<1$. This is a triangle with area $1/2$, so $$ \mathbb{P}(A>1/2,R>1/2)=\frac12\cdot\frac12=\frac14. $$ Check against the marginals: for $0\le s\le2$, $\overline F(s)=1-s/2$. Thus $$ \mathbb{P}(A>1/2)=\int_{1/2}^2(1-u/2)\,du =1-\left(\frac12-\frac{1}{16}\right)=\frac{9}{16}. $$ The residual-life tail is also $9/16$. Their product is $81/256$, not $1/4=64/256$, so the two quantities are not independent.

Worked Example: A Region Defined by the Total Interval Length

For the same uniform interarrival law on $(0,2)$, find the limiting probability that $A+R>1$.

At total length $z=a+r$, the possible ages $a$ range from $0$ to $z$, a segment of length $z$. The joint density is $1/2$ for $0<z<2$. Hence $$ \mathbb{P}(A+R>1)=\frac12\int_1^2 z\,dz =\frac12\left(\frac{4}{2}-\frac{1}{2}\right)=\frac34. $$ Check: the total interval length has density $z f(z)/\mu=z/2$ on $(0,2)$; its integral is $\int_0^2 z/2\,dz=1$, and its tail above $1$ is $3/4$, matching the joint-region calculation.

A Common Mistake and Its Correction

A tempting shortcut is to calculate the equilibrium marginal tails and multiply them, even when the interarrival law is not exponential. For the uniform $(0,2)$ example, that shortcut gives $(9/16)^2=81/256$ for the event $A>1/2,\ R>1/2$. The joint density calculation gives $1/4=64/256$. The error is the assumption of independence: both age and residual life are constrained by the same containing interval, whose length is $A+R$. Independence holds for exponential interarrivals, but not for this uniform law.

For a general region $D$, the safe calculation is to integrate $f(a+r)/\mu$ over the specified part of the nonnegative quadrant. Boundaries may be included or excluded without changing the answer when they have zero probability under this density. In particular, a line such as $a+r=c$ has zero probability, so strict and non-strict inequalities at that boundary give the same probability.

Check Your Understanding

Use the joint density and the assumptions stated above to answer these questions.

For interarrivals uniform on $(0,2)$, what is the equilibrium joint density, and on which region is it nonzero?
Show from the joint density that the equilibrium total interval length has density $x f(x)/\mu$.
For exponential interarrivals with rate $\lambda>0$, write the joint density and identify the marginal law of each coordinate.
For uniform $(0,2)$ interarrivals, calculate the equilibrium probability of $A\le1,\ R\le1$ by integrating over the rectangle.
Why does a finite mean matter in the normalization of the proposed joint density?