Inspection Paradox Examples Across Interarrival Laws

What an Inspector Sees

When observation occurs at a typical time rather than at a renewal epoch, long interarrival intervals are more likely to contain the observation. The interval containing the observation is therefore generally not distributed like a fresh interarrival time. This tutorial calculates that effect for exponential, deterministic, and Pareto interarrivals, and compares the interval length with the time still left in it.

Let $(X_k)_{k\ge1}$ be independent, identically distributed, strictly positive interarrival times on a probability space $(\Omega,\mathcal{F},\mathbb{P})$, with common distribution $F$ and finite mean $\mu=\mathbb{E}[X_1]$. The associated renewal process is $(N(t))_{t\ge0}$, with time index $t\ge0$ and countable state space $\mathbb{N}_0$: set $S_0=0$, $S_n=\sum_{k=1}^n X_k$, and $N(t)=\max\{n\ge0:S_n\le t\}$. We consider an observation at a typical time in the equilibrium renewal setting, not an observation forced to occur at a renewal epoch.

Definition: The length-biased interarrival time $L$ has distribution $\mathbb{P}(L\in dx)=\frac{xF(dx)}{\mu}$. Given $L$, the position $U$ of the observation within its interval is uniform on $[0,L]$. The forward residual time is $R=L-U$. This is the standard equilibrium inspection identity from renewal theory: an interval is sampled in proportion to its length, and, conditional on that interval, the observation is uniform within it.

The formula is normalized because $\int xF(dx)=\mu$. It also shows why the interval seen by an inspector can differ from a fresh interval. For any nonnegative measurable function $g$ for which the expectation exists, the definition gives

\mathbb{E}[g(L)]=\frac{\mathbb{E}[X_1g(X_1)]}{\mu}.

In particular, when $\mathbb{E}[X_1^2]$ is finite, the mean inspected interval is $\mathbb{E}[L]=\mathbb{E}[X_1^2]/\mu$. Conditional on $L=x$, the mean remaining time is $x/2$, so the equilibrium mean residual time is half the mean inspected interval. The next result also gives its distribution.

Theorem: Under the equilibrium inspection identity, for $t\ge0$, $$ \mathbb{P}(R>t)=\frac{1}{\mu}\int_t^\infty \mathbb{P}(X_1>u)\,du. $$ Consequently, $\mathbb{E}[R]=\mathbb{E}[X_1^2]/(2\mu)$, with the expectation allowed to be infinite.

Proof. Conditional on $L=x$, the uniform position $U$ gives $\mathbb{P}(R>t\mid L=x)=(x-t)_+/x$ for $x>0$, where $(x-t)_+=\max\{x-t,0\}$. Averaging under the length-biased law yields $$ \mathbb{P}(R>t)=\frac{1}{\mu}\int (x-t)_+F(dx). $$ For each $x>0$, $(x-t)_+=\int_t^\infty \mathbf{1}_{\{x>u\}}\,du$; the value at the single endpoint $u=x$ does not affect the integral. Since the integrand is nonnegative, interchanging the integrals gives the stated tail formula. For any nonnegative random variable, its expectation equals the integral of its tail, including when that integral is infinite. Thus $$ \mathbb{E}[R]=\frac{1}{\mu}\int_0^\infty\int_t^\infty\mathbb{P}(X_1>u)\,du\,dt =\frac{1}{\mu}\int_0^\infty u\,\mathbb{P}(X_1>u)\,du. $$ Finally, the tail integral identity $\mathbb{E}[X_1^2]=2\int_0^\infty u\,\mathbb{P}(X_1>u)\,du$ follows by writing $x^2=\int_0^x 2u\,du$ and integrating over $X_1$. This proves the formula, including the case of an infinite second moment. $\square$

There are two different residual-life questions worth keeping separate. The equilibrium residual $R$ asks how much time remains at a typical observation time. By contrast, the conditional mean residual life at age $a$ asks for $\mathbb{E}[X_1-a\mid X_1>a]$. This conditional quantity is defined only when $\mathbb{P}(X_1>a)>0$. The equilibrium formula averages over the random age and interval seen at inspection; it is not generally the conditional mean residual life at age zero.

Exponential Interarrivals: The Memoryless Case

Worked Example: Exponential Interarrivals

Suppose each interarrival time is exponential with rate $\lambda=2$, so its density is $f(x)=2e^{-2x}$ for $x>0$. Its mean is $\mu=1/2$. Find the inspected-interval law, its mean, and the equilibrium residual law.

The length-biased density is $x f(x)/\mu$, so substitution gives $$ f_L(x)=\frac{x(2e^{-2x})}{1/2}=4xe^{-2x},\qquad x>0. $$ This is a Gamma distribution with shape $2$ and rate $2$. Its mean can be checked directly: $$ \mathbb{E}[L]=\int_0^\infty x(4xe^{-2x})\,dx =4\frac{2!}{2^3}=1. $$ Thus an inspected interval has mean $1$, twice the fresh interarrival mean $1/2$.

The exponential survival function is $\mathbb{P}(X_1>u)=e^{-2u}$. The theorem gives, for $t\ge0$, $$ \mathbb{P}(R>t)=\frac{1}{1/2}\int_t^\infty e^{-2u}\,du =2\frac{e^{-2t}}{2}=e^{-2t}. $$ So the equilibrium residual is exponential with rate $2$, and its mean is $1/2$. The check $\mathbb{E}[R]=\mathbb{E}[X_1^2]/(2\mu)= (1/2)/(1)=1/2$ gives the same result, since $\mathbb{E}[X_1^2]=2/2^2=1/2$.

At any age $a\ge0$, $\mathbb{P}(X_1-a>t\mid X_1>a)=e^{-2t}$ for $t\ge0$: the numerator is $e^{-2(a+t)}$ and the conditioning probability is $e^{-2a}>0$. Hence the conditional mean residual life is $1/2$ at every age. The equilibrium residual and the conditional residual happen to have the same law here, a consequence of memorylessness.

Deterministic Interarrivals: No Variation to Bias

Worked Example: Deterministic Interarrivals

Suppose every interarrival time equals $3$, so $F$ puts all its mass at $3$, and $\mu=3$. Find the inspected-interval and residual distributions, and the conditional mean residual life.

Length bias multiplies the mass at $3$ by $3/\mu=1$, so $L=3$ with probability $1$. Its mean is $3$, unchanged from a fresh interval. Given $L=3$, $U$ is uniform on $[0,3]$, so $R=3-U$ is also uniform on $[0,3]$. In particular, $$ \mathbb{P}(R>t)=\frac{3-t}{3}\quad(0\le t\le3),\qquad \mathbb{E}[R]=\frac{3}{2}. $$ The general moment formula agrees: $\mathbb{E}[X_1^2]/(2\mu)=9/6=3/2$.

For the conditional mean residual life, if $0\le a<3$, then $X_1>a$ with probability $1$, and the remaining time is exactly $3-a$. Thus $\mathbb{E}[X_1-a\mid X_1>a]=3-a$. At $a\ge3$, the conditioning event has probability zero, so this conditional expectation is undefined. The equilibrium mean $3/2$ averages over the uniform location in the interval; it is not the remaining time at every age.

Pareto Interarrivals: A Heavy-Tail Contrast

For a Pareto interarrival with scale $x_m>0$ and shape $\alpha>1$, the survival function is $1$ for $0\le t<x_m$, and $(x_m/t)^\alpha$ for $t\ge x_m$. The mean is $\mu=\alpha x_m/(\alpha-1)$, so the finite-mean assumption required for the equilibrium inspection law holds. The second moment is finite only when $\alpha>2$, a distinction that will determine whether the equilibrium mean residual is finite.

Worked Example: Pareto Interarrivals with Finite Residual Mean

Take $x_m=1$ and $\alpha=3$. Then $\mu=3/2$. Find the length-biased law, its mean, the equilibrium residual tail and mean, and the conditional mean residual life.

The original density is $3x^{-4}$ for $x\ge1$. Therefore the length-biased density is $$ f_L(x)=\frac{x(3x^{-4})}{3/2}=2x^{-3},\qquad x\ge1. $$ This is Pareto with scale $1$ and shape $2$. Its mean is $$ \mathbb{E}[L]=\int_1^\infty x(2x^{-3})\,dx =2\int_1^\infty x^{-2}\,dx=2. $$ Directly, $\mathbb{E}[X_1^2]=3\int_1^\infty x^{-2}\,dx=3$, so the general formula gives $\mathbb{E}[L]=3/(3/2)=2$.

For $0\le t<1$, the integrated survival function is $$ \int_t^\infty\mathbb{P}(X_1>u)\,du =\int_t^1 1\,du+\int_1^\infty u^{-3}\,du =(1-t)+\frac12=\frac32-t. $$ Dividing by $\mu=3/2$ gives $\mathbb{P}(R>t)=1-\frac{2t}{3}$. For $t\ge1$, the integral is $\int_t^\infty u^{-3}\,du=1/(2t^2)$; dividing by $3/2$ gives $\mathbb{P}(R>t)=1/(3t^2)$. The two expressions agree at $t=1$, where both equal $1/3$.

The residual mean is $\mathbb{E}[R]=\mathbb{E}[X_1^2]/(2\mu)=3/3=1$, which is also half the inspected-interval mean $2$. For the conditional mean residual life, if $0\le a<1$, then $X_1>a$ surely and the mean remaining time is $3/2-a$. If $a\ge1$, the conditional density is Pareto above $a$; integrating its survival tail gives $\mathbb{E}[X_1-a\mid X_1>a]=a/2$. At $a=1$, the two formulas both give $1/2$.

The Pareto example also shows why a finite mean interarrival time is not enough to ensure a finite mean wait at inspection. Keep $x_m=1$ but take $1<\alpha\le2$. The mean interval remains finite, while $\mathbb{E}[X_1^2]=\infty$, since its tail integral $2\int_0^\infty u\,\mathbb{P}(X_1>u)\,du$ includes a constant multiple of $\int_1^\infty u^{1-\alpha}\,du$, which diverges for $\alpha\le2$. The theorem then gives $\mathbb{E}[R]=\infty$. For example, at $\alpha=3/2$ the length-biased density is $\tfrac12 x^{-3/2}$ for $x\ge1$; its integral is $1$, but its mean $\tfrac12\int_1^\infty x^{-1/2}\,dx$ is infinite.

Comparing the Three Laws

Fresh interarrival law	Mean inspected interval	Equilibrium mean residual	Conditional mean residual
Exponential, rate $\lambda$	$2/\lambda$	$1/\lambda$	$1/\lambda$ at every age $a\ge0$
Deterministic, length $c$	$c$	$c/2$	$c-a$ for $0\le a<c$
Pareto, scale $x_m$, shape $\alpha>2$	$x_m(\alpha-1)/(\alpha-2)$	$x_m(\alpha-1)/(2(\alpha-2))$	$\mu-a$ for $a<x_m$; $a/(\alpha-1)$ for $a\ge x_m$

The table’s Pareto conditional formula uses $\mu=\alpha x_m/(\alpha-1)$. For $1<\alpha\le2$, the mean inspected interval and equilibrium mean residual are infinite, although the fresh mean $\mu$ is finite. The law of $L$ is still well-defined because it is normalized by $\mu$; it is its mean that diverges.

Common mistake: applying memorylessness to every inspection problem. For exponential interarrivals with rate $2$, one might claim that the inspected interval still has mean $1/2$, reasoning that the residual waiting time is memoryless. That confuses the interval containing the observation with the remaining time. The actual inspected-interval density is $4xe^{-2x}$, whose mean is $1$; the residual time has mean $1/2$. Memorylessness preserves the residual law, not the length-biased interval law. For deterministic intervals, there is no length bias in the interval length, but the residual still varies with the observation’s position.

Check Your Understanding

For each question, distinguish the inspected interval from the remaining time, and state any moment or conditioning requirements.

If exponential interarrivals have rate $4$, what are the mean fresh interval, mean inspected interval, and mean equilibrium residual time?
For deterministic interarrivals of length $5$, find the equilibrium residual tail for $0\le t\le5$ and the conditional mean residual life at age $a=2$.
For a Pareto law with scale $2$ and shape $3$, calculate the mean interarrival and use the moment formula to find the equilibrium mean residual time.
For Pareto interarrivals with $1<\alpha\le2$, explain why the equilibrium mean residual time is infinite even though the equilibrium inspection law is defined.
In one sentence, explain why the conditional mean residual life at age zero need not equal the equilibrium mean residual time.

Fresh interarrival law	Mean inspected interval	Equilibrium mean residual	Conditional mean residual
Exponential, rate \(\lambda\)	\(2/\lambda\)	\(1/\lambda\)	\(1/\lambda\) at every age \(a\ge0\)
Deterministic, length \(c\)	\(c\)	\(c/2\)	\(c-a\) for \(0\le a<c\)
Pareto, scale \(x_m\), shape \(\alpha>2\)	\(x_m(\alpha-1)/(\alpha-2)\)	\(x_m(\alpha-1)/(2(\alpha-2))\)	\(\mu-a\) for \(a<x_m\); \(a/(\alpha-1)\) for \(a\ge x_m\)