The Inspection Paradox and Length-Biased Intervals

Why an Interval Seen in Progress Is Biased

Suppose renewals occur at times \(S_0,S_1,\ldots\), and an observer chooses a time uniformly from a long observation window. The interval containing that time is not generally distributed like the interval between two successive renewals. An interval twice as long occupies twice as much time in which the observer could land. Consequently, time sampling favors long intervals. This is the inspection paradox: the intervals encountered by observation can be longer, on average, than the intervals generated at renewals.

To make the comparison precise, let \(X_1,X_2,\ldots\) be independent, identically distributed, strictly positive interarrival times, with common distribution \(F\) and finite mean \(\mu=\mathbb{E}[X_1]\), where \(0<\mu<\infty\). Set \(S_0=0\) and \(S_n=\sum_{k=1}^n X_k\). For a time \(u\geq 0\), define its containing interval to be \([S_{n-1},S_n)\) when \(S_{n-1}\leq u<S_n\). The endpoint convention has no effect on sampling with an independent continuous uniform time.

The formula weights each possible interval length \(x\) by \(x\). It defines a probability distribution because \(F_L((0,\infty))=\mathbb{E}[X_1]/\mu=1\). If \(X_1\) has probability mass function \(p(x)\), then the length-biased mass at \(x\) is \(x p(x)/\mu\). If it has density \(f(x)\), the length-biased density is \(f_L(x)=x f(x)/\mu\). The finite-mean assumption is essential for this normalization.

Why Uniform Time Sampling Produces Length Bias

We use the standard strong law of large numbers for independent, identically distributed integrable random variables: their sample averages converge almost surely to their common expectation. This standard probability result is not proved here. Its application below is to interval lengths and to lengths multiplied by an indicator.

Proof. Fix a Borel set \(A\), and let \(N(T)=\max\{n:S_n\leq T\}\), the number of completed intervals by time \(T\). Conditional on the renewal intervals, the probability that \(U_T\) falls in a portion of total length \(r\) is \(r/T\). The completed intervals whose lengths belong to \(A\) contribute total time \(\sum_{k=1}^{N(T)}X_k\mathbf{1}_{\{X_k\in A\}}\). The only additional contribution can come from the final, incomplete interval, and its length is at most \(X_{N(T)+1}\). Therefore the conditional probability of selecting an interval with length in \(A\) differs from $$ \frac{1}{T}\sum_{k=1}^{N(T)}X_k\mathbf{1}_{\{X_k\in A\}} $$ by at most \(X_{N(T)+1}/T\).

By the strong law, \(S_n/n\to\mu\) almost surely. Since \(S_{N(T)}\leq T<S_{N(T)+1}\) and \(N(T)\to\infty\), this implies \(N(T)/T\to1/\mu\) almost surely. Also, \(X_n/n=S_n/n-((n-1)/n)(S_{n-1}/(n-1))\to0\) as \(n\to\infty\). Hence \(X_{N(T)+1}/T\to0\), since \(X_{N(T)+1}/(N(T)+1)\to0\) and \((N(T)+1)/T\to1/\mu\).

The variables \(X_k\mathbf{1}_{\{X_k\in A\}}\) are independent, identically distributed, and integrable, with mean \(\mathbb{E}[X_1\mathbf{1}_{\{X_1\in A\}}]\). Applying the strong law to them gives \(\frac{1}{N(T)}\sum_{k=1}^{N(T)}X_k\mathbf{1}_{\{X_k\in A\}}\to\mathbb{E}[X_1\mathbf{1}_{\{X_1\in A\}}]\). Multiplying by \(N(T)/T\) shows that the displayed time proportion converges almost surely to the claimed ratio. The final incomplete interval contributes a vanishing proportion, so the conditional probability converges to that ratio. These conditional probabilities lie between zero and one; bounded convergence therefore gives the same limit for their expectations, which are \(\mathbb{P}(L_T\in A)\). \(\square\)

This argument compares intervals by the time they occupy, not by how often they occur. It needs no assumption that the interarrival law is continuous or nonarithmetic. The theorem concerns the interval’s length; it does not by itself describe the observer’s position within that interval.

Worked Distributions

In What Sense Is the Observed Interval Longer?

Length bias gives more than a comparison of means. For every threshold \(c\geq0\), the probability that the time-sampled interval exceeds \(c\) is at least the corresponding ordinary interarrival probability. Indeed, the length-biased tail is \(\mathbb{E}[X_1\mathbf{1}_{\{X_1>c\}}]/\mu\). The next result proves the comparison directly.

Proof. Let \(X_1'\) be an independent copy of \(X_1\), and put \(g(x)=\mathbf{1}_{\{x>c\}}\), which is nondecreasing. Independence and identical distributions give \(\mathbb{E}[(X_1-X_1')(g(X_1)-g(X_1'))]=2\operatorname{Cov}(X_1,g(X_1))\). The product inside the expectation is nonnegative: if \(X_1\geq X_1'\), then \(g(X_1)\geq g(X_1')\), and if \(X_1<X_1'\), both differences are nonpositive. Thus \(\operatorname{Cov}(X_1,g(X_1))\geq0\). Dividing the identity \(\mathbb{E}[X_1g(X_1)]-\mu\mathbb{E}[g(X_1)]=\operatorname{Cov}(X_1,g(X_1))\geq0\) by \(\mu>0\) proves the tail inequality using the definition of \(F_L\).

When the second moment is finite, the mean under \(F_L\) is \(\int x\,F_L(dx)=\mathbb{E}[X_1^2]/\mu\). Subtracting \(\mu\) gives \((\mathbb{E}[X_1^2]-\mu^2)/\mu=\operatorname{Var}(X_1)/\mu\). This difference is strictly positive exactly when the interarrival time is nonconstant. If the second moment is infinite, the length-biased mean is infinite, even though the ordinary mean is finite. \(\square\)

A common mistake is to assume that observing an interval in progress leaves its distribution unchanged. In the two-length example, that mistake assigns probability \(\tfrac12\) to length \(3\), and therefore gives mean \(1\cdot\tfrac12+3\cdot\tfrac12=2\). But an interval of length \(3\) contributes three times as much sampling time per occurrence as an interval of length \(1\). The correct probabilities are \(\tfrac14\) and \(\tfrac34\), giving mean \(\tfrac52\). The correction is to weight each interarrival probability by its length and then divide by the original mean.

The size bias disappears when the interval length is constant: if \(X_1=a\) almost surely for some \(a>0\), then \(F_L\) also assigns probability one to \(a\). Thus “longer on average” is a strict statement only when the ordinary interarrival distribution is nonconstant and has finite second moment. The tail comparison remains valid under the finite-mean assumption, whether or not the second moment exists.