Why Compare Expected Widths?
Two confidence procedures can have the same coverage probability but produce intervals of different sizes. A procedure with smaller intervals is often more informative, so it is natural to compare procedures by their widths. But a width is calculated from a random sample and can itself be random. Comparing the average width is therefore a comparison across repeated samples, not a guarantee about which interval will be shorter for the particular sample in hand.
The comparison is meaningful only after specifying the model, the parameter, and the coverage being compared. In this tutorial, coverage refers to the repeated-sampling probability that the interval contains the fixed parameter. Expected width is the average of the random interval length under that same sampling model. It is not the probability that one procedure is narrower than another.
Expected width is measured in the units of the parameter. A comparison may depend on the unknown parameter, and some width expectations may be infinite even when each realized interval has finite endpoints. These details matter: a numerical average is useful only if it is defined under the stated model.
Expected Width From a Uniform Sample Maximum
A uniform model gives a direct example of how to compare equal-coverage procedures. Suppose \(X_1,\ldots,X_n\) are independent observations from the uniform distribution on \((0,\theta)\), where \(\theta>0\) is unknown, and let \(M=\max(X_1,\ldots,X_n)\). For \(0\le r\le1\), the pivot \(M/\theta\) has distribution function \(P_\theta(M/\theta\le r)=r^n\). Thus its \(p\)-quantile is \(p^{1/n}\).
For pivot quantiles \(a\) and \(b\), where \(0<a<b<1\), the event \(a\le M/\theta\le b\) is equivalent to \(M/b\le\theta\le M/a\). The resulting interval has coverage \(b^n-a^n\), and its width is \(M(1/a-1/b)\). Since the factor multiplying \(M\) is fixed, its expected width follows from \(E_\theta[M]\).
The expectation of the maximum can be derived from its survival probability. For any nonnegative random variable \(Z\), \(E[Z]=\int_0^\infty P(Z>m)\,dm\). Here \(P(M>m)=1-(m/\theta)^n\) for \(0\le m\le\theta\), and it is zero for \(m>\theta\). Therefore,
The interval-width result follows by taking the expectation of \(M(1/a-1/b)\). This derivation also shows why expected width depends on \(\theta\): in this scale model, the observations and the interval widths scale with the unknown endpoint.
Worked Example: Equal Coverage With Different Pivot Tails
Suppose \(X_1,X_2\) are independent uniform \((0,\theta)\) observations, and \(M=\max(X_1,X_2)\). Compare a central 90% interval, with 5% in each pivot tail, to an interval with lower pivot-tail probability 9% and upper pivot-tail probability 1%. Calculate both expected widths and identify the narrower procedure on average.
Method and coverage. Here \(n=2\), so the pivot is \(M/\theta\) with distribution function \(r^2\). For a central 90% interval, \(a=\sqrt{0.05}=0.2236068\) and \(b=\sqrt{0.95}=0.9746794\). Its coverage is \(b^2-a^2=0.95-0.05=0.90\).
For the second procedure, the 9% lower tail gives \(a=\sqrt{0.09}=0.3\), and the 1% upper tail gives \(b=\sqrt{0.99}=0.9949874\). Its coverage is \(b^2-a^2=0.99-0.09=0.90\) as well.
Central interval width. The interval is \([M/b,M/a]\), so its width is \(M(1/a-1/b)\). Substitution gives \(1/a=4.472136\) and \(1/b=1.025978\), so the width is \((4.472136-1.025978)M=3.446158M\). Since \(E_\theta[M]=2\theta/3\), its expected width is \(3.446158(2\theta/3)=2.297439\theta\).
Unequal-tail interval width. Here \(1/a=3.333333\) and \(1/b=1.005038\). Its width is \((3.333333-1.005038)M=2.328295M\), and its expected width is \(2.328295(2\theta/3)=1.552197\theta\).
Both procedures have exact 90% coverage, but the unequal-tail interval has the smaller expected width: approximately \(1.5522\theta\), compared with \(2.2974\theta\). The comparison is about average length under the model, not a claim that every possible interval from the second procedure must be shorter.
Common mistake to avoid: Reversing the quantile endpoints or using \(M(1/b-1/a)\) gives a negative width. Since \(a<b\), \(1/a>1/b\), and the correct width is \(M(1/a-1/b)\).
An Exact Rate Interval and Its Expected Width
Expected width can also be calculated for a confidence interval whose width is not proportional to a sample maximum. Suppose \(X_1,\ldots,X_n\) are independent exponential observations with rate \(\lambda>0\), using density \(f(x)=\lambda e^{-\lambda x}\) for \(x>0\), and let \(T=\sum_{j=1}^n X_j\). The pivot \(2\lambda T\) has a chi-square distribution with \(2n\) degrees of freedom. If \(q_p\) denotes the \(p\)-quantile of that distribution, inversion of the central pivot interval gives the exact interval \([q_{\alpha/2}/(2T),q_{1-\alpha/2}/(2T)]\) for \(\lambda\).
Its width is \((q_{1-\alpha/2}-q_{\alpha/2})/(2T)\). To obtain an expected width, we need \(E_\lambda[1/T]\), which exists only when \(n>1\). Since \(T\) has a gamma distribution with shape \(n\) and rate \(\lambda\), direct integration gives \(E_\lambda[1/T]=\lambda/(n-1)\) for \(n>1\). Thus
The restriction \(n>1\) is substantive. With just one exponential observation, the interval has finite width for every observed \(T>0\), but \(E_\lambda[1/T]\) is infinite. It would be incorrect to report a finite expected width in that case.
Worked Example: Expected Width for an Exponential-Rate Interval
A hypothetical waiting-time study records two independent times \(X_1=0.50\) years and \(X_2=0.75\) years. Assume both follow an exponential distribution with a common unknown rate \(\lambda\) per year. Construct the central 90% chi-square interval for \(\lambda\), calculate its observed width, and find the procedure’s expected width under the model.
Method and assumptions. The observations are independent exponential variables with rate \(\lambda\), and \(T=X_1+X_2\). Thus \(T=0.50+0.75=1.25\) years and \(2\lambda T\) has a chi-square distribution with 4 degrees of freedom. For this distribution, \(q_{0.05}=0.7107\) and \(q_{0.95}=9.4877\).
Interval endpoints. Inverting \(q_{0.05}\le2\lambda T\le q_{0.95}\) gives \(\lambda\in[q_{0.05}/(2T),q_{0.95}/(2T)]\). Since \(2T=2.50\) years, the lower endpoint is \(0.7107/2.50=0.28428\) per year, and the upper endpoint is \(9.4877/2.50=3.79508\) per year. The observed interval is approximately \([0.2843,3.7951]\) per year.
Observed and expected widths. The observed width is \((9.4877-0.7107)/2.50=8.7770/2.50=3.5108\) per year. For \(n=2\), \(E_\lambda[1/T]=\lambda/(2-1)=\lambda\). Therefore the expected width is \((9.4877-0.7107)\lambda/2=8.7770\lambda/2=4.3885\lambda\) per year.
The observed width is a number obtained from the particular total \(T=1.25\); the expected width depends on the true rate \(\lambda\). They are different quantities and should not be confused.
Common mistake to avoid: From \(q_{0.05}\le2\lambda T\le q_{0.95}\), divide each endpoint by \(2T\) to obtain the interval for \(\lambda\). The endpoints are \(q_{0.05}/(2T)\) and \(q_{0.95}/(2T)\), not \(T/q_{0.95}\) and \(T/q_{0.05}\). The correct width is \((q_{0.95}-q_{0.05})/(2T)\), approximately \(4.3885/T\), not a difference of reciprocal quantiles multiplied by \(T\).
Why Smaller Expected Width Does Not Mean Smaller Every Time
An expected width summarizes a procedure over repeated samples. It does not order the intervals sample by sample. The following construction makes the distinction explicit using a normal model and an ancillary event: an event determined by the data whose probability does not depend on the parameter.
Worked Example: Equal Coverage but Crossing Widths
Let \(X_1,X_2\) be independent observations from \(N(\mu,4)\), where \(\mu\) is unknown and the variance \(4\) is known. Set \(D=X_1-X_2\). Compare procedure A, which uses a central 90% interval for \(\mu\) when \(D\ge0\) and a central 70% interval when \(D<0\), with procedure B, which always uses a central 80% interval. Determine their coverages and expected widths, and check whether one is narrower for every sample.
Coverage. The sample mean \(\bar X=(X_1+X_2)/2\) has distribution \(N(\mu,2)\), so its standard error is \(\sqrt{2}\). Also, \(D\sim N(0,8)\), so \(P(D\ge0)=P(D<0)=0.5\). Under the joint normal model, \(\bar X\) and \(D\) are independent because their covariance is zero. Therefore the conditional coverage of each central interval, given either sign of \(D\), is its stated normal coverage.
Procedure A has unconditional coverage \(0.5(0.90)+0.5(0.70)=0.80\). Procedure B has coverage \(0.80\) for every sample condition on \(D\), and hence unconditional coverage \(0.80\). The procedures have equal coverage.
Expected widths. A central interval with coverage \(c\) has width \(2z_{(1+c)/2}\sqrt{2}\), where \(z_p\) is the \(p\)-quantile of the standard normal distribution. Use \(z_{0.95}=1.6449\), \(z_{0.85}=1.0364\), and \(z_{0.90}=1.2816\). Procedure A has width \(2(1.6449)\sqrt2\approx4.6525\) when \(D\ge0\), and width \(2(1.0364)\sqrt2\approx2.9314\) when \(D<0\). Its expected width is \(0.5(4.6525)+0.5(2.9314)=3.7920\), to four decimal places.
Procedure B always has width \(2(1.2816)\sqrt2\approx3.6249\). Thus B has the smaller expected width. Nevertheless, when \(D<0\), procedure A’s width is about \(2.9314\), less than B’s \(3.6249\); when \(D\ge0\), A’s width is about \(4.6525\), greater than B’s. Neither procedure is narrower for every sample.
For example, if the observed values are \(X_1=10\) and \(X_2=14\), then \(D=-4\), so A uses its 70% interval and is narrower than B. Reversing the observations gives \(D=4\), so A uses its 90% interval and is wider than B, even though the sample mean remains 12.
Common mistake to avoid: A smaller expected width does not imply that the procedure has a shorter interval for every realized sample. Here B has the smaller average, while A is shorter on the event \(D<0\).
How to Make a Fair Width Comparison
A sound comparison reports both the coverage and the model used to calculate widths. If coverage differs, a narrower average interval may simply be paying for less frequent coverage. If the model changes, the expected width may change as well. When an exact expectation is difficult to calculate, it can be estimated by simulation: generate many samples under a stated parameter value, calculate each interval width, and average those widths. The estimate describes that model and parameter setting; it is not a guarantee for any one sample.
When reviewing a comparison, check the following points:
- Same target and coverage: The procedures should estimate the same parameter and attain the same stated coverage under the model being used.
- Correct width: For an interval with ordered endpoints, subtract the lower endpoint from the upper endpoint.
- Expectation under the right model: Average the random width over the sampling distribution, and verify that the expectation is finite.
- Interpretation: Expected width describes average interval length, not a probability of being narrower on a particular sample.
Check Your Understanding
For each problem, state the comparison or calculation requested and check that the coverage and model assumptions are respected.
- For two independent uniform \((0,\theta)\) observations, let \(M\) be their maximum. Compare the expected widths of the 80% central pivot interval, using \(a=\sqrt{0.10}\), \(b=\sqrt{0.90}\), and an interval using \(a=0.2\), \(b=\sqrt{0.84}\). (Answer: The central expected width is approximately \(1.4055\theta\); the second is approximately \(2.6059\theta\), so the central interval is narrower on average.)
- Two independent exponential observations with unknown rate \(\lambda\) have total \(T\). For a central 95% interval, use chi-square 4-degree-of-freedom quantiles \(q_{0.025}=0.4844\) and \(q_{0.975}=11.1433\). Find the expected width. (Answer: \((11.1433-0.4844)\lambda/2=5.3295\lambda\) per unit of time.)
- For the exponential-rate model in question 2, the observed total is \(T=2\) time units. Calculate the observed interval width using the given quantiles. (Answer: \((11.1433-0.4844)/(2\cdot2)=2.6647\) per unit of time.)
- For one exponential observation \(X\), the same chi-square pivot gives a finite interval width for every observed \(X>0\). Is its expected width finite under the model? (Answer: No. With \(n=1\), \(E_\lambda[1/T]\) is infinite, so the expected interval width is infinite.)
- In the normal ancillary-event example, suppose the two procedures retain central coverages of 90% and 70% on equally likely events for procedure A, while B always uses 80%. Explain why their coverage is equal and why the smaller expected width of either procedure would not by itself establish that it is narrower on every sample. (Answer: Both have unconditional coverage \(0.5(0.90)+0.5(0.70)=0.80\); their widths cross between the two events, so average width does not imply sample-by-sample ordering.)