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Approximate Pivots and Large-Sample Coverage

Use large-sample distributions to construct approximate confidence intervals, and judge when their coverage may be unreliable.

Intermediate 12 min read

What You'll Learn

  • Distinguish an exact pivot from a statistic that is pivotal only asymptotically.
  • Derive an approximate interval by inverting a large-sample normal approximation.
  • Construct approximate intervals for Poisson rates and binomial proportions.
  • Check whether sample size and model conditions support a normal approximation.
  • Calculate exact repeated-sampling coverage to assess finite-sample error.

Why Large-Sample Pivots Are Useful

An exact pivot has a distribution that does not depend on the unknown parameter, so its quantiles can produce a confidence interval with exact coverage under the model. In many problems, however, no convenient exact pivot is available. A statistic may instead become approximately parameter-free as the sample size grows. This can still support a useful interval, provided the approximation is understood and its limitations are checked.

The key distinction is between an interval whose coverage is exactly its stated level and one whose coverage approaches that level as the sample size increases. A large sample can make the approximation accurate, but it does not make the result exact for every finite sample. This tutorial develops the general construction and examines that difference.

Definition: A sequence of statistics \(T_n(\theta)\) is an approximate pivot when its distribution approaches a fixed distribution that does not depend on the unknown parameter \(\theta\) as the sample size \(n\) increases. In the common normal case, \(T_n(\theta)\) converges in distribution to \(N(0,1)\). An interval obtained by treating this limiting distribution as an approximation has large-sample (asymptotic) coverage: under stated regularity conditions, its repeated-sampling coverage approaches the nominal confidence level as \(n\) grows.

From an Approximate Pivot to an Interval

Suppose an estimator \(\hat{\theta}\) is approximately normal for large \(n\), centered at the true parameter \(\theta\), with standard error estimated by \(\widehat{\operatorname{se}}(\hat{\theta})\). If the standardized estimator is an approximate pivot, then

$$ T_n(\theta)=\frac{\hat{\theta}-\theta}{\widehat{\operatorname{se}}(\hat{\theta})} \ \xrightarrow{d}\ N(0,1). $$

Here \(\xrightarrow{d}\) means convergence in distribution. For a two-sided confidence level \(1-\alpha\), let \(z_{1-\alpha/2}\) be the \(1-\alpha/2\) quantile of the standard normal distribution. The normal approximation gives

$$ P\left(-z_{1-\alpha/2}\leq T_n(\theta)\leq z_{1-\alpha/2}\right) \ \approx\ 1-\alpha. $$

Solving these inequalities for \(\theta\) gives the approximate interval

$$ \left[ \hat{\theta}-z_{1-\alpha/2}\widehat{\operatorname{se}}(\hat{\theta}), \quad \hat{\theta}+z_{1-\alpha/2}\widehat{\operatorname{se}}(\hat{\theta}) \right]. $$

Why the Construction Works

The justification combines a limiting distribution with a consistent estimate of the standard error. For example, suppose the estimator has an asymptotic normal distribution with variance depending on \(\theta\). If a plug-in estimate of that variance converges to the true variance, replacing the unknown variance by that estimate does not change the limiting standardized distribution. This is an application of Slutsky’s theorem.

1
Start with a limiting result.
Assume that a suitable standardized version of \(\hat{\theta}-\theta\) converges in distribution to \(N(0,1)\). This is often established using the central limit theorem and an estimator’s variance.
2
Estimate any unknown scale.
If the standardization involves an unknown parameter, replace it with a consistent estimate. Under the required regularity conditions, Slutsky’s theorem preserves the \(N(0,1)\) limiting distribution.
3
Use central quantiles and invert.
The standard normal probability between \(-z_{1-\alpha/2}\) and \(z_{1-\alpha/2}\) is \(1-\alpha\). Applying that central range to the approximate pivot and rearranging yields the interval for \(\theta\).

Thus the coverage statement is a limiting one: as \(n\) increases, the probability that the interval contains the fixed true parameter approaches \(1-\alpha\). For a particular finite \(n\), the actual coverage may differ from \(1-\alpha\), sometimes substantially. The argument also requires a valid large-sample approximation and a suitable standard error; a large sample alone does not repair a wrong model or a badly chosen statistic.

Worked Example: An Approximate Interval for a Poisson Rate

Worked Example: An Approximate Interval for a Poisson Rate

Model and assumptions. Suppose daily incident counts \(X_1,\ldots,X_n\) are independent and identically distributed as \(\operatorname{Poisson}(\lambda)\), where \(\lambda\) is the unknown mean number of incidents per day. Assume the recording process and underlying rate are stable across the observed days. For a hypothetical 100-day period, the total recorded count is 256. Construct an approximate 95% interval for \(\lambda\).

Method. The estimator is the sample mean, \(\hat{\lambda}=\bar{X}\). Under the Poisson model, its variance is \(\lambda/n\), and the central limit theorem gives an approximately normal standardized estimator for large \(n\). Estimate the unknown \(\lambda\) in the standard error by \(\hat{\lambda}\).

Step 1: Calculate the estimate. The total is 256 over 100 days, so

$$ \hat{\lambda}=\bar{X}=\frac{256}{100}=2.56 \quad\text{incidents per day}. $$

Step 2: Estimate the standard error. The plug-in standard error is

$$ \widehat{\operatorname{se}}(\bar{X}) =\sqrt{\frac{\hat{\lambda}}{n}} =\sqrt{\frac{2.56}{100}} =\sqrt{0.0256} =0.16. $$

Step 3: Apply the 95% normal critical value. Using \(z_{0.975}=1.96\), the margin of error is

$$ 1.96(0.16)=0.3136. $$

The endpoints are \(2.56-0.3136=2.2464\) and \(2.56+0.3136=2.8736\). Rounded to two decimals, the approximate interval is \((2.25,\,2.87)\) incidents per day.

Check and trap avoided. The midpoint of the unrounded interval is \((2.2464+2.8736)/2=2.56\), the observed rate. A common arithmetic error is to miscalculate \(256/100\) as 6.25; the correct rate is 2.56, and both the standard error and interval must be calculated from 2.56. This is an approximate interval, not an exact finite-sample Poisson interval.

Conditions and Approximation Error

A large-sample interval is only as dependable as the conditions supporting its approximate pivot. Before using one, check the model, the estimator’s limiting distribution, and the estimated standard error. In particular, observations should meet the independence or dependence conditions used in the limiting result, and the estimator should have the stated variance behavior.

The approximation may be poor when the sample is small, the distribution is strongly skewed, or the parameter is near a boundary. Plug-in standard errors can also behave poorly in small samples. For instance, a normal interval for a probability can extend below zero or above one. Truncating its endpoints to the parameter space may make the interval easier to interpret, but it does not turn the original approximation into an exact procedure.

Worked Example: A Large-Sample Interval for a Proportion

Worked Example: A Large-Sample Interval for a Proportion

Model and assumptions. Consider 400 independent records from a stable hypothetical review process. Each record is flagged with probability \(p\), and each flag indicator has a Bernoulli distribution with the same unknown probability \(p\). Suppose 92 of the records are flagged. Construct an approximate 95% interval for \(p\).

Method. The estimator is the sample proportion \(\hat{p}\). For independent Bernoulli observations, its variance is \(p(1-p)/n\). For a sufficiently large sample, the standardized proportion is approximately normal. Use \(\hat{p}\) in place of the unknown \(p\) to estimate its standard error.

Step 1: Calculate the observed proportion and check the counts.

$$ \hat{p}=\frac{92}{400}=0.23,\qquad n\hat{p}=92,\qquad n(1-\hat{p})=308. $$

Both observed category counts are large, supporting use of the normal approximation in this example.

Step 2: Calculate the estimated standard error.

$$ \widehat{\operatorname{se}}(\hat{p}) =\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} =\sqrt{\frac{(0.23)(0.77)}{400}} =\sqrt{0.00044275} \approx 0.0210416. $$

Step 3: Calculate the margin and endpoints. The 95% margin is \(1.96(0.0210416)\approx 0.0412415\). Therefore the endpoints are approximately \(0.23-0.0412415=0.1887585\) and \(0.23+0.0412415=0.2712415\). To four decimal places, the approximate interval is \((0.1888,\,0.2712)\).

Check and trap avoided. The interval is centered on 0.23 and lies within the possible range from zero to one. Do not substitute the total number of unflagged records for the number of flagged records: the estimate is \(92/400\), not \(308/400\). The favorable observed counts support the approximation, but they do not guarantee exactly 95% coverage.

Nominal Confidence Is Not Guaranteed Coverage

The nominal level is the probability targeted by the quantile calculation under the approximation. Actual coverage at a given sample size is the repeated-sampling probability that the computed interval contains the true parameter under the actual model. These probabilities need not match. A direct finite-sample calculation can show the size of the discrepancy.

Worked Example: Checking Finite-Sample Coverage

Worked Example: Checking Finite-Sample Coverage

Model and assumptions. Let \(X\) be the number of successes in \(n=10\) independent Bernoulli trials with true success probability \(p=0.5\), so \(X\sim\operatorname{Binomial}(10,0.5)\). Consider the approximate 95% Wald interval formed from \(\hat{p}=X/10\) and estimated standard error \(\sqrt{\hat{p}(1-\hat{p})/10}\). First calculate the interval when the observed count is \(X=3\), then find the procedure’s exact coverage under this model.

Method. Calculate the observed Wald interval, then enumerate which possible values of \(X\) produce an interval containing the true \(p=0.5\). This distinguishes the interval’s nominal level from its actual finite-sample coverage.

Step 1: Calculate the interval for the observed data. With \(X=3\), \(\hat{p}=3/10=0.3\). The estimated standard error is

$$ \sqrt{\frac{(0.3)(0.7)}{10}} =\sqrt{0.021} \approx 0.1449. $$

The margin is \(1.96(0.1449)\approx 0.2840\), giving endpoints \(0.3-0.2840=0.0160\) and \(0.3+0.2840=0.5840\). This interval contains 0.5.

Step 2: Determine all counts whose intervals contain 0.5. The interval contains 0.5 exactly when

$$ \left|\frac{x}{10}-0.5\right| \leq 1.96\sqrt{\frac{(x/10)(1-x/10)}{10}}. $$

For \(x=2\), the left side is \(0.3\), while the right side is \(1.96\sqrt{0.016}\approx 0.2479\); the interval misses 0.5. For \(x=3\), the sides are \(0.2\) and \(1.96\sqrt{0.021}\approx 0.2840\), so it covers. By symmetry around \(x=5\), the covering counts are \(x=3,4,5,6,7\); counts 0, 1, 2, 8, 9, and 10 do not cover.

Step 3: Sum their probabilities. Under \(X\sim\operatorname{Binomial}(10,0.5)\), the probabilities for \(x=3,4,5,6,7\) sum to

$$ \frac{\binom{10}{3}+\binom{10}{4}+\binom{10}{5} +\binom{10}{6}+\binom{10}{7}}{2^{10}} =\frac{120+210+252+210+120}{1024} =\frac{912}{1024} =0.890625. $$

Thus the exact coverage is approximately 0.8906, or 89.06%, rather than the nominal 95%. The count calculation checks the result directly under the stated model. The trap is to treat a normal critical value as a guarantee of exact coverage; it only supplies the nominal level for this approximate procedure.

Practical Checks Before Reporting an Interval

For an exam calculation or an applied analysis, a short checklist helps keep the interpretation aligned with the mathematics:

  • State the model. Identify the sampling assumptions and parameter before selecting a standard error.
  • Identify the approximate pivot. Explain why its distribution is approximately parameter-free, and name the limiting distribution being used.
  • Check the data scale. Look for small expected counts, boundary estimates, strong skewness, or dependence that could weaken the approximation.
  • Use the correct standard error and critical value. Keep the estimated standard error distinct from the estimate itself, and match the quantile to the intended confidence level.
  • Describe the result accurately. Call it an approximate interval with asymptotic coverage, not an exact interval unless exact coverage has been established.
Key takeaway: A large-sample confidence interval comes from inverting a statistic whose distribution approaches a parameter-free limit. Its nominal confidence level describes the target of that approximation; actual finite-sample coverage depends on how well the limiting distribution represents the sampling distribution at the sample size used.

Check Your Understanding

For each problem, state the model-based method and keep the interpretation tied to approximate coverage.

  1. Independent daily counts follow a stable \(\operatorname{Poisson}(\lambda)\) model. Across 64 days, 144 events are observed. Use the plug-in normal approximation to construct a 90% interval for \(\lambda\), with \(z_{0.95}=1.645\). Give endpoints to four decimals. (Answer: \((1.9416,\,2.5584)\) events per day)
  2. In 250 independent trials with a common success probability \(p\), 40 successes are observed. Construct the approximate 95% Wald interval using \(z_{0.975}=1.96\). Give endpoints to four decimals. (Answer: \((0.1146,\,0.2054)\))
  3. An estimator of a probability has observed value \(0.84\) from \(n=100\) independent Bernoulli observations. Use the plug-in standard error and \(z_{0.95}=1.645\) for an approximate 90% interval. Give endpoints to four decimals. (Answer: \((0.7797,\,0.9003)\))
  4. For \(n=8\) independent Bernoulli trials with true \(p=0.5\), consider the 95% Wald interval. Which observed counts \(x\) yield an interval containing 0.5, and what is the exact coverage? (Answer: \(x=2,3,4,5,6\); coverage \(238/256=0.9297\))

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