From a Pivot to an Interval
A pivotal quantity has a known distribution that does not depend on the unknown parameter. That known distribution supplies probability bounds; inversion then turns those bounds into an interval for the parameter. The essential step is algebra, but the direction of the inequalities matters: solving for a parameter can preserve or reverse their order.
A confidence interval is a random interval before the data are observed. Its endpoints are functions of the sample, while the parameter being estimated is fixed. A confidence procedure with coverage \(1-\alpha\) satisfies \(P_\theta\{L(X)\leq\theta\leq U(X)\}=1-\alpha\), under the stated model. After observing the sample, the endpoints become numbers; the coverage statement describes the procedure’s repeated-sampling behavior.
The quantiles need not be symmetric around zero, and a pivot need not increase with the parameter. What matters is to use bounds whose probability is known and to solve the inequalities without changing their meaning. In the examples, the selected bounds are specified so the algebra can be followed directly.
Why Inversion Gives the Coverage
The justification is an equivalence of events. Start with an event involving the pivot whose probability is known. If solving its inequalities gives exactly the event that the parameter lies between two sample-dependent endpoints, the two events have the same probability. That probability is the coverage of the interval.
Find fixed \(q_L,q_U\) for which the pivot lies between them with probability \(1-\alpha\).
Keep the sample statistic and unknown parameter visible in \(q_L\leq Q(X,\theta)\leq q_U\).
Use the model’s parameter restrictions and track whether each operation preserves or reverses inequality order.
Express the result as \(L(X)\leq\theta\leq U(X)\). Event equivalence gives the claimed coverage.
For example, if \(Q(X,\theta)=(\bar X-\theta)/c\), where \(c>0\), then \(q_L\leq(\bar X-\theta)/c\leq q_U\) is equivalent to \(\bar X-cq_U\leq\theta\leq\bar X-cq_L\). The endpoint order reverses relative to the pivot bounds because \(\theta\) appears with a negative sign. This algebraic step is the basis of the first two examples.
Inverting a Standard Normal Pivot
Worked Example: A Mean with Known Standard Deviation
Model and task. A hypothetical laboratory measures the output of 36 independent sensors. Assume the measurements are identically distributed as \(N(\mu,12^2)\), with known standard deviation 12 units and unknown mean \(\mu\). The observed sample mean is 84 units. Derive a 95% confidence interval using the standard normal pivot. State the method and the common mistake it avoids.
Method: invert the standard normal pivot. Under the stated independent normal model, \(\bar X\sim N(\mu,12^2/36)\). Thus the standard error is \(12/\sqrt{36}=2\), and \(Z=(\bar X-\mu)/2\sim N(0,1)\). The standard normal quantiles \(-1.96\) and \(1.96\) enclose probability approximately 0.95.
Write the pivot event and substitute the observed sample mean and standard error:
Multiply all three parts by 2, which is positive, so the inequality directions do not change: \( -3.92\leq84-\mu\leq3.92\). Subtract 84 throughout: \( -87.92\leq-\mu\leq-80.08\). Multiplying by \(-1\) reverses both inequalities, giving \(80.08\leq\mu\leq87.92\).
The 95% interval is therefore \([80.08,87.92]\) units. The endpoints are symmetric around the observed mean: each is 3.92 units from 84. Substituting either endpoint into the pivot gives \(-1.96\) or \(1.96\), which checks the inversion.
Common mistake avoided. The pivot contains \(-\mu\), so multiplying by \(-1\) requires reversing inequality signs. Also, the standard error is \(12/\sqrt{36}=2\), not 12; using the population standard deviation directly as the standard error would produce an interval far too wide.
The same reasoning applies when the pivot has a t distribution, but its critical value depends on degrees of freedom. The distributional result is specific to a random sample from a normal population with unknown variance; it is not an exact t result for every population distribution.
When the Variance Is Unknown: A t Pivot
Worked Example: A Normal Mean with Unknown Variance
Model and task. A hypothetical sample of 12 independent measurements is drawn from \(N(\mu,\sigma^2)\), where both \(\mu\) and \(\sigma^2\) are unknown. The sample mean is 31.4 and the sample standard deviation is \(s=4.8\). Derive a 95% confidence interval for \(\mu\). Use \(t_{0.975,11}=2.201\), where \(t_{p,\nu}\) denotes the \(p\)-quantile of the t distribution with \(\nu\) degrees of freedom.
Method: invert the t pivot. For an independent normal sample, \(T=(\bar X-\mu)/(S/\sqrt n)\) has a t distribution with \(n-1\) degrees of freedom. Here \(n=12\), so the degrees of freedom are 11. The chosen bounds \(-2.201\) and \(2.201\) give probability 0.95.
First calculate the observed standard error: \(s/\sqrt n=4.8/\sqrt{12}=1.3856\), to four decimal places. The pivot inequality is
Solving in the same way as for the normal pivot gives \(\bar X-t_{0.975,11}s/\sqrt n\leq\mu\leq\bar X+t_{0.975,11}s/\sqrt n\). The margin is \(2.201(1.3856406)=3.0498\), using unrounded standard error in the multiplication. Therefore the endpoints are \(31.4-3.0498=28.3502\) and \(31.4+3.0498=34.4498\). Rounded to three decimal places, the interval is \([28.350,34.450]\).
The interval is centered at 31.4, as expected from the symmetric t bounds. Substituting either endpoint into the pivot yields one of the two critical values, up to rounding.
Common mistake avoided. The t quantile uses 11 degrees of freedom, not 12: estimating the sample mean leaves \(n-1\) degrees of freedom for the sample variance. The exact t pivot also depends on the normal-sampling assumption stated in the problem.
Inverting a Decreasing Pivot
Not every pivot increases with the parameter. A particularly important case is the chi-square pivot for a normal variance. For an independent sample from \(N(\mu,\sigma^2)\), with \(\mu\) unknown, define \(S^2=\sum_{i=1}^n(X_i-\bar X)^2/(n-1)\). Then \((n-1)S^2/\sigma^2\) has a chi-square distribution with \(n-1\) degrees of freedom. Because \(\sigma^2\) is in the denominator, solving for it reverses the ordering of the pivot quantiles.
Worked Example: An Interval for a Normal Variance
Model and task. A hypothetical sample of 10 independent observations comes from a normal distribution with unknown mean and variance \(\sigma^2\). Its sample variance, calculated with denominator \(n-1\), is \(S^2=16\) squared units. Find a 90% confidence interval for \(\sigma^2\), using the chi-square quantiles \(q_{0.05}=3.325\) and \(q_{0.95}=16.919\) for 9 degrees of freedom. Here \(q_p\) is the \(p\)-quantile of that chi-square distribution.
Method: invert the chi-square pivot and reverse the endpoint order. The pivot has 9 degrees of freedom, and the specified quantiles enclose probability 0.90:
Since \(9S^2=9(16)=144\), this becomes \(3.325\leq144/\sigma^2\leq16.919\). The variance is positive, so multiply by \(\sigma^2\) without changing the inequality directions. Then divide by each positive quantile: \(144/16.919\leq\sigma^2\leq144/3.325\). The order has reversed because the smaller value of \(144/\sigma^2\) corresponds to the larger value of \(\sigma^2\).
The calculations are \(144/16.919=8.5106\) and \(144/3.325=43.3083\). Rounded to three decimal places, the 90% interval for the variance is \([8.511,43.308]\) squared units. Substitution of the lower endpoint gives a pivot of approximately 16.919; substitution of the upper endpoint gives approximately 3.325, confirming the reversed pairing.
Common mistake avoided. Writing the endpoints as \(144/3.325\) followed by \(144/16.919\) produces \(43.308\) followed by \(8.511\), an incorrectly ordered interval. The reciprocal operation reverses the order; the lower variance limit must use the larger chi-square quantile.
What to Check Before Reporting an Interval
An interval is not justified merely because its endpoints look plausible. Check that the pivot’s distribution applies under the stated model, the chosen fixed bounds have the claimed probability, and the algebra produces an event equivalent to the original pivot event. Then verify that the lower endpoint does not exceed the upper endpoint and that the units match the parameter.
| Feature of the pivot | What to inspect during inversion |
|---|---|
| Parameter appears with a minus sign | Multiplication by \(-1\) reverses the inequality signs. |
| Positive parameter appears in a denominator | Taking reciprocals reverses the ordering; check positivity before doing so. |
| Standard error is estimated | Use the pivot’s correct reference distribution and degrees of freedom. |
| Sample statistic is observed | Insert it only after writing the general inequality, then retain units. |
These inversion steps are aligned with the SOA VEE Mathematical Statistics syllabus. A useful final check is to start from the reported interval and reverse the algebra: its endpoints should correspond to the selected pivot bounds, and the probability statement should follow from the assumed sampling model.
Check Your Understanding
For each problem, write the pivot inequality first, then solve it for the parameter. Quantiles are provided where needed.
- A sample of 25 independent observations comes from \(N(\mu,10^2)\), with known standard deviation 10. The sample mean is 42. Use standard normal bounds \(-1.645\) and \(1.645\) to find the interval. (Answer: \([38.710,45.290]\))
- A normal sample has size 16 and sample variance \(S^2=9\), with unknown mean and variance. For 15 degrees of freedom, use chi-square bounds 6.262 and 27.488 to find the interval for \(\sigma^2\). (Answer: \([4.911,21.559]\), to three decimal places)
- Five independent waiting times have an exponential distribution with rate \(\lambda\) per hour, and their observed sum is 12.5 hours. The pivot \(2\lambda\sum X_i\) has a chi-square distribution with 10 degrees of freedom. Using bounds 3.940 and 18.307, find the interval for \(\lambda\). (Answer: \([0.1576,0.7323]\) per hour, to four decimal places)
- A normal sample has size 9, sample mean 17.2, and sample standard deviation 6. Use the t pivot with 8 degrees of freedom and the symmetric 90% bounds \(-1.860\) and \(1.860\) to find the interval for the population mean. (Answer: \([13.480,20.920]\), to three decimal places)
- For a positive parameter \(\theta\), a pivot is \(Q=20/\theta\). Its fixed probability bounds are 2 and 5. Invert \(2\leq Q\leq5\) and give the interval for \(\theta\). (Answer: \([4,10]\))