One-Sided Confidence Bounds For A Normal Variance

Why One-Sided Bounds Are Useful

A two-sided interval for a variance describes a range of plausible values. Some questions instead ask for a limit in only one direction: for example, how large could the variability reasonably be, or how small can it reasonably be? A one-sided confidence bound answers that question while assigning the full significance level to a single tail of a reference distribution.

The calculation relies on the exact sampling distribution of the sample variance from a normal population. The central practical issue is choosing the correct chi-square quantile: an upper bound uses a lower-tail quantile, while a lower bound uses an upper-tail quantile. These formulas are aligned with the SOA VEE Mathematical Statistics syllabus.

Model and notation: Let \(X_1,\ldots,X_n\) be an independent random sample from \(N(\mu,\sigma^2)\), where both \(\mu\) and \(\sigma^2\) are unknown and \(n\ge 2\). Define the unbiased sample variance by \(S^2=\sum_{i=1}^{n}(X_i-\bar X)^2/(n-1)\), and let \(\nu=n-1\). Under this model, \(Q=\nu S^2/\sigma^2\) has a chi-square distribution with \(\nu\) degrees of freedom. Write \(q_{p,\nu}\) for the lower-tail \(p\)-quantile, so that \(P(\chi^2_\nu\le q_{p,\nu})=p\).

The normality and independence assumptions matter. The chi-square result is exact for an independent sample from a normal population; it is not generally an exact result for a nonnormal population. The variance \(S^2\) here uses denominator \(n-1\), not \(n\). A different definition of sample variance would require adjusting the statistic before applying these formulas.

The examples below concern hypothetical measurement studies. No interest-rate or mortality assumptions are involved: the probability model is the stated normal sampling model, with its mean and variance parameters specified or estimated as described.

Upper Confidence Bound for the Variance

An upper confidence bound \(U\) is constructed so that the procedure covers the population variance from above with probability \(1-\alpha\). The confidence level is \(100(1-\alpha)\%\). Use the lower-tail chi-square quantile \(q_{\alpha,\nu}\), not \(q_{1-\alpha,\nu}\).

U=\frac{\nu S^2}{q_{\alpha,\nu}}

The bound is one-sided: the confidence set is \((0,U]\), rather than an interval with both a lower and an upper endpoint. Its coverage statement is \(P(\sigma^2\le U)=1-\alpha\), where the probability is calculated over repeated samples under the model.

Derivation: Since \(Q=\nu S^2/\sigma^2\sim\chi^2_\nu\), its lower-tail probability at \(q_{\alpha,\nu}\) is \(\alpha\). Therefore \(P(Q\ge q_{\alpha,\nu})=1-\alpha\). On the event \(S^2>0\), all quantities used in the rearrangement are positive, so \(Q\ge q_{\alpha,\nu}\) is equivalent to \(\nu S^2/\sigma^2\ge q_{\alpha,\nu}\), and hence to \(\sigma^2\le\nu S^2/q_{\alpha,\nu}=U\). Thus \(P(\sigma^2\le U)=1-\alpha\). For \(n\ge2\), \(S^2>0\) with probability one under the continuous normal model; the event of all sample values being equal has probability zero and does not affect coverage.

Worked Example: A 90% Upper Bound for Measurement Variability

A hypothetical laboratory takes an independent sample of \(n=4\) readings from a normal population with unknown mean and variance. The reported sample variance, calculated with denominator \(n-1\), is \(S^2=3.00\) measurement-units squared. Find a 90% upper confidence bound for \(\sigma^2\). The model is normal sampling; no interest or mortality assumptions apply.

Method. For a 90% bound, \(\alpha=0.10\), and the degrees of freedom are \(\nu=n-1=3\). The required lower-tail quantile is \(q_{0.10,3}=0.5844\) (rounded to four decimals).

Substitute into the upper-bound formula:

U=\frac{\nu S^2}{q_{\alpha,\nu}} =\frac{3(3.00)}{0.5844} =\frac{9.00}{0.5844} \approx 15.40

Thus the 90% upper confidence bound is \(15.40\) measurement-units squared. Over repeated samples from this normal population, the procedure produces an upper bound at least as large as the true variance 90% of the time. The common mistake to avoid is using \(q_{0.90,3}\) in the denominator; that is the wrong tail for an upper bound.

Lower Confidence Bound for the Variance

A lower confidence bound \(L\) is constructed so that the procedure places the true variance above the bound with probability \(1-\alpha\). In this case use \(q_{1-\alpha,\nu}\), the upper-tail cutoff whose lower-tail probability is \(1-\alpha\).

L=\frac{\nu S^2}{q_{1-\alpha,\nu}}

The confidence set is \([L,\infty)\). This result follows from the other side of the same chi-square pivot. Since \(P(Q\le q_{1-\alpha,\nu})=1-\alpha\), the event \(Q\le q_{1-\alpha,\nu}\) can be rearranged, when \(S^2>0\), to \(\sigma^2\ge \nu S^2/q_{1-\alpha,\nu}=L\). The zero-variance sample event has probability zero under the stated normal model, as above.

Worked Example: A 95% Lower Bound from a Sample Variance

A hypothetical sensor study samples \(n=3\) independent readings from a normal population with unknown mean and variance. Its unbiased sample variance is \(S^2=2.00\) units squared. Find a 95% lower confidence bound for the population variance. The distributional assumption is \(N(\mu,\sigma^2)\); no interest or mortality assumptions apply.

Method. Here \(\alpha=0.05\) and \(\nu=3-1=2\). A lower bound uses \(q_{0.95,2}\). For two degrees of freedom, the chi-square lower-tail quantile is \(q_{p,2}=-2\ln(1-p)\), so

q_{0.95,2}=-2\ln(0.05)=5.9915

Now calculate the bound:

L=\frac{2(2.00)}{5.9915} =\frac{4.00}{5.9915} \approx 0.6676

The 95% lower confidence bound is \(0.67\) units squared when rounded to two decimal places. The statement is that this procedure gives a lower endpoint no greater than the true variance in 95% of repeated samples. It does not mean that, after observing this sample, the fixed parameter has a 95% probability of exceeding \(0.67\).

Calculating the Sample Variance Before Bounding

A problem may provide the observations rather than \(S^2\). In that case, first calculate the sample mean, then the sum of squared deviations, and then divide by \(n-1\). Keep the degrees of freedom visible: the numerator \(\nu S^2\) in either bound is exactly the sum of squared deviations.

Worked Example: An Upper Bound from Four Raw Readings

A hypothetical calibration exercise records \(8, 10, 12,\) and \(14\) units from an independent sample of a normal population with unknown mean and variance. Find a 95% upper confidence bound for the population variance. Assume the sample readings are independent and normally distributed; no interest or mortality assumptions apply.

Method. First find the sample mean:

\bar X=\frac{8+10+12+14}{4}=\frac{44}{4}=11

The deviations from 11 are \(-3,-1,1,3\), so the squared deviations are \(9,1,1,9\). Their sum is \(20\), giving

S^2=\frac{20}{4-1}=\frac{20}{3}=6.6667,\qquad \nu=3

For a 95% upper bound, \(\alpha=0.05\), so use the lower-tail quantile \(q_{0.05,3}=0.3518\) (rounded to four decimals). The numerator is \(\nu S^2=3(20/3)=20\). Therefore,

U=\frac{20}{0.3518}\approx 56.85

The 95% upper confidence bound is \(56.85\) units squared, rounded to two decimal places. As a check, the calculation used \(n-1=3\) degrees of freedom and the sum of squared deviations \(20\) in the numerator. Using \(S^2\) alone as the numerator would omit the degrees-of-freedom factor.

How the Two Bounds Differ

Both bounds use the same pivot and the same observed sample variance. They differ in which tail probability is set to \(1-\alpha\). For a fixed sample, a lower bound is usually a positive lower endpoint, while an upper bound is finite; neither is a two-sided interval.

Desired bound	Confidence set	Chi-square quantile	Formula
Upper \(100(1-\alpha)\%\) bound	\((0,U]\)	\(q_{\alpha,\nu}\)	\(U=\nu S^2/q_{\alpha,\nu}\)
Lower \(100(1-\alpha)\%\) bound	\([L,\infty)\)	\(q_{1-\alpha,\nu}\)	\(L=\nu S^2/q_{1-\alpha,\nu}\)

The quantile convention is essential. Here \(q_{p,\nu}\) always means a lower-tail quantile, so \(P(\chi^2_\nu\le q_{p,\nu})=p\). For an upper variance bound, the event used in the derivation is \(Q\ge q_{\alpha,\nu}\), which has probability \(1-\alpha\). For a lower variance bound, it is \(Q\le q_{1-\alpha,\nu}\), also with probability \(1-\alpha\).

Common Mistakes and Conditions

Using the wrong tail. An upper bound uses \(q_{\alpha,\nu}\), the small lower-tail quantile. A lower bound uses \(q_{1-\alpha,\nu}\), the large lower-tail quantile.
Using \(n\) degrees of freedom. When the population mean is unknown and \(S^2\) is defined with denominator \(n-1\), use \(\nu=n-1\).
Interpreting confidence as a probability about the fixed variance. The parameter is fixed; the 95% coverage statement describes the long-run performance of the procedure.
Applying the exact formula without normality. The chi-square sampling distribution stated here depends on normal sampling. Without that assumption, these bounds are not generally exact.
Dividing by zero in a rearrangement. The algebraic equivalences use \(S^2>0\). Under a continuous normal model with \(n\ge2\), this holds with probability one, because an exactly constant sample has probability zero.

Key takeaway: For an independent normal sample with unknown mean, set \(\nu=n-1\) and use the unbiased variance \(S^2\). The upper bound is \(\nu S^2/q_{\alpha,\nu}\); the lower bound is \(\nu S^2/q_{1-\alpha,\nu}\). Always identify the quantile convention before substituting.

Check Your Understanding

For each problem, use the stated normal-sampling assumption and the lower-tail quantile convention \(P(\chi^2_\nu\le q_{p,\nu})=p\). Round variance bounds to two decimal places.

A hypothetical independent normal sample has \(n=4\) and \(S^2=5.00\). For a 90% upper bound, use \(q_{0.10,3}=0.5844\). (Answer: \(15(1)/0.5844=71.19\) units squared.)
A hypothetical independent normal sample has \(n=3\) and \(S^2=6.00\). For a 90% lower bound, use \(q_{0.90,2}=4.6052\). (Answer: \(12/4.6052=2.61\) units squared.)
A hypothetical normal sample consists of \(5,7,9,11\). Find the 95% upper bound using \(q_{0.05,3}=0.3518\). (Answer: \(\bar X=8\), \(S^2=20/3\), and \(U=20/0.3518=56.85\) units squared.)
A hypothetical independent normal sample has \(n=6\) and \(S^2=4.00\). Find the 95% lower bound using \(q_{0.95,5}=11.0705\). (Answer: \(20/11.0705=1.81\) units squared.)
A hypothetical independent normal sample has \(n=6\) and \(S^2=9.00\). Find the 90% upper bound using \(q_{0.10,5}=1.6103\). (Answer: \(45/1.6103=27.94\) units squared.)

What You'll Learn