Why Independent Samples Are Required for the Exact F Test

Why the Exact F Test Needs Independent Samples

A ratio of two sample variances may look like an F statistic, but its distribution is not determined by the ratio alone. For the classical exact F reference law, each sample must come from a normal population, and the two samples must be independent. Pairing observations or sharing a source of variation can make the two sample variances dependent. In that case, the usual F table does not give an exact reference distribution for their ratio.

The distinction matters in variance comparisons because a test’s reference distribution is what calibrates its tail probabilities. A statistic can be calculated perfectly and still be compared with the wrong distribution if the sampling design violates independence. The results below are aligned with the SOA VEE Mathematical Statistics syllabus.

How Independence Produces the F Distribution

Let \(X_1,\ldots,X_{n_1}\) be an independent random sample from \(N(\mu_1,\sigma_1^2)\), and let \(Y_1,\ldots,Y_{n_2}\) be an independent random sample from \(N(\mu_2,\sigma_2^2)\). Assume, in addition, that the entire \(X\) sample is independent of the entire \(Y\) sample. Write \(S_X^2\) and \(S_Y^2\) for their usual unbiased sample variances, each calculated with its own sample mean and divisor \(n_j-1\).

For a normal sample, the scaled sample variance has a chi-square distribution. Specifically, \((n_1-1)S_X^2/\sigma_1^2\) has a chi-square distribution with \(n_1-1\) degrees of freedom, and the corresponding result holds for the \(Y\) sample. Since each statistic is a function of its own sample, independence of the two samples makes these two chi-square statistics independent.

The F distribution is defined as the ratio of two independent chi-square variables after each is divided by its degrees of freedom. Applying that definition gives the variance-scaled ratio below. Under the null hypothesis of equal population variances, the population-variance ratio is one, so the unscaled sample-variance ratio itself has the stated F law.

If \(H_0:\sigma_1^2=\sigma_2^2\), this reduces to \(S_X^2/S_Y^2\sim F_{n_1-1,n_2-1}\). If the population variances are not equal, the raw ratio \(S_X^2/S_Y^2\) does not have this standard F law; it is the ratio scaled by \(\sigma_1^2/\sigma_2^2\) that has the stated distribution.

Justification: Why the Two Variance Estimates Are Independent

The reasoning has two separate parts. Normality supplies a chi-square distribution for each scaled sample variance. Independence of the samples supplies independence between those two statistics. For the second part, no special distribution calculation is needed: a statistic computed only from the \(X\) sample is independent of a statistic computed only from the independent \(Y\) sample.

This also shows precisely where a paired design can break the argument. If \(X_i\) and \(Y_i\) are linked within each pair, then \(S_X^2\) and \(S_Y^2\) are functions of dependent data. Their individual scaled distributions may still be chi-square when the marginal samples are normal, but that fact alone does not make the two chi-square variables independent. The defining independent-ratio step for the exact F law is then unavailable.

What Pairing Does to Sample Variances

A simple normal model makes the dependence visible. Suppose each pair has observations \(X_i=U_i+\epsilon_i\) and \(Y_i=U_i+\eta_i\), where all components are normal, the components are independent, and their variances are \(\operatorname{Var}(U_i)=2\), \(\operatorname{Var}(\epsilon_i)=2\), and \(\operatorname{Var}(\eta_i)=2\). The common component \(U_i\) is shared by the two observations in pair \(i\). Consequently, \(\operatorname{Var}(X_i)=4\), \(\operatorname{Var}(Y_i)=4\), and \(\operatorname{Cov}(X_i,Y_i)=2\), so the within-pair correlation is \(2/\sqrt{4\cdot4}=0.5\).

For independent pairs from a bivariate normal distribution with marginal variances \(\sigma_X^2,\sigma_Y^2\) and within-pair covariance \(\sigma_{XY}\), the covariance of the usual sample variances is \(\operatorname{Cov}(S_X^2,S_Y^2)=2\sigma_{XY}^2/(n-1)\). One way to see the factor is to express each centered sample as \(n-1\) orthogonal normal contrasts. Corresponding contrasts in the two samples have covariance \(\sigma_{XY}\); for a jointly normal pair of contrasts \(A,B\), \(\operatorname{Cov}(A^2,B^2)=2\operatorname{Cov}(A,B)^2\). Summing over the \(n-1\) contrasts and dividing by \((n-1)^2\) gives the formula.

When a Numerical Ratio Is Not an Exact F Statistic

A variance ratio can always be calculated from observed data. Whether it can be assessed against an exact F distribution depends on the sampling model, not on the arithmetic. For example, two treatments measured on the same subjects produce paired observations. If treatment responses are dependent within subjects, the two sample variances are generally dependent as well. Merely inserting the two sample sizes into F degrees of freedom does not fix that problem.

Common Exam Trap: Normal Does Not Mean Independent

The incorrect argument is: “Both populations are normal, so \(S_X^2/\sigma_X^2\) and \(S_Y^2/\sigma_Y^2\) are scaled chi-square variables; therefore their ratio is F.” The final inference does not follow unless the two chi-square variables are independent. In a paired design, each marginal variance result can be valid while the independence condition fails.

The correction is to check the design before selecting the reference distribution. Confirm that the samples are separately collected and mutually independent, and that each population is normal. Only then use the exact classical F law. In particular, a paired comparison is not converted into an independent-samples comparison by treating the two columns as separate lists.