Deriving the F Law for Two Normal Sample Variances

Why Ratios of Sample Variances Lead to an F Distribution

A sample variance is random: a new sample from the same population will generally produce a different value. When two independent samples come from normal populations, the variation in each sample variance has a precise chi-square form. Dividing the two appropriately scaled chi-square quantities produces an F variable. This connection explains when a ratio of sample variances has an exact F distribution, rather than merely an approximate one.

The distinction between a raw ratio and a standardized ratio matters. If the two populations have the same positive variance, the raw ratio has an F distribution. If their variances differ, it is the ratio after scaling each estimate by its own population variance that has the standard F distribution. This derivation is aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: Let $X_{11},\ldots,X_{1n_1}$ be an independent sample from $N(\mu_1,\sigma_1^2)$, and let $X_{21},\ldots,X_{2n_2}$ be an independent sample from $N(\mu_2,\sigma_2^2)$. Assume the two samples are independent of one another, $\sigma_1^2>0$, $\sigma_2^2>0$, and $n_1,n_2\ge 2$. For sample $i$, define $$ S_i^2=\frac{1}{n_i-1}\sum_{j=1}^{n_i}(X_{ij}-\bar X_i)^2, \qquad \bar X_i=\frac{1}{n_i}\sum_{j=1}^{n_i}X_{ij}. $$ The sample variance uses $n_i-1$ in its denominator. A variable with an F distribution with $\nu_1$ numerator degrees of freedom and $\nu_2$ denominator degrees of freedom is denoted $F_{\nu_1,\nu_2}$.

From Normal Sample Variances to an F Variable

For a normal sample of size $n_i$, the standard sample-variance result is $(n_i-1)S_i^2/\sigma_i^2\sim\chi^2_{n_i-1}$. This result applies because the observations are independent and normally distributed, and the sample variance is calculated about that sample's own mean. The two chi-square variables formed from the two samples are independent because the samples are independent.

Write $\nu_i=n_i-1$ for the degrees of freedom in sample $i$. The F distribution is defined by taking two independent chi-square variables, dividing each by its degrees of freedom, and forming their ratio. Substitution of the sample-variance chi-square results gives the standardized variance-ratio law.

U=\frac{\nu_1 S_1^2}{\sigma_1^2}\sim\chi^2_{\nu_1}, \qquad V=\frac{\nu_2 S_2^2}{\sigma_2^2}\sim\chi^2_{\nu_2}, \qquad U\ \text{and}\ V\ \text{independent}.

Theorem: Under the independent normal-sample assumptions above, $$ \frac{S_1^2/\sigma_1^2}{S_2^2/\sigma_2^2} = \frac{U/\nu_1}{V/\nu_2} \sim F_{\nu_1,\nu_2}, \qquad \nu_i=n_i-1. $$ If the population variances are equal and positive, so that $\sigma_1^2=\sigma_2^2=\sigma^2>0$, then $$ \frac{S_1^2}{S_2^2}\sim F_{n_1-1,n_2-1}. $$ Both sample sizes must be at least 2 for these sample variances and their ratio to be defined.

The result follows directly from the definition of the F distribution. In detail, $$ \frac{U/\nu_1}{V/\nu_2} = \frac{(S_1^2/\sigma_1^2)}{(S_2^2/\sigma_2^2)}. $$ The chi-square quantities $U$ and $V$ are independent, so their ratio after division by their respective degrees of freedom is $F_{\nu_1,\nu_2}$. When the population variances are equal, the factors $\sigma^2$ cancel, leaving the raw sample-variance ratio. When they are unequal, they do not cancel: the raw ratio is scaled by the population variance ratio.

\frac{S_1^2}{S_2^2} = \frac{\sigma_1^2}{\sigma_2^2} \left(\frac{S_1^2/\sigma_1^2}{S_2^2/\sigma_2^2}\right).

Thus, for unequal positive variances, $S_1^2/S_2^2$ has the distribution of $(\sigma_1^2/\sigma_2^2)F_{\nu_1,\nu_2}$, not the standard $F_{\nu_1,\nu_2}$ distribution itself. The scaling factor is fixed, while the F variable is random. Knowing this distinction is essential whenever a variance ratio is interpreted probabilistically.

Worked Applications

Worked Example: Equal Population Variances and Sample Summaries

Two independent hypothetical normal populations have the same unknown positive variance $\sigma^2$. A sample of $n_1=4$ observations from the first population has corrected sum of squares $\sum_{j=1}^{4}(X_{1j}-\bar X_1)^2=18$. A sample of $n_2=6$ observations from the second has corrected sum of squares $\sum_{j=1}^{6}(X_{2j}-\bar X_2)^2=25$. Calculate the sample-variance ratio and identify its distribution under the model.

Method. Divide each corrected sum of squares by its sample's degrees of freedom. Since both samples are normal and independent, and their population variances are equal and positive, the ratio follows the equal-variance result.

For the first sample, the degrees of freedom are $n_1-1=4-1=3$, so $$ S_1^2=\frac{18}{3}=6. $$ For the second sample, the degrees of freedom are $n_2-1=6-1=5$, so $$ S_2^2=\frac{25}{5}=5. $$ Therefore the observed ratio is $$ \frac{S_1^2}{S_2^2}=\frac{6}{5}=1.2. $$ Under the stated model, the random ratio $S_1^2/S_2^2$ has an $F_{3,5}$ distribution. The observed value is $1.2$; it is not itself a distribution, but one realization of that F variable.

Check. Both sample sizes are at least 2, the degrees of freedom are 3 and 5, and the variances were computed using $n_i-1$. The common positive population variance cancels from the ratio. The common mistake avoided here is using $n_i$, rather than $n_i-1$, to calculate sample variance or degrees of freedom.

Worked Example: Standardizing When Population Variances Differ

Suppose two independent hypothetical normal populations have known variances $\sigma_1^2=9$ and $\sigma_2^2=4$. Independent samples have sizes $n_1=6$ and $n_2=5$, and their corrected sums of squares are 45 and 16, respectively. Find the raw sample-variance ratio and the standardized ratio, then state which one has a standard F distribution.

Method. First calculate each sample variance with its own degrees of freedom. Then divide each sample variance by its population variance before taking the standardized ratio.

The first sample has $6-1=5$ degrees of freedom, giving $$ S_1^2=\frac{45}{5}=9. $$ The second sample has $5-1=4$ degrees of freedom, giving $$ S_2^2=\frac{16}{4}=4. $$ The raw ratio is $$ \frac{S_1^2}{S_2^2}=\frac{9}{4}=2.25. $$ The standardized ratio is $$ \frac{S_1^2/\sigma_1^2}{S_2^2/\sigma_2^2} = \frac{9/9}{4/4} = \frac{1}{1} = 1. $$ By the theorem, the standardized random ratio has an $F_{5,4}$ distribution. The raw ratio is the standardized ratio multiplied by $\sigma_1^2/\sigma_2^2=9/4$, so it is not itself a standard $F_{5,4}$ variable.

Check. The sample variances equal the given population variances in this particular set of summaries, so each standardized estimate is 1. The degrees of freedom remain 5 and 4, not 6 and 5. The common mistake avoided is labeling the raw value $2.25$ as an $F_{5,4}$ observation despite the unequal population variances.

Worked Example: A Probability for a Ratio of Sample Variances

Two independent samples, each of size 3, are drawn from normal populations with the same unknown positive variance. What is the probability that the first sample variance is more than three times the second sample variance?

Method. Under equal positive population variances, the raw sample-variance ratio has an F distribution. Identify its degrees of freedom, then evaluate the resulting F tail probability.

Each sample has $3-1=2$ degrees of freedom, so $$ \frac{S_1^2}{S_2^2}\sim F_{2,2}. $$ For $W\sim F_{2,2}$, the density is $f_W(w)=(1+w)^{-2}$ for $w>0$. Integrating gives $$ P(W\leq x)=\int_0^x\frac{1}{(1+w)^2}\,dw =1-\frac{1}{1+x} =\frac{x}{1+x}, \qquad x>0. $$ Consequently, $$ P\left(\frac{S_1^2}{S_2^2}>3\right) =P(W>3) =1-\frac{3}{1+3} =1-\frac{3}{4} =\frac{1}{4} =0.25. $$ The probability is $0.25$, or $0.2500$ to four decimal places.

Check. The degrees of freedom are 2 and 2 because each sample has size 3. The probability is an upper-tail probability: the event is a ratio greater than 3, so the CDF at 3 must be subtracted from 1. The result lies between 0 and 1 as required.

Conditions and Common Traps

The exact result relies on the model, not just on the fact that two sample variances have been calculated. Each sample must consist of independent observations from a normal population, the two samples must be independent of one another, and both sample sizes must be at least 2. The population variances used for standardization must be positive. If a sample has size 1, its sample-variance formula divides by zero; there is no defined sample variance and no variance ratio of the stated form.

The numerator and denominator degrees of freedom follow the order of the sample variances in the ratio. For $S_1^2/S_2^2$, they are $n_1-1$ and $n_2-1$. Reversing the sample-variance ratio also reverses the order of the degrees of freedom. In either order, the values come from sample sizes minus one, not from the sample sizes themselves.

Using the result without normality. The exact chi-square sample-variance result used in the derivation is for normal samples. A ratio of variances does not automatically have an exact F distribution for arbitrary populations.
Ignoring dependence. The ratio construction uses independent chi-square quantities. If the samples are dependent, the stated derivation does not establish an F law.
Forgetting to standardize. With unequal population variances, $S_1^2/S_2^2$ carries the factor $\sigma_1^2/\sigma_2^2$. The standard F variable is the ratio after dividing each sample variance by its own population variance.
Using the wrong denominator. Sample variance is computed with $n-1$, and its chi-square degrees of freedom are $n-1$. Using $n$ changes both the variance estimate and the claimed F degrees of freedom.

Key takeaway: For independent normal samples with $n_1,n_2\ge2$ and positive population variances, the ratio $(S_1^2/\sigma_1^2)/(S_2^2/\sigma_2^2)$ is $F_{n_1-1,n_2-1}$. Only when the population variances are equal does this reduce to the unscaled ratio $S_1^2/S_2^2$.

Check Your Understanding

For each problem, identify the relevant degrees of freedom and apply the variance-ratio result under the stated assumptions.

Independent normal samples have sizes 7 and 9 from populations with a common positive variance. State the distribution of the first sample variance divided by the second. (Answer: $F_{6,8}$.)
Independent normal samples have sizes 5 and 8, with population variances 16 and 4. State the distribution of $(S_1^2/16)/(S_2^2/4)$. (Answer: $F_{4,7}$.)
For the samples in question 2, express the raw ratio $S_1^2/S_2^2$ as a constant times a standard F variable, and identify the degrees of freedom. (Answer: $4F_{4,7}$.)
Two independent normal samples of size 3 come from populations with equal positive variances. Find the probability that the second sample variance is at least twice the first. (Answer: $1/3$.)
An analyst has one observation from the first population and four from the second and proposes $S_1^2/S_2^2\sim F_{0,3}$. Explain the defect. (Answer: The first sample variance is undefined because its denominator is $1-1=0$; the ratio and the proposed F distribution do not apply.)

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