Scaling F for Unequal Population Variances

Why the Population Variances Matter

The ratio of two sample variances reflects two sources of variation: the random variation from sampling, and the difference in spread between the two populations. The F distribution describes the sampling part after each sample variance has been standardized by its own population variance. If the population variances are unequal, the raw ratio retains their ratio as a scale factor.

Keeping that factor visible is important in probability calculations. Treating the unstandardized ratio as an ordinary F variable effectively assumes the population variances are equal. This tutorial derives the scaled distribution and shows how to use it; it is aligned with the SOA VEE Mathematical Statistics syllabus.

Definition: 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 the two samples are independent, $\sigma_1^2>0$, $\sigma_2^2>0$, and $n_1,n_2\ge 2$. Define $S_1^2$ and $S_2^2$ as the usual sample variances, each calculated about its own sample mean and with denominator $n_i-1$. Then $$ \frac{S_1^2}{S_2^2}\ \sim\ \frac{\sigma_1^2}{\sigma_2^2}F_{n_1-1,n_2-1}. $$ Here $F_{\nu_1,\nu_2}$ denotes an F random variable with numerator degrees of freedom $\nu_1$ and denominator degrees of freedom $\nu_2$. The notation on the right means a positive constant multiplied by an F random variable; it does not mean the raw ratio itself has an F distribution unless $\sigma_1^2=\sigma_2^2$.

Deriving the Scale Factor

For a normal sample of size $n_i$, the standard sample-variance result is $(n_i-1)S_i^2/\sigma_i^2\sim\chi^2_{n_i-1}$. In words, the sample variance divided by its own population variance, and multiplied by its degrees of freedom, has a chi-square distribution. The two such chi-square variables are independent because the samples are independent.

The derivation makes clear where the scale factor enters. Write each sample variance as its population variance times a standardized random quantity, then form their ratio:

\begin{aligned} \frac{S_1^2}{S_2^2} &= \frac{\sigma_1^2}{\sigma_2^2} \frac{S_1^2/\sigma_1^2}{S_2^2/\sigma_2^2}\\ &= \frac{\sigma_1^2}{\sigma_2^2} \frac{\left((n_1-1)S_1^2/\sigma_1^2\right)/(n_1-1)} {\left((n_2-1)S_2^2/\sigma_2^2\right)/(n_2-1)}\\ &= \frac{\sigma_1^2}{\sigma_2^2} \frac{U_1/(n_1-1)}{U_2/(n_2-1)}, \end{aligned}

where $U_1=(n_1-1)S_1^2/\sigma_1^2$ and $U_2=(n_2-1)S_2^2/\sigma_2^2$. By the normal-sample variance result, $U_1\sim\chi^2_{n_1-1}$ and $U_2\sim\chi^2_{n_2-1}$, independently. The ratio of these independent chi-square variables, each divided by its degrees of freedom, has the $F_{n_1-1,n_2-1}$ distribution. Therefore the remaining factor $\sigma_1^2/\sigma_2^2$ scales that F variable.

Theorem: Under the assumptions in the definition, set $$ F=\frac{S_1^2/\sigma_1^2}{S_2^2/\sigma_2^2} =\frac{(n_1-1)S_1^2/\sigma_1^2}{n_1-1}\left/\frac{(n_2-1)S_2^2/\sigma_2^2}{n_2-1}\right. . $$ Then $F\sim F_{n_1-1,n_2-1}$, and $S_1^2/S_2^2=(\sigma_1^2/\sigma_2^2)F$. The degrees of freedom come from the sample sizes; the scale comes from the population variances.

Using the Distribution in Calculations

Write $c=\sigma_1^2/\sigma_2^2$, so that $S_1^2/S_2^2=cF$. For a positive threshold $r$, divide by the positive constant $c$ to translate an event about the sample-variance ratio into an event about an F variable:

P\left(\frac{S_1^2}{S_2^2}\le r\right) =P\left(F_{n_1-1,n_2-1}\le \frac{r}{c}\right), \qquad P\left(\frac{S_1^2}{S_2^2}>r\right) =P\left(F_{n_1-1,n_2-1}>\frac{r}{c}\right).

The same scaling applies to quantiles. If $q_p$ is the $p$-quantile of $F_{n_1-1,n_2-1}$, then the $p$-quantile of $S_1^2/S_2^2$ is $cq_p$. In practical work, F table entries or software can provide the F probability or quantile after the threshold has been divided by $c$.

Worked Example: A Probability for Unequal Population Variances

Suppose, hypothetically, that two independent groups each consist of $n_1=n_2=3$ normally distributed measurements. The first population is $N(5,4)$ and the second is $N(8,9)$, where the second parameter is the variance. Find the probability that the first sample variance is no larger than the second.

Method and assumptions. The samples are independent normal samples, both sizes are at least 2, and the population variances are $4$ and $9$. Thus the sample-variance ratio has scale $c=4/9$ and degrees of freedom $2$ and $2$: $$ \frac{S_1^2}{S_2^2}\sim \frac{4}{9}F_{2,2}. $$

For $F_{2,2}$, the cumulative distribution function at $x>0$ is $P(F_{2,2}\le x)=x/(1+x)$. The event in the question is $S_1^2/S_2^2\le1$. Divide its threshold by $4/9$: $$ P\left(\frac{S_1^2}{S_2^2}\le1\right) =P\left(F_{2,2}\le\frac{1}{4/9}\right) =P(F_{2,2}\le 9/4). $$

Evaluate the probability: $$ \frac{9/4}{1+9/4} =\frac{9/4}{13/4} =\frac{9}{13} =0.6923. $$ The probability is $0.6923$ to four decimal places. The check is that a threshold of 1 for the raw ratio becomes the larger threshold $9/4$ for $F$, because the raw ratio is scaled down by $4/9$. The method avoids confusing the population scale with the F degrees of freedom.

Worked Example: Translating an Upper-Tail Event

In a hypothetical quality study, two independent samples of size 3 come from normal populations with variances $\sigma_1^2=12$ and $\sigma_2^2=3$. Find the probability that $S_1^2/S_2^2>8$.

Method and assumptions. Both samples are independent, normal, and have $n_i=3$, so the degrees of freedom are $2$ and $2$. The scale factor is $$ c=\frac{12}{3}=4, \qquad \frac{S_1^2}{S_2^2}\sim4F_{2,2}. $$

Divide the raw-ratio threshold 8 by 4. Then use the $F_{2,2}$ upper-tail probability: $$ P\left(\frac{S_1^2}{S_2^2}>8\right) =P(F_{2,2}>2) =1-P(F_{2,2}\le2). $$ Since $P(F_{2,2}\le2)=2/(1+2)=2/3$, the requested probability is $$ 1-\frac{2}{3}=\frac{1}{3}=0.3333. $$

The probability is $0.3333$ to four decimal places. As a check, the scale factor is 4, so a raw ratio greater than 8 corresponds to the standardized F variable being greater than 2, not greater than 8. The method avoids applying the threshold to the wrong random variable.

Worked Example: Finding a Percentile of the Raw Ratio

Suppose two independent normal samples each have size 3, with population variances $\sigma_1^2=6$ and $\sigma_2^2=4$. Find the 90th percentile of $S_1^2/S_2^2$.

Method and assumptions. The samples satisfy the independence and normality assumptions, and both sample sizes are at least 2. The scale is $c=6/4=3/2$, and the degrees of freedom are $2,2$. If $q_{0.90}$ is the 90th percentile of $F_{2,2}$, then the desired percentile is $(3/2)q_{0.90}$.

For $F_{2,2}$, its cumulative distribution function is $x/(1+x)$. Solve $x/(1+x)=0.90$: $$ x=0.90(1+x),\qquad 0.10x=0.90,\qquad q_{0.90}=9. $$ Therefore the 90th percentile of the sample-variance ratio is $$ \frac{3}{2}\times9=13.5. $$

The 90th percentile is $13.5$. As a check, the probability that the raw ratio is at most $13.5$ is the probability that $F_{2,2}\le13.5/(3/2)=9$, which is $9/(1+9)=0.90$. The method avoids reporting the F percentile without applying the population-variance scale.

Assumptions, Interpretation, and a Common Trap

The exact result depends on specific conditions. The observations within each sample must be independent and identically distributed normal random variables, and the two samples must be independent of each other. Each sample size must be at least 2, so both sample variances and both positive degrees of freedom are defined. The population variances must be positive. The population means may differ; they do not appear in the ratio distribution.

The result does not assert that all ratios of variances follow a scaled F distribution. In particular, for non-normal populations or dependent samples, the chi-square result used in the derivation does not generally hold, so this exact distribution need not apply. Also, $S_1^2/S_2^2$ is not itself an F variable when $\sigma_1^2\ne\sigma_2^2$. Only after dividing each sample variance by its own population variance does the ratio have the standard F distribution.

Quantity	Distribution or role	What to check
$(n_1-1)S_1^2/\sigma_1^2$	$\chi^2_{n_1-1}$	First sample is normal; $n_1\ge2$
$(n_2-1)S_2^2/\sigma_2^2$	$\chi^2_{n_2-1}$	Second sample is normal; $n_2\ge2$
$(S_1^2/\sigma_1^2)/(S_2^2/\sigma_2^2)$	$F_{n_1-1,n_2-1}$	Samples are independent
$S_1^2/S_2^2$	$(\sigma_1^2/\sigma_2^2)F_{n_1-1,n_2-1}$	Retain the population-variance ratio as scale

Common mistake and correction. Suppose $\sigma_1^2=12$, $\sigma_2^2=3$, and both sample sizes are 3. A mistaken calculation treats $S_1^2/S_2^2$ as $F_{2,2}$ and obtains $$ P\left(\frac{S_1^2}{S_2^2}>8\right)\ \text{(incorrectly)}=P(F_{2,2}>8)=\frac{1}{9}=0.1111. $$ The correction is to retain the scale $12/3=4$: $$ P\left(\frac{S_1^2}{S_2^2}>8\right) =P(F_{2,2}>8/4) =P(F_{2,2}>2) =\frac{1}{3}=0.3333. $$ The incorrect calculation uses the right F degrees of freedom but the wrong threshold. Degrees of freedom alone do not determine the distribution of the unstandardized ratio when the population variances differ.

Key takeaway: For independent normal samples with $n_1,n_2\ge2$, standardize each sample variance by its own population variance. The resulting ratio is $F_{n_1-1,n_2-1}$; the raw sample-variance ratio is that F variable multiplied by $\sigma_1^2/\sigma_2^2$.

Check Your Understanding

For each problem, identify the scale and degrees of freedom before calculating or translating the requested quantity.

Two independent normal samples each have size 3 and population variances 8 and 4, respectively. Find $P(S_1^2/S_2^2\le3)$. (Answer: $0.6000$)
Two independent normal samples each have size 3 and population variances 2 and 8, respectively. Find $P(S_1^2/S_2^2>1)$. (Answer: $0.2000$)
Two independent normal samples each have size 3 and population variances 10 and 2, respectively. Find the 75th percentile of $S_1^2/S_2^2$. (Answer: $15$)
Two independent normal samples each have size 3 and population variances 6 and 4, respectively. Find $P(S_1^2/S_2^2>9)$. (Answer: $0.1429$)
A proposed calculation uses $S_1^2/S_2^2\sim F_{n_1-1,n_2-1}$ for independent normal samples of sizes 5 and 4 with population variances 15 and 5. State the correct distribution of the raw ratio and explain the error. (Answer: $(S_1^2/S_2^2)\sim3F_{4,3}$; the proposed calculation omits the population-variance scale factor 3.)

Quantity	Distribution or role	What to check
\((n_1-1)S_1^2/\sigma_1^2\)	\(\chi^2_{n_1-1}\)	First sample is normal; \(n_1\ge2\)
\((n_2-1)S_2^2/\sigma_2^2\)	\(\chi^2_{n_2-1}\)	Second sample is normal; \(n_2\ge2\)
\((S_1^2/\sigma_1^2)/(S_2^2/\sigma_2^2)\)	\(F_{n_1-1,n_2-1}\)	Samples are independent
\(S_1^2/S_2^2\)	\((\sigma_1^2/\sigma_2^2)F_{n_1-1,n_2-1}\)	Retain the population-variance ratio as scale

What You'll Learn