When the Sample-Variance Chi-Square Law Is Exact

Why Exactness Depends on the Population Model

A formula involving a chi-square distribution can look usable whenever a problem gives a sample variance and a sample size. The crucial question is whether the sampling model justifies that distribution. For the usual exact result, the observations must be an independent random sample from a normal population. A large sample from a nonnormal population may permit an approximation, but that is a different claim.

This distinction matters whenever a calculation relies on the shape or tail probabilities of a chi-square distribution. The sample size determines the degrees of freedom only after the distributional assumptions are checked; it cannot make an unjustified exact law valid.

Definition: Let $X_1,\ldots,X_n$ be an independent random sample from a population with mean $\mu$ and variance $\sigma^2>0$. The unbiased sample variance is $$ S^2=\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\bar X)^2, \qquad \bar X=\frac{1}{n}\sum_{i=1}^{n}X_i. $$ If the population is $N(\mu,\sigma^2)$ and $n\geq 2$, then $$ \frac{(n-1)S^2}{\sigma^2}\sim\chi^2_{n-1}. $$ This is an exact distributional statement under the stated normal random-sample model, not a general law for sample variances.

The notation $N(\mu,\sigma^2)$ uses $\sigma^2$ as the population variance. The divisor $n-1$ in $S^2$ is part of the result: if a problem instead defines a variance using divisor $n$, its scaling must be adjusted. Neither changing the divisor nor increasing the sample size removes the need to check the population model.

Why Normality Gives the Exact Law

The normal assumption is what allows the residual sum of squares to be expressed as a sum of $n-1$ independent squared standard normal variables. Here is a justification of the result. Standardize each observation by setting $Z_i=(X_i-\mu)/\sigma$. Under the assumptions, the vector of these standardized observations has independent standard normal components.

Apply an orthogonal transformation whose first row is $(1/\sqrt n,\ldots,1/\sqrt n)$. Orthogonal transformations preserve the sum of squares, and the joint standard normal density depends only on that sum of squares. Consequently, the transformed components are again independent standard normal variables. The first transformed component is $\sqrt n(\bar X-\mu)/\sigma$; the other $n-1$ components describe deviations orthogonal to the sample-mean direction.

\sum_{i=1}^{n}\left(\frac{X_i-\mu}{\sigma}\right)^2 = \frac{n(\bar X-\mu)^2}{\sigma^2} + \frac{\sum_{i=1}^{n}(X_i-\bar X)^2}{\sigma^2}

The left side is the sum of squares of all $n$ transformed standard normal components. The first term on the right is the square of the first component. The remaining term is therefore the sum of squares of the other $n-1$ independent standard normal components. By the definition of a chi-square distribution, that residual sum of squares has a $\chi^2_{n-1}$ distribution. Since the residual sum of squares is $(n-1)S^2$, dividing by $\sigma^2$ gives the stated result.

For a nonnormal population, the standardized observation vector is not generally jointly normal. An orthogonal transformation still preserves the sum of squares as an algebraic identity, but it does not generally produce independent standard normal components. The key distributional step in the derivation therefore fails. Having independent observations and a finite variance alone is not enough to establish the exact chi-square law.

Worked Example: Checking an Exact Normal-Sample Claim

A hypothetical laboratory models repeated readings of a stable instrument as independent observations from $N(42,25)$, with readings measured in the same units. Nine readings are taken, and their unbiased sample variance is $s^2=20$. Identify the exact distribution of the random sample-variance statistic and calculate its observed value.

Method. Check the model, sample size, and variance convention before applying the exact result.

The stated population is normal with mean $\mu=42$ and variance $\sigma^2=25$. The nine readings are independent, so $n=9$, and the given sample variance uses divisor $n-1=8$. All assumptions for the exact law are met.

\frac{(n-1)S^2}{\sigma^2} = \frac{8S^2}{25} \sim\chi^2_8

For the observed sample variance $s^2=20$, the realized statistic is $$ \frac{8(20)}{25}=\frac{160}{25}=6.4. $$ Thus the random statistic has an exact $\chi^2_8$ distribution, and this sample produced the value $6.4$.

If the same data were summarized using $T^2=\frac{1}{9}\sum_{i=1}^{9}(X_i-\bar X)^2$, then $T^2=\frac{8}{9}S^2$. Its corresponding statistic is $$ \frac{9T^2}{25}=\frac{8S^2}{25}, $$ so it is still the same chi-square statistic. The common mistake is to use $9S^2/25$ when $S^2$ has divisor $n-1$; the correct multiplier is $8$, not $9$.

What Changes for a Nonnormal Population?

If the population is nonnormal, do not label $(n-1)S^2/\sigma^2$ as chi-square merely because $S^2$ is a sample variance. The statistic may have a distribution with a different shape, different support, or both. The next example gives a direct check using a population with a finite variance.

Worked Example: A Discrete Sample Variance Is Not Chi-Square

A hypothetical inspection records whether each item has a minor surface mark: $X_i=1$ if marked and $X_i=0$ otherwise. Suppose the observations are independent with $X_i\sim\operatorname{Bernoulli}(1/2)$, and a sample of three items is taken. Find the possible values and probabilities of $(n-1)S^2/\sigma^2$, then assess the exact chi-square claim.

Method. Count the number of marked items and calculate the sample variance for each possible count. The population is not normal, so the normal-sample theorem cannot be invoked.

The population mean is $\mu=1/2$, and its variance is $\sigma^2=p(1-p)=1/4$. Let $K$ be the number of marked items among the three. Then $K\sim\operatorname{Binomial}(3,1/2)$, and the sample mean is $K/3$. For this zero-one sample, the sum of squared deviations from the sample mean is $K(3-K)/3$. Therefore, $$ S^2=\frac{1}{2}\frac{K(3-K)}{3}=\frac{K(3-K)}{6}. $$

When $K=0$ or $K=3$, the sample variance is $0$, and the combined probability is $$ P(K=0)+P(K=3)=\frac18+\frac18=\frac14. $$ When $K=1$ or $K=2$, the sample variance is $2/6=1/3$, and the combined probability is $3/8+3/8=3/4$. Since $n-1=2$, the scaled statistic takes the values $$ \frac{2(0)}{1/4}=0 \quad\text{with probability } \frac14, \qquad \frac{2(1/3)}{1/4}=\frac83 \quad\text{with probability } \frac34. $$

This is a discrete distribution with probability $1/4$ at zero. A $\chi^2_2$ variable is continuous and has probability zero of being exactly zero. Thus the scaled sample variance in this model is not $\chi^2_2$. The trap is to use $n-1=2$ degrees of freedom without first checking normality; the degrees-of-freedom count does not establish the distribution.

Approximation Is a Separate Claim

For some nonnormal populations, large-sample theory can provide an approximation for $S^2$. A standard version applies to independent, identically distributed observations with mean $\mu$, variance $\sigma^2>0$, and finite fourth central moment $\mu_4=E[(X_i-\mu)^4]$. Under these conditions,

Theorem: For an independent, identically distributed sample with finite fourth central moment, $$ \sqrt n(S^2-\sigma^2)\ \xrightarrow{d}\ N(0,\mu_4-\sigma^4). $$ Thus, for large $n$, a normal approximation for $S^2$ has mean approximately $\sigma^2$ and variance approximately $(\mu_4-\sigma^4)/n$. This is an asymptotic approximation, not an exact chi-square law. Its conditions and quality must be considered for the population in question.

One way to understand the approximation is to write the sample variance in terms of averages of $X_i$ and $X_i^2$. To first order, its fluctuation around $\sigma^2$ is the fluctuation of the average of $(X_i-\mu)^2$. This quantity has mean $\sigma^2$ and variance $\mu_4-\sigma^4$, which explains the asymptotic variance in the result. The approximation depends on the fourth moment, not just on the population variance.

In the special normal case, $\mu_4=3\sigma^4$, so this asymptotic variance is $2\sigma^4/n$. That large-sample normal approximation is still not the same statement as the exact finite-sample chi-square law. For a nonnormal population, both the approximation's variance and its adequacy can differ from the normal case.

Worked Example: An Exact Support Check and a Large-Sample Approximation

Suppose hypothetical measurements are an independent random sample from the uniform distribution on $[-\sqrt3,\sqrt3]$. Assess an exact chi-square claim when $n=2$, then calculate the large-sample normal approximation's variance for $S^2$ when $n=100$.

Method. For exactness, compare the possible values of the statistic with the support of a chi-square distribution. For the approximation, calculate the population variance and fourth central moment, then apply the asymptotic result.

For a uniform distribution on $[-a,a]$, the mean is zero and the variance is $a^2/3$. Here $a=\sqrt3$, so $\mu=0$ and $\sigma^2=3/3=1$. With $n=2$, $$ S^2=\frac{(X_1-\bar X)^2+(X_2-\bar X)^2}{1} =\frac{(X_1-X_2)^2}{2}. $$ Both observations lie between $-\sqrt3$ and $\sqrt3$, so $|X_1-X_2|\leq2\sqrt3$. It follows that $0\leq S^2\leq6$. The proposed scaled statistic is $(n-1)S^2/\sigma^2=S^2$, and therefore it cannot exceed $6$. A $\chi^2_1$ variable has positive probability above $6$, so this statistic cannot have that exact distribution.

For the large-sample approximation, a uniform variable on $[-a,a]$ has fourth central moment $a^4/5$. Thus $\mu_4=9/5$, while $\sigma^4=1$. For $n=100$, the approximate variance of $S^2$ is $$ \frac{\mu_4-\sigma^4}{n} = \frac{9/5-1}{100} = \frac{4/5}{100} = 0.008. $$ The resulting asymptotic approximation is $S^2$ approximately $N(1,0.008)$, with standard deviation $\sqrt{0.008}\approx0.0894$. This is an approximation for a large sample, not an exact distributional statement for the uniform population. The mistake to avoid is treating a large sample as proof of the exact chi-square law.

Exam Checks and Common Traps

Before using the exact result, check the model and the definition of the variance in the question. The following distinctions are useful:

Exact result: independent observations from $N(\mu,\sigma^2)$, with $S^2$ defined using divisor $n-1$, give $(n-1)S^2/\sigma^2\sim\chi^2_{n-1}$.
Not enough for exactness: independence, a finite variance, a large sample, or a bell-shaped sketch does not by itself establish the exact law.
Approximation: under suitable moment conditions, $S^2$ may be approximately normal for large $n$; the approximation's variance involves the fourth central moment.
Scaling trap: distinguish the unbiased sample variance's divisor $n-1$ from a variance defined using divisor $n$.
Model trap: knowing the sample mean and variance, or observing data that look roughly symmetric, does not establish that the population is normal.

For an exam-style question, state the assumption that justifies the distribution before using it. If normality is specified, apply the exact theorem with the correct degrees of freedom. If the population is nonnormal, explain why exactness is not guaranteed and identify any separate approximation assumptions rather than silently substituting a chi-square distribution.

Check Your Understanding

For each problem, check both the distributional assumption and the sample-variance convention before deciding what conclusion is justified.

A sample of 12 independent observations comes from $N(18,9)$, and $S^2$ uses divisor $n-1$. State the exact distribution of the scaled statistic. (Answer: $11S^2/9\sim\chi^2_{11}$.)
For five independent observations from a normal population with variance $16$, the variance is defined as $T^2=\frac{1}{5}\sum_{i=1}^{5}(X_i-\bar X)^2$. Identify a chi-square statistic and its degrees of freedom. (Answer: $5T^2/16\sim\chi^2_4$.)
Independent observations are sampled from a Bernoulli population with parameter $p=0.2$. Does the normal-sample theorem establish that $(n-1)S^2/\sigma^2$ is exactly chi-square? (Answer: No; the population is not normal, so the exact chi-square result is not established.)
A nonnormal population has finite variance but no finite fourth central moment. Does the finite-fourth-moment asymptotic normal result stated here apply? (Answer: No; its finite fourth central moment condition is not met.)
A sample of size 40 is independent and identically distributed from a nonnormal population with finite fourth central moment. Is $(n-1)S^2/\sigma^2$ exactly chi-square, and what kind of approximation may be available for $S^2$? (Answer: Exact chi-square is not guaranteed; for sufficiently large samples, an asymptotic normal approximation for $S^2$ may be available, with variance approximately $(\mu_4-\sigma^4)/40$.)

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