The Sample Mean and Sample Variance

Suppose a factory produces bolts whose true diameter has some unknown mean $\mu$ and unknown variance $\sigma^2$ . You cannot measure every bolt ever made. Instead, you measure a sample of $n$ of them and compute two numbers. These are their average, and how spread out they are around that average. These two numbers — the sample mean and sample variance — are the workhorses of statistical inference. Almost every downstream method (confidence intervals, hypothesis tests, regression diagnostics) is built by asking how these two quantities behave as random variables in their own right. The central subtlety is this: the sample mean and sample variance are not fixed numbers waiting to be discovered. They are functions of a random sample. Therefore they are themselves random variables with their own distributions, expectations, and variances. Understanding how they fluctuate from sample to sample — not just what they equal for one particular dataset — is the entire content of statistical inference. This article answers the question precisely: what are $\mathbb{E}[\bar X]$ , $\operatorname{Var}(\bar X)$ , $\mathbb{E}[S^2]$ , and (for normal populations) $\operatorname{Var}(S^2)$ ? The big picture is a bridge between two worlds. On one side sits the population, described by an unknown distribution with parameters $\mu$ and $\sigma^2$ . On the other side sits data, described by empirical summaries $\bar X$ and $S^2$ . The theorems here are the precise dictionary translating properties of the population into properties of these summaries. They explain a famous fact that surprises many students on first sight. Dividing by $n-1$ , rather than $n$ , in the sample variance is not an arbitrary convention. It is a mathematically forced correction. This article gives rigorous definitions. It proves the key expectation and variance identities without any distributional assumption beyond independence. It then specializes to the normal population case to compute $\operatorname{Var}(S^2)$ exactly. It closes the loop toward the machinery — $t$ , $\chi^2$ , and $F$ distributions — that make these facts usable for confidence intervals and tests. It also points toward the general theory of Point Estimation and Estimators.

Key takeaways

The sample mean $\bar X$ and sample variance $S^2$ are statistics: functions of a random sample, hence random variables.
For any i.i.d. sample with finite mean $\mu$ and variance $\sigma^2$ : $\mathbb{E}[\bar X]=\mu$ and $\operatorname{Var}(\bar X)=\sigma^2/n$ .
Dividing by $n-1$ instead of $n$ in $S^2$ is exactly what makes $\mathbb{E}[S^2]=\sigma^2$ — this is Bessel's correction.
These two facts hold for any population distribution with two finite moments; no normality is needed.
Under normality, $\bar X$ and $S^2$ are independent, and $(n-1)S^2/\sigma^2 \sim \chi^2_{n-1}$ , giving $\operatorname{Var}(S^2)=2\sigma^4/(n-1)$ .
Key idea: the sample mean and sample variance are random variables whose own mean and variance can be computed exactly from those of the parent population — this computation is the foundation of statistical inference.

Where this lives today

The sample mean and sample variance are the entry point to essentially all of applied statistics and much of applied mathematics beyond it. In statistical quality control, control charts (like the classical $\bar X$ -chart and $S$ -chart used on factory floors) are built directly from the sampling distributions derived here, flagging processes whose observed mean or variance drifts beyond the natural fluctuation predicted by $\sigma^2/n$ and $2\sigma^4/(n-1)$ . In experimental physics, reporting a measurement as " $\bar x \pm \text{SE}$ " is a direct application of $\operatorname{Var}(\bar X)=\sigma^2/n$ , quantifying the precision gained by repeated trials. In machine learning, mini-batch statistics (batch normalization layers in neural networks) compute running sample means and variances whose bias and variance properties determine training stability. In finance, estimating volatility (variance of returns) from historical price data uses exactly $S^2$ , and the $\operatorname{Var}(S^2)$ formula quantifies how much estimation error to expect in a volatility estimate from a finite historical window. In genomics and clinical trials, sample-size calculations for detecting a treatment effect are built by combining $\operatorname{Var}(\bar X)=\sigma^2/n$ with hypothesis-testing theory to decide how many subjects are needed. And in pure probability, this article's identities are the seed from which the t, Chi-Squared, and F Distributions grow — the $t$ -distribution arises precisely from studentizing $\bar X$ by $S/\sqrt n$ using the independence and chi-squared results proved here, and the $F$ -distribution compares two independent sample variances built the same way, both central to the general theory of Point Estimation and Estimators and to confidence-interval construction built on Sampling Distributions.

Ask the tutor: How does the independence of $\bar X$ and $S^2$ under normality get used to derive the Student $t$ -distribution?

X_1, X_2, \dots, X_n

Intuition

Picture the population as an urn with an unknown but fixed "chemical composition" — a mean $\mu$ around which values cluster, and a spread $\sigma^2$ describing how loosely they cluster. Each time you draw a sample of size $n$ , you get a slightly different snapshot: a different collection of $n$ numbers, hence a different $\bar X$ and a different $S^2$ . If you repeated the whole experiment many times — drawing sample after sample of size $n$ — the resulting sequence of $\bar X$ values would itself trace out a distribution, and likewise for $S^2$ . The theorem below identifies the center and spread of that distribution, in terms of the population's $\mu$ and $\sigma^2$ . The key geometric fact behind the $n-1$ divisor is that the sample mean $\bar X$ is itself estimated from the data, so the deviations $X_i - \bar X$ are not free to vary independently: they are constrained to sum to zero. This costs exactly one degree of freedom, which is why $n-1$ rather than $n$ appears in the denominator of $S^2$ — using $n$ would systematically underestimate $\sigma^2$ .

Figure

Repeated sampling: each sample of size n gives a different (x̄, s²) pair, scattering around the true (μ, σ²)

Central results

X_1,\dots,X_n

\mathbb{E}[\bar X] = \mathbb{E}\!\left[\frac{1}{n}\sum_{i=1}^n X_i\right] = \frac{1}{n}\sum_{i=1}^n \mathbb{E}[X_i] = \frac{1}{n}\sum_{i=1}^n \mu = \frac{1}{n}\cdot n\mu = \mu.

n-1

X_1,\dots,X_n

\mathbf X = (X_1,\dots,X_n)^\top

X_1,\dots,X_n

W := (n-1)S^2/\sigma^2 \sim \chi^2_{n-1}

Worked examples

Example 1 (asymmetric numeric sample; illustrating $\bar X$ , $S^2$ , and the bias correction directly). Consider the sample of $n=5$ measurements (in some unit), deliberately asymmetric and non-round:

x_1=2.1,\quad x_2=5.4,\quad x_3=3.0,\quad x_4=9.7,\quad x_5=4.8.

Step 1: sample mean.

\bar x = \frac{2.1+5.4+3.0+9.7+4.8}{5} = \frac{25.0}{5} = 5.0.

Step 2: deviations and squared deviations.

$x_i$	$x_i - \bar x$	$(x_i-\bar x)^2$
2.1	$-2.9$	$8.41$
5.4	$0.4$	$0.16$
3.0	$-2.0$	$4.00$
9.7	$4.7$	$22.09$
4.8	$-0.2$	$0.04$

Sum of squared deviations: $8.41+0.16+4.00+22.09+0.04 = 34.70$ . (Check: the deviations sum to $-2.9+0.4-2.0+4.7-0.2 = 0.0$ , confirming $\sum (x_i-\bar x)=0$ as required by the algebra in the proof.) Step 3: sample variance with $n-1=4$ .

s^2 = \frac{34.70}{4} = 8.675.

Step 4: compare to the naive divide-by- $n$ estimate.

\tilde s^2 = \frac{34.70}{5} = 6.94.

Indeed $\tilde s^2 = \frac{n-1}{n}s^2 = \frac{4}{5}(8.675) = 6.94$ , matching the theorem's bias factor $\frac{n-1}{n}$ exactly. This single sample cannot show unbiasedness (that is a statement about the average over repeated sampling, not about one dataset) — but it verifies the algebraic relationship between $S^2$ and $\tilde S^2$ predicted by the corollary. Step 5: standard deviation and standard error of the mean. $s = \sqrt{8.675} \approx 2.945$ . If this population had a known variance $\sigma^2 = 8.675$ (treating the sample value as if it were the truth, purely for illustration of the formula), the theorem gives

\operatorname{Var}(\bar X) = \frac{\sigma^2}{n} = \frac{8.675}{5} = 1.735, \qquad \text{so } \sqrt{\operatorname{Var}(\bar X)} \approx 1.317.

This quantity, the standard deviation of $\bar X$ , is called the standard error of the mean and shrinks like $1/\sqrt n$ as sample size grows — the quantitative content of Part 2 of the theorem.

Example 2 (using the theorems to predict sampling behavior before collecting data — normal population). A quality-control engineer knows from decades of production data that bolt diameters (in mm) are very well modeled as $N(\mu, \sigma^2)$ with $\sigma^2 = 0.09$ (so $\sigma = 0.3$ mm), though $\mu$ itself is unknown and will be estimated from a fresh sample of $n = 10$ bolts. Question A: What is the standard error of $\bar X$ for this sample size? By Part 2 of the theorem,

\operatorname{Var}(\bar X) = \frac{\sigma^2}{n} = \frac{0.09}{10} = 0.009, \qquad \text{SE}(\bar X) = \sqrt{0.009} \approx 0.0949 \text{ mm}.

So even though individual bolts vary with standard deviation $0.3$ mm, the average of 10 bolts varies with standard deviation only about $0.095$ mm — roughly $\sqrt{10}\approx 3.16$ times tighter, exactly as $\sigma/\sqrt n$ predicts. Question B: What is $\mathbb{E}[S^2]$ for the sample variance the engineer will compute? By Part 3, regardless of what particular $10$ bolts are drawn,

\mathbb{E}[S^2] = \sigma^2 = 0.09,

so on average, across many hypothetical batches of 10 bolts, the computed $S^2$ centers exactly on the true variance $0.09$ — this is the precise sense in which $S^2$ is the "right" estimator, not the naive $\tilde S^2$ (which would average to $\frac{9}{10}(0.09)=0.081$ , a $10\%$ downward bias). Question C: How much does $S^2$ itself fluctuate from batch to batch? Using the normal-population theorem with $n-1=9$ ,

\operatorname{Var}(S^2) = \frac{2\sigma^4}{n-1} = \frac{2(0.09)^2}{9} = \frac{2(0.0081)}{9} = \frac{0.0162}{9} = 0.0018.

So the standard deviation of $S^2$ itself is $\sqrt{0.0018} \approx 0.0424$ , meaning a single computed $S^2$ can quite plausibly land anywhere in roughly $0.09 \pm 0.08$ (two standard deviations) purely from sampling noise with only $n=10$ — a concrete illustration of why $S^2$ from small samples is a noisy estimate of $\sigma^2$ , even though it is unbiased. Increasing $n$ to $100$ would shrink $\operatorname{Var}(S^2)$ to $2(0.0081)/99 \approx 0.000164$ , an order of magnitude tighter, illustrating the $1/(n-1)$ rate of concentration.

\operatorname{Var}(\bar X)

Scope, edge cases & common traps

Situation	What holds	What can fail
I.i.d. sample, any distribution with finite $\mu,\sigma^2$	$\mathbb E[\bar X]=\mu$ , $\operatorname{Var}(\bar X)=\sigma^2/n$ , $\mathbb E[S^2]=\sigma^2$	Nothing — these three hold universally under the stated hypotheses
I.i.d. normal sample	Additionally: $\bar X \perp S^2$ ; $(n-1)S^2/\sigma^2\sim\chi^2_{n-1}$ ; $\operatorname{Var}(S^2)=2\sigma^4/(n-1)$	The exact independence of $\bar X$ and $S^2$ is special to normal populations — it can fail for other distributions
Non-independent sample (e.g. time series, clustered data)	$\mathbb{E}[\bar X]=\mu$ still holds by linearity	$\operatorname{Var}(\bar X)=\sigma^2/n$ generally fails; extra covariance terms appear
Infinite variance population (e.g. Cauchy, or heavy-tailed with $\sigma^2=\infty$ )	$\bar X$ may still be defined as a statistic	$\operatorname{Var}(\bar X)=\sigma^2/n$ is meaningless/undefined since $\sigma^2=\infty$ ; $\bar X$ need not even concentrate as $n\to\infty$
$n=1$	$\bar X = X_1$ , so $\operatorname{Var}(\bar X)=\sigma^2$ trivially	$S^2$ is undefined ( $n-1=0$ in the denominator); at least $n=2$ observations are needed to estimate spread

Common trapThe classic error: treating "unbiased" as "close to the truth for any single sample"

$\mathbb{E}[S^2]=\sigma^2$ is a statement about averaging over infinitely many hypothetical repetitions of the sampling process — it says nothing about how close any one computed $S^2$ will be to $\sigma^2$ . Students often conflate unbiasedness with accuracy for a single dataset. Worked Example 2 shows a concrete case where $\operatorname{Var}(S^2)\approx 0.0018$ with $n=10$ : individual values of $S^2$ can deviate substantially from $\sigma^2$ even while their long-run average is exactly correct. Unbiasedness controls the center of the sampling distribution of $S^2$ , not its spread — the spread is a separate quantity, quantified by $\operatorname{Var}(S^2)$ , and it is this quantity (together with the shape of the sampling distribution) that governs how trustworthy a single sample's $S^2$ actually is.

The mathematician behind it

Origins · 1 minFriedrich Wilhelm Bessel and how this was discovered

Friedrich Wilhelm Bessel

1784–1846 · Astronomy and error theory, Prussia

Bessel was a self-taught astronomer and mathematician who became director of the Königsberg Observatory. Though best known for Bessel functions (arising from his work on planetary perturbations) and for the first reliable measurement of stellar parallax (61 Cygni, 1838), his name is permanently attached to the "Bessel correction" — the $n/(n-1)$ factor that converts a biased mean-squared-deviation into an unbiased estimator of variance. His work sat at the intersection of practical observational astronomy, where instruments produced noisy repeated measurements, and the emerging theory of errors developed alongside Gauss and Laplace; the drive to correctly combine and summarize many imperfect telescope readings is precisely the setting in which the distinction between a population variance and its unbiased sample estimate first became practically indispensable.

History

- 1809 — Gauss develops the theory of errors and the normal distribution in the context of astronomical observation, laying groundwork for treating repeated measurements probabilistically.

- 1823 — Gauss formulates the Gauss–Markov framework for least-squares estimation, implicitly using sums of squared deviations.

- 1830s — Bessel, working on precision astronomical measurement, identifies and popularizes the

n-1

correction for estimating variance from a finite sample of observations.

- 1900s — Karl Pearson and R.A. Fisher formalize the modern vocabulary of "statistic," "sampling distribution," and "degrees of freedom," giving the

n-1

correction its rigorous justification via the chi-squared distribution.

- 1925 — Fisher's Statistical Methods for Research Workers cements

S^2

with divisor

n-1

as the standard unbiased estimator of variance in applied statistics.

Why isn't the sample variance just defined with divisor n, like a plain average of squared deviations?

Dividing by $n$ gives $\tilde S^2 = \frac1n\sum_i(X_i-\bar X)^2$ , whose expectation is $\frac{n-1}{n}\sigma^2$ , strictly less than $\sigma^2$ for every finite $n$ . The reason is that $\bar X$ is itself estimated from the same data, so the deviations $X_i - \bar X$ are constrained (they must sum to zero), giving effectively only $n-1$ "free" pieces of information about spread rather than $n$ . Dividing by $n-1$ exactly compensates for this constraint, yielding an unbiased estimator $\mathbb{E}[S^2]=\sigma^2$ . This is Bessel's correction, and it is a mathematical necessity for unbiasedness, not a stylistic choice.

Does Var(X̄) = σ²/n require the population to be normally distributed?

No. The identities $\mathbb{E}[\bar X]=\mu$ , $\operatorname{Var}(\bar X)=\sigma^2/n$ , and $\mathbb{E}[S^2]=\sigma^2$ hold for any i.i.d. sample from any population with finite mean and variance — Bernoulli, exponential, uniform, or otherwise. Normality is only needed for the stronger results: that $\bar X$ and $S^2$ are exactly independent, that $(n-1)S^2/\sigma^2$ is exactly chi-squared, and hence that $\operatorname{Var}(S^2)=2\sigma^4/(n-1)$ exactly. For non-normal populations these last facts generally fail or hold only approximately for large $n$ .

Is S² the only reasonable estimator of σ², or just the unbiased one?

$S^2$ is the standard unbiased estimator, but it is not the only reasonable one and not always the "best" by every criterion. For instance, under normality, the maximum likelihood estimator of $\sigma^2$ is exactly $\tilde S^2 = \frac{n-1}{n}S^2$ (the divide-by- $n$ version), which is biased but has smaller variance and in fact smaller mean-squared error than $S^2$ for finite $n$ . Choosing between estimators involves trading off bias and variance, a theme developed fully in Point Estimation and Estimators; unbiasedness is a desirable property but not the sole criterion for a "good" estimator.

Why does Var(S²) matter if S² is already unbiased?

Unbiasedness only guarantees that $S^2$ is correct on average across hypothetical repeated sampling; it says nothing about how far a single computed $S^2$ might land from $\sigma^2$ in any particular dataset. $\operatorname{Var}(S^2)=2\sigma^4/(n-1)$ quantifies exactly that spread for normal populations, showing it shrinks like $1/(n-1)$ — so small samples give noisy variance estimates even though they are unbiased. This variance is exactly what feeds into building valid confidence intervals for $\sigma^2$ using the chi-squared distribution, and into deriving the $t$ -distribution for confidence intervals on $\mu$ when $\sigma$ is unknown.

What happens to these formulas when n = 1?

With $n=1$ , $\bar X = X_1$ exactly, so $\mathbb{E}[\bar X]=\mu$ and $\operatorname{Var}(\bar X)=\sigma^2$ still hold trivially (consistent with $\sigma^2/n = \sigma^2/1$ ). But $S^2$ is undefined, since its denominator $n-1=0$ : with only one observation there is no way to estimate spread around an unknown mean, because a single point offers zero information about variability. At least two observations are the minimum requirement for the sample variance to exist.

Can the sample mean and sample variance be dependent for non-normal populations?

Yes, and in general they typically are. The proof that $\bar X \perp S^2$ relies specifically on properties of the multivariate normal distribution (that zero covariance implies independence for jointly Gaussian vectors, and that projections of an isotropic Gaussian onto orthogonal subspaces are independent). For a non-normal population — say, exponential or uniform — $\bar X$ and $S^2$ are generally correlated, sometimes strongly so, and no analogous clean chi-squared result for $(n-1)S^2/\sigma^2$ holds exactly (only asymptotically, via the Central Limit Theorem-type approximations)

How do these results connect to hypothesis testing and confidence intervals?

The confidence interval for $\mu$ when $\sigma$ is unknown is built from the statistic $T = \dfrac{\bar X - \mu}{S/\sqrt n}$ , whose distribution is derived precisely from the facts proved here: numerator behavior from $\operatorname{Var}(\bar X)=\sigma^2/n$ , and denominator behavior from $(n-1)S^2/\sigma^2 \sim \chi^2_{n-1}$ together with the independence of $\bar X$ and $S^2$ . Combining an $N(0,1)$ numerator with an independent $\chi^2_{n-1}$ denominator (suitably scaled) is exactly the construction of the $t$ -distribution with $n-1$ degrees of freedom, covered in The t, Chi-Squared, and F Distributions. Every $t$ -test and $t$ -confidence-interval formula ultimately traces back to the identities proved in this article.