sample statistic

If we don’t have access to the entire dataset like with the population - statistic, but only a sample, we need to apply Bessel’s correction, e.g. for the variance or the std, by dividing by $(n-1)$ instead of $n$.

$$\begin{array}{rl}{s}^2& =\frac{1}{n-1}\sum _{i=1}^n(X_i-\bar{X}{)}^2\\ s& =\sqrt{s^2}\end{array}$$

Why we need Bessel's correction

The sample mean $\bar{X}$ is the value that minimizes the sum of squared deviations from the sample. For any other value $c$:

$$\sum (X_i-c{)}^2>\sum (X_i-\bar{X}{)}^2$$

This means $\sum (X_i-\bar{X}{)}^2<\sum (X_i-\mu {)}^2$ where $\mu$ is the true population mean.
The sample always clusters tightly around its own mean compared to the population mean “overfit”, systematically underestimating variance.

Additionally, the deviations must sum to zero: $\sum (X_i-\bar{X})=0$ (since $\bar{X}=\frac{1}{n}\sum X_i$). This constraint means only $(n-1)$ deviations are free to vary - if you know $(n-1)$ of them, the last is determined.

Dividing by $(n-1)$ exactly compensates for both effects, making $E[s^2]={\sigma }^2$ (unbiased).

Practical impact

Small samples (n < 30): correction is crucial (10% difference at n=10)
Medium samples (n = 30-100): still noticeable (3.3% at n=30, 1% at n=100)
Large samples (n > 100): essentially negligible
As $n\to \infty$: $\frac{n}{n-1}\to 1$, correction vanishes