표본분산을 구할 때 (n-1)로 나누는 이유

$\text{표본분산을 구할 때 n으로 나누지 않고 n-1로 나누는 이유를 설명합니다.}$ $\\ $

$s^2 = \frac{ \sum_{i=1}^n (x_i-\overline{x})^2} {n-1}$ $$\begin{align*} E[s_n^2] &=E[\frac{1}{n} \sum_{i=1}^n(xi- \overline{x})^2]\\ &=\frac{1}{n} E[\sum_{i=1}^n ((x_i-u)-(\overline{x}-u))^2]\\ &=\frac{1}{n} E[\sum_{i=1}^n ( (x_i-u)^2- 2(x_i-u)(\overline{x}-u)+(\overline{x}-u)^2)]\\ &=\frac{1}{n} E[\sum_{i=1}^n ( x_i-u)^2- 2(\overline{x}-u) \sum_{i=1}^n(x_i-u)+\sum_{i=1}^n(\overline{x}-u)^2]\\ &=\frac{1}{n} E[\sum_{i=1}^n ( x_i-u)^2- 2(\overline{x}-u)( \sum_{i=1}^nx_i-\sum_{i=1}^nu)+n(\overline{x}-u)^2]\\ &=\frac{1}{n} E[\sum_{i=1}^n ( x_i-u)^2- 2(\overline{x}-u) (n\frac{\sum_{i=1}^nx_i}{n}-nu)+n(\overline{x}-u)^2]\\ &=\frac{1}{n} E[\sum_{i=1}^n ( x_i-u)^2- 2(\overline{x}-u) n(\overline{x}-u)+n(\overline{x}-u)^2]\\ &=\frac{1}{n} E[\sum_{i=1}^n ( x_i-u)^2- 2n(\overline{x}-u)^2 +n(\overline{x}-u)^2]\\ &=\frac{1}{n} E[\sum_{i=1}^n ( x_i-u)^2- n(\overline{x}-u)^2]\\ &=\frac{1}{n}([\sum_{i=1}^n E[( x_i-u)^2]- nE[(\overline{x}-u)^2]\\ &=\frac{1}{n}(\sum_{i=1}^n \sigma^2- n\frac{\sigma^2}{n})\\ &=\frac{1}{n}(n\sigma^2- \sigma^2)\\ &=\frac{n-1}{n}\sigma^2 \lt \sigma^2\\ \\ & s^2 = \frac{ \sum_{i=1}^n (x_i-\overline{x})^2} {n-1}\\ \end{align*}$$