Math - 样本方差（sample variance）的分母为什么是 n - 1
按照定义，方差的 estimator 应该是： $$ \begin{equation} S^2 = \frac{1}{n}\sum_{i = 1}^n\left(X_i - \overline{X}\right)^2 \label{eq:sample-variance} \end{equation} $$ 但是这个 estimator 有 bias，因为： $$ \begin{align*} E(S^2) &= \frac{1}{n}\sum_{i = 1}^nE\left[\left(X_i - \overline{X}\right)^2\right] \\ &= \frac{1}{n}E\left[\sum_{i = 1}^n\left(X_i - \mu + \mu - \overline{X}\right)^2\right]\\ &= \frac{1}{n}E\left[\sum_{i = 1}^n\left(X_i - \mu\right)^2 - n \left(\overline{X} - \mu \right)^2\right] \\ &= \frac{1}{n} \left[\sum_{i = 1}^nE\left(\left(X_i - \mu\right)^2\right) - nE\left(\left(\overline{X} - \mu\right)^2\right)\right] \\ &= \frac{1}{n}\left[n\operatorname{Var}(X) - n\operatorname{Var}(\overline{X})\right] \\ &= \operatorname{Var}(X) - \operatorname{Var}(\overline{X}) \\ &= \sigma^2 - \frac{\sigma^2}{n} = \frac{n - 1}{n}\sigma^2 \end{align*} $$ 关于 $\operatorname{Var}(\overline{X}) = \frac{\sigma^2}{n}$ 的证明参考： Prove that $E (\overline{X} - \mu)^2 = \frac{1}{n}\sigma^2$ ...