命题 6.3.1 (样本均值与样本方差的数字特征)

命题 6.3.1 (样本均值与样本方差的数字特征)

设 $\left(X_1,X_2,\cdots ,X_n\right)$ 是来自总体 $X$ 的样本, $E\left[X\right]=\mu$, $\mathrm{Var}\left[X\right]={\sigma }^2$. 则

$$E\left[\bar{X}\right]=\mu ,\mathrm{Var}\left[\bar{X}\right]=\frac{{\sigma }^2}{n};\text{\hspace{1em}(6.3.1)}$$ $$E\left[S_n^2\right]=\frac{n-1}{n}{\sigma }^2,E\left[S_n^{\ast 2}\right]={\sigma }^2.\text{\hspace{1em}(6.3.2)}$$

证明

1

$$\begin{array}{rl}E\left[\bar{X}\right]& =E\left[\frac{1}{n}\sum _{i=1}^n{X}_i\right]\\ & =\frac{1}{n}\sum _{i=1}^{n}E\left[X_i\right]\\ & =\frac{1}{n}\sum _{i=1}^n\mu \\ & =\mu \end{array}$$ $$\begin{array}{rl}\mathrm{Var}\left[\bar{X}\right]& =\mathrm{Var}\left[\frac{1}{n}\sum _{i=1}^n{X}_i\right]\\ & =\frac{1}{n^2}\sum _{i=1}^n\mathrm{Var}\left[X_i\right]\\ & =\frac{1}{n^2}\sum _{i=1}^n{\sigma }^2\\ & =\frac{{\sigma }^2}{n}\end{array}$$

2

由于

$$\begin{array}{rl}{S}_n^2& =\frac{1}{n}\sum _{i=1}^n{\left(X_i-\bar{X}\right)}^2\\ & =\frac{1}{n}\sum _{i=1}^n\left(X_i^2-2\bar{X}\cdot X_i+{\bar{X}}^2\right)\\ & =\frac{1}{n}\left(\sum _{i=1}^n{X}_i^2-2\bar{X}\cdot \sum _{i=1}^n{X}_i+n{\bar{X}}^2\right)\\ & =\frac{1}{n}\sum _{i=1}^n{X}_i^2-{\bar{X}}^2\end{array}$$

所以

$$\begin{array}{rl}E\left[S_n^2\right]& =E\left[\frac{1}{n}\sum _{i=1}^n{X}_i^2-{\bar{X}}^2\right]\\ & =\frac{1}{n}\sum _{i=1}^{n}E\left[X_i^2\right]-E\left[{\bar{X}}^2\right]\\ & =\frac{1}{n}\sum _{i=1}^n\left(\mathrm{Var}\left[X_i\right]+{\left(E\left[X_i\right]\right)}^2\right)-\left(\mathrm{Var}\left[\bar{X}\right]+E{\left[\bar{X}\right]}^2\right)\\ & =\frac{1}{n}\left(n{\sigma }^2+n{\mu }^2\right)-\frac{{\sigma }^2}{n}-{\mu }^2\\ & =\frac{n-1}{n}{\sigma }^2.\end{array}$$ $$\begin{array}{rl}E\left[S_n^{\ast 2}\right]& =E\left[\frac{n}{n-1}{S}_n^2\right]\\ & =\frac{n}{n-1}E\left[S_n^2\right]\\ & ={\sigma }^2.\end{array}$$