> For the complete documentation index, see [llms.txt](https://zhjunqin.gitbook.io/machine-learning/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://zhjunqin.gitbook.io/machine-learning/ji-qi-xue-xi/shu-xue-ji-chu/gai-lv-tong-ji/yang-ben-tong-ji-liang.md).

# 样本统计量

设$$X\_1$$，$$X\_2$$，...，$$X\_n$$是来自总体$$X$$（随机变量）的一个样本，它们相互独立，$$g(X\_1,X\_2,...,X\_n)$$是$$X\_1$$，$$X\_2$$，...，$$X\_n$$的函数，若$$g$$中不含未知参数，则称$$g(X\_1,X\_2,...,X\_n)$$是一**统计量**。

因为$$X\_1$$，$$X\_2$$，...，$$X\_n$$都是随机变量，而统计量是随机变量的函数，因此统计量是一个随机变量。设$$x\_1,x\_2,...,x\_n$$是相应于样本$$X\_1$$，$$X\_2$$，...，$$X\_n$$的样本值，则称$$g(x\_1,x\_2,...,x\_3)$$是$$g(X\_1,X\_2,...,X\_n)$$的**观察值**。

**样本均值**：

$$
\overline{X}=\frac{1}{n}\displaystyle\sum\_{i=1}^{n} X\_i
$$

**样本方差（无偏估计）**：

$$
S^2=\frac{1}{n-1}\displaystyle\sum\_{i=1}^{n} (X\_i-\overline{X})^2=\frac{1}{n-1}(\displaystyle\sum\_{i=1}^{n} X\_i^2-n\overline{X}^2)
$$

**样本标准差**：

$$
S=\sqrt{S^2}=\sqrt{\frac{1}{n-1}\displaystyle\sum\_{i=1}^{n} (X\_i-\overline{X})^2}
$$

**样本**$$k$$**阶（原点）距**：

$$
A\_k=\frac{1}{n}\displaystyle\sum\_{i=1}^{n} X\_i^k
$$

**样本**$$k$$**阶中心距**：

$$
A\_k=\frac{1}{n}\displaystyle\sum\_{i=1}^{n} (X\_i-\overline{X})^k
$$

**样本的协方差：**

$$
Cov(X,Y)=\frac{1}{n-1}\displaystyle\sum\_{i=1}^{n} (X\_i-\overline{X})(Y\_i-\overline{Y})
$$

其中$$X\_1$$，$$X\_2$$，...，$$X\_n$$是来自总体$$X$$的一个样本，$$Y\_1$$，$$Y\_2$$，...，$$Y\_n$$是来自总体$$Y$$的一个样本。

**样本协方差矩阵：**

假定$$X\_1$$，$$X\_2$$，...，$$X\_n$$是多维随机变量

$$
c\_{ij}=Cov(X\_{i},X\_{j})=\frac{1}{n-1}\displaystyle\sum\_{k=1}^{n} (X\_{ik}-\overline{X\_i})(X\_{jk}-\overline{X\_j})
$$

$$
C=\begin{bmatrix}
c\_{11} & c\_{12} & ... & c\_{1n}  \\
c\_{21} & c\_{22} & ... & c\_{2n} \\
\vdots & \vdots & & \vdots \\
c\_{n1} & c\_{n2} & ... & c\_{nn}
\end{bmatrix}
$$

> 为什么样本方差是除以$$n-1$$，而不是$$n$$？
>
> “均值已经用了$$n$$个数的平均来做估计，在求方差时，只有$$n-1$$个数和均值信息是不相关的。而第$$n$$个数已经可以由前$$n-1$$个数和均值来唯一确定，实际上没有信息量，所以在计算方差时，只除以$$n-1$$“
>
> （详细请参考 <https://www.zhihu.com/question/20099757）>


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