一、样本方差

设样本均值为$\bar x$，样本方差为S²，总体均值为${\rm{\mu }}$，总体方差为${

{\rm{\sigma }}^2}$，那么样本方差

${S^2} = \frac{1}{

{n - 1}}\mathop \sum \limits_{i = 1}^n {\left( {

{x_i} - \bar x} \right)^2}$

推导：假设样本数量等于总体数量，应有

 ${S^2} = \frac{1}{n}\mathop \sum \limits_{i = 1}^n {\left( {

{x_i} - \bar x} \right)^2}$

在多次重复抽取样本过程中，样本方差会逐渐接近总体方差，假设每次抽取的样本方差为

（S₁²,S₂²,S₃²…），然后对这些样本方差求平均值记为E(S²)，则

 

${\rm{E}}\left( {

{

{\rm{S}}^2}} \right) = {\rm{E}}\left( {\frac{1}{n}\mathop \sum \limits_{i = 1}^n {

{\left( {

{x_i} - \bar x} \right)}^2}} \right)$

$ = {\rm{E}}\left( {\frac{1}{n}\mathop \sum \limits_{i = 1}^n {

{\left( {\left( {

{x_i} - \mu } \right) - \left( {\bar x - \mu } \right)} \right)}^2}} \right)$

因为

$\frac{1}{n}\mathop \sum \limits_{i = 1}^n \left( {

{x_i} - \mu } \right) = \frac{1}{n}\mathop \sum \limits_{i = 1}^n {x_i} - \mu  = \bar x - \mu $

接上式

${\rm{E}}\left( {\frac{1}{n}\mathop \sum \limits_{i = 1}^n {

{\left( {\left( {

{x_i} - \mu } \right) - \left( {\bar x - \mu } \right)} \right)}^2}} \right) = {\rm{E}}\left( {\frac{1}{n}\mathop \sum \limits_{i = 1}^n {

{\left( {

{x_i} - \mu } \right)}^2} - \frac{1}{n}\mathop \sum \limits_{i = 1}^n 2({x_i} - \mu )\left( {\bar x - \mu } \right) + \frac{1}{n}\mathop \sum \limits_{i = 1}^n {

{\left( {

{x_i} - \mu } \right)}^2}} \right)$

$ = {\rm{E}}\left( {\frac{1}{n}\mathop \sum \limits_{i = 1}^n {

{\left( {

{x_i} - \mu } \right)}^2} - 2\left( {\bar x - \mu } \right)\left( {\bar x - \mu } \right) + {

{\left( {\bar x - \mu } \right)}^2}} \right)$

$ = {\rm{\;E}}\left( {\frac{1}{n}\mathop \sum \limits_{i = 1}^n {

{\left( {

{x_i} - \mu } \right)}^2} - {

{\left( {\bar x - \mu } \right)}^2}} \right)$

$ = {\rm{E}}\left( {\frac{1}{n}\mathop \sum \limits_{i = 1}^n {

{\left( {

{x_i} - \mu } \right)}^2}} \right) - E({\left( {\bar x - \mu } \right)^2}) \le {\sigma ^2}$

所以样本方差除以n会小于总体方差

${\rm{E}}\left( {\frac{1}{n}\mathop \sum \limits_{i = 1}^n {

{\left( {

{x_i} - \mu } \right)}^2}} \right) - E({\left( {\bar x - \mu } \right)^2}) = {\sigma ^2} - \frac{1}{n}{\sigma ^2} = \frac{

{n - 1}}{n}{\sigma ^2}$

所以样本方差与总体方差差(n-1)/n倍。

 

二、协方差

协方差是对两个随机变量联合分布线性相关程度的一种度量。两个随机变量越线性相关，协方差越大，完全线性无关，协方差为零。

Cov(x,y) = E[(x-E(x))(y-E(y))]

特殊的当只存在一个变量x，x与自身的协方差等于方差，记作Var(x)

Cov(x,x) =Var(x)= E[(x-E(x))(x-E(x))]

样本协方差

对于多维随机变量Q(x₁,x₂,x₃,…,x_n)，样本集合为x_ij=[x_1j,x_2j,…,x_nj](j=1,2,…,m)，m为样本数量，在a,b（a,b=1,2…n）两个维度内

${\rm{cov}}\left( {

{

{\rm{x}}_{\rm{a}}},{

{\rm{x}}_{\rm{b}}}} \right) = \frac{

{\mathop \sum \nolimits_{j = 1}^m \left( {

{x_{aj}} - {

{\bar x}_a}} \right)\left( {

{x_{bj}} - {

{\bar x}_b}} \right)}}{

{m - 1}}$

 

三、协方差矩阵

对于多维随机变量Q(x₁,x₂,x₃,…,x_n)我们需要对任意两个变量(x_i,x_j)求线性关系，即需要对任意两个变量求协方差矩阵

Cov(x_i,x_j)= E[(x_i-E(x_i))(x_j-E(x_j))]

\[{\rm{cov}}\left( {

{x_i},{x_j}} \right) = \left[ {\begin{array}{*{20}{c}}

{

{\rm{cov}}\left( {

{x_1},{x_1}} \right)}&{

{\rm{cov}}\left( {

{x_1},{x_2}} \right)}&{

{\rm{cov}}\left( {

{x_1},{x_3}} \right)}& \cdots &{

{\rm{cov}}\left( {

{x_1},{x_{\rm{n}}}} \right)}\\

{

{\rm{cov}}\left( {

{x_2},{x_1}} \right)}&{

{\rm{cov}}\left( {

{x_2},{x_2}} \right)}&{

{\rm{cov}}\left( {

{x_2},{x_3}} \right)}& \cdots &{

{\rm{cov}}\left( {

{x_2},{x_n}} \right)}\\

{

{\rm{cov}}\left( {

{x_3},{x_1}} \right)}&{

{\rm{cov}}\left( {

{x_3},{x_2}} \right)}&{

{\rm{cov}}\left( {

{x_3},{x_3}} \right)}& \cdots &{

{\rm{cov}}\left( {

{x_3},{x_n}} \right)}\\

\vdots & \vdots & \vdots & \ddots & \vdots \\

{

{\rm{cov}}\left( {

{x_m}{x_1}} \right)}&{

{\rm{cov}}\left( {

{x_m},{x_2}} \right)}&{

{\rm{cov}}\left( {

{x_m},{x_3}} \right)}& \cdots &{

{\rm{cov}}\left( {

{x_m},{x_n}} \right)}

\end{array}} \right]\]

 

【 结束 】