樣本方差，協方差，協方差矩陣

一、樣本方差

設樣本均值為$\bar x$，樣本方差為S²，總體均值為${\rm{\mu }}$，總體方差為${{\rm{\sigma }}^2}$，那么樣本方差

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

推導：假設樣本數量等於總體數量，應有

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

在多次重復抽取樣本過程中，樣本方差會逐漸接近總體方差，假設每次抽取的樣本方差為

（S₁²,S₂²,S₃²…），然后對這些樣本方差求平均值記為E(S²)，則

 

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

$ = {\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)$

因為

$\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}$

所以樣本方差除以n會小於總體方差

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

所以樣本方差與總體方差差(n-1)/n倍。

 

二、協方差

協方差是對兩個隨機變量聯合分布線性相關程度的一種度量。兩個隨機變量越線性相關，協方差越大，完全線性無關，協方差為零。

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

特殊的當只存在一個變量x，x與自身的協方差等於方差，記作Var(x)

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

樣本協方差

對於多維隨機變量Q(x₁,x₂,x₃,…,x_n)，樣本集合為x_ij=[x_1j,x_2j,…,x_nj](j=1,2,…,m)，m為樣本數量，在a,b（a,b=1,2…n）兩個維度內

${\rm{cov}}\left( {{{\rm{x}}_{\rm{a}}},{{\rm{x}}_{\rm{b}}}} \right) = \frac{{\mathop \sum \nolimits_{j = 1}^m \left( {{x_{aj}} - {{\bar x}_a}} \right)\left( {{x_{bj}} - {{\bar x}_b}} \right)}}{{m - 1}}$

 

三、協方差矩陣

對於多維隨機變量Q(x₁,x₂,x₃,…,x_n)我們需要對任意兩個變量(x_i,x_j)求線性關系，即需要對任意兩個變量求協方差矩陣

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

\[{\rm{cov}}\left( {{x_i},{x_j}} \right) = \left[ {\begin{array}{*{20}{c}}
{{\rm{cov}}\left( {{x_1},{x_1}} \right)}&{{\rm{cov}}\left( {{x_1},{x_2}} \right)}&{{\rm{cov}}\left( {{x_1},{x_3}} \right)}& \cdots &{{\rm{cov}}\left( {{x_1},{x_{\rm{n}}}} \right)}\\
{{\rm{cov}}\left( {{x_2},{x_1}} \right)}&{{\rm{cov}}\left( {{x_2},{x_2}} \right)}&{{\rm{cov}}\left( {{x_2},{x_3}} \right)}& \cdots &{{\rm{cov}}\left( {{x_2},{x_n}} \right)}\\
{{\rm{cov}}\left( {{x_3},{x_1}} \right)}&{{\rm{cov}}\left( {{x_3},{x_2}} \right)}&{{\rm{cov}}\left( {{x_3},{x_3}} \right)}& \cdots &{{\rm{cov}}\left( {{x_3},{x_n}} \right)}\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
{{\rm{cov}}\left( {{x_m}{x_1}} \right)}&{{\rm{cov}}\left( {{x_m},{x_2}} \right)}&{{\rm{cov}}\left( {{x_m},{x_3}} \right)}& \cdots &{{\rm{cov}}\left( {{x_m},{x_n}} \right)}
\end{array}} \right]\]

 

【 結束 】