样本方差公式推导--为什么样本方差的分母是n-1

概要

因为使用n作为分母会导致方差被低估，将分母替换为n-1可以保证样本方差是一种无偏估计

理想情况

首先，我们假定随机变量 X X X的数学期望 μ \mu μ是已知的，然而方差 σ 2 {{\sigma }^{2}} σ2未知。如果我们得到一组随机变量 X X X的样本 { X i , i = 1 , 2 , 3... n } \left\{ {{X}_{i}},i=1,2,3...n \right\} {Xi​,i=1,2,3...n}。

在这个条件下，根据方差的定义我们有：

E [ ( X i − μ ) 2 ] = σ 2 , ∀ i = 1 , … , n E\left[ {{\left( {{X}_{i}}-\mu \right)}^{2}} \right]={{\sigma }^{2}},\quad \forall i=1,\ldots ,n E[(Xi​−μ)2]=σ2,∀i=1,…,n

由此可得：

E [ 1 n ∑ i = 1 n ( X i − μ ) 2 ] = σ 2 E\left[ \frac{1}{n}\sum\limits_{i=1}^{n}{{{\left( {{X}_{i}}-\mu \right)}^{2}}} \right]={{\sigma }^{2}} E[n1​i=1∑n​(Xi​−μ)2]=σ2

因此， 1 n ∑ i = 1 n ( X i − μ ) 2 \frac{1}{n}\sum\limits_{i=1}^{n}{{{\left( {{X}_{i}}-\mu \right)}^{2}}} n1​i=1∑n​(Xi​−μ)2是方差 σ 2 {{\sigma }^{2}} σ2的一个无偏估计。此时，除的分母仍然是 n n n。

使用样本均值代替数学期望

现在，假定随机变量 X X X的数学期望 μ \mu μ是未知的，我们使用样本数据来估计数学期望 μ \mu μ：

X ˉ = 1 n ∑ i = 1 n X i \bar{X}=\frac{1}{n}\sum\limits_{i=1}^{n}{{{X}_{i}}} Xˉ=n1​i=1∑n​Xi​

如果我们直接使用上式，代替数学期望 μ \mu μ，则会导致低估方差，如下所示：

E ( 1 n ∑ i = 1 n ( X i − X ˉ ) 2 ) = E ( 1 n ∑ i = 1 n [ ( X i − μ ) + ( μ − X ˉ ) ] 2 ) = E ( 1 n ∑ i = 1 n ( X i − μ ) 2 + 2 n ∑ i = 1 n ( X i − μ ) ( μ − X ˉ ) + 1 n ∑ i = 1 n ( μ − X ˉ ) 2 ) = E ( 1 n ∑ i = 1 n ( X i − μ ) 2 + 2 ( X ˉ − μ ) ( μ − X ˉ ) + ( μ − X ˉ ) 2 ) = E ( 1 n ∑ i = 1 n ( X i − μ ) 2 − ( μ − X ˉ ) 2 ) ≤ E ( 1 n ∑ i = 1 n ( X i − μ ) 2 ) = σ 2 \begin{array}{l} E\left(\frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}\right)=E\left(\frac{1}{n} \sum_{i=1}^{n}\left[\left(X_{i}-\mu\right)+(\mu-\bar{X})\right]^{2}\right) \\ =E\left(\frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2}+\frac{2}{n} \sum_{i=1}^{n}\left(X_{i}-\mu\right)(\mu-\bar{X})+\frac{1}{n} \sum_{i=1}^{n}(\mu-\bar{X})^{2}\right) \\ =E\left(\frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2}+2(\bar{X}-\mu)(\mu-\bar{X})+(\mu-\bar{X})^{2}\right) \\ =E\left(\frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2}-(\mu-\bar{X})^{2}\right) \\ \leq E\left(\frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2}\right)=\sigma^{2} \end{array} E(n1​∑i=1n​(Xi​−Xˉ)2)=E(n1​∑i=1n​[(Xi​−μ)+(μ−Xˉ)]2)=E(n1​∑i=1n​(Xi​−μ)2+n2​∑i=1n​(Xi​−μ)(μ−Xˉ)+n1​∑i=1n​(μ−Xˉ)2)=E(n1​∑i=1n​(Xi​−μ)2+2(Xˉ−μ)(μ−Xˉ)+(μ−Xˉ)2)=E(n1​∑i=1n​(Xi​−μ)2−(μ−Xˉ)2)≤E(n1​∑i=1n​(Xi​−μ)2)=σ2​

对 ( μ − X ˉ ) 2 {{(\mu -\bar{X})}^{2}} (μ−Xˉ)2项进行分析:

E ( ( μ − X ˉ ) 2 ) = E ( ( X ˉ − μ ) 2 ) = E ( ( 1 n ∑ i = 1 n X i − μ ) 2 ) = E ( ( 1 n ∑ i = 1 n ( X i − μ ) ) 2 ) \begin{array}{l} E\left((\mu-\bar{X})^{2}\right)=E\left((\bar{X}-\mu)^{2}\right) \\ =E\left(\left(\frac{1}{n} \sum_{i=1}^{n} X_{i}-\mu\right)^{2}\right) \\ =E\left(\left(\frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\mu\right)\right)^{2}\right) \end{array} E((μ−Xˉ)2)=E((Xˉ−μ)2)=E((n1​∑i=1n​Xi​−μ)2)=E((n1​∑i=1n​(Xi​−μ))2)​

对多个独立随机变量，存在下述公式：

方差计算公式：

D ( X ) = E ( X 2 ) − [ E ( X ) ] 2 D(X)=E\left( {{X}^{2}} \right)-{{[E(X)]}^{2}} D(X)=E(X2)−[E(X)]2

均值的均值：

E ( X ) = E ( 1 n ∑ i = 1 n X i ) = 1 n E ( ∑ i = 1 n X i ) = E ( X i ) = X ˉ \begin{aligned} & E(X)=E\left( \frac{1}{n}\sum\limits_{i=1}^{n}{{{X}_{i}}} \right) \\ & =\frac{1}{n}E\left( \sum\limits_{i=1}^{n}{{{X}_{i}}} \right) \\ & =E\left( {{X}_{i}} \right) \\ & =\bar{X} \end{aligned} ​E(X)=E(n1​i=1∑n​Xi​)=n1​E(i=1∑n​Xi​)=E(Xi​)=Xˉ​

均值的方差：

D ( X ˉ ) = D ( 1 n ∑ i = 1 n X i ) = 1 n 2 D ( ∑ i = 1 n X i ) = 1 n D ( X i ) \begin{aligned} D(\bar{X}) &=D\left(\frac{1}{n} \sum_{i=1}^{n} X_{i}\right) \\ &=\frac{1}{n^{2}} D\left(\sum_{i=1}^{n} X_{i}\right) \\ &=\frac{1}{n} D\left(X_{i}\right) \end{aligned} D(Xˉ)​=D(n1​i=1∑n​Xi​)=n21​D(i=1∑n​Xi​)=n1​D(Xi​)​

所以：

E ( ( μ − X ˉ ) 2 ) = E ( ( 1 n ∑ i = 1 n ( X i − μ ) ) 2 ) → A = 1 n ∑ i = 1 n ( X i − μ ) E ( A 2 ) = D ( A ) − E ( A ) 2 → E ( A ) = 0 1 n D ( X i − μ ) = 1 n D ( X i ) = 1 n σ 2 \begin{aligned} & E\left( {{(\mu -\bar{X})}^{2}} \right)=E\left( {{\left( \frac{1}{n}\sum\limits_{i=1}^{n}{\left( {{X}_{i}}-\mu \right)} \right)}^{2}} \right) \\ & \xrightarrow{A=\frac{1}{n}\sum\limits_{i=1}^{n}{\left( {{X}_{i}}-\mu \right)}}E\left( {{A}^{2}} \right) \\ & =D\left( A \right)-E{{\left( A \right)}^{2}} \\ & \xrightarrow{E(A)=0}\frac{1}{n}D\left( {{X}_{i}}-\mu \right) \\ & =\frac{1}{n}D\left( {{X}_{i}} \right) \\ & =\frac{1}{n}{{\sigma }^{2}} \end{aligned} ​E((μ−Xˉ)2)=E⎝⎛​(n1​i=1∑n​(Xi​−μ))2⎠⎞​A=n1​i=1∑n​(Xi​−μ) ​E(A2)=D(A)−E(A)2E(A)=0 ​n1​D(Xi​−μ)=n1​D(Xi​)=n1​σ2​

结合以上结果，可以知道：

E ( 1 n ∑ i = 1 n ( X i − X ˉ ) 2 ) = E ( 1 n ∑ i = 1 n ( X i − μ ) 2 − ( μ − X ˉ ) 2 ) = E ( 1 n ∑ i = 1 n ( X i − μ ) 2 ) − E ( ( μ − X ˉ ) 2 ) = σ 2 − 1 n σ 2 = n − 1 n σ 2 \begin{aligned} & E\left( \frac{1}{n}\sum\limits_{i=1}^{n}{{{\left( {{X}_{i}}-\bar{X} \right)}^{2}}} \right)=E\left( \frac{1}{n}\sum\limits_{i=1}^{n}{{{\left( {{X}_{i}}-\mu \right)}^{2}}}-{{(\mu -\bar{X})}^{2}} \right) \\ & =E\left( \frac{1}{n}\sum\limits_{i=1}^{n}{{{\left( {{X}_{i}}-\mu \right)}^{2}}} \right)-E\left( {{(\mu -\bar{X})}^{2}} \right) \\ & ={{\sigma }^{2}}-\frac{1}{n}{{\sigma }^{2}} \\ & =\frac{n-1}{n}{{\sigma }^{2}} \end{aligned} ​E(n1​i=1∑n​(Xi​−Xˉ)2)=E(n1​i=1∑n​(Xi​−μ)2−(μ−Xˉ)2)=E(n1​i=1∑n​(Xi​−μ)2)−E((μ−Xˉ)2)=σ2−n1​σ2=nn−1​σ2​

要使样本方差的期望等于总体方差，就需要进行修正，也即给样本方差乘上 n n − 1 \frac{n}{n-1} n−1n​。

所以得到样本方差为：

n n − 1 ⋅ 1 n ∑ i = 1 n ( X i − X ˉ ) 2 = 1 n − 1 ∑ i = 1 n ( X i − X ˉ ) 2 \frac{n}{n-1}\cdot \frac{1}{n}\sum\limits_{i=1}^{n}{{{\left( {{X}_{i}}-\bar{X} \right)}^{2}}}=\frac{1}{n-1}\sum\limits_{i=1}^{n}{{{\left( {{X}_{i}}-\bar{X} \right)}^{2}}} n−1n​⋅n1​i=1∑n​(Xi​−Xˉ)2=n−11​i=1∑n​(Xi​−Xˉ)2

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