欢迎光临散文网 会员登陆 & 注册

为什么样本方差(sample variance)的分母是 n-1?

2021-06-18 10:40 作者:马同学图解数学  | 我要投稿

先把问题完整的描述下。

如果已知随机变量X的期望为%5Cmu,那么可以如下计算方差%5Csigma%5E2

%5Csigma%5E2%3DE%5B(X-%5Cmu)%5E2%5D

上面的式子需要知道X的具体分布是什么(在现实应用中往往不知道准确分布),计算起来也比较复杂。

所以实践中常常采样之后,用下面这个S%5E2来近似%5Csigma%5E2

 S%5E2%3D%5Cfrac%7B1%7D%7Bn%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7D(X_i-%5Cmu)%5E2

其实现实中,往往连X的期望%5Cmu也不清楚,只知道样本的均值:

%5Coverline%7BX%7D%3D%5Cfrac%7B1%7D%7Bn%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7DX_i

那么可以这么来计算S%5E2

S%5E2%3D%5Cfrac%7B1%7D%7Bn-1%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7D(X_i-%5Coverline%7BX%7D)%5E2

那这里就有两个问题了:

  • 为什么可以用S%5E2来近似%5Csigma%5E2

  • 为什么使用%5Coverline%7BX%7D替代%5Cmu之后,分母是%5Cdisplaystyle%5Cfrac%7B1%7D%7Bn-1%7D

我们来仔细分析下细节,就可以弄清楚这两个问题。

1 为什么可以用S%5E2来近似%5Csigma%5E2

举个例子,假设X服从这么一个正态分布:

X%20%5Csim%20N(145%2C%201.4%5E2)

%5Cmu%3D145%2C%5Csigma%5E2%3D1.4%5E2%3D1.96 图形如下:

当然,现实中往往并不清楚X服从的分布是什么,具体参数又是什么?所以我用虚线来表明我们并不是真正知道X的分布:

很幸运的,我们知道%5Cmu%3D145,因此对X采样,并通过:

S%5E2%3D%5Cfrac%7B1%7D%7Bn%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7D(X_i-%5Cmu)%5E2

来估计%5Csigma%5E2。某次采样计算出来的S%5E2

看起来比%5Csigma%5E2%3D1.96要小。采样具有随机性,我们多采样几次,S%5E2会围绕%5Csigma%5E2上下波动:

S%5E2作为%5Csigma%5E2的一个估计量,算是可以接受的选择。 

很容易算出:

E%5B%5Cfrac%7B1%7D%7Bn%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7D(X_i-%5Cmu)%5E2%5D%3D%5Csigma%5E2

这也就是所谓的无偏估计量。从这个分布来看,选择S%5E2作为估计量确实可以接受。

2 为什么使用%5Coverline%7BX%7D替代%5Cmu之后,分母是%5Cdisplaystyle%5Cfrac%7B1%7D%7Bn-1%7D

更多的情况,我们不知道%5Cmu是多少的,只能计算出%5Coverline%7BX%7D。不同的采样对应不同的%5Coverline%7BX%7D

对于某次采样而言,当%5Cmu%3D%5Coverline%7BX%7D时,下式取得最小值:

%5Csum_%7Bi%3D1%7D%5E%7Bn%7D(X_i-%5Cmu)%5E2

我们也是比较容易从图像中观察出这一点,只要%5Cmu偏离%5Coverline%7BX%7D,该值就会增大:

所以可知:

%5Csum_%7Bi%3D1%7D%5E%7Bn%7D(X_i-%5Coverline%7BX%7D)%5E2%20%5Cleq%20%5Csum_%7Bi%3D1%7D%5E%7Bn%7D(X_i-%5Cmu)%5E2

可推出:

%5Cfrac%7B1%7D%7Bn%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7D(X_i-%5Coverline%7BX%7D)%5E2%20%5Cleq%20%5Cfrac%7B1%7D%7Bn%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7D(X_i-%5Cmu)%5E2

进而推出:

E%5B%5Cfrac%7B1%7D%7Bn%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7D(X_i-%5Coverline%7BX%7D)%5E2%5D%20%5Cleq%20E%5B%5Cfrac%7B1%7D%7Bn%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7D(X_i-%5Cmu)%5E2%5D%3D%5Csigma%5E2

如果用下面这个式子来估计:

S%5E2%3D%5Cfrac%7B1%7D%7Bn%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7D(X_i-%5Coverline%7BX%7D)%5E2

那么S%5E2采样均值会服从一个偏离1.4%5E2的正态分布:

可见,此分布倾向于低估%5Csigma%5E2

具体小了多少,我们可以来算下:

%7B%5Cdisplaystyle%20%7B%5Cbegin%7Baligned%7D%5Coperatorname%20%7BE%7D%20%5BS%5E%7B2%7D%5D%26%3D%5Coperatorname%20%7BE%7D%20%5Cleft%5B%7B%5Cfrac%20%7B1%7D%7Bn%7D%7D%5Csum%20_%7Bi%3D1%7D%5E%7Bn%7D%7B%5Cbig%20(%7DX_%7Bi%7D-%7B%5Coverline%20%7BX%7D%7D%7B%5Cbig%20)%7D%5E%7B2%7D%5Cright%5D%3D%5Coperatorname%20%7BE%7D%20%7B%5Cbigg%20%5B%7D%7B%5Cfrac%20%7B1%7D%7Bn%7D%7D%5Csum%20_%7Bi%3D1%7D%5E%7Bn%7D%7B%5Cbigg%20(%7D(X_%7Bi%7D-%5Cmu%20)-(%7B%5Coverline%20%7BX%7D%7D-%5Cmu%20)%7B%5Cbigg%20)%7D%5E%7B2%7D%7B%5Cbigg%20%5D%7D%5C%5C%5B8pt%5D%26%3D%5Coperatorname%20%7BE%7D%20%7B%5Cbigg%20%5B%7D%7B%5Cfrac%20%7B1%7D%7Bn%7D%7D%5Csum%20_%7Bi%3D1%7D%5E%7Bn%7D%7B%5Cbigg%20(%7D(X_%7Bi%7D-%5Cmu%20)%5E%7B2%7D-2(%7B%5Coverline%20%7BX%7D%7D-%5Cmu%20)(X_%7Bi%7D-%5Cmu%20)%2B(%7B%5Coverline%20%7BX%7D%7D-%5Cmu%20)%5E%7B2%7D%7B%5Cbigg%20)%7D%7B%5Cbigg%20%5D%7D%5C%5C%5B8pt%5D%26%3D%5Coperatorname%20%7BE%7D%20%7B%5Cbigg%20%5B%7D%7B%5Cfrac%20%7B1%7D%7Bn%7D%7D%5Csum%20_%7Bi%3D1%7D%5E%7Bn%7D(X_%7Bi%7D-%5Cmu%20)%5E%7B2%7D-%7B%5Cfrac%20%7B2%7D%7Bn%7D%7D(%7B%5Coverline%20%7BX%7D%7D-%5Cmu%20)%5Csum%20_%7Bi%3D1%7D%5E%7Bn%7D(X_%7Bi%7D-%5Cmu%20)%2B%7B%5Cfrac%20%7B1%7D%7Bn%7D%7D(%7B%5Coverline%20%7BX%7D%7D-%5Cmu%20)%5E%7B2%7D%5Csum%20_%7Bi%3D1%7D%5E%7Bn%7D1%7B%5Cbigg%20%5D%7D%5C%5C%5B8pt%5D%26%3D%5Coperatorname%20%7BE%7D%20%7B%5Cbigg%20%5B%7D%7B%5Cfrac%20%7B1%7D%7Bn%7D%7D%5Csum%20_%7Bi%3D1%7D%5E%7Bn%7D(X_%7Bi%7D-%5Cmu%20)%5E%7B2%7D-%7B%5Cfrac%20%7B2%7D%7Bn%7D%7D(%7B%5Coverline%20%7BX%7D%7D-%5Cmu%20)%5Csum%20_%7Bi%3D1%7D%5E%7Bn%7D(X_%7Bi%7D-%5Cmu%20)%2B%7B%5Cfrac%20%7B1%7D%7Bn%7D%7D(%7B%5Coverline%20%7BX%7D%7D-%5Cmu%20)%5E%7B2%7D%5Ccdot%20n%7B%5Cbigg%20%5D%7D%5C%5C%5B8pt%5D%26%3D%5Coperatorname%20%7BE%7D%20%7B%5Cbigg%20%5B%7D%7B%5Cfrac%20%7B1%7D%7Bn%7D%7D%5Csum%20_%7Bi%3D1%7D%5E%7Bn%7D(X_%7Bi%7D-%5Cmu%20)%5E%7B2%7D-%7B%5Cfrac%20%7B2%7D%7Bn%7D%7D(%7B%5Coverline%20%7BX%7D%7D-%5Cmu%20)%5Csum%20_%7Bi%3D1%7D%5E%7Bn%7D(X_%7Bi%7D-%5Cmu%20)%2B(%7B%5Coverline%20%7BX%7D%7D-%5Cmu%20)%5E%7B2%7D%7B%5Cbigg%20%5D%7D%5C%5C%5B8pt%5D%5Cend%7Baligned%7D%7D%7D%0A

其中:

%7B%5Cdisplaystyle%20%7B%5Cbegin%7Baligned%7D%7B%5Coverline%20%7BX%7D%7D-%5Cmu%20%3D%7B%5Cfrac%20%7B1%7D%7Bn%7D%7D%5Csum%20_%7Bi%3D1%7D%5E%7Bn%7DX_%7Bi%7D-%5Cmu%20%3D%7B%5Cfrac%20%7B1%7D%7Bn%7D%7D%5Csum%20_%7Bi%3D1%7D%5E%7Bn%7DX_%7Bi%7D-%7B%5Cfrac%20%7B1%7D%7Bn%7D%7D%5Csum%20_%7Bi%3D1%7D%5E%7Bn%7D%5Cmu%20%5C%20%3D%7B%5Cfrac%20%7B1%7D%7Bn%7D%7D%5Csum%20_%7Bi%3D1%7D%5E%7Bn%7D(X_%7Bi%7D-%5Cmu%20).%5C%5C%5B8pt%5D%5Cend%7Baligned%7D%7D%7D%0A

所以我们接着算下去:

%7B%5Cdisplaystyle%20%7B%5Cbegin%7Baligned%7D%5Coperatorname%20%7BE%7D%20%5BS%5E%7B2%7D%5D%26%3D%5Coperatorname%20%7BE%7D%20%7B%5Cbigg%20%5B%7D%7B%5Cfrac%20%7B1%7D%7Bn%7D%7D%5Csum%20_%7Bi%3D1%7D%5E%7Bn%7D(X_%7Bi%7D-%5Cmu%20)%5E%7B2%7D-%7B%5Cfrac%20%7B2%7D%7Bn%7D%7D(%7B%5Coverline%20%7BX%7D%7D-%5Cmu%20)%5Csum%20_%7Bi%3D1%7D%5E%7Bn%7D(X_%7Bi%7D-%5Cmu%20)%2B(%7B%5Coverline%20%7BX%7D%7D-%5Cmu%20)%5E%7B2%7D%7B%5Cbigg%20%5D%7D%5C%5C%5B8pt%5D%26%3D%5Coperatorname%20%7BE%7D%20%7B%5Cbigg%20%5B%7D%7B%5Cfrac%20%7B1%7D%7Bn%7D%7D%5Csum%20_%7Bi%3D1%7D%5E%7Bn%7D(X_%7Bi%7D-%5Cmu%20)%5E%7B2%7D-%7B%5Cfrac%20%7B2%7D%7Bn%7D%7D(%7B%5Coverline%20%7BX%7D%7D-%5Cmu%20)%5Ccdot%20n%5Ccdot%20(%7B%5Coverline%20%7BX%7D%7D-%5Cmu%20)%2B(%7B%5Coverline%20%7BX%7D%7D-%5Cmu%20)%5E%7B2%7D%7B%5Cbigg%20%5D%7D%5C%5C%5B8pt%5D%26%3D%5Coperatorname%20%7BE%7D%20%7B%5Cbigg%20%5B%7D%7B%5Cfrac%20%7B1%7D%7Bn%7D%7D%5Csum%20_%7Bi%3D1%7D%5E%7Bn%7D(X_%7Bi%7D-%5Cmu%20)%5E%7B2%7D-2(%7B%5Coverline%20%7BX%7D%7D-%5Cmu%20)%5E%7B2%7D%2B(%7B%5Coverline%20%7BX%7D%7D-%5Cmu%20)%5E%7B2%7D%7B%5Cbigg%20%5D%7D%5C%5C%5B8pt%5D%26%3D%5Coperatorname%20%7BE%7D%20%7B%5Cbigg%20%5B%7D%7B%5Cfrac%20%7B1%7D%7Bn%7D%7D%5Csum%20_%7Bi%3D1%7D%5E%7Bn%7D(X_%7Bi%7D-%5Cmu%20)%5E%7B2%7D-(%7B%5Coverline%20%7BX%7D%7D-%5Cmu%20)%5E%7B2%7D%7B%5Cbigg%20%5D%7D%5C%5C%5B8pt%5D%26%3D%5Coperatorname%20%7BE%7D%20%7B%5Cbigg%20%5B%7D%7B%5Cfrac%20%7B1%7D%7Bn%7D%7D%5Csum%20_%7Bi%3D1%7D%5E%7Bn%7D(X_%7Bi%7D-%5Cmu%20)%5E%7B2%7D%7B%5Cbigg%20%5D%7D-%5Coperatorname%20%7BE%7D%20%7B%5Cbigg%20%5B%7D(%7B%5Coverline%20%7BX%7D%7D-%5Cmu%20)%5E%7B2%7D%7B%5Cbigg%20%5D%7D%5C%5C%5B8pt%5D%26%3D%5Csigma%20%5E%7B2%7D-%5Coperatorname%20%7BE%7D%20%5Cleft%5B(%7B%5Coverline%20%7BX%7D%7D-%5Cmu%20)%5E%7B2%7D%5Cright%5D%5Cend%7Baligned%7D%7D%7D

其中:

%0A%7B%5Cdisplaystyle%20%5Coperatorname%20%7BE%7D%20%7B%5Cbig%20%5B%7D(%7B%5Coverline%20%7BX%7D%7D-%5Cmu%20)%5E%7B2%7D%7B%5Cbig%20%5D%7D%3D%7B%5Cfrac%20%7B1%7D%7Bn%7D%7D%5Csigma%20%5E%7B2%7D.%7D%0A

所以:

E%5B%5Cfrac%7B1%7D%7Bn%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7D(X_i-%5Coverline%7BX%7D)%5E2%5D%3D%5Csigma%5E%7B2%7D-%7B%5Cfrac%20%7B1%7D%7Bn%7D%7D%5Csigma%5E%7B2%7D%3D%5Cfrac%7Bn-1%7D%7Bn%7D%5Csigma%5E%7B2%7D

也就是说,低估了%5Cdisplaystyle%7B%5Cfrac%20%7B1%7D%7Bn%7D%7D%5Csigma%5E%7B2%7D,进行一下调整:

%5Cfrac%7Bn%7D%7Bn-1%7DE%5B%5Cfrac%7B1%7D%7Bn%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7D(X_i-%5Coverline%7BX%7D)%5E2%5D%3DE%5B%5Cfrac%7B1%7D%7Bn-1%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7D(X_i-%5Coverline%7BX%7D)%5E2%5D%3D%5Csigma%5E%7B2%7D

因此使用下面这个式子进行估计,得到的就是无偏估计:

S%5E2%3D%5Cfrac%7B1%7D%7Bn-1%7D%5Csum_%7Bi%3D1%7D%5E%7Bn%7D(X_i-%5Coverline%7BX%7D)%5E2


更多内容请关注:马同学图解数学





为什么样本方差(sample variance)的分母是 n-1?的评论 (共 条)

分享到微博请遵守国家法律