参考链接
/s/blog_c96053d60101n24f.html总体均值和方差
设 X X X服从某一分布 X ∼ F X\sim F X∼F,则 X X X的总体均值为 E ( X ) = μ E\left ( X\right )=\mu E(X)=μ,总体方差为 D ( X ) ∼ σ 2 D\left ( X\right )\sim \sigma ^{2} D(X)∼σ2。
样本的均值
我们不能获取到分布中的所有点,只能从中随机采样一部分样本,以估计整体分布情况。
设 n n n个样本为 x 1 , x 2 , ⋅ ⋅ ⋅ , x n x_{1},x_{2},\cdot \cdot \cdot ,x_{n} x1,x2,⋅⋅⋅,xn,则样本的均值为 x ˉ = 1 n ∑ i = 1 n x i \bar{x}=\frac{1}{n}\sum_{i=1}^{n}x_{i} xˉ=n1∑i=1nxi。由于样本是随机采样的,所以样本的均值实际上也是随机的,同样有一个期望和方差:
E ( x ˉ ) = E ( 1 n ∑ i = 1 n x i ) = 1 n ∑ i = 1 n E ( x i ) = 1 n × n E ( x i ) = E ( x i ) = μ E\left ( \bar{x}\right )=E\left ( \frac{1}{n}\sum_{i=1}^{n}x_{i}\right )=\frac{1}{n}\sum_{i=1}^{n}E\left ( x_{i}\right )=\frac{1}{n}\times nE\left ( x_{i}\right )=E\left ( x_{i}\right )=\mu E(xˉ)=E(n1∑i=1nxi)=n1∑i=1nE(xi)=n1×nE(xi)=E(xi)=μ
D ( x ˉ ) = D ( 1 n ∑ i = 1 n x i ) = 1 n 2 ∑ i = 1 n D ( x i ) = σ 2 n D\left ( \bar{x}\right )=D\left ( \frac{1}{n}\sum_{i=1}^{n}x_{i}\right )=\frac{1}{n^{2}}\sum_{i=1}^{n}D\left ( x_{i}\right )=\frac{\sigma ^{2}}{n} D(xˉ)=D(n1∑i=1nxi)=n21∑i=1nD(xi)=nσ2
可以看出,样本均值的期望就是总体均值,因此可以说均值是无偏的。
样本的方差
样本方差的期望为:
E ( S 2 ) = E [ 1 n ∑ i = 1 n ( x i − x ˉ ) 2 ] = 1 n E [ ∑ i = 1 n ( x i 2 − 2 x i x ˉ + x ˉ 2 ) ] = 1 n E [ ∑ i = 1 n ( x i 2 ) ] − 1 n E ( 2 x ˉ × n x ˉ − n x ˉ 2 ) = 1 n ∑ i = 1 n E ( x i 2 ) − E ( x ˉ 2 ) E\left ( S^{2}\right )=E\left [ \frac{1}{n}\sum_{i=1}^{n}\left ( x_{i}-\bar{x}\right )^{2}\right ]=\frac{1}{n}E\left [ \sum_{i=1}^{n}\left ( x_{i}^{2}-2x_{i}\bar{x}+\bar{x}^{2}\right )\right ]\\ =\frac{1}{n}E\left [ \sum_{i=1}^{n}\left ( x_{i}^{2}\right )\right ]-\frac{1}{n}E\left ( 2\bar{x}\times n\bar{x}-n\bar{x}^{2}\right )=\frac{1}{n}\sum_{i=1}^{n}E\left ( x_{i}^{2}\right )-E\left ( \bar{x}^{2}\right ) E(S2)=E[n1∑i=1n(xi−xˉ)2]=n1E[∑i=1n(xi2−2xixˉ+xˉ2)]=n1E[∑i=1n(xi2)]−n1E(2xˉ×nxˉ−nxˉ2)=n1∑i=1nE(xi2)−E(xˉ2)
由方差公式,有:
E ( x i 2 ) = D ( x i ) + E 2 ( x i ) = σ 2 + μ 2 E\left ( x_{i}^{2}\right )=D\left ( x_{i}\right )+E^{2}\left ( x_{i}\right )=\sigma ^{2}+\mu ^{2} E(xi2)=D(xi)+E2(xi)=σ2+μ2
E ( x ˉ 2 ) = D ( x ˉ ) + E 2 ( x ˉ ) = σ 2 n + μ 2 E\left ( \bar{x}^{2}\right )=D\left ( \bar{x}\right )+E^{2}\left ( \bar{x}\right )=\frac{\sigma ^{2}}{n}+\mu ^{2} E(xˉ2)=D(xˉ)+E2(xˉ)=nσ2+μ2
因此, E ( S 2 ) = n − 1 n σ 2 E\left ( S^{2}\right )=\frac{n-1}{n}\sigma ^{2} E(S2)=nn−1σ2
无偏估计的样本方差
从上面得到的 E ( S 2 ) = n − 1 n σ 2 E\left ( S^{2}\right )=\frac{n-1}{n}\sigma ^{2} E(S2)=nn−1σ2,可以看出,样本方差的期望不是无偏的,需要乘上一个系数,于是:
n − 1 n S 2 = 1 n − 1 ∑ i = 1 n ( x i − x ˉ ) 2 \frac{n-1}{n}S^{2}=\frac{1}{n-1}\sum_{i=1}^{n}\left ( x_{i}-\bar{x}\right )^{2} nn−1S2=n−11∑i=1n(xi−xˉ)2
n-1即为自由度,就是说,在一个容量为n的样本里,当确定了n-1个变量以后,第n个变量就确定了,因为样本均值是无偏的。
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