【抽样技术系列02】简单随机抽样SRS PART1


这一个系列的笔记和整理希望可以帮助到正在学习抽样技术的同学。我会慢慢更新各个章节的内容。

Chap 2 简单随机抽样

2.1 概念,表示以及基本知识

抽样技术有限总体的方差定义与数理统计的不同!
σ 2 = 1 N ∑ i = 1 N ( Y i − Y ˉ ) 2 S 2 = 1 N − 1 ∑ i = 1 N ( Y i − Y ˉ ) 2 \sigma^2 = \frac{1}{N}\sum_{i=1}^N(Y_i-\bar Y)^2 \\ S^2 = \frac{1}{N-1}\sum_{i=1}^N(Y_i-\bar Y)^2σ2=N1i=1N(YiYˉ)2S2=N11i=1N(YiYˉ)2
SRSWOR下,关于α i \alpha_iαi 的常识
E ( α i ) = π i = f = n N V ( α i ) = f ( 1 − f ) = n N ( 1 − n N ) E ( α i α j ) = 1 ∗ π i j + 0 = n ( n − 1 ) N ( N − 1 ) C o v ( α i , α j ) = E ( α i α j ) − E ( α i ) E ( α j ) = − f ( 1 − f ) N − 1 \begin{aligned} E(\alpha_i) &= \pi_i = f = \frac{n}{N}\\ V(\alpha_i) &= f(1-f) = \frac{n}{N} (1- \frac{n}{N})\\ E(\alpha_i \alpha_j) &= 1*\pi_{ij}+0 =\frac{n(n-1)}{N(N-1)} \\ Cov(\alpha_i, \alpha_j) &= E(\alpha_i \alpha_j) - E(\alpha_i)E(\alpha_j) \\ &= \frac{-f(1-f)}{N-1} \end{aligned}E(αi)V(αi)E(αiαj)Cov(αi,αj)=πi=f=Nn=f(1f)=Nn(1Nn)=1πij+0=N(N1)n(n1)=E(αiαj)E(αi)E(αj)=N1f(1f)

2.2 估计

2.2.1 总体均值的估计

总体均值的估计:样本均值y ˉ = ∑ i = 1 n y i \bar y = \sum_{i = 1}^{n} y_iyˉ=i=1nyi

  • 估计的无偏性
    E ( y ˉ ) = ∑ i = 1 C N n y ˉ i P ( S i ) = ∑ i = 1 C N n y ˉ i 1 C N n = ∑ i = 1 C N n ( Y i 1 + Y i 2 + . . . + Y i n ) 1 n C N n = 1 n C N n ( ∑ i = 1 N C 1 1 C N − 1 n − 1 Y i ) = ( N − 1 ) ! ( n − 1 ) ! ( N − n ) ! n N ! n ! ( N − n ) ! ∑ i = 1 N Y i = 1 N ∑ i = 1 N Y i = Y ˉ \begin{aligned} E(\bar y) &= \sum_{i=1}^{C_N^n} \bar y_i P(S_i) \\ &= \sum_{i=1}^{C_N^n} \bar y_i \frac{1}{C_N^n} \\ &= \sum_{i=1}^{C_N^n} (Y_{i1} + Y_{i2} + ... + Y_{in}) \frac{1}{nC_N^n} \\ &= \frac{1}{nC_N^n} (\sum_{i=1}^{N}C_1^1C_{N-1}^{n-1}Y_i) \\ &= \frac{\frac{(N-1)!}{(n-1)!(N-n)!}}{\frac{n N!}{n!(N-n)!}}\sum_{i=1}^{N}Y_i \\ &= \frac{1}{N}\sum_{i=1}^{N}Y_i \\ &= \bar Y \end{aligned}E(yˉ)=i=1CNnyˉiP(Si)=i=1CNnyˉiCNn1=i=1CNn(Yi1+Yi2+...+Yin)nCNn1=nCNn1(i=1NC11CN1n1Yi)=n!(Nn)!nN!(n1)!(Nn)!(N1)!i=1NYi=N1i=1NYi=Yˉ
  • 估计的方差
    V ( y ˉ ) = V ( 1 n ∑ i = 1 n y i ) = V ( 1 n ∑ i = 1 N α i Y i ) = 1 n 2 ( ∑ i = 1 N Y i 2 V ( α i ) + ∑ i ≠ j Y i Y j C o v ( α i , α j ) ) = f ( 1 − f ) n 2 ( ∑ i = 1 N Y i 2 − ∑ i ≠ j Y i Y j 1 N − 1 ) = ( 1 − f ) n N ( N − 1 ) ( N N − 1 ∑ i = 1 N Y i 2 − 1 N − 1 ( ∑ i = 1 N Y i 2 + ∑ i ≠ j Y i Y j ) ) = ( 1 − f ) n N ( N − 1 ) ( N ∑ i = 1 N Y i 2 − ( ∑ i = 1 N Y i ) 2 ) = ( 1 − f ) n N ( N − 1 ) ( N ∑ i = 1 N Y i 2 − ( N Y ˉ ) 2 ) = ( 1 − f ) n 1 N − 1 ( ∑ i = 1 N Y i 2 − N Y ˉ 2 ) = ( 1 − f ) n 1 N − 1 ( ∑ i = 1 N Y i 2 − 2 ∑ i = 1 N Y i Y ˉ + ∑ i = 1 N Y ˉ 2 ) = ( 1 − f ) n 1 N − 1 ∑ i = 1 N ( Y i − Y ˉ ) 2 = 1 − f n S 2 \begin{aligned} V(\bar y) &= V(\frac{1}{n} \sum_{i=1}^{n}y_i) \\ &= V(\frac{1}{n} \sum_{i=1}^{N}\alpha_iY_i) \\ &= \frac{1}{n^2}( \sum_{i=1}^{N}Y_i^2V(\alpha_i)+\sum_{i\neq j}Y_iY_jCov(\alpha_i, \alpha_j)) \\ &= \frac{f(1-f)}{n^2}( \sum_{i=1}^{N}Y_i^2-\sum_{i\neq j}Y_iY_j\frac{1}{N-1}) \\ &= \frac{(1-f)}{nN(N-1)}( \frac{N}{N-1}\sum_{i=1}^{N}Y_i^2-\frac{1}{N-1}(\sum_{i=1}^{N}Y_i^2+\sum_{i\neq j}Y_iY_j)) \\ &= \frac{(1-f)}{nN(N-1)}( N\sum_{i=1}^{N}Y_i^2-(\sum_{i=1}^{N}Y_i)^2) \\ &= \frac{(1-f)}{nN(N-1)}( N\sum_{i=1}^{N}Y_i^2-(N\bar Y)^2) \\ &= \frac{(1-f)}{n}\frac{1}{N-1}( \sum_{i=1}^{N}Y_i^2-N\bar Y^2) \\ &= \frac{(1-f)}{n}\frac{1}{N-1}( \sum_{i=1}^{N}Y_i^2-2\sum_{i=1}^{N}Y_i\bar Y+\sum_{i=1}^N\bar Y^2) \\ &= \frac{(1-f)}{n}\frac{1}{N-1}\sum_{i=1}^{N}(Y_i-\bar Y)^2 \\ &= \frac{1-f}{n}S^2 \\ \end{aligned}V(yˉ)=V(n1i=1nyi)=V(n1i=1NαiYi)=n21(i=1NYi2V(αi)+i=jYiYjCov(αi,αj))=n2f(1f)(i=1NYi2i=jYiYjN11)=nN(N1)(1f)(N1Ni=1NYi2N11(i=1NYi2+i=jYiYj))=nN(N1)(1f)(Ni=1NYi2(i=1NYi)2)=nN(N1)(1f)(Ni=1NYi2(NYˉ)2)=n(1f)N11(i=1NYi2NYˉ2)=n(1f)N11(i=1NYi22i=1NYiYˉ+i=1NYˉ2)=n(1f)N11i=1N(YiYˉ)2=n1fS2
  • 方差的样本估计的无偏性
    E ( v ( y ˉ ) ) = E ( 1 − f n s 2 ) = 1 − f n S 2 = E ( V ( y ˉ ) ) E(v(\bar y)) = E(\frac{1-f}{n}s^2) = \frac{1-f}{n}S^2=E(V(\bar y))E(v(yˉ))=E(n1fs2)=n1fS2=E(V(yˉ))
    只需要证明(证明的关键还是要把y i y_iyi找到合适的形式转到α i Y i \alpha_i Y_iαiYi
    ∵ E ( s 2 ) = E ( 1 n − 1 ∑ i = 1 n ( y i − y ˉ ) 2 ) = 1 n − 1 E ( ∑ i = 1 n y i 2 − n y ˉ 2 ) = 1 n − 1 [ E ( ∑ i = 1 N α i Y i 2 ) − n E y ˉ 2 ] = 1 n − 1 [ ∑ i = 1 N E ( α i ) Y i 2 − n ( V ( y ˉ ) + E ( y ˉ ) 2 ) ] = 1 n − 1 [ n N ∑ i = 1 N Y i 2 − n ( 1 − f n S 2 + Y ˉ 2 ) ] = n N ( n − 1 ) [ ∑ i = 1 N Y i 2 − N Y ˉ 2 − N n ( 1 − n N ) S 2 ) ] = n N ( n − 1 ) [ ∑ i = 1 N ( Y i − Y ˉ ) 2 − N n S 2 + S 2 ) ] = n N ( n − 1 ) ( N − 1 + 1 − N n ) S 2 = n N ( n − 1 ) ( N n − N n ) S 2 = S 2 ∴ E ( v ( y ˉ ) ) = E ( 1 − f n s 2 ) = 1 − f n S 2 = E ( V ( y ˉ ) ) \begin{aligned} \because E(s^2) &= E(\frac{1}{n-1}\sum_{i = 1}^n(y_i-\bar y)^2) \\ &= \frac{1}{n-1}E(\sum_{i = 1}^ny_i^2 - n\bar y^2)\\ &= \frac{1}{n-1}[E(\sum_{i = 1}^N\alpha_i Y_i^2) - nE\bar y^2]\\ &= \frac{1}{n-1}[\sum_{i = 1}^NE(\alpha_i) Y_i^2 - n(V(\bar y)+E(\bar y)^2)]\\ &= \frac{1}{n-1}[\frac{n}{N}\sum_{i = 1}^N Y_i^2 - n(\frac{1-f}{n}S^2+\bar Y^2)]\\ &= \frac{n}{N(n-1)}[\sum_{i = 1}^N Y_i^2 -N\bar Y^2 -\frac{N}{n}(1-\frac{n}{N})S^2)]\\ &= \frac{n}{N(n-1)}[\sum_{i = 1}^N(Y_i-\bar Y)^2 -\frac{N}{n}S^2+S^2)]\\ &= \frac{n}{N(n-1)}(N-1+1-\frac{N}{n})S^2\\ &= \frac{n}{N(n-1)}(\frac{Nn-N}{n})S^2 = S^2\\ \therefore E(v(\bar y)) &= E(\frac{1-f}{n}s^2) = \frac{1-f}{n}S^2=E(V(\bar y)) \end{aligned}E(s2)E(v(yˉ))=E(n11i=1n(yiyˉ)2)=n11E(i=1nyi2nyˉ2)=n11[E(i=1NαiYi2)nEyˉ2]=n11[i=1NE(αi)Yi2n(V(yˉ)+E(yˉ)2)]=n11[Nni=1NYi2n(n1fS2+Yˉ2)]=N(n1)n[i=1NYi2NYˉ2nN(1Nn)S2)]=N(n1)n[i=1N(YiYˉ)2nNS2+S2)]=N(n1)n(N1+1nN)S2=N(n1)n(nNnN)S2=S2=E(n1fs2)=n1fS2=E(V(yˉ))

2.2.2 总体总量的估计

总体总量的估计:总体均值的乘上N即可

2.2.3 总体比例的估计

总体比例的估计:Y i = { 0 , o t h e r w i s e 1 , 第 i 个 总 体 单 元 含 有 某 考 虑 的 特 征 i = 1 , 2 , . . . N Y_i = \left\{ _{0,otherwise}^{1,第i个总体单元含有某考虑的特征}\right.i=1,2,...NYi={0,otherwise1,ii=1,2,...N只是总体均值估计的特殊情况

比例P = A N = ∑ i = 1 N Y i N P=\frac{A}{N}=\frac{\sum_{i=1}^N Y_i}{N}P=NA=Ni=1NYi

2.2.4 两个估计的协方差

c o v ( y ˉ , x ˉ ) = E ( y ˉ − Y ˉ ) ( x ˉ − X ˉ ) cov(\bar y, \bar x) = E(\bar y - \bar Y)(\bar x - \bar X)cov(yˉ,xˉ)=E(yˉYˉ)(xˉXˉ)
可以证明
c o v ( y ˉ , x ˉ ) = 1 − f n S y x 其 中 , S y x = 1 N − 1 ∑ i = 1 N ( Y i − Y ˉ ) ( X i − X ˉ ) 为 总 体 协 方 差 cov(\bar y, \bar x) = \frac{1-f}{n}S_{yx} \\ 其中,S_{yx} = \frac{1}{N-1}\sum_{i=1}^N (Y_i-\bar Y)(X_i - \bar X)为总体协方差cov(yˉ,xˉ)=n1fSyxSyx=N11i=1N(YiYˉ)(XiXˉ)

此外,正如 2.2.1-2.2.3节中涉及到的估计量方差及其方差估计一样,y ˉ \bar yyˉx ˉ \bar xxˉ之间的协方差也有无偏估计——简单随机样本协方差

s y x = 1 n − 1 ∑ i = 1 n ( y i − y ˉ ) ( x i − x ˉ ) s a t i s f i e s , E ( s y x ) = S y x s_{yx}=\frac{1}{n-1}\sum_{i=1}^n(y_i-\bar y)(x_i - \bar x) \\ satisfies, E(s_{yx})=S_{yx}syx=n11i=1n(yiyˉ)(xixˉ)satisfies,E(syx)=Syx

2.3 比率估计

2.4 回归估计

2.5 Carry out a SRS

2.5.1 样本量的确定


版权声明:本文为weixin_45632492原创文章,遵循CC 4.0 BY-SA版权协议,转载请附上原文出处链接和本声明。