统计量由来与定义
如前所述,在进行统计推断时,构造样本的适当的函数是关键,为此,我们给出下列定义.
定义:设 ( X 1 , X 2 , . . . , X n ) ({X_1},{X_2},...,{X_n}) (X1,X2,...,Xn)为总体 X {X} X的一个样本,称此样本的任一不含总体分布未知参数的函数为该样本的统计量。
举例说明:设总体 X {X} X服从正态分布, E ( X ) = 5 {E(X)=5} E(X)=5, D ( X ) = σ 2 D(X) = {\sigma ^2} D(X)=σ2, σ {\sigma} σ未知。 ( X 1 , X 2 , . . . , X n ) ({X_1},{X_2},...,{X_n}) (X1,X2,...,Xn)为总体 X {X} X的一个样本,令:
X ˉ = X 1 + X 2 + . . . + X n n \bar X = {
{
{X_1} + {X_2} + ... + {X_n}} \over n} Xˉ=nX1+X2+...+Xn
U = X 1 + X 2 + . . . + X n n σ U = {
{
{X_1} + {X_2} + ... + {X_n}} \over {n\sigma }} U=nσX1+X2+...+Xn
那么 X ˉ \bar X Xˉ为该样本的统计量,而 U {U} U不是,因为其表达式中含有了 X {X} X的未知参数 σ \sigma σ
常用的统计量
设 ( X 1 , X 2 , . . . , X n ) ({X_1},{X_2},...,{X_n}) (X1,X2,...,Xn)为总体 X {X} X的一个样本
样本均值
称样本的算术平均值为样本均值,记为 X ˉ \bar X Xˉ,即
X ˉ = X 1 + X 2 + . . . + X n n \bar X = {
{
{X_1} + {X_2} + ... + {X_n}} \over n} Xˉ=nX1+X2+...+Xn
样本方差
样本方差是用来描述样本中诸分量与样本均值的均方差异的,它有两种定义方式。
未修正样本方差
S 0 2 = 1 n ∑ i = 1 n ( X i − X ˉ ) 2 S_0^2 = {1 \over n}\sum\limits_{i = 1}^n { { {\left( { {X_i} - \bar X} \right)}^2}} S02=n1i=1∑n(Xi−Xˉ)2
修正样本方差
S 0 2 = 1 n − 1 ∑ i = 1 n ( X i − X ˉ ) 2 S_0^2 = {1 \over {n - 1}}\sum\limits_{i = 1}^n { { {\left( { {X_i} - \bar X} \right)}^2}} S02=n−11i=1∑n(Xi−Xˉ)2
由于存在数学关系 E ( S 2 ) = σ 2 E({S^2}) = {\sigma ^2} E(S2)=σ2(后面证明),修正样本方差具有更好的统计性质,故样本方差均采用修正样本方差。
样本标准差
样本标准差定义为样本方差的算术平方根
S = 1 n − 1 ∑ i = 1 n ( X i − X ˉ ) 2 S = \sqrt { {1 \over {n - 1}}\sum\limits_{i = 1}^n { { {\left( { {X_i} - \bar X} \right)}^2}} } S=n−11i=1∑n(Xi−Xˉ)2
样本原点矩
k阶原点矩为
A k = 1 n ∑ i = 1 n X i k {A_k} = {1 \over n}\sum\limits_{i = 1}^n {X_i^k} Ak=n1i=1∑nXik
样本中心距
k阶中心距为
B k = 1 n ∑ i = 1 n ( X i − X ˉ ) k {B_k} = {1 \over n}\sum\limits_{i = 1}^n { { {\left( { {X_i} - \bar X} \right)}^k}} Bk=n1i=1∑n(Xi−Xˉ)k
样本均值与方差性质讨论
均值性质
如果总体 X {X} X具有数学期望$E(X) = \mu $,则
E ( X ˉ ) = E ( X ) = σ E(\bar X) = E(X) = \sigma E(Xˉ)=E(X)=σ
结论显然,证明从略。
方差性质
如果总体 X {X} X具有方差 D ( X ) = σ 2 D(X) = {\sigma ^2} D(X)=σ2,则
D ( X ˉ ) = 1 n D ( X ) = σ 2 n D(\bar X) = {1 \over n}D(X) = {
{
{\sigma ^2}} \over n} D(Xˉ)=n1D(X)=nσ2
E ( S 2 ) = D ( X ) = σ 2 E({S^2}) = D(X) = {\sigma ^2} E(S2)=D(X)=σ2
证明:
对于 D ( X ˉ ) = 1 n D ( X ) = σ 2 n D(\bar X) = {1 \over n}D(X) = { { {\sigma ^2}} \over n} D(Xˉ)=n1D(X)=nσ2:
D ( X ˉ ) = D ( X 1 + X 2 + . . . + X n n ) D(\bar X) = D({ { {X_1} + {X_2} + ... + {X_n}} \over n}) D(Xˉ)=D(nX1+X2+...+Xn)
= D ( X 1 n ) + D ( X 2 n ) + . . . + D ( X n n ) = D({ { {X_1}} \over n}) + D({ { {X_2}} \over n}) + ... + D({ { {X_n}} \over n}) =D(nX1)+D(nX2)+...+D(nXn)
= 1 n 2 D ( X 1 ) + 1 n 2 D ( X 2 ) + . . . + 1 n 2 D ( X n ) = {1 \over { {n^2}}}D({X_1}) + {1 \over { {n^2}}}D({X_2}) + ... + {1 \over { {n^2}}}D({X_n}) =n21D(X1)+n21D(X2)+...+n21D(Xn)
= 1 n 2 D ( X ) × n = D ( X ) n = σ 2 n = {1 \over { {n^2}}}D(X) \times n = { {D(X)} \over n} = { { {\sigma ^2}} \over n} =n21D(X)×n=nD(X)=nσ2
对于 E ( S 2 ) = D ( X ) = σ 2 E({S^2}) = D(X) = {\sigma ^2} E(S2)=D(X)=σ2:
E ( S 2 ) = E ( ∑ i = 1 n ( X i − X ˉ ) 2 n − 1 ) E({S^2}) = E\left( { { {\sum\limits_{i = 1}^n { { {\left( { {X_i} - \bar X} \right)}^2}} } \over {n - 1}}} \right) E(S2)=E⎝⎜⎜⎛n−1i=1∑n(Xi−Xˉ)2⎠⎟⎟⎞
= 1 n − 1 E ( ∑ i = 1 n ( X i − X ˉ ) 2 ) = {1 \over {n - 1}}E(\sum\limits_{i = 1}^n { { {\left( { {X_i} - \bar X} \right)}^2}} ) =n−11E(i=1∑n(Xi−Xˉ)2)
= 1 n − 1 E ( ∑ i = 1 n ( X i 2 − 2 X i X ˉ + X ˉ 2 ) ) = {1 \over {n - 1}}E(\sum\limits_{i = 1}^n {\left( {X_i^2 - 2{X_i}\bar X + { {\bar X}^2}} \right)} ) =n−11E(i=1∑n(Xi2−2XiXˉ+Xˉ2))
= 1 n − 1 E ( ∑ i = 1 n X i 2 − 2 ∑ i = 1 n X i X ˉ + ∑ i = 1 n X ˉ 2 ) = {1 \over {n - 1}}E(\sum\limits_{i = 1}^n {X_i^2} - 2\sum\limits_{i = 1}^n { {X_i}\bar X} + \sum\limits_{i = 1}^n { { {\bar X}^2}} ) =n−11E(i=1∑nXi2−2i=1∑nXiXˉ+i=1∑nXˉ2)
= 1 n − 1 E ( ∑ i = 1 n X i 2 − 2 n X ˉ 2 + ∑ i = 1 n X ˉ 2 ) = {1 \over {n - 1}}E(\sum\limits_{i = 1}^n {X_i^2} - 2n{\bar X^2} + \sum\limits_{i = 1}^n { { {\bar X}^2}} ) =n−11E(i=1∑nXi2−2nXˉ2+i=1∑nXˉ2)
= 1 n − 1 E ( ∑ i = 1 n X i 2 − n X ˉ 2 ) = {1 \over {n - 1}}E(\sum\limits_{i = 1}^n {X_i^2} - n{\bar X^2}) =n−11E(i=1∑nXi2−nXˉ2)
= 1 n − 1 ( ∑ i = 1 n E ( X i 2 ) − n E ( X ˉ 2 ) ) = {1 \over {n - 1}}\left( {\sum\limits_{i = 1}^n {E(X_i^2)} - nE({ {\bar X}^2})} \right) =n−11(i=1∑nE(Xi2)−nE(Xˉ2))
由方差公式,有: E ( X 2 ) = D ( X ) + ( E ( X ) ) 2 E({X^2}) = D(X) + {(E(X))^2} E(X2)=D(X)+(E(X))2,代入即得上式:
= 1 n − 1 ( ∑ i = 1 n D ( X i ) + ( E ( X i ) ) 2 − n ( D ( X ˉ ) + ( E ( X ˉ ) ) 2 ) ) = {1 \over {n - 1}}\left( {\sum\limits_{i = 1}^n {D({X_i}) + (E({X_i})} {)^2} - n\left( {D(\bar X) + { {(E(\bar X))}^2}} \right)} \right) =n−11(i=1∑nD(Xi)+(E(Xi))2−n(D(Xˉ)+(E(Xˉ))2))
= 1 n − 1 ( ∑ i = 1 n σ 2 + μ 2 − n ( σ 2 n + μ 2 ) ) = {1 \over {n - 1}}\left( {\sum\limits_{i = 1}^n { {\sigma ^2} + {\mu ^2}} - n\left( { { { {\sigma ^2}} \over n} + {\mu ^2}} \right)} \right) =n−11(i=1∑nσ2+μ2−n(nσ2+μ2))
= 1 n − 1 ( n σ 2 + n μ 2 − n ( σ 2 n + μ 2 ) ) = {1 \over {n - 1}}\left( {n{\sigma ^2} + n{\mu ^2} - n\left( { { { {\sigma ^2}} \over n} + {\mu ^2}} \right)} \right) =n−11(nσ2+nμ2−n(nσ2+μ2))
= σ 2 = {\sigma ^2} =σ2