UA MATH564 概率論 樣本均值的偏度與峰度

• 偏度
• 峰度

假設$X_1,?,X_n$是一組簡單隨機樣本，$\bar{X}$是樣本均值，總體均值為$\mu$，總體方差為${\sigma }^2$，總體的偏度為${\gamma }_1$，峰度為${\gamma }_2$，下面計算樣本均值的偏度${\gamma }_{1,n}$與峰度${\gamma }_{2,n}$，先給出結論：
$${\gamma }_{1,n}=\frac{{\gamma }_1}{\sqrt{n}},{\gamma }_{2,n}=\frac{{\gamma }_2}{n}+\frac{3n(n?1)}{n^2}$$

偏度

偏度的定義是
$$\begin{gathered} {\gamma }_1=E{\left[\frac{X?\mu }{\sigma }\right]}^3=\frac{1}{{\sigma }^3}(E{X}^3?3\mu E{X}^2+3{\mu }^{2}EX?{\mu }^3) \\ =\frac{1}{{\sigma }^3}(E{X}^3?3\mu ({\mu }^2+{\sigma }^2)+3{\mu }^3?{\mu }^3)=\frac{E{X}^3?3\mu {\sigma }^2?{\mu }^3}{{\sigma }^3} \end{gathered}$$

我們知道
$$E\bar{X}=\mu ,Var(\bar{X})=\frac{{\sigma }^2}{n}$$

因此
$$\begin{gathered} {\gamma }_{1,n}=\frac{E{\bar{X}}^3?3\mu {\sigma }^2/n?{\mu }^3}{{\sigma }^3/n^{3/2}} \\ E{\bar{X}}^3=\frac{1}{n^3}E{\left(\sum _{i=1}^n{X}_i\right)}^3 \end{gathered}$$

注意到${\left({\sum }_{i=1}^n{X}_i\right)}^3$的展開式一共有$n^3$項，其中有$A_n^1$項是$X_i^3$，有$C_3^1{C}_2^2{A}_n^2$項是$X_i^2{X}_j,i=j$，有$A_n^3$項是$X_i{X}_j{X}_k,i=j=k$
$$\begin{gathered} E[X_i{X}_j{X}_k]=E{X}_{i}E{X}_{j}E{X}_k={\mu }^3 \\ E[X_i^2{X}_j]=E{X}_i^{2}E{X}_j=({\mu }^2+{\sigma }^2)\mu ={\mu }^3+\mu {\sigma }^2 \\ E[X_i^3]={\mu }^3+3\mu {\sigma }^2+{\gamma }_1{\sigma }^3 \end{gathered}$$

因此
$$\begin{gathered} E{\bar{X}}^3=\frac{n({\mu }^3+3\mu {\sigma }^2+{\gamma }_1{\sigma }^3)+3n(n?1)({\mu }^3+\mu {\sigma }^2)+n(n?1)(n?2){\mu }^3}{n^3} \\ =\frac{{\sigma }^3{\gamma }_1}{n^2}+\frac{3\mu {\sigma }^2}{n}+{\mu }^3 \\ E{\bar{X}}^3?3\mu {\sigma }^2/n?{\mu }^3=\frac{{\sigma }^3{\gamma }_1}{n^2} \end{gathered}$$

所以樣本均值的偏度為
$${\gamma }_{1,n}=\frac{{\gamma }_1}{\sqrt{n}}$$

峰度

偏度的定義是（另一種定義是在這個基礎上減3，3是標準正態分布的峰度，用來作參考）
$$\begin{gathered} {\gamma }_2=E{\left[\frac{X?\mu }{\sigma }\right]}^4=\frac{1}{{\sigma }^4}(E{X}^4?4\mu E{X}^3+6{\mu }^{2}E{X}^2?4{\mu }^{3}EX+{\mu }^4) \\ =\frac{1}{{\sigma }^4}(E{X}^4?4\mu ({\mu }^3+3\mu {\sigma }^2+{\gamma }_1{\sigma }^3)+6{\mu }^2({\mu }^2+{\sigma }^2)?4{\mu }^4+{\mu }^4) \\ =\frac{E{X}^4?4\mu {\gamma }_1{\sigma }^3?6{\mu }^2{\sigma }^2?{\mu }^4}{{\sigma }^4} \end{gathered}$$

我們知道
$$E\bar{X}=\mu ,Var(\bar{X})=\frac{{\sigma }^2}{n},{\gamma }_{1,n}=\frac{{\gamma }_1}{\sqrt{n}}$$

因此
$$\begin{gathered} {\gamma }_{2,n}=\frac{E{\bar{X}}^4?4\mu {\gamma }_1{\sigma }^3/n^2?6{\mu }^2{\sigma }^2/n?{\mu }^4}{{\sigma }^4/n^2} \\ E{\bar{X}}^4=\frac{1}{n^4}E{\left(\sum _{i=1}^n{X}_i\right)}^4 \end{gathered}$$

注意到${\left({\sum }_{i=1}^n{X}_i\right)}^4$的展開式一共有$n^4$項，其中有$A_n^1$項是$X_i^4$，有$C_4^2{C}_n^2$項是$X_i^2{X}_j^2,i=j$，有$C_4^1{C}_3^3{A}_n^2$項是$X_i^3{X}_j,i=j$，有$C_4^2{C}_2^1{C}_1^1{A}_n^3$項是$X_i^2{X}_j{X}_k,i=j=k$，有$A_n^4$項是$X_i{X}_j{X}_k{X}_l,i=j=k=l$
$$\begin{gathered} E[X_i{X}_j{X}_k{X}_l]=E{X}_{i}E{X}_{j}E{X}_{k}E{X}_l={\mu }^4 \\ E[X_i^2{X}_j{X}_k]=E{X}_i^{2}E{X}_{j}E{X}_k=({\mu }^2+{\sigma }^2){\mu }^2={\mu }^4+{\mu }^2{\sigma }^2 \\ E[X_i^3{X}_j]=E{X}_i^{3}E{X}_j=({\mu }^3+3\mu {\sigma }^2+{\gamma }_1{\sigma }^3)\mu ={\mu }^4+3{\mu }^2{\sigma }^2+\mu {\gamma }_1{\sigma }^3 \\ E[X_i^2{X}_j^2]=E{X}_i^{2}E{X}_j^2=({\mu }^2+{\sigma }^2{)}^2={\mu }^4+2{\mu }^2{\sigma }^2+{\sigma }^4 \\ E{X}_i^4={\sigma }^4{\gamma }_2+4\mu {\gamma }_1{\sigma }^3+6{\mu }^2{\sigma }^2+{\mu }^4 \end{gathered}$$

計算
$$\begin{gathered} n({\sigma }^4{\gamma }_2+4\mu {\gamma }_1{\sigma }^3+6{\mu }^2{\sigma }^2+{\mu }^4)+3n(n?1)({\mu }^4+2{\mu }^2{\sigma }^2+{\sigma }^4) \\ +4n(n?1)({\mu }^4+3{\mu }^2{\sigma }^2+\mu {\gamma }_1{\sigma }^3)+6n(n?1)(n?2)({\mu }^4+{\mu }^2{\sigma }^2) \\ +n(n?1)(n?2)(n?3){\mu }^4=n{\sigma }^4{\gamma }_2+4n^2\mu {\gamma }_1{\sigma }^2+3n(n?1){\sigma }^4+n^4{\mu }^4+6n^3{\mu }^2{\sigma }^2 \end{gathered}$$

也就是
$$\begin{gathered} E{\bar{X}}^4=\frac{n{\sigma }^4{\gamma }_2+4n^2\mu {\gamma }_1{\sigma }^2+3n(n?1){\sigma }^4+n^4{\mu }^4+6n^3{\mu }^2{\sigma }^2}{n^4} \\ =\frac{{\sigma }^4}{n^3}{\gamma }_2+\frac{4\mu {\gamma }_1{\sigma }^2}{n^2}+{\mu }^4+\frac{6}{n}{\mu }^2{\sigma }^2+\frac{3n(n?1)}{n^4}{\sigma }^4 \end{gathered}$$

因此
$${\gamma }_{2,n}=\frac{{\gamma }_2}{n}+\frac{3n(n?1)}{n^2}$$

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