UA MATH564 概率论 样本均值的偏度与峰度
偏度峰度假设X1,⋯,XnX_1,\cdots,X_nX1,⋯,Xn是一组简单随机样本，Xˉ\bar{X}Xˉ是样本均值，总体均值为μ\muμ，总体方差为σ2\sigma^2σ2，总体的偏度为γ1\gamma_1γ1，峰度为γ2\gamma_2γ2，下面计算样本均值的偏度γ1,n\gamma_{1,n}γ1,n与峰度γ2,n\gamma_{2,n}γ2,n，先给出结论：
γ1,n=γ1n,γ2,n=γ2n+3n(n−1)n2\gamma_{1,n} = \frac{\gamma_1}{\sqrt{n}},\ \ \gamma_{2,n} = \frac{\gamma_2}{n} + \frac{3n(n-1)}{n^2}γ1,n=nγ1,γ2,n=nγ2+n23n(n−1)
偏度
偏度的定义是
γ1=E[X−μσ]3=1σ3(EX3−3μEX2+3μ2EX−μ3)=1σ3(EX3−3μ(μ2+σ2)+3μ3−μ3)=EX3−3μσ2−μ3σ3\gamma_1 = E \left[ \frac{X-\mu}{\sigma} \right]^3 = \frac{1}{\sigma^3}( EX^3 - 3\mu EX^2 + 3\mu^2 EX - \mu^3) \\ = \frac{1}{\sigma^3}( EX^3 - 3\mu (\mu^2 + \sigma^2) + 3\mu^3 - \mu^3) = \frac{EX^3-3\mu\sigma^2-\mu^3}{\sigma^3}γ1=E[σX−μ]3=σ31(EX3−3μEX2+3μ2EX−μ3)=σ31(EX3−3μ(μ2+σ2)+3μ3−μ3)=σ3EX3−3μσ2−μ3
我们知道
EXˉ=μ,Var(Xˉ)=σ2nE\bar{X} = \mu,\ \ Var(\bar{X}) = \frac{\sigma^2}{n}EXˉ=μ,Var(Xˉ)=nσ2
因此
γ1,n=EXˉ3−3μσ2/n−μ3σ3/n3/2EXˉ3=1n3E(∑i=1nXi)3\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)^3γ1,n=σ3/n3/2EXˉ3−3μσ2/n−μ3EXˉ3=n31E(i=1∑nXi)3
注意到(∑i=1nXi)3\left( \sum_{i=1}^n X_i \right)^3(∑i=1nXi)3的展开式一共有n3n^3n3项，其中有An1A_n^1An1项是Xi3X_i^3Xi3，有C31C22An2C_3^1C_2^2A_n^2C31C22An2项是Xi2Xj,i≠jX_i^2X_j,i \ne jXi2Xj,i=j，有An3A_n^3An3项是XiXjXk,i≠j≠kX_iX_jX_k,i \ne j \ne kXiXjXk,i=j=k
E[XiXjXk]=EXiEXjEXk=μ3E[Xi2Xj]=EXi2EXj=(μ2+σ2)μ=μ3+μσ2E[Xi3]=μ3+3μσ2+γ1σ3E[X_iX_jX_k] = EX_iEX_jEX_k = \mu^3 \\ E[X_i^2X_j] = EX_i^2EX_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 E[XiXjXk]=EXiEXjEXk=μ3E[Xi2Xj]=EXi2EXj=(μ2+σ2)μ=μ3+μσ2E[Xi3]=μ3+3μσ2+γ1σ3
因此
EXˉ3=n(μ3+3μσ2+γ1σ3)+3n(n−1)(μ3+μσ2)+n(n−1)(n−2)μ3n3=σ3γ1n2+3μσ2n+μ3EXˉ3−3μσ2/n−μ3=σ3γ1n2E\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} EXˉ3=n3n(μ3+3μσ2+γ1σ3)+3n(n−1)(μ3+μσ2)+n(n−1)(n−2)μ3=n2σ3γ1+n3μσ2+μ3EXˉ3−3μσ2/n−μ3=n2σ3γ1
所以样本均值的偏度为
γ1,n=γ1n\gamma_{1,n} = \frac{\gamma_1}{\sqrt{n}}γ1,n=nγ1
峰度
偏度的定义是（另一种定义是在这个基础上减3，3是标准正态分布的峰度，用来作参考）
γ2=E[X−μσ]4=1σ4(EX4−4μEX3+6μ2EX2−4μ3EX+μ4)=1σ4(EX4−4μ(μ3+3μσ2+γ1σ3)+6μ2(μ2+σ2)−4μ4+μ4)=EX4−4μγ1σ3−6μ2σ2−μ4σ4\gamma_2 = E \left[ \frac{X-\mu}{\sigma} \right]^4 = \frac{1}{\sigma^4}( EX^4 - 4\mu EX^3 + 6\mu^2 EX^2 - 4\mu^3EX + \mu^4) \\ = \frac{1}{\sigma^4}( EX^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{EX^4 - 4\mu\gamma_1 \sigma^3 - 6\mu^2\sigma^2 - \mu^4}{\sigma^4}γ2=E[σX−μ]4=σ41(EX4−4μEX3+6μ2EX2−4μ3EX+μ4)=σ41(EX4−4μ(μ3+3μσ2+γ1σ3)+6μ2(μ2+σ2)−4μ4+μ4)=σ4EX4−4μγ1σ3−6μ2σ2−μ4
我们知道
EXˉ=μ,Var(Xˉ)=σ2n,γ1,n=γ1nE\bar{X} = \mu,\ \ Var(\bar{X}) = \frac{\sigma^2}{n}, \ \ \gamma_{1,n} = \frac{\gamma_1}{\sqrt{n}}EXˉ=μ,Var(Xˉ)=nσ2,γ1,n=nγ1
因此
γ2,n=EXˉ4−4μγ1σ3/n2−6μ2σ2/n−μ4σ4/n2EXˉ4=1n4E(∑i=1nXi)4\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)^4γ2,n=σ4/n2EXˉ4−4μγ1σ3/n2−6μ2σ2/n−μ4EXˉ4=n41E(i=1∑nXi)4
注意到(∑i=1nXi)4\left( \sum_{i=1}^n X_i \right)^4(∑i=1nXi)4的展开式一共有n4n^4n4项，其中有An1A_n^1An1项是Xi4X_i^4Xi4，有C42Cn2C_4^2C_n^2C42Cn2项是Xi2Xj2,i≠jX_i^2X_j^2,i \ne jXi2Xj2,i=j，有C41C33An2C_4^1C_3^3A_n^2C41C33An2项是Xi3Xj,i≠jX_i^3X_j,i \ne jXi3Xj,i=j，有C42C21C11An3C_4^2C_2^1C_1^1A_n^3C42C21C11An3项是Xi2XjXk,i≠j≠kX_i^2X_jX_k,i \ne j \ne kXi2XjXk,i=j=k，有An4A_n^4An4项是XiXjXkXl,i≠j≠k≠lX_iX_jX_kX_l,i\ne j \ne k \ne lXiXjXkXl,i=j=k=l
E[XiXjXkXl]=EXiEXjEXkEXl=μ4E[Xi2XjXk]=EXi2EXjEXk=(μ2+σ2)μ2=μ4+μ2σ2E[Xi3Xj]=EXi3EXj=(μ3+3μσ2+γ1σ3)μ=μ4+3μ2σ2+μγ1σ3E[Xi2Xj2]=EXi2EXj2=(μ2+σ2)2=μ4+2μ2σ2+σ4EXi4=σ4γ2+4μγ1σ3+6μ2σ2+μ4E[X_iX_jX_kX_l] = EX_iEX_jEX_k EX_l= \mu^4\\ E[X_i^2X_jX_k] = EX_i^2EX_jEX_k = (\mu^2 + \sigma^2)\mu^2 = \mu^4 + \mu^2 \sigma^2 \\ E[X_i^3X_j] = EX_i^3EX_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^2X_j^2] = EX_i^2EX_j^2 = (\mu^2 + \sigma^2)^2 = \mu^4 + 2\mu^2 \sigma^2 + \sigma^4 \\ EX_i^4 = \sigma^4 \gamma_2+ 4\mu\gamma_1 \sigma^3 + 6\mu^2\sigma^2 + \mu^4E[XiXjXkXl]=EXiEXjEXkEXl=μ4E[Xi2XjXk]=EXi2EXjEXk=(μ2+σ2)μ2=μ4+μ2σ2E[Xi3Xj]=EXi3EXj=(μ3+3μσ2+γ1σ3)μ=μ4+3μ2σ2+μγ1σ3E[Xi2Xj2]=EXi2EXj2=(μ2+σ2)2=μ4+2μ2σ2+σ4EXi4=σ4γ2+4μγ1σ3+6μ2σ2+μ4
计算
n(σ4γ2+4μγ1σ3+6μ2σ2+μ4)+3n(n−1)(μ4+2μ2σ2+σ4)+4n(n−1)(μ4+3μ2σ2+μγ1σ3)+6n(n−1)(n−2)(μ4+μ2σ2)+n(n−1)(n−2)(n−3)μ4=nσ4γ2+4n2μγ1σ2+3n(n−1)σ4+n4μ4+6n3μ2σ2n(\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^2n(σ4γ2+4μγ1σ3+6μ2σ2+μ4)+3n(n−1)(μ4+2μ2σ2+σ4)+4n(n−1)(μ4+3μ2σ2+μγ1σ3)+6n(n−1)(n−2)(μ4+μ2σ2)+n(n−1)(n−2)(n−3)μ4=nσ4γ2+4n2μγ1σ2+3n(n−1)σ4+n4μ4+6n3μ2σ2
也就是
EXˉ4=nσ4γ2+4n2μγ1σ2+3n(n−1)σ4+n4μ4+6n3μ2σ2n4=σ4n3γ2+4μγ1σ2n2+μ4+6nμ2σ2+3n(n−1)n4σ4E\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^4EXˉ4=n4nσ4γ2+4n2μγ1σ2+3n(n−1)σ4+n4μ4+6n3μ2σ2=n3σ4γ2+n24μγ1σ2+μ4+n6μ2σ2+n43n(n−1)σ4
因此
γ2,n=γ2n+3n(n−1)n2\gamma_{2,n} = \frac{\gamma_2}{n} + \frac{3n(n-1)}{n^2}γ2,n=nγ2+n23n(n−1)
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