【概率论基础进阶】参数估计-点估计

烧灯续昼2002
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定义：用样本X1,X2,⋯ ,XnX_{1},X_{2},\cdots,X_{n}构造的统计量θ^(X1,X2,⋯ ,Xn)\hat{\theta }(X_{1},X_{2},\cdots,X_{n})来估计参数θ\theta称为点估计，统计量θ^(X1,X2,⋯ ,Xn)\hat{\theta }(X_{1},X_{2},\cdots,X_{n})称为估计量

 

估计量是随机变量，它所取得的观测值θ^(x1,x2,⋯ ,xn)\hat{\theta }(x_{1},x_{2},\cdots,x_{n})称为估计量，有时将θ\theta的估计量和估计值统称为θ\theta的估计

 

定义：设θ^\hat{\theta}是θ\theta的估计量，如果E(θ^)E(\hat{\theta})，则称θ^=θ^(X1,X2,⋯ ,Xn)\hat{\theta}=\hat{\theta}(X_{1},X_{2},\cdots,X_{n})是未知参数θ\theta的无偏估计量

 

例1：设总体XX的E(X)=μ,D(X)=σ2E(X)=\mu,D(X)=\sigma^{2}，从来自总体XX的样本X1,X2,⋯ ,XnX_{1},X_{2},\cdots,X_{n}得样本均值Xˉ=1n∑i=1nXi\bar{X}=\frac{1}{n}\sum\limits_{i=1}^{n}X_{i}和样本方差S2=1n−1∑i=1n(Xi−Xˉ)2S^{2}=\frac{1}{n-1}\sum\limits_{i=1}^{n}(X_{i}-\bar{X})^{2}，试证S2S^{2}是σ2\sigma^{2}的无偏估计

 

E(S2)=1n−1E{∑i=1n[(Xi−μ)−(Xˉ−μ)]2}=1n−1E{∑i=1n[(Xi−μ)2−2(Xi−μ)(Xˉ−μ)+(Xˉ−μ)2]}=1n−1E[∑i=1n(Xi−μ)2−n(Xˉ−μ)2]=1n−1[∑i=1nE(Xi−μ)2−nE(Xˉ−μ)2]=1n−1[nσ2−nD(Xˉ)]=σ2 \begin{aligned} E(S^{2})&=\frac{1}{n-1}E \left\{\sum\limits_{i=1}^{n}[(X_{i}-\mu)-(\bar{X}-\mu)]^{2}\right\}\\ &=\frac{1}{n-1}E \left\{\sum\limits_{i=1}^{n}[(X_{i}-\mu)^{2}-2(X_{i}-\mu)(\bar{X}-\mu)+(\bar{X}-\mu)^{2}]\right\}\\ &=\frac{1}{n-1}E\left[\sum\limits_{i=1}^{n}(X_{i}-\mu)^{2}-n(\bar{X}-\mu)^{2}\right]\\ &=\frac{1}{n-1}\left[\sum\limits_{i=1}^{n}E(X_{i}-\mu)^{2}-nE(\bar{X}-\mu)^{2}\right]\\ &=\frac{1}{n-1}[n \sigma^{2}-n D(\bar{X})]\\ &=\sigma^{2} \end{aligned}

 

之前做过用∑i=1n(Xi−Xˉ)2=∑i=1nXi2−nXˉ2\sum\limits_{i=1}^{n}(X_{i}-\bar{X})^{2}=\sum\limits_{i=1}^{n}X_{i}^{2}-n \bar{X}^{2}做的，这里不再赘述

∑i=1n(Xi−Xˉ)2=∑i=1nXi2−nXˉ2∑i=1n(Xi−Xˉ)2=∑i=1n(Xi−μ)2−n(Xˉ−μ)2\begin{aligned} \sum\limits_{i=1}^{n}(X_{i}-\bar{X})^{2}&=\sum\limits_{i=1}^{n}X_{i}^{2}-n \bar{X}^{2}\\ \sum\limits_{i=1}^{n}(X_{i}-\bar{X})^{2}&=\sum\limits_{i=1}^{n}(X_{i}-\mu)^{2}-n(\bar{X}-\mu)^{2}\end{aligned}

这两个公式都很重要

 

定义：设θ1^\hat{\theta_{1}}和θ2^\hat{\theta_{2}}都是θ\theta的无偏估计量，且D(θ1^)≤D(θ2^)D(\hat{\theta_{1}})\leq D(\hat{\theta_{2}})，则称θ1^\hat{\theta_{1}}比θ2^\hat{\theta_{2}}更有效，或θ1^\hat{\theta_{1}}比θ2^\hat{\theta_{2}}更有效估计量

 

例2：设总体的数学期望和方差分别为μ\mu和σ2\sigma^{2}，X1,X2X_{1},X_{2}是来自总体XX的样本。记X~=(1−a)X1+aX2\tilde{X}=(1-a)X_{1}+aX_{2}，确定aa，使D(X~)D(\tilde{X})最小

 

E(X~)=(1−a)E(X1)+aE(X2)=μD(X~)=(1−a)2D(X1)+a2D(X2)=[(1−a)2+a2]σ2 \begin{aligned} E(\tilde{X})&=(1-a)E(X_{1})+aE(X_{2})=\mu\\ D(\tilde{X})&=(1-a)^{2}D(X_{1})+a^{2}D(X_{2})\\ &=[(1-a)^{2}+a^{2}]\sigma^{2} \end{aligned}

显然当a=12\begin{aligned} a=\frac{1}{2}\end{aligned}时D(X~)D(\tilde{X})最小

 

可以证明如果样本为X1,X2,⋯ ,XnX_{1},X_{2},\cdots,X_{n}时，μ\mu的估计Xˉ=1n∑i=1nXi\begin{aligned} \bar{X}=\frac{1}{n}\sum\limits_{i=1}^{n}X_{i}\end{aligned}是形如

X~=∑i=1naiXi(其中∑i=1nai=1)\tilde{X}=\sum\limits_{i=1}^{n}a_{i}X_{i}(其中\sum\limits_{i=1}^{n}a_{i}=1)

的估计中最有效的

 

定义：设θ~(X1,X2,⋯ ,Xn)\tilde{\theta}(X_{1},X_{2},\cdots,X_{n})是θ\theta的估计，如果θ~\tilde{\theta}依概率收敛于θ\theta，则称θ~(X1,X2,⋯ ,Xn)\tilde{\theta}(X_{1},X_{2},\cdots,X_{n})为θ\theta的一致估计量

 

例3：设总体XX的概率密度为

f(x;θ)={2x3θθ<x<2θ0其他 f(x;\theta)=\left\{\begin{aligned}& \frac{2x}{3\theta}&\theta<x<2\theta\\&0&其他\end{aligned}\right.

其中θ\theta是未知参数，X1,X2,⋯ ,XnX_{1},X_{2},\cdots,X_{n}为来自总体的简单随机样本，已知E(c∑i=1nXi2)=θ2E(c \sum\limits_{i=1}^{n}X_{i}^{2})=\theta^{2}，则c=()c=()

 

E(c∑i=1nXi2)=c∑i=1nE(Xi2)=cnE(X2)=cn∫θ2θx2f(x;θ)dx=cn∫θ2θ2x33θ2dx=cn⋅52θ2=θ2⇒c=25n \begin{aligned} E(c \sum\limits_{i=1}^{n}X_{i}^{2})&=c \sum\limits_{i=1}^{n}E(X_{i}^{2})\\ &=cn E(X^{2})\\ &=cn \int_{\theta}^{2\theta}x^{2}f(x;\theta)dx\\ &=cn \int_{\theta}^{2\theta} \frac{2x^{3}}{3\theta^{2}}dx\\ &=cn \cdot \frac{5}{2}\theta^{2}=\theta^{2}\Rightarrow c=\frac{2}{5n} \end{aligned}

 

例4：设总体XX的概率密度为

f(x)={2e−x(x−θ)x>θ0x≤θ f(x)=\left\{\begin{aligned}&2e^{-x(x-\theta)}&x>\theta\\&0&x \leq \theta\end{aligned}\right.

其中θ>0\theta>0是未知参数，从总体XX中抽取简单随机样本X1,X2,⋯ ,XnX_{1},X_{2},\cdots,X_{n}，记θ^=min⁡{X1,X2,⋯ ,Xn}\hat{\theta}=\min \left\{X_{1},X_{2},\cdots,X_{n}\right\}

  • 求总体XX的分布函数F(x)F(x)

 

F(x)=P{X≤x}=∫−∞xf(x)dx={∫0x2e−2(x−θ)dx=1−e−2(x−θ)x>θ0x≤θ \begin{aligned} F(x)&=P \left\{X \leq x\right\}\\ &=\int_{-\infty}^{x}f(x)dx\\ &=\left\{\begin{aligned}&\int_{0}^{x}2e^{-2(x-\theta)}dx=1-e^{-2(x-\theta)}&x>\theta\\&0&x \leq \theta\end{aligned}\right. \end{aligned}

 

  • 求统计量θ^\hat{\theta}的分布函数Fθ^(x)F_{\hat{\theta}}(x)

 

Fθ^(X)=P{θ^≤x}=P{min⁡(X1,X2,⋯ ,Xn)≤x}=1−P{min⁡(X1,X2,⋯ ,Xn)>x}=1−P{X1>x,X2>x,⋯ ,Xn>x}=1−P{X1>x}P{X2>x}⋯P{Xn>x}=1−[1−F(x)]n={1−e2n(x−θ)x>θ0x≤θ \begin{aligned} F_{\hat{\theta}}(X)&=P \left\{\hat{\theta}\leq x\right\}\\ &=P \left\{\min (X_{1},X_{2},\cdots,X_{n})\leq x\right\}\\ &=1-P \left\{\min (X_{1},X_{2},\cdots,X_{n})>x\right\}\\ &=1-P \left\{X_{1}>x,X_{2}>x,\cdots ,X_{n}>x\right\}\\ &=1-P \left\{X_{1}>x\right\}P \left\{X_{2}>x\right\}\cdots P \left\{X_{n}>x\right\}\\ &=1-[1-F(x)]^{n}\\ &=\left\{\begin{aligned}&1-e^{2n(x-\theta)}&x>\theta\\&0&x \leq \theta\end{aligned}\right. \end{aligned}

 

  • 如果用θ^\hat{\theta}作为θ\theta的估计量，讨论它是否具有性质Eθ^=θE \hat{\theta}=\theta

 

fθ^(x)=Fθ^′(x)={2ne−2n(x−θ)x>θ0x≤θEθ^=∫−∞+∞xfθ^(x)dx=∫0+∞x⋅2ne−2n(x−θ)dx=θ+12θ≠θ \begin{aligned} f_{\hat{\theta}}(x)&=F'_{\hat{\theta}}(x)=\left\{\begin{aligned}&2ne^{-2n(x-\theta)}&x>\theta\\&0&x \leq \theta\end{aligned}\right.\\ E \hat{\theta}&=\int_{-\infty}^{+\infty}xf_{\hat{\theta}}(x)dx\\ &=\int_{0}^{+\infty}x \cdot 2ne^{-2n(x-\theta)}dx\\ &=\theta+ \frac{1}{2\theta}\ne \theta \end{aligned}

所以不具有性质Eθ^=θE \hat{\theta}=\theta

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