题目
18.设x1,x2,···,xn是总体N(μ,σ^2)的简单随机样本.记-|||-overline (X)=dfrac (1)(n)sum _(i=1)^nX , ^2=dfrac (1)(n-1)sum _(i=1)^n(({X)_(i)-overline (X))}^2 , =(overline {X)}^2-dfrac (1)(n)(S)^2-|||-(1)证明T是μ^2的无偏估计量;-|||-(2)当 mu =0 , =1 时,求D(T).
题目解答
答案
解析
步骤 1：证明T是μ^2的无偏估计量
根据无偏估计的定义，我们需要证明E(T) = μ^2。首先，我们计算E(T)。
步骤 2：计算E(T)
$E(T)=E({\overline {X}}^{2}-\dfrac {1}{n}{S}^{2})=E({\overline {X}}^{2})-\dfrac {1}{n}E({S}^{2})$
$={[ E(\overline {X})] }^{2}+D(\overline {X})-\dfrac {1}{n}E({S}^{2})$
$={\mu }^{2}+\dfrac {{\sigma }^{2}}{n}-\dfrac {{\sigma }^{2}}{n}={\mu }^{2}$
步骤 3：计算D(T)当 $\mu =0$ , $0=1$ 时
方法一:当 $\mu =0$ , $0=1$ 时,有 $D(T)=D({x}^{2}-\dfrac {1}{n}{S}^{2})$
$=D({x}^{2})+\dfrac {1}{{n}^{2}}D({s}^{2})=\dfrac {1}{{n}^{2}}D[ {(\sqrt {n}x)}^{2}] +\dfrac {1}{{n}^{2}}\cdot \dfrac {1}{{(n-1)}^{2}}D[ $
$=1/22+ \dfrac {1}{{n}^{2}}\cdot \dfrac {1}{{(n-1)}^{2}}\cdot 2(n-1)=\dfrac {2}{{n}^{2}}(1+\dfrac {1}{n-1})=\dfrac {2}{n(n-1)}$
方法二: $D(T)=E({T}^{2})-{[ E(T)] }^{2}$ , E(T)=0 , $E({S}^{2})={\sigma }^{2}=1$ ,
根据无偏估计的定义，我们需要证明E(T) = μ^2。首先，我们计算E(T)。
步骤 2：计算E(T)
$E(T)=E({\overline {X}}^{2}-\dfrac {1}{n}{S}^{2})=E({\overline {X}}^{2})-\dfrac {1}{n}E({S}^{2})$
$={[ E(\overline {X})] }^{2}+D(\overline {X})-\dfrac {1}{n}E({S}^{2})$
$={\mu }^{2}+\dfrac {{\sigma }^{2}}{n}-\dfrac {{\sigma }^{2}}{n}={\mu }^{2}$
步骤 3：计算D(T)当 $\mu =0$ , $0=1$ 时
方法一:当 $\mu =0$ , $0=1$ 时,有 $D(T)=D({x}^{2}-\dfrac {1}{n}{S}^{2})$
$=D({x}^{2})+\dfrac {1}{{n}^{2}}D({s}^{2})=\dfrac {1}{{n}^{2}}D[ {(\sqrt {n}x)}^{2}] +\dfrac {1}{{n}^{2}}\cdot \dfrac {1}{{(n-1)}^{2}}D[ $
$=1/22+ \dfrac {1}{{n}^{2}}\cdot \dfrac {1}{{(n-1)}^{2}}\cdot 2(n-1)=\dfrac {2}{{n}^{2}}(1+\dfrac {1}{n-1})=\dfrac {2}{n(n-1)}$
方法二: $D(T)=E({T}^{2})-{[ E(T)] }^{2}$ , E(T)=0 , $E({S}^{2})={\sigma }^{2}=1$ ,