${\displaystyle X\sim B(n,p)}$, ${\displaystyle p}$是未知参数. ${\displaystyle g(p)=\sin p}$. 因为${\displaystyle X\in \{0,1,\cdots ,n\}}$, 故要给出估计${\displaystyle \hat{g}}$, 只需指出${\displaystyle a_i=\hat{g}(i)}$即可. 若${\displaystyle \hat{g}}$无偏, 则$$E_p[\hat{g}(X)]=\sum _{i=0}^n{a}_i(\genfrac{}{}{0ex}{}{n}{i})p^i(1-p{)}^{n-i}=g(p)=\sin p,$$
但是多项式一定不是超越函数, 因此${\displaystyle \sin p}$没有无偏估计.

${\displaystyle T=T(X)}$是一个 充分统计量, ${\displaystyle \hat{g}}$是${\displaystyle g(\theta )}$的一个无偏估计, 则存在${\displaystyle g(\theta )}$的无偏估计${\displaystyle h(T(X))}$, 使得$$\begin{array}{}\text{(1.3)}& {\mathrm{Var}}_{\theta }(h(T(X)))\le {\mathrm{Var}}_{\theta }(\hat{g}(X)),\end{array}$$等号当且仅当${\displaystyle \hat{g}(X)}$能表为${\displaystyle T(X)}$的函数.

${\displaystyle T}$是充分统计量, 故按定义, 给定${\displaystyle T}$时, ${\displaystyle X}$的条件分布与${\displaystyle \theta }$无关, 故${\displaystyle E_{\theta }(\hat{g}(X)|T)}$与${\displaystyle \theta }$无关, 记为${\displaystyle h(T)}$. 又${\displaystyle \hat{g}}$为无偏估计, 故$$\begin{array}{}\text{(塔式法则)}& E_{\theta }(h(T(X)))=E_{\theta }{E}_{\theta }(\hat{g}(X)|T)=E_{\theta }(\hat{g}(X))=g(\theta )\end{array}$$故${\displaystyle h(T(X))}$是${\displaystyle g(\theta )}$的无偏估计. 现任取${\displaystyle {\theta }_0\in \mathrm{\Theta }}$, 有$$h^2(T(X))=[E_{{\theta }_0}(\hat{g}(X)|T){]}^2\le E_{{\theta }_0}({\hat{g}}^2(X)|T).$$套期望: $$E_{{\theta }_0}[h^2(T(X))]\le E_{{\theta }_0}{E}_{{\theta }_0}({\hat{g}}^2(X)|T)=E_{{\theta }_0}({\hat{g}}^2(X)),$$进而$$\begin{array}{rl}{\mathrm{Var}}_{{\theta }_0}(h(T(X)))=& E_{{\theta }_0}(h^2(T(X)))-g^2({\theta }_0)\\ \le & E_{{\theta }_0}({\hat{g}}^2(X))-g^2({\theta }_0)\\ =& {\mathrm{Var}}_{{\theta }_0}(\hat{g}(X)),\end{array}$$
等号成立时${\displaystyle {\mathrm{Var}}_{{\theta }_0}(\hat{g}(X)|T)=0}$, 也即给定${\displaystyle T}$时${\displaystyle \hat{g}(X)}$是常数, 也即${\displaystyle \hat{g}(X)}$可表为${\displaystyle T}$的函数.

取任一无偏估计${\displaystyle {\hat{g}}_1(X)}$, 记${\displaystyle l(X)={\hat{g}}_1(X)-\hat{g}(X)}$, 则 (2.1) 成立, 故$$\begin{array}{rl}{\mathrm{Var}}_{\theta }({\hat{g}}_1(X))=& {\mathrm{Var}}_{\theta }(\hat{g}(X)+l(X))\\ =& {\mathrm{Var}}_{\theta }(\hat{g}(X))+{\mathrm{Var}}_{\theta }(l(X))+2{\mathrm{Cov}}_{\theta }(\hat{g}(X),l(X))\\ =& {\mathrm{Var}}_{\theta }(\hat{g}(X))+{\mathrm{Var}}_{\theta }(l(X))\ge {\mathrm{Var}}_{\theta }(\hat{g}(X)).\end{array}$$

记${\displaystyle T(X)=X_1+\cdots +X_n}$, 依据 这个例子, ${\displaystyle T}$是充分统计量; 依据 引理1.1, UMVUE 的寻找范围可以缩小到可表为${\displaystyle T}$的函数的无偏估计类. 现取${\displaystyle \hat{g}=\frac{T}{n}}$, 来验证它满足 定理2.1.
首先, ${\displaystyle \hat{g}}$是${\displaystyle p}$的无偏估计, ${\displaystyle {\mathrm{Var}}_p(\hat{g})<\mathrm{\infty }(0\le p\le 1)}$.
然后, 设${\displaystyle l=l(T)}$是零的无偏估计, ${\displaystyle a_i=l(i)}$, 则由于${\displaystyle T\sim B(n,p)}$, 记${\displaystyle s=\frac{p}{1-p}}$, 则$$\begin{array}{rl}& E_{p}l=\sum _{i=0}^n{a}_i(\genfrac{}{}{0ex}{}{n}{i})p^i(1-p{)}^{n-i}=0\\ \\ \Rightarrow & \sum _{i=1}^n{a}_i(\genfrac{}{}{0ex}{}{n}{i})s^i=0,\mathrm{\forall }s\in (0,\mathrm{\infty })\Rightarrow a_i=0,\end{array}$$因此${\displaystyle l}$只能取${\displaystyle 0}$值, 满足定理 2.1. 这样, ${\displaystyle \frac{T}{n}}$确实是${\displaystyle p}$的 UMVUE.

再次利用${\displaystyle T=X_1+\cdots X_n}$是充分统计量. 取${\displaystyle \hat{g}=\frac{T}{n}}$是 ${\displaystyle \frac{1}{\lambda }}$的无偏估计, 且${\displaystyle {\mathrm{Var}}_{\lambda }(\hat{g})<\mathrm{\infty }}$. 现设${\displaystyle l=l(T)}$为零的无偏估计, 则由 这里的(2.4), 有$$0=E_{\lambda }(l(T))=\frac{1}{(n-1)!}{\int }_0^{\mathrm{\infty }}{\lambda }^n{\mathrm{e}}^{-\lambda x}{x}^{n-1}l(x)\mathrm{d}x,(\lambda >0)$$
即$${\int }_0^{\mathrm{\infty }}l(x){\mathrm{e}}^{-\lambda x}{x}^{n-1}\mathrm{d}x=0.$$两边对${\displaystyle \lambda }$求导数:$${\int }_0^{\mathrm{\infty }}xl(x){\mathrm{e}}^{-\lambda x}{x}^{n-1}\mathrm{d}x=0\Rightarrow E_{\lambda }(Tl(T))=0,$$这样${\displaystyle \hat{g}}$满足定理 2.1.

该例子已经证明${\displaystyle T=max\{X_1,\cdots ,X_n\}}$是充分的. 由于$$P_{\theta }(T<t)=P_{\theta }(X_1<t,\cdots ,X_n<t)=\prod _{i=1}^n{P}_{\theta }(X_i<t)=\frac{t^n}{{\theta }^n},0<t<\theta ,$$故${\displaystyle T}$的概率密度为$$\begin{array}{}\text{(求导)}& \{\begin{array}{rl}& \frac{n{t}^{n-1}}{{\theta }^n},0<t<\theta ,\\ & 0,\text{其他情形}.\end{array}\end{array}$$取${\displaystyle \hat{g}=\hat{g}(T)=\frac{n+1}{n}T}$, 则$$E_{\theta }\hat{g}={\int }_0^{\theta }{\theta }^{-n}n{t}^{n-1}\frac{n+1}{n}t\mathrm{d}t=\theta ,\mathrm{\forall }\theta >0.$$即${\displaystyle \hat{g}}$是${\displaystyle \theta }$的无偏估计. 现设${\displaystyle l=l(T)}$是零的无偏估计, 则$${\int }_0^{\mathrm{\infty }}l(x)\frac{n{x}^{n-1}}{{\theta }^n}\mathrm{d}x=0\Rightarrow {\int }_0^{\mathrm{\infty }}l(x)x^{n-1}\mathrm{d}x=0.$$对${\displaystyle \theta }$求导, 得${\displaystyle l(\theta ){\theta }^{n-1}=0\Rightarrow l(\theta )=0(\theta >0)}$, 故满足定理 2.1.

记${\displaystyle T=(T_1,T_2)}$, 其中${\displaystyle T_1=\bar{X},T_2=(n-1)S^2=\sum _{i=1}^n(X_i-\bar{X}{)}^2}$. 在 #？ 中证明了${\displaystyle T}$是充分统计量. 又根据 这里的定理3.1, 得到${\displaystyle T}$的密度函数$$\begin{array}{}\text{(?)}& f_{\theta }(t_1,t_2)=\frac{\sqrt{n}}{\sqrt{2\pi }\sigma }\exp (-\frac{n(t_1-a{)}^2}{2{\sigma }^2}){\left(2^{\frac{n-1}{2}}\mathrm{\Gamma }\left(\frac{n-1}{2}\right){\sigma }^{n-1}\right)}^{-1}{t}_2^{\frac{n-3}{2}}\exp (-\frac{t_2}{2{\sigma }^2})\end{array}$$
(在非${\displaystyle -\mathrm{\infty }<t_1<\mathrm{\infty },t_2>0}$时为${\displaystyle 0}$).
先估计${\displaystyle a}$. 令${\displaystyle \hat{g}=\hat{g}(T)=T_1}$, 则${\displaystyle \hat{g}}$为${\displaystyle g(\theta )=a}$的无偏估计, ${\displaystyle {\mathrm{Var}}_{\theta }(\hat{g})<\mathrm{\infty }}$. 设${\displaystyle l=l(T)}$为零的无偏估计, ${\displaystyle E_{\theta }(l(T))=0}$, 也即$${\int }_0^{\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}l(t_1,t_2)t_2^{\frac{n-3}{2}}\exp (-\frac{1}{2{\sigma }^2}[n(t_1-a{)}^2+t_2])\mathrm{d}{t}_1\mathrm{d}{t}_2=0.$$对${\displaystyle a}$求导: $${\int }_0^{\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}l(t_1,t_2)(t_1-a)t_2^{\frac{n-3}{2}}\exp (-\frac{1}{2{\sigma }^2}[n(t_1-a{)}^2+t_2])\mathrm{d}{t}_1\mathrm{d}{t}_2=0$$从而$${\mathrm{Cov}}_{\theta }(l,\hat{g})=E_{\theta }[l(T)(T_1-a)]=0,-\mathrm{\infty }<a<\mathrm{\infty },\sigma >0.$$满足定理 2.1.
类似地证明: ${\displaystyle \frac{T_2}{n-1}=S^2}$是${\displaystyle {\sigma }^2}$的 UMVUE.

回到 这个例子, 考虑废品率${\displaystyle \frac{M}{N}}$的估计 (${\displaystyle N}$已知, ${\displaystyle M}$为参数). 令${\displaystyle T(X)=X_1+\cdots +X_n}$, 这个例子 已经证明它是完备统计量, ${\displaystyle \frac{T}{n}}$又是${\displaystyle \frac{M}{N}}$的无偏估计, 因此根据 定理3.1, 需要说明${\displaystyle T}$是完备统计量. 现取${\displaystyle l(T)}$满足$$E_M(l(T))=\sum _{i=0}^{M}l(i)(\genfrac{}{}{0ex}{}{M}{i})(\genfrac{}{}{0ex}{}{N-M}{n-i})/(\genfrac{}{}{0ex}{}{N}{n})=0,0\le M\le N.$$令${\displaystyle M=0}$, 代入得${\displaystyle l(0)=0}$. 再令${\displaystyle M=1}$, 得$$(\genfrac{}{}{0ex}{}{N-M}{n})l(0)+(\genfrac{}{}{0ex}{}{M}{1})(\genfrac{}{}{0ex}{}{N-M}{n-1})l(1)=0\Rightarrow l(1)=0,$$如此往复得${\displaystyle l(i)=0,0\le i\le n}$. 这样${\displaystyle T}$是完备统计量.

回到 这个例子, 证明${\displaystyle T=X_1+\cdots +X_n}$是完备的. 事实上, 若${\displaystyle E_{\lambda }(l(T))=0,\mathrm{\forall }\lambda >0}$, 则$${\int }_0^{\mathrm{\infty }}l(x)x^{n-1}{\mathrm{e}}^{-\lambda x}\mathrm{d}x=0,\lambda >0,$$则${\displaystyle l(x)x^{n-1}}$的 #Laplace变换 为${\displaystyle 0}$. 依据 Laplace 变换的唯一性, ${\displaystyle l(x)x^{n-1}=0\Rightarrow l(x)=0}$.

${\displaystyle X_1,\cdots ,X_n\sim N(a,{\sigma }^2),\theta =(a,{\sigma }^2)}$, 则${\displaystyle X}$的概率密度满足 (3.4) 的形式. 其中$$\begin{array}{rl}& k=2,T_1(x)=\sum _{i=1}^n{x}_i,T_2(x)=\sum _{i=1}^n{x}_i^2,Q_1(\theta )=\frac{a}{{\sigma }^2},Q_2(\theta )=-\frac{1}{2{\sigma }^2},\\ & C(\theta )=(\sqrt{2\pi }\sigma {)}^{-n}\exp (-\frac{n{a}^2}{2{\sigma }^2}).\end{array}$$而${\displaystyle \{(\frac{a}{{\sigma }^2},-\frac{1}{2{\sigma }^2})|-\mathrm{\infty }<a<\mathrm{\infty },\sigma >0\}}$是${\displaystyle {\mathbb{R}}^2}$的半平面, 存在内点, 故依据 定理3.2, ${\displaystyle T=(T_1,T_2)}$是充分完备统计量. 这样, 基于${\displaystyle T}$的无偏估计量$$\bar{X}=\frac{T_1}{n},S^2=\frac{1}{n-1}(T_2-\frac{T_1^2}{n})$$是${\displaystyle a,{\sigma }^2}$的唯一 UMVUE (依据 定理3.1)

根据无偏估计的定义$${\int }_{\mathcal{X}}\hat{g}(x)f(x,\theta )\mathrm{d}x=g(\theta ),$$
对${\displaystyle \theta }$求导:$${\int }_{\mathcal{X}}\hat{g}(x)\frac{\partial f(x,\theta )}{\partial \theta }\mathrm{d}x=g^{\prime }(\theta ).$$又$$\int f(x,\theta )\mathrm{d}x=1\Rightarrow \int \frac{\partial f(x,\theta )}{\partial \theta }\mathrm{d}x=0,$$
因此$$\int [\hat{g}(x)-g(\theta )]\frac{\partial f(x,\theta )}{\partial \theta }\mathrm{d}x=g^{\prime }(\theta ).$$将它改写为$$\int \{[\hat{g}(x)-g(\theta )]\sqrt{f(x,\theta )}\}\left(\frac{1}{\sqrt{f(x,\theta )}}\frac{\partial f(x,\theta )}{\partial \theta }\right)\mathrm{d}x=g^{\prime }(\theta ).$$利用 Cauchy-Schwarz 不等式:$$\int [\hat{g}(x)-g(\theta ){]}^{2}f(x,\theta )\mathrm{d}x\cdot \int {\left(\frac{1}{f(x,\theta )}\frac{\partial f(x,\theta )}{\partial \theta }\right)}^{2}f(x,\theta )\mathrm{d}x\ge [g^{\prime }(\theta ){]}^2,$$
也即$${\mathrm{Var}}_{\theta }(\hat{g}(X))\cdot E_{\theta }{\left(\frac{\partial \ln f(X,\theta )}{\partial \theta }\right)}^2\ge [g^{\prime }(\theta ){]}^2.$$这就完成了证明.

注意到$$\begin{array}{}\text{(*)}& \frac{\partial \ln f(x,\theta )}{\partial \theta }=\sum _{i=1}^n\frac{\partial \ln f_{\theta }(x_i)}{\partial \theta }.\end{array}$$
当${\displaystyle i\ne j}$时, ${\displaystyle X_i,X_j}$独立, 故$$E_{\theta }\left(\frac{\partial \ln f_{\theta }(X_i)}{\partial \theta }\frac{\partial \ln f_{\theta }(X_j)}{\partial \theta }\right)=E_{\theta }\left(\frac{\partial \ln f_{\theta }(X_i)}{\partial \theta }\right)E_{\theta }\left(\frac{\partial \ln f_{\theta }(X_j)}{\partial \theta }\right).$$
但$$\begin{array}{rl}{E}_{\theta }\left(\frac{\partial \ln f_{\theta }(X_i)}{\partial \theta }\right)=& \int \frac{1}{f_{\theta }(x_i)}\frac{\partial f_{\theta }(x_i)}{\partial x_i}{f}_{\theta }(x_i)\mathrm{d}{x}_i\\ =& \int \frac{\partial f_{\theta }(x_i)}{\partial \theta }\mathrm{d}{x}_i=\frac{\partial }{\partial \theta }\int f_{\theta }(x_i)\mathrm{d}{x}_i=\frac{\partial }{\partial \theta }(1)=0,\end{array}$$故$$E_{\theta }\left(\frac{\partial \ln f_{\theta }(X_i)}{\partial \theta }\frac{\partial \ln f_{\theta }(X_j)}{\partial \theta }\right)=0.$$ 这样，对 (*) 两边平方再求期望:$$E_{\theta }{\left(\frac{\partial \ln f(X,\theta )}{\partial \theta }\right)}^2=\sum _{i=1}^n{E}_{\theta }{\left(\frac{\partial \ln f_{\theta }(X_i)}{\partial \theta }\right)}^2=n{E}_{\theta }{\left(\frac{\partial \ln f_{\theta }(X_1)}{\partial \theta }\right)}^2.$$

回到 这个例子, 总体概率函数为${\displaystyle f_p(1)=p,f_p(0)=1-p}$, 也即${\displaystyle f_p(t)=p^t(1-p{)}^{1-t}(t=0,1)}$. 故$$\frac{\partial \ln f_p(X_1)}{\partial p}=\frac{X_1-p}{p(1-p)}\Rightarrow I(p)=\frac{1}{p(1-p)}.$$从而代入 (4.3), 得${\displaystyle {\mathrm{Var}}_{\theta }(\hat{g}(X))\ge \frac{p(1-p)}{n}}$. 另一方面, ${\displaystyle \bar{X}}$的方差恰是这个 (参考 这里的推导 ), 因此就是${\displaystyle p}$的 UMVUE.

在 Poisson 分布中估计${\displaystyle g(\theta )=\theta }$. 注意到${\displaystyle \frac{\partial f_{\theta }(X_1)}{\partial \theta }=\frac{X_1-\theta }{\theta }}$(对上面的${\displaystyle P_{\theta }}$求导, 去除不必要的系数), 得${\displaystyle I(\theta )=\frac{1}{\theta }}$. 当样本大小为${\displaystyle n}$, ${\displaystyle \theta }$的无偏估计的方差下降到${\displaystyle \frac{\theta }{n}}$, 而${\displaystyle \bar{X}}$的方差达到了这个下限, 因此是${\displaystyle \theta }$的 UMVUE.

再次回到 这个例子, 总体密度为${\displaystyle \lambda {\mathrm{e}}^{-\lambda x}}$, 因此$$\frac{\partial \ln f_{\lambda }(X_1)}{\partial \lambda }=\frac{1-\lambda X_1}{\lambda }\Rightarrow I(\lambda )=\frac{1}{{\lambda }^2}.$$待估函数为${\displaystyle g(\lambda )=\frac{1}{\lambda }}$, 得 C-R 下界${\displaystyle {\left(\frac{1}{{\lambda }^2}\right)}^2/\left(n\frac{1}{{\lambda }^2}\right)=\frac{1}{n{\lambda }^2}}$, ${\displaystyle \bar{X}}$的方差恰好达到这个下界.

回到 这个例子, ${\displaystyle f_{\theta }(t)=\frac{1}{\theta }(0<t<\theta )}$, 在其他地方为${\displaystyle 0}$. 固定${\displaystyle t>0}$, 则${\displaystyle f_{\theta }(t)}$在${\displaystyle \theta =t}$处不连续, 不存在偏导数. 如果形式地按照公式计算, ${\displaystyle \frac{\partial \ln f_{\theta }(X_1)}{\partial \theta }=-\frac{1}{\theta },I(\theta )=\frac{1}{{\theta }^2}}$, 故 C-R 不等式提供的下界为${\displaystyle \frac{{\theta }^2}{n}}$. 但从例子中的无偏估计${\displaystyle \frac{n+1}{n}T=\frac{n+1}{n}max(X_1,\cdots ,X_n)}$来看, 根据给出的概率密度函数, 可得${\displaystyle {\mathrm{Var}}_{\theta }\left(\frac{n+1}{n}T\right)=\frac{{\theta }^2}{n(n+2)}<\frac{{\theta }^2}{n}}$矛盾.

设${\displaystyle X_1,\cdots ,X_n\sim N(a,{\sigma }^2),\theta =(a,{\sigma }^2),f_{\theta }=(2\pi {\theta }_2{)}^{-1/2}\exp (-\frac{(t-{\theta }_1{)}^2}{2{\theta }_2})}$. 因此$$\frac{\partial \ln f_{\theta }(X_1)}{\partial {\theta }_1}=\frac{X_1-{\theta }_1}{{\theta }_2},\frac{\partial \ln f_{\theta }(X_1)}{\partial {\theta }_2}=\frac{{\theta }_2-(X_1-{\theta }_1{)}^2}{2{\theta }_2^2},$$可以算出$$\begin{array}{rl}& I_{11}(\theta )=\frac{1}{{\sigma }^2},I_{22}(\theta )=\frac{1}{2{\sigma }^4},I_{12}(\theta )=I_{21}(\theta )=0;\\ & I_{11}^{\ast }(\theta )=\frac{1}{{\sigma }^2},I_{22}^{\ast }(\theta )=2{\sigma }^4,I_{12}^{\ast }(\theta )=I_{21}^{\ast }(\theta )=0.\end{array}$$
这给出了${\displaystyle a,{\sigma }^2}$无偏估计方差的下限, 分别是${\displaystyle {\sigma }^2/n,2{\sigma }^4/n}$. 前一个下限${\displaystyle \bar{X}}$可以达到, 后一个无法达到; 在 这个例子中我们证明了${\displaystyle S^2}$是${\displaystyle {\sigma }^2}$的 UMVUE. 利用${\displaystyle (n-1)S^2/{\sigma }^2\sim {\chi }_{n-1}^2,\mathrm{Var}({\chi }_{n-1}^2)=2(n-1)}$的事实, ${\displaystyle {\mathrm{Var}}_{\theta }(S^2)=\frac{2{\sigma }^4}{n-1}>\frac{2{\sigma }^4}{n}}$.