某商店每天进货 ${\displaystyle N}$ 件商品, 抽取 ${\displaystyle n}$ 件检验, ${\displaystyle X}$ 为其中的废品数, 用来决定是否接受这批商品. 如果接受 (行动 ${\displaystyle d_1}$), 则每件废品损失 ${\displaystyle 10}$ 元; 若拒绝 (${\displaystyle d_2}$), 则当日无货出售, 每件损失利润 ${\displaystyle 2}$ 元.
可以认为 ${\displaystyle X\sim \mathrm{Binomial}(n,p)}$, ${\displaystyle \mathrm{\Theta }=[0,1]}$, ${\displaystyle \mathcal{D}=\{d_1,d_2\}}$. 设废品率 ${\displaystyle p}$, 则该批商品有废品 ${\displaystyle Np}$, 从而损失函数为 $$L(d_i,p)=\{\begin{array}{rl}& 10Np,i=1,\\ & 2N(1-p),i=2.\end{array}$$

考虑任意判决函数 ${\displaystyle \delta }$. 根据定义, Bayes 风险为 ${\displaystyle R_H(\delta )={\mathbb{E}}_{(X,\theta )}[L(\delta (X),\theta )]}$. 把它分成两步计算. 第一步是条件期望 $${\int }_{\mathrm{\Theta }}L(\delta (x),\theta )H(\mathrm{d}\theta |x)=R(\delta (x)|x).$$ 第二步是对 ${\displaystyle x}$ 求期望. 记 ${\displaystyle X}$ 的边缘分布为 ${\displaystyle Q}$, 则 $$R_H(\delta )={\int }_{\mathcal{F}}R(\delta (x)|x)\mathrm{d}Q(x).$$ 根据条件, ${\displaystyle \mathrm{\forall }x\in \mathcal{X}}$, 有 $$R({\delta }_H(x)|x)=\underset{d\in \mathcal{D}}{min}R(d|x)\le R(\delta (x)|x).$$ 于是 $$R_h({\delta }_H)={\int }_{\mathcal{F}}R({\delta }_H(x)|x)\mathrm{d}Q(x)\le {\int }_{\mathcal{F}}R(\delta (x)|x)\mathrm{d}Q(x)=R_H(\delta ).$$

记 ${\displaystyle m(x),{\sigma }^2(x)}$ 为 ${\displaystyle H(\theta |x)}$ 的均值、方差, 则 $$R(d|x)=[d-m(x){]}^2+{\sigma }^2(x).$$ 因此 Bayes 判决函数为 ${\displaystyle {\delta }_H(x)=m(x)}$, Bayes 风险为 $$R_H({\delta }_H)={\int }_{\mathcal{F}}{\sigma }^2(x)\mathrm{d}Q(x).$$

则得到样本 ${\displaystyle x}$ 后, 后验分布为$$\begin{array}{rl}H({\theta }_1|x)& =\frac{pf(x,{\theta }_1)}{pf(x,{\theta }_1)+(1-p)f(x,{\theta }_2)},\\ H({\theta }_2|x)& =\frac{(1-p)f(x,{\theta }_2)}{pf(x,{\theta }_1)+(1-p)f(x,{\theta }_2)}.\end{array}$$ 于是行动的后验风险分别为$$\begin{array}{rl}R(d_1|x)& =\frac{a(1-p)f(x,{\theta }_2)}{pf(x,{\theta }_1)+(1-p)f(x,{\theta }_2)},\\ R(d_2|x)& =\frac{bpf(x,{\theta }_1)}{pf(x,{\theta }_1)+(1-p)f(x,{\theta }_2)}.\end{array}$$
根据 后验风险最小原则, Bayes 解为 $${\delta }_H(x)=\{\begin{array}{rl}& d_1,\frac{f(x,{\theta }_2)}{f(x,{\theta }_1)}\le \frac{bp}{a(1-p)},\\ & d_2,\frac{f(x,{\theta }_2)}{f(x,{\theta }_1)}>\frac{bp}{a(1-p)}.\end{array}$$ 这个解和 NP基本引理 提供的 UMP 检验符合, 不同的是 NP 中临界值由 ${\displaystyle \alpha }$ 确定, 而这里则取决于 ${\displaystyle a,b,p}$. ${\displaystyle a,b}$ 的取值正好对应了 ${\displaystyle \alpha }$ 的作用, 差异反映了两类错误后果的不同.

根据 这个结果, ${\displaystyle \theta }$ 的"后验分布"^[2]为 ${\displaystyle \mathcal{N}(\bar{x},\frac{1}{n})}$. 从而 Bayes 解为均值 ${\displaystyle {\delta }_H(x)=\bar{X}}$. 它的风险函数为 ${\displaystyle {\mathbb{E}}_{\theta }(\bar{X}-\theta {)}^2=\frac{1}{n}}$, 从而 Bayes 风险为 $$R_H({\delta }_H)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{n}\mathrm{d}\theta =\mathrm{\infty }.$$ 从而 ${\displaystyle \mathrm{\forall }\delta }$, ${\displaystyle R_H(\delta )=\mathrm{\infty }}$.

如果 ${\displaystyle {\delta }^{\ast }}$ 不是的话, ${\displaystyle \mathrm{\exists }\delta :M(\delta )=c^{\prime }<c}$, 则 $$R_H(\delta )={\int }_{\mathrm{\Theta }}R(\delta ,\theta )\mathrm{d}H(\theta )\le {\int }_{\mathrm{\Theta }}M(\delta )\mathrm{d}H(\theta )=c^{\prime }<c=R_H({\delta }^{\ast }).$$ 这和 ${\displaystyle {\delta }^{\ast }}$ 为 Bayes 解相矛盾.

取 ${\displaystyle p}$ 的 共轭先验分布 ${\displaystyle \mathrm{Beta}(a,b)}$. 当 ${\displaystyle X=x}$, 后验分布为 ${\displaystyle \mathrm{Beta}(x+a,n+b-x)}$. 根据 这个例子, Bayes 估计就是 Beta 分布的期望值 ${\displaystyle {\delta }_{ab}(x)=\frac{a+x}{n+a+b}}$, 风险函数为 $$\begin{array}{rl}R({\delta }_{ab},p)& ={\mathbb{E}}_p{(\frac{X+a}{n+a+b}-p)}^2\\ & ={\mathrm{Var}}_p\left(\frac{X}{n+a+b}\right)+{(\frac{{\mathbb{E}}_p(X)+a}{n+a+b}-p)}^2.\end{array}$$ 考虑到 ${\displaystyle {\mathbb{E}}_p(X)=np,{\mathrm{Var}}_p(X)=np(1-p)}$, 故 $$R({\delta }_{ab},p)=\frac{np(1-p)+[a-(a+b)p{]}^2}{(n+a+b{)}^2}.$$ 若取 ${\displaystyle a=b=\frac{\sqrt{n}}{2}}$, 则上式右边为 ${\displaystyle \frac{n}{4(n+\sqrt{n}{)}^2}}$. 于是根据 定理5.1, $${\delta }_{\frac{\sqrt{n}}{2},\frac{\sqrt{n}}{2}}(x)=\frac{x+\frac{\sqrt{n}}{2}}{n+\sqrt{n}}$$ 是 ${\displaystyle p}$ 的 Minimax 估计.

反设 ${\displaystyle {\delta }^{\ast }}$ 不是 Minimax 解, 则 ${\displaystyle \mathrm{\exists }\delta :M(\delta )<M({\delta }^{\ast })}$. 则根据定理 5.2 的条件, 当 ${\displaystyle k}$ 充分大, ${\displaystyle M(\delta )<r_k}$. 于是 ${\displaystyle \delta }$ 的 Bayes 风险 ${\displaystyle R_{H_k}(\delta )}$ 满足 $$R_{H_k}(\delta )\le M(\delta )<r_k=R_{H_k}({\delta }_k).$$ 这和 ${\displaystyle {\delta }_k}$ 是 ${\displaystyle H_k}$ 下的 Bayes 解矛盾.

作正交变换 ${\displaystyle Y=(Y_1,\cdots ,Y_n{)}^{\mathrm{T}}=Q(X_1,\cdots ,X_n{)}^{\mathrm{T}}}$, ${\displaystyle Q}$ 的第一行为 ${\displaystyle (\frac{1}{\sqrt{n}},\cdots ,\frac{1}{\sqrt{n}})}$, 且 ${\displaystyle \sum _{j=1}^n{q}_{ij}=0}$, ${\displaystyle i=2,\cdots ,n}$. 根据 引理3.1, ${\displaystyle Y_1,\cdots ,Y_n}$ 独立, ${\displaystyle \bar{X}=\frac{Y_1}{\sqrt{n}}}$ 只和 ${\displaystyle Y_1}$ 有关.
设 ${\displaystyle f(X_1,\cdots ,X_n)=g(Y_1,\cdots ,Y_n)}$. 当 ${\displaystyle (X_1,\cdots ,X_n)\to (X_1+c,\cdots ,X_n+c)}$, ${\displaystyle Y_1,\cdots ,Y_n}$ 变换为$$\begin{array}{rl}{Y}_1^{\prime }& =\frac{1}{\sqrt{n}}\sum _{i=1}^n(X_i+c)=Y_1+\sqrt{n}c,\\ Y_i^{\prime }& =\sum _{j=1}^n{q}_{ij}(X_i+c)=Y_i,i=2,\cdots ,n.\end{array}$$
于是由 ${\displaystyle f(X_1+c,\cdots ,X_n+c)=f(X_1,\cdots ,X_n)}$, 有 $$g(Y_1+\sqrt{n}c,Y_2,\cdots ,Y_n)=g(Y_1,\cdots ,Y_n),\mathrm{\forall }c\in \mathbb{R}.$$ 这说明 ${\displaystyle g(Y_1,\cdots ,Y_n)}$ 其实和 ${\displaystyle Y_1}$ 无关, 即有 ${\displaystyle h(Y_2,\cdots ,Y_n)}$ 的形式. 故 $$\bar{X}=\frac{Y_1}{\sqrt{n}},f(X_1,\cdots ,X_n)=h(Y_2,\cdots ,Y_n).$$ 考虑到 ${\displaystyle Y_1,\cdots ,Y_n}$ 相互独立, ${\displaystyle \bar{X}\perp \perp f(X_1,\cdots ,X_n)}$.

否则 ${\displaystyle \mathrm{\exists }\delta }$: ${\displaystyle R(\delta ,\theta )\le R({\delta }_H,\theta )}$, ${\displaystyle \mathrm{\forall }\theta \in \mathrm{\Theta }}$, 且 ${\displaystyle \mathrm{\exists }{\theta }_0\in \mathrm{\Theta }}$, 使 ${\displaystyle R({\delta }_H,{\theta }_0)-R(\delta ,{\theta }_0)=2\epsilon >0}$. 因为 ${\displaystyle R(\delta ,\theta )}$, ${\displaystyle R({\delta }_H,\theta )}$ 关于 ${\displaystyle \theta }$ 连续, 故 ${\displaystyle \mathrm{\exists }\rho >0}$, ${\displaystyle \mathrm{\forall }\theta \in {\mathrm{B}}_{\rho }({\theta }_0)}$, 有 ${\displaystyle R({\delta }_H,\theta )-R(\delta ,\theta )\ge \epsilon }$. 此时$$\begin{array}{rl}{R}_H(\delta )& ={\int }_{\mathrm{\Theta }}R(\delta ,\theta )\mathrm{d}H(\theta )\\ & ={\int }_{{\mathrm{\Theta }}_1}R(\delta ,\theta )\mathrm{d}H(\theta )+{\int }_{{\mathrm{\Theta }}_2}R(\delta ,\theta )\mathrm{d}H(\theta ),\end{array}$$ 这里 ${\displaystyle {\mathrm{\Theta }}_1=\mathrm{\Theta }\cap S_{{\theta }_0}(\rho )}$, ${\displaystyle {\mathrm{\Theta }}_2=\mathrm{\Theta }-{\mathrm{\Theta }}_1}$. 因为 ${\displaystyle {\mathrm{\Theta }}_1}$ 上 ${\displaystyle R({\delta }_H,\theta )\ge R(\delta ,\theta )+\epsilon }$, ${\displaystyle {\mathrm{\Theta }}_2}$ 上 ${\displaystyle R({\delta }_H,\theta )\ge R(\delta ,\theta )}$, 从而$$\begin{array}{rl}{R}_h({\delta }_H)& ={\int }_{{\mathrm{\Theta }}_1}R({\delta }_H,\theta )\mathrm{d}H(\theta )+{\int }_{{\mathrm{\Theta }}_2}R(\delta ,\theta )\mathrm{d}H(\theta )\\ & \ge {\int }_{\mathrm{\Theta }}R(\delta ,\theta )\mathrm{d}H(\theta )+\epsilon H(\mathrm{\Theta }\cap S_{{\theta }_0}(\rho ))+{\int }_{{\mathrm{\Theta }}_2}R(\delta ,\theta )\mathrm{d}H(\theta )\\ & =R_H(\delta )+\epsilon H(\mathrm{\Theta }\cap S_{{\theta }_0}(\rho ))>R_H(\delta ).\end{array}$$ 这与 ${\displaystyle {\delta }_H}$ 是 ${\displaystyle H}$ 下的 Bayes 解矛盾.

1. 考虑 这个例子. ${\displaystyle H}$ 在 ${\displaystyle 0<p<1}$ 上有处处大于 ${\displaystyle 0}$ 的密度, 故条件 2 满足; 条件 3 显然满足; 对条件 1, 注意到任一 ${\displaystyle \delta }$ 的风险函数: $$R(\delta ,p)=\sum _{i=0}^n[\delta (i)-p{]}^2(\genfrac{}{}{0ex}{}{n}{i})p^i(1-p{)}^{n-i}$$ 关于 ${\displaystyle p}$ 连续. 因此根据 定理7.1, Minimax 估计是可容许的.

2. 设 ${\displaystyle X_1,\cdots ,X_n\sim \mathcal{N}(\theta ,1)}$. 平方损失. ${\displaystyle c\in (0,1)}$ 为常数, 则 ${\displaystyle c\bar{X}}$ 是 ${\displaystyle \theta }$ 的可容许估计. 事实上取 ${\displaystyle \theta \sim \mathcal{N}(0,{\tau }^2)}$, 选择 ${\displaystyle \tau }$: ${\displaystyle \frac{n{\tau }^2}{1+n{\tau }^2}=c}$. 则由 这个例子 和 这个结论, ${\displaystyle c\bar{X}}$ 是这个先验分布下的 Bayes 解. 不难验证定理 7.1 的 1～3 都满足, 因此 ${\displaystyle c\bar{X}}$ 是容许估计. 其中 3: $$\begin{array}{rl}R(\delta ,\theta )& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\cdots {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}[\delta (x_1,\cdots ,x_n)-\theta {]}^2{\left(\frac{1}{\sqrt{2\pi }}\right)}^n\\ & \cdot \exp [-\frac{1}{2}\sum _{i=1}^n(x_i-\theta {)}^2]\mathrm{d}{x}_1\cdots \mathrm{d}{x}_n,\end{array}$$ 需要验证它处处有限且可在积分下对 ${\displaystyle \theta }$ 求导.
   令 ${\displaystyle c\to 1}$ 就有 ${\displaystyle c\bar{X}\to \bar{X}}$, 因此 ${\displaystyle \bar{X}}$ 是容许估计的极限, 但它并不一定就容许. 下面的方法基于 C-R 不等式, 可以处理有关单参数指数族的参数在平方损失下的某些容许估计问题.
   设 ${\displaystyle \delta }$ 为 ${\displaystyle \theta }$ 任意估计, 满足: $$R(\delta ,\theta )\le R(\bar{X},\theta )=\frac{1}{n},\mathrm{\forall }\theta .$$ 记 ${\displaystyle {\mathbb{E}}_{\theta }(\delta (X_1,\cdots ,X_n))=\theta +b(\theta )}$, ${\displaystyle b^{\prime }(\theta )}$ 存在. 根据 C-R不等式, $$\begin{array}{}\text{(*)}& \frac{[1+b^{\prime }(\theta ){]}^2}{n}+b^2(\theta )\le \frac{1}{n},\mathrm{\forall }\theta \in \mathbb{R}.\end{array}$$ 由此 ${\displaystyle b^{\prime }(\theta )\le -l{b}^2(\theta )}$, 其中 ${\displaystyle l=\frac{n}{2}>0}$. 则 ${\displaystyle \frac{b^{\prime }(\theta )}{b^2(\theta )}\le -l}$. 由上式 ${\displaystyle b^{\prime }(\theta )\le 0}$, 故 ${\displaystyle b(\theta )}$ 单减. 现在证 ${\displaystyle b(\theta )\ge 0}$: 否则 ${\displaystyle \mathrm{\exists }{\theta }_0:b({\theta }_0)<0}$. 根据单调性, ${\displaystyle b(\theta )<0,\mathrm{\forall }\theta >{\theta }_0}$. 从而由 ${\displaystyle \frac{b^{\prime }(\theta )}{b^2(\theta )}\le -l}$, 当 ${\displaystyle \theta >{\theta }_0}$, 有 $$\begin{array}{}\text{(**)}& -l(\theta -{\theta }_0)\ge {\int }_{{\theta }_0}^{\theta }\frac{b^{\prime }(\theta )}{b^2(\theta )}\mathrm{d}\theta =\frac{1}{b({\theta }_0)}-\frac{1}{b(\theta )}.\end{array}$$ 令 ${\displaystyle \theta \to \mathrm{\infty }}$, 则 ${\displaystyle b(\theta )\to 0}$, 这和 ${\displaystyle b({\theta }_0)<0}$、${\displaystyle b(\theta )}$ 单减矛盾, 从而 ${\displaystyle b(\theta )\ge 0}$.
   这样有两种情况:

   • ${\displaystyle b(\theta )\equiv 0}$. 此时 ${\displaystyle \delta }$ 为 ${\displaystyle \theta }$ 的无偏估计, 而 ${\displaystyle \bar{X}}$ 为 ${\displaystyle \theta }$ 的 UMVUE, 故 $$R(\delta ,\theta )={\mathrm{Var}}_{\theta }(\delta )\ge {\mathrm{Var}}_{\theta }(\bar{X})=R(\bar{X},\theta ),$$ 此时 ${\displaystyle \delta }$ 不一致优于 ${\displaystyle \bar{X}}$.
   • 存在 ${\displaystyle {\theta }_0}$: ${\displaystyle b({\theta }_0)>0}$. 此时 ${\displaystyle \mathrm{\forall }\theta <{\theta }_0}$: ${\displaystyle b(\theta )\ge b({\theta }_0)>0}$. 但 ${\displaystyle \theta \to -\mathrm{\infty }}$ 时, ${\displaystyle b(\theta )\nrightarrow \mathrm{\infty }}$. 否则 (*) 不对一切 ${\displaystyle \theta }$ 成立. 故 ${\displaystyle \underset{\theta \to -\mathrm{\infty }}{lim}b(\theta )=c}$, ${\displaystyle 0<c<\mathrm{\infty }}$. 因此在 (**) 中令 ${\displaystyle {\theta }_0=\theta -1\to -\mathrm{\infty }}$, 有 $$-l\ge \frac{1}{b(\theta -1)}-\frac{1}{b(\theta )}\to \frac{1}{c}-\frac{1}{c}=0.$$

   因此只有情况 1 成立. 而一致优于 ${\displaystyle \bar{X}}$ 的 ${\displaystyle \delta }$ 不存在, 这就证明了 ${\displaystyle \bar{X}}$ 的可容许性.