以${\displaystyle h}$记${\displaystyle \theta ={\theta }_0}$时${\displaystyle f(X,{\theta }_1)/f(X,{\theta }_0)}$的分布函数, 它是左连续的. 由于${\displaystyle 0<\alpha <1}$, ${\displaystyle \mathrm{\exists }C:h(C)\le 1-\alpha \le h(C^{+})}$. 取$$\gamma =\{\begin{array}{rl}& 1,h(C)=1-\alpha ,\\ & 0,h(C^{+})=1-\alpha ,\\ & \frac{h(C^{+})-(1-\alpha )}{h(C^{+})-h(C)},h(C)<1-\alpha <h(C^{+}).\end{array}$$这样的${\displaystyle C,\gamma }$满足$$\begin{array}{}\text{(1.4)}& E_{{\theta }_0}({\phi }_{\alpha }(X))=1-h(C^{+})+\gamma [h(C^{+})-h(C)]=\alpha ,\end{array}$$即${\displaystyle {\phi }_{\alpha }}$有水平${\displaystyle \alpha }$. 设${\displaystyle \phi }$为检验问题 (1.3) 的任一水平${\displaystyle \alpha }$检验, 设${\displaystyle f(x,\theta )}$是${\displaystyle X}$的密度. 定义${\displaystyle S^{+}=\{x|{\phi }_{\alpha }(x)>\phi (x)\}\subset \mathcal{X},S^{-}=\{x|{\phi }_{\alpha }(x)<\phi (x)\}}$. 在${\displaystyle x\in S^{+}}$时${\displaystyle {\phi }_{\alpha }(x)>0\Rightarrow \frac{f(x,{\theta }_1)}{f(x,{\theta }_0)}\ge C}$; ${\displaystyle x\in S^{-}}$时${\displaystyle {\phi }_{\alpha }(x)<1\Rightarrow \frac{f(x,{\theta }_1)}{f(x,{\theta }_0)}\le C}$. 因此总有$$[{\phi }_{\alpha }(x)-\phi (x)][f(x,{\theta }_1)-f(x,{\theta }_0)]\ge 0,\mathrm{\forall }x\in \mathcal{X}.$$从而$$\begin{array}{rl}& {\int }_{\mathcal{X}}[{\phi }_{\alpha }(x)-\phi (x)][f(x,{\theta }_1)-f(x,{\theta }_0)]\mathrm{d}x\ge 0\\ \Rightarrow & {\int }_{\mathcal{X}}{\phi }_{\alpha }(x)f(x,{\theta }_1)\mathrm{d}x-{\int }_{\mathcal{X}}\phi (x)f(x,{\theta }_1)\mathrm{d}x\\ \text{(1.5)}& & \ge C[{\int }_{\mathcal{X}}{\phi }_{\alpha }(x)f(x,{\theta }_0)\mathrm{d}x-{\int }_{\mathcal{X}}\phi (x)f(x,{\theta }_0)\mathrm{d}x].\end{array}$$易知${\displaystyle C\ge 0}$. 由 (1.4),$${\int }_{\mathcal{X}}{\phi }_{\alpha }(x)f(x,{\theta }_0)=\alpha \Rightarrow {\int }_{\mathcal{X}}\phi (x)f(x,{\theta }_0)\mathrm{d}x\le \alpha$$

因此 (1.5) 右边非负, 从而$${\beta }_{{\phi }_{\alpha }}({\theta }_1)={\int }_{\mathcal{X}}{\phi }_{\alpha }(x)f(x,{\theta }_1)\mathrm{d}x\ge {\int }_{\mathcal{X}}\phi (x)f(x,{\theta }_1)\mathrm{d}x={\beta }_{\phi }({\theta }_1).$$这就说明${\displaystyle {\phi }_{\alpha }}$是水平${\displaystyle \alpha }$的 UMP 检验.

任取${\displaystyle {\theta }_1>{\theta }_0}$构成检验问题 (2.2), 有$$\frac{f(x,{\theta }_1)}{f(x,{\theta }_0)}=\frac{C({\theta }_1)}{C({\theta }_0)}\exp [T(x)(Q({\theta }_1)-Q({\theta }_0))].$$上式关于${\displaystyle T(x)}$严格单增. 因此, ${\displaystyle \frac{f(x,{\theta }_1)}{f(x,{\theta }_0)}}$和某个常数${\displaystyle C^{\prime }}$的大小关系对应于${\displaystyle T(x)}$和${\displaystyle C}$的大小关系. 由此和 NP 引理, (2.2) 的水平 UMP 检验式为 (2.3), 其中${\displaystyle C,\gamma }$由 (2.3) 确定. 现在要证这样确定的${\displaystyle \phi }$是 (2.2) 的水平${\displaystyle \alpha }$的 UMP检验. 由于${\displaystyle C}$与${\displaystyle \gamma ,{\theta }_1}$无关, 只需证${\displaystyle \phi }$是 (2.2) 的水平${\displaystyle \alpha }$检验. 只需证: ${\displaystyle {\beta }_{\phi }(\theta )}$是${\displaystyle \theta }$的非降函数.
为此, 任取${\displaystyle {\theta }^{\prime }<{\theta }^{″}}$, 则$$\frac{f(x,{\theta }^{\prime })}{f(x,{\theta }^{″})}=\frac{C({\theta }^{\prime })}{C({\theta }^{″})}\exp [(Q({\theta }^{\prime })-Q({\theta }^{″}))T(x)].$$由于${\displaystyle Q}$为${\displaystyle \theta }$的严格单增函数, 故${\displaystyle Q({\theta }^{\prime })-Q({\theta }^{″})<0}$. 结合${\displaystyle C({\theta }^{\prime }),C({\theta }^{″})>0}$, 知${\displaystyle \frac{f(x,{\theta }^{\prime })}{f(x,{\theta }^{″})}}$关于${\displaystyle T(x)}$严格单减. 找${\displaystyle t_0}$(${\displaystyle t_0}$一定存在), 使得$$\frac{C({\theta }^{\prime })}{C({\theta }^{″})}\exp [(Q({\theta }^{\prime })-Q({\theta }^{″}))t_0]=1.$$

令${\displaystyle S_1=\{x|T(x)>t_0\},S_2=\{x|T(x)<t_0\},S_3=\{x|T(x)=t_0\}}$, 则$$0\le {\int }_{S_1}[f(x,{\theta }^{″})-f(x,{\theta }^{\prime })]\mathrm{d}x=-{\int }_{S_2}[f(x,{\theta }^{″})-f(x,{\theta }^{\prime })]\mathrm{d}x.$$而 (2.3) 看出${\displaystyle \phi }$只与${\displaystyle T(x)}$有关 (且非减), 故${\displaystyle \underset{x\in S_1}{inf}\phi (x)\ge \underset{x\in S_2}{sup}\phi (x)}$. 因此$$\begin{array}{rl}{\beta }_{\phi }({\theta }^{″})-{\beta }_{\phi }({\theta }^{\prime })=& {\int }_X\phi (x)[f(x,{\theta }^{″})-f(x,{\theta }^{\prime })]\mathrm{d}x\\ =& ({\int }_{S_1}+{\int }_{S_2})\phi (x)[f(x,{\theta }^{″})-f(x,{\theta }^{\prime })]\mathrm{d}x\\ \ge & [\underset{x\in S_1}{inf}\phi (x)-\underset{x\in S_2}{sup}\phi (x)]{\int }_{S_1}[f(x,{\theta }^{″})-f(x,{\theta }^{\prime })]\mathrm{d}x\ge 0,\end{array}$$因此证明结束.

由于$$\begin{array}{rl}f(x_1,\cdots ,x_n,\theta )=& \frac{1}{(2\pi {\sigma }_0^2{)}^{\frac{n}{2}}}\exp [-\frac{1}{2{\sigma }_0^2}\sum _{i=1}^n(x_i-\theta {)}^2]\\ =& \frac{1}{(2\pi {\sigma }_0^2{)}^{\frac{n}{2}}}\exp (-\frac{n{\theta }^2}{2{\sigma }_0^2})\exp \left(\frac{n}{{\sigma }_0^2}\theta \bar{x}\right)\exp (-\frac{1}{2{\sigma }_0^2}\sum _{i=1}^n{x}_i^2).\end{array}$$与 (2.1) 比对, ${\displaystyle Q(\theta )=\frac{n\theta }{{\sigma }_0^2}}$严格单增, ${\displaystyle T(x)=\bar{x}=\frac{1}{n}\sum _{i=1}^n{x}_i}$, 则根据定理得水平${\displaystyle \alpha }$的 UMP 检验存在, 有否定域${\displaystyle \{\bar{x}>C\}}$. 要计算常数${\displaystyle C}$, 需要注意到${\displaystyle P_0(\bar{X}>C)=\alpha }$, 则$$C=\frac{{\sigma }_0{u}_{\alpha }}{\sqrt{n}},\frac{1}{\sqrt{2\pi }}{\int }_{u_{\alpha }}^{\mathrm{\infty }}{\mathrm{e}}^{-\frac{t^2}{2}}\mathrm{d}t=\alpha .$$ ${\displaystyle u_{\alpha }}$的取值可以查表得到.