${\displaystyle X_1,\cdots ,X_n\sim N(a,{\sigma }^2)}$. 考虑统计量${\displaystyle T=\sum _{i=1}^n{a}_i{X}_i}$, 则${\displaystyle T\sim N(a\sum _{i=1}^n{a}_i,{\sigma }^2\sum _{i=1}^n{a}_i^2)}$. 特别地${\displaystyle \bar{X}\sim N(a,{\sigma }^2/n)}$.

如果${\displaystyle X_1,\cdots ,X_n\sim f(x,\lambda )}$, ${\displaystyle f(x,\lambda )}$由 (6) 给出，求${\displaystyle \bar{X}}$的样本分布.
以${\displaystyle f_n(x,\lambda )}$记统计量${\displaystyle T_n=X_1+\cdots +X_n}$的密度函数(特别地, ${\displaystyle f_n(x,\lambda )=f(x,\lambda )}$), 则${\displaystyle T_n=T_{n-1}+X_n}$. 则$$f_n(x,\lambda )={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}f(x-y,\lambda )f_{n-1}(y,\lambda )\mathrm{d}y={\int }_{-\mathrm{\infty }}^x\lambda {\mathrm{e}}^{-\lambda (x-y)}{f}_{n-1}(y,\lambda )\mathrm{d}y.$$
不难用归纳法证明:$$f_n(x,\lambda )=\{\begin{array}{rl}& \frac{1}{(n-1)!}{\lambda }^n{\mathrm{e}}^{-\lambda x}{x}^{n-1},x>0,\\ & 0,x\le 0.\end{array}$$
记${\displaystyle g_n(x,\lambda )}$为${\displaystyle \bar{X}}$的概率密度函数. 因为$$\begin{array}{rl}& P(\frac{T_n}{n}\le x)=P(T_n\le nx)\\ \Rightarrow & {\int }_{-\mathrm{\infty }}^x{g}_n(y,\lambda )\mathrm{d}y={\int }_{-\mathrm{\infty }}^{nx}{f}_n(y,\lambda )\mathrm{d}y=n{\int }_{-\mathrm{\infty }}^x{f}_n(nt,\lambda )\mathrm{d}t\\ \Rightarrow & g_n(x,\lambda )=n{f}_n(nx,\lambda ),\end{array}$$因此得到最后的答案$$\begin{array}{}\text{(2.4)}& g_n(x,\lambda )=\{\begin{array}{rl}& \frac{1}{(n-1)!}{n}^n{\lambda }^n{\mathrm{e}}^{-\lambda nx}{x}^{n-1},x>0,\\ & 0,x\le 0.\end{array}\end{array}$$

${\displaystyle X_1,\cdots ,X_n\sim F}$, ${\displaystyle F}$有密度${\displaystyle f}$, 求次序统计量${\displaystyle X_{(m)}}$的抽样分布.
事件${\displaystyle \{X_{(m)}<x\}}$发生等价于至少有${\displaystyle m}$个不同的${\displaystyle i}$使得${\displaystyle \{X_i<x\}}$. 这${\displaystyle n}$个事件相互独立，且都有概率${\displaystyle F(x)}$, 故$$\begin{array}{rl}P(X_{(m)}<x)=& \sum _{i=m}^n(\genfrac{}{}{0ex}{}{n}{i})F^i(x)[1-F(x){]}^{n-i}\\ =& m(\genfrac{}{}{0ex}{}{n}{m}){\int }_0^{F(x)}{t}^{m-1}(1-t{)}^{n-m}\mathrm{d}t.\end{array}$$
(为了证明这个式子，记${\displaystyle F(x)=p\in (0,1)}$, 欲证$$\sum _{i=m}^n(\genfrac{}{}{0ex}{}{n}{i})p^i(1-p{)}^{n-i}=m(\genfrac{}{}{0ex}{}{n}{m}){\int }_0^p{t}^{m-1}(1-t{)}^{n-m}\mathrm{d}t.$$注意到等式对${\displaystyle p=0}$成立，只需要证明两边对${\displaystyle p}$求导相等. 事实上, $$\begin{array}{rl}{\text{左边}}^{\prime }=& \sum _{i=m}^n(\genfrac{}{}{0ex}{}{n}{i})[i{p}^{i-1}(1-p{)}^{n-i}-(n-i)(1-p{)}^{n-i-1}{p}^i]\\ =& \sum _{i=m}^{n}i(\genfrac{}{}{0ex}{}{n}{i})p^{i-1}(1-p{)}^{n-i}-\sum _{i=m}^{n-1}(n-i)(\genfrac{}{}{0ex}{}{n}{n-i})(1-p{)}^{n-i-1}{p}^i\\ =& n\sum _{i=m}^n(\genfrac{}{}{0ex}{}{n-1}{i-1})p^{i-1}(1-p{)}^{n-i}-n\sum _{i=m}^{i-1}(\genfrac{}{}{0ex}{}{n-1}{n-i-1})(1-p{)}^{n-i-1}{p}^i\\ =& n\sum _{i=m}^n(\genfrac{}{}{0ex}{}{n-1}{i-1})p^{i-1}(1-p{)}^{n-i}-n\sum _{t=m+1}^n(\genfrac{}{}{0ex}{}{n-1}{n-t})(1-p{)}^{n-t}{p}^{t-1}\\ =& n(\genfrac{}{}{0ex}{}{n-1}{m-1})p^{m-1}(1-p{)}^{n-m}=m(\genfrac{}{}{0ex}{}{n}{m})p^{m-1}(1-p{)}^{n-m}={\text{右边}}^{\prime }.\end{array}$$因此$$f_m(x)=\frac{\mathrm{d}}{\mathrm{d}x}P(X_{(m)}<x)=m(\genfrac{}{}{0ex}{}{n}{m})F^{(m-1)}(x)[1-F(x){]}^{n-m}f(x).$$
进一步地, 次序统计量的联合密度为$$\begin{array}{}\text{(2.4)}& g(x_{(1)},\cdots ,x_{(n)})=\{\begin{array}{rl}& n!f(x_{(1)})\cdots f(x_{(n)}),x_{(1)}<\cdots <x_{(n)},\\ & 0,\text{其他情况}.\end{array}\end{array}$$

也即要计算${\displaystyle P(\xi <x)}$. ${\displaystyle x\le 0}$时显然${\displaystyle P(\xi <x)=0}$. ${\displaystyle x>0}$时, 记${\displaystyle B=\{(x_1,\cdots ,x_n)|\sum _{i=1}^n{x}_i^2<x\}}$. 则依据 (2.1), $$\begin{array}{rl}P(\xi <x)=& (2\pi {)}^{-\frac{n}{2}}\underset{B}{\iiint }\exp (-\frac{1}{2}\sum _{i=1}^n{x}_i^2)\mathrm{d}{x}_1\cdots \mathrm{d}{x}_n\\ =& c_n{\int }_0^{\sqrt{x}}{\mathrm{e}}^{-\frac{r^2}{2}}{r}^{n-1}\mathrm{d}r.\end{array}$$
为了求出系数${\displaystyle c_n}$, 首先注意到$$1=c_n{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\frac{r^2}{2}}{r}^{n-1}\mathrm{d}r,$$然后作代换 ${\displaystyle r=\sqrt{2t},\mathrm{d}r=\frac{\mathrm{d}t}{\sqrt{2t}}}$:$$\begin{array}{rl}1=& c_n{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-t}(2t{)}^{\frac{n-1}{2}}\frac{1}{\sqrt{2t}}\mathrm{d}t=c_n{2}^{\frac{n}{2}-1}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-t}{t}^{\frac{n}{2}-1}\mathrm{d}t=c_n{2}^{\frac{n}{2}-1}\mathrm{\Gamma }\left(\frac{n}{2}\right).\end{array}$$

#GammaFunction 表达式为 $$\mathrm{\Gamma }(x)={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-t}{t}^{x-1}\mathrm{d}t.$$

${\displaystyle X_1,\cdots ,X_n}$相互独立，${\displaystyle X_i\sim N(a_i,{\sigma }^2)}$. ${\displaystyle A=(a_{ij})}$为${\displaystyle n}$阶正交方阵, ${\displaystyle Y_i=\sum _{j=1}^n{a}_{ij}{X}_j(1\le i\le n)}$, 则${\displaystyle Y_1,\cdots ,Y_n}$相互独立, ${\displaystyle Y_i\sim N(b_i,{\sigma }^2)}$, 其中${\displaystyle b_i=\sum _{j=1}^n{a}_{ij}{a}_j}$.

证明 ${\displaystyle A}$ 的正交性保证了变换的 Jacobi 行列式的绝对值为 ${\displaystyle 1}$, 以及 ${\displaystyle \sum _{i=1}^n(y_i-b_i{)}^2=\sum _{i=1}^n(x_i-a_i{)}^2}$. ${\displaystyle (X_1,\cdots ,X_n)}$ 有联合密度 $$(2\pi {\sigma }^2{)}^{-\frac{n}{2}}\exp (-\frac{1}{2{\sigma }^2}\sum _{i=1}^n(x_i-a_i{)}^2),$$ 从而运用密度变换公式 有 ${\displaystyle (Y_1,\cdots ,Y_N)}$ 的概率密度 $$(2\pi {\sigma }^2{)}^{-\frac{n}{2}}\exp (-\frac{1}{2{\sigma }^2}\sum _{i=1}^n(y_i-b_i{)}^2).$$

取正交方阵${\displaystyle A}$(${\displaystyle n}$阶), 使第一行$$(a_{11},\cdots ,a_{1n})=(\frac{1}{\sqrt{n}},\cdots ,\frac{1}{\sqrt{n}}).$$
作变换${\displaystyle Y=AX}$, 其中${\displaystyle Y=(Y_1,\cdots ,Y_n{)}^T,X=(X_1,\cdots ,X_n{)}^T}$(这说明${\displaystyle \bar{X}=Y_1/\sqrt{n}}$), 则 引理1.1 指出${\displaystyle Y_1,\cdots ,Y_n}$相互独立,${\displaystyle Y_i\sim N(b_i,{\sigma }^2)}$, 其中${\displaystyle b_i=a\sum _{j=1}^n{a}_{ij}}$. 由于${\displaystyle A}$正交, ${\displaystyle \sum _{j=1}^n{a}_{ij}=0,i\ge 2}$, 因此$$b_i=\{\begin{array}{rl}& \sqrt{n}a,i=1,\\ & 0,i\ge 2.\end{array}$$
因此$$\begin{array}{rl}(n-1)S^2=& \sum _{i=1}^n(X_i-\bar{X}{)}^2=\sum _{i=1}^n{X}_i^2-n{\bar{X}}^2\\ =& \sum _{i=1}^n{Y}_i^2-Y_1^2=\sum _{i=2}^n{Y}_i^2.\end{array}$$
而${\displaystyle Y_2/\sigma ,\cdots ,Y_n/\sigma \sim N(0,1)}$且独立同分布, 故由${\displaystyle {\chi }^2}$分布的定义知, ${\displaystyle (n-1)S^2/{\sigma }^2\sim {\chi }_{n-1}^2}$.
最后, 由于${\displaystyle \bar{X}}$只依赖于${\displaystyle Y_1}$, ${\displaystyle S^2}$只依赖于${\displaystyle (Y_2,\cdots ,Y_n)}$, 而${\displaystyle Y_1,\cdots ,Y_n}$独立, 故${\displaystyle \bar{X}}$与${\displaystyle S}$独立.

由矩阵论，存在${\displaystyle B_{i_{n\times n_i}}:A_i=B_i{B}_i^T}$. 记${\displaystyle B=(B_1,\cdots ,B_m)}$为${\displaystyle n}$阶方阵, 作变换 ${\displaystyle Y=(Y_1,\cdots ,Y_n{)}^T=B^{T}X}$. 则$$\begin{array}{rl}{Y}^{T}Y=& X^{T}B{B}^{T}X=X^T\sum _{i=1}^m{B}_i{B}_i^{T}X\\ =& X^T\sum _{i=1}^m{A}_{i}X=X^{T}X.\end{array}$$由此, ${\displaystyle B}$为正交阵, 由 定理1.1 得${\displaystyle Y_1,\cdots ,Y_n}$相互独立, ${\displaystyle Y_i\sim N(b_i,{\sigma }^2)}$. 记${\displaystyle b=(b_1,\cdots ,b_n)=B^{T}a}$, 注意到$$B_i^{T}X=(Y_{n_1+\cdots +n_{i-1}+1},\cdots ,Y_{n_1+\cdots +n_i}),$$
知${\displaystyle {\xi }_i=\sum _{j=n_1+\cdots +n_{i-1}+1}^{n_1+\cdots +n_i}{Y}_j^2(1\le i\le m)}$. 于是由${\displaystyle Y_1,\dots ,Y_n}$独立推出${\displaystyle {\xi }_1,\cdots ,{\xi }_m}$独立.
当${\displaystyle a^T{A}_{1}a=0}$, 则${\displaystyle a^T{B}_1{B}_1^{T}a=0\Rightarrow B_1^T=0\Rightarrow b_1=\cdots =b_{n_1}=0}$, 于是${\displaystyle {\xi }_1/{\sigma }^2=\sum _{j=1}^{n_1}(Y_j/\sigma {)}^2}$, 推出${\displaystyle {\xi }_1/{\sigma }^2\sim {\chi }_{n_1}^2}$.

根据 (3.1), 得到${\displaystyle \sqrt{\frac{1}{n}Y}}$的密度$$h_2(x)=\frac{2n^{\frac{n}{2}}{x}^{n-1}{\mathrm{e}}^{-\frac{n{x}^2}{2}}}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}.$$
再结合${\displaystyle X}$的密度$$h_1(x)=\frac{1}{\sqrt{2\pi }}{\mathrm{e}}^{-\frac{x^2}{2}},$$
和概率论中的密度变换公式, $$g(x)={\int }_0^{\mathrm{\infty }}t{h}_1(xt)h_2(t)\mathrm{d}t,$$
因此$$\begin{array}{rl}g(x)& ={\int }_0^{\mathrm{\infty }}t\frac{1}{\sqrt{2\pi }}{\mathrm{e}}^{-\frac{x^2{t}^2}{2}}\frac{2n^{\frac{n}{2}}{t}^{n-1}{\mathrm{e}}^{-\frac{n{t}^2}{2}}}{2^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}\mathrm{d}t\\ & =\frac{2n^{\frac{n}{2}}}{\sqrt{2\pi }{2}^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\frac{(x^2+n)t^2}{2}}{t}^n\mathrm{d}t.\end{array}$$
进行换元${\displaystyle \frac{(x^2+n)t^2}{2}=u}$, 得$$\begin{array}{rl}g(x)& =\frac{2n^{\frac{n}{2}}}{\sqrt{2\pi }{2}^{\frac{n}{2}}\mathrm{\Gamma }\left(\frac{n}{2}\right)}\frac{1}{2}{\left(\frac{2}{x^2+n}\right)}^{\frac{n+1}{2}}\mathrm{\Gamma }\left(\frac{n+1}{2}\right)\\ & =\frac{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}{\sqrt{n\pi }\mathrm{\Gamma }\left(\frac{n}{2}\right)}{(1+\frac{x^2}{n})}^{-\frac{n+1}{2}}.\end{array}$$

根据 这个公式, 左边的函数可以写成 ${\displaystyle g_m(x)=A_1{A}_2}$, 其中 $$\begin{array}{rl}{A}_1& =\sqrt{pq}\sqrt{n}(\genfrac{}{}{0ex}{}{n}{m})F^m(t)[1-F(t){]}^{n-m}\frac{m}{nF(t)},\\ A_2& =\frac{f(t)}{f({\xi }_p)}.\end{array}$$ 其中 ${\displaystyle q=1-p}$, ${\displaystyle t={\xi }_p+\sqrt{\frac{pq}{n}}\frac{x}{f({\xi }_p)}}$. 由于 ${\displaystyle f}$ 在 ${\displaystyle {\xi }_p}$ 点连续且 ${\displaystyle f({\xi }_p)>0}$, 知 ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{f(t)}{f({\xi }_p)}=1}$. 而 ${\displaystyle (\genfrac{}{}{0ex}{}{n}{m})F^m(t)[1-F(t){]}^{n-m}}$ 为二项分布概率, 依据概率论局部极限定理, 如果能证明 $$\underset{n\to \mathrm{\infty }}{lim}\frac{m-nF(t)}{\sqrt{npq}}=-x,$$ 就有 ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{A}_1=\frac{1}{\sqrt{2\pi }}{\mathrm{e}}^{-\frac{x^2}{2}}}$, 从而 ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{g}_m(x)=\frac{1}{\sqrt{2\pi }}{\mathrm{e}}^{-\frac{x^2}{2}}}$.
对于要证的式子, 只要把 ${\displaystyle F}$ 在 ${\displaystyle {\xi }_p}$ 处 Taylor展开, 就有 $$F(t)=p+\sqrt{\frac{pq}{n}}x+o\left(n^{-\frac{1}{2}}\right).$$ (注意 ${\displaystyle F({\xi }_p)=p}$). 则 $$\frac{m-nF(t)}{\sqrt{npq}}=\frac{m-np}{npq}-x-\frac{1}{\sqrt{pq}}\sqrt{n}o(n^{-\frac{1}{2}}).$$ 再由假定 ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{m-np}{\sqrt{n}}=0}$, 证毕.

${\displaystyle X_1,\cdots ,X_n\sim N(\theta ,1)}$, ${\displaystyle T=\bar{X}=\frac{1}{n}\sum _{i=1}^n{X}_i}$. 类似这个证明, 作正交变换$$Y_1=\sum _{j=1}^n\frac{1}{\sqrt{n}}{X}_j=\sqrt{n}\bar{X}=\sqrt{n}T,Y_i=\sum _{j=1}^n{a}_{ij}{X}_j,2\le i\le n.$$
由于${\displaystyle Y_1,\cdots ,Y_n}$独立, 则给定${\displaystyle Y_1}$(即给定${\displaystyle T}$) 时, ${\displaystyle (Y_2,\cdots ,Y_n)}$的条件分布就是它的无条件分布, 且${\displaystyle Y_i\sim N(0,1),2\le i\le n}$, 与${\displaystyle \theta }$无关, 因此${\displaystyle \bar{X}}$是充分统计量.

依据这个例子, 考虑统计量${\displaystyle T=X_1+\cdots +X_n}$, 即抽出的${\displaystyle n}$个产品中的废品个数, 有$$\begin{array}{rl}P(X_i=x_i,1\le i\le n|T=a)=& \frac{P(X_i=x_i,T=a)}{P(T=a)}\\ =& \frac{P(X_i=x_i)}{P(T=a)}={(\genfrac{}{}{0ex}{}{n}{a})}^{-1}.\end{array}$$

回到 这个例子, 考虑统计量${\displaystyle T=X_1}$, 此时${\displaystyle (X_2,\cdots ,X_n)}$是无条件分布, 概率密度${\displaystyle \prod _{i=2}^n\left(\frac{1}{\sqrt{2\pi }}{\mathrm{e}}^{-\frac{(x_i-\theta {)}^2}{2}}\right)}$与${\displaystyle \theta }$有关, 故${\displaystyle T}$不是充分统计量.

• 充分性 考虑 ${\displaystyle X}$ 有概率密度, ${\displaystyle T}$ 有形式 ${\displaystyle (T_1,\cdots ,T_k)}$ 且可以补充 ${\displaystyle W=(W_1,\cdots ,W_{n-k})}$ 满足 这段评注, 记 ${\displaystyle J}$ 为对应的 Jacobi 行列式的绝对值, 记 ${\displaystyle T=h(x_1,\cdots ,x_n)/J}$ 和 ${\displaystyle H(T,W)}$, 则依据 密度变换公式 和条件式 (4.1), 知 ${\displaystyle (T,W)}$ 的密度 ${\displaystyle g_{\theta }(T)H(T,W)}$, 故 $$W|T\sim g_{\theta }(T)H(T,W)/{\int }_{{\mathbb{R}}^{n-k}}H(T,W)\mathrm{d}W=H(T,W)/{\int }_{{\mathbb{R}}^{n-k}}H(T,W)\mathrm{d}W$$ 与 ${\displaystyle \theta }$ 无关, 故 ${\displaystyle T}$ 为充分统计量.
• 必要性 以 ${\displaystyle G(T,W)}$ 记 ${\displaystyle W|T}$ 的密度, 由 ${\displaystyle T}$ 充分知 ${\displaystyle G}$ 与 ${\displaystyle \theta }$ 无关, 故记 ${\displaystyle (T,W)}$ 的联合密度为 ${\displaystyle g_{\theta }(T)G(T,W)}$. 再记 ${\displaystyle G(T,W)=h_1(x_1,\cdots ,x_n)}$, 则再次利用密度变换公式, 得 ${\displaystyle (X_1,\cdots ,X_n)}$ 的密度为 $$g_{\theta }(T(x_1,\cdots ,x_n))h_1(x_1,\cdots ,x_n)J=g_{\theta }(T(x_1,\cdots ,x_n))h(x_1,\cdots ,x_n),$$
  其中${\displaystyle h=h_{1}J}$与${\displaystyle \theta }$无关, 因此必要性得证.

${\displaystyle X_1,\cdots ,X_n\sim \mathrm{Uniform}(0,\theta )(\theta >0)}$, 记 $$\begin{array}{}\text{(4.2)}& T=max(X_1,\cdots ,X_n),\end{array}$$
则${\displaystyle (X_1,\cdots ,X_n)}$的联合密度为$$f_{\theta }(x_1,\cdots ,x_n)=\{\begin{array}{rl}& \frac{1}{{\theta }^n},max(x_1,\cdots ,x_n)<\theta ,\\ & 0,\text{其他情况}.\end{array}$$本例中样本只能取正值, 故${\displaystyle f_{\theta }(x_1,\cdots ,x_n)}$只能在${\displaystyle B=\{(x_1,\cdots ,x_n)|x_i>0\}}$上定义. 定义${\displaystyle g_{\theta }(t)=\frac{1}{{\theta }^n}}$(${\displaystyle 0<t<\theta }$, 在其他地方为${\displaystyle 0}$), ${\displaystyle h\equiv 1}$, 则$$f_{\theta }(x_1,\cdots ,x_n)=g_{\theta }(t)h(x_1,\cdots ,x_n).$$因此${\displaystyle T}$是充分统计量.