定义 ${\displaystyle X}$ 的特征函数为 ${\displaystyle {\mathrm{\Phi }}_X(t)=\mathbb{E}{\mathrm{e}}^{\mathrm{i}{t}^{\mathrm{T}}X}}$, 它是 ${\displaystyle t=(t_1,\cdots ,t_p{)}^{\mathrm{T}}}$ 的函数, 这里如果 ${\displaystyle X\sim {\mathcal{N}}_1(\mu ,{\sigma }^2)}$, 特征函数为 ${\displaystyle \exp \{\mathrm{i}\mu t-\frac{1}{2}{\sigma }^2{t}^2\}}$. 这样 ${\displaystyle \mathrm{\forall }t\in {\mathbb{R}}^p}$, ${\displaystyle t^{\mathrm{T}}X}$ 是正态随机变量, 它的特征函数是 $$\mathrm{\Phi }(\theta )=\mathbb{E}{\mathrm{e}}^{\mathrm{i}\theta t^{\mathrm{T}}X}=\exp \{\mathrm{i}\theta \mathbb{E}(t^{\mathrm{T}}X)-\frac{1}{2}\mathrm{Var}(t^{\mathrm{T}}X){\theta }^2\}.$$ 注意到 $$\mathbb{E}(t^{\mathrm{T}}X)=t^{\mathrm{T}}\mu ,\mathrm{Var}(t^{\mathrm{T}}X)=t^{\mathrm{T}}\mathrm{Cov}(Xt)=t^{\mathrm{T}}\mathrm{\Sigma }t,$$ 于是在 ${\displaystyle \mathrm{\Phi }(\theta )}$ 中令 ${\displaystyle \theta =1}$, 有 $$\mathbb{E}{\mathrm{e}}^{\mathrm{i}{t}^{\mathrm{T}}X}=\exp \{\mathrm{i}{t}^{\mathrm{T}}\mu -\frac{1}{2}{t}^{\mathrm{T}}\mathrm{\Sigma }t\}.$$ 因此可以说明 ${\displaystyle X}$ 的分布由 ${\displaystyle \mu ,\mathrm{\Sigma }}$ 唯一决定.

1. ${\displaystyle \mathrm{\forall }a\in {\mathbb{R}}^q}$, 有 ${\displaystyle a^{\mathrm{T}}Y=a^{\mathrm{T}}AX+a^{\mathrm{T}}b}$. 因为 ${\displaystyle a^{\mathrm{T}}AX}$ 正态, ${\displaystyle a^{\mathrm{T}}b}$ 常数, 从而 ${\displaystyle Y}$ 正态.
2. 根据 这里第三条, ${\displaystyle \mathrm{\Sigma }}$ 可以分解为 ${\displaystyle \mathrm{\Sigma }=A{A}^{\mathrm{T}}}$. 取 ${\displaystyle Y_1,\cdots ,Y_p\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }\mathcal{N}(0,1)}$. 记 ${\displaystyle Y=(Y_1,\cdots ,Y_p{)}^{\mathrm{T}}}$, 则 ${\displaystyle Y\sim {\mathcal{N}}_p(0,I_p)}$. 令 ${\displaystyle X=AY+\mu }$, 则由性质 1: $$X\sim {\mathcal{N}}_p(\mu ,A{A}^{\mathrm{T}})={\mathcal{N}}_p(\mu ,\mathrm{\Sigma }).$$
3. 由上面的证明蕴含.
4. 由性质 3: ${\displaystyle X={\mathrm{\Sigma }}^{\frac{1}{2}}Y+\mu }$, ${\displaystyle Y\sim {\mathcal{N}}_p(0,I_p)}$. 则 $$f(y)=(2\pi {)}^{-\frac{p}{2}}\exp \{-\frac{1}{2}{y}^{\mathrm{T}}y\}.$$ 作变换 ${\displaystyle y={\mathrm{\Sigma }}^{-\frac{1}{2}}(x-\mu )}$, 则 Jacobi 行列式是 ${\displaystyle det{\mathrm{\Sigma }}^{-\frac{1}{2}}}$. 这样根据 密度变换公式 得证.
5. 只证明 2. 令 ${\displaystyle P=\left(\begin{array}{cc}{I}_{p_1}& 0\\ -{\mathrm{\Sigma }}_{21}{\mathrm{\Sigma }}_{11}^{-1}& I_{p_2}\end{array}\right)}$, 变换 ${\displaystyle Y=PX}$, 则$$\begin{array}{rl}Y& =\left(\begin{array}{c}{X}_{(1)}\\ X_{(2)}-{\mathrm{\Sigma }}_{21}{\mathrm{\Sigma }}_{11}^{-1}{X}_{(1)}\end{array}\right)=\left(\begin{array}{c}{Y}_{(1)}\\ Y_{(2)}\end{array}\right),\\ \mathrm{Cov}Y& =P\mathrm{Cov}(X{P}^{\mathrm{T}})=\left(\begin{array}{cc}{\mathrm{\Sigma }}_{11}& 0\\ 0& {\mathrm{\Sigma }}_{22}-{\mathrm{\Sigma }}_{21}{\mathrm{\Sigma }}_{11}^{-1}{\mathrm{\Sigma }}_{12}\end{array}\right),\end{array}$$ 从而 ${\displaystyle Y_{(1)}\perp \perp Y_{(2)}}$, 因此给定 ${\displaystyle Y_{(1)}=X_{(1)}}$ 时, $$Y_{(2)}\sim {\mathcal{N}}_{p_2}({\mu }_{(2)}-{\mathrm{\Sigma }}_{21}{\mathrm{\Sigma }}_{11}^{-1}{\mu }_{(1)},{\mathrm{\Sigma }}_{22}-{\mathrm{\Sigma }}_{21}{\mathrm{\Sigma }}_{11}^{-1}{\mathrm{\Sigma }}_{12}).$$ 但 ${\displaystyle X_{(2)}=Y_{(2)}+{\mathrm{\Sigma }}_{21}{\mathrm{\Sigma }}_{11}^{-1}{X}_{(1)}}$, 从而得证.

只证明 2. 记 ${\displaystyle Y^{\mathrm{T}}=A{X}^{\mathrm{T}}=(y_1,\cdots ,y_n)}$. 易见 ${\displaystyle y_{\alpha }\sim {\mathcal{N}}_k(A{\mu }_{\alpha },A\mathrm{\Sigma }{A}^{\mathrm{T}})}$, 且相互独立. 因此 ${\displaystyle AW{A}^{\mathrm{T}}=Y^{\mathrm{T}}Y\sim W_k(n,A\mathrm{\Sigma }{A}^{\mathrm{T}},A\mathrm{\Delta }{A}^{\mathrm{T}})}$.

1. • "${\displaystyle \Rightarrow }$": ${\displaystyle \mathrm{\forall }a\in {\mathbb{R}}^p}$, ${\displaystyle a^{\mathrm{T}}Wa\sim {\chi }_k^2(\delta )}$, 从而 ${\displaystyle C}$ 为正投影阵.
   • "${\displaystyle \Leftarrow }$": ${\displaystyle \mathrm{\exists }{U}_{n\times k}}$: ${\displaystyle C=U{U}^{\mathrm{T}}}$, ${\displaystyle U^{\mathrm{T}}U=I_k}$. 令 ${\displaystyle Y=U^{\mathrm{T}}X}$. 记 ${\displaystyle U=(u_1,\cdots ,u_k)}$, 有 $$\mathrm{Cov}(X^{\mathrm{T}}{u}_i,X^{\mathrm{T}}{u}_j)=\mathrm{Cov}(\sum _{\alpha }{x}_{\alpha }{u}_{i\alpha },\sum _{\alpha }{x}_{\alpha }{u}_{j\alpha })={\delta }_{ij}\mathrm{\Sigma }.$$ 于是 ${\displaystyle X^{\mathrm{T}}CX=Y^{\mathrm{T}}Y}$, 而 ${\displaystyle Y\sim {\mathcal{N}}_{kp}(U^{\mathrm{T}}M,I,\mathrm{\Sigma })}$.
2. ${\displaystyle \mathrm{\forall }i,j}$: $$\mathrm{Cov}(\sum _{\alpha }{x}_{\alpha }{a}_{i\alpha },\sum _{\alpha }{x}_{\alpha }{b}_{j\alpha })=\sum _{\alpha }{a}_{i\alpha }{b}_{j\alpha }\cdot \mathrm{\Sigma }=0⟺A{B}^{\mathrm{T}}=0.$$
   从而得证.

按定义, ${\displaystyle W=X^{\mathrm{T}}X}$, ${\displaystyle X\sim {\mathcal{N}}_{np}(M,I,\mathrm{\Sigma })}$, ${\displaystyle M^{\mathrm{T}}M=\mathrm{\Delta }}$. 记 ${\displaystyle X=(x_{(1)},\cdots ,x_{(p)})}$. 对 ${\displaystyle k<p}$, 由 性质5: ${\displaystyle x_{(k+1)}|x_{(1)},\cdots ,x_{(k)}\sim \mathcal{N}(v_{k+1},{\sigma }_{k+1}^{2}I)}$. 这里 ${\displaystyle {\sigma }_{k+1}^2=e_1^{\mathrm{T}}({\mathrm{\Sigma }}_{22}-{\mathrm{\Sigma }}_{21}{\mathrm{\Sigma }}_{11}^{-1}{\mathrm{\Sigma }}_{12})e_1}$, 这里 ${\displaystyle e_1=(1,0,\cdots ,0)}$. 因此条件分布非退化, 故有 $$\mathbb{P}(x_{(k+1)}\in \mathrm{Im}(x_{(1)},\cdots ,x_{(k)})|x_{(1)},\cdots ,x_{(k)})=0.$$ 这是由于 ${\displaystyle \mathrm{Im}(x_{(1)},\cdots ,x_{(k)})}$ 至多 ${\displaystyle k}$ 维, 而 ${\displaystyle x_{(k+1)}}$ 是 ${\displaystyle n}$ 维的, 从而$$\begin{array}{rl}& \mathbb{P}(x_{(k+1)}\in \mathrm{Im}(x_{(1)},\cdots ,x_{(k)}))\\ =& \mathbb{E}\{\mathbb{P}(x_{(k+1)}\in \mathrm{Im}(x_{(1)},\cdots ,x_{(k)})|x_{(1)},\cdots ,x_{(k)})\}=0.\end{array}$$ 于是 $$\begin{array}{rl}\mathbb{P}(\mathrm{rank}X<p)& =\mathbb{P}(x_{(1)},\cdots ,x_{(p)}\text{线性相关})\\ & \le \sum _{k=0}^{p-1}\mathbb{P}(x_{(k+1)}\in \mathrm{Im}(x_{(1)},\cdots ,x_{(k)}))=0,\end{array}$$ 因此 ${\displaystyle \mathbb{P}(X^{\mathrm{T}}X>0)=1}$, 从而得证.

在 定理2.2(2) 中, 取 ${\displaystyle A=\left(\begin{array}{cc}0& 0\\ 0& I_{p-q}\end{array}\right)}$, 得 ${\displaystyle W_{22}\sim W_{p-q}(n,{\mathrm{\Sigma }}_{22})}$.
设 ${\displaystyle W=X^{\mathrm{T}}X}$, ${\displaystyle X\sim {\mathcal{N}}_{np}(0,I,\mathrm{\Sigma })}$, 记 ${\displaystyle X=(X_1,X_2)}$, ${\displaystyle X_1}$ 有 ${\displaystyle q}$ 列. 因 ${\displaystyle X}$ 的行独立同分布, 由 性质5(2): $$(X_1|X_2)\sim {\mathcal{N}}_{nq}(X_2{\mathrm{\Sigma }}_{22}^{-1}{\mathrm{\Sigma }}_{21},I,{\mathrm{\Sigma }}_{11,2}).$$
注意到 ${\displaystyle W_{11}=X_1^{\mathrm{T}}{X}_1}$, ${\displaystyle W_{21}=X_2^{\mathrm{T}}{X}_1}$, ${\displaystyle W_{22}=X_2^{\mathrm{T}}{X}_2}$. 由 定理2.1: $$W_{22}^{-1}{W}_{21}=(X_2^{\mathrm{T}}{X}_2{)}^{-1}{X}_2^{\mathrm{T}}{X}_1|X_2\sim {\mathcal{N}}_{p-q,q}({\mathrm{\Sigma }}_{22}^{-1}{\mathrm{\Sigma }}_{21},(X_2^{\mathrm{T}}{X}_2{)}^{-1},{\mathrm{\Sigma }}_{11,2}),$$ 知它仅依赖 ${\displaystyle X_2^{\mathrm{T}}{X}_2=W_{22}}$. 故 $$W_{22}^{-1}{W}_{21}|W_{22}\sim {\mathcal{N}}_{p-q,q}({\mathrm{\Sigma }}_{22}^{-1}{\mathrm{\Sigma }}_{21},W_{22}^{-1},{\mathrm{\Sigma }}_{11,2}).$$
而 $$W_{11,2}=X_1^{\mathrm{T}}(I-X_2(X_2^{\mathrm{T}}{X}_2{)}^{-1}{X}_2^{\mathrm{T}})X_1=X_1^{\mathrm{T}}{P}_{X_2^{\perp }}{X}_1,$$ 由 定理2.3(1): $$W_{11,2}|X_2\sim W_q(n-p+q,{\mathrm{\Sigma }}_{11,2}).$$ (此时 ${\displaystyle \mathrm{\Delta }={\mathrm{\Sigma }}_{12}{\mathrm{\Sigma }}_{22}^{-1}{X}_2^{\mathrm{T}}{P}_{X_2^{\perp }}{X}_2{\mathrm{\Sigma }}_{22}^{-1}{\mathrm{\Sigma }}_{21}=0}$). 又由 定理2.3(2): 由 ${\displaystyle (X_2^{\mathrm{T}}{X}_2{)}^{-1}{X}_2^{\mathrm{T}}{P}_{X_2}=0}$ 得 ${\displaystyle W_{11,2}|X_2\perp \perp W_{22}^{-1}{W}_{21}|X_2}$. 而 ${\displaystyle W_{11,2}}$ 和 ${\displaystyle X_2}$ 无关, ${\displaystyle W_{22}^{-1}{W}_{21}}$ 仅依赖于 ${\displaystyle W_{22}}$, 故 $$W_{11,2}|(W_{22}^{-1}{W}_{21},W_{22})\sim W_q(n-p+q,{\mathrm{\Sigma }}_{11,2}).$$

1. 先考虑 ${\displaystyle a=e_1}$, 有 $$\frac{a^{\mathrm{T}}{\mathrm{\Sigma }}^{-1}a}{a^{\mathrm{T}}{W}^{-1}a}=\frac{({\mathrm{\Sigma }}^{-1}{)}_{11}}{(W^{-1}{)}_{11}}=\frac{({\mathrm{\Sigma }}_{11}-{\mathrm{\Sigma }}_{12}{\mathrm{\Sigma }}_{22}^{-1}{\mathrm{\Sigma }}_{21}{)}^{-1}}{(W_{11}-W_{12}{W}_{22}^{-1}{W}_{21}{)}^{-1}}=\frac{W_{11}-W_{12}{W}_{22}^{-1}{W}_{21}}{{\mathrm{\Sigma }}_{11}-{\mathrm{\Sigma }}_{12}{\mathrm{\Sigma }}_{22}^{-1}{\mathrm{\Sigma }}_{21}}.$$ 由 引理2.1 中 ${\displaystyle q=1}$ 的情形: $$\frac{e_1^{\mathrm{T}}{\mathrm{\Sigma }}^{-1}{e}_1}{e_1^{\mathrm{T}}{W}^{-1}{e}_1}\sim {\chi }_{n-p+1}^2.$$ 一般地对 ${\displaystyle a\in {\mathbb{R}}^p}$, 令 ${\displaystyle u_1=\frac{a}{||a||}}$, 作正交阵 ${\displaystyle U=(u_1,\cdots ,u_p)}$, 有 ${\displaystyle a=U{e}_1||a||}$: $$\frac{a^{\mathrm{T}}{\mathrm{\Sigma }}^{-1}a}{a^{\mathrm{T}}{W}^{-1}a}=\frac{e_1^{\mathrm{T}}{U}^{\mathrm{T}}{\mathrm{\Sigma }}^{-1}U{e}_1}{e_1^{\mathrm{T}}{U}^{\mathrm{T}}{W}^{-1}U{e}_1}=\frac{e_1^{\mathrm{T}}(U^{\mathrm{T}}\mathrm{\Sigma }U{)}^{-1}{e}_1}{e_1^{\mathrm{T}}(U^{\mathrm{T}}WU{)}^{-1}{e}_1}.$$ 由 定理2.2: ${\displaystyle U^{\mathrm{T}}WU\sim W_p(n,U^{\mathrm{T}}\mathrm{\Sigma }U)}$, 故得证.
2. 注意到 $$x^{\mathrm{T}}{W}^{-1}x=\frac{x^{\mathrm{T}}{\mathrm{\Sigma }}^{-1}x}{x^{\mathrm{T}}{\mathrm{\Sigma }}^{-1}x/(x^{\mathrm{T}}{W}^{-1}x)}.$$ 有 ${\displaystyle \left(\frac{x^{\mathrm{T}}{\mathrm{\Sigma }}^{-1}x}{x^{\mathrm{T}}{W}^{-1}x}|x\right)\sim {\chi }_{n-p+1}^2}$. 因此实际上条件分布不依赖 ${\displaystyle x}$, 从而 ${\displaystyle \frac{x^{\mathrm{T}}{\mathrm{\Sigma }}^{-1}x}{x^{\mathrm{T}}{W}^{-1}x}\sim {\chi }_{n-p+1}^2}$, 与 ${\displaystyle x}$ 独立, 从而与 ${\displaystyle x^{\mathrm{T}}{\mathrm{\Sigma }}^{-1}x}$ 独立.

容易验算 $$\mathbb{E}\bar{y}=\mathbb{E}\left(Y^{\mathrm{T}}{\mathbf{1}}_n\frac{1}{n}\right)=M^{\mathrm{T}}{\mathbf{1}}_n\frac{1}{n}=\mu .$$
设 ${\displaystyle Y^{\mathrm{T}}a}$ 是 ${\displaystyle \mu }$ 的任一无偏估计, 则有 $$\mathbb{E}(Y^{\mathrm{T}}a)=M^{\mathrm{T}}a=\sum _{i=1}^n{a}_i\mu =\mu ⟺\sum _{i=1}^n{a}_i=1.$$ 又注意到$$\begin{array}{rl}\mathrm{Cov}(Y^{\mathrm{T}}a-Y^{\mathrm{T}}\mathbf{1}\frac{1}{n},Y^{\mathrm{T}}\mathbf{1}\frac{1}{n})& =\mathrm{Cov}(\sum _{\alpha =1}^n(a_{\alpha }-\frac{1}{n})y_{\alpha },\sum _{\alpha =1}^n\frac{1}{n}{y}_{\alpha })\\ & =\sum _{\alpha =1}^n\frac{1}{n}(a_{\alpha }-\frac{1}{n})\mathrm{\Sigma }=0,\end{array}$$
从而 $$\mathrm{Cov}(Y^{\mathrm{T}}a)-\mathrm{Cov}(\bar{y})=\mathrm{Cov}(Y^{\mathrm{T}}a-\bar{y})\ge 0.$$
顺便注意到 ${\displaystyle \mathrm{Cov}\bar{y}=\frac{1}{n}\mathrm{\Sigma }}$, 由于$$\begin{array}{rl}(n-1)\mathbb{E}S& =\mathbb{E}(\sum _{\alpha =1}^n(y_{\alpha }-\bar{y})(y_{\alpha }-\bar{y}{)}^{\mathrm{T}})=\mathbb{E}(\sum _{\alpha =1}^n{y}_{\alpha }{y}_{\alpha }^{\mathrm{T}}-n\bar{y}{\bar{y}}^{\mathrm{T}})\\ & =\sum _{\alpha =1}^n(\mathrm{\Sigma }+\mu {\mu }^{\mathrm{T}})-n(\frac{1}{n}\mathrm{\Sigma }+\mu {\mu }^{\mathrm{T}})=(n-1)\mathrm{\Sigma },\end{array}$$ 得 ${\displaystyle S}$ 是 ${\displaystyle \mathrm{\Sigma }}$ 的无偏估计.

${\displaystyle Y}$ 的联合密度是$$\begin{array}{rl}f(Y)& =C(det\mathrm{\Sigma }{)}^{-\frac{n}{2}}\exp \{-\frac{1}{2}\sum _{\alpha =1}^n(y_{\alpha }-{\mu }^{\mathrm{T}}){\mathrm{\Sigma }}^{-1}(y_{\alpha }-\mu )\}\\ & =C(det\mathrm{\Sigma }{)}^{-\frac{n}{2}}\exp \{-\frac{1}{2}\mathrm{tr}({\mathrm{\Sigma }}^{-1}\sum _{\alpha =1}^n(y_{\alpha }-\mu )(y_{\alpha }-\mu {)}^{\mathrm{T}})\}.\end{array}$$
由于 $$\begin{array}{rl}\sum _{\alpha =1}^n(y_{\alpha }-\mu )(y_{\alpha }-\mu {)}^{\mathrm{T}}& =\sum _{\alpha =1}^n{y}_{\alpha }{y}_{\alpha }^{\mathrm{T}}-n\bar{y}{\mu }^{\mathrm{T}}-n\mu {\bar{y}}^{\mathrm{T}}+n\mu {\mu }^{\mathrm{T}}\\ \text{(3.3)}& & =Y^{\mathrm{T}}Y-n\bar{y}{\mu }^{\mathrm{T}}-n\mu {\bar{y}}^{\mathrm{T}}+n\mu {\mu }^{\mathrm{T}},\end{array}$$ 得 $$f(Y)=C(det\mathrm{\Sigma }{)}^{-\frac{n}{2}}\exp \{-\frac{1}{2}\mathrm{tr}({\mathrm{\Sigma }}^{-1}{Y}^{\mathrm{T}}Y)+n{\mu }^{\mathrm{T}}{\mathrm{\Sigma }}^{-1}\bar{y}\}\exp \{-\frac{n}{2}{\mu }^{\mathrm{T}}{\mathrm{\Sigma }}^{-1}\mu \},$$ 可见 ${\displaystyle (\bar{y},Y^{\mathrm{T}}Y)}$ 是充分完备统计量 (因为是指数型分布族). 又因为是无偏估计, 所以是 UMVUE.

由 (3.3), 对给定的 ${\displaystyle \mathrm{\Sigma }}$, 为了让 ${\displaystyle f(Y;\mu ,\mathrm{\Sigma })}$ 达到极大, 只需要 ${\displaystyle \hat{\mu }=\bar{y}}$, 从而 $$f(Y;\hat{\mu },\mathrm{\Sigma })=C(det\mathrm{\Sigma }{)}^{-\frac{n}{2}}\exp \{-\frac{1}{2}\mathrm{tr}({\mathrm{\Sigma }}^{-1}W)\}.$$ 故$$\begin{array}{rl}\ln f(Y;\hat{\mu },\mathrm{\Sigma })& =\ln C+\frac{n}{2}\ln (det{\mathrm{\Sigma }}^{-1})-\frac{n}{2}\mathrm{tr}\left({\mathrm{\Sigma }}^{-1}\frac{W}{n}\right)\\ & =\ln C-\frac{n}{2}\ln det\frac{W}{n}+\frac{n}{2}(\ln det\frac{{\mathrm{\Sigma }}^{-1}W}{n}-\mathrm{tr}\frac{{\mathrm{\Sigma }}^{-1}W}{n}).\end{array}$$
记 ${\displaystyle \frac{{\mathrm{\Sigma }}^{-1}W}{n}}$ 的特征值为 ${\displaystyle {\lambda }_1,\cdots ,{\lambda }_p}$, 则$$\begin{array}{rl}& \ln det\frac{{\mathrm{\Sigma }}^{-1}W}{n}-\mathrm{tr}\frac{{\mathrm{\Sigma }}^{-1}W}{n}=\sum _{i=1}^p(\ln {\lambda }_i-{\lambda }_i)\\ =& \sum _{i=1}^p[\ln (1+{\lambda }_i-1)-{\lambda }_i]\le \sum _{i=1}^p({\lambda }_i-1-{\lambda }_i)=-p.\end{array}$$ 且上述不等式当且仅当 ${\displaystyle {\lambda }_i=1}$ 时取等号. 因此, ${\displaystyle f(Y;\hat{\mu },\mathrm{\Sigma })}$ 的极大值点是 ${\displaystyle \hat{\epsilon }=\frac{W}{n}}$.