设\(M_{n \times p}\)是正态总体\(\text{N}(\boldsymbol 0, \Sigma)\)的观测数据集， \(C\)为对称阵，则

(1) \(M^T C M\)服从加权Wishart分布，即 \[\begin{aligned} M^T C M = \sum_{i=1}^n \lambda_i W_p(\Sigma, 1) , \end{aligned}\] 其中\(\lambda_i, i=1,\dots, n\)是\(C\)的各个特征值， 右式中各个Wishart分布相互独立。

(2) \(M^T C M\)服从Wishart分布当且仅当\(C^2=C\)，这时 \[\begin{aligned} M^T C M \sim W_p(\Sigma, r) , \end{aligned}\] \(r = \text{rank}(C) = \text{tr}(C)\)。

定理4.1 设\(\boldsymbol x^{(1)}, \boldsymbol x^{(2)}, \dots, \boldsymbol x^{(n)}\) iid \(\sim \text{N}(\boldsymbol\mu, \Sigma)\), \(M^T = (\boldsymbol x^{(1)}, \boldsymbol x^{(2)}, \dots, \boldsymbol x^{(n)})\), 即\(M_{n \times p}\)是正态总体\(\text{N}(\boldsymbol\mu, \Sigma)\)的观测数据集， 样本统计量 \[\begin{aligned} \bar{\boldsymbol x} =& \frac{1}{n} \sum_{i=1}^n \boldsymbol x^{(i)}, \\ S =& \frac{1}{n} M_C^T M_C = \frac{1}{n} M^T H M \\ =& \frac{1}{n} \sum_{i=1}^n (\boldsymbol x^{(i)} - \bar{\boldsymbol x}) (\boldsymbol x^{(i)} - \bar{\boldsymbol x})^T , \end{aligned}\] 则有

(1) \(\bar{\boldsymbol x} \sim \text{N}(\boldsymbol\mu, \frac{1}{n} \Sigma)\)。

(2) \(n S = M_C^T M_C = M^T H M \sim W_p(\Sigma, n-1)\)。

(3) \(\bar{\boldsymbol x}\)与\(S\)相互独立。

证明： (1) 前面已经证明。

(2)和(3)的证明： 取\(n \times n\)正交阵\(\Gamma\)， 设其各个列向量为\(\boldsymbol\gamma_i\)， \(i=1,2,\dots,n\)， 取\(\boldsymbol\gamma_n = \frac{1}{\sqrt{n}} \boldsymbol 1_n\)。令 \[ N = \Gamma^T M = (\boldsymbol z^{(1)}, \boldsymbol z^{(2)}, \dots, \boldsymbol z^{(n)})^T , \ N^T = M^T \Gamma, \] 最后一行 \[ \boldsymbol z^{(n)} = M \boldsymbol\gamma_n = \frac{1}{\sqrt{n}} M^T \boldsymbol 1_n = \sqrt{n} \bar{\boldsymbol x} . \]

由随机矩阵性质和正交阵性质， \[ E(N) = \Gamma^T E(M) = \Gamma^T \boldsymbol 1_n \boldsymbol\mu^T = \begin{pmatrix} \boldsymbol 0_{(n-1) \times n} \\ \sqrt{n} \boldsymbol\mu^T \end{pmatrix} . \] 所以 \[\begin{aligned} E(\boldsymbol z^{(i)}) =& \boldsymbol 0_p, \ i=1,2,\dots, n-1, \\ E(\boldsymbol z^{(n)}) =& \sqrt{n} \boldsymbol\mu . \end{aligned}\]

由随机矩阵性质(3.5)式， 可得 \[\begin{aligned} & \text{Var}(\text{vec}(N^T)) \\ =& (\Gamma^T \Gamma) \otimes (I \Sigma I) = I_n \otimes \Sigma , \end{aligned}\] 从而\(\boldsymbol z^{(i)}\)， \(i=1,2,\dots,n\)相互不相关， 由多元正态性质可知其相互独立， 都服从多元正态分布， 且 \[\begin{aligned} \boldsymbol z^{(i)} \sim& \text{N}_p(\boldsymbol 0, \Sigma), \; n=1,2,\dots, n-1, \\ \boldsymbol z^{(n)} \sim& \text{N}_p(\sqrt{n} \boldsymbol\mu, \Sigma) . \end{aligned}\]

注意 \[ N^T N = M^T \Gamma \Gamma^T M = M^T M, \] 而 \[\begin{aligned} M^T M =& (\boldsymbol x^{(1)}, \boldsymbol x^{(2)}, \dots, \boldsymbol x^{(n)}) \begin{pmatrix} [\boldsymbol x^{(1)}]^T \\ [\boldsymbol x^{(2)}]^T \\ \vdots \\ [\boldsymbol x^{(n)}]^T \end{pmatrix} \\ =& \sum_{i=1}^n \boldsymbol x^{(i)} [\boldsymbol x^{(i)}]^T, \\ N^T N =& \sum_{i=1}^n \boldsymbol z^{(i)} [\boldsymbol z^{(i)}]^T, \end{aligned}\] 注意到\(\boldsymbol z^{(n)} = \sqrt{n} \bar{\boldsymbol x}\)， 于是 \[\begin{aligned} & \sum_{i=1}^{n-1} \boldsymbol z^{(i)} [\boldsymbol z^{(i)}]^T \\ =& \sum_{i=1}^n \boldsymbol x^{(i)} [\boldsymbol x^{(i)}]^T - n \bar{\boldsymbol x} \bar{\boldsymbol x}^T \\ =& \sum_{i=1}^n (\boldsymbol x^{(i)} - \bar{\boldsymbol x}) (\boldsymbol x^{(i)} - \bar{\boldsymbol x})^T \\ =& n S = n M^T H M . \end{aligned}\] 因为\(\sum_{i=1}^{n-1} \boldsymbol z^{(i)} [\boldsymbol z^{(i)}]^T\)服从\(W_p(\Sigma, n-1)\)， 所以\(n S\)服从\(W_p(\Sigma, n-1)\)， 由\(\boldsymbol z^{(n)}\)与\(\boldsymbol z^{(1)}, \dots, \boldsymbol z^{(n-1)}\)的独立性可知\(\bar{\boldsymbol x}\)与\(S\)独立。

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