3 多元分布

设\(\boldsymbol M\)为\(n \times p\)随机矩阵， 则\(E \boldsymbol M\)也是\(n \times p\)矩阵， 其\((i,j)\)元素等于\(\boldsymbol M\)的第\((i,j)\)元素的期望值。

若\(A, B, C\)为普通矩阵， 有 \[ E(A \boldsymbol M B^T + C) = A E(\boldsymbol M) B^T + C . \]

设\(\boldsymbol X\)为\(p \times 1\)随机向量， \(\boldsymbol a\)为\(n \times 1\)向量， 则\(\boldsymbol a \boldsymbol X^T\)为\(n \times p\)随机矩阵。 将\(\boldsymbol a \boldsymbol X^T\)按行拉直，有 \[\begin{aligned} & \text{Var}\left\{ \text{vec}( [\boldsymbol a \boldsymbol X^T]^T) \right\} = \text{Var}\left\{ \text{vec}( \boldsymbol X \boldsymbol a^T) \right\} \\ =& \text{Var}\left\{ \begin{pmatrix} a_1 \boldsymbol X \\ a_2 \boldsymbol X \\ \vdots \\ a_n \boldsymbol X \end{pmatrix} \right\} \\ =& \begin{pmatrix} a_1^2 \Sigma & a_1 a_2 \Sigma & \cdots & a_1 a_n \Sigma \\ a_2 a_1 \Sigma & a_2^2 \Sigma & \cdots & a_2 a_n \Sigma \\ \vdots & \vdots & \ddots & \vdots \\ a_n a_1 \Sigma & a_n a_2 \Sigma & \cdots & a_n^2 \Sigma \end{pmatrix} \\ =& (\boldsymbol a \boldsymbol a^T) \otimes \Sigma . \end{aligned}\] 即： \[\begin{align} \text{Var}\left\{ \text{vec}( [\boldsymbol a \boldsymbol X^T]^T) \right\} = \text{Var}\left\{ \text{vec}( \boldsymbol X \boldsymbol a^T) \right\} = (\boldsymbol a \boldsymbol a^T) \otimes \Sigma . \tag{3.1} \end{align}\]

若\(\boldsymbol M\)的各行\([\boldsymbol x^{(i)}]^T\), \(i=1,2,\dots,n\)相互独立同分布， \[ E(\boldsymbol x^{(i)}) = \boldsymbol\mu, \quad \text{Var}(\boldsymbol x^{(i)}) = \Sigma, \] 则\(\boldsymbol M\)按行拉直后，有 \[\begin{aligned} E[\text{vec}(\boldsymbol M^T)] =& \boldsymbol 1_n \otimes \boldsymbol \mu, \\ \text{Var}[\text{vec}(\boldsymbol M^T)] =& I_n \otimes \boldsymbol \Sigma . \end{aligned}\] 这时令随机矩阵\(N\)定义为 \[\begin{aligned} \boldsymbol N = \boldsymbol M B^T + C, \end{aligned}\] 其中\(B\)为\(q \times p\)矩阵， 则\(N\)的第\(i\)行\([\boldsymbol y^{(i)}]^T\)完全由\([\boldsymbol x^{(i)}]^T\)决定： \[\begin{aligned} \; [\boldsymbol y^{(i)}]^T =& [\boldsymbol x^{(i)}]^T B^T + [\boldsymbol c^{(i)}]^T, \\ \boldsymbol y^{(i)} =& B \boldsymbol x^{(i)} + \boldsymbol c^{(i)} . \end{aligned}\] 其中\([\boldsymbol c^{(i)}]^T\)是\(C\)的第\(i\)行。 于是， 各\(\boldsymbol y^{(i)}\), \(i=1,2,\dots,n\)独立同分布， \[ \text{Var}(\boldsymbol y^{(i)}) = \text{Var}(B \boldsymbol x^{(i)}) = B \Sigma B^T, \] 于是 \[\begin{align} E(\boldsymbol N) =& E(\boldsymbol M) B^T + C, \tag{3.2} \\ \text{Var}(\text{vec}(\boldsymbol N^T)) =& I_n \otimes (B \Sigma B^T) . \tag{3.3} \end{align}\]

进一步地， 仍设\(\boldsymbol M\)的各行独立同分布， 令 \[\begin{aligned} \boldsymbol N = A \boldsymbol M B^T + C, \end{aligned}\] 则有 \[\begin{align} E(\boldsymbol N) =& A E(\boldsymbol M) B^T + C, \tag{3.4} \\ \text{Var}(\text{vec}(\boldsymbol N^T)) =& (A A^T) \otimes (B \Sigma B^T) . \tag{3.5} \end{align}\] (3.4)由期望的线性性质可得。 下面证明(3.5)。

因为\(C\)不影响方差和协方差计算， 在证明(3.5)时设\(C = 0\)。 令\(\boldsymbol z^{(i)} = B \boldsymbol x^{(i)}\)， 则各\(\boldsymbol z^{(i)}\)独立同分布， \[ \text{Var}(\boldsymbol z^{(i)}) = B \Sigma B . \] 且\(M B^T\)的各行为\([\boldsymbol z^{(i)}]^T\)。 设\(A\)为\(m \times n\)矩阵， 各个列向量为\(\boldsymbol a_i\), \(i=1,2,\dots, n\)， 则 \[\begin{aligned} \boldsymbol N =& A \boldsymbol M B^T \\ =& (\boldsymbol a_1, \boldsymbol a_2, \dots, \boldsymbol a_n) \begin{pmatrix} [\boldsymbol z^{(1)}]^T \\ [\boldsymbol z^{(2)}]^T \\ \vdots \\ [\boldsymbol z^{(n)}]^T \\ \end{pmatrix} \\ =& \sum_{i=1}^n \boldsymbol a_i [\boldsymbol z^{(i)}]^T , \end{aligned}\] 其按行拉直的协方差阵为 \[\begin{aligned} & \text{Var}\left\{ \text{vec}( \boldsymbol N^T) \right\} = \text{Var}\left\{ \text{vec} \left[ \left( \sum_{i=1}^n \boldsymbol a_i [\boldsymbol z^{(i)}]^T \right)^T \right] \right\} \\ =& \text{Var}\left\{ \text{vec} \left[ \sum_{i=1}^n \boldsymbol z^{(i)} \boldsymbol a_i^T \right] \right\} = \text{Var}\left\{ \sum_{i=1}^n \text{vec} \left[ \boldsymbol z^{(i)} \boldsymbol a_i^T \right] \right\} \\ =& \sum_{i=1}^n \text{Var}\left\{ \boldsymbol z^{(i)} \boldsymbol a_i^T \right\} \\ =& \sum_{i=1}^n (\boldsymbol a_i \boldsymbol a_i^T) \otimes (B \Sigma B) \\ =& (A A^T) \otimes (B \Sigma B) , \end{aligned}\] 证明中用到不同的\(\boldsymbol z^{(i)}\)的不相关性， 拉直运算的线性性质， Kronecker乘积的线性性质， 和式(3.1)。 (3.5)式证毕。