假设二维空间中有8个样本点分成以下两类。第1类：02-2035-3-5第2类：35138602。请计算两类样本的均值与协方差矩阵计算02与35分别在两类中的马氏距离。给出计算过程。

第1类样本均值： $(\overline{x_1},\overline{x_2})=\frac{1}{4}[(0+(-2)+3+(-3)),(2+0+5+(-5))]=(0,0)$

第2类样本均值： $(\overline{x_1},\overline{x_2})=\frac{1}{4}[(3+1+8+0),(5+3+6+2)]=(3,4)$

第1类样本协方差矩阵： $\begin{bmatrix} D_{11} & D_{12}\ D_{21} & D_{22} \end{bmatrix}=\begin{bmatrix} \frac{1}{4}[(0-0)^2+((-2)-0)^2+(3-0)^2+((-3)-0)^2] & \frac{1}{4}[(0-0)(2-0)+((-2)-0)(0-0)+(3-0)(5-0)+((-3)-0)((-5)-0)]\ \frac{1}{4}[(2-0)(0-0)+((0)-0)(2-0)+(5-0)(-5-0)+((-5)-0)(5-0)] & \frac{1}{4}[(2-0)^2+((0)-0)^2+(5-0)^2+((-5)-0)^2] \end{bmatrix}=\begin{bmatrix} \frac{29}{4} & -\frac{15}{4}\ -\frac{15}{4} & \frac{89}{4} \end{bmatrix}$

第2类样本协方差矩阵： $\begin{bmatrix} D_{11} & D_{12}\ D_{21} & D_{22} \end{bmatrix}=\begin{bmatrix} \frac{1}{4}[(3-3)^2+(1-3)^2+(8-3)^2+(0-3)^2] & \frac{1}{4}[(3-3)(5-4)+(1-3)(3-4)+(8-3)(6-4)+(0-3)(2-4)]\ \frac{1}{4}[(5-4)(3-3)+(3-4)(1-3)+(6-4)(8-3)+(2-4)(0-3)] & \frac{1}{4}[(5-4)^2+(3-4)^2+(6-4)^2+(2-4)^2] \end{bmatrix}=\begin{bmatrix} \frac{31}{4} & -\frac{5}{4}\ -\frac{5}{4} & \frac{7}{4} \end{bmatrix}$

(0,2)到第1类样本的马氏距离： $d_M^2=(0,2)\begin{bmatrix} \frac{29}{4} & -\frac{15}{4}\ -\frac{15}{4} & \frac{89}{4} \end{bmatrix}^{-1}(0,2)^T=(0,2)\begin{bmatrix} \frac{89}{149} & \frac{15}{149}\ \frac{15}{149} & \frac{29}{149} \end{bmatrix}(0,2)^T=\frac{2}{149}$

(0,2)到第2类样本的马氏距离： $d_M^2=(0,2)\begin{bmatrix} \frac{31}{4} & -\frac{5}{4}\ -\frac{5}{4} & \frac{7}{4} \end{bmatrix}^{-1}(0,2)^T=(0,2)\begin{bmatrix} \frac{7}{29} & \frac{5}{29}\ \frac{5}{29} & \frac{31}{29} \end{bmatrix}(0,2)^T=\frac{10}{29}$

(3,5)到第1类样本的马氏距离： $d_M^2=(3,5)\begin{bmatrix} \frac{29}{4} & -\frac{15}{4}\ -\frac{15}{4} & \frac{89}{4} \end{bmatrix}^{-1}(3,5)^T=(3,5)\begin{bmatrix} \frac{89}{149} & \frac{15}{149}\ \frac{15}{149} & \frac{29}{149} \end{bmatrix}(3,5)^T=\frac{170}{149}$

(3,5)到第2类样本的马氏距离： $d_M^2=(3,5)\begin{bmatrix} \frac{31}{4} & -\frac{5}{4}\ -\frac{5}{4} & \frac{7}{4} \end{bmatrix}^{-1}(3,5)^T=(3,5)\begin{bmatrix} \frac{7}{29} & \frac{5}{29}\ \frac{5}{29} & \frac{31}{29} \end{bmatrix}(3,5)^T=\frac{10}{29}