Fisher准则例题
设两类样本集
C
1
:
(
0.2
0.7
)
,
(
0.3
0.8
)
,
(
0.4
0.5
)
,
(
0.6
0.5
)
,
(
0.1
0.4
)
C_1: \begin{pmatrix} 0.2 \\ 0.7 \end{pmatrix}, \begin{pmatrix} 0.3 \\ 0.8 \end{pmatrix}, \begin{pmatrix} 0.4 \\ 0.5 \end{pmatrix}, \begin{pmatrix} 0.6 \\ 0.5 \end{pmatrix}, \begin{pmatrix} 0.1 \\ 0.4 \end{pmatrix}
C1:(0.20.7),(0.30.8),(0.40.5),(0.60.5),(0.10.4)
C
2
:
(
0.4
0.6
)
,
(
0.6
0.2
)
,
(
0.7
0.4
)
,
(
0.8
0.6
)
,
(
0.7
0.5
)
C_2: \begin{pmatrix} 0.4 \\ 0.6 \end{pmatrix}, \begin{pmatrix} 0.6 \\ 0.2 \end{pmatrix}, \begin{pmatrix} 0.7 \\ 0.4 \end{pmatrix}, \begin{pmatrix} 0.8 \\ 0.6 \end{pmatrix}, \begin{pmatrix} 0.7 \\ 0.5 \end{pmatrix}
C2:(0.40.6),(0.60.2),(0.70.4),(0.80.6),(0.70.5)
设计Fisher判别分析（Linear Discriminant Analysis, LDA）线性分类器，将
C
1
C_1
C1和
C
2
C_2
C2的数据分开。
设计线性分类器
f ( x ) = w 0 + w 1 x 1 + w 2 x 2 = 0 f({\bm x}) = {w}_0 + { w}_1 x_1 + { w}_2 x_2 = 0 f(x)=w0+w1x1+w2x2=0
均值向量
x ˉ 1 = [ 0.32 0.48 ] \bar{\bm x}_1 = \begin{bmatrix} 0.32 \\ 0.48 \end{bmatrix} xˉ1=[0.320.48]
x ˉ 2 = [ 0.64 0.46 ] \bar{\bm x}_2 = \begin{bmatrix} 0.64 \\ 0.46 \end{bmatrix} xˉ2=[0.640.46]
协方差矩阵
S 1 = [ 0.148 0.002 0.002 0.088 ] {\bm S}_1 = \begin{bmatrix} 0.148 & 0.002 \\ 0.002 & 0.088 \end{bmatrix} S1=[0.1480.0020.0020.088]
S 2 = [ 0.092 − 0.002 − 0.002 0.112 ] {\bm S}_2 = \begin{bmatrix} 0.092 & -0.002 \\ -0.002 & 0.112 \end{bmatrix} S2=[0.092−0.002−0.0020.112]
总协方差矩阵
S W = [ 0.24 0 0 0.2 ] {\bm S}_W = \begin{bmatrix} 0.24 & 0 \\ 0 & 0.2 \end{bmatrix} SW=[0.24000.2]
总协方差矩阵的逆矩阵
S W − 1 = [ 4.1667 0 0 5 ] {\bm S}_W^{-1} = \begin{bmatrix} 4.1667 & 0 \\ 0 & 5 \end{bmatrix} SW−1=[4.1667005]
权重向量
w = [ − 1.3333 0.1 ] {\bm w} = \begin{bmatrix} -1.3333 \\ 0.1 \end{bmatrix} w=[−1.33330.1]
归一化的权重向量
w ^ = w ∥ w ∥ = [ − 0.9972 0.0748 ] \hat{{\bm w}} = \frac{{\bm w}}{\|{\bm w}\|} = \begin{bmatrix} -0.9972 \\ 0.0748 \end{bmatrix} w^=∥w∥w=[−0.99720.0748]
阈值
y ˉ 1 = − 0.2832 , y ˉ 2 = − 0.6038 , w 0 = − 0.4435 \bar{y}_1 = -0.2832, \quad \bar{y}_2 = -0.6038, \quad {w}_0 = -0.4435 yˉ1=−0.2832,yˉ2=−0.6038,w0=−0.4435
线性判别函数
f ( x ) = − 0.4435 − 0.9972 x 1 + 0.0748 x 2 = 0 f({\bm x}) = -0.4435 -0.9972 x_1 + 0.0748 x_2 = 0 f(x)=−0.4435−0.9972x1+0.0748x2=0
