Fisher准则例题——给定类内散度矩阵和类样本均值
设有两类样本，两类样本的类内散度矩阵分别为
S 1 = ( 1 1 / 2 1 / 2 1 ) , S 2 = ( 1 − 1 / 2 − 1 / 2 1 ) S_1 = \begin{pmatrix} 1 & 1/2 \\ 1/2 & 1 \end{pmatrix}, \quad S_2 = \begin{pmatrix} 1 & -1/2 \\ -1/2 & 1 \end{pmatrix} S1=(11/21/21),S2=(1−1/2−1/21)
各类样本均值分别为
μ 1 = ( 2 , 0 ) ⊤ 和 μ 2 = ( 2 , 2 ) ⊤ \mu_1 = (2, 0)^\top \text{ 和 } \mu_2 = (2, 2)^\top μ1=(2,0)⊤ 和 μ2=(2,2)⊤
利用 Fisher 准则求其决策面方程（假定分类阈值点为均值），并求新样本 ( 1 , 1 ) ⊤ (1, 1)^\top (1,1)⊤ 属于哪类？
解：
S w = S 1 + S 2 = ( 2 0 0 2 ) S_{\bm w} = S_1 + S_2 = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} Sw=S1+S2=(2002)
S w − 1 = ( 1 / 2 0 0 1 / 2 ) S_{\bm w}^{-1} = \begin{pmatrix} 1/2 & 0 \\ 0 & 1/2 \end{pmatrix} Sw−1=(1/2001/2)
w = S w − 1 ( μ 1 − μ 2 ) = ( 0 , − 1 ) ⊤ {\bm w} = S_{\bm w}^{-1} (\mu_1 - \mu_2) = (0, -1)^\top w=Sw−1(μ1−μ2)=(0,−1)⊤
y 0 ∗ = w ⊤ μ 1 + μ 2 2 = ( 0 , − 1 ) ( 2 , 1 ) ⊤ = − 1 y_0^* = {\bm w}^\top \frac{\mu_1 + \mu_2}{2} = (0, -1)(2, 1)^\top = -1 y0∗=w⊤2μ1+μ2=(0,−1)(2,1)⊤=−1
w ⊤ ( 1 , 1 ) ⊤ = − 1 = y 0 ∗ {\bm w}^\top (1, 1)^\top = -1 = y_0^* w⊤(1,1)⊤=−1=y0∗