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S1[11/21/21] 和 S2[1−1/2−1/21] {\boldsymbol S}_1 \begin{bmatrix} 1 1/2 \\ 1/2 1 \end{bmatrix} \text{ 和 } {\boldsymbol S}_2 \begin{bmatrix} 1 -1/2 \\ -1/2 1 \end{bmatrix}S1​[11/2​1/21​]和S2​[1−1/2​−1/21​]两类样本均值分别为μ1[2,0]⊤ 和 μ2[2,2]⊤ \boldsymbol {\mu}_1 [2, 0]^\top \text{ 和 } \boldsymbol {\mu}_2 [2, 2]^\topμ1​[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[1,2]⊤[1, 2]^\top[1,2]⊤属于哪类解答SwS1S2[2002] {\boldsymbol S}_{\mathrm w} {\boldsymbol S}_1 {\boldsymbol S}_2 \begin{bmatrix} 2 0 \\ 0 2 \end{bmatrix}Sw​S1​S2​[20​02​]Sw−1[1/2001/2] {\boldsymbol S}_{\mathrm w}^{-1} \begin{bmatrix} 1/2 0 \\ 0 1/2 \end{bmatrix}Sw−1​[1/20​01/2​]wSw−1(μ1−μ2)[0,−1]⊤ {\bm w} {\boldsymbol S}_{\mathrm w}^{-1} (\boldsymbol {\mu}_1 - \boldsymbol {\mu}_2) [0, -1]^\topwSw−1​(μ1​−μ2​)[0,−1]⊤y0w⊤μ1μ22[0,−1][2,1]⊤−1 y_0 {\bm w}^\top \frac{\boldsymbol {\mu}_1 \boldsymbol {\mu}_2}{2} [0, -1][2, 1]^\top -1y0​w⊤2μ1​μ2​​[0,−1][2,1]⊤−1w⊤[1,2]⊤−2 {\bm w}^\top [1, 2]^\top -2w⊤[1,2]⊤−2第2类