第6章支持向量机

文档摘要

6.9 $$\boldsymbol{w} = \sum{i=1}^m\alphaiyi\boldsymbol{x}i$$ [推导]：公式(6.8)可作如下展开 $$\begin{aligned} L(\boldsymbol{w},b,\boldsymbol{\alpha}) &= \frac{1}{2}||\boldsymbol{w}||^2+\sum{i=1}^m\alphai(1-yi(\boldsymbol{w}^T\boldsymbol{x}i+b)) \\ & = \frac{1}{2}||\boldsymbol{w}||^2+\sum{i=1}^m(\alphai-\alphaiyi \boldsymbol{w}^T\boldsymbol{x}i-\alphaiyib)\\ &

6.9

\boldsymbol{w} = \sum_{i=1}^m\alpha_iy_i\boldsymbol{x}_i

[推导]：公式(6.8)可作如下展开

\begin{aligned} L(\boldsymbol{w},b,\boldsymbol{\alpha}) &= \frac{1}{2}||\boldsymbol{w}||^2+\sum_{i=1}^m\alpha_i(1-y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b)) \\ & = \frac{1}{2}||\boldsymbol{w}||^2+\sum_{i=1}^m(\alpha_i-\alpha_iy_i \boldsymbol{w}^T\boldsymbol{x}_i-\alpha_iy_ib)\\ & =\frac{1}{2}\boldsymbol{w}^T\boldsymbol{w}+\sum_{i=1}^m\alpha_i -\sum_{i=1}^m\alpha_iy_i\boldsymbol{w}^T\boldsymbol{x}_i-\sum_{i=1}^m\alpha_iy_ib \end{aligned}

对\boldsymbol{w}和b分别求偏导数并令其等于0

\frac {\partial L}{\partial \boldsymbol{w}}=\frac{1}{2}\times2\times\boldsymbol{w} + 0 - \sum_{i=1}^{m}\alpha_iy_i \boldsymbol{x}_i-0= 0 \Longrightarrow \boldsymbol{w}=\sum_{i=1}^{m}\alpha_iy_i \boldsymbol{x}_i

\frac {\partial L}{\partial b}=0+0-0-\sum_{i=1}^{m}\alpha_iy_i=0 \Longrightarrow \sum_{i=1}^{m}\alpha_iy_i=0

值得一提的是，上述求解过程遵循的是西瓜书附录B中公式(B.7)左边的那段话“在推导对偶问题时，常通过将拉格朗日函数L(\boldsymbol{x},\boldsymbol{\lambda},\boldsymbol{\mu})对\boldsymbol{x}求导并令导数为0，来获得对偶函数的表达形式”。那么这段话背后的缘由是啥呢？在这里我认为有两种说法可以进行解释：

对于强对偶性成立的优化问题，其主问题的最优解\boldsymbol{x}^*一定满足附录①给出的KKT条件（证明参见参考文献[3]的§ 5.5），而KKT条件中的条件(1)就要求最优解\boldsymbol{x}^*能使得拉格朗日函数L(\boldsymbol{x},\boldsymbol{\lambda},\boldsymbol{\mu})关于\boldsymbol{x}的一阶导数等于0；
对于任意优化问题，若拉格朗日函数L(\boldsymbol{x},\boldsymbol{\lambda},\boldsymbol{\mu})是关于\boldsymbol{x}的凸函数，那么此时对L(\boldsymbol{x},\boldsymbol{\lambda},\boldsymbol{\mu})关于\boldsymbol{x}求导并令导数等于0解出来的点一定是最小值点。根据对偶函数的定义可知，将最小值点代回L(\boldsymbol{x},\boldsymbol{\lambda},\boldsymbol{\mu})即可得到对偶函数。

显然，对于SVM来说，从以上任意一种说法都能解释得通。

6.10

0=\sum_{i=1}^m\alpha_iy_i

[解析]：参见公式(6.9)

6.11

\begin{aligned} \max_{\boldsymbol{\alpha}} & \sum_{i=1}^m\alpha_i - \frac{1}{2}\sum_{i = 1}^m\sum_{j=1}^m\alpha_i \alpha_j y_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j \\ \text { s.t. } & \sum_{i=1}^m \alpha_i y_i =0 \\ & \alpha_i \geq 0 \quad i=1,2,\dots ,m \end{aligned}$$ [推导]：将公式(6.9)和公式(6.10)代入公式(6.8)即可将$L(\boldsymbol{w},b,\boldsymbol{\alpha})$中的$\boldsymbol{w}$和$b$消去，再考虑公式(6.10)的约束，就得到了公式(6.6)的对偶问题 $$\begin{aligned} \inf_{\boldsymbol{w},b} L(\boldsymbol{w},b,\boldsymbol{\alpha}) &=\frac{1}{2}\boldsymbol{w}^T\boldsymbol{w}+\sum_{i=1}^m\alpha_i -\sum_{i=1}^m\alpha_iy_i\boldsymbol{w}^T\boldsymbol{x}_i-\sum_{i=1}^m\alpha_iy_ib \\ &=\frac {1}{2}\boldsymbol{w}^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i-\boldsymbol{w}^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alpha_ i -b\sum _{i=1}^m\alpha_iy_i \\ & = -\frac {1}{2}\boldsymbol{w}^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alpha_i -b\sum _{i=1}^m\alpha_iy_i \end{aligned}

由于\sum\limits_{i=1}^{m}\alpha_iy_i=0，所以上式最后一项可化为0，于是得

\begin{aligned} \inf_{\boldsymbol{w},b} L(\boldsymbol{w},b,\boldsymbol{\alpha}) &= -\frac {1}{2}\boldsymbol{w}^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alpha_i \\ &=-\frac {1}{2}(\sum_{i=1}^{m}\alpha_iy_i\boldsymbol{x}_i)^T(\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i)+\sum _{i=1}^m\alpha_i \\ &=-\frac {1}{2}\sum_{i=1}^{m}\alpha_iy_i\boldsymbol{x}_i^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alpha_i \\ &=\sum _{i=1}^m\alpha_i-\frac {1}{2}\sum_{i=1 }^{m}\sum_{j=1}^{m}\alpha_i\alpha_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j \end{aligned}

所以

\max_{\boldsymbol{\alpha}}\inf_{\boldsymbol{w},b} L(\boldsymbol{w},b,\boldsymbol{\alpha})=\max_{\boldsymbol{\alpha}} \sum_{i=1}^m\alpha_i - \frac{1}{2}\sum_{i = 1}^m\sum_{j=1}^m\alpha_i \alpha_j y_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j

6.13

\left\{\begin{array}{l}\alpha_{i} \geqslant 0 \\ y_{i} f\left(\boldsymbol{x}_{i}\right)-1 \geqslant 0 \\ \alpha_{i}\left(y_{i} f\left(\boldsymbol{x}_{i}\right)-1\right)=0\end{array}\right.

[解析]：参见公式(6.9)中给出的第1点理由

6.35

\begin{aligned} \min _{\boldsymbol{w}, b, \xi_{i}} & \frac{1}{2}\|\boldsymbol{w}\|^{2}+C \sum_{i=1}^{m} \xi_{i} \\ \text { s.t. } & y_{i}\left(\boldsymbol{w}^{\mathrm{T}} \boldsymbol{x}_{i}+b\right) \geqslant 1-\xi_{i} \\ & \xi_{i} \geqslant 0, i=1,2, \ldots, m \end{aligned}

[解析]：令

\max \left(0,1-y_{i}\left(\boldsymbol{w}^{\mathrm{T}} \boldsymbol{x}_{i}+b\right)\right)=\xi_{i}

显然\xi_i\geq 0，而且当1-y_{i}\left(\boldsymbol{w}^{\mathrm{T}} \boldsymbol{x}_{i}+b\right)>0时

1-y_{i}\left(\boldsymbol{w}^{\mathrm{T}} \boldsymbol{x}_{i}+b\right)=\xi_i

当1-y_{i}\left(\boldsymbol{w}^{\mathrm{T}} \boldsymbol{x}_{i}+b\right)\leq 0时

\xi_i = 0

所以综上可得

1-y_{i}\left(\boldsymbol{w}^{\mathrm{T}} \boldsymbol{x}_{i}+b\right)\leq\xi_i\Rightarrow y_{i}\left(\boldsymbol{w}^{\mathrm{T}} \boldsymbol{x}_{i}+b\right) \geqslant 1-\xi_{i}

6.37

\boldsymbol{w}=\sum_{i=1}^{m}\alpha_{i}y_{i}\boldsymbol{x}_{i}

[解析]：参见公式(6.9)

6.38

0=\sum_{i=1}^{m}\alpha_{i}y_{i}

[解析]：参见公式(6.10)

6.39

C=\alpha_i +\mu_i

[推导]：对公式(6.36)关于\xi_i求偏导并令其等于0可得：

\frac{\partial L}{\partial \xi_i}=0+C \times 1 - \alpha_i \times 1-\mu_i \times 1 =0\Longrightarrow C=\alpha_i +\mu_i

6.40

\begin{aligned} \max_{\boldsymbol{\alpha}}&\sum _{i=1}^m\alpha_i-\frac {1}{2}\sum_{i=1 }^{m}\sum_{j=1}^{m}\alpha_i\alpha_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j \\ s.t. &\sum_{i=1}^m \alpha_i y_i=0 \\ & 0 \leq\alpha_i \leq C \quad i=1,2,\dots ,m \end{aligned}

将公式(6.37)-(6.39)代入公式(6.36)可以得到公式(6.35)的对偶问题：

\begin{aligned} &\frac{1}{2}||\boldsymbol{w}||^2+C\sum_{i=1}^m \xi_i+\sum_{i=1}^m \alpha_i(1-\xi_i-y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b))-\sum_{i=1}^m\mu_i \xi_i \\ =&\frac{1}{2}||\boldsymbol{w}||^2+\sum_{i=1}^m\alpha_i(1-y_i(\boldsymbol{w}^T\boldsymbol{x}_i+b))+C\sum_{i=1}^m \xi_i-\sum_{i=1}^m \alpha_i \xi_i-\sum_{i=1}^m\mu_i \xi_i \\ =&-\frac {1}{2}\sum_{i=1}^{m}\alpha_iy_i\boldsymbol{x}_i^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alpha_i +\sum_{i=1}^m C\xi_i-\sum_{i=1}^m \alpha_i \xi_i-\sum_{i=1}^m\mu_i \xi_i \\ =&-\frac {1}{2}\sum_{i=1}^{m}\alpha_iy_i\boldsymbol{x}_i^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alpha_i +\sum_{i=1}^m (C-\alpha_i-\mu_i)\xi_i \\ =&\sum _{i=1}^m\alpha_i-\frac {1}{2}\sum_{i=1 }^{m}\sum_{j=1}^{m}\alpha_i\alpha_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j\\ =&\min_{\boldsymbol{w},b,\boldsymbol{\xi}}L(\boldsymbol{w},b,\boldsymbol{\alpha},\boldsymbol{\xi},\boldsymbol{\mu}) \end{aligned}

所以

\begin{aligned} \max_{\boldsymbol{\alpha},\boldsymbol{\mu}} \min_{\boldsymbol{w},b,\boldsymbol{\xi}}L(\boldsymbol{w},b,\boldsymbol{\alpha},\boldsymbol{\xi},\boldsymbol{\mu})&=\max_{\boldsymbol{\alpha},\boldsymbol{\mu}}\sum _{i=1}^m\alpha_i-\frac {1}{2}\sum_{i=1 }^{m}\sum_{j=1}^{m}\alpha_i\alpha_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j \\ &=\max_{\boldsymbol{\alpha}}\sum _{i=1}^m\alpha_i-\frac {1}{2}\sum_{i=1 }^{m}\sum_{j=1}^{m}\alpha_i\alpha_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j \end{aligned}

又

\begin{aligned} \alpha_i &\geq 0 \\ \mu_i &\geq 0 \\ C &= \alpha_i+\mu_i \end{aligned}

消去\mu_i可得等价约束条件为：

0 \leq\alpha_i \leq C \quad i=1,2,\dots ,m

6.41

\left\{\begin{array}{l}\alpha_{i} \geqslant 0, \quad \mu_{i} \geqslant 0 \\ y_{i} f\left(\boldsymbol{x}_{i}\right)-1+\xi_{i} \geqslant 0 \\ \alpha_{i}\left(y_{i} f\left(\boldsymbol{x}_{i}\right)-1+\xi_{i}\right)=0 \\ \xi_{i} \geqslant 0, \mu_{i} \xi_{i}=0\end{array}\right.

[解析]：参见公式(6.13)

6.52

\left\{\begin{array}{l} {\alpha_{i}\left(f\left(\boldsymbol{x}_{i}\right)-y_{i}-\epsilon-\xi_{i}\right)=0} \\ {\hat{\alpha}_{i}\left(y_{i}-f\left(\boldsymbol{x}_{i}\right)-\epsilon-\hat{\xi}_{i}\right)=0} \\ {\alpha_{i} \hat{\alpha}_{i}=0, \xi_{i} \hat{\xi}_{i}=0} \\ {\left(C-\alpha_{i}\right) \xi_{i}=0,\left(C-\hat{\alpha}_{i}\right) \hat{\xi}_{i}=0} \end{array}\right.

[推导]：将公式(6.45)的约束条件全部恒等变形为小于等于0的形式可得：

\left\{\begin{array}{l} {f\left(\boldsymbol{x}_{i}\right)-y_{i}-\epsilon-\xi_{i} \leq 0 } \\ {y_{i}-f\left(\boldsymbol{x}_{i}\right)-\epsilon-\hat{\xi}_{i} \leq 0 } \\ {-\xi_{i} \leq 0} \\ {-\hat{\xi}_{i} \leq 0} \end{array}\right.

由于以上四个约束条件的拉格朗日乘子分别为\alpha_i,\hat{\alpha}_i,\mu_i,\hat{\mu}_i，所以由附录①可知，以上四个约束条件可相应转化为以下KKT条件：

\left\{\begin{array}{l} {\alpha_i\left(f\left(\boldsymbol{x}_{i}\right)-y_{i}-\epsilon-\xi_{i} \right) = 0 } \\ {\hat{\alpha}_i\left(y_{i}-f\left(\boldsymbol{x}_{i}\right)-\epsilon-\hat{\xi}_{i} \right) = 0 } \\ {-\mu_i\xi_{i} = 0 \Rightarrow \mu_i\xi_{i} = 0 } \\ {-\hat{\mu}_i \hat{\xi}_{i} = 0 \Rightarrow \hat{\mu}_i \hat{\xi}_{i} = 0 } \end{array}\right.

由公式(6.49)和公式(6.50)可知：

\begin{aligned} \mu_i=C-\alpha_i \\ \hat{\mu}_i=C-\hat{\alpha}_i \end{aligned}

所以上述KKT条件可以进一步变形为：

\left\{\begin{array}{l} {\alpha_i\left(f\left(\boldsymbol{x}_{i}\right)-y_{i}-\epsilon-\xi_{i} \right) = 0 } \\ {\hat{\alpha}_i\left(y_{i}-f\left(\boldsymbol{x}_{i}\right)-\epsilon-\hat{\xi}_{i} \right) = 0 } \\ {(C-\alpha_i)\xi_{i} = 0 } \\ {(C-\hat{\alpha}_i) \hat{\xi}_{i} = 0 } \end{array}\right.

又因为样本(\boldsymbol{x}_i,y_i)只可能处在间隔带的某一侧，那么约束条件f\left(\boldsymbol{x}_{i}\right)-y_{i}-\epsilon-\xi_{i}=0和y_{i}-f\left(\boldsymbol{x}_{i}\right)-\epsilon-\hat{\xi}_{i}=0不可能同时成立，所以\alpha_i和\hat{\alpha}_i中至少有一个为0，也即\alpha_i\hat{\alpha}_i=0。在此基础上再进一步分析可知，如果\alpha_i=0的话，那么根据约束(C-\alpha_i)\xi_{i} = 0可知此时\xi_i=0，同理，如果\hat{\alpha}_i=0的话，那么根据约束(C-\hat{\alpha}_i)\hat{\xi}_{i} = 0可知此时\hat{\xi}_i=0，所以\xi_i和\hat{\xi}_i中也是至少有一个为0，也即\xi_{i} \hat{\xi}_{i}=0。将\alpha_i\hat{\alpha}_i=0,\xi_{i} \hat{\xi}_{i}=0整合进上述KKT条件中即可得到公式(6.52)。

6.60

\max _{\boldsymbol{w}} J(\boldsymbol{w})=\frac{\boldsymbol{w}^{\mathrm{T}} \mathbf{S}_{b}^{\phi} \boldsymbol{w}}{\boldsymbol{w}^{\mathrm{T}} \mathbf{S}_{w}^{\phi} \boldsymbol{w}}

[解析]：类似于第3章的公式(3.35)。

6.62

\mathbf{S}_{b}^{\phi}=\left(\boldsymbol{\mu}_{1}^{\phi}-\boldsymbol{\mu}_{0}^{\phi}\right)\left(\boldsymbol{\mu}_{1}^{\phi}-\boldsymbol{\mu}_{0}^{\phi}\right)^{\mathrm{T}}

[解析]：类似于第3章的公式(3.34)。

6.63

\mathbf{S}_{w}^{\phi}=\sum_{i=0}^{1} \sum_{\boldsymbol{x} \in X_{i}}\left(\phi(\boldsymbol{x})-\boldsymbol{\mu}_{i}^{\phi}\right)\left(\phi(\boldsymbol{x})-\boldsymbol{\mu}_{i}^{\phi}\right)^{\mathrm{T}}

[解析]：类似于第3章的公式(3.33)。

6.65

\boldsymbol{w}=\sum_{i=1}^{m} \alpha_{i} \phi\left(\boldsymbol{x}_{i}\right)

[推导]：由表示定理可知，此时二分类KLDA最终求得的投影直线方程总可以写成如下形式

h(\boldsymbol{x})=\sum_{i=1}^{m} \alpha_{i} \kappa\left(\boldsymbol{x}, \boldsymbol{x}_{i}\right)

又因为直线方程的固定形式为

h(\boldsymbol{x})=\boldsymbol{w}^{\mathrm{T}}\phi(\boldsymbol{x})

所以

\boldsymbol{w}^{\mathrm{T}}\phi(\boldsymbol{x})=\sum_{i=1}^{m} \alpha_{i} \kappa\left(\boldsymbol{x}, \boldsymbol{x}_{i}\right)

将\kappa\left(\boldsymbol{x}, \boldsymbol{x}_{i}\right)=\phi(\boldsymbol{x})^{\mathrm{T}}\phi(\boldsymbol{x}_i)代入可得

\boldsymbol{w}^{\mathrm{T}}\phi(\boldsymbol{x})=\sum_{i=1}^{m} \alpha_{i} \phi(\boldsymbol{x})^{\mathrm{T}}\phi(\boldsymbol{x}_i)

\boldsymbol{w}^{\mathrm{T}}\phi(\boldsymbol{x})=\phi(\boldsymbol{x})^{\mathrm{T}}\cdot\sum_{i=1}^{m} \alpha_{i} \phi(\boldsymbol{x}_i)

由于\boldsymbol{w}^{\mathrm{T}}\phi(\boldsymbol{x})的计算结果为标量，而标量的转置等于其本身，所以

\boldsymbol{w}^{\mathrm{T}}\phi(\boldsymbol{x})=\left(\boldsymbol{w}^{\mathrm{T}}\phi(\boldsymbol{x})\right)^{\mathrm{T}}=\phi(\boldsymbol{x})^{\mathrm{T}}\cdot\sum_{i=1}^{m} \alpha_{i} \phi(\boldsymbol{x}_i)

\boldsymbol{w}^{\mathrm{T}}\phi(\boldsymbol{x})=\phi(\boldsymbol{x})^{\mathrm{T}}\boldsymbol{w}=\phi(\boldsymbol{x})^{\mathrm{T}}\cdot\sum_{i=1}^{m} \alpha_{i} \phi(\boldsymbol{x}_i)

\boldsymbol{w}=\sum_{i=1}^{m} \alpha_{i} \phi(\boldsymbol{x}_i)

6.66

\hat{\boldsymbol{\mu}}_{0}=\frac{1}{m_{0}} \mathbf{K} \mathbf{1}_{0}

[解析]：为了详细地说明此公式的计算原理，下面首先先举例说明，然后再在例子的基础上延展出其一般形式。假设此时仅有4个样本，其中第1和第3个样本的标记为0，第2和第4个样本的标记为1，那么此时：

m=4

m_0=2,m_1=2

X_0=\{\boldsymbol{x}_1,\boldsymbol{x}_3\},X_1=\{\boldsymbol{x}_2,\boldsymbol{x}_4\}

\mathbf{K}=\left[ \begin{array}{cccc} \kappa\left(\boldsymbol{x}_1, \boldsymbol{x}_1\right) & \kappa\left(\boldsymbol{x}_1, \boldsymbol{x}_2\right) & \kappa\left(\boldsymbol{x}_1, \boldsymbol{x}_3\right) & \kappa\left(\boldsymbol{x}_1, \boldsymbol{x}_4\right)\\ \kappa\left(\boldsymbol{x}_2, \boldsymbol{x}_1\right) & \kappa\left(\boldsymbol{x}_2, \boldsymbol{x}_2\right) & \kappa\left(\boldsymbol{x}_2, \boldsymbol{x}_3\right) & \kappa\left(\boldsymbol{x}_2, \boldsymbol{x}_4\right)\\ \kappa\left(\boldsymbol{x}_3, \boldsymbol{x}_1\right) & \kappa\left(\boldsymbol{x}_3, \boldsymbol{x}_2\right) & \kappa\left(\boldsymbol{x}_3, \boldsymbol{x}_3\right) & \kappa\left(\boldsymbol{x}_3, \boldsymbol{x}_4\right)\\ \kappa\left(\boldsymbol{x}_4, \boldsymbol{x}_1\right) & \kappa\left(\boldsymbol{x}_4, \boldsymbol{x}_2\right) & \kappa\left(\boldsymbol{x}_4, \boldsymbol{x}_3\right) & \kappa\left(\boldsymbol{x}_4, \boldsymbol{x}_4\right)\\ \end{array} \right]\in \mathbb{R}^{4\times 4}

\mathbf{1}_{0}=\left[ \begin{array}{c} 1\\ 0\\ 1\\ 0\\ \end{array} \right]\in \mathbb{R}^{4\times 1}

\mathbf{1}_{1}=\left[ \begin{array}{c} 0\\ 1\\ 0\\ 1\\ \end{array} \right]\in \mathbb{R}^{4\times 1}

所以

\hat{\boldsymbol{\mu}}_{0}=\frac{1}{m_{0}} \mathbf{K} \mathbf{1}_{0}=\frac{1}{2}\left[ \begin{array}{c} \kappa\left(\boldsymbol{x}_1, \boldsymbol{x}_1\right)+\kappa\left(\boldsymbol{x}_1, \boldsymbol{x}_3\right)\\ \kappa\left(\boldsymbol{x}_2, \boldsymbol{x}_1\right)+\kappa\left(\boldsymbol{x}_2, \boldsymbol{x}_3\right)\\ \kappa\left(\boldsymbol{x}_3, \boldsymbol{x}_1\right)+\kappa\left(\boldsymbol{x}_3, \boldsymbol{x}_3\right)\\ \kappa\left(\boldsymbol{x}_4, \boldsymbol{x}_1\right)+\kappa\left(\boldsymbol{x}_4, \boldsymbol{x}_3\right)\\ \end{array} \right]\in \mathbb{R}^{4\times 1}

\hat{\boldsymbol{\mu}}_{1}=\frac{1}{m_{1}} \mathbf{K} \mathbf{1}_{1}=\frac{1}{2}\left[ \begin{array}{c} \kappa\left(\boldsymbol{x}_1, \boldsymbol{x}_2\right)+\kappa\left(\boldsymbol{x}_1, \boldsymbol{x}_4\right)\\ \kappa\left(\boldsymbol{x}_2, \boldsymbol{x}_2\right)+\kappa\left(\boldsymbol{x}_2, \boldsymbol{x}_4\right)\\ \kappa\left(\boldsymbol{x}_3, \boldsymbol{x}_2\right)+\kappa\left(\boldsymbol{x}_3, \boldsymbol{x}_4\right)\\ \kappa\left(\boldsymbol{x}_4, \boldsymbol{x}_2\right)+\kappa\left(\boldsymbol{x}_4, \boldsymbol{x}_4\right)\\ \end{array} \right]\in \mathbb{R}^{4\times 1}

根据此结果易得\hat{\boldsymbol{\mu}}_{0},\hat{\boldsymbol{\mu}}_{1}的一般形式为

\hat{\boldsymbol{\mu}}_{0}=\frac{1}{m_{0}} \mathbf{K} \mathbf{1}_{0}=\frac{1}{m_{0}}\left[ \begin{array}{c} \sum_{\boldsymbol{x} \in X_{0}}\kappa\left(\boldsymbol{x}_1, \boldsymbol{x}\right)\\ \sum_{\boldsymbol{x} \in X_{0}}\kappa\left(\boldsymbol{x}_2, \boldsymbol{x}\right)\\ \vdots\\ \sum_{\boldsymbol{x} \in X_{0}}\kappa\left(\boldsymbol{x}_m, \boldsymbol{x}\right)\\ \end{array} \right]\in \mathbb{R}^{m\times 1}

\hat{\boldsymbol{\mu}}_{1}=\frac{1}{m_{1}} \mathbf{K} \mathbf{1}_{1}=\frac{1}{m_{1}}\left[ \begin{array}{c} \sum_{\boldsymbol{x} \in X_{1}}\kappa\left(\boldsymbol{x}_1, \boldsymbol{x}\right)\\ \sum_{\boldsymbol{x} \in X_{1}}\kappa\left(\boldsymbol{x}_2, \boldsymbol{x}\right)\\ \vdots\\ \sum_{\boldsymbol{x} \in X_{1}}\kappa\left(\boldsymbol{x}_m, \boldsymbol{x}\right)\\ \end{array} \right]\in \mathbb{R}^{m\times 1}

6.67

\hat{\boldsymbol{\mu}}_{1}=\frac{1}{m_{1}} \mathbf{K} \mathbf{1}_{1}

[解析]：参见公式(6.66)的解析。

6.70

\max _{\boldsymbol{\alpha}} J(\boldsymbol{\alpha})=\frac{\boldsymbol{\alpha}^{\mathrm{T}} \mathbf{M} \boldsymbol{\alpha}}{\boldsymbol{\alpha}^{\mathrm{T}} \mathbf{N} \boldsymbol{\alpha}}

[推导]：此公式是将公式(6.65)代入公式(6.60)后推得而来的，下面给出详细的推导过程。首先将公式(6.65)代入公式(6.60)的分子可得：

\begin{aligned} \boldsymbol{w}^{\mathrm{T}} \mathbf{S}_{b}^{\phi} \boldsymbol{w}&=\left(\sum_{i=1}^{m} \alpha_{i} \phi\left(\boldsymbol{x}_{i}\right)\right)^{\mathrm{T}}\cdot\mathbf{S}_{b}^{\phi}\cdot \sum_{i=1}^{m} \alpha_{i} \phi\left(\boldsymbol{x}_{i}\right) \\ &=\sum_{i=1}^{m} \alpha_{i} \phi\left(\boldsymbol{x}_{i}\right)^{\mathrm{T}}\cdot\mathbf{S}_{b}^{\phi}\cdot \sum_{i=1}^{m} \alpha_{i} \phi\left(\boldsymbol{x}_{i}\right) \\ \end{aligned}

其中

\begin{aligned} \mathbf{S}_{b}^{\phi} &=\left(\boldsymbol{\mu}_{1}^{\phi}-\boldsymbol{\mu}_{0}^{\phi}\right)\left(\boldsymbol{\mu}_{1}^{\phi}-\boldsymbol{\mu}_{0}^{\phi}\right)^{\mathrm{T}} \\ &=\left(\frac{1}{m_{1}} \sum_{\boldsymbol{x} \in X_{1}} \phi(\boldsymbol{x})-\frac{1}{m_{0}} \sum_{\boldsymbol{x} \in X_{0}} \phi(\boldsymbol{x})\right)\left(\frac{1}{m_{1}} \sum_{\boldsymbol{x} \in X_{1}} \phi(\boldsymbol{x})-\frac{1}{m_{0}} \sum_{\boldsymbol{x} \in X_{0}} \phi(\boldsymbol{x})\right)^{\mathrm{T}} \\ &=\left(\frac{1}{m_{1}} \sum_{\boldsymbol{x} \in X_{1}} \phi(\boldsymbol{x})-\frac{1}{m_{0}} \sum_{\boldsymbol{x} \in X_{0}} \phi(\boldsymbol{x})\right)\left(\frac{1}{m_{1}} \sum_{\boldsymbol{x} \in X_{1}} \phi(\boldsymbol{x})^{\mathrm{T}}-\frac{1}{m_{0}} \sum_{\boldsymbol{x} \in X_{0}} \phi(\boldsymbol{x})^{\mathrm{T}}\right) \\ \end{aligned}

将其代入上式可得

\begin{aligned} \boldsymbol{w}^{\mathrm{T}} \mathbf{S}_{b}^{\phi} \boldsymbol{w}=&\sum_{i=1}^{m} \alpha_{i} \phi\left(\boldsymbol{x}_{i}\right)^{\mathrm{T}}\cdot\left(\frac{1}{m_{1}} \sum_{\boldsymbol{x} \in X_{1}} \phi(\boldsymbol{x})-\frac{1}{m_{0}} \sum_{\boldsymbol{x} \in X_{0}} \phi(\boldsymbol{x})\right)\cdot\left(\frac{1}{m_{1}} \sum_{\boldsymbol{x} \in X_{1}} \phi(\boldsymbol{x})^{\mathrm{T}}-\frac{1}{m_{0}} \sum_{\boldsymbol{x} \in X_{0}} \phi(\boldsymbol{x})^{\mathrm{T}}\right)\cdot \sum_{i=1}^{m} \alpha_{i} \phi\left(\boldsymbol{x}_{i}\right) \\ =&\left(\frac{1}{m_{1}} \sum_{\boldsymbol{x} \in X_{1}}\sum_{i=1}^{m} \alpha_{i} \phi\left(\boldsymbol{x}_{i}\right)^{\mathrm{T}} \phi(\boldsymbol{x})-\frac{1}{m_{0}} \sum_{\boldsymbol{x} \in X_{0}} \sum_{i=1}^{m} \alpha_{i} \phi\left(\boldsymbol{x}_{i}\right)^{\mathrm{T}}\phi(\boldsymbol{x})\right)\\ &\cdot\left(\frac{1}{m_{1}} \sum_{\boldsymbol{x} \in X_{1}} \sum_{i=1}^{m} \alpha_{i} \phi(\boldsymbol{x})^{\mathrm{T}}\phi\left(\boldsymbol{x}_{i}\right)-\frac{1}{m_{0}} \sum_{\boldsymbol{x} \in X_{0}} \sum_{i=1}^{m} \alpha_{i} \phi(\boldsymbol{x})^{\mathrm{T}}\phi\left(\boldsymbol{x}_{i}\right)\right) \\ \end{aligned}

由于\kappa\left(\boldsymbol{x}_i, \boldsymbol{x}\right)=\phi(\boldsymbol{x}_i)^{\mathrm{T}}\phi(\boldsymbol{x})为标量，所以其转置等于本身，也即\kappa\left(\boldsymbol{x}_i, \boldsymbol{x}\right)=\phi(\boldsymbol{x}_i)^{\mathrm{T}}\phi(\boldsymbol{x})=\left(\phi(\boldsymbol{x}_i)^{\mathrm{T}}\phi(\boldsymbol{x})\right)^{\mathrm{T}}=\phi(\boldsymbol{x})^{\mathrm{T}}\phi(\boldsymbol{x}_i)=\kappa\left(\boldsymbol{x}_i, \boldsymbol{x}\right)^{\mathrm{T}}，将其代入上式可得

\begin{aligned} \boldsymbol{w}^{\mathrm{T}} \mathbf{S}_{b}^{\phi} \boldsymbol{w}=&\left(\frac{1}{m_{1}} \sum_{i=1}^{m}\sum_{\boldsymbol{x} \in X_{1}}\alpha_{i} \kappa\left(\boldsymbol{x}_i, \boldsymbol{x}\right)-\frac{1}{m_{0}} \sum_{i=1}^{m} \sum_{\boldsymbol{x} \in X_{0}} \alpha_{i} \kappa\left(\boldsymbol{x}_i, \boldsymbol{x}\right)\right)\\ &\cdot\left(\frac{1}{m_{1}} \sum_{i=1}^{m}\sum_{\boldsymbol{x} \in X_{1}} \alpha_{i} \kappa\left(\boldsymbol{x}_i, \boldsymbol{x}\right)-\frac{1}{m_{0}}\sum_{i=1}^{m} \sum_{\boldsymbol{x} \in X_{0}} \alpha_{i} \kappa\left(\boldsymbol{x}_i, \boldsymbol{x}\right)\right) \end{aligned}

令\boldsymbol{\alpha}=(\alpha_1;\alpha_2;...;\alpha_m)^{\mathrm{T}}\in \mathbb{R}^{m\times 1}，同时结合公式(6.66)的解析中得到的\hat{\boldsymbol{\mu}}_{0},\hat{\boldsymbol{\mu}}_{1}的一般形式，上式可以化简为

\begin{aligned} \boldsymbol{w}^{\mathrm{T}} \mathbf{S}_{b}^{\phi} \boldsymbol{w}&=\left(\boldsymbol{\alpha}^{\mathrm{T}}\hat{\boldsymbol{\mu}}_{1}-\boldsymbol{\alpha}^{\mathrm{T}}\hat{\boldsymbol{\mu}}_{0}\right)\cdot\left(\hat{\boldsymbol{\mu}}_{1}^{\mathrm{T}}\boldsymbol{\alpha}-\hat{\boldsymbol{\mu}}_{0}^{\mathrm{T}}\boldsymbol{\alpha}\right)\\ &=\boldsymbol{\alpha}^{\mathrm{T}}\cdot\left(\hat{\boldsymbol{\mu}}_{1}-\hat{\boldsymbol{\mu}}_{0}\right)\cdot\left(\hat{\boldsymbol{\mu}}_{1}^{\mathrm{T}}-\hat{\boldsymbol{\mu}}_{0}^{\mathrm{T}}\right)\cdot\boldsymbol{\alpha}\\ &=\boldsymbol{\alpha}^{\mathrm{T}}\cdot\left(\hat{\boldsymbol{\mu}}_{1}-\hat{\boldsymbol{\mu}}_{0}\right)\cdot\left(\hat{\boldsymbol{\mu}}_{1}-\hat{\boldsymbol{\mu}}_{0}\right)^{\mathrm{T}}\cdot\boldsymbol{\alpha}\\ &=\boldsymbol{\alpha}^{\mathrm{T}} \mathbf{M} \boldsymbol{\alpha}\\ \end{aligned}

以上便是公式(6.70)分子部分的推导，下面继续推导公式(6.70)的分母部分。将公式(6.65)代入公式(6.60)的分母可得：

\begin{aligned} \boldsymbol{w}^{\mathrm{T}} \mathbf{S}_{w}^{\phi} \boldsymbol{w}&=\left(\sum_{i=1}^{m} \alpha_{i} \phi\left(\boldsymbol{x}_{i}\right)\right)^{\mathrm{T}}\cdot\mathbf{S}_{w}^{\phi}\cdot \sum_{i=1}^{m} \alpha_{i} \phi\left(\boldsymbol{x}_{i}\right) \\ &=\sum_{i=1}^{m} \alpha_{i} \phi\left(\boldsymbol{x}_{i}\right)^{\mathrm{T}}\cdot\mathbf{S}_{w}^{\phi}\cdot \sum_{i=1}^{m} \alpha_{i} \phi\left(\boldsymbol{x}_{i}\right) \\ \end{aligned}

其中

\begin{aligned} \mathbf{S}_{w}^{\phi}&=\sum_{i=0}^{1} \sum_{\boldsymbol{x} \in X_{i}}\left(\phi(\boldsymbol{x})-\boldsymbol{\mu}_{i}^{\phi}\right)\left(\phi(\boldsymbol{x})-\boldsymbol{\mu}_{i}^{\phi}\right)^{\mathrm{T}} \\ &=\sum_{i=0}^{1} \sum_{\boldsymbol{x} \in X_{i}}\left(\phi(\boldsymbol{x})-\boldsymbol{\mu}_{i}^{\phi}\right)\left(\phi(\boldsymbol{x})^{\mathrm{T}}-\left(\boldsymbol{\mu}_{i}^{\phi}\right)^{\mathrm{T}}\right) \\ &=\sum_{i=0}^{1} \sum_{\boldsymbol{x} \in X_{i}}\left(\phi(\boldsymbol{x})\phi(\boldsymbol{x})^{\mathrm{T}}-\phi(\boldsymbol{x})\left(\boldsymbol{\mu}_{i}^{\phi}\right)^{\mathrm{T}}-\boldsymbol{\mu}_{i}^{\phi}\phi(\boldsymbol{x})^{\mathrm{T}}+\boldsymbol{\mu}_{i}^{\phi}\left(\boldsymbol{\mu}_{i}^{\phi}\right)^{\mathrm{T}}\right) \\ &=\sum_{i=0}^{1} \sum_{\boldsymbol{x} \in X_{i}}\phi(\boldsymbol{x})\phi(\boldsymbol{x})^{\mathrm{T}}-\sum_{i=0}^{1} \sum_{\boldsymbol{x} \in X_{i}}\phi(\boldsymbol{x})\left(\boldsymbol{\mu}_{i}^{\phi}\right)^{\mathrm{T}}-\sum_{i=0}^{1} \sum_{\boldsymbol{x} \in X_{i}}\boldsymbol{\mu}_{i}^{\phi}\phi(\boldsymbol{x})^{\mathrm{T}}+\sum_{i=0}^{1} \sum_{\boldsymbol{x} \in X_{i}}\boldsymbol{\mu}_{i}^{\phi}\left(\boldsymbol{\mu}_{i}^{\phi}\right)^{\mathrm{T}} \\ \end{aligned}

由于

\begin{aligned} \sum_{i=0}^{1} \sum_{\boldsymbol{x} \in X_{i}} \phi(\boldsymbol{x})\left(\boldsymbol{\mu}_{i}^{\phi}\right)^{\mathrm{T}} &=\sum_{\boldsymbol{x} \in X_{0}} \phi(\boldsymbol{x})\left(\boldsymbol{\mu}_{0}^{\phi}\right)^{\mathrm{T}}+\sum_{\boldsymbol{x} \in X_{1}} \phi(\boldsymbol{x})\left(\boldsymbol{\mu}_{1}^{\phi}\right)^{\mathrm{T}} \\ &=m_{0} \boldsymbol{\mu}_{0}^{\phi}\left(\boldsymbol{\mu}_{0}^{\phi}\right)^{\mathrm{T}}+m_{1} \boldsymbol{\mu}_{1}^{\phi}\left(\boldsymbol{\mu}_{1}^{\phi}\right)^{\mathrm{T}} \\ \sum_{i=0}^{1} \sum_{\boldsymbol{x} \in X_{i}} \boldsymbol{\mu}_{i}^{\phi} \phi(\boldsymbol{x})^{\mathrm{T}} &=\sum_{i=0}^{1} \boldsymbol{\mu}_{i}^{\phi} \sum_{\boldsymbol{x} \in X_{i}} \phi(\boldsymbol{x})^{\mathrm{T}} \\ &=\boldsymbol{\mu}_{0}^{\phi} \sum_{\boldsymbol{x} \in X_{0}} \phi(\boldsymbol{x})^{\mathrm{T}}+\boldsymbol{\mu}_{1}^{\phi} \sum_{\boldsymbol{x} \in X_{1}} \phi(\boldsymbol{x})^{\mathrm{T}} \\ &=m_{0} \boldsymbol{\mu}_{0}^{\phi}\left(\boldsymbol{\mu}_{0}^{\phi}\right)^{\mathrm{T}}+m_{1} \boldsymbol{\mu}_{1}^{\phi}\left(\boldsymbol{\mu}_{1}^{\phi}\right)^{\mathrm{T}} \end{aligned}

所以

\begin{aligned} \mathbf{S}_{w}^{\phi}&=\sum_{\boldsymbol{x} \in D}\phi(\boldsymbol{x})\phi(\boldsymbol{x})^{\mathrm{T}}-2\left[m_0\boldsymbol{\mu}_{0}^{\phi}\left(\boldsymbol{\mu}_{0}^{\phi}\right)^{\mathrm{T}}+m_1\boldsymbol{\mu}_{1}^{\phi}\left(\boldsymbol{\mu}_{1}^{\phi}\right)^{\mathrm{T}}\right]+m_0 \boldsymbol{\mu}_{0}^{\phi}\left(\boldsymbol{\mu}_{0}^{\phi}\right)^{\mathrm{T}}+m_1 \boldsymbol{\mu}_{1}^{\phi}\left(\boldsymbol{\mu}_{1}^{\phi}\right)^{\mathrm{T}} \\ &=\sum_{\boldsymbol{x} \in D}\phi(\boldsymbol{x})\phi(\boldsymbol{x})^{\mathrm{T}}-m_0\boldsymbol{\mu}_{0}^{\phi}\left(\boldsymbol{\mu}_{0}^{\phi}\right)^{\mathrm{T}}-m_1\boldsymbol{\mu}_{1}^{\phi}\left(\boldsymbol{\mu}_{1}^{\phi}\right)^{\mathrm{T}}\\ \end{aligned}

再将此式代回\boldsymbol{w}^{\mathrm{T}} \mathbf{S}_{b}^{\phi} \boldsymbol{w}可得

\begin{aligned} \boldsymbol{w}^{\mathrm{T}} \mathbf{S}_{w}^{\phi} \boldsymbol{w}=&\sum_{i=1}^{m} \alpha_{i} \phi\left(\boldsymbol{x}_{i}\right)^{\mathrm{T}}\cdot\mathbf{S}_{w}^{\phi}\cdot \sum_{i=1}^{m} \alpha_{i} \phi\left(\boldsymbol{x}_{i}\right) \\ =&\sum_{i=1}^{m} \alpha_{i} \phi\left(\boldsymbol{x}_{i}\right)^{\mathrm{T}}\cdot\left(\sum_{\boldsymbol{x} \in D}\phi(\boldsymbol{x})\phi(\boldsymbol{x})^{\mathrm{T}}-m_0\boldsymbol{\mu}_{0}^{\phi}\left(\boldsymbol{\mu}_{0}^{\phi}\right)^{\mathrm{T}}-m_1\boldsymbol{\mu}_{1}^{\phi}\left(\boldsymbol{\mu}_{1}^{\phi}\right)^{\mathrm{T}}\right)\cdot \sum_{i=1}^{m} \alpha_{i} \phi\left(\boldsymbol{x}_{i}\right) \\ =&\sum_{i=1}^{m}\sum_{j=1}^{m}\sum_{\boldsymbol{x} \in D}\alpha_{i} \phi\left(\boldsymbol{x}_{i}\right)^{\mathrm{T}}\phi(\boldsymbol{x})\phi(\boldsymbol{x})^{\mathrm{T}}\alpha_{j} \phi\left(\boldsymbol{x}_{j}\right)-\sum_{i=1}^{m}\sum_{j=1}^{m}\alpha_{i} \phi\left(\boldsymbol{x}_{i}\right)^{\mathrm{T}}m_0\boldsymbol{\mu}_{0}^{\phi}\left(\boldsymbol{\mu}_{0}^{\phi}\right)^{\mathrm{T}}\alpha_{j} \phi\left(\boldsymbol{x}_{j}\right)\\ &-\sum_{i=1}^{m}\sum_{j=1}^{m}\alpha_{i} \phi\left(\boldsymbol{x}_{i}\right)^{\mathrm{T}}m_1\boldsymbol{\mu}_{1}^{\phi}\left(\boldsymbol{\mu}_{1}^{\phi}\right)^{\mathrm{T}}\alpha_{j} \phi\left(\boldsymbol{x}_{j}\right) \\ \end{aligned}

其中，第1项可化简为

\begin{aligned} \sum_{i=1}^{m}\sum_{j=1}^{m}\sum_{\boldsymbol{x} \in D}\alpha_{i} \phi\left(\boldsymbol{x}_{i}\right)^{\mathrm{T}}\phi(\boldsymbol{x})\phi(\boldsymbol{x})^{\mathrm{T}}\alpha_{j} \phi\left(\boldsymbol{x}_{j}\right)&=\sum_{i=1}^{m}\sum_{j=1}^{m}\sum_{\boldsymbol{x} \in D}\alpha_{i} \alpha_{j}\kappa\left(\boldsymbol{x}_i, \boldsymbol{x}\right)\kappa\left(\boldsymbol{x}_j, \boldsymbol{x}\right)\\ &=\boldsymbol{\alpha}^{\mathrm{T}} \mathbf{K} \mathbf{K}^{\mathrm{T}} \boldsymbol{\alpha} \end{aligned}

第2项可化简为

\begin{aligned} \sum_{i=1}^{m}\sum_{j=1}^{m}\alpha_{i} \phi\left(\boldsymbol{x}_{i}\right)^{\mathrm{T}}m_0\boldsymbol{\mu}_{0}^{\phi}\left(\boldsymbol{\mu}_{0}^{\phi}\right)^{\mathrm{T}}\alpha_{j} \phi\left(\boldsymbol{x}_{j}\right)&=m_0\sum_{i=1}^{m}\sum_{j=1}^{m}\alpha_{i}\alpha_{j}\phi\left(\boldsymbol{x}_{i}\right)^{\mathrm{T}}\boldsymbol{\mu}_{0}^{\phi}\left(\boldsymbol{\mu}_{0}^{\phi}\right)^{\mathrm{T}} \phi\left(\boldsymbol{x}_{j}\right)\\ &=m_0\sum_{i=1}^{m}\sum_{j=1}^{m}\alpha_{i}\alpha_{j}\phi\left(\boldsymbol{x}_{i}\right)^{\mathrm{T}}\left[\frac{1}{m_{0}} \sum_{\boldsymbol{x} \in X_{0}} \phi(\boldsymbol{x})\right]\left[\frac{1}{m_{0}} \sum_{\boldsymbol{x} \in X_{0}} \phi(\boldsymbol{x})\right]^{\mathrm{T}} \phi\left(\boldsymbol{x}_{j}\right)\\ &=m_0\sum_{i=1}^{m}\sum_{j=1}^{m}\alpha_{i}\alpha_{j}\left[\frac{1}{m_{0}} \sum_{\boldsymbol{x} \in X_{0}} \phi\left(\boldsymbol{x}_{i}\right)^{\mathrm{T}}\phi(\boldsymbol{x})\right]\left[\frac{1}{m_{0}} \sum_{\boldsymbol{x} \in X_{0}} \phi(\boldsymbol{x})^{\mathrm{T}}\phi\left(\boldsymbol{x}_{j}\right)\right] \\ &=m_0\sum_{i=1}^{m}\sum_{j=1}^{m}\alpha_{i}\alpha_{j}\left[\frac{1}{m_{0}} \sum_{\boldsymbol{x} \in X_{0}} \kappa\left(\boldsymbol{x}_i, \boldsymbol{x}\right)\right]\left[\frac{1}{m_{0}} \sum_{\boldsymbol{x} \in X_{0}} \kappa\left(\boldsymbol{x}_j, \boldsymbol{x}\right)\right] \\ &=m_0\boldsymbol{\alpha}^{\mathrm{T}} \hat{\boldsymbol{\mu}}_{0} \hat{\boldsymbol{\mu}}_{0}^{\mathrm{T}} \boldsymbol{\alpha} \end{aligned}

同理可得，第3项可化简为

\sum_{i=1}^{m}\sum_{j=1}^{m}\alpha_{i} \phi\left(\boldsymbol{x}_{i}\right)^{\mathrm{T}}m_1\boldsymbol{\mu}_{1}^{\phi}\left(\boldsymbol{\mu}_{1}^{\phi}\right)^{\mathrm{T}}\alpha_{j} \phi\left(\boldsymbol{x}_{j}\right)=m_1\boldsymbol{\alpha}^{\mathrm{T}} \hat{\boldsymbol{\mu}}_{1} \hat{\boldsymbol{\mu}}_{1}^{\mathrm{T}} \boldsymbol{\alpha}

将上述三项的化简结果代回再将此式代回\boldsymbol{w}^{\mathrm{T}} \mathbf{S}_{b}^{\phi} \boldsymbol{w}可得

\begin{aligned} \boldsymbol{w}^{\mathrm{T}} \mathbf{S}_{b}^{\phi} \boldsymbol{w}&=\boldsymbol{\alpha}^{\mathrm{T}} \mathbf{K} \mathbf{K}^{\mathrm{T}} \boldsymbol{\alpha}-m_0\boldsymbol{\alpha}^{\mathrm{T}} \hat{\boldsymbol{\mu}}_{0} \hat{\boldsymbol{\mu}}_{0}^{\mathrm{T}} \boldsymbol{\alpha}-m_1\boldsymbol{\alpha}^{\mathrm{T}} \hat{\boldsymbol{\mu}}_{1} \hat{\boldsymbol{\mu}}_{1}^{\mathrm{T}} \boldsymbol{\alpha}\\ &=\boldsymbol{\alpha}^{\mathrm{T}} \cdot\left(\mathbf{K} \mathbf{K}^{\mathrm{T}} -m_0\hat{\boldsymbol{\mu}}_{0} \hat{\boldsymbol{\mu}}_{0}^{\mathrm{T}} -m_1\hat{\boldsymbol{\mu}}_{1} \hat{\boldsymbol{\mu}}_{1}^{\mathrm{T}} \right)\cdot\boldsymbol{\alpha}\\ &=\boldsymbol{\alpha}^{\mathrm{T}} \cdot\left(\mathbf{K} \mathbf{K}^{\mathrm{T}}-\sum_{i=0}^{1} m_{i} \hat{\boldsymbol{\mu}}_{i} \hat{\boldsymbol{\mu}}_{i}^{\mathrm{T}} \right)\cdot\boldsymbol{\alpha}\\ &=\boldsymbol{\alpha}^{\mathrm{T}} \mathbf{N}\boldsymbol{\alpha}\\ \end{aligned}

附录

①KKT条件^[1]

对于一般地约束优化问题

\begin{array}{ll} {\min } & {f(\boldsymbol x)} \\ {\text {s.t.}} & {g_{i}(\boldsymbol x) \leq 0 \quad(i=1, \ldots, m)} \\ {} & {h_{j}(\boldsymbol x)=0 \quad(j=1, \ldots, n)} \end{array}

其中，自变量\boldsymbol x\in \mathbb{R}^n。设f(\boldsymbol x),g_i(\boldsymbol x),h_j(\boldsymbol x)具有连续的一阶偏导数，\boldsymbol x^*是优化问题的局部可行解。若该优化问题满足任意一个约束限制条件（constraint qualifications or regularity conditions）^[2]，则一定存在\boldsymbol \mu^*=(\mu_1^*,\mu_2^*,...,\mu_m^*)^T,\boldsymbol \lambda^*=(\lambda_1^*,\lambda_2^*,...,\lambda_n^*)^T,使得

\left\{ \begin{aligned} & \nabla_{\boldsymbol x} L(\boldsymbol x^* ,\boldsymbol \mu^* ,\boldsymbol \lambda^* )=\nabla f(\boldsymbol x^* )+\sum_{i=1}^{m}\mu_i^* \nabla g_i(\boldsymbol x^* )+\sum_{j=1}^{n}\lambda_j^* \nabla h_j(\boldsymbol x^*)=0 &(1) \\ & h_j(\boldsymbol x^*)=0 &(2) \\ & g_i(\boldsymbol x^*) \leq 0 &(3) \\ & \mu_i^* \geq 0 &(4)\\ & \mu_i^* g_i(\boldsymbol x^*)=0 &(5) \end{aligned} \right.

其中L(\boldsymbol x,\boldsymbol \mu,\boldsymbol \lambda)为拉格朗日函数

L(\boldsymbol x,\boldsymbol \mu,\boldsymbol \lambda)=f(\boldsymbol x)+\sum_{i=1}^{m}\mu_i g_i(\boldsymbol x)+\sum_{j=1}^{n}\lambda_j h_j(\boldsymbol x)

以上5条即为KKT条件，严格数学证明参见参考文献[1]的§ 4.2.1。

参考文献

[1] 王燕军. 《最优化基础理论与方法》

[2] https://en.wikipedia.org/wiki/Karush%E2%80%93Kuhn%E2%80%93Tucker_conditions#Regularity_conditions_(or_constraint_qualifications)

[3] 王书宁译.《凸优化》

第6章 支持向量机