chapter6.md 31 KB
6.3
$$ \left{\begin{array}{ll}{\boldsymbol{w}^{\mathrm{T}} \boldsymbol{x}{i}+b \geqslant+1,} & {y{i}=+1} \ {\boldsymbol{w}^{\mathrm{T}} \boldsymbol{x}{i}+b \leqslant-1,} & {y{i}=-1}\end{array}\right. $$ [推导]：假设这个超平面是$\left(\boldsymbol{w}^{\prime}\right)^{\top} \boldsymbol{x}+b^{\prime}=0$，对于$\left(\boldsymbol{x}{i}, y{i}\right) \in D$，有： $$ \left{\begin{array}{ll}{\left(\boldsymbol{w}^{\prime}\right)^{\top} \boldsymbol{x}{i}+b^{\prime}>0,} & {y{i}=+1} \ {\left(\boldsymbol{w}^{\prime}\right)^{\top} \boldsymbol{x}{i}+b^{\prime}<0,} & {y{i}=-1}\end{array}\right. $$ 根据几何间隔，将以上关系修正为： $$ \left{\begin{array}{ll}{\left(\boldsymbol{w}^{\prime}\right)^{\top} \boldsymbol{x}{i}+b^{\prime} \geq+\zeta,} & {y{i}=+1} \ {\left(\boldsymbol{w}^{\prime}\right)^{\top} \boldsymbol{x}{i}+b^{\prime} \leq-\zeta,} & {y{i}=-1}\end{array}\right. $$ 其中$\zeta$为某个大于零的常数，两边同除以$\zeta$，再次修正以上关系为： $$ \left{\begin{array}{ll}{\left(\frac{1}{\zeta} \boldsymbol{w}^{\prime}\right)^{\top} \boldsymbol{x}{i}+\frac{b^{\prime}}{\zeta} \geq+1,} & {y{i}=+1} \ {\left(\frac{1}{\zeta} \boldsymbol{w}^{\prime}\right)^{\top} \boldsymbol{x}{i}+\frac{b^{\prime}}{\zeta} \leq-1,} & {y{i}=-1}\end{array}\right. $$ 令：$\boldsymbol{w}=\frac{1}{\zeta} \boldsymbol{w}^{\prime}, b=\frac{b^{\prime}}{\zeta}$，则以上关系可写为： $$ \left{\begin{array}{ll}{\boldsymbol{w}^{\top} \boldsymbol{x}{i}+b \geq+1,} & {y{i}=+1} \ {\boldsymbol{w}^{\top} \boldsymbol{x}{i}+b \leq-1,} & {y{i}=-1}\end{array}\right. $$
6.8
$$ L(\boldsymbol{w}, b, \boldsymbol{\alpha})=\frac{1}{2}|\boldsymbol{w}|^{2}+\sum{i=1}^{m} \alpha{i}\left(1-y{i}\left(\boldsymbol{w}^{\top} \boldsymbol{x}{i}+b\right)\right) $$ [推导]： 待求目标: $$\begin{aligned} \min_{\boldsymbol{x}}\quad f(\boldsymbol{x})\ s.t.\quad h(\boldsymbol{x})&=0\ g(\boldsymbol{x}) &\leq 0 \end{aligned}$$
等式约束和不等式约束：$h(\boldsymbol{x})=0, g(\boldsymbol{x}) \leq 0$分别是由一个等式方程和一个不等式方程组成的方程组。
拉格朗日乘子：$\boldsymbol{\lambda}=\left(\lambda{1}, \lambda{2}, \ldots, \lambda{m}\right)$ $\qquad\boldsymbol{\mu}=\left(\mu{1}, \mu{2}, \ldots, \mu{n}\right)$
拉格朗日函数：$L(\boldsymbol{x}, \boldsymbol{\lambda}, \boldsymbol{\mu})=f(\boldsymbol{x})+\boldsymbol{\lambda} h(\boldsymbol{x})+\boldsymbol{\mu} g(\boldsymbol{x})$
6.9-6.10
$$\begin{aligned} w &= \sum_{i=1}^m\alpha_iy_i\boldsymbol{x}i \ 0 &=\sum{i=1}^m\alpha_iyi \end{aligned}$$ [推导]：式（6.8）可作如下展开： $$\begin{aligned} L(\boldsymbol{w},b,\boldsymbol{\alpha}) &= \frac{1}{2}||\boldsymbol{w}||^2+\sum{i=1}^m\alpha_i(1-y_i(\boldsymbol{w}^T\boldsymbol{x}i+b)) \ & = \frac{1}{2}||\boldsymbol{w}||^2+\sum{i=1}^m(\alpha_i-\alpha_iy_i \boldsymbol{w}^T\boldsymbol{x}_i-\alpha_iyib)\ & =\frac{1}{2}\boldsymbol{w}^T\boldsymbol{w}+\sum{i=1}^m\alphai -\sum{i=1}^m\alpha_iy_i\boldsymbol{w}^T\boldsymbol{x}i-\sum{i=1}^m\alpha_iy_ib \end{aligned}$$ 对$\boldsymbol{w}$和$b$分别求偏导数并令其等于0：
$$\frac {\partial L}{\partial \boldsymbol{w}}=\frac{1}{2}\times2\times\boldsymbol{w} + 0 - \sum_{i=1}^{m}\alpha_iy_i \boldsymbol{x}i-0= 0 \Longrightarrow \boldsymbol{w}=\sum{i=1}^{m}\alpha_iy_i \boldsymbol{x}_i$$
$$\frac {\partial L}{\partial b}=0+0-0-\sum_{i=1}^{m}\alpha_iyi=0 \Longrightarrow \sum{i=1}^{m}\alpha_iy_i=0$$
6.11
$$\begin{aligned}
\max{\boldsymbol{\alpha}} & \sum{i=1}^m\alphai - \frac{1}{2}\sum{i = 1}^m\sum_{j=1}^m\alpha_i \alpha_j y_iy_j\boldsymbol{x}_i^T\boldsymbol{x}j \
s.t. & \sum{i=1}^m \alpha_i y_i =0 \
& \alphai \geq 0 \quad i=1,2,\dots ,m
\end{aligned}$$
[推导]：将式 (6.9)代入 (6.8) ，即可将$L(\boldsymbol{w},b,\boldsymbol{\alpha})$ 中的 $\boldsymbol{w}$ 和 $b$ 消去,再考虑式 (6.10) 的约束,就得到式 (6.6) 的对偶问题：
$$\begin{aligned}
\min{\boldsymbol{w},b} L(\boldsymbol{w},b,\boldsymbol{\alpha}) &=\frac{1}{2}\boldsymbol{w}^T\boldsymbol{w}+\sum_{i=1}^m\alphai -\sum{i=1}^m\alpha_iy_i\boldsymbol{w}^T\boldsymbol{x}i-\sum{i=1}^m\alpha_iy_ib \
&=\frac {1}{2}\boldsymbol{w}^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i-\boldsymbol{w}^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum {i=1}^m\alpha
i -b\sum _{i=1}^m\alpha_iy_i \
& = -\frac {1}{2}\boldsymbol{w}^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alpha_i -b\sum _{i=1}^m\alpha_iyi
\end{aligned}$$
又$\sum\limits{i=1}^{m}\alpha_iyi=0$，所以上式最后一项可化为0，于是得：
$$\begin{aligned}
\min{\boldsymbol{w},b} L(\boldsymbol{w},b,\boldsymbol{\alpha}) &= -\frac {1}{2}\boldsymbol{w}^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alphai \
&=-\frac {1}{2}(\sum{i=1}^{m}\alpha_iy_i\boldsymbol{x}_i)^T(\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i)+\sum _{i=1}^m\alphai \
&=-\frac {1}{2}\sum{i=1}^{m}\alpha_iy_i\boldsymbol{x}_i^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alpha_i \
&=\sum _{i=1}^m\alphai-\frac {1}{2}\sum{i=1 }^{m}\sum_{j=1}^{m}\alpha_i\alpha_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}j
\end{aligned}$$
所以
$$\max{\boldsymbol{\alpha}}\min{\boldsymbol{w},b} L(\boldsymbol{w},b,\boldsymbol{\alpha}) =\max{\boldsymbol{\alpha}} \sum_{i=1}^m\alphai - \frac{1}{2}\sum{i = 1}^m\sum_{j=1}^m\alpha_i \alpha_j y_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j $$
6.39
$$ C=\alpha_i +\mu_i $$
[推导]：对式（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
$$\begin{aligned}
\max_{\boldsymbol{\alpha}}&\sum _{i=1}^m\alphai-\frac {1}{2}\sum{i=1 }^{m}\sum_{j=1}^{m}\alpha_i\alpha_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}j \
s.t. &\sum{i=1}^m \alpha_i y_i=0 \
& 0 \leq\alphai \leq C \quad i=1,2,\dots ,m
\end{aligned}$$
将式6.37-6.39代入6.36可以得到6.35的对偶问题：
$$\begin{aligned}
\min{\boldsymbol{w},b,\boldsymbol{\xi}}L(\boldsymbol{w},b,\boldsymbol{\alpha},\boldsymbol{\xi},\boldsymbol{\mu}) &= \frac{1}{2}||\boldsymbol{w}||^2+C\sum_{i=1}^m \xii+\sum{i=1}^m \alpha_i(1-\xi_i-y_i(\boldsymbol{w}^T\boldsymbol{x}i+b))-\sum{i=1}^m\mu_i \xii \
&=\frac{1}{2}||\boldsymbol{w}||^2+\sum{i=1}^m\alpha_i(1-y_i(\boldsymbol{w}^T\boldsymbol{x}i+b))+C\sum{i=1}^m \xii-\sum{i=1}^m \alpha_i \xii-\sum{i=1}^m\mu_i \xii \
& = -\frac {1}{2}\sum{i=1}^{m}\alpha_iy_i\boldsymbol{x}_i^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alphai +\sum{i=1}^m C\xii-\sum{i=1}^m \alpha_i \xii-\sum{i=1}^m\mu_i \xii \
& = -\frac {1}{2}\sum{i=1}^{m}\alpha_iy_i\boldsymbol{x}_i^T\sum _{i=1}^m\alpha_iy_i\boldsymbol{x}_i+\sum _{i=1}^m\alphai +\sum{i=1}^m (C-\alpha_i-\mu_i)\xi_i \
&=\sum _{i=1}^m\alphai-\frac {1}{2}\sum{i=1 }^{m}\sum_{j=1}^{m}\alpha_i\alpha_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}j
\end{aligned}$$
所以
$$\begin{aligned}
\max{\boldsymbol{\alpha},\boldsymbol{\mu}} \min{\boldsymbol{w},b,\boldsymbol{\xi}}L(\boldsymbol{w},b,\boldsymbol{\alpha},\boldsymbol{\xi},\boldsymbol{\mu})&=\max{\boldsymbol{\alpha},\boldsymbol{\mu}}\sum _{i=1}^m\alphai-\frac {1}{2}\sum{i=1 }^{m}\sum_{j=1}^{m}\alpha_i\alpha_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}j \
&=\max{\boldsymbol{\alpha}}\sum _{i=1}^m\alphai-\frac {1}{2}\sum{i=1 }^{m}\sum_{j=1}^{m}\alpha_i\alpha_jy_iy_j\boldsymbol{x}_i^T\boldsymbol{x}_j
\end{aligned}$$
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：
$$0 \leq\alpha_i \leq C \quad i=1,2,\dots ,m$$
6.52
$$ \left{\begin{array}{l} {\alpha{i}\left(f\left(\boldsymbol{x}{i}\right)-y{i}-\epsilon-\xi{i}\right)=0} \ {\hat{\alpha}{i}\left(y{i}-f\left(\boldsymbol{x}{i}\right)-\epsilon-\hat{\xi}{i}\right)=0} \ {\alpha{i} \hat{\alpha}{i}=0, \xi{i} \hat{\xi}{i}=0} \ {\left(C-\alpha{i}\right) \xi{i}=0,\left(C-\hat{\alpha}{i}\right) \hat{\xi}{i}=0} \end{array}\right. $$ [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$\alpha_i,\hat{\alpha}_i,\mu_i,\hat{\mu}_i$，所以由西瓜书附录式（B.3）可知，以上四个约束条件可相应转化为以下KKT条件： $$ \left{\begin{array}{l} {\alphai\left(f\left(\boldsymbol{x}{i}\right)-y{i}-\epsilon-\xi{i} \right) = 0 } \ {\hat{\alpha}i\left(y{i}-f\left(\boldsymbol{x}{i}\right)-\epsilon-\hat{\xi}{i} \right) = 0 } \ {-\mui\xi{i} = 0 \Rightarrow \mui\xi{i} = 0 } \ {-\hat{\mu}i \hat{\xi}{i} = 0 \Rightarrow \hat{\mu}i \hat{\xi}{i} = 0 } \end{array}\right. $$ 由式（6.49）和式（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： $$ \left{\begin{array}{l} {\alphai\left(f\left(\boldsymbol{x}{i}\right)-y{i}-\epsilon-\xi{i} \right) = 0 } \ {\hat{\alpha}i\left(y{i}-f\left(\boldsymbol{x}{i}\right)-\epsilon-\hat{\xi}{i} \right) = 0 } \ {(C-\alphai)\xi{i} = 0 } \ {(C-\hat{\alpha}i) \hat{\xi}{i} = 0 } \end{array}\right. $$ 又因为样本$(\boldsymbol{x}_i,yi)$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$(C-\alphai)\xi{i} = 0$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条件中即可得到式（6.52）。
6.57
$$\min {h \in \mathbb{H}} F(h)=\Omega\left(|h|{\mathbb{H}}\right)+\ell\left(h\left(\boldsymbol{x}{1}\right), h\left(\boldsymbol{x}{2}\right), \ldots, h\left(\boldsymbol{x}_{m}\right)\right)$$ [解析]：略
6.58
$$h^{*}(\boldsymbol{x})=\sum{i=1}^{m} \alpha{i} \kappa\left(\boldsymbol{x}, \boldsymbol{x}_{i}\right)$$ [解析]：略
6.59
$$h(\boldsymbol{x})=\boldsymbol{w}^{\mathrm{T}}\phi(\boldsymbol{x})$$ [解析]：由于书上已经交代公式(6.60)是公式(3.35)引入核函数后的形式，而公式(3.35)是二分类LDA的损失函数，并且此式为直线方程，所以此时讨论的KLDA应当也是二分类KLDA。那么此公式就类似于第3章图3.3里的$y=\boldsymbol{w}^{\mathrm{T}}\boldsymbol{x}$，表示的是二分类KLDA中所要求解的那条投影直线。
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)。
6.61
$$\boldsymbol{\mu}{i}^{\phi}=\frac{1}{m{i}} \sum{\boldsymbol{x} \in X{i}} \phi(\boldsymbol{x})$$ [解析]：略
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)。
6.64
$$h(\boldsymbol{x})=\sum{i=1}^{m} \alpha{i} \kappa\left(\boldsymbol{x}, \boldsymbol{x}_{i}\right)$$ [解析]：略
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。
6.68
$$\mathbf{M}=\left(\hat{\boldsymbol{\mu}}{0}-\hat{\boldsymbol{\mu}}{1}\right)\left(\hat{\boldsymbol{\mu}}{0}-\hat{\boldsymbol{\mu}}{1}\right)^{\mathrm{T}}$$ [解析]：略
6.69
$$\mathbf{N}=\mathbf{K} \mathbf{K}^{\mathrm{T}}-\sum{i=0}^{1} m{i} \hat{\boldsymbol{\mu}}{i} \hat{\boldsymbol{\mu}}{i}^{\mathrm{T}}$$ [解析]：略
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，下面给出详细地推导过程。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 $$\boldsymbol{w}^{\mathrm{T}} \mathbf{S}{b}^{\phi} \boldsymbol{w}=\left(\frac{1}{m{1}} \sum{i=1}^{m}\sum{\boldsymbol{x} \in X{1}}\alpha_{i} \kappa\left(\boldsymbol{x}i, \boldsymbol{x}\right)-\frac{1}{m{0}} \sum{i=1}^{m} \sum{\boldsymbol{x} \in X{0}} \alpha{i} \kappa\left(\boldsymbol{x}i, \boldsymbol{x}\right)\right)\cdot\left(\frac{1}{m{1}} \sum{i=1}^{m}\sum{\boldsymbol{x} \in X{1}} \alpha{i} \kappa\left(\boldsymbol{x}i, \boldsymbol{x}\right)-\frac{1}{m{0}}\sum{i=1}^{m} \sum{\boldsymbol{x} \in X{0}} \alpha{i} \kappa\left(\boldsymbol{x}_i, \boldsymbol{x}\right)\right)$$ 令$\boldsymbol{\alpha}=(\alpha_1;\alpha_2;...;\alpham)^{\mathrm{T}}\in \mathbb{R}^{m\times 1}$，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 $$\begin{aligned} \mathbf{S}{w}^{\phi}&=\sum{i=0}^{1} \sum{\boldsymbol{x} \in X{i}}\left(\phi(\boldsymbol{x})-\boldsymbol{\mu}{i}^{\phi}\right)\left(\phi(\boldsymbol{x})-\boldsymbol{\mu}{i}^{\phi}\right)^{\mathrm{T}} \ &=\sum{i=0}^{1} \sum{\boldsymbol{x} \in X{i}}\left(\phi(\boldsymbol{x})-\boldsymbol{\mu}{i}^{\phi}\right)\left(\phi(\boldsymbol{x})^{\mathrm{T}}-\left(\boldsymbol{\mu}{i}^{\phi}\right)^{\mathrm{T}}\right) \ &=\sum{i=0}^{1} \sum{\boldsymbol{x} \in X{i}}\left(\phi(\boldsymbol{x})\phi(\boldsymbol{x})^{\mathrm{T}}-\phi(\boldsymbol{x})\left(\boldsymbol{\mu}{i}^{\phi}\right)^{\mathrm{T}}-\boldsymbol{\mu}{i}^{\phi}\phi(\boldsymbol{x})^{\mathrm{T}}+\boldsymbol{\mu}{i}^{\phi}\left(\boldsymbol{\mu}{i}^{\phi}\right)^{\mathrm{T}}\right) \ \end{aligned}$$ 由于$\phi(\boldsymbol{x})\left(\boldsymbol{\mu}{i}^{\phi}\right)^{\mathrm{T}}$的计算结果为标量，所以$\phi(\boldsymbol{x})\left(\boldsymbol{\mu}{i}^{\phi}\right)^{\mathrm{T}}=\left[\phi(\boldsymbol{x})\left(\boldsymbol{\mu}{i}^{\phi}\right)^{\mathrm{T}}\right]^{\mathrm{T}}=\boldsymbol{\mu}{i}^{\phi}\phi(\boldsymbol{x})^{\mathrm{T}}$，将其代回上式可得 $$\begin{aligned} \mathbf{S}{w}^{\phi}&=\sum{i=0}^{1} \sum{\boldsymbol{x} \in X{i}}\left(\phi(\boldsymbol{x})\phi(\boldsymbol{x})^{\mathrm{T}}-2\boldsymbol{\mu}{i}^{\phi}\phi(\boldsymbol{x})^{\mathrm{T}}+\boldsymbol{\mu}{i}^{\phi}\left(\boldsymbol{\mu}{i}^{\phi}\right)^{\mathrm{T}}\right) \ &=\sum{i=0}^{1} \sum{\boldsymbol{x} \in X{i}}\phi(\boldsymbol{x})\phi(\boldsymbol{x})^{\mathrm{T}}-\sum{i=0}^{1} \sum{\boldsymbol{x} \in X{i}}2\boldsymbol{\mu}{i}^{\phi}\phi(\boldsymbol{x})^{\mathrm{T}}+\sum{i=0}^{1} \sum{\boldsymbol{x} \in X{i}}\boldsymbol{\mu}{i}^{\phi}\left(\boldsymbol{\mu}{i}^{\phi}\right)^{\mathrm{T}} \ &=\sum{\boldsymbol{x} \in D}\phi(\boldsymbol{x})\phi(\boldsymbol{x})^{\mathrm{T}}-2\boldsymbol{\mu}{0}^{\phi}\sum{\boldsymbol{x} \in X{0}}\phi(\boldsymbol{x})^{\mathrm{T}}-2\boldsymbol{\mu}{1}^{\phi}\sum{\boldsymbol{x} \in X{1}}\phi(\boldsymbol{x})^{\mathrm{T}}+\sum{\boldsymbol{x} \in X{0}}\boldsymbol{\mu}{0}^{\phi}\left(\boldsymbol{\mu}{0}^{\phi}\right)^{\mathrm{T}}+\sum{\boldsymbol{x} \in X{1}}\boldsymbol{\mu}{1}^{\phi}\left(\boldsymbol{\mu}{1}^{\phi}\right)^{\mathrm{T}} \ &=\sum{\boldsymbol{x} \in D}\phi(\boldsymbol{x})\phi(\boldsymbol{x})^{\mathrm{T}}-2m0\boldsymbol{\mu}{0}^{\phi}\left(\boldsymbol{\mu}_{0}^{\phi}\right)^{\mathrm{T}}-2m1\boldsymbol{\mu}{1}^{\phi}\left(\boldsymbol{\mu}_{1}^{\phi}\right)^{\mathrm{T}}+m0 \boldsymbol{\mu}{0}^{\phi}\left(\boldsymbol{\mu}_{0}^{\phi}\right)^{\mathrm{T}}+m1 \boldsymbol{\mu}{1}^{\phi}\left(\boldsymbol{\mu}{1}^{\phi}\right)^{\mathrm{T}} \ &=\sum{\boldsymbol{x} \in D}\phi(\boldsymbol{x})\phi(\boldsymbol{x})^{\mathrm{T}}-m0\boldsymbol{\mu}{0}^{\phi}\left(\boldsymbol{\mu}_{0}^{\phi}\right)^{\mathrm{T}}-m1\boldsymbol{\mu}{1}^{\phi}\left(\boldsymbol{\mu}{1}^{\phi}\right)^{\mathrm{T}}\ \end{aligned}$$ 再将此式代回$\boldsymbol{w}^{\mathrm{T}} \mathbf{S}{b}^{\phi} \boldsymbol{w}$可得 $$\begin{aligned} \boldsymbol{w}^{\mathrm{T}} \mathbf{S}{w}^{\phi} \boldsymbol{w}&=\sum{i=1}^{m} \alpha{i} \phi\left(\boldsymbol{x}{i}\right)^{\mathrm{T}}\cdot\mathbf{S}{w}^{\phi}\cdot \sum{i=1}^{m} \alpha{i} \phi\left(\boldsymbol{x}{i}\right) \ &=\sum{i=1}^{m} \alpha{i} \phi\left(\boldsymbol{x}{i}\right)^{\mathrm{T}}\cdot\left(\sum{\boldsymbol{x} \in D}\phi(\boldsymbol{x})\phi(\boldsymbol{x})^{\mathrm{T}}-m0\boldsymbol{\mu}{0}^{\phi}\left(\boldsymbol{\mu}_{0}^{\phi}\right)^{\mathrm{T}}-m1\boldsymbol{\mu}{1}^{\phi}\left(\boldsymbol{\mu}{1}^{\phi}\right)^{\mathrm{T}}\right)\cdot \sum{i=1}^{m} \alpha{i} \phi\left(\boldsymbol{x}{i}\right) \ &=\sum{i=1}^{m}\sum{j=1}^{m}\sum{\boldsymbol{x} \in D}\alpha{i} \phi\left(\boldsymbol{x}{i}\right)^{\mathrm{T}}\phi(\boldsymbol{x})\phi(\boldsymbol{x})^{\mathrm{T}}\alpha{j} \phi\left(\boldsymbol{x}{j}\right)-\sum{i=1}^{m}\sum{j=1}^{m}\alpha{i} \phi\left(\boldsymbol{x}_{i}\right)^{\mathrm{T}}m0\boldsymbol{\mu}{0}^{\phi}\left(\boldsymbol{\mu}{0}^{\phi}\right)^{\mathrm{T}}\alpha{j} \phi\left(\boldsymbol{x}{j}\right)-\sum{i=1}^{m}\sum{j=1}^{m}\alpha{i} \phi\left(\boldsymbol{x}_{i}\right)^{\mathrm{T}}m1\boldsymbol{\mu}{1}^{\phi}\left(\boldsymbol{\mu}{1}^{\phi}\right)^{\mathrm{T}}\alpha{j} \phi\left(\boldsymbol{x}{j}\right) \ \end{aligned}$$ 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 $$\begin{aligned} \sum{i=1}^{m}\sum{j=1}^{m}\alpha{i} \phi\left(\boldsymbol{x}_{i}\right)^{\mathrm{T}}m0\boldsymbol{\mu}{0}^{\phi}\left(\boldsymbol{\mu}{0}^{\phi}\right)^{\mathrm{T}}\alpha{j} \phi\left(\boldsymbol{x}_{j}\right)&=m0\sum{i=1}^{m}\sum{j=1}^{m}\alpha{i}\alpha{j}\phi\left(\boldsymbol{x}{i}\right)^{\mathrm{T}}\boldsymbol{\mu}{0}^{\phi}\left(\boldsymbol{\mu}{0}^{\phi}\right)^{\mathrm{T}} \phi\left(\boldsymbol{x}_{j}\right)\ &=m0\sum{i=1}^{m}\sum{j=1}^{m}\alpha{i}\alpha{j}\phi\left(\boldsymbol{x}{i}\right)^{\mathrm{T}}\left[\frac{1}{m{0}} \sum{\boldsymbol{x} \in X{0}} \phi(\boldsymbol{x})\right]\left[\frac{1}{m{0}} \sum{\boldsymbol{x} \in X{0}} \phi(\boldsymbol{x})\right]^{\mathrm{T}} \phi\left(\boldsymbol{x}_{j}\right)\ &=m0\sum{i=1}^{m}\sum{j=1}^{m}\alpha{i}\alpha{j}\left[\frac{1}{m{0}} \sum{\boldsymbol{x} \in X{0}} \phi\left(\boldsymbol{x}{i}\right)^{\mathrm{T}}\phi(\boldsymbol{x})\right]\left[\frac{1}{m{0}} \sum{\boldsymbol{x} \in X{0}} \phi(\boldsymbol{x})^{\mathrm{T}}\phi\left(\boldsymbol{x}_{j}\right)\right] \ &=m0\sum{i=1}^{m}\sum{j=1}^{m}\alpha{i}\alpha{j}\left[\frac{1}{m{0}} \sum{\boldsymbol{x} \in X{0}} \kappa\left(\boldsymbol{x}i, \boldsymbol{x}\right)\right]\left[\frac{1}{m{0}} \sum{\boldsymbol{x} \in X{0}} \kappa\left(\boldsymbol{x}_j, \boldsymbol{x}\right)\right] \ &=m0\boldsymbol{\alpha}^{\mathrm{T}} \hat{\boldsymbol{\mu}}{0} \hat{\boldsymbol{\mu}}{0}^{\mathrm{T}} \boldsymbol{\alpha} \end{aligned}$$ 同理可得，第3项可化简为 $$\sum{i=1}^{m}\sum{j=1}^{m}\alpha{i} \phi\left(\boldsymbol{x}_{i}\right)^{\mathrm{T}}m1\boldsymbol{\mu}{1}^{\phi}\left(\boldsymbol{\mu}{1}^{\phi}\right)^{\mathrm{T}}\alpha{j} \phi\left(\boldsymbol{x}_{j}\right)=m1\boldsymbol{\alpha}^{\mathrm{T}} \hat{\boldsymbol{\mu}}{1} \hat{\boldsymbol{\mu}}{1}^{\mathrm{T}} \boldsymbol{\alpha}$$ 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 $$\begin{aligned} \boldsymbol{w}^{\mathrm{T}} \mathbf{S}_{b}^{\phi} \boldsymbol{w}&=\boldsymbol{\alpha}^{\mathrm{T}} \mathbf{K} \mathbf{K}^{\mathrm{T}} \boldsymbol{\alpha}-m0\boldsymbol{\alpha}^{\mathrm{T}} \hat{\boldsymbol{\mu}}{0} \hat{\boldsymbol{\mu}}_{0}^{\mathrm{T}} \boldsymbol{\alpha}-m1\boldsymbol{\alpha}^{\mathrm{T}} \hat{\boldsymbol{\mu}}{1} \hat{\boldsymbol{\mu}}_{1}^{\mathrm{T}} \boldsymbol{\alpha}\ &=\boldsymbol{\alpha}^{\mathrm{T}} \cdot\left(\mathbf{K} \mathbf{K}^{\mathrm{T}} -m0\hat{\boldsymbol{\mu}}{0} \hat{\boldsymbol{\mu}}_{0}^{\mathrm{T}} -m1\hat{\boldsymbol{\mu}}{1} \hat{\boldsymbol{\mu}}{1}^{\mathrm{T}} \right)\cdot\boldsymbol{\alpha}\ &=\boldsymbol{\alpha}^{\mathrm{T}} \cdot\left(\mathbf{K} \mathbf{K}^{\mathrm{T}}-\sum{i=0}^{1} m{i} \hat{\boldsymbol{\mu}}{i} \hat{\boldsymbol{\mu}}_{i}^{\mathrm{T}} \right)\cdot\boldsymbol{\alpha}\ &=\boldsymbol{\alpha}^{\mathrm{T}} \mathbf{N}\boldsymbol{\alpha}\ \end{aligned}$$