## 13.1

$$
p(\boldsymbol{x})=\sum_{i=1}^{N} \alpha_{i} \cdot p\left(\boldsymbol{x} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)
$$

[解析]： 高斯混合分布的定义式。

## 13.2

$$
\begin{aligned} f(\boldsymbol{x}) &=\underset{j \in \mathcal{Y}}{\arg \max } p(y=j | \boldsymbol{x}) \\ &=\underset{j \in \mathcal{Y}}{\arg \max } \sum_{i=1}^{N} p(y=j, \Theta=i | \boldsymbol{x}) \\ &=\underset{j \in \mathcal{Y}}{\arg \max } \sum_{i=1}^{N} p(y=j | \Theta=i, \boldsymbol{x}) \cdot p(\Theta=i | \boldsymbol{x}) \end{aligned}
$$

[解析]：从公式第 1 行到第 2 行是对概率进行边缘化(marginalization)；通过引入$\Theta$并对其求和 $\sum_{i=1}^N$以抵消引入的影响。从公式第 2 行到第 3 行推导如下
$$
\begin{aligned}p(y=j, \Theta=i | \boldsymbol{x}) &=\frac{p(y=j, \Theta=i, \boldsymbol{x})}{p(\boldsymbol{x})} \\&=\frac{p(y=j, \Theta=i, \boldsymbol{x})}{p(\Theta=i, \boldsymbol{x})} \cdot \frac{p(\Theta=i, \boldsymbol{x})}{p(\boldsymbol{x})} \\&=p(y=j | \Theta=i, \boldsymbol{x}) \cdot p(\Theta=i | \boldsymbol{x})\end{aligned}
$$

## 13.3

$$
p(\Theta=i | \boldsymbol{x})=\frac{\alpha_{i} \cdot p\left(\boldsymbol{x} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)}{\sum_{i=1}^{N} \alpha_{i} \cdot p\left(\boldsymbol{x} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)}
$$

[解析]：根据 13.1 
$$
p(\boldsymbol{x})=\sum_{i=1}^{N} \alpha_{i} \cdot p\left(\boldsymbol{x} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)
$$
因此
$$
\begin{aligned}p(\Theta=i | \boldsymbol{x})&=\frac{p(\Theta=i , \boldsymbol{x})}{P(x)}\\&=\frac{\alpha_{i} \cdot p\left(\boldsymbol{x} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)}{\sum_{i=1}^{N} \alpha_{i} \cdot p\left(\boldsymbol{x} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)}\end{aligned}
$$

## 13.4

$$
\begin{aligned} L L\left(D_{l} \cup D_{u}\right)=& \sum_{\left(x_{j}, y_{j}\right) \in D_{l}} \ln \left(\sum_{i=1}^{N} \alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right) \cdot p\left(y_{j} | \Theta=i, \boldsymbol{x}_{j}\right)\right) \\ &+\sum_{x_{j} \in D_{u}} \ln \left(\sum_{i=1}^{N} \alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)\right) \end{aligned}
$$

[解析]：第二项很好解释，当不知道类别信息的时候，样本$x_j$的概率可以用式 13.1 表示，所有无类别信息的样本$D_u$的似然是所有样本的乘积，因为$\ln$函数是单调的，所以也可以将$\ln$函数作用于这个乘积消除因为连乘产生的数值计算问题。第一项引入了样本的标签信息，由
$$
p(y=j | \Theta=i, \boldsymbol{x})=\left\{\begin{array}{ll}1, & i=j \\0, & i \neq j\end{array}\right.
$$
可知，这项限定了样本$x_j$只可能来自于$y_j$所对应的高斯分布。

## 13.5

$$
\gamma_{j i}=\frac{\alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)}{\sum_{i=1}^{N} \alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)}
$$

[解析]：参见式 13.3，这项可以理解成样本$x_j$属于类别标签$i$(或者说由第$i$个高斯分布生成)的后验概率。其中$\alpha_i,\boldsymbol{\mu}_{i}\boldsymbol{\Sigma}_i$可以通过有标记样本预先计算出来。即：
$$
\begin{array}{l}\alpha_{i}=\frac{l_{i}}{\left|D_{l}\right|}, \text { where }\left|D_{l}\right|=\sum_{i=1}^{N} l_{i} \\\boldsymbol{\mu}_{i}=\frac{1}{l_{i}} \sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} \boldsymbol{x}_{j} \\\boldsymbol{\Sigma}_{i}=\frac{1}{l_{i}} \sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top}\end{array}
$$


## 13.6

$$
\boldsymbol{\mu}_{i}=\frac{1}{\sum_{\boldsymbol{x}_{j} \in D_{u}} \gamma_{j i}+l_{i}}\left(\sum_{\boldsymbol{x}_{j} \in D_{u}} \gamma_{j i} \boldsymbol{x}_{j}+\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} \boldsymbol{x}_{j}\right)
$$

[推导]：这项可以由$$\cfrac{\partial LL(D_l \cup D_u) }{\partial \mu_i}=0$$而得，将式 13.4 的两项分别记为：
$$
\begin{aligned}LL(D_l)&=\sum_{(\boldsymbol{x_j},y_j \in D_l)}\ln\left(\sum_{s=1}^{N}\alpha_s \cdot p(\boldsymbol{x_j}\vert \boldsymbol{\mu}_s,\boldsymbol{\Sigma}_s) \cdot p(y_i|\Theta = s,\boldsymbol{x_j})\right)\\&=\sum_{(\boldsymbol{x_j},y_j \in D_l)}\ln\left(\sum_{s=1}^{N}\alpha_{y_j} \cdot p(\boldsymbol{x_j} \vert \boldsymbol{\mu}_{y_j},\boldsymbol{\Sigma}_{y_j})\right)\\LL(D_u)&=\sum_{\boldsymbol{x_j} \in D_u} \ln\left(\alpha_s \cdot p(\boldsymbol{x_j} | \boldsymbol{\mu}_s,\boldsymbol{\Sigma}_s)\right)\end{aligned}
$$
首先，$LL(D_l)$对$\boldsymbol{\mu_i}$求偏导，$LL(D_l)$求和号中只有$y_j=i$ 的项能留下来，即
$$
\begin{aligned}\frac{\partial L L\left(D_{l}\right)}{\partial \boldsymbol{\mu}_{i}} &=\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} \frac{\partial \ln \left(\alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)\right)}{\partial \boldsymbol{\mu}_{i}} \\&=\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} \frac{1}{p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)} \cdot \frac{\partial p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)}{\partial \boldsymbol{\mu}_{i}} \\&=\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} \frac{1}{p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right) \cdot \boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right) \\&=\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} \boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\end{aligned}
$$
$LL(D_u)$对$\boldsymbol{\mu_i}$求导，参考 9.33 的推导：
$$
\begin{aligned}
\frac{\partial L L\left(D_{u}\right)}{\partial \boldsymbol{\mu}_{i}} &=\sum_{\boldsymbol{x}_{j} \in D_{u}} \frac{\alpha_{i}}{\sum_{s=1}^{N} \alpha_{s} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{s}, \boldsymbol{\Sigma}_{s}\right)} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right) \cdot \boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right) \\
&=\sum_{\boldsymbol{x}_{j} \in D_{u}} \gamma_{j i} \cdot \boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)
\end{aligned}
$$


综上，
$$
\begin{aligned}\frac{\partial L L\left(D_{l} \cup D_{u}\right)}{\partial \boldsymbol{\mu}_{i}} &=\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} \boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)+\sum_{\boldsymbol{x}_{j} \in D_{u}} \gamma_{j i} \cdot \boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right) \\&=\boldsymbol{\Sigma}_{i}^{-1}\left(\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)+\sum_{\boldsymbol{x}_{j} \in D_{u}} \gamma_{j i} \cdot\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\right) \\&=\boldsymbol{\Sigma}_{i}^{-1}\left(\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} \boldsymbol{x}_{j}+\sum_{\boldsymbol{x}_{j} \in D_{u}} \gamma_{j i} \cdot \boldsymbol{x}_{j}-\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} \boldsymbol{\mu}_{i}-\sum_{\boldsymbol{x}_{j} \in D_{u}} \gamma_{j i} \cdot \boldsymbol{\mu}_{i}\right)\end{aligned}
$$
令$\frac{\partial L L\left(D_{l} \cup D_{u}\right)}{\partial \boldsymbol{\mu}_{i}}=0$，两边同时左乘$\Sigma_i$并移项：
$$
\sum_{\boldsymbol{x}_{j} \in D_{u}} \gamma_{j i} \cdot \boldsymbol{\mu}_{i}+\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} \boldsymbol{\mu}_{i}=\sum_{\boldsymbol{x}_{j} \in D_{u}} \gamma_{j i} \cdot \boldsymbol{x}_{j}+\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} \boldsymbol{x}_{j}
$$
上式中，$\boldsymbol{\mu_i}$ 可以作为常量提到求和号外面，而$\sum_{\left(x_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} 1=l_{i}$，即第$i$类样本的有标记 样本数目，因此

$$
\left(\sum_{x_{j} \in D_{u}} \gamma_{j i}+\sum_{\left(x_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} 1\right) \boldsymbol{\mu}_{i}=\sum_{x_{j} \in D_{u}} \gamma_{j i} \cdot \boldsymbol{x}_{j}+\sum_{\left(x_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} \boldsymbol{x}_{j}
$$


即得式 13.6。

## 13.7

$$
\begin{aligned}\boldsymbol{\Sigma}_{i}=& \frac{1}{\sum_{\boldsymbol{x}_{j} \in D_{u}} \gamma_{j i}+l_{i}}\left(\sum_{\boldsymbol{x}_{j} \in D_{u}} \gamma_{j i} \cdot\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top}\right.\\&\left.+\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top}\right)\end{aligned}
$$

[推导]：首先$LL(D_l)$对$\boldsymbol{\Sigma_i}$求偏导 ，类似于 13.6
$$
\begin{aligned} \frac{\partial L L\left(D_{l}\right)}{\partial \boldsymbol{\Sigma}_{i}} &=\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} \frac{\partial \ln \left(\alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)\right)}{\partial \boldsymbol{\Sigma}_{i}} \\ &=\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} \frac{1}{p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)} \cdot \frac{\partial p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)}{\partial \boldsymbol{\Sigma}_{i}} \\
&=\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} \frac{1}{p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right) \cdot\left(\boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top}-\boldsymbol{I}\right) \cdot \frac{1}{2} \boldsymbol{\Sigma}_{i}^{-1}\\
&=\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i}\left(\boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top}-\boldsymbol{I}\right) \cdot \frac{1}{2} \boldsymbol{\Sigma}_{i}^{-1}
\end{aligned}
$$
然后$LL(D_u)$ 对$\boldsymbol{\Sigma_i}$求偏导，类似于 9.35
$$
\frac{\partial L L\left(D_{u}\right)}{\partial \boldsymbol{\Sigma}_{i}}=\sum_{\boldsymbol{x}_{j} \in D_{u}} \gamma_{j i} \cdot\left(\boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top}-\boldsymbol{I}\right) \cdot \frac{1}{2} \boldsymbol{\Sigma}_{i}^{-1}
$$
综合可得：
$$
\begin{aligned} \frac{\partial L L\left(D_{l} \cup D_{u}\right)}{\partial \boldsymbol{\Sigma}_{i}}=& \sum_{\boldsymbol{x}_{j} \in D_{u}} \gamma_{j i} \cdot\left(\boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top}-\boldsymbol{I}\right) \cdot \frac{1}{2} \boldsymbol{\Sigma}_{i}^{-1} \\ &+\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i}\left(\boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top}-\boldsymbol{I}\right) \cdot \frac{1}{2} \boldsymbol{\Sigma}_{i}^{-1} \\=&\left(\sum_{\boldsymbol{x}_{j} \in D_{u}} \gamma_{j i} \cdot\left(\boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top}-\boldsymbol{I}\right)\right.\\ &\left.+\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i}\left(\boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top}-\boldsymbol{I}\right)\right) \cdot \frac{1}{2} \boldsymbol{\Sigma}_{i}^{-1} \end{aligned}
$$
令$\frac{\partial L L\left(D_{l} \cup D_{u}\right)}{\partial \boldsymbol{\Sigma}_{i}}=0$，两边同时右乘$2\Sigma_i$并移项：
$$
\begin{aligned} \sum_{\boldsymbol{x}_{j} \in D_{u}} \gamma_{j i} \cdot \boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top}+& \sum_{\left(\boldsymbol{x}_{j}, y_{j} \in D_{l} \wedge y_{j}=i\right.} \boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top} \\=& \sum_{\boldsymbol{x}_{j} \in D_{u}} \gamma_{j i} \cdot \boldsymbol{I}+\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} \boldsymbol{I} \\ &=\left(\sum_{\boldsymbol{x}_{j} \in D_{u}} \gamma_{j i}+l_{i}\right) \boldsymbol{I} \end{aligned}
$$
两边同时左乘以$\Sigma_i$：
$$
\sum_{\boldsymbol{x}_{j} \in D_{u}} \gamma_{j i} \cdot\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top}+\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top}=\left(\sum_{\boldsymbol{x}_{j} \in D_{u}} \gamma_{j i}+l_{i}\right) \boldsymbol{\Sigma}_{i}
$$
即得式 13.7。

## 13.8

$$
\alpha_{i}=\frac{1}{m}\left(\sum_{\boldsymbol{x}_{j} \in D_{u}} \gamma_{j i}+l_{i}\right)
$$

[推导]：类似于式 9.36，写出$LL(D_l \cup D_u)$的拉格朗日形式
$$
\begin{aligned}\mathcal{L}\left(D_{l} \cup D_{u}, \lambda\right) &=L L\left(D_{l} \cup D_{u}\right)+\lambda\left(\sum_{s=1}^{N} \alpha_{s}-1\right) \\&=L L\left(D_{l}\right)+L L\left(D_{u}\right)+\lambda\left(\sum_{s=1}^{N} \alpha_{s}-1\right)\end{aligned}
$$


类似于式 9.37，对$\alpha_i$求偏导。对于$LL(D_u)$，求导结果与式 9.37 的推导过程一样
$$
\frac{\partial L L\left(D_{u}\right)}{\partial \alpha_{i}}=\sum_{\boldsymbol{x}_{j} \in D_{u}} \frac{1}{\sum_{s=1}^{N} \alpha_{s} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{s}, \boldsymbol{\Sigma}_{s}\right)} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)
$$


对于$LL(D_l)$，类似于 13.6 和 13.7 的推导过程
$$
\begin{aligned}\frac{\partial L L\left(D_{l}\right)}{\partial \alpha_{i}} &=\sum_{\left(x_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} \frac{\partial \ln \left(\alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)\right)}{\partial \alpha_{i}} \\&=\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} \frac{1}{\alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)} \cdot \frac{\partial\left(\alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)\right)}{\partial \alpha_{i}} \\&=\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} \frac{1}{\alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right) \\&=\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} \frac{1}{\alpha_{i}}=\frac{1}{\alpha_{i}} \cdot \sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} 1=\frac{l_{i}}{\alpha_{i}}\end{aligned}
$$


上式推导过程中，重点注意变量是$\alpha_i$ ，$p(x_j|\mu_i,\Sigma_i)$是常量；最后一行$\alpha_i$相对于求和变量为常量，因此作为公因子提到求和号外面； $l_i$ 为第$i$类样本的有标记样本数目。

综合两项结果：
$$
\frac{\partial \mathcal{L}\left(D_{l} \cup D_{u}, \lambda\right)}{\partial \alpha_{i}}=\frac{l_{i}}{\alpha_{i}}+\sum_{\boldsymbol{x}_{j} \in D_{u}} \frac{p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)}{\sum_{s=1}^{N} \alpha_{s} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{s}, \boldsymbol{\Sigma}_{s}\right)}+\lambda
$$


令$\cfrac{\partial LL(D_l \cup D_u) }{\partial \alpha_i}=0$ 并且两边同乘以$\alpha_i$，得
$$
\alpha_{i} \cdot \frac{l_{i}}{\alpha_{i}}+\sum_{\boldsymbol{x}_{j} \in D_{u}} \frac{\alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)}{\sum_{s=1}^{N} \alpha_{s} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{s}, \boldsymbol{\Sigma}_{s}\right)}+\lambda \cdot \alpha_{i}=0
$$


结合式 9.30 发现，求和号内即为后验概率$\gamma_{ji}$,即
$$
l_i+\sum_{x_i \in D_u} \gamma_{ji}+\lambda \alpha_i = 0
$$
对所有混合成分求和，得
$$
\sum_{i=1}^N l_i+\sum_{i=1}^N  \sum_{x_i \in D_u} \gamma_{ji}+\sum_{i=1}^N \lambda \alpha_i = 0
$$


这里$\sum_{i=1}^N \alpha_i =1$ ，因此$\sum_{i=1}^N \lambda \alpha_i=\lambda\sum_{i=1}^N \alpha_i=\lambda$，根据 9.30 中$\gamma_{ji}$表达式可知
$$
\sum_{i=1}^N \gamma_{ji} =  \sum_{i =1}^{N} \cfrac{\alpha_i \cdot  p(x_j|\mu_i,\Sigma_i)}{\Sigma_{s=1}^N \alpha_s \cdot p(x_j| \mu_s, \Sigma_s)}=  \cfrac{\sum_{i =1}^{N}\alpha_i \cdot  p(x_j|\mu_i,\Sigma_i)}{\sum_{s=1}^N \alpha_s \cdot p(x_j| \mu_s, \Sigma_s)}=1
$$


再结合加法满足交换律，所以
$$
\sum_{i=1}^N  \sum_{x_i \in D_u} \gamma_{ji}=\sum_{x_i \in D_u} \sum_{i=1}^N  \gamma_{ji} =\sum_{x_i \in D_u} 1=u
$$


以上分析过程中，$\sum_{x_j\in D_u}$ 形式与$\sum_{j=1}^u$等价，其中u为未标记样本集的样本个数； $\sum_{i=1}^Nl_i=l$其中$l$为有标记样本集的样本个数；将这些结果代入
$$
\sum_{i=1}^N l_i+\sum_{i=1}^N  \sum_{x_i \in D_u} \gamma_{ji}+\sum_{i=1}^N \lambda \alpha_i = 0
$$


解出$l+u+\lambda = 0$  且$l+u =m$ 其中$m$为样本总个数，移项即得$\lambda = -m$
最后代入整理解得
$$
l_i + \sum_{x_j \in{D_u}} \gamma_{ji}-\lambda \alpha_i = 0
$$
整理即得式 13.8。

## 13.9

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

[解析]：这个公式和公式 6.35 基本一致，除了引入了无标记样本的松弛变量$\xi_i, i=l+1,\cdots m$和对应的权重系数$C_u$和无标记样本的标记指派$\hat{y}_i$。

## 13.12

$$
\begin{aligned}
E(f) &=\frac{1}{2} \sum_{i=1}^{m} \sum_{j=1}^{m}(\mathbf{W})_{i j}\left(f\left(\boldsymbol{x}_{i}\right)-f\left(\boldsymbol{x}_{j}\right)\right)^{2} \\
&=\frac{1}{2}\left(\sum_{i=1}^{m} d_{i} f^{2}\left(\boldsymbol{x}_{i}\right)+\sum_{j=1}^{m} d_{j} f^{2}\left(\boldsymbol{x}_{j}\right)-2 \sum_{i=1}^{m} \sum_{j=1}^{m}(\mathbf{W})_{i j} f\left(\boldsymbol{x}_{i}\right) f\left(\boldsymbol{x}_{j}\right)\right) \\
&=\sum_{i=1}^{m} d_{i} f^{2}\left(\boldsymbol{x}_{i}\right)-\sum_{i=1}^{m} \sum_{j=1}^{m}(\mathbf{W})_{i j} f\left(\boldsymbol{x}_{i}\right) f\left(\boldsymbol{x}_{j}\right) \\
&=\boldsymbol{f}^{\mathrm{T}}(\mathbf{D}-\mathbf{W}) \boldsymbol{f}
\end{aligned}
$$

[解析]：首先解释下这个能量函数的定义。原则上，我们希望能量函数$E(f)$越小越好，对于节点$i,j$，如果它们不相邻，则$(\mathbf{W})_{i j}=0$，如果它们相邻，则最小化能量函数要求$f(x_i)$和$f(x_j)$尽量相似，和逻辑相符。下面进行公式的推导，首先由二项展开可得：
$$
\begin{aligned}
E(f) &=\frac{1}{2} \sum_{i=1}^{m} \sum_{j=1}^{m}(\mathbf{W})_{i j}\left(f\left(\boldsymbol{x}_{i}\right)-f\left(\boldsymbol{x}_{j}\right)\right)^{2} \\
&=\frac{1}{2} \sum_{i=1}^{m} \sum_{j=1}^{m}(\mathbf{W})_{i j}\left(f^{2}\left(\boldsymbol{x}_{i}\right)-2 f\left(\boldsymbol{x}_{i}\right) f\left(\boldsymbol{x}_{j}\right)+f^{2}\left(\boldsymbol{x}_{j}\right)\right) \\
&=\frac{1}{2}\left( \sum_{i=1}^{m} \sum_{j=1}^{m}(\mathbf{W})_{i j} f^{2}\left(\boldsymbol{x}_{i}\right)+ \sum_{i=1}^{m} \sum_{j=1}^{m}(\mathbf{W})_{i j} f^{2}\left(\boldsymbol{x}_{j}\right)-2\sum_{i=1}^{m} \sum_{j=1}^{m}(\mathbf{W})_{i j} f\left(\boldsymbol{x}_{i}\right) f\left(\boldsymbol{x}_{j}\right)\right)
\end{aligned}
$$
由于$\mathbf{W}$是一个对称矩阵，可以通过变量替换得到
$$
\begin{aligned}
\sum_{i=1}^{m} \sum_{j=1}^{m}(\mathbf{W})_{i j} f^{2}\left(\boldsymbol{x}_{j}\right)&=\sum_{j=1}^{m} \sum_{i=1}^{m}(\mathbf{W})_{j i} f^{2}\left(\boldsymbol{x}_{i}\right)\\
&=\sum_{i=1}^{m} \sum_{j=1}^{m}(\mathbf{W})_{i j} f^{2}\left(\boldsymbol{x}_{i}\right)\\
&=
\sum_{i=1}^{m} \sum_{j=1}^{m}(\mathbf{W})_{i j} f^{2}\left(\boldsymbol{x}_{j}\right)
\end{aligned}
$$
因此$E(f)$可化简为
$$
\begin{aligned}
E(f) &=  \sum_{i=1}^{m} \sum_{j=1}^{m}(\mathbf{W})_{i j} f^{2}\left(\boldsymbol{x}_{i}\right)-\sum_{i=1}^{m} \sum_{j=1}^{m}(\mathbf{W})_{i j} f\left(\boldsymbol{x}_{i}\right) f\left(\boldsymbol{x}_{j}\right)
\end{aligned}
$$
根据定义 $d_i=\sum_{j=1}^{l+u}\left(\mathbf{W}\right)_{ij}$，且$m=l+u$则
$$
\begin{aligned}
E(f)&=\sum_{i=1}^{m} d_{i} f^{2}\left(\boldsymbol{x}_{i}\right)-\sum_{i=1}^{m} \sum_{j=1}^{m}(\mathbf{W})_{i j} f\left(\boldsymbol{x}_{i}\right) f\left(\boldsymbol{x}_{j}\right)\\
&=\boldsymbol{f}^{\mathrm{T}}\mathbf{D}\boldsymbol{f}-\boldsymbol{f}^{\mathrm{T}}\mathbf{W}\boldsymbol{f}\\
&=\boldsymbol{f}^{\mathrm{T}}(\mathbf{D}-\mathbf{W}) \boldsymbol{f}
\end{aligned}
$$

## 13.13

$$
\begin{aligned}
E(f) &=\left(\boldsymbol{f}_{l}^{\mathrm{T}} \boldsymbol{f}_{u}^{\mathrm{T}}\right)\left(\left[\begin{array}{ll}
\mathbf{D}_{l l} & \mathbf{0}_{l u} \\
\mathbf{0}_{u l} & \mathbf{D}_{u u}
\end{array}\right]-\left[\begin{array}{ll}
\mathbf{W}_{l l} & \mathbf{W}_{l u} \\
\mathbf{W}_{u l} & \mathbf{W}_{u u}
\end{array}\right]\right)\left[\begin{array}{l}
\boldsymbol{f}_{l} \\
\boldsymbol{f}_{u}
\end{array}\right] \\
&=\boldsymbol{f}_{l}^{\mathrm{T}}\left(\mathbf{D}_{l l}-\mathbf{W}_{l l}\right) \boldsymbol{f}_{l}-2 \boldsymbol{f}_{u}^{\mathrm{T}} \mathbf{W}_{u l} \boldsymbol{f}_{l}+\boldsymbol{f}_{u}^{\mathrm{T}}\left(\mathbf{D}_{u u}-\mathbf{W}_{u u}\right) \boldsymbol{f}_{u}
\end{aligned}
$$

[解析]：这里第一项西瓜书中的符号有歧义，应该表示成$\left[\begin{array}{ll}
\boldsymbol{f}_{l}^{\mathrm{T}} & \boldsymbol{f}_{u}^{\mathrm{T}}
\end{array}\right]$即一个$\mathbb{R}^{1\times(l+u)}$的行向量，根据矩阵乘法的定义，有：
$$
\begin{aligned}
E(f) &=\left[\begin{array}{ll}
\boldsymbol{f}_{l}^{\mathrm{T}} & \boldsymbol{f}_{u}^{\mathrm{T}}
\end{array}\right]\left[\begin{array}{cc}
\boldsymbol{D}_{l l}-\boldsymbol{W}_{l l} & -\boldsymbol{W}_{l u} \\
-\boldsymbol{W}_{u l} & \boldsymbol{D}_{u u}-\boldsymbol{W}_{u u}
\end{array}\right]\left[\begin{array}{l}
\boldsymbol{f}_{l} \\
\boldsymbol{f}_{u}
\end{array}\right] \\
&=\left[\begin{array}{ll}\boldsymbol{f}_{l}^{\mathrm{T}}\left(\boldsymbol{D}_{l l}-\boldsymbol{W}_{l l}\right)-\boldsymbol{f}_{u}^{\mathrm{T}} \boldsymbol{W}_{u l} & -\boldsymbol{f}_{l}^{\mathrm{T}} \boldsymbol{W}_{l u}+\boldsymbol{f}_{u}^{\mathrm{T}}\left(\boldsymbol{D}_{u u}-\boldsymbol{W}_{u u}\right)\end{array}\right]\left[\begin{array}{l}
\boldsymbol{f}_{l} \\
\boldsymbol{f}_{u}
\end{array}\right] \\
&=\left(\boldsymbol{f}_{l}^{\mathrm{T}}\left(\boldsymbol{D}_{l l}-\boldsymbol{W}_{l l}\right)-\boldsymbol{f}_{u}^{\mathrm{T}} \boldsymbol{W}_{u l}\right) \boldsymbol{f}_{l}+\left(-\boldsymbol{f}_{l}^{\mathrm{T}} \boldsymbol{W}_{l u}+\boldsymbol{f}_{u}^{\mathrm{T}}\left(\boldsymbol{D}_{u u}-\boldsymbol{W}_{u u}\right)\right) \boldsymbol{f}_{u} \\
&=\boldsymbol{f}_{l}^{\mathrm{T}}\left(\boldsymbol{D}_{l l}-\boldsymbol{W}_{l l}\right) \boldsymbol{f}_{l}-\boldsymbol{f}_{u}^{\mathrm{T}} \boldsymbol{W}_{u l} \boldsymbol{f}_{l}-\boldsymbol{f}_{l}^{\mathrm{T}} \boldsymbol{W}_{l u} \boldsymbol{f}_{u}+\boldsymbol{f}_{u}^{\mathrm{T}}\left(\boldsymbol{D}_{u u}-\boldsymbol{W}_{u u}\right) \boldsymbol{f}_{u} \\
&=\boldsymbol{f}_{l}^{\mathrm{T}}\left(\boldsymbol{D}_{l l}-\boldsymbol{W}_{l l}\right) \boldsymbol{f}_{l}-2 \boldsymbol{f}_{u}^{\mathrm{T}} \boldsymbol{W}_{u l} \boldsymbol{f}_{l}+\boldsymbol{f}_{u}^{\mathrm{T}}\left(\boldsymbol{D}_{u u}-\boldsymbol{W}_{u u}\right) \boldsymbol{f}_{u}
\end{aligned}
$$
其中最后一步，$\boldsymbol{f}_{l}^{\mathrm{T}} \boldsymbol{W}_{l u} \boldsymbol{f}_{u}=\left(\boldsymbol{f}_{l}^{\mathrm{T}} \boldsymbol{W}_{l u} \boldsymbol{f}_{u}\right)^{\mathrm{T}}=f_{u}^{\mathrm{T}} \boldsymbol{W}_{u l} \boldsymbol{f}_{l}$，因为这个式子的结果是一个标量。

## 13.14

$$
\begin{aligned}
E(f) &=\left(\boldsymbol{f}_{l}^{\mathrm{T}} \boldsymbol{f}_{u}^{\mathrm{T}}\right)\left(\left[\begin{array}{ll}
\mathbf{D}_{l l} & \mathbf{0}_{l u} \\
\mathbf{0}_{u l} & \mathbf{D}_{u u}
\end{array}\right]-\left[\begin{array}{ll}
\mathbf{W}_{l l} & \mathbf{W}_{l u} \\
\mathbf{W}_{u l} & \mathbf{W}_{u u}
\end{array}\right]\right)\left[\begin{array}{l}
\boldsymbol{f}_{l} \\
\boldsymbol{f}_{u}
\end{array}\right] \\
&=\boldsymbol{f}_{l}^{\mathrm{T}}\left(\mathbf{D}_{l l}-\mathbf{W}_{l l}\right) \boldsymbol{f}_{l}-2 \boldsymbol{f}_{u}^{\mathrm{T}} \mathbf{W}_{u l} \boldsymbol{f}_{l}+\boldsymbol{f}_{u}^{\mathrm{T}}\left(\mathbf{D}_{u u}-\mathbf{W}_{u u}\right) \boldsymbol{f}_{u}
\end{aligned}
$$



[解析]：参考 13.13

## 13.15

$$
\boldsymbol{f}_{u}=\left(\mathbf{D}_{u u}-\mathbf{W}_{u u}\right)^{-1} \mathbf{W}_{u l} \boldsymbol{f}_{l}
$$

[解析]：由 13.13，有
$$
\begin{aligned}
\frac{\partial E(f)}{\partial \boldsymbol{f}_{u}} &=\frac{\partial \boldsymbol{f}_{l}^{\mathrm{T}}\left(\boldsymbol{D}_{l l}-\boldsymbol{W}_{l l}\right) \boldsymbol{f}_{l}-2 \boldsymbol{f}_{u}^{\mathrm{T}} \boldsymbol{W}_{u l} \boldsymbol{f}_{l}+\boldsymbol{f}_{u}^{\mathrm{T}}\left(\boldsymbol{D}_{u u}-\boldsymbol{W}_{u u}\right) \boldsymbol{f}_{u}}{\partial \boldsymbol{f}_{u}} \\
&=-2 \boldsymbol{W}_{u l} \boldsymbol{f}_{l}+2\left(\boldsymbol{D}_{u u}-\boldsymbol{W}_{u u}\right) \boldsymbol{f}_{u}
\end{aligned}
$$
令结果等于 0 即得 13.15。

## 13.16

$$
\begin{aligned}
\mathbf{P} &=\mathbf{D}^{-1} \mathbf{W}=\left[\begin{array}{cc}
\mathbf{D}_{l l}^{-1} & \mathbf{0}_{l u} \\
\mathbf{0}_{u l} & \mathbf{D}_{u u}^{-1}
\end{array}\right]\left[\begin{array}{ll}
\mathbf{W}_{l l} & \mathbf{W}_{l u} \\
\mathbf{W}_{u l} & \mathbf{W}_{u u}
\end{array}\right] \\
&=\left[\begin{array}{ll}
\mathbf{D}_{l l}^{-1} \mathbf{W}_{l l} & \mathbf{D}_{l l}^{-1} \mathbf{W}_{l u} \\
\mathbf{D}_{u u}^{-1} \mathbf{W}_{u l} & \mathbf{D}_{u u}^{-1} \mathbf{W}_{u u}
\end{array}\right]
\end{aligned}
$$

[解析]：根据矩阵乘法的定义计算可得该式，其中需要注意的是，对角矩阵$\mathbf{D}$的拟等于其各个对角元素的倒数。

## 13.17

$$
\begin{aligned}
\boldsymbol{f}_{u} &=\left(\mathbf{D}_{u u}\left(\mathbf{I}-\mathbf{D}_{u u}^{-1} \mathbf{W}_{u u}\right)\right)^{-1} \mathbf{W}_{u l} \boldsymbol{f}_{l} \\
&=\left(\mathbf{I}-\mathbf{D}_{u u}^{-1} \mathbf{W}_{u u}\right)^{-1} \mathbf{D}_{u u}^{-1} \mathbf{W}_{u l} \boldsymbol{f}_{l} \\
&=\left(\mathbf{I}-\mathbf{P}_{u u}\right)^{-1} \mathbf{P}_{u l} \boldsymbol{f}_{l}
\end{aligned}
$$

[解析]：第一项到第二项是根据矩阵乘法逆的定义：$(\mathbf{A}\mathbf{B})^{-1}=\mathbf{B}^{-1}\mathbf{A}^{-1}$，在这个式子中​
$$
\begin{aligned}
\mathbf{P}_{u u}&=\mathbf{D}_{u u}^{-1} \mathbf{W}_{u u}\\
\mathbf{P}_{ul}&=\mathbf{D}_{u u}^{-1} \mathbf{W}_{u l}
\end{aligned}
$$
均可以根据$\mathbf{W}_{ij}$计算得到，因此可以通过标记$\mathbf{f}_l$计算未标记数据的标签$\mathbf{f}_u$。

## 13.20

$$
\mathbf{F}^{*}=\lim _{t \rightarrow \infty} \mathbf{F}(t)=(1-\alpha)(\mathbf{I}-\alpha \mathbf{S})^{-1} \mathbf{Y}
$$

[解析]：由 13.19
$$
\mathbf{F}(t+1)=\alpha \mathbf{S} \mathbf{F}(t)+(1-\alpha) \mathbf{Y}
$$
当 t取不同的值时，有：
$$
\begin{aligned}
t=0: \mathbf{F}(1) &=\alpha \mathbf{S F}(0)+(1-\alpha) \mathbf{Y}\\
&=\alpha \mathbf{S} \mathbf{Y}+(1-\alpha) \mathbf{Y} \\
t=1: \mathbf{F}(2) &=\alpha \mathbf{S F}(1)+(1-\alpha) \mathbf{Y}=\alpha \mathbf{S}(\alpha \mathbf{S} \mathbf{Y}+(1-\alpha) \mathbf{Y})+(1-\alpha) \mathbf{Y} \\
&=(\alpha \mathbf{S})^{2} \mathbf{Y}+(1-\alpha)\left(\sum_{i=0}^{1}(\alpha \mathbf{S})^{i}\right) \mathbf{Y} \\
t=2:\mathbf{F}(3)&=\alpha\mathbf{S}\mathbf{F}(2)+(1-\alpha)\mathbf{Y}\\&=\alpha \mathbf{S}\left((\alpha \mathbf{S})^{2} \mathbf{Y}+(1-\alpha)\left(\sum_{i=0}^{1}(\alpha \mathbf{S})^{i}\right) \mathbf{Y}\right)+(1-\alpha) \mathbf{Y} \\
&=(\alpha \mathbf{S})^{3} \mathbf{Y}+(1-\alpha)\left(\sum_{i=0}^{2}(\alpha \mathbf{S})^{i}\right) \mathbf{Y}\\
\end{aligned}
$$
可以观察到规律
$$
\mathbf{F}(t)=(\alpha \mathbf{S})^{t} \mathbf{Y}+(1-\alpha)\left(\sum_{i=0}^{t-1}(\alpha \mathbf{S})^{i}\right) \mathbf{Y}
$$
则
$$
\mathbf{F}^{*}=\lim _{t \rightarrow \infty}\mathbf{F}(t)=\lim _{t \rightarrow \infty}(\alpha \mathbf{S})^{t} \mathbf{Y}+\lim _{t \rightarrow \infty}(1-\alpha)\left(\sum_{i=0}^{t-1}(\alpha \mathbf{S})^{i}\right) \mathbf{Y}
$$
其中第一项由于$\mathbf{S}=\mathbf{D}^{-\frac{1}{2}} \mathbf{W} \mathbf{D}^{-\frac{1}{2}}$的特征值介于[-1, 1]之间(这里省略详细推导，可以参见 https://en.wikipedia.org/wiki/Laplacian_matrix 其中对称拉普拉斯矩阵的特征值介于 0 和 2 之间)，而$\alpha\in(0,1)$，所以$\lim _{t \rightarrow \infty}(\alpha \mathbf{S})^{t}=0$，第二项由等比数列公式
$$
\lim _{t \rightarrow \infty} \sum_{i=0}^{t-1}(\alpha \mathbf{S})^{i}=\frac{\mathbf{I}-\lim _{t \rightarrow \infty}(\alpha \mathbf{S})^{t}}{\mathbf{I}-\alpha \mathbf{S}}=\frac{\mathbf{I}}{\mathbf{I}-\alpha \mathbf{S}}=(\mathbf{I}-\alpha \mathbf{S})^{-1}
$$
综合可得式 13.20。



