update chapter 8,9,11,13

docs/chapter11/chapter11.md

@@ -28,7 +28,7 @@ $$
 \min _{\boldsymbol{w}} \sum_{i=1}^{m}\left(y_{i}-\boldsymbol{w}^{\mathrm{T}} \boldsymbol{x}_{i}\right)^{2}+\lambda\|\boldsymbol{w}\|_{2}^{2}
 $$
 
-[解析]：该式为加入了$\mathrm{L}_2$正规化项的优化目标，也叫”岭回归“，$\lambda$用来调节误差项和正规化项的相对重要性，引入正规化项的目的是为了防止$w$的分量过太而导致过拟合的风险。
+[解析]：该式为加入了$\mathrm{L}_2$正规化项的优化目标，也叫"岭回归"，$\lambda$用来调节误差项和正规化项的相对重要性，引入正规化项的目的是为了防止$w$的分量过太而导致过拟合的风险。
 
 ## 11.7
 

docs/chapter13/chapter13.md

@@ -1,7 +1,7 @@
 ## 13.1
 
 $$
-p(\boldsymbol{x})=\sum_{i=1}^{N} \alpha_{i} \cdot p\left(\boldsymbol{x} | \boldsymbol{\mu}_{i}, \mathbf{\Sigma}_{i}\right)
+p(\boldsymbol{x})=\sum_{i=1}^{N} \alpha_{i} \cdot p\left(\boldsymbol{x} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)
 $$
 
 [解析]： 高斯混合分布的定义式。
@@ -20,34 +20,34 @@ $$
 ## 13.3
 
 $$
-p(\Theta=i | \boldsymbol{x})=\frac{\alpha_{i} \cdot p\left(\boldsymbol{x} | \boldsymbol{\mu}_{i}, \mathbf{\Sigma}_{i}\right)}{\sum_{i=1}^{N} \alpha_{i} \cdot p\left(\boldsymbol{x} | \boldsymbol{\mu}_{i}, \mathbf{\Sigma}_{i}\right)}
+p(\Theta=i | \boldsymbol{x})=\frac{\alpha_{i} \cdot p\left(\boldsymbol{x} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)}{\sum_{i=1}^{N} \alpha_{i} \cdot p\left(\boldsymbol{x} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)}
 $$
 
 [解析]：根据 13.1 
 $$
-p(\boldsymbol{x})=\sum_{i=1}^{N} \alpha_{i} \cdot p\left(\boldsymbol{x} | \boldsymbol{\mu}_{i}, \mathbf{\Sigma}_{i}\right)
+p(\boldsymbol{x})=\sum_{i=1}^{N} \alpha_{i} \cdot p\left(\boldsymbol{x} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)
 $$
 因此
 $$
-\begin{aligned}p(\Theta=i | \boldsymbol{x})&=\frac{p(\Theta=i , \boldsymbol{x})}{P(x)}\\&=\frac{\alpha_{i} \cdot p\left(\boldsymbol{x} | \boldsymbol{\mu}_{i}, \mathbf{\Sigma}_{i}\right)}{\sum_{i=1}^{N} \alpha_{i} \cdot p\left(\boldsymbol{x} | \boldsymbol{\mu}_{i}, \mathbf{\Sigma}_{i}\right)}\end{aligned}
+\begin{aligned}p(\Theta=i | \boldsymbol{x})&=\frac{p(\Theta=i , \boldsymbol{x})}{P(x)}\\&=\frac{\alpha_{i} \cdot p\left(\boldsymbol{x} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)}{\sum_{i=1}^{N} \alpha_{i} \cdot p\left(\boldsymbol{x} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)}\end{aligned}
 $$
 
 ## 13.4
 
 $$
-\begin{aligned} L L\left(D_{l} \cup D_{u}\right)=& \sum_{\left(x_{j}, y_{j}\right) \in D_{l}} \ln \left(\sum_{i=1}^{N} \alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \mathbf{\Sigma}_{i}\right) \cdot p\left(y_{j} | \Theta=i, \boldsymbol{x}_{j}\right)\right) \\ &+\sum_{x_{j} \in D_{u}} \ln \left(\sum_{i=1}^{N} \alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \mathbf{\Sigma}_{i}\right)\right) \end{aligned}
+\begin{aligned} L L\left(D_{l} \cup D_{u}\right)=& \sum_{\left(x_{j}, y_{j}\right) \in D_{l}} \ln \left(\sum_{i=1}^{N} \alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right) \cdot p\left(y_{j} | \Theta=i, \boldsymbol{x}_{j}\right)\right) \\ &+\sum_{x_{j} \in D_{u}} \ln \left(\sum_{i=1}^{N} \alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)\right) \end{aligned}
 $$
 
 [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$所对应的高斯分布。
+可知，这项限定了样本$x_j$只可能来自于$y_j$所对应的高斯分布。
 
 ## 13.5
 
 $$
-\gamma_{j i}=\frac{\alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \mathbf{\Sigma}_{i}\right)}{\sum_{i=1}^{N} \alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \mathbf{\Sigma}_{i}\right)}
+\gamma_{j i}=\frac{\alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)}{\sum_{i=1}^{N} \alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)}
 $$
 
 [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：
@@ -105,9 +105,9 @@ $$
 [推导]：类似于13.6 由$\cfrac{\partial LL(D_l \cup D_u) }{\partial \Sigma_i}=0$得，化简过程同13.6过程类似
 首先$LL(D_l)$对$\boldsymbol{\Sigma_i}$求偏导 ，类似于 13.6 
 $$
-\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}, \mathbf{\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}, \mathbf{\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(\mathbf{\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}
+\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}
 $$
 然后$LL(D_u)$ 对$\boldsymbol{\Sigma_i}$求偏导，类似于 9.35
@@ -156,7 +156,7 @@ $$
 
 综合两项结果：
 $$
-\frac{\partial \mathcal{L}\left(D_{l} \cup D_{u}, \lambda\right)}{\partial \alpha_{i}}=\frac{l_{i}}{\alpha_{i}}+\sum_{\boldsymbol{x}_{j} \in D_{u}} \frac{p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)}{\sum_{s=1}^{N} \alpha_{s} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{s}, \mathbf{\Sigma}_{s}\right)}+\lambda
+\frac{\partial \mathcal{L}\left(D_{l} \cup D_{u}, \lambda\right)}{\partial \alpha_{i}}=\frac{l_{i}}{\alpha_{i}}+\sum_{\boldsymbol{x}_{j} \in D_{u}} \frac{p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)}{\sum_{s=1}^{N} \alpha_{s} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{s}, \boldsymbol{\Sigma}_{s}\right)}+\lambda
 $$
 
 

docs/chapter8/chapter8.md

@@ -117,7 +117,7 @@ $$
 \end{aligned}
 $$
 
-[解析]：第一行到第二行显然成立，第二行到第三行是利用了$\arg\max$函数的定义。$\underset{y \in\{-1,1\}}{\arg \max } P(f(x)=y | \boldsymbol{x})$表示使得函数$P(f(x)=y | \boldsymbol{x}$取得最大值的$y$的值，展开刚好和第二行的式子。
+[解析]：第一行到第二行显然成立，第二行到第三行是利用了$\arg\max$函数的定义。$\underset{y \in\{-1,1\}}{\arg \max } P(f(x)=y | \boldsymbol{x})$表示使得函数$P(f(x)=y | \boldsymbol{x}$取得最大值的$y$的值，展开刚好是第二行的式子。
 
 ## 8.9
 

docs/chapter9/chapter9.md

@@ -71,11 +71,11 @@ $$
 
 ## 9.33
 $$
-\sum_{j=1}^m \cfrac{\alpha_{i}\cdot p\left(\boldsymbol x_{j}|\boldsymbol\mu _{i},\mathbf\Sigma_{i}\right)}{\sum_{l=1}^k \alpha_{l}\cdot p(\boldsymbol x_{j}|\boldsymbol\mu_{l},\mathbf\Sigma_{l})}(\boldsymbol x_{j}-\boldsymbol\mu_{i})=0
+\sum_{j=1}^{m} \frac{\alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)}{\sum_{l=1}^{k} \alpha_{l} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{l}, \boldsymbol{\Sigma}_{l}\right)}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)=0
 $$
 [推导]：根据公式(9.28)可知：
 $$
-p(\boldsymbol x_{j}|\boldsymbol\mu_{i},\mathbf\Sigma_{i})=\cfrac{1}{(2\pi)^\frac{n}{2}\left| \mathbf\Sigma_{i}\right |^\frac{1}{2}}\exp\left({-\frac{1}{2}(\boldsymbol x_{j}-\boldsymbol\mu_{i})^T\mathbf\Sigma_{i}^{-1}(\boldsymbol x_{j}-\boldsymbol\mu_{i})}\right)
+p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)=\frac{1}{(2 \pi)^{\frac{n}{2}}\left|\boldsymbol{\Sigma}_{i}\right|^{\frac{1}{2}}} \exp \left(-\frac{1}{2}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{T} \boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\right)
 $$
 又根据公式(9.32)，由
 $$
@@ -83,12 +83,12 @@ $$
 $$
 可得
 $$\begin{aligned}
-\cfrac {\partial LL(D)}{\partial\boldsymbol\mu_{i}}&=\cfrac {\partial}{\partial \boldsymbol\mu_{i}}\left[\sum_{j=1}^m\ln\Bigg(\sum_{i=1}^k \alpha_{i}\cdot p(\boldsymbol x_{j}|\boldsymbol\mu_{i},\mathbf\Sigma_{i})\Bigg)\right] \\
-&=\sum_{j=1}^m\frac{\partial}{\partial\boldsymbol\mu_{i}}\left[\ln\Bigg(\sum_{i=1}^k \alpha_{i}\cdot p(\boldsymbol x_{j}|\boldsymbol\mu_{i},\mathbf\Sigma_{i})\Bigg)\right] \\
-&=\sum_{j=1}^m\cfrac{\alpha_{i}\cdot \cfrac{\partial}{\partial\boldsymbol\mu_{i}}\left(p(\boldsymbol x_{j}|\boldsymbol\mu_{i},\mathbf\Sigma_{i})\right)}{\sum_{l=1}^k\alpha_{l}\cdot p(\boldsymbol x_{j}|\boldsymbol\mu_{l},\mathbf\Sigma_{l})} \\
-&=\sum_{j=1}^m\cfrac{\alpha_{i}\cdot \cfrac{1}{(2\pi)^\frac{n}{2}\left| \mathbf\Sigma_{i}\right |^\frac{1}{2}}\exp\left({-\frac{1}{2}(\boldsymbol x_{j}-\boldsymbol\mu_{i})^T\mathbf\Sigma_{i}^{-1}(\boldsymbol x_{j}-\boldsymbol\mu_{i})}\right)}{\sum_{l=1}^k\alpha_{l}\cdot p(\boldsymbol x_{j}|\boldsymbol\mu_{l},\mathbf\Sigma_{l})}\cfrac{\partial}{\partial \boldsymbol\mu_{i}}\left(-\frac{1}{2}\left(\boldsymbol x_{j}-\boldsymbol\mu_{i}\right)^T\mathbf\Sigma_{i}^{-1}\left(\boldsymbol x_{j}-\boldsymbol\mu_{i}\right)\right) \\
-&=\sum_{j=1}^m\cfrac{\alpha_{i}\cdot p(\boldsymbol x_{j}|\boldsymbol\mu_{i},\mathbf\Sigma_{i})}{\sum_{l=1}^k\alpha_{l}\cdot p(\boldsymbol x_{j}|\boldsymbol\mu_{l},\mathbf\Sigma_{l})}\cdot(-\cfrac{1}{2})\cdot\cfrac{\partial}{\partial \boldsymbol\mu_{i}}\left(\boldsymbol x_j^T\mathbf{\Sigma}_i^{-1}\boldsymbol x_j-\boldsymbol x_j^T\mathbf{\Sigma}_i^{-1}\boldsymbol\mu_i-\boldsymbol\mu_i^T\mathbf{\Sigma}_i^{-1}\boldsymbol x_j+\boldsymbol\mu_i^T\mathbf{\Sigma}_i^{-1}\boldsymbol\mu_i\right) \\
-&=\sum_{j=1}^m\cfrac{\alpha_{i}\cdot p(\boldsymbol x_{j}|\boldsymbol\mu_{i},\mathbf\Sigma_{i})}{\sum_{l=1}^k\alpha_{l}\cdot p(\boldsymbol x_{j}|\boldsymbol\mu_{l},\mathbf\Sigma_{l})}\cdot(-\cfrac{1}{2})\cdot\cfrac{\partial}{\partial \boldsymbol\mu_{i}}\left(-\boldsymbol x_j^T\mathbf{\Sigma}_i^{-1}\boldsymbol\mu_i-\boldsymbol\mu_i^T\mathbf{\Sigma}_i^{-1}\boldsymbol x_j+\boldsymbol\mu_i^T\mathbf{\Sigma}_i^{-1}\boldsymbol\mu_i\right) \\
+\frac{\partial L L(D)}{\partial \boldsymbol{\mu}_{i}} &=\frac{\partial}{\partial \boldsymbol{\mu}_{i}}\left[\sum_{j=1}^{m} \ln \left(\sum_{i=1}^{k} \alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)\right)\right] \\
+&=\sum_{j=1}^{m} \frac{\partial}{\partial \boldsymbol{\mu}_{i}}\left[\ln \left(\sum_{i=1}^{k} \alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)\right)\right] \\
+&=\sum_{j=1}^{m} \frac{\alpha_{i} \cdot \frac{\partial}{\partial \boldsymbol{\mu}_{i}}\left(p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)\right)}{\sum_{l=1}^{k} \alpha_{l} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{l}, \boldsymbol{\Sigma}_{l}\right)} \\
+&=\sum_{j=1}^{m} \frac{1}{(2 \pi)^{\frac{n}{2}}\left|\boldsymbol{\Sigma}_{i}\right|^{\frac{1}{2}} \exp \left(-\frac{1}{2}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{T} \boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\right)} \frac{\partial}{\partial \boldsymbol{\mu}_{l=1} \alpha_{l} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{l}, \boldsymbol{\Sigma}_{l}\right)}\left(-\frac{1}{2}\left(\boldsymbol{x}_{i}\right)^{m}\right) \cdot \\ &\qquad\frac{\partial}{\partial \boldsymbol{\mu}_{i}}\left(\boldsymbol{x}_{j}^{T} \boldsymbol{\Sigma}_{i}^{-1} \boldsymbol{x}_{j}-\boldsymbol{x}_{j}^{T} \boldsymbol{\Sigma}_{i}^{-1} \boldsymbol{\mu}_{i}-\boldsymbol{\mu}_{i}^{T} \boldsymbol{\Sigma}_{i}^{-1} \boldsymbol{x}_{j}+\boldsymbol{\mu}_{i}^{T} \boldsymbol{\Sigma}_{i}^{-1} \boldsymbol{\mu}_{i}\right) \\
+&=\sum_{j=1}^{m} \frac{\alpha_{i} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)}{\sum_{l=1}^{k} \alpha_{l} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{l}, \boldsymbol{\Sigma}_{l}\right)} \cdot\left(-\frac{1}{2}\right) \cdot \frac{\partial_{i}}{2}\left(\boldsymbol{x}_{i}, \boldsymbol{\Sigma}_{i}\right) \\
+&=\sum_{j=1}^{m} \frac{\left.\alpha_{i} \cdot p\left(\boldsymbol{x}_{i}\right)^{T} \boldsymbol{\Sigma}_{l=1}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\right)}{\left(x_{l} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{l}, \boldsymbol{\Sigma}_{l}\right)\right.} \cdot\left(-\frac{1}{2}\right) \cdot \frac{\partial}{\partial \boldsymbol{\mu}_{i}}\left(-\boldsymbol{x}_{j}^{T} \boldsymbol{\Sigma}_{i}^{-1} \boldsymbol{\mu}_{i}-\boldsymbol{\mu}_{i}^{T} \boldsymbol{\Sigma}_{i}^{-1} \boldsymbol{x}_{j}+\boldsymbol{\mu}_{i}^{T} \boldsymbol{\Sigma}_{i}^{-1} \boldsymbol{\mu}_{i}\right)
 \end{aligned}$$
 由于$\boldsymbol x_j^T\mathbf{\Sigma}_i^{-1}\boldsymbol\mu_i$和$\boldsymbol\mu_i^T\mathbf{\Sigma}_i^{-1}\boldsymbol x_j$均为标量且$\mathbf{\Sigma}_i$为对称矩阵，所以
 $$(\boldsymbol x_j^T\mathbf{\Sigma}_i^{-1}\boldsymbol\mu_i)^T=\boldsymbol\mu_i^T({\mathbf{\Sigma}_i^{-1}})^T\boldsymbol x_j=\boldsymbol\mu_i^T({\mathbf{\Sigma}_i^T})^{-1}\boldsymbol x_j=\boldsymbol\mu_i^T\mathbf{\Sigma}_i^{-1}\boldsymbol x_j=\boldsymbol x_j^T\mathbf{\Sigma}_i^{-1}\boldsymbol\mu_i$$