代修订13.6/7/8

docs/chapter13/chapter13.md

@@ -74,7 +74,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 对于式(13.4)中的第 2 项$LL(D_u)$，求导结果与式(9.33)的推导过程一样
-$$\cfrac{\partial LL(D_l \cup D_u) }{\partial \mu_i}=\sum_{x_j \in {D_u}} \cfrac{\alpha_i}{\sum_{s=1}^N \alpha_s \cdotp(x_j|\mu_s,\Sigma_s)} \cdot p(x_j|\mu_i,\Sigma_i )\cdot \Sigma_i^{-1}(x_j-\mu_i)$$
+$$\cfrac{\partial LL(D_u) }{\partial \mu_i}=\sum_{x_j \in {D_u}} \cfrac{\alpha_i}{\sum_{s=1}^N \alpha_s \cdotp(x_j|\mu_s,\Sigma_s)} \cdot p(x_j|\mu_i,\Sigma_i )\cdot \Sigma_i^{-1}(x_j-\mu_i)$$
 $$=\sum_{x_j \in D_u }\gamma_{ji} \cdot  \Sigma_i^{-1}(x_j-\mu_i)$$
 综合两项结果，则$\cfrac{\partial LL(D_l \cup D_u) }{\partial \mu_i}$为
 $$
@@ -106,9 +106,8 @@ $$
 \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}
 $$
 综合两项结果，则$\cfrac{\partial LL(D_l \cup D_u) }{\partial \Sigma_i}$为
-$$\begin{aligned} \frac{\partial L L\left(D_{l} \cup D_{u}\right)}{\partial \boldsymbol{\mu}_{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.\\ &+\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}
+$$
+\begin{aligned} \frac{\partial L L\left(D_{l} \cup D_{u}\right)}{\partial \boldsymbol{\mu}_{i}}=& \sum_{\boldsymbol{x}_{j} \in D_{u}} \gamma_{j i} \cdot\left(\boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top}-\boldsymbol{I}\right) \cdot \frac{1}{2} \boldsymbol{\Sigma}_{i}^{-1} \\ &+\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i}\left(\boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top}-\boldsymbol{I}\right) \cdot \frac{1}{2} \boldsymbol{\Sigma}_{i}^{-1} \\=&\left(\sum_{\boldsymbol{x}_{j} \in D_{u}} \gamma_{j i} \cdot\left(\boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top}-\boldsymbol{I}\right)\right.\\ &\left.+\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i}\left(\boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top}-\boldsymbol{I}\right)\right) \cdot \frac{1}{2} \boldsymbol{\Sigma}_{i}^{-1} \end{aligned}
 $$
 令$\frac{\partial L L\left(D_{l} \cup D_{u}\right)}{\partial \boldsymbol{\Sigma}_{i}}=0$，两边同时右乘$2\Sigma_i$可将 $\cfrac{1}{2}\Sigma_i^{-1}$消掉，移项即得
 $$
@@ -130,7 +129,7 @@ $$\begin{aligned}
 \end{aligned}$$
 类似于式(9.37)，对$\alpha_i$求偏导。对于LL(D_u)，求导结果与式(9.37)的推导过程一样：
 $$\cfrac{\partial LL(D_u)}{\partial\alpha_i} = \sum_{x_j \in D_u} \cfrac{1}{\Sigma_{s=1}^N \alpha_s \cdot p(x_j|\mu_s,\Sigma_s)} \cdot p(x_j|\mu_i,\Sigma_i)$$
-对于$LL(D_l)$，类似于类似于(13.6)和(13.7)的推导过程
+对于$LL(D_l)$，类似于(13.6)和(13.7)的推导过程
 $$\begin{aligned}
 \cfrac{\partial LL(D_l)}{\partial\alpha_i} &= \sum_{(x_i,y_i)\in D_l \wedge y_j=i} \cfrac{\partial ln(\alpha_i \cdot p(x_j| \mu_i,\Sigma_i))}{\partial\alpha_i}\\
 &=\sum_{(x_i,y_i)\in D_l \wedge y_j=i}\cfrac{1}{ \alpha_i \cdot p(x_j|\mu_i,\Sigma_i) }\cdot  \cfrac{\partial (\alpha_i \cdot  p(x_j|\mu_i,\Sigma_i))}{\partial \alpha_i}\\
@@ -157,6 +156,4 @@ $$\sum_{i=1}^N l_i+\sum_{i=1}^N  \sum_{x_i \in D_u} \gamma_{ji}+\sum_{i=1}^N \la
 解出$l+u+\lambda = 0$  且$l+u =m$ 其中$m$为样本总个数，移项即得$\lambda = -m$
 最后带入整理解得
 $$l_i + \Sigma_{X_j \in{D_u}} \gamma_{ji}-m \alpha_i = 0$$
-整理即得式(13.8)；
-
-
+整理即得式(13.8)；