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@@ -69,7 +69,8 @@ $$\begin{aligned}
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\sum_{(x_i,y_i)\in D_l \wedge y_j=i} \cfrac{\partial ln(\alpha_i \cdot p(x_j| \mu_i,\Sigma_i))}{\partial\mu_i}\\
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\sum_{(x_i,y_i)\in D_l \wedge y_j=i} \cfrac{\partial ln(\alpha_i \cdot p(x_j| \mu_i,\Sigma_i))}{\partial\mu_i}\\
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&=\sum_{(x_i,y_i)\in D_l \wedge y_j=i}\cfrac{1}{p(x_j|\mu_i,\Sigma_i) }\cdot \cfrac{\partial p(x_j|\mu_i,\Sigma_i)}{\partial\mu_i}\\
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&=\sum_{(x_i,y_i)\in D_l \wedge y_j=i}\cfrac{1}{p(x_j|\mu_i,\Sigma_i) }\cdot \cfrac{\partial p(x_j|\mu_i,\Sigma_i)}{\partial\mu_i}\\
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&=\sum_{(x_i,y_i)\in D_l \wedge y_j=i}\cfrac{1}{p(x_j|\mu_i,\Sigma_i) }\cdot p(x_j|\mu_i,\Sigma_i) \cdot \Sigma_i^{-1}(x_j-\mu_i)\\
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&=\sum_{(x_i,y_i)\in D_l \wedge y_j=i}\cfrac{1}{p(x_j|\mu_i,\Sigma_i) }\cdot p(x_j|\mu_i,\Sigma_i) \cdot \Sigma_i^{-1}(x_j-\mu_i)\\
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-&=\sum_{x_j \in D_u } \Sigma_i^{-1}(x_j-\mu_i)
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+&=\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i}
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+\boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)
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\end{aligned}$$
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\end{aligned}$$
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对于式(13.4)中的第 2 项$LL(D_u)$,求导结果与式(9.33)的推导过程一样
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对于式(13.4)中的第 2 项$LL(D_u)$,求导结果与式(9.33)的推导过程一样
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@@ -85,7 +86,7 @@ $$
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$$
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$$
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上式中, 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 类样本的有标记 样本数目,因此
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上式中, 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 类样本的有标记 样本数目,因此
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$$
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$$
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-\left(\sum_{x_{j} \in D_{u}} \gamma_{j i}+\sum_{\left(x_{j}, y_{j}\right) \in D_{l} \backslash y_{j}=i} 1\right) \mu_{i}=\sum_{x_{j} \in D_{u}} \gamma_{j i} \cdot x_{j}+\sum_{\left(x_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} x_{j}
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+\left(\sum_{x_{j} \in D_{u}} \gamma_{j i}+\sum_{\left(x_{j}, y_{j}\right) \in D_{l} \backslash y_{j}=i} \mu_{i} \right) =\sum_{x_{j} \in D_{u}} \gamma_{j i} \cdot x_{j}+\sum_{\left(x_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} x_{j}
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$$
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$$
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即得式(13.6);
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即得式(13.6);
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## 13.7
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## 13.7
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