|
|
@@ -69,7 +69,8 @@ $$\begin{aligned}
|
|
|
\sum_{(x_i,y_i)\in D_l \wedge y_j=i} \cfrac{\partial ln(\alpha_i \cdot p(x_j| \mu_i,\Sigma_i))}{\partial\mu_i}\\
|
|
|
&=\sum_{(x_i,y_i)\in D_l \wedge y_j=i}\cfrac{1}{p(x_j|\mu_i,\Sigma_i) }\cdot \cfrac{\partial p(x_j|\mu_i,\Sigma_i)}{\partial\mu_i}\\
|
|
|
&=\sum_{(x_i,y_i)\in D_l \wedge y_j=i}\cfrac{1}{p(x_j|\mu_i,\Sigma_i) }\cdot p(x_j|\mu_i,\Sigma_i) \cdot \Sigma_i^{-1}(x_j-\mu_i)\\
|
|
|
-&=\sum_{x_j \in D_u } \Sigma_i^{-1}(x_j-\mu_i)
|
|
|
+&=\sum_{\left(\boldsymbol{x}_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i}
|
|
|
+\boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)
|
|
|
\end{aligned}$$
|
|
|
|
|
|
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
|
|
|
@@ -85,7 +86,7 @@ $$
|
|
|
$$
|
|
|
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
|
|
|
$$
|
|
|
-\left(\sum_{x_{j} \in D_{u}} \gamma_{j i}+\sum_{\left(x_{j}, y_{j}\right) \in D_{l} \backslash y_{j}=i} 1\right) \mu_{i}=\sum_{x_{j} \in D_{u}} \gamma_{j i} \cdot x_{j}+\sum_{\left(x_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} x_{j}
|
|
|
+\left(\sum_{x_{j} \in D_{u}} \gamma_{j i}+\sum_{\left(x_{j}, y_{j}\right) \in D_{l} \backslash y_{j}=i} \mu_{i} \right) =\sum_{x_{j} \in D_{u}} \gamma_{j i} \cdot x_{j}+\sum_{\left(x_{j}, y_{j}\right) \in D_{l} \wedge y_{j}=i} x_{j}
|
|
|
$$
|
|
|
即得式(13.6)；
|
|
|
## 13.7
|
Sm1les