修订9.38下标

docs/chapter9/chapter9.md

@@ -140,20 +140,21 @@ $$\sum_{j=1}^m\gamma_{ji}(\boldsymbol x_j-\boldsymbol\mu_i)(\boldsymbol x_j-\bol
 $$\mathbf{\Sigma}_i^{-1}\cdot\sum_{j=1}^m\gamma_{ji}(\boldsymbol x_j-\boldsymbol\mu_i)(\boldsymbol x_j-\boldsymbol\mu_i)^T=\sum_{j=1}^m\gamma_{ji}$$
 $$\mathbf{\Sigma}_i=\cfrac{\sum_{j=1}^m\gamma_{ji}(\boldsymbol x_j-\boldsymbol\mu_i)(\boldsymbol x_j-\boldsymbol\mu_i)^T}{\sum_{j=1}^m\gamma_{ji}}$$
 此即为公式(9.35)
+
 ## 9.38
 $$
 \alpha_{i}=\frac{1}{m}\sum_{j=1}^m\gamma_{ji}
 $$
 [推导]：对公式(9.37)两边同时乘以$\alpha_{i}$可得
-$$\sum_{j=1}^m\frac{\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_{i},\mathbf\Sigma_{i})}+\lambda\alpha_{i}=0$$
-$$\sum_{j=1}^m\frac{\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_{i},\mathbf\Sigma_{i})}=-\lambda\alpha_{i}$$
+$$\sum_{j=1}^m\frac{\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})}+\lambda\alpha_{i}=0$$
+$$\sum_{j=1}^m\frac{\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})}=-\lambda\alpha_{i}$$
 两边对所有混合成分求和可得
-$$\sum_{i=1}^k\sum_{j=1}^m\frac{\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_{i},\mathbf\Sigma_{i})}=-\lambda\sum_{i=1}^k\alpha_{i}$$
-$$\sum_{j=1}^m\sum_{i=1}^k\frac{\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_{i},\mathbf\Sigma_{i})}=-\lambda\sum_{i=1}^k\alpha_{i}$$
+$$\sum_{i=1}^k\sum_{j=1}^m\frac{\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})}=-\lambda\sum_{i=1}^k\alpha_{i}$$
+$$\sum_{j=1}^m\sum_{i=1}^k\frac{\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})}=-\lambda\sum_{i=1}^k\alpha_{i}$$
 $$m=-\lambda$$
 所以
-$$\sum_{j=1}^m\frac{\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_{i},\mathbf\Sigma_{i})}=-\lambda\alpha_{i}=m\alpha_{i}$$
-$$\alpha_{i}=\cfrac{1}{m}\sum_{j=1}^m\frac{\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_{i},\mathbf\Sigma_{i})}$$
+$$\sum_{j=1}^m\frac{\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})}=-\lambda\alpha_{i}=m\alpha_{i}$$
+$$\alpha_{i}=\cfrac{1}{m}\sum_{j=1}^m\frac{\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})}$$
 又由公式(9.30)可知$\cfrac{\alpha_{i}\cdot p(\boldsymbol x_{j}|\boldsymbol\mu_{i},\mathbf\Sigma_{i})}{\sum_{l=1}^k\alpha_{l}\cdot p(\boldsymbol x_{j}|\boldsymbol\mu_{l},\mathbf\Sigma_{l})}=\gamma_{ji}$，所以上式可进一步化简为
 $$\alpha_{i}=\cfrac{1}{m}\sum_{j=1}^m\gamma_{ji}$$
 此即为公式(9.38)