update 7.13

docs/chapter7/chapter7.md

@@ -37,7 +37,21 @@ $$\begin{aligned}
 \hat{\boldsymbol{\mu}}_{c}, \hat{\boldsymbol{\Sigma}}_{c}&=\underset{\boldsymbol{\mu}_{c},\boldsymbol{\Sigma}_c}{\arg \min }\sum_{i=1}^{n} \left[\frac{1}{2}\log|\boldsymbol{\Sigma}_c|+\frac{1}{2}(\boldsymbol{x}_{i}-\boldsymbol{\mu}_c)^{\mathrm{T}} \boldsymbol{\Sigma}_c^{-1}(\boldsymbol{x}_{i}-\boldsymbol{\mu}_c)\right]\\
 &=\underset{\boldsymbol{\mu}_{c},\boldsymbol{\Sigma}_c}{\arg \min }\frac{n}{2}\log|\boldsymbol{\Sigma}_c|+\sum_{i=1}^{n}\frac{1}{2}(\boldsymbol{x}_i-\boldsymbol{\mu}_c)^{\mathrm{T}} \boldsymbol{\Sigma}_c^{-1}(\boldsymbol{x}_i-\boldsymbol{\mu}_c)\\
 \end{aligned}$$
-为了便于分别求解$\hat{\boldsymbol{\mu}}_{c}$和$\hat{\boldsymbol{\Sigma}}_{c}$，在这里我们根据公式$\boldsymbol{x}^{\mathrm{T}}\mathbf{A}\boldsymbol{x}=\operatorname{tr}(\mathbf{A}\boldsymbol{x}\boldsymbol{x}^{\mathrm{T}}),\bar{\boldsymbol{x}}=\frac{1}{n}\sum_{i=1}^{n}\boldsymbol{x}_i$将上式恒等变形为
+为了便于分别求解$\hat{\boldsymbol{\mu}}_{c}$和$\hat{\boldsymbol{\Sigma}}_{c}$，在这里我们根据公式$\boldsymbol{x}^{\mathrm{T}}\mathbf{A}\boldsymbol{x}=\operatorname{tr}(\mathbf{A}\boldsymbol{x}\boldsymbol{x}^{\mathrm{T}}),\bar{\boldsymbol{x}}=\frac{1}{n}\sum_{i=1}^{n}\boldsymbol{x}_i$将上式中的最后一项作如下恒等变形
+$$\begin{aligned}
+&\sum_{i=1}^{n}\frac{1}{2}(\boldsymbol{x}_i-\boldsymbol{\mu}_c)^{\mathrm{T}} \boldsymbol{\Sigma}_c^{-1}(\boldsymbol{x}_i-\boldsymbol{\mu}_c)\\
+=&\frac{1}{2}\operatorname{tr}\left[\boldsymbol{\Sigma}_c^{-1}\sum_{i=1}^{n}(\boldsymbol{x}_i-\boldsymbol{\mu}_c)(\boldsymbol{x}_i-\boldsymbol{\mu}_c)^{\mathrm{T}}\right]\\
+=&\frac{1}{2}\operatorname{tr}\left[\boldsymbol{\Sigma}_c^{-1}\sum_{i=1}^{n}\left(\boldsymbol{x}_i\boldsymbol{x}_i^{\mathrm{T}}-\boldsymbol{x}_i\boldsymbol{\mu}_c^{\mathrm{T}}-\boldsymbol{\mu}_c\boldsymbol{x}_i^{\mathrm{T}}+\boldsymbol{\mu}_c\boldsymbol{\mu}_c^{\mathrm{T}}\right)\right]\\
+=&\frac{1}{2}\operatorname{tr}\left[\boldsymbol{\Sigma}_c^{-1}\left(\sum_{i=1}^{n}\boldsymbol{x}_i\boldsymbol{x}_i^{\mathrm{T}}-n\bar{\boldsymbol{x}}\boldsymbol{\mu}_c^{\mathrm{T}}-n\boldsymbol{\mu}_c\bar{\boldsymbol{x}}^{\mathrm{T}}+n\boldsymbol{\mu}_c\boldsymbol{\mu}_c^{\mathrm{T}}\right)\right]\\
+=&\frac{1}{2}\operatorname{tr}\left[\boldsymbol{\Sigma}_c^{-1}\left(\sum_{i=1}^{n}\boldsymbol{x}_i\boldsymbol{x}_i^{\mathrm{T}}-2n\bar{\boldsymbol{x}}\boldsymbol{\mu}_c^{\mathrm{T}}+n\boldsymbol{\mu}_c\boldsymbol{\mu}_c^{\mathrm{T}}+2n\bar{\boldsymbol{x}}\bar{\boldsymbol{x}}^{\mathrm{T}}-2n\bar{\boldsymbol{x}}\bar{\boldsymbol{x}}^{\mathrm{T}}\right)\right]\\
+=&\frac{1}{2}\operatorname{tr}\left[\boldsymbol{\Sigma}_c^{-1}\left(\left(\sum_{i=1}^{n}\boldsymbol{x}_i\boldsymbol{x}_i^{\mathrm{T}}-2n\bar{\boldsymbol{x}}\bar{\boldsymbol{x}}^{\mathrm{T}}+n\bar{\boldsymbol{x}}\bar{\boldsymbol{x}}^{\mathrm{T}}\right)+\left(n\boldsymbol{\mu}_c\boldsymbol{\mu}_c^{\mathrm{T}}-2n\bar{\boldsymbol{x}}\boldsymbol{\mu}_c^{\mathrm{T}}+n\bar{\boldsymbol{x}}\bar{\boldsymbol{x}}^{\mathrm{T}}\right)\right)\right]\\
+=&\frac{1}{2}\operatorname{tr}\left[\boldsymbol{\Sigma}_c^{-1}\left(\sum_{i=1}^{n}(\boldsymbol{x}_i-\bar{\boldsymbol{x}})(\boldsymbol{x}_i-\bar{\boldsymbol{x}})^{\mathrm{T}}+\sum_{i=1}^{n}(\boldsymbol{\mu}_c-\bar{\boldsymbol{x}})(\boldsymbol{\mu}_c-\bar{\boldsymbol{x}})^{\mathrm{T}}\right)\right]\\
+=&\frac{1}{2}\operatorname{tr}\left[\boldsymbol{\Sigma}_c^{-1}\sum_{i=1}^{n}(\boldsymbol{x}_i-\bar{\boldsymbol{x}})(\boldsymbol{x}_i-\bar{\boldsymbol{x}})^{\mathrm{T}}\right]+\frac{1}{2}\operatorname{tr}\left[\boldsymbol{\Sigma}_c^{-1}\sum_{i=1}^{n}(\boldsymbol{\mu}_c-\bar{\boldsymbol{x}})(\boldsymbol{\mu}_c-\bar{\boldsymbol{x}})^{\mathrm{T}}\right]\\
+=&\frac{1}{2}\operatorname{tr}\left[\boldsymbol{\Sigma}_c^{-1}\sum_{i=1}^{n}(\boldsymbol{x}_i-\bar{\boldsymbol{x}})(\boldsymbol{x}_i-\bar{\boldsymbol{x}})^{\mathrm{T}}\right]+\frac{1}{2}\operatorname{tr}\left[n\cdot\boldsymbol{\Sigma}_c^{-1}(\boldsymbol{\mu}_c-\bar{\boldsymbol{x}})(\boldsymbol{\mu}_c-\bar{\boldsymbol{x}})^{\mathrm{T}}\right]\\
+=&\frac{1}{2}\operatorname{tr}\left[\boldsymbol{\Sigma}_c^{-1}\sum_{i=1}^{n}(\boldsymbol{x}_i-\bar{\boldsymbol{x}})(\boldsymbol{x}_i-\bar{\boldsymbol{x}})^{\mathrm{T}}\right]+\frac{n}{2}\operatorname{tr}\left[\boldsymbol{\Sigma}_c^{-1}(\boldsymbol{\mu}_c-\bar{\boldsymbol{x}})(\boldsymbol{\mu}_c-\bar{\boldsymbol{x}})^{\mathrm{T}}\right]\\
+=&\frac{1}{2}\operatorname{tr}\left[\boldsymbol{\Sigma}_c^{-1}\sum_{i=1}^{n}(\boldsymbol{x}_i-\bar{\boldsymbol{x}})(\boldsymbol{x}_i-\bar{\boldsymbol{x}})^{\mathrm{T}}\right]+\frac{n}{2}(\boldsymbol{\mu}_c-\bar{\boldsymbol{x}})^{\mathrm{T}} \boldsymbol{\Sigma}_c^{-1}(\boldsymbol{\mu}_c-\bar{\boldsymbol{x}})
+\end{aligned}$$
+所以
 $$\hat{\boldsymbol{\mu}}_{c}, \hat{\boldsymbol{\Sigma}}_{c}=\underset{\boldsymbol{\mu}_{c},\boldsymbol{\Sigma}_c}{\arg \min }\frac{n}{2}\log|\boldsymbol{\Sigma}_c|+\frac{1}{2}\operatorname{tr}\left[\boldsymbol{\Sigma}_{c}^{-1}\sum_{i=1}^{n}(\boldsymbol{x}_i-\bar{\boldsymbol{x}})(\boldsymbol{x}_i-\bar{\boldsymbol{x}})^{\mathrm{T}}\right]+\frac{n}{2}(\boldsymbol{\mu}_c-\bar{\boldsymbol{x}})^{\mathrm{T}} \boldsymbol{\Sigma}_c^{-1}(\boldsymbol{\mu}_c-\bar{\boldsymbol{x}})$$
 观察上式可知，由于此时$\boldsymbol{\Sigma}_c^{-1}$和$\boldsymbol{\Sigma}_c$一样均为正定矩阵，所以当$\boldsymbol{\mu}_c-\bar{\boldsymbol{x}}\neq\boldsymbol{0}$时，上式最后一项为正定二次型。根据正定二次型的性质可知，上式最后一项取值的大小此时仅与$\boldsymbol{\mu}_c-\bar{\boldsymbol{x}}$相关，而且当且仅当$\boldsymbol{\mu}_c-\bar{\boldsymbol{x}}=\boldsymbol{0}$时，上式最后一项取到最小值0，此时可以解得
 $$\hat{\boldsymbol{\mu}}_{c}=\bar{\boldsymbol{x}}=\frac{1}{n}\sum_{i=1}^{n}\boldsymbol{x}_i$$