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@@ -95,11 +95,20 @@ $$
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\frac{\partial p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right)}{\partial \boldsymbol{\mu}_{i}} &=\frac{\partial \frac{1}{(2 \pi)^{\frac{n}{2}}\left|\Sigma_{i}\right|^{\frac{1}{2}}} \exp\left({-\frac{1}{2}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top}\boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)}\right)}{\partial \boldsymbol{\mu}_{i}} \\
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&=\frac{1}{(2 \pi)^{\frac{n}{2}}\left|\boldsymbol{\Sigma}_{i}\right|^{\frac{1}{2}}} \cdot \frac{\partial \exp\left({-\frac{1}{2}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top} \boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)}\right)}{\partial \boldsymbol{\mu}_{i}}\\
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&=\frac{1}{(2 \pi)^{\frac{n}{2}}\left|\boldsymbol{\Sigma}_{i}\right|^{\frac{1}{2}}}\cdot \exp\left({-\frac{1}{2}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top} \boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)}\right) \cdot-\frac{1}{2} \frac{\partial\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top} \boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)}{\partial \boldsymbol{\mu}_{i}}\\
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-&=\frac{1}{(2 \pi)^{\frac{n}{2}}\left|\boldsymbol{\Sigma}_{i}\right|^{\frac{1}{2}}}\cdot \exp\left({-\frac{1}{2}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top} \boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)}\right) \cdot-\boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{i}-\boldsymbol{\mu}_{j}\right)\\
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+&=\frac{1}{(2 \pi)^{\frac{n}{2}}\left|\boldsymbol{\Sigma}_{i}\right|^{\frac{1}{2}}}\cdot \exp\left({-\frac{1}{2}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top} \boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)}\right) \cdot\boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)\\
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&=p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right) \cdot \boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)
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\end{aligned}
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$$
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+其中，由矩阵求导的法则$\frac{\partial \mathbf{a}^{T} \mathbf{X} \mathbf{a}}{\partial \mathbf{a}}=2\mathbf{X} \mathbf{a}$可得：
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+$$
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+\begin{aligned}
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+-\frac{1}{2} \frac{\partial\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)^{\top} \boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)}{\partial \boldsymbol{\mu}_{i}} &=-\frac{1}{2} \cdot 2 \boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{\mu}_{i}-\boldsymbol{x}_{j}\right) \\
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+&=\boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)
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+\end{aligned}
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+$$
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+
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+
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因此有：
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$$
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\frac{\partial L L(D)}{\partial \boldsymbol{\mu}_{i}}=\sum_{j=1}^{m} \frac{\alpha_{i}}{\sum_{l=1}^{k} \alpha_{l} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{l}, \mathbf{\Sigma}_{l}\right)} \cdot p\left(\boldsymbol{x}_{j} | \boldsymbol{\mu}_{i}, \boldsymbol{\Sigma}_{i}\right) \cdot \boldsymbol{\Sigma}_{i}^{-1}\left(\boldsymbol{x}_{j}-\boldsymbol{\mu}_{i}\right)=0
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archwalker