\[ \newcommand\comment[1]{\color{red}{{\Tiny\mbox{#1}}}} \newcommand\commentb[1]{\color{red}{\mbox{#1}}} \newcommand\b[1]{\pmb{\mathrm{#1}}} \newcommand\T{\mathsf{T}} \newcommand\bx{\b{x}} \newcommand\bX{\b{X}} \newcommand\bZ{\b{Z}} \newcommand\bmu{\b{\mu}} \newcommand\bpi{\b{\pi}} \newcommand\Ex{\mathbb{E}} \newcommand\pdiff[2]{ \frac{\partial #1}{\partial #2} } \newcommand\S[1]{ \sum_{#1}} \newcommand\l{\left} \newcommand\r{\right} \newcommand\mean[1]{\overline{#1}} \newcommand\mbx{\mean{\bx}} \] \[ \leqalignno{ &\Ex_{\bZ}[\ln p(\bX,\bZ)] = \S{n}\S{j}\gamma(z_{nj})\l\{ \ln \pi_j + \S{i}\l[x_{ni}\mu_{ji}+(1-x_{ni})\ln(1-\mu_{ji})\r] \r\} &(9.55) } \] これより \[ \leqalignno{ \pdiff{\Ex_{\bZ}}{\mu_{kl}} &= \S{n}\pdiff{}{\mu_{kl}}\gamma(z_{nk}) \l\{ \ln \pi_k + \S{i}[ x_{ni}\ln\mu_{ki}+(1-x_{ni})\ln(1-\mu_{ki}) ] \r\} ~~~\comment{※1} \\ &= \S{n}\gamma(z_{nk})\pdiff{}{\mu_{kl}}\S{i}[ x_{ni}\ln\mu_{ki}+(1-x_{ni})\ln(1-\mu_{ki}) ] \\ &= \S{n}\gamma(z_{nk})\pdiff{}{\mu_{kl}}[ x_{nl}\ln\mu_{kl}+(1-x_{nl})\ln(1-\mu_{kl}))] ~~~\comment{※2} \\ &= \S{n}\gamma(z_{nk})\l[x_{nl}\frac{1}{\mu_{kl}} + (1-x_{nl})\frac{-1}{1-\mu_{kl}} \r] } \] を得る。\(これを\ ０\ とおいて\ \mu_{kl}\ について解く\) \[ \leqalignno{ &\S{n}\gamma(z_{nk})\l[x_{nl}\frac{1}{\mu_{kl}}+(1-x_{nl})\frac{-1}{1-\mu_{kl}} \r] = 0 \\ &\therefore \frac{1}{\mu_{kl}}\S{n}\gamma(z_{nk})x_{nl} - \frac{1}{1-\mu_{kl}}\S{n}\gamma(z_{nk})(1-x_{nl}) = 0 \\ &\therefore \frac{1}{\mu_{kl}} N_k \mbx_{kl} - \frac{1}{1-\mu_{kl}}(N_k - N_k \mbx_{kl}) = 0 ~~~\comment{※３} } \] \(両辺に\ \mu_{kl}(1-\mu_{kl})\ を掛けると\) \[ \leqalignno{ &N_k\mbx_{ｋｌ} - \mu_{kl}N_k\mbx_{kl} - N_k\mu_{kl}+\mu_{kl}N_k\mbx_{kl} = 0 \\ &\therefore \mu_{kl} = \mbx_{kl} } \] を得る。 \( \mu_{kl}\ は\ \bmu_k\ の成分なのでベクトル表示すると \) \[ \leqalignno{ &\bmu_k = \mbx_k &(9.59) } \] を得る。