PRML 第1章 1.11(基本)

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$$
\begin{array}{ll}
(1.54) & \ln p(\mathbf{x} | \mu,\sigma^2) = -\dfrac{1}{2\sigma^2} {\displaystyle \sum_{n=1}^N} (x_n - \mu)^2 - \dfrac{N}{2} \ln\sigma^2 - \dfrac{N}{2} \ln(2\pi)\\[1em]
(1.55) & \mu_{\scriptscriptstyle\mathrm{ML}} = \dfrac{1}{N} {\displaystyle \sum_{n=1}^N} x_n\\[1em]
(1.56) & \sigma_{\scriptscriptstyle\mathrm{ML}}^2 = \dfrac{1}{N} {\displaystyle \sum_{n=1}^N} (x_n - \mu_{\scriptscriptstyle\mathrm{ML}})^2
\end{array}
$$

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解答

　(1.54)を$${\mu}$$で微分すると，

$$
\begin{array}{l}
\dfrac{\partial}{\partial\mu} \ln p(\mathbf{x} | \mu,\sigma^2) = -\dfrac{1}{2\sigma^2} \dfrac{\partial}{\partial\mu} {\displaystyle \sum_{n=1}^N} (x_n - \mu)^2 = \dfrac{1}{\sigma^2} {\displaystyle \sum_{n=1}^N} (x_n - \mu)
\end{array}
$$

となる．これが0であることから，

$$
\begin{array}{l}
{\displaystyle \sum_{n=1}^N} (x_n - \mu_{\scriptscriptstyle\mathrm{ML}}) = 0 \quad\Longleftrightarrow\quad {\displaystyle \sum_{n=1}^N} x_n - N\mu_{\scriptscriptstyle\mathrm{ML}} = 0 \qquad\therefore \mu_{\scriptscriptstyle\mathrm{ML}} = \dfrac{1}{N} {\displaystyle \sum_{n=1}^N} x_n
\end{array}
$$

となり，(1.55)が成り立つ．

　また，(1.54)で$${\mu = \mu_{\scriptscriptstyle\mathrm{ML}}}$$として$${\sigma^2}$$で微分すると，

$$
\begin{array}{l}
\dfrac{\partial}{\partial\sigma^2} \ln p(\mathbf{x} | \mu_{\scriptscriptstyle\mathrm{ML}},\sigma^2) = \dfrac{1}{2\sigma^4} {\displaystyle \sum_{n=1}^N} (x_n - \mu_{\scriptscriptstyle\mathrm{ML}})^2 - \dfrac{N}{2\sigma^2}
\end{array}
$$

となる．これが0であることから，

$$
\begin{array}{l}
\dfrac{1}{2\sigma_{\scriptscriptstyle\mathrm{ML}}^4} \left( {\displaystyle \sum_{n=1}^N} (x_n - \mu)^2 - N\sigma_{\scriptscriptstyle\mathrm{ML}}^2 \right) = 0
\qquad\therefore \sigma_{\scriptscriptstyle\mathrm{ML}}^2 = \dfrac{1}{N} {\displaystyle \sum_{n=1}^N} (x_n - \mu_{\scriptscriptstyle\mathrm{ML}})^2
\end{array}
$$

となり，(1.56)が成り立つ．