Exercise 3.6 Solution Example - Hoff, A First Course in Bayesian Statistical Methods
標準ベイズ統計学 演習問題 3.6 解答例

Taking derivatives with respect to \(\phi\) of both sides of the equation \( \int p(y \mid \theta) dy = 1 \), The left-hand side is

\begin{align*} \frac{d}{d \phi } \int p(y \mid \theta) dy &= \int \frac{d}{d \phi } p(y \mid \theta) dy \\ &= \int \frac{d}{d \phi } c(\phi) h(y) \exp\{\phi t(y) \} dy \\ &= \int h(y) \left( c(\phi) \frac{d \exp \{\phi t(y) \} }{ d \phi } + \frac{d c(\phi)}{d \phi } \exp \{\phi t(y) \} \right) dy \\ &= \int h(y) \left( c(\phi) t(y) \exp \{\phi t(y) \} + c'(\phi) \exp \{\phi t(y) \} \right) dy \\ &= \int h(y) c(\phi) t(y) \exp \{\phi t(y) \} dy + \int h(y) c'(\phi) \exp \{\phi t(y) \} dy \\ &= E[ t(y) \mid \phi] + \frac{c'(\phi)}{c(\phi)} \int c(\phi) h(y) \exp \{\phi t(y) \} dy \\ &\quad (\because c(\phi) \neq 0 \text{ because } p(y \mid \theta) \text{ is a probability density}) \\ &= E[ t(y) \mid \phi] + \frac{c'(\phi)}{c(\phi)} \int p(y \mid \theta) dy \\ &= E[ t(y) \mid \phi] + \frac{c'(\phi)}{c(\phi)}. \\ \end{align*}

Since the right-hand side is 0, \[ E[ t(y) \mid \phi] = - \frac{c'(\phi)}{c(\phi)}. \]

\begin{align*} \frac{d p(\phi)}{d \phi} &= \frac{d}{d \phi} \left( \kappa(n_0, t_0) c(\phi)^{n_0} \exp \{ n_0 t_0 \phi \} \right) \\ &= \kappa(n_0, t_0) \left( \frac{d \ c(\phi)^{n_0} }{d \phi} \exp \{ n_0 t_0 \phi \} + c(\phi)^{n_0} \frac{d}{d \phi} \exp \{ n_0 t_0 \phi \} \right) \\ &= \kappa(n_0, t_0) \left( n_0 c(\phi)^{n_0 - 1} c'(\phi) \exp \{ n_0 t_0 \phi \} + c(\phi)^{n_0} n_0 t_0 \exp \{ n_0 t_0 \phi \} \right) \\ &= \frac{c'(\phi)}{c(\phi)} n_0 \kappa(n_0, t_0) c(\phi)^{n_0} \exp \{ n_0 t_0 \phi \} + n_0 t_0 \kappa(n_0, t_0) c(\phi)^{n_0} \exp \{ n_0 t_0 \phi \} \\ &= \frac{c'(\phi)}{c(\phi)} n_0 p(\phi) + n_0 t_0 p(\phi) \\ \end{align*}

Taking the integral of both sides with respect to \(\phi\), the left-hand side is

\begin{align*} \int \frac{d p(\phi)}{d \phi} d \phi &= \frac{d}{d \phi} \int p(\phi) d \phi \\ &= \frac{d}{d \phi} 1 \\ &= 0. \\ \end{align*}

The right-hand side is

\begin{align*} n_0 \int \frac{c'(\phi)}{c(\phi)} p(\phi) d \phi + n_0 t_0 \int p(\phi) d \phi & = n_0 E \left[ \frac{c'(\phi)}{c(\phi)} \right] + n_0 t_0. \end{align*}

Thus, we have \[ E \left[- \frac{c'(\phi)}{c(\phi)} \right] = t_0. \]