Exercise 5.5 Solution Example - Hoff, A First Course in Bayesian Statistical Methods
標準ベイズ統計学 演習問題 5.5 解答例

\begin{align*} l(\theta, \psi | \boldsymbol{y}) &= \sum_{i=1}^n \log p(y_i | \theta, \psi) \\ &= \sum_{i=1}^n \log \left( (2\pi)^{ - \frac{1}{2} } \psi^{ \frac{1}{2} } \exp \left( - \frac{\psi}{2} (y_i - \theta )^2 \right) \right) \\ &= \sum_{i=1}^n \left\{ \log (2\pi)^{ - \frac{1}{2} } + \log \psi^{ \frac{1}{2} } - \frac{\psi}{2} (y_i - \theta )^2 \right\} \\ &= - \frac{n}{2} \log (2\pi) + \frac{n}{2} \log \psi - \frac{\psi}{2} \sum_{i=1}^n (y_i - \theta )^2 \\ \end{align*}

\begin{align*} l(\theta, \psi | \boldsymbol{y}) &= - \frac{n}{2} \log (2\pi) + \frac{n}{2} \log \psi - \frac{\psi}{2} \sum_{i=1}^n (y_i - \theta )^2 \\ &= - \frac{n}{2} \log (2\pi) + \frac{n}{2} \log \psi - \frac{\psi}{2} \left( \sum_{i=1}^n (y_i - \bar{y} )^2 + n (\bar{y} - \theta )^2 \right) \\ &= - \frac{n}{2} \log (2\pi) + \frac{n}{2} \log \psi - \frac{\psi}{2} n s^2 - \frac{\psi}{2} n (\bar{y} - \theta )^2 \\ \end{align*}

従って、

\begin{align*} \log p_U(\theta, \psi) &= \frac{l(\theta, \psi | \boldsymbol{y})}{n} + c \\ &= - \frac{1}{2} \log (2\pi) + \frac{1}{2} \log \psi - \frac{\psi}{2} s^2 - \frac{\psi}{2} (\bar{y} - \theta )^2 + c \\ \end{align*}

となるので、

\begin{align*} p_U(\theta, \psi) &= (2 \pi)^{- \frac{1}{2} } \psi^{ \frac{1}{2} } \exp \left( - \frac{s^2}{2} \psi \right) \exp \left( - \frac{\psi}{2} (\bar{y} - \theta )^2 \right) \exp \left( c \right) \\ &\propto \psi^{ \frac{1}{2} } \exp \left( - \frac{\psi}{2} ( \theta - \bar{y} )^2 \right) \times \psi^{ 1 - 1} \exp \left( - \frac{s^2}{2} \psi \right) \\ &\propto \text{dnormal}(\theta, \bar{y}, \psi^{-1}) \times \text{dgamma}(\psi, 1, \frac{s^2}{2}) \end{align*}

\begin{align*} p_U(\theta, \psi | \boldsymbol{y}) &\propto p_U(\theta, \psi) \times p( y_1, \ldots, y_n | \theta, \psi) \\ &\propto \exp \left( - \frac{\psi}{2} ( \theta - \bar{y} )^2 \right) \times \psi^{ \frac{3}{2} - 1} \exp \left( - \frac{s^2}{2} \psi \right) \times \psi^{ \frac{n}{2} } \exp \left( - \frac{\psi}{2} \sum_{i=1}^n (y_i - \theta )^2 \right) \\ &= \exp \left( - \frac{\psi}{2} ( \theta - \bar{y} )^2 \right) \times \psi^{ \frac{3}{2} - 1} \exp \left( - \frac{s^2}{2} \psi \right) \times \psi^{ \frac{n}{2} } \exp \left( - \frac{\psi}{2} \left( \sum_{i=1}^n (y_i - \bar{y} )^2 + n (\bar{y} - \theta )^2 \right) \right) \\ &= \exp \left( - \frac{\psi}{2} ( \theta - \bar{y} )^2 \right) \times \psi^{ \frac{n + 3}{2} - 1} \exp \left( - \frac{s^2}{2} \psi \right) \times \exp \left( - \frac{n s^2}{2} \psi \right) \times \exp \left( - \frac{\psi}{2} n (\bar{y} - \theta )^2 \right) \\ &= \exp \left( - \frac{ (n + 1) \psi}{2} ( \theta - \bar{y} )^2 \right) \times \psi^{ \frac{n + 3}{2} - 1} \exp \left( - \frac{(n + 1) s^2}{2} \psi \right) \\ &\propto \text{dnormal}(\theta, \bar{y}, \frac{1}{(n + 1)\psi}) \times \text{dgamma}(\psi, \frac{n + 3}{2}, \frac{(n + 1) s^2}{2}) \end{align*}

上式より、joint density は posterior density になっていることが確認できた。