Exercise 3.5 Solution Example - Hoff, A First Course in Bayesian Statistical Methods
標準ベイズ統計学 演習問題 3.5 解答例

Table of Contents

Answer

a)

\begin{align*} p(\theta \mid y_1, \dots, y_n) &\propto \tilde{p}(\theta) p(y_1, \dots, y_n \mid \theta) \\ &= \sum_{k=1}^K w_k p_k(\theta \mid n_k, n_k t_k) \prod_{i=1}^n p(y_i \mid \theta) \\ &= \sum_{k=1}^K w_k \kappa_k(n_k, t_k) c(\theta)^{n_k} e^{n_k t_k \theta} \times \prod_{i=1}^n c(\theta) h(y_i) \exp \{\theta t(y_i)\} \\ &= \sum_{k=1}^K w_k \kappa_k(n_k, t_k) c(\theta)^{n_k} e^{n_k t_k \theta} \times \left( c(\theta)^n \times \prod_{i=1}^n h(y_i) \times \exp \{\theta \sum_{i=1}^n t(y_i)\} \right) \\ &\propto \sum_{k=1}^K w_k \kappa_k(n_k, t_k) c(\theta)^{n_k} e^{n_k t_k \theta} \times \left( c(\theta)^n \times \exp \{\theta \sum_{i=1}^n t(y_i)\} \right) \\ &= \sum_{k=1}^K w_k \kappa_k(n_k, t_k) c(\theta)^{n_k + n} \exp \left\{ \theta \times \left[ n_k t_k + \sum_{i=1}^n t(y_i) \right] \right\} \\ &\propto \sum_{k=1}^K w_k p_k (\theta \mid n_k + n, \frac{n_k t_k + n \bar{t}(\boldsymbol{y})}{n_k + n}) \end{align*}

where \(\bar{t}(\boldsymbol{y}) = \frac{\sum t(y_i)}{n}\)

b)

\begin{align*} p(\theta \mid y_1, \dots, y_n) &\propto \tilde{p}(\theta) p(y_1, \dots, y_n \mid \theta) \\ &= \sum_{k=1}^K w_k p_k(\theta \mid a_k, b_k) \times \prod_{i=1}^n p(y_i \mid \theta) \\ &= \sum_{k=1}^K w_k \frac{b_k^{a_k}}{\Gamma(a_k)} \theta^{a_k-1} \exp \left( - b_k \theta \right) \times \prod_{i=1}^n \frac{\theta^{y_i} e^{- \theta } }{y_i!} \\ &\propto \sum_{k=1}^K w_k \frac{b_k^{a_k}}{\Gamma(a_k)} \theta^{a_k-1} \exp \left( - b_k \theta \right) \times \theta^{\sum_{i=1}^n y_i} \exp \left( - n \theta \right) \\ &= \sum_{k=1}^K w_k \frac{b_k^{a_k}}{\Gamma(a_k)} \theta^{a_k + \sum y_i - 1} \exp \left( - (b_k + n) \theta \right) \\ &\propto \sum_{k=1}^K w_k \text{dgamma}(\theta \mid a_k + \sum y_i, b_k + n) \end{align*}

Author: Kaoru Babasaki

Email: [email protected]

Last Updated: 2025-05-02 金 16:29

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