Exercise 6.3 Solution Example - Hoff, A First Course in Bayesian Statistical Methods
標準ベイズ統計学 演習問題 6.3 解答例

\begin{align*} p(\beta | \boldsymbol{y}, \boldsymbol{x}, \boldsymbol{z}, c) &\propto p(\boldsymbol{y}, \boldsymbol{x}, \boldsymbol{z}, c | \beta) p(\beta) \\ &= p(\boldsymbol{y} | \boldsymbol{x}, \boldsymbol{z}, c, \beta) p(\boldsymbol{x}, \boldsymbol{z}, c | \beta) p(\beta) \\ &= p(\boldsymbol{y} | \boldsymbol{x}, \boldsymbol{z}, c, \beta) p(\boldsymbol{z} | \boldsymbol{x}, c, \beta) p(\boldsymbol{x}, c | \beta) p(\beta) \\ &= p(\boldsymbol{y} | \boldsymbol{z}, c) p(\boldsymbol{z} | \boldsymbol{x}, \beta) p(\boldsymbol{x}) p(c) p(\beta) \\ &\propto p(\boldsymbol{z} | \boldsymbol{x},\beta) p(\beta) \\ &\propto \prod_{i=1}^n \exp \left( -\frac{1}{2} (z_i - \beta x_i)^2 \right) \exp \left( -\frac{1}{2 \tau_{\beta}^2} \beta^2 \right) \\ &= \exp \left( -\frac{1}{2} \sum_{i=1}^n (z_i - \beta x_i)^2 \right) \exp \left( -\frac{1}{2 \tau_{\beta}^2} \beta^2 \right) \\ &= \exp \left( -\frac{1}{2} \left( \sum_{i=1}^n z_i^2 - 2 \beta \sum_{i=1}^n x_i z_i + \beta^2 \sum_{i=1}^n x_i^2 \right) \right) \exp \left( -\frac{1}{2 \tau_{\beta}^2} \beta^2 \right) \\ \end{align*}

ここで、

\begin{align*} \sum_{i=1}^n (z_i - \beta x_i)^2 + \frac{1}{\tau_{\beta}^2} \beta^2 &= \sum_{i=1}^n z_i^2 - 2 \beta \sum_{i=1}^n x_i z_i + \beta^2 \sum_{i=1}^n x_i^2 + \frac{1}{\tau_{\beta}^2} \beta^2 \\ &= a \beta^2 - 2 b \beta + \sum_{i=1}^n z_i^2 \\ \mathrm{where} \quad a &= \sum_{i=1}^n x_i^2 + \frac{1}{\tau_{\beta}^2} \\ b &= \sum_{i=1}^n x_i z_i \\ \end{align*}

となるので、

\begin{align*} p(\beta | \boldsymbol{y}, \boldsymbol{x}, \boldsymbol{z}, c) &\propto \exp \left\{ -\frac{1}{2} \left( a \beta^2 - 2 b \beta \right) \right\} \\ &= \exp \left\{ -\frac{1}{2} a \left( \beta^2 - \frac{2 b}{a} \beta + \frac{b^2}{a^2} \right) + \frac{1}{2} \frac{b^2}{a} \right\} \\ &\propto \exp \left\{ -\frac{1}{2} a \left( \beta - \frac{b}{a} \right)^2 \right\} \\ &= \exp \left\{ - \frac{1}{2} \left( \frac{\beta - b/a}{1/ \sqrt{a} } \right) \right\} \\ &= \mathrm{dnorm}(\beta, b/a, 1/ \sqrt{a} ) \\ \end{align*}

よって、

\begin{align*} \beta | \boldsymbol{y}, \boldsymbol{x}, \boldsymbol{z}, c \sim \mathrm{Normal} \left( \frac{ \sum_{i=1}^n x_i z_i }{ \sum_{i=1}^n x_i^2 + \frac{1}{\tau_{\beta}^2} }, \frac{1}{ \sum_{i=1}^n x_i^2 + \frac{1}{\tau_{\beta}^2} } \right) \end{align*}

From \[ p(y_i \mid z_i, c) = \mathbf{1}\{y_i=1, z_i>c\}\,\mathbf{1}\{y_i=0, z_i\le c\}, \] we obtain

\begin{align*} p(c | \boldsymbol{y}, \boldsymbol{x}, \boldsymbol{z}, \beta) &\propto p(\boldsymbol{y} | \boldsymbol{z}, c) \, p(c) \\ &= \left( \prod_{i=1}^n \mathbf{1}\{y_i=1, z_i>c\} \, \mathbf{1}\{y_i=0, z_i \le c\} \right) p(c) \\ &= \mathbf{1}\left\{ \max_{i:y_i=0} z_i \le c < \min_{i:y_i=1} z_i \right\} \mathrm{dnorm}(c, 0, \tau_c) \\ &= \mathrm{dtruncatednorm}\left(c, 0, \tau_c, \max_{i:y_i=0} z_i,\ \min_{i:y_i=1} z_i \right) \\ \end{align*}

Similarly,

\begin{align*} p(z_i | \boldsymbol{y}, \boldsymbol{x}, \boldsymbol{z}_{-i}, \beta, c) &\propto p(y_i | z_i, c) \, p(z_i | x_i, \beta) \\ &= \mathbf{1}\{y_i=1, z_i>c\} \, \mathbf{1}\{y_i=0, z_i \le c\} \, \mathrm{dnorm}(z_i, \beta x_i, 1) \\ &= \begin{cases} \mathrm{dtruncatednorm}(z_i, \beta x_i, 1, c, \infty) & \text{if } y_i = 1 \\ \mathrm{dtruncatednorm}(z_i, \beta x_i, 1, -\infty, c) & \text{if } y_i = 0 \end{cases} \\ \end{align*}