Exercise 5.3 Solution Example - Hoff, A First Course in Bayesian Statistical Methods
標準ベイズ統計学 演習問題 5.3 解答例

\begin{align*} p(\theta | y_1, \dots, y_n) &= \int_0^{\infty} p(\theta, \sigma^2 | y_1, \dots, y_n) d\sigma^2 \\ &= \int_0^{\infty} p(\theta | \sigma^2, y_1, \dots, y_n) p(\sigma^2 | y_1, \dots, y_n) d\sigma^2 \\ &= \int_0^{\infty} \sqrt{ \frac{\kappa_n}{2 \pi \sigma^2 } } \exp \left( - \frac{\kappa_n (\theta - \mu_n)^2}{2 \sigma^2} \right) \frac{ (\frac{\nu_n \sigma_n^2}{2})^{\frac{\nu_n}{2} } }{\Gamma(\frac{\nu_n}{2})} (\sigma^2)^{-(\frac{\nu_n}{2} + 1)} \exp \left( - \frac{\nu_n \sigma_n^2}{2 \sigma^2} \right) d\sigma^2 \\ &= \sqrt{ \frac{\kappa_n}{2 \pi} } \frac{ (\frac{\nu_n \sigma_n^2}{2})^{\frac{\nu_n}{2} } }{\Gamma(\frac{\nu_n}{2})} \int_0^{\infty} (\sigma^2)^{-(\frac{\nu_n + 1}{2} + 1)} \exp \left( - \frac{\kappa_n (\theta - \mu_n)^2}{2 \sigma^2} - \frac{\nu_n \sigma_n^2}{2 \sigma^2} \right) d\sigma^2 \\ &= \sqrt{ \frac{\kappa_n}{2 \pi} } \frac{ (\frac{\nu_n \sigma_n^2}{2})^{\frac{\nu_n}{2} } }{\Gamma(\frac{\nu_n}{2})} \int_0^{\infty} (\sigma^2)^{-(\frac{\nu_n + 1}{2} + 1)} \exp \left( - \frac{1}{2 \sigma^2} (\kappa_n (\theta - \mu_n)^2 + \nu_n \sigma_n^2) \right) d\sigma^2 \\ &= \sqrt{ \frac{\kappa_n}{2 \pi} } \frac{ (\frac{\nu_n \sigma_n^2}{2})^{\frac{\nu_n}{2} } }{\Gamma(\frac{\nu_n}{2})} \Gamma(\frac{\nu_n + 1}{2}) \times (\frac{\kappa_n (\theta - \mu_n)^2 + \nu_n \sigma_n^2}{2})^{-\frac{\nu_n + 1}{2}} \\ &= \frac{ \Gamma(\frac{\nu_n + 1}{2}) }{ \Gamma(\frac{\nu_n}{2}) } \sqrt{ \frac{\kappa_n}{2 \pi} } \left(\frac{2}{\nu_n \sigma_n^2}\right)^{\frac{1}{2} } \left(\frac{2}{\nu_n \sigma_n^2}\right)^{-\frac{\nu_n + 1}{2} } \times (\frac{\kappa_n (\theta - \mu_n)^2 + \nu_n \sigma_n^2}{2})^{-\frac{\nu_n + 1}{2}} \\ &= \frac{ \Gamma(\frac{\nu_n + 1}{2}) }{ \Gamma(\frac{\nu_n}{2}) } \sqrt{ \frac{\kappa_n}{\pi \nu_n \sigma_n^2} } \left( 1 + \frac{\kappa_n (\theta - \mu_n)^2}{\nu_n \sigma_n^2} \right)^{-\frac{\nu_n + 1}{2}} \\ &= \text{d-Student t}(\theta | \nu_n, \mu_n, \tau_n^2) \end{align*}

where

\begin{align*} \kappa_n &= \kappa_0 + n \\ \mu_n &= \frac{\kappa_0 \mu_0 + n \bar{y}}{\kappa_n} \\ \nu_n &= \nu_0 + n \\ \sigma_n^2 &= \frac{1}{\nu_n} \left( \nu_0 \sigma_0^2 + (n-1)s^2 + \frac{\kappa_0 n}{\kappa_n} (\bar{y} - \mu_0)^2 \right) \\ \tau_n^2 &= \frac{\nu_n \sigma_n^2}{\kappa_n} \\ \end{align*}

\(\sigma^2\)の marginal posterior distribution は、

\begin{align*} p(\sigma^2 | y_1, \dots, y_n) & \propto p(\sigma^2) p(y_1, \dots, y_n | \sigma^2) \\ &= p(\sigma^2) \int p(y_1, \dots, y_n | \theta, \sigma^2) p(\theta | \sigma^2) d\theta \\ \end{align*}

となる。

ここで、

\begin{equation} \label{eq:ex5-3.1} \begin{aligned}[b] \sum_{i=1}^n (y_i - \theta)^2 &= \sum_{i=1}^n (y_i - \bar{y} + \bar{y} - \theta)^2 \\ &= \sum_{i=1}^n (y_i - \bar{y})^2 + n (\bar{y} - \theta)^2 + 2 (\bar{y} - \theta) \sum_{i=1}^n (y_i - \bar{y}) \\ &= \sum_{i=1}^n (y_i - \bar{y})^2 + n (\bar{y} - \theta)^2 \\ &= n (\theta - \bar{y})^2 + (n-1)s^2 \\ \end{aligned} \end{equation} \begin{equation} \label{eq:ex5-3.2} \begin{aligned} A(x - a)^2 + B(x - b)^2 &= (A + B)(x - c)^2 + \frac{AB}{A + B}(a - b)^2 \\ \text{where } c &= \frac{aA + bB}{A + B} \\ \end{aligned} \end{equation}

を用いると、

\begin{align*} p(y_1, \dots, y_n | \theta, \sigma^2) p(\theta | \sigma^2) &= \prod_{i=1}^n p(y_i | \theta, \sigma^2) p(\theta | \sigma^2) \\ &= \prod_{i=1}^n \left( \frac{1}{\sqrt{2 \pi \sigma^2}} \exp \left( - \frac{(y_i - \theta)^2}{2 \sigma^2} \right) \right) \left( \frac{\sqrt{\kappa_0} }{\sqrt{2 \pi \sigma^2}} \exp \left( - \frac{ \kappa_0 (\theta - \mu_0)^2}{2 \sigma^2} \right) \right) \\ &= (2 \pi \sigma^2)^{-\frac{n + 1}{2} } \sqrt{\kappa_0} \exp \left( - \frac{1}{2 \sigma^2} \sum_{i=1}^n (y_i - \theta)^2 \right) \exp \left( - \frac{\kappa_0}{2 \sigma^2} (\theta - \mu_0)^2 \right) \\ &= (2 \pi \sigma^2)^{-\frac{n + 1}{2} } \sqrt{\kappa_0} \exp \left( - \frac{1}{2 \sigma^2} ( n (\theta - \bar{y})^2 + (n-1)s^2 ) \right) \\ & \qquad \times \exp \left( - \frac{\kappa_0}{2 \sigma^2} (\theta - \mu_0)^2 \right) \quad (\because \eqref{eq:ex5-3.1}) \\ &= (2 \pi \sigma^2)^{-\frac{n + 1}{2} } \sqrt{\kappa_0} \exp \left( - \frac{1}{2 \sigma^2} ( n(\theta - \bar{y})^2 + \kappa_0 (\theta - \mu_0)^2 ) \right) \exp \left( - \frac{1}{2 \sigma^2} (n-1)s^2 \right) \\ &= (2 \pi \sigma^2)^{-\frac{n + 1}{2} } \sqrt{\kappa_0} \exp \left( - \frac{1}{2 \sigma^2} ( (n + \kappa_0)(\theta - \theta^{\ast})^2 + r ) \right) \exp \left( - \frac{1}{2 \sigma^2} (n-1)s^2 \right) \\ & \qquad \text{ where } \theta^{\ast} = \frac{n \bar{y} + \kappa_0 \mu_0}{n + \kappa_0}, \quad r = \frac{n \kappa_0}{n + \kappa_0} (\bar{y} - \mu_0)^2 \qquad \qquad \qquad \qquad \qquad (\because \eqref{eq:ex5-3.2}) \\ &= (2 \pi \sigma^2)^{-\frac{n + 1}{2} } \sqrt{\kappa_0} \exp \left( - \frac{ n + \kappa_0}{2 \sigma^2} (\theta - \theta^{\ast})^2 \right) \exp \left( - \frac{r}{2 \sigma^2} \right) \exp \left( - \frac{1}{2 \sigma^2} (n-1)s^2 \right) \\ \end{align*}

となるので、

\begin{align*} p(\sigma^2 | y_1, \dots, y_n) & \propto p(\sigma^2) p(y_1, \dots, y_n | \sigma^2) \\ &= p(\sigma^2) \int p(y_1, \dots, y_n | \theta, \sigma^2) p(\theta | \sigma^2) d\theta \\ &= p(\sigma^2) \int (2 \pi \sigma^2)^{-\frac{n + 1}{2} } \sqrt{\kappa_0} \exp \left( - \frac{ n + \kappa_0}{2 \sigma^2} (\theta - \theta^{\ast})^2 \right) \exp \left( - \frac{r}{2 \sigma^2} \right) \exp \left( - \frac{1}{2 \sigma^2} (n-1)s^2 \right) d\theta \\ &= p(\sigma^2) (2 \pi \sigma^2)^{-\frac{n + 1}{2} } \sqrt{\kappa_0} \exp \left( - \frac{1}{2 \sigma^2} ( (n-1)s^2 + r) \right) \\ & \qquad \times \left(2 \pi \frac{\sigma^2}{n + \kappa_0} \right)^{\frac{1}{2} } \int \left(2 \pi \frac{\sigma^2}{n + \kappa_0} \right)^{-\frac{1}{2} } \exp \left( - \frac{ n + \kappa_0}{2 \sigma^2} (\theta - \theta^{\ast})^2 \right) d\theta \\ &= p(\sigma^2) (2 \pi \sigma^2)^{-\frac{n + 1}{2} } \sqrt{\kappa_0} \exp \left( - \frac{1}{2 \sigma^2} ( (n-1)s^2 + r) \right) \left(2 \pi \frac{\sigma^2}{n + \kappa_0} \right)^{\frac{1}{2} } \\ & \propto p(\sigma^2) (\sigma^2)^{- \frac{n}{2} } \exp \left( - \frac{1}{2 \sigma^2} ( (n-1)s^2 + r) \right) \\ & \propto (\sigma^2)^{- (\frac{ \nu_0 }{2} + 1 )} \exp \left( - \frac{\nu_0 \sigma_0^2}{2 \sigma^2} \right) (\sigma^2)^{- \frac{n}{2} } \exp \left( - \frac{1}{2 \sigma^2} ( (n-1)s^2 + r) \right) \\ &= (\sigma^2)^{- (\frac{ \nu_0 + n }{2} + 1 )} \exp \left( - \frac{\nu_0 \sigma_0^2 + (n-1)s^2 + r}{2 \sigma^2} \right) \\ &= (\sigma^2)^{- (\frac{ \nu_0 + n }{2} + 1 )} \exp \left( - \frac{\nu_0 + n}{2 \sigma^2} \left( \frac{\nu_0 \sigma_0^2 + (n-1)s^2 + \frac{\kappa_0 n}{\kappa_0 + n} (\bar{y} - \mu_0)^2 }{\nu_0 + n} \right) \right) \\ &\propto \text{dinverse-gamma}(\sigma^2, \frac{\nu_n}{2}, \frac{\nu_n \sigma_n^2}{2}) \\ & \text{ where } \nu_n = \nu_0 + n, \quad \sigma_n^2 = \frac{1}{\nu_n}\left[ \nu_0 \sigma_0^2 + (n-1)s^2 + \frac{\kappa_0 n}{\kappa_n} (\bar{y} - \mu_0)^2 \right] \end{align*}

よって、

\begin{equation*} \left| \frac{d \sigma^2}{d \tilde{\sigma}^2} \right| = (\tilde{\sigma}^2)^{-2} \end{equation*}

より

\begin{align*} p( \tilde{\sigma}^2 | y_1, \dots, y_n) &= p( (\tilde{\sigma}^2)^{-1} | y_1, \dots, y_n) \left| \frac{d \sigma^2}{d \tilde{\sigma}^2} \right| \\ &= \text{dinverse-gamma}(\tilde{\sigma}^{-2} | \frac{\nu_n}{2}, \frac{\nu_n \sigma_n^2}{2}) \times (\tilde{\sigma}^2)^{-2} \\ &\propto (\tilde{\sigma}^2)^{(\frac{ \nu_0 + n }{2} + 1 )} \exp \left( - \frac{\nu_n \sigma_n^2}{2 (\tilde{\sigma}^2)^{-1} } \right) \times (\tilde{\sigma}^2)^{-2} \\ &= (\tilde{\sigma}^2)^{(\frac{ \nu_0 + n }{2} - 1 )} \exp \left( - \tilde{\sigma}^2 \frac{\nu_n \sigma_n^2}{2} \right) \\ &= \text{dgamma}(\tilde{\sigma}^2 | \frac{\nu_n}{2}, \frac{\nu_n \sigma_n^2}{2}) \end{align*}