Exercise 8.1 Solution Example - Hoff, A First Course in Bayesian Statistical Methods
標準ベイズ統計学 演習問題 8.1 解答例

\( \text{Var}[y_{i,j}|\mu, \tau^2] \) is bigger because it includes both between-group and within-group variability while \( \text{Var}[y_{i,j}|\theta_j, \sigma^2] \) only includes within-group variability.

I think \(\text{Cov}[y_{i_1, j}, y_{i_2,j}|\theta_j, \sigma^2]\) is zero because when \(\theta_j\) is known, \(y_{i_2,j}\) tells us nothing about \(y_{i_1,j}\) more than what we already know. However, \(\text{Cov}[y_{i_1, j}, y_{i_2,j}|\mu, \tau^2]\) will be positive because \(y_{i_2,j}\) tells us something about group-specific information \(\theta_j\) which we do not know.

\begin{align*} \text{Var}[y_{i,j}|\theta_j, \sigma^2] &= \sigma^2 \\ \text{Var}[\bar{y}_{\cdot, j}|\theta_j, \sigma^2] &= \text{Var}[\frac{1}{n_j} \sum_{i=1}^{n_j} y_{i,j} | \theta_j, \sigma^2] \\ &= \frac{1}{n_j^2} \text{Var}[\sum_{i=1}^{n_j} y_{i,j} | \theta_j, \sigma^2] \\ &= \frac{ \sigma^2 }{n_j} \\ \text{Cov}[y_{i_1, j}, y_{i_2,j}|\theta_j, \sigma^2] &= E[(y_{i_1, j} - \theta_j)(y_{i_2, j} - \theta_j) | \theta_j, \sigma^2] \\ &= E[y_{i_1, j}y_{i_2, j} - \theta_j y_{i_1, j} - \theta_j y_{i_2, j}) + \theta_j^2 | \theta_j, \sigma^2] \\ &= E[y_{i_1, j}y_{i_2, j} | \theta_j, \sigma^2] - \theta_j E[y_{i_1, j} | \theta_j, \sigma^2] - \theta_j E[y_{i_2, j} | \theta_j, \sigma^2] + \theta_j^2 \\ &= E[y_{i_1, j} | \theta_j, \sigma^2] E[y_{i_2, j} | \theta_j, \sigma^2] - \theta_j^2 - \theta_j^2 + \theta_j^2 \\ &\qquad (\because y_{i_1, j} \text{ and } y_{i_2, j} \text{ are conditionally independent given} \theta_j, \sigma^2) \\ &= \theta_j^2 - \theta_j^2 \\ &= 0 \\ \end{align*} \begin{align*} \text{Var}[y_{i,j}|\mu, \tau^2] &= E \left[\text{Var}[y_{i,j}|\theta_j, \sigma^2, \mu, \tau^2] | \mu, \tau^2\right] + \text{Var} \left[ E[y_{i,j}|\theta_j, \sigma^2, \mu, \tau^2] | \mu, \tau^2\right] \\ &= E[\text{Var}[y_{i,j}|\theta_j, \sigma^2] | \mu, \tau^2] + \text{Var}[E[y_{i,j}|\theta_j, \sigma^2] | \mu, \tau^2] \\ &= E[\sigma^2 | \mu, \tau^2] + \text{Var}[\theta_j | \mu, \tau^2] \\ &= \sigma^2 + \tau^2 \\ \text{Var}[\bar{y}_{\cdot, j}|\mu, \tau^2] &= E \left[ \text{Var}[\bar{y}_{\cdot, j}|\theta_j, \sigma^2, \mu, \tau^2] | \mu, \tau^2 \right] + \text{Var} \left[ E[\bar{y}_{\cdot, j}|\theta_j, \sigma^2, \mu, \tau^2] | \mu, \tau^2 \right] \\ &= E[\text{Var}[\bar{y}_{\cdot, j}|\theta_j, \sigma^2] | \mu, \tau^2] + \text{Var}[E[\bar{y}_{\cdot, j}|\theta_j, \sigma^2] | \mu, \tau^2] \\ &= E[ \frac{ \sigma^2 }{n_j} | \mu, \tau^2] + \text{Var}[\theta_j | \mu, \tau^2] \\ &= \frac{ \sigma^2 }{n_j} + \tau^2 \\ \text{Cov}[y_{i_1, j}, y_{i_2,j}|\mu, \tau^2] &= E \left[ \text{Cov}[y_{i_1, j}, y_{i_2,j}|\theta_j, \sigma^2, \mu, \tau^2] | \mu, \tau^2 \right] + \text{Cov} \left[ E[y_{i_1, j}|\theta_j, \sigma^2, \mu, \tau^2], E[y_{i_2, j}|\theta_j, \sigma^2, \mu, \tau^2] | \mu, \tau^2 \right] \\ &= E[\text{Cov}[y_{i_1, j}, y_{i_2,j}|\theta_j, \sigma^2] | \mu, \tau^2] + \text{Cov}[E[y_{i_1, j}|\theta_j, \sigma^2], E[y_{i_2, j}|\theta_j, \sigma^2] | \mu, \tau^2] \\ &= E[0 | \mu, \tau^2] + \text{Cov}[\theta_j, \theta_j | \mu, \tau^2] \\ &= 0 + \text{Var}[\theta_j | \mu, \tau^2] \\ &= \tau^2 \\ \end{align*}

\begin{align*} p(\mu | \theta_1, \ldots, \theta_m, \sigma^2, \tau^2, \boldsymbol{y}_1, \ldots, \boldsymbol{y}_m) &= \frac{ p(\mu, \theta_1, \ldots, \theta_m, \sigma^2, \tau^2, \boldsymbol{y}_1, \ldots, \boldsymbol{y}_m) }{ p(\theta_1, \ldots, \theta_m, \sigma^2, \tau^2, \boldsymbol{y}_1, \ldots, \boldsymbol{y}_m) } \\ &= \frac{ p(\mu, \theta_1, \ldots, \theta_m, \sigma^2, \tau^2, \boldsymbol{y}_1, \ldots, \boldsymbol{y}_m) }{ \int p( \mu, \theta_1, \ldots, \theta_m, \sigma^2, \tau^2, \boldsymbol{y}_1, \ldots, \boldsymbol{y}_m) d \mu } \\ &= \frac{ p(\boldsymbol{y}_1, \dots, \boldsymbol{y}_m | \theta_1, \ldots, \theta_m, \sigma^2) p(\theta_1, \ldots, \theta_m | \mu, \tau^2) p(\mu) p(\sigma^2) p(\tau^2) }{ \int p(\boldsymbol{y}_1, \dots, \boldsymbol{y}_m | \theta_1, \ldots, \theta_m, \sigma^2) p(\theta_1, \ldots, \theta_m | \mu, \tau^2) p(\mu) p(\sigma^2) p(\tau^2) d \mu } \\ &= \frac{ p(\boldsymbol{y}_1, \dots, \boldsymbol{y}_m | \theta_1, \ldots, \theta_m, \sigma^2) p(\theta_1, \ldots, \theta_m | \mu, \tau^2) p(\mu) p(\sigma^2) p(\tau^2) }{ p(\boldsymbol{y}_1, \dots, \boldsymbol{y}_m | \theta_1, \ldots, \theta_m, \sigma^2) p(\sigma^2) \int p(\theta_1, \ldots, \theta_m | \mu, \tau^2) p(\mu) p(\tau^2) d \mu } \\ &= \frac{ p(\theta_1, \ldots, \theta_m | \mu, \tau^2) p(\mu) p(\tau^2)}{ \int p(\theta_1, \ldots, \theta_m | \mu, \tau^2) p(\mu) p(\tau^2) d \mu } \\ &= \frac{ p(\mu, \theta_1, \ldots, \theta_m, \tau^2) }{ \int p(\mu, \theta_1, \ldots, \theta_m, \tau^2) d \mu } \\ &= \frac{ p(\mu, \theta_1, \ldots, \theta, \tau^2) }{ p(\theta_1, \ldots, \theta_m, \tau^2) } \\ &= p(\mu | \theta_1, \ldots, \theta_m, \tau^2) \\ \end{align*}

This means that the \(\sigma^2, \boldsymbol{y}_1, \dots, \boldsymbol{y}_2\) give us no additional information about \(\mu\) given \(\theta_1, \ldots, \theta_m, \tau^2\).