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I am reading the paper by Willemsen et al (2015), "A multivariate Bayesian model for embryonic growth", Statistics in Medicine, 34:8, 1351–1365

where they define the posterior distribution as, \begin{multline} p(\beta,\sigma^{2},\gamma,\Sigma_{\gamma}|y) \propto \prod\limits_{ij} N(y_{ij}|\gamma_{i2} + z^{T}_{ij}\beta_{j},\sigma^{2}) \prod\limits_{i}N(\gamma_{i} \mid 0,\Sigma_{\gamma})\\ \times \prod\limits_{j}N(\beta_{j}|0,\sigma^{2}_{\beta}\mathbf{I}_{5}) \times \prod\limits_{i}N(\sigma^{2}_{i}|\alpha_{\sigma},\beta_{\sigma}) IW (\Sigma_{\gamma}|\delta,\psi) \end{multline}

where, $y_{ij} = \gamma_{i2} + z^{T}_{ij}\beta + \epsilon_{ij}$

where, $z_{ij} = B(\exp(\gamma_{i3})(t_{ij} + \gamma_{i1} )) , i= 1, \ldots N; j= 1, \ldots n $, $\gamma_{i} = (\gamma_{i1},\gamma_{i2}\gamma_{i3})$, $\gamma_{i} \sim N_{3}(0, \Sigma_{(3*3)})$, $\epsilon_{ij} \sim N(0, \sigma^{2}) $

where, $z^{T}_{ij}$ is a spline function. I am trying to figure out the full conditional distribution(Gibbs sampling) for $\beta$.

They said the $\beta \sim N(\bar \mu,\bar\Sigma_{\beta})$, where $\bar\mu = (\bar\Sigma_{\beta}\bar Z \tilde y)/\sigma^{2}$ and $\bar\Sigma_{\beta}=(\bar\Sigma^{-1}_{\beta}+ \bar Z^{T}\bar Z / \sigma^{2})^{-1}$

My question is how did they get that distribution of $\beta$?

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    $\begingroup$ What exactly is your question? You have the full conditional distribution for $\beta$, what more do you need to know? $\endgroup$ – jbowman Jul 12 '17 at 16:39
  • $\begingroup$ @jbowman: My question is how they got that distribution of $\beta$ $\endgroup$ – Benzamin Jul 12 '17 at 16:55
  • $\begingroup$ @Xi'an: i tried that. I took the full conditional distribution of $y_{ij}$|rest of the parameter , and the prior of $\beta$. but my answer doesn't match with their answer. $\endgroup$ – Benzamin Jul 12 '17 at 17:03
  • $\begingroup$ @Xi'an: can you please recheck your work ? $\endgroup$ – Benzamin Jul 12 '17 at 18:26
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    $\begingroup$ You have a notational error in your formula for $\bar{\Sigma}_{\beta}$. To see this, try raising both sides to the $-1$ power; you will see that you get an "$=$" where there is no equality. $\endgroup$ – jbowman Jul 12 '17 at 20:56
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\begin{align*} p(\beta|\sigma^{2},\gamma,\Sigma_{\gamma},y) &\propto \prod\limits_{ij} N(y_{ij}|\gamma_{i2} + z^{T}_{ij}\beta_{j},\sigma^{2}) \prod\limits_{i}N(\gamma \mid 0,\Sigma_{\gamma})\\ &\qquad\times \prod\limits_{i}N(\beta_{i}|0,\sigma^{2}_{\beta}I_{p}) \times \prod\limits_{i}N(\sigma^{2}_{i}|\alpha_{\sigma},\beta_{\sigma}) \times IW (\Sigma_{\gamma}|\delta,\psi)\\ &\propto\prod\limits_{ij} N(y_{ij}|\gamma_{i2} + z^{T}_{ij}\beta_{j},\sigma^{2}) \prod\limits_{i}N(\beta_{i}|0,\sigma^{2}_{\beta}I_{p})\\ &\propto\exp-\left\{||\mathbf{y}-\mathbf{\gamma}-\mathbf{Z}\mathbf{\beta}||^2 \right\}/2\sigma^2\times\exp-\left\{||\mathbf{\beta}||^2\right\}/2\sigma_\beta^2\\ &\propto\exp-\left\{\sigma^{-2}\mathbf{\beta}^\text{T}\mathbf{Z}^\text{T}\mathbf{Z}\mathbf{\beta}-2\sigma^{-2}(\mathbf{y}-\mathbf{\gamma})^\text{T}\mathbf{Z}\mathbf{\beta}+\sigma^{-2}_{\beta}\mathbf{\beta}^\text{T}\mathbf{\beta} \right\}/2\\ &\propto\exp-\big\{\mathbf{\beta}^\text{T}\underbrace{[\sigma^{-2}\mathbf{Z}^\text{T}\mathbf{Z}+\sigma^{-2}_{\beta}I_p]}_{\bar\Sigma_{\beta}^{-1}}\mathbf{\beta}-2\sigma^{-2}(\mathbf{y}-\mathbf{\gamma})^\text{T}\mathbf{Z}\mathbf{\beta} \big\}/2\\ &\propto\exp-\left\{\mathbf{\beta}^\text{T}\bar\Sigma_{\beta}^{-1}\mathbf{\beta}-2\sigma^{-2}\mathbf{\beta}^\text{T}\bar\Sigma_{\beta}^{-1}\bar\Sigma_{\beta}\mathbf{Z}^\text{T}(\mathbf{y}-\mathbf{\gamma}) \right\}/2\\ &\propto\exp-\left\{[\mathbf{\beta}-\sigma^{-2}\bar\Sigma_{\beta}\mathbf{Z}(\mathbf{y}-\mathbf{\gamma})]^\text{T}\bar\Sigma_{\beta}^{-1}\bar[\mathbf{\beta}-\sigma^{-2}\bar\Sigma_{\beta}\mathbf{Z}^\text{T}(\mathbf{y}-\mathbf{\gamma})] \right\}/2\\ \end{align*}

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