# Formulating posterior predictive distribution from hierarchical model

I have been reading a couple related papers using Bayesian inference in hierarchical models1,2,3 but am struggling to bridge the gap in one aspect of the papers. I think the struggle is in relation to the posterior predictive distribution. The model is described as

$$log(y_{i,t}) \sim \mathcal{N}(\beta_{0,i} + \beta_{1,i}a_{i,t} + \eta_t, \sigma_y^2)$$

$$\eta_t \sim \mathcal{N}(\beta_2 x_t, \sigma_{\eta}^2)$$

$$x_t \sim \mathcal{N}(\mu_x, \sigma_x^2)$$

In this case, $$y_{i,t}$$, $$a_{i,t}$$, $$x_t$$ are measured but the goal will be to predict new values of $$x_t$$ (climate) for which we have measures of $$y_{i,t}$$ and $$a_{i,t}$$. They state that the posterior predictive distribution can be sampled from

$$x_t^{(j)} \sim \mathcal{N}\left(\frac{\sigma_{\eta}^{2(j)}\mu_x^{(j)} + \sigma_x^{2(j)}\beta_2^{(j)}\eta_t^{(j)}}{\sigma_{\eta}^{2(j)} + \sigma_x^{2(j)}\beta_2^{2(j)}}, \left[\frac{1}{\sigma_x^{2(j)}} + \frac{\beta_2^{2(j)}}{\sigma_{\eta}^{2(j)}} \right] \right)$$

where $$(j)$$ represents the $$j^{th}$$ MCMC sample. I know that the posterior predictive distribution is defined as

$$p(\tilde{x} \mid x) = \int_\theta p(\tilde{x} \mid \theta)p(\theta \mid x)d\theta$$

However, I am unable to get from the model description to the posterior using this equation. Could anyone walk me through the probability/integration steps necessary to come up with this specific posterior predictive distribution?

• Let me know if this would be better posted on math.stackexchange.com – djhocking May 6 '20 at 18:23

It appears that the stated distribution is for $$x_t^{(j)} | \eta_t^{(j)}$$ and the random variable $$y_{i,t}$$ is being ignored for now. It also appears that the authors are being a bit loose in their notation for the normal distribution, using the variance parameter in some statements and the precision parameter in others. (I will parameterise with the variance unless otherwise stated.) To obtain the conditional density we take the joint density kernel and "complete the square" to simplify. Taking proportionality with respect to the $$x$$ variable gives:
\begin{aligned} p(x| \eta) &\equiv p(x_t^{(j)} = x| \eta_t^{(j)} = \eta) \\[12pt] &\overset{x}{\propto} p(x_t^{(j)} = x, \eta_t^{(j)} = \eta) \\[12pt] &= p(\eta_t^{(j)} = \eta | x_t^{(j)} = x) \cdot p(x_t^{(j)} = x) \\[12pt] &= \text{N}(\eta | \beta_2 x, \sigma_\eta^2) \cdot \text{N}(x | \mu_x, \sigma_x^2) \\[6pt] &\overset{x}{\propto} \exp \bigg( - \frac{1}{2 \sigma_\eta^2} (\eta - \beta_2 x)^2 \bigg) \cdot \exp \bigg( - \frac{1}{2 \sigma_x^2} (x - \mu_x)^2 \bigg) \\[6pt] &= \exp \bigg( - \frac{1}{2} \bigg[ \frac{1}{\sigma_\eta^2} (\eta - \beta_2 x)^2 + \frac{1}{\sigma_x^2} (x - \mu_x)^2 \bigg] \bigg) \\[6pt] &= \exp \bigg( - \frac{1}{2} \bigg[ \frac{1}{\sigma_\eta^2} (\eta^2 - 2 \eta \beta_2 x + \beta_2^2 x^2) + \frac{1}{\sigma_x^2} (x^2 - 2 \mu_x x + \mu_x^2) \bigg] \bigg) \\[6pt] &= \exp \bigg( - \frac{1}{2} \bigg[ \Big( \frac{1}{\sigma_x^2} + \frac{\beta_2^2}{\sigma_\eta^2} \Big) x^2 -2 \Big( \frac{\mu_x}{\sigma_x^2} + \frac{\eta \beta_2}{\sigma_\eta^2} \Big) x + \Big( \frac{\eta^2}{\sigma_\eta^2} + \frac{\mu_x^2}{\sigma_x^2} \Big) \bigg] \bigg) \\[6pt] &= \exp \bigg( - \frac{1}{2} \Big( \frac{1}{\sigma_x^2} + \frac{\beta_2^2}{\sigma_\eta^2} \Big) \bigg[ x^2 -2 \Big( \frac{\eta \sigma_x^2 \beta_2^2 + \mu_x \sigma_\eta^2}{\sigma_x^2 \beta_2^2 + \sigma_\eta^2} \Big) x + \text{const} \bigg] \bigg) \\[6pt] &\overset{x}{\propto} \exp \bigg( - \frac{1}{2} \Big( \frac{1}{\sigma_x^2} + \frac{\beta_2^2}{\sigma_\eta^2} \Big) \bigg( x - \frac{\eta \sigma_x^2 \beta_2^2 + \mu_x \sigma_\eta^2}{\sigma_x^2 \beta_2^2 + \sigma_\eta^2} \bigg)^2 \bigg) \\[6pt] &\overset{x}{\propto} \text{N}\bigg( x \Bigg| \text{Mean} = \frac{\eta \sigma_x^2 \beta_2^2 + \mu_x \sigma_\eta^2}{\sigma_x^2 \beta_2^2 + \sigma_\eta^2}, \text{Precision} = \frac{1}{\sigma_x^2} + \frac{\beta_2^2}{\sigma_\eta^2} \bigg). \\[6pt] \end{aligned}
$$x_t^{(j)} | \eta_t^{(j)} \sim \text{N}\bigg(\text{Mean} = \frac{\eta_t^{(j)} \sigma_x^2 \beta_2^2 + \mu_x \sigma_\eta^2}{\sigma_x^2 \beta_2^2 + \sigma_\eta^2}, \text{Precision} = \frac{1}{\sigma_x^2} + \frac{\beta_2^2}{\sigma_\eta^2} \bigg).$$