why do average fitted values have less uncertainty than the intercept in a Bayesian random effects model? Consider a linear, random intercept model:
$$
y_{ij} \sim \mathcal{N}(\mu_i, \sigma_e^2)
$$
where
$$
\mu_i \sim \mathcal{N}(\gamma_0, \sigma_{mu}^2)
$$
Estimated in a Bayesian framework using MCMC sampling, I would expect that extracting fitted values for each participant in sample for each MCMC draw of the posterior would give me:
$$
\mu_{i,draw}
$$
If I were to average across the $i$ participants, I would get samples from a posterior distribution like:
$$
\bar{\mu}_{draw}
$$
which I would expect to have the same value and same uncertainty as $\gamma_0$.
In most software, the model is separated as:
$$
\mu_i = \gamma_0 + u_i
$$
where
$$
\gamma_0 \sim \mathcal{N}(0, \sigma_{\mu}^2)
$$
When I actually try this, my observation is that the posterior draws for $\gamma_0$ and the average of the fitted values $\mu_i$ have very similar average values, but $\gamma_0$ consistently has a wider uncertainty interval than the uncertainty interval for the average of $\mu_i$, the fitted values for each participant.
Now why I care about this. In the linear model case, I could just use the parameter estimates for $\gamma_0$. However, when fitting say a logistic mixed effects model, the estimate for $\gamma_0$ is on the log odds scale and back transformed to a probability is not the same as getting the averaged probability taking into account the random intercept.
My best guess is that in taking the fitted values:
$$
\hat{\mu_i} = \gamma_0 + u_i
$$
and then averaging across all participants and then summarizing that posterior distribution is somehow missing a source of variance, but I've been over this a dozen times and tried to look for articles or books discussing it and have come up empty, so am hoping for some insight here.

In that bayesian model, $\gamma_0$ posterior depends on $\vec\mu$, while $\vec\mu$ posterior depends on data (you may think of it as in a DAG). $\gamma_0$ is conditionally independent on data (conditioned on $\vec\mu$).
In particular, drafted $\gamma_0$ will vary around $\bar\mu$, and this variation is independent on data, so it just sums up to $\bar\mu$'s variance.
You may think it this way: information streams up from data to $\vec\mu$ (low variance), and then to $\gamma_0$ (higher variance), because you set a low informative prior. If you set a prior more informative than data, $\gamma_0$ would have been less variable than $\bar\mu$.
