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Suppose I have unstandardized data and estimate a 2-level Bayesian hierarchical model with varying intercepts and no covariates. The grand mean $\alpha_i$ and within-$i$ variance of the outcome $\sigma_y$ are estimated since I assign them priors.

$$y_{it} \sim N(\alpha_0 + \alpha_{i}, \sigma_y) \\ \alpha_{i} \sim N(0, \sigma_{\alpha})\\ \alpha_0 \sim N(0,1)\\ \sigma_{\alpha} \sim Exp(0.1)\\ \sigma_y \sim Exp(0.1)\\ $$

Suppose I standardized the outcome vector $y_{it}$ to mean 0 and variance of 1. In this case, $\alpha_0$ and $\sigma_y$ are known. Given this information, my inclination is to fix these parameters. Accordingly, I could write the model as:

$$y_{it} \sim N(\alpha_{i},1) \\ \alpha_{i} \sim N(0, \sigma_{\alpha})\\ \sigma_{\alpha} \sim Exp(0.1)\\ $$

While removing $\alpha_0$ seems warranted, can I do this for $\sigma_y$?

The decomposition of the variance in $y$ into a within ($\sigma_y$)and between ($\sigma_{\alpha}$) seems to be lost in the process.

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  • $\begingroup$ You call $\alpha_i$ the grand mean, not $\alpha_0$? $\endgroup$
    – frank
    Oct 6, 2022 at 9:50
  • $\begingroup$ Whoops meant to just have the intercepts centered at 0. Updated. $\endgroup$ Oct 6, 2022 at 10:03

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If you standardize your data, you would expect the total mean to be somewhat near zero, and the total variance to be somewhat near one. It doesn't mean that the within-group variance should be one.

You could follow your inclination and use the model you have described. It mainly depends on what you are interested in. However, presuming $y_{it} = \alpha_i + \epsilon_{it}$ and independency, the variances would add up, thus $1 = \operatorname{Var}(y) = \operatorname{Var}(\alpha) + \operatorname{Var}(\epsilon)$. So I would suggest: $$ \begin{align} y_{it}|\alpha_i, \sigma_y &\sim N(\alpha_i, \sigma_y)\\ \alpha_i|\sigma_\alpha &\sim N(0, \sigma_\alpha)\\ \sigma_y &\sim Exp(0.1)\\ \sigma_\alpha &\sim Exp(0.1). \end{align} $$ I don't see a reason to somehow force the optimization to create variances that add up to one. That would add complexity and in addition, the fact that the data has sample variance one doesn't mean that the real marginal variance of $Y$ has to be zero, too.

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  • $\begingroup$ The motivation for "fixing" some parameters like the marginal mean of $Y$ to 0 (i.e. marginal variance set to 0) is that it makes identifying some latent variable models possible. For example, if $y$ is actually a latent outcome. Of course, I didn't mention that... $\endgroup$ Oct 6, 2022 at 11:12

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