I am thinking of a null model (also known as a random-effects ANOVA, or intercept-only model)

$$ y_{ij} = \gamma_{00} + u_j + e_{ij} $$

where $u_j$ are the level-2 residuals and $e_{ij}$ are the level-1 residuals.

I see $\gamma_{00}$ often referred to as an "overall intercept" or "overall mean". I can see that if all the units J are the same size, then $\gamma_{00}$ is exactly equal to the grand mean of y.

However, if the units J are not all the same size, then $\gamma_{00}$ is not quite equal to the overall grand overall mean of y anymore. I noticed that if I take $\gamma_{00}$ and add the conditional modes of each group from it, and then take a weighted average of the resulting values, then it is exactly equal to the overall grand mean. But I don't really have any intuition for how that would relate to an interpretation.

So what is the right interpretation of $\gamma_{00}$? I'm looking for an explanation that amongst other things can explain why it's equal to the overall grand mean when sample sizes are even, but isn't when they are not.

Also, how does the interpretation change when a level-1 predictor is added, e.g.

$$ y_{ij} = \gamma_{00} + \gamma_{10}x_1{_{ij}} + u_{0j} + e_{ij} $$

Or when a level-2 predictor is added?


1 Answer 1


$\gamma_{00}$ is a weighted grand mean when you have unequal group sample sizes. The formula for calculating $\gamma_{00}$, from Rabe-Hesketh & Skrondal (2012):

$\hat\gamma_{00} = {\sum^J_{j=1} w_j \overline y_{.j}\over \sum^J_{j=1} w_j}$ where $w_j = {1\over \hat\psi + \hat\theta/ n_j}$, $n_j$ is a cluster sample size, $\hat\theta$ the within cluster (level 1) variance, and $\hat\psi$ is the between cluster (level 2) variance.*

When you add a predictor, whether level 1 or level 2, $\gamma_{00}$ is equivalent to the weighted grand mean when all other predictors are equal to 0. This is the exact same interpretation as in OLS.

*It is interesting to note that the empirical Bayes correction factor ($\hat R_j$) for a cluster has just a single substitution relative of the formula for $w_j$:

$\hat R_j ={\hat\psi\over \hat\psi + \hat\theta/ n_j}$

To get the empirical Bayes estimate for a cluster, you multiply the correction factor $\hat R_j$ by the maximum likelihood value of the group deviation from the grand mean $\hat u^{ML}_{0j}$.


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