# Assumptions of the random-effects ANOVA

Wilcox (1994) writes on p.289 that

The usual random effects model is that for $$J$$ randomly sampled groups,

$$X_{ij} = \mu + a_j + \epsilon_{ij},$$

where $$X_{ij}$$ is the $$i$$-th observation randomly sampled from the $$j$$-th group ($$i$$ = $$1$$, ... , $$n_j$$; $$j = 1$$, ..., $$J$$), $$\mu$$ is the grand mean, $$a_j = \mu_j - \mu$$, $$\mu_j$$ is the mean associated with the $$j$$-th group, $$a_j$$ is normally distributed with mean 0 and variance $$\sigma^2_a$$, and $$\epsilon_{ij}$$ is normally distributed with a mean of $$0$$ and variance $$\sigma^2$$ with $$a_j$$ and $$\epsilon_{ij}$$ independent.

My questions are:

• Why does the random-effects ANOVA assume that $$a_j$$ and $$\epsilon_{ij}$$ will be independent? How would we test this assumption? Does this assumption have a "counterpart" in fixed-effects ANOVA?
• Is it the case, that as per this question with no answer, independent observations are not an assumption of the random-effects ANOVA? Why is that?
• Are there any other assumptions made by the random-effects ANOVA but not mentioned by Wilcox in the text above? I'm aware that other conditions may be helpful, e.g. that inferences might be suspect if we don't have equal variances. But I wonder if there are any other strict assumptions that underpin the model.

Wilcox, R. R. (1994). A one-way random effects model for trimmed means. Psychometrika, 59(3), 289-306.

• Correct me if I am wrong, but isn't $\epsilon_{ij}$ your within-group variance? The ANOVA assumes that there are no other factors - outside the ones you have modeled explicitly - that systematically affect the groups, hence $\epsilon_{ij}$ must be independent of the other variables, including (but not limited to) $a_j$. This is - likely - wrong in the real world, but probably good enough for practical purposes, just like assuming all effects are linear. Jun 4, 2021 at 6:19
• That makes sense to me, and unless I'm mistaken that would also be true of fixed-effects ANOVA. So that seems to take care of most of the first dot point. Jun 4, 2021 at 10:56
• I'm not sure that fixed-effects ANOVA has such a counterpart because in a random-effects model, the $a_{j}$ are (as noted above) normally distributed random variables as are the $\epsilon_{ij}$. However, in a fixed-effects model, parameterized this way, we would not have $a_{j}$ but, say, $\tau_{j}$ which would be a (set of) constant(s) we would want to estimate. The difference is covered, albeit densely in Milliken & Johnson's Analysis of Messy Data Volume 1 Jun 4, 2021 at 21:03

Why does the random-effects ANOVA assume that $$a_j$$ and $$\epsilon_{ij}$$ will be independent? How would we test this assumption? Does this assumption have a "counterpart" in fixed-effects ANOVA?
This assumption is there to handle omitted variable bias, that is, to exclude the possibility of other variable(s) not in the model being associated with the outcome, so from that point of view there is a parallel with regular ANOVA (and any other regression model for that matter). As far as testing it goes, I suppose we could just plot them and hope to see no pattern, as we would for a plot of residuals vs fitted values. We could maybe go a bit further and collapse the $$\epsilon_{ij}$$ to, say $$\gamma_{j}$$ in order that it can have the same dimension as $$a_j$$ and simply compute their correlation, although that would not rule out some nonlinear dependence.
That is because with repeated measures we expect observations within one group to be more similar to observations in the same group, than to observations in other groups and that is what $$a_j$$ is there to capture. The more similar they are within each group, that is, the more they are correlated, the larger $$\sigma^2_a$$ will be.