Suppose $$X_{ij} = \mu_j + \varepsilon_{ij}, \quad j = 1, \cdots, J, \quad i = 1, \cdots, N_j$$ ANOVA can allow us to test whether $\mu_1 = \cdots = \mu_J$.
In traditional ANOVA, however, the number of groups $J$ is fixed. But I want to know the proof technique in the case that $J$ goes to infinity.
Does anyone know some papers containing this question with detailed proof? Thanks so much!
The following article is what you are looking for:
Akritas, M., Arnold, S. (2000). Asymptotics for Analysis of Variance When the Number of Levels is Large, Journal of the American Statistical Association, 95:449, 212-226.
Set $J$ the number of groups, and $N_j$ the number of observations for the group $j$, $1\le j\le N$. The test statistic is the ratio $$ T_{n,J}:=\frac{A}{B}, $$ with $$ A:=\frac{\sum_{j=1}^J N_j (\bar X- \bar{X}_{\cdot j})^2}{J-1} $$ and $$ B:=\frac{\sum_{j=1}^J \sum_{i=1}^{N_j}(X_{ij}- \bar{X}_{\cdot j})}{N-J}, $$ where $\bar{X}_{\cdot j}$ is the mean of the $j$th group and $N:= \sum_{j=1}^JN_j$.
Under the classical assumptions of normality of the residuals and homoscedasticity, $T_{n,J}$ has a Fisher distribution with d.f. $J-1,N-J$ under the null hypothesis that all group means are equal.
In your case, if you consider a sequence of $(J_k)_{k\in \mathbb{N}}$ with $J_k \to \infty$ as $k\to \infty$, the classical result still holds true for each $J_k$.
Now, if you want to go further and study the properties of the test, you should first make hypotheses about the sample sizes per group. If you want to study effect sizes, for instance, you will again require hypotheses on the differences of the means and so on.
