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I'm having a bit of trouble with the following problem in statistics (particularly ANOVA): Consider a two-way layout in which the effects of the factors are additive; suppose also that there are $K_{ij}$ observations $Y_{ijk}$ per cell, where $K_{ij} > 0$ for $i = 1, \dots, I$ and $j=1, \dots, J$. Let $v_i = K_{i+} = \sum_{j=1}^J K_{ij}$ for $i = 1, \dots, I$, and let $w_j = K_{+j} = \sum_{i=1}^I K_{ij}$. Assume that $E(Y_{ijk}) = \mu + \alpha_i + \beta_j$ for $k=1, \dots, K_{ij}$, $i = 1, \dots, I$, and $j = 1, \dots, J$, where $\sum_{i=1}^I v_i \alpha_i = \sum_{j=1}^J w_j \beta_j = 0$. (It can be shown that such $\mu, \alpha_i, \beta_j$ exist and are unique given the assumption of additivity). We want to verify that the least-squares estimators of $\mu, \alpha_i, \beta_j$ are as follows: $$\hat{\mu} = \overline{Y}_{+++} \\ \hat{\alpha}_i = \frac{1}{K_{i+}} Y_{i++} - \overline{Y}_{+++} \\ \hat{\beta}_j = \frac{1}{K_{+j}} Y_{+j+} - \overline{Y}_{+++} $$ where $\overline{Y}_{+++} = \frac{1}{\sum_{i=1}^I \sum_{j=1}^J K_{ij}} \sum_{i=1}^I \sum_{j=1}^J \sum_{k=1}^{K_{ij}} Y_{ijk}$, $Y_{i++} = \sum_{j=1}^J \sum_{k=1}^{K_{ij}} Y_{ijk}$ and likewise for $Y_{+j+}$.

It is easy enough to verify that $\hat{\mu} = \overline{Y}_{+++}$ is correct, but trying to solve $\frac{\partial Q}{\partial \alpha_i} = 0 $ where $$Q = \sum_{i=1}^I \sum_{j=1}^J \sum_{k=1}^{K_{ij}} (Y_{ijk} - \mu - \alpha_i - \beta_j)^2$$ is the quantity we would like to minimize, has some $\beta_j$ terms that I don't know how to get rid of.

Thanks for the help! Please let me know if I can clarify anything.

Chase Uyeda
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$\hat{\alpha_i}$ and $\hat{\beta_j}$ should be solved jointly through the following equations. $$ \left\{ \begin{aligned} &\frac{\partial Q}{\partial \alpha_i}=0 \\ &\frac{\partial Q}{\partial \beta_j}=0 \\ \end{aligned}\right. $$

MoonKnight
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  • Thanks for the response! Your suggestion makes sense; mind walking me through a few of the first steps? If I try to make this into a matrix equation, it doesn't look particularly easy to solve; and trying to solve for $\alpha_i$ and substituting into $\frac{\partial Q}{\partial \beta_j}$ doesn't look too nice either. Am I missing something? – Chase Uyeda Dec 25 '13 at 20:52
  • One observation you can make is that since $\sum_{i=1}^I v_i \alpha_i = 0$, you can solve $\alpha_I = \frac{-1}{v_I} \sum_{i=1}^{I-1} v_i \alpha_i \Rightarrow \frac{\partial \alpha_I}{\partial \alpha_i} = -\frac{v_i}{v_I}$, hence by the chain rule you can deduce $\frac{1}{v_i} \sum_{j=1}^J \sum_{k=1}^{K_{ij}} (Y_{ijk} - \mu - \alpha_i - \beta_j)$ has the same value for all $i = 1, \dots, I$. Would this help us solve? Unfortunately I don't see how to use that to actually minimize $Q$... – Chase Uyeda Dec 25 '13 at 21:00