Estimability of Contrasts in Two-Way ANOVA

While I personally think it’s time to retire estimability from statistics, the following problem interested me:

Let $Y_{ij} = \mu + \tau_i + \theta_j + \epsilon_{ij}$, where $\epsilon_{ij} \sim \mathcal{N}(0,\sigma^2)$, $i=1,\ldots,a$, $j=1,\ldots,b$. Let $\boldsymbol{\beta} = (\mu,\tau_1,\ldots,\tau_a,\theta_1,\ldots,\theta_b)^\prime$. For any $\ell=1,\ldots,a-1$, define $$\boldsymbol{c}_{\ell}^\prime \boldsymbol{\beta} = \left(\sum_{i=1}^\ell \tau_i - \ell \tau_{\ell+1} \right)/\sqrt{\ell(\ell+1)}.$$ Verify that $\boldsymbol{c}_{\ell}^\prime \boldsymbol{\beta}$ is estimable.

To begin with, note that any linear combination of $\mu + \tau_i +\theta_j$ is estimable since $\boldsymbol{X}\boldsymbol{\beta}$ is always estimable. So rewriting the contrast as $$\boldsymbol{c}_\ell^\prime\boldsymbol{\beta} = \left(\sum_{i=1}^\ell(\mu+\tau_i + \theta_j)-\ell(\mu+\tau_i+\theta_j) \right)/\sqrt{\ell(\ell+1)}$$ clearly reveals that it is of the form $\boldsymbol{t}^\prime \boldsymbol{X}\boldsymbol{\beta}$.

Another interesting, and more general, way of approaching this problem is to write the model in a matrix-vector form. That is, $$ \boldsymbol{X}\boldsymbol{\beta} = \begin{pmatrix}\boldsymbol{1}_b & \boldsymbol{1}_b & \cdots & \cdots & \cdots & \boldsymbol{I}_b\\ \boldsymbol{1}_b & \cdots & \boldsymbol{1}_b & \cdots & & \boldsymbol{I}_b \\ \vdots & & & \ddots & & \vdots \\ \boldsymbol{1}_b & \cdots & \cdots & \cdots & \boldsymbol{1}_b & \boldsymbol{I}_b \end{pmatrix}\begin{pmatrix}\mu \\ \tau_1 \\ \vdots \\ \tau_a \\ \theta_1 \\ \vdots \\ \theta_b \end{pmatrix}.$$ Note that $\boldsymbol{X}$ has rank $a + b - 1$ and therefore $\mathrm{dim}(\mathcal{N}(\boldsymbol{X}))=2$. Due to the estimability condition which states $\boldsymbol{c}$ must be in $\mathcal{R}(\boldsymbol{X})$, it implies that $\boldsymbol{c} \in \mathcal{N}^\perp(\boldsymbol{X})$. Thus, we can find a basis $(\boldsymbol{b}_1,\boldsymbol{b}_2)$ for $\mathcal{N}(\boldsymbol{X})$ and show that $\boldsymbol{c}^\prime \boldsymbol{b}_j=0$ for $j=1,2$.

An obvious basis for $\mathcal{N}(\boldsymbol{X})$ is $$\left\{\begin{pmatrix}1 \\ -\boldsymbol{1}_a \\ \boldsymbol{0}_b \end{pmatrix}, \begin{pmatrix}1 \\ \boldsymbol{0}_a \\ -\boldsymbol{1}_b \end{pmatrix} \right\}.$$ Thus, $\boldsymbol{c}^\prime \boldsymbol{\beta}$ is estimable if and only if $c_0 - \sum_{i=1}^a c_i = 0$ and $c_0 - \sum_{j=1}^b c_{a+j}=0$. This includes

  • grand mean - $\mu + \bar{\tau}_\cdot + \bar{\theta}_\cdot$
  • cell means - $\mu + \tau_i + \theta_j$
  • first-factor contrasts - $\sum_{i=1}^a c_i \tau_i$ where $\sum_{i=1}^a c_i = 0$
  • second-factor contrasts - $\sum_{j=1}^b d_j \theta_j$ where $\sum_{j=1}^b d_j = 0$.

The contrast we have as $\boldsymbol{c}_\ell^\prime\boldsymbol{\beta}$ falls into the third category, and thus is estimable.

Daeyoung Lim
Daeyoung Lim
Statistics PhD Candidate

My research interests include Bayesian statistics, biostatistics, and computational statistics. I’m an English grammar fiend and a staunch proponent of plain language.