Corrigenda to “Proportional subclass numbers in two-factor ANOVA”

This article refers to:

Proportional subclass numbers in two-factor ANOVA

This note is to provide proofs of sufficiency for Propositions A.3 and A.5 in LaMotte (2018).

1. Formulation of psn

The incidence matrix for a setting with cell sample sizes $n_{ij}$, $i=1,\dots ,a,j=1,\dots ,b$ is a matrix $\mathbb{K}$ with $n_{\cdot \cdot }=\sum _{i,j}{n}_{ij}$ rows and ab columns, where the i,j,s-th row has a 1 in the i,j column and zeroes in all the other columns. This means that the i,j-th column has $n_{ij}$ ones, and every row has exactly one 1.

Models for effects are built with matrices defined as follows. For each positive integer m, define $U_m=\left(1/m\right){\mathbf{1}}_m{\mathbf{1}}_m{}^{\mathrm{\prime }}$ and $S_m={\mathrm{I}}_m-U_m$. For dummy variables, define $E_{00}={\mathbf{1}}_a\otimes {\mathbf{1}}_b$, $E_{10}={\mathrm{I}}_a\otimes 1_b$, $E_{01}={\mathbf{1}}_a\otimes {\mathrm{I}}_b$, and $E_{11}={\mathrm{I}}_a\otimes {\mathrm{I}}_b$. For definitions of ANOVA effects, define $H_{00}=U_a\otimes U_b$, $H_{10}=S_a\otimes U_b$, $H_{01}=U_a\otimes S_b$, and $H_{11}=S_a\otimes S_b$.

The property ‘proportional subclass numbers’ (psn) is defined here to mean that there exist positive integers $r_i$ and $c_j$ such that $n_{ij}=r_i{c}_j$, $i=1,\dots ,a,j=1,\dots ,b$. Let $r_{\cdot }=\sum _i{r}_i$ and $c_{\cdot }=\sum _j{c}_j$. Then $n_{i\cdot }=\sum _j{n}_{ij}=c_{\cdot }{r}_i$, $n_{\cdot j}=r_{\cdot }{c}_j$, and $n_{\cdot \cdot }=r_{\cdot }{c}_{\cdot }$. Matrices defined in LaMotte (2018) are $D_a=\mathrm{Diag}\left(n_{i\cdot }\right)=c_{\cdot }\mathrm{Diag}\left(r_i\right)=c_{\cdot }{D}_r$ and $D_b=\mathrm{Diag}\left(n_{\cdot j}\right)=r_{\cdot }\mathrm{Diag}\left(c_j\right)=r_{\cdot }{D}_c$

Let ${\mathbb{K}}_a=\mathrm{Diag}\left({\mathbf{1}}_{r_i}\right)$ and ${\mathbb{K}}_b=\mathrm{Diag}\left({\mathbf{1}}_{c_j}\right)$. Then, if $n_{ij}=r_i{c}_j$ for all i,j, $\mathbb{K}=\mathrm{Diag}\left({\mathbf{1}}_{n_{ij}}\right)$ can be formed by permuting the rows of ${\mathbb{K}}_a\otimes {\mathbb{K}}_b$, so that there exists an $n_{\cdot \cdot }$ by $n_{\cdot \cdot }$ orthogonal matrix R such that $\mathbb{K}=R({\mathbb{K}}_a\otimes {\mathbb{K}}_b)$, or ${\mathbb{K}}_a\otimes {\mathbb{K}}_b=R{}^{\mathrm{\prime }}\mathbb{K}$.

Recall that $Q_A={\mathbf{P}}_{\mathbb{K}{E}_{10}}-{\mathbf{P}}_{{\mathbf{1}}_{n_{\cdot \cdot }}}={\mathbf{P}}_{\mathbb{K}{E}_{10}|\mathbb{K}{E}_{00}}.$ and, from (A2) in LaMotte (2018), (1) $\begin{array}{rl}\mathbb{K}{E}_{10}|\mathbb{K}{E}_{00}& =\mathbb{K}\left[\right({\mathrm{I}}_a-\left(1/n_{\cdot \cdot }\right){\mathbf{1}}_a{\mathbf{1}}_a{}^{\mathrm{\prime }}{D}_a)\otimes {\mathbf{1}}_b]\\ & =R\left\{\right({\mathbb{K}}_a\otimes {\mathbb{K}}_b\left)\right[({\mathrm{I}}_a-(1/n_{\cdot \cdot }\left){\mathbf{1}}_a{\mathbf{1}}_a{}^{\mathrm{\prime }}{D}_a\right)\otimes {\mathbf{1}}_b\left]\right\}\\ & =R\left\{\right[{\mathbb{K}}_a({\mathrm{I}}_a-(1/n_{\cdot \cdot }\left){\mathbf{1}}_a{\mathbf{1}}_a{}^{\mathrm{\prime }}{D}_a\right)]\otimes ({\mathbb{K}}_b{\mathbf{1}}_b\left)\right\}.\end{array}$(1)

2. Proposition A.3

The proof of necessity is correct: that is, $Q_A={\mathbf{P}}_{\mathbb{K}}-{\mathbf{P}}_{\mathbb{K}(\mathrm{I}-H_{10})}$ $⟹$ $n_{ij}=n_{i\cdot }/b$, $i=1,\dots ,a,j=1,\dots ,b$. While it is true that $Q_A={\mathbf{P}}_{\mathbb{K}}-{\mathbf{P}}_{\mathbb{K}(\mathrm{I}-H_{10})}$ $⟹$ $Q_A{\mathbf{P}}_{\mathbb{K}(\mathrm{I}-H_{10})}=0$, the implication in the other direction is not true. Following is a proof of sufficiency.

Proposition A.3, sufficiency. If $n_{ij}=n_{i\cdot }/b=r_i$, $i=1,\dots ,a,j=1,\dots ,b$, then $Q_A={\mathbf{P}}_{\mathbb{K}}-{\mathbf{P}}_{\mathbb{K}(\mathrm{I}-H_{10})}$.

Proof. That $n_{ij}=r_i$, $i=1,\dots ,a,j=1,\dots ,b$, means that $c_j=1$, $j=1,\dots ,b$, and hence that ${\mathbb{K}}_a=\mathrm{Diag}\left({\mathbf{1}}_{r_i}\right)$ and ${\mathbb{K}}_b={\mathrm{I}}_b$.

Note that $\mathrm{I}-H_{10}=H_{00}+H_{01}+H_{11}=H_{00}+{\mathrm{I}}_a\otimes S_b$. Then $\begin{array}{rl}R{}^{\mathrm{\prime }}\mathbb{K}(H_{00}+{\mathrm{I}}_a\otimes S_b)& =({\mathbb{K}}_a\otimes {\mathbb{K}}_b)(U_a\otimes U_b+{\mathrm{I}}_a\otimes S_b)\\ & =\left({\mathbb{K}}_a{U}_a\right)\otimes U_b+{\mathbb{K}}_a\otimes S_b.\end{array}$ Then $({\mathbb{K}}_a{U}_a\otimes U_b){}^{\mathrm{\prime }}({\mathbb{K}}_a\otimes S_b)=\left(U_a{}^{\mathrm{\prime }}{\mathbb{K}}_a{}^{\mathrm{\prime }}\right){\mathbb{K}}_a\otimes \left(U_b{S}_b\right)=0$, which implies that ${\mathbf{P}}_{({\mathbb{K}}_a\otimes {\mathbb{K}}_b)(\mathrm{I}-H_{10})}={\mathbf{P}}_{{\mathbb{K}}_a{U}_a\otimes U_b}+{\mathbf{P}}_{{\mathbb{K}}_a\otimes S_b}$ and hence $\begin{array}{rl}& R{}^{\mathrm{\prime }}[{\mathbf{P}}_{\mathbb{K}}-{\mathbf{P}}_{\mathbb{K}(\mathrm{I}-H_{10})}]R\\ & ={\mathbf{P}}_{{\mathbb{K}}_a\otimes {\mathbb{K}}_b}-{\mathbf{P}}_{({\mathbb{K}}_a\otimes {\mathbb{K}}_b)(\mathrm{I}-H_{10})}\\ & ={\mathbf{P}}_{{\mathbb{K}}_a}\otimes {\mathrm{I}}_b-{\mathbf{P}}_{{\mathbb{K}}_a{U}_a\otimes U_b}-{\mathbf{P}}_{{\mathbb{K}}_a}\otimes S_b\\ & ={\mathbf{P}}_{{\mathbb{K}}_a}\otimes U_b-{\mathbf{P}}_{\mathbf{1}}\\ & ={\mathbf{P}}_{({\mathbb{K}}_a\otimes {\mathbb{K}}_b)E_{10}}-{\mathbf{P}}_{\mathbf{1}}\\ & =R{}^{\mathrm{\prime }}({\mathbf{P}}_{\mathbb{K}{E}_{10}}-{\mathbf{P}}_{\mathbf{1}})R,\end{array}$ which, since R is orthogonal, implies that $Q_A={\mathbf{P}}_{\mathbb{K}}-{\mathbf{P}}_{\mathbb{K}(\mathrm{I}-H_{10})}$.

3. Proposition A.5

The additive model is $\mathrm{sp}\left(\mathbb{K}\right(E_{10},E_{01}\left)\right)$. Within it, the model that includes no A effects is $\mathrm{sp}\left(\mathbb{K}{E}_{01}\right)$. Then the extra SSE due to omitting possible A effects is $\mathit{y}{}^{\mathrm{\prime }}({\mathbf{P}}_{\mathbb{K}(E_{10},E_{01})}-{\mathbf{P}}_{\mathbb{K}{E}_{01}})\mathit{y}$. This is otherwise known as Type II SS for A effects. The argument given for Proposition A.5 proves necessity, that if $Q_A=Q_{IIA}:={\mathbf{P}}_{\mathbb{K}(E_{10},E_{01})}-{\mathbf{P}}_{\mathbb{K}{E}_{01}}$ then $n_{ij}=n_{i\cdot }{n}_{\cdot j}/n_{\cdot \cdot }$ for all i,j. The purpose of this section is to provide a proof of sufficiency.

Proposition A.5, sufficiency. If $n_{ij}=n_{i\cdot }{n}_{\cdot j}/n_{\cdot \cdot }$ for all i,j, then ${\mathbf{P}}_{\mathbb{K}{E}_{10}|\mathbb{K}{E}_{00}}={\mathbf{P}}_{\mathbb{K}(E_{10},E_{01})|\mathbb{K}{E}_{01}}.$

Proof. We shall show that $\mathbb{K}(E_{10},E_{01})|\mathbb{K}{E}_{01}=\left[\right(\mathbb{K}{E}_{10}|\mathbb{K}{E}_{00}),0],$ and hence the result, by showing that $\mathbb{K}{E}_{10}|\mathbb{K}{E}_{01}:=(\mathrm{I}-{\mathbf{P}}_{\mathbb{K}{E}_{01}})\mathbb{K}{E}_{10}=\mathbb{K}{E}_{10}|\mathbb{K}{E}_{00}.$ First note that $\begin{array}{rl}R{}^{\mathrm{\prime }}\mathbb{K}{E}_{01}& =({\mathbb{K}}_a\otimes {\mathbb{K}}_b)({\mathbf{1}}_a\otimes {\mathrm{I}}_b)\\ & =\left({\mathbb{K}}_a{\mathbf{1}}_a\right)\otimes {\mathbb{K}}_b={\mathbf{1}}_{r_{\cdot }}\otimes {\mathbb{K}}_b.\end{array}$ Then $\begin{array}{rl}{\mathbf{P}}_{R{}^{\mathrm{\prime }}\mathbb{K}{E}_{01}}& =R{}^{\mathrm{\prime }}{\mathbf{P}}_{\mathbb{K}{E}_{01}}R\\ & ={\mathbf{P}}_{{\mathbf{1}}_{r_{\cdot }}}\otimes {\mathbf{P}}_{{\mathbb{K}}_b}\\ & =U_{r_{\cdot }}\otimes {\mathbf{P}}_{{\mathbb{K}}_b},\end{array}$ and so $\begin{array}{rl}R{}^{\mathrm{\prime }}\mathbb{K}{E}_{10}|\mathbb{K}{E}_{01}& =(\mathrm{I}-U_{r_{\cdot }}\otimes {\mathbf{P}}_{{\mathbb{K}}_b})({\mathbb{K}}_a\otimes {\mathbb{K}}_b{\mathbf{1}}_b)\\ & ={\mathbb{K}}_a\otimes {\mathbb{K}}_b{\mathbf{1}}_b-U_{r_{\cdot }}{\mathbb{K}}_a\otimes {\mathbb{K}}_b{\mathbf{1}}_b\\ & ={\mathbb{K}}_a\otimes {\mathbf{1}}_{c_{\cdot }}-\left(1/r_{\cdot }\right){\mathbb{K}}_a{\mathbf{1}}_a{\mathbf{1}}_a{}^{\mathrm{\prime }}{\mathbb{K}}_a{}^{\mathrm{\prime }}{\mathbb{K}}_a\otimes {\mathbf{1}}_{c_{\cdot }}\\ & ={\mathbb{K}}_a({\mathrm{I}}_a-(1/n_{\cdot \cdot }\left){\mathbf{1}}_a{\mathbf{1}}_a{}^{\mathrm{\prime }}{D}_a\right)\otimes {\mathbb{K}}_b{\mathbf{1}}_b\\ & =({\mathbb{K}}_a\otimes {\mathbb{K}}_b)\left[\right({\mathrm{I}}_a-\left(1/n_{\cdot \cdot }\right){\mathbf{1}}_a{\mathbf{1}}_a{}^{\mathrm{\prime }}{D}_a)\otimes {\mathbf{1}}_b]\\ & =R{}^{\mathrm{\prime }}\mathbb{K}{E}_{10}|\mathbb{K}{E}_{00},\end{array}$ from (1(1) $\begin{array}{rl}\mathbb{K}{E}_{10}|\mathbb{K}{E}_{00}& =\mathbb{K}\left[\right({\mathrm{I}}_a-\left(1/n_{\cdot \cdot }\right){\mathbf{1}}_a{\mathbf{1}}_a{}^{\mathrm{\prime }}{D}_a)\otimes {\mathbf{1}}_b]\\ & =R\left\{\right({\mathbb{K}}_a\otimes {\mathbb{K}}_b\left)\right[({\mathrm{I}}_a-(1/n_{\cdot \cdot }\left){\mathbf{1}}_a{\mathbf{1}}_a{}^{\mathrm{\prime }}{D}_a\right)\otimes {\mathbf{1}}_b\left]\right\}\\ & =R\left\{\right[{\mathbb{K}}_a({\mathrm{I}}_a-(1/n_{\cdot \cdot }\left){\mathbf{1}}_a{\mathbf{1}}_a{}^{\mathrm{\prime }}{D}_a\right)]\otimes ({\mathbb{K}}_b{\mathbf{1}}_b\left)\right\}.\end{array}$(1) ). It follows that $Q_A=Q_{IIA}$.