Blocking, pairs, and Latin squares

Let $H=\{a\in {\mathbb{R}}^t:{\mathbf{1}}_t^{T}a=0\}$. The coordinate-sum map ${\mathbb{R}}^t\to \mathbb{R}$ is onto, so [F2] gives $\dim H=t-1$. By [F6], $\dim \mathrm{row}(C)=\dim \mathrm{im}C=\mathrm{rank}C=q$. For every $a\in {\mathbb{R}}^t$, the vector $h=a-({\mathbf{1}}_t^{T}a\mathrm{/}t){\mathbf{1}}_t$ lies in $H$ and satisfies $Ch=Ca$, because $C{\mathbf{1}}_t=0$. Thus $\mathrm{im}(C{\mathrm{\mid }}_H)=\mathrm{im}C$ has dimension $q$, and [F2] gives $\dim (H\cap \ker C)=t-1-q$. The row space of $C$ lies in $H$ and is orthogonal to $H\cap \ker C$. The orthogonal complement of $H\cap \ker C$ within $H$ has dimension $q$: both subspaces are finite-dimensional by [F4], and [F3]–[F5] subtract dimensions from $\dim H=t-1$. Since this complement contains the $q$-dimensional row space, the equality case of [F4] identifies it with that row space.

In $V$, the constant arrays form a one-dimensional space $G$, the arrays $(a_i{)}_{ij}$ for $a\in H$ form a $(t-1)$-dimensional treatment space $T$, and the arrays $(d_j{)}_{ij}$ for $d\in {\mathbb{R}}^b$ with ${\sum }_j{d}_j=0$ form a $(b-1)$-dimensional block space $B$. The replication maps from the two zero-sum coordinate spaces to $T$ and $B$ are linear bijections: their repeated rows or columns recover the original coordinates, so they transport bases and give the stated dimensions; the block zero-sum space has dimension $b-1$ by [F2] applied to its onto coordinate-sum map. These spaces are pairwise orthogonal: the treatment-block inner product is ${\sum }_{i,j}{a}_i{d}_j=({\sum }_i{a}_i)({\sum }_j{d}_j)=0$, and either zero-sum condition also makes the relevant inner product with $G$ zero. Their sum is direct: if three mutually orthogonal vectors sum to zero, taking the inner product with each one gives its squared norm zero. Thus $W_U=G\oplus T\oplus B$, and the finite-dimensional direct-sum formula [F5] gives $\dim W_U=1+(t-1)+(b-1)=t+b-1$. The reduced space replaces $T$ by the replicated arrays from $H\cap \ker C$, whose dimension is $t-1-q$ by step 1.1; the same formula gives $\dim W_R(C)=1+(t-1-q)+(b-1)=t+b-1-q$.

For any $y$, set $r_{ij}=y_{ij}-{\bar{y}}_{i\cdot }-{\bar{y}}_{\cdot j}+{\bar{y}}_{\cdot \cdot }$. Summing $r$ over any row gives zero because ${\sum }_j{y}_{ij}=b{\bar{y}}_{i\cdot }$ and ${\sum }_j{\bar{y}}_{\cdot j}=b{\bar{y}}_{\cdot \cdot }$; summing over any column likewise gives zero. Conversely, if $v$ is orthogonal to $T$, then its row-sum vector is orthogonal to every $a\in H$, so all row sums are equal; orthogonality to $G$ makes their total zero, so every row sum is zero. The same argument with $B$ gives zero column sums. Thus $W_U^{\perp }=E$. Also $(y-r{)}_{ij}={\bar{y}}_{i\cdot }+{\bar{y}}_{\cdot j}-{\bar{y}}_{\cdot \cdot }$ belongs to $W_U$. By [F3], this is the orthogonal decomposition of $y$ relative to $W_U$, proving the formula for $RS{S}_U$; the subspace $E$ is finite-dimensional by [F4], and [F3] gives $V=W_U\oplus E$. Since the standard cell-coordinate basis makes $\dim V=tb$, [F5] gives $\dim E=tb-(t+b-1)=(t-1)(b-1)$.

Put $P_C=C^T(C{C}^T{)}^{-1}C$. The matrix $C{C}^T$ is invertible since $C$ has full row rank. Direct multiplication gives $P_C^T=P_C$, $P_C^2=P_C$, and image $P_C=\mathrm{row}(C)$. For any $x$ and $a\in {\mathbb{R}}^q$, $⟨x-P_{C}x,C^{T}a⟩=a^{T}Cx-a^{T}C{C}^T(C{C}^T{)}^{-1}Cx=0$, so [F3] identifies $P_C$ as the orthogonal projection onto $\mathrm{row}(C)$. Consequently $T$ is the orthogonal direct sum of the reduced treatment space and the treatment arrays formed from $\mathrm{row}(C)$. Therefore $W_U=W_R(C)\oplus D_C$, where $D_C=\{(a_i{)}_{ij}:a\in \mathrm{row}(C)\}$.

The projection of $y$ onto $D_C$ is the array repeating $P_C\bar{y}$ across all $b$ blocks: its residual is orthogonal to every repeated $a\in \mathrm{row}(C)$ because the inner product is $b{a}^T(\bar{y}-P_C\bar{y})=0$. Its squared norm is $b\mathrm{\parallel }{P}_C\bar{y}{\mathrm{\parallel }}^2=b(C\bar{y}{)}^T(C{C}^T{)}^{-1}(C\bar{y})$. In the orthogonal direct sum of step 3.2, $y-P_{W_R}y$ is this projection plus $y-P_{W_U}y$, and the two parts are orthogonal; expanding the squared norm gives the claimed residual-sum reduction.

Embed the pair's design law in the finite binary assignment model on $\{0,1{\}}^2$ by giving weight $1\mathrm{/}2$ to $(1,0)$ and $(0,1)$ and weight zero to $(0,0)$ and $(1,1)$. Zero-probability assignments are allowed by [F1]. The positive-weight assignments give the displayed $D_1,D_2$, with average $({\delta }_1+{\delta }_2)\mathrm{/}2$ and difference $({\delta }_1-{\delta }_2)+2(Y_1(0)-Y_2(0))$.

Define $D$ by the treated-minus-control formula on the two positive-weight assignments and, for example, by zero on the two zero-weight assignments. This is a function on the full-power-set data space [F7], hence a measurable statistic and estimator of $\tau$ by [F6]; its arbitrary zero-weight values do not affect design moments.

By [F4], ${\mathbb{E}}_d[D]=(D_1+D_2)\mathrm{/}2=({\delta }_1+{\delta }_2)\mathrm{/}2=\tau$, so the contrast estimates the pair's finite-population average effect without bias.

Let $X$ indicate that unit $1$ is assigned treatment $1$, so $D=D_2+(D_1-D_2)X$. Since the assignment space is finite, the variables are square-integrable. By [F5] and [F8], covariance with a constant is zero and ${\mathrm{Var}}_d(D)=(D_1-D_2{)}^2{\mathrm{Var}}_d(X)$.

The two equally weighted assignments give ${\mathbb{E}}_d[X]=1\mathrm{/}2$ by [F4], hence ${\mathrm{Var}}_d(X)=1\mathrm{/}4$ by [F5]. Therefore ${\mathrm{Var}}_d(D)=(D_1-D_2{)}^2\mathrm{/}4$, which is the formula in the Statement.

Let $S={\delta }_1$ when unit $1$ receives treatment $1$ and $S={\delta }_2$ when unit $2$ receives it. These outcomes have weights $1\mathrm{/}2$ by [F3]. Under assignment $(1,0)$, fix $Y_1(1)$ and $Y_2(0)$ but change $Y_1(0)$; the observed assignment and response data stay the same while $S=Y_1(1)-Y_1(0)$ changes. By [F7], no statistic of the observed data can equal $S$ over the model.

By [F4], ${\mathbb{E}}_d[S]=({\delta }_1+{\delta }_2)\mathrm{/}2=\tau$, agreeing with the pair's finite-population average effect [F2].

By [F5], ${\mathrm{Var}}_d(S)=\frac{1}{2}({\delta }_1-\tau {)}^2+\frac{1}{2}({\delta }_2-\tau {)}^2=({\delta }_1-{\delta }_2{)}^2\mathrm{/}4$.

Write the cell array as $Y=m+\gamma +ϵ$, where $m\in W_U$ is additive and $\gamma \in E$. By [F1], the residual from projection onto $W_U$ is $(I-P_{W_U})Y=\gamma +(I-P_{W_U})ϵ$, and the latter error projection lies in $E=W_U^{\perp }$.

For any $h\in E$, the identity $\gamma +(I-P_{W_U})ϵ=(\gamma +h)+((I-P_{W_U})ϵ-h)$ gives two decompositions of the same observed residual into vectors in $E$. Thus the cell data alone do not distinguish the unrestricted interaction component from the projected error component.

The additive space has dimension $t+b-1$ and $E$ has dimension $(t-1)(b-1)$, so their orthogonal direct sum has dimension $tb$, the full cell-space dimension. Therefore the model with a free interaction uses all cell-space directions and leaves zero residual degrees of freedom for a separate pure-error estimate.

By [F2], $W_R(C)\subset W_U$, $p=\dim W_U=t+b-1$, $q=\dim W_U-\dim W_R(C)$, and $n=tb$. Since $t,b\ge 2$, $n-p=tb-t-b+1=(t-1)(b-1)=\nu >0$, and $q\le t-1\le p$. Under $C\tau =0$, the mean array $(\mu +{\tau }_i+{\beta }_j{)}_{ij}$ belongs to $W_R(C)$. Thus the nested-space, positive-degree, and null-mean hypotheses of [F1] hold.

Applying [F1] gives the piecewise nested-model statistic $\{(RS{S}_R-RS{S}_U)\mathrm{/}q\}\mathrm{/}\{RS{S}_U\mathrm{/}\nu \}$ the law $F_{q,\nu }$, with its value set to zero when $RS{S}_U=0$. The finite assignment law [F5] supplies no normal-error assumption; Gaussianity here is separately imposed in the Given. AC is used exactly through [A1] and [F1].

By [F2], the numerator reduction is $b(C\bar{Y}{)}^T(C{C}^T{)}^{-1}(C\bar{Y})$, so the statistic in the Statement is the same piecewise ratio. Its numerator and residual sum of squares are continuous functions of the data, and the ratio with the stated zero-denominator branch is Borel measurable; hence it is a statistic by [F4].

Since $q,\nu >0$, [F3] defines the named central law and step 3.1 is the piecewise nested-model ratio from [F1]. It therefore has law $F_{q,\nu }$, as claimed.