Consider allocation of *v* treatments with replications *t*_{1}, …, *t*_{v} in *k* blocks of sizes *b*_{1}, …, *b*_{k}, where ∑_{i}*t*_{i} = ∑_{j}*b*_{j} = *n* . Let us introduce matrices **B** = *diag*(**1**_{b1}, …, **1**_{bk}) and **D** = (*d*_{ij}), where
$$\begin{array}{}{d}_{ij}=\left\{\begin{array}{}1,& \text{if the\hspace{0.17em}}i\text{-th observation refers to the\hspace{0.17em}}j\text{-th treatment},\\ 0,& \text{otherwise}.\end{array}\right.\end{array}$$

These matrices indicate allocation of treatments in blocks. For this reason **D** is sometimes identified with block design.

To each pair (**B**, **D**) corresponds a linear experiment 𝓝 = 𝓝(**D ***α* + [**1**_{n}, **B**] *β*, *σ* **I**_{n}), where *α* = (*α*_{1}, …, *α*_{υ})^{T} refers to the treatment effects, while *β* = (*μ*, *β*_{1}, …, *β*_{k})^{T} refers to the general mean and block effects. In this case the reduced information matrix (4), called also **C**-matrix (see [18, 19, 20]), may be presented in the form
$$\begin{array}{}\mathbf{C}={\mathbf{D}}^{T}\mathbf{D}-{\mathbf{D}}^{T}diag({b}_{1}^{-1}{\mathbf{1}}_{{b}_{1}}{\mathbf{1}}_{{b}_{1}}^{T},...,{b}_{k}^{-1}{\mathbf{1}}_{{b}_{k}}{\mathbf{1}}_{{b}_{k}}^{T})\mathbf{D}\\ \phantom{\rule{1em}{0ex}}=diag({t}_{1},...,{t}_{v})-\mathbf{N}diag({b}_{1}^{-1},...,{b}_{k}^{-1}){\mathbf{N}}^{T},\end{array}$$

where **N** = (*n*_{ij}) is the incidence matrix defined as **N** = **DB**^{T}. It is clear that **N1**_{k} = **t** and $\begin{array}{}{\mathbf{1}}_{v}^{T}{\mathbf{N}\mathbf{=}\mathbf{b}}^{T},\end{array}$ where **t** = (*t*_{1}, …, *t*_{v})^{T} and **b** = (*b*_{1}, …, *b*_{k})^{T}. A design **D** is said to be orthogonal if $\begin{array}{}\mathbf{N}=\frac{1}{n}{\mathbf{t}\mathbf{b}}^{T}.\end{array}$

One can verify that
$$\begin{array}{}\mathbf{N}diag({b}_{1}^{-1},...,{b}_{k}^{-1}){\mathbf{N}}^{T}=\left[\begin{array}{cccc}{\sum}_{j}\frac{{n}_{1j}^{2}}{{b}_{j}}& {\sum}_{j}\frac{{n}_{1j}{n}_{2j}}{{b}_{j}}& ...& {\sum}_{j}\frac{{n}_{1j}{n}_{vj}}{{b}_{j}}\\ {\sum}_{j}\frac{{n}_{2j}{n}_{1j}}{{b}_{j}}& {\sum}_{j}\frac{{n}_{2j}^{2}}{{b}_{j}}& ...& {\sum}_{j}\frac{{n}_{2j}{n}_{vj}}{{b}_{j}}\\ ...& ...& ...& ...\\ {\sum}_{j}\frac{{n}_{vj}{n}_{1j}}{{b}_{j}}& {\sum}_{j}\frac{{n}_{vj}{n}_{2j}}{{b}_{j}}& ...& {\sum}_{j}\frac{{n}_{vj}^{2}}{{b}_{j}}\end{array}\right].\end{array}$$

In particular, for the orthogonal design,
$\begin{array}{}\mathbf{N}diag({b}_{1}^{-1},...,{b}_{k}^{-1}){\mathbf{N}}^{T}=\frac{1}{n}{\mathbf{t}\mathbf{t}}^{T}.\end{array}$

Denote by 𝓓 = 𝓓(**t**; **b**) the class of all possible allocations of *v* treatments with replications *t*_{1}, …, *t*_{v} in *k* blocks of sizes *b*_{1}, …, *b*_{k} for *v*, *k* ≥ 2. Such class contains or does not contain an orthogonal design. If it does then by Stępniak [8] this design is optimal in 𝓓 w.r.t. invariant linear estimation, i.e. it is at least as good as any other design in the class.

It is natural to ask whether the orthogonal design is also optimal w.r.t. invariant quadratic estimation. In the light of the results presented in Section 3 we are strongly convinced that the answer is negative, but for formal reasons we are ready to provide a rigorous proof of this fact. By Corollary 3.8 we only need to show that for any incidence matrix **N** = (*n*_{ij}) corresponding to the orthogonal design there exists an incidence matrix **M** = (*m*_{ij}) such that
$\begin{array}{}\mathbf{M}diag({b}_{1}^{-1},...,{b}_{k}^{-1}){\mathbf{M}}^{T}\ne \mathbf{N}diag({b}_{1}^{-1},...,{b}_{k}^{-1}){\mathbf{N}}^{T}\end{array}$ Define
$$\begin{array}{}{m}_{ij}=\left\{\begin{array}{}{n}_{ij}+1,& \text{if\hspace{0.17em}}i=1\text{\hspace{0.17em}and\hspace{0.17em}}j=1,\text{\hspace{0.17em}or\hspace{0.17em}}i=2\text{\hspace{0.17em}and\hspace{0.17em}}j=2,\\ {n}_{ij}-1,& \text{if\hspace{0.17em}}i=1\text{\hspace{0.17em}and\hspace{0.17em}}j=2,\text{\hspace{0.17em}or\hspace{0.17em}}i=2\text{\hspace{0.17em}and\hspace{0.17em}}j=1,\\ {n}_{ij},& \text{otherwise}.\end{array}\right.\end{array}$$

We note that **M1**_{k} = **N1**_{k} and
$\begin{array}{}{\mathbf{1}}_{v}^{T}{\mathbf{M}\mathbf{=}\mathbf{1}}_{v}^{T}\mathbf{N}.\end{array}$ Therefore, the designs represented by **M** and **N** belong to the same class. To show the desired inequality we only need, for instance, to compare the left upper entries, say *u*_{11} and
$\begin{array}{}{u}_{11}^{0},\end{array}$ of the matrices
$\begin{array}{}\mathbf{M}diag({b}_{1}^{-1},...,{b}_{k}^{-1}){\mathbf{M}}^{T}\text{\hspace{0.17em}and\hspace{0.17em}}\mathbf{N}diag({b}_{1}^{-1},...,{b}_{k}^{-1}){\mathbf{N}}^{T}.\end{array}$

Since $\begin{array}{}{n}_{ij}=\frac{1}{n}{t}_{i}{b}_{j}\end{array}$ we have
$$\begin{array}{}{\displaystyle {u}_{11}-{u}_{11}^{0}=\frac{{m}_{11}^{2}}{{b}_{1}}+\frac{{m}_{12}^{2}}{{b}_{2}}-(\frac{{n}_{11}^{2}}{{b}_{1}}+\frac{{n}_{12}^{2}}{{b}_{2}})}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle =\frac{({n}_{11}+1{)}^{2}}{{b}_{1}}+\frac{({n}_{12}-1{)}^{2}}{{b}_{2}}-(\frac{{n}_{11}^{2}}{{b}_{1}}+\frac{{n}_{12}^{2}}{{b}_{2}})}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle =2(\frac{{n}_{11}}{{b}_{1}}-\frac{{n}_{12}}{{b}_{2}})+\frac{1}{{b}_{1}}+\frac{1}{{b}_{2}}}\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}{\displaystyle =\frac{2}{n}(\frac{{t}_{1}{b}_{1}}{{b}_{1}}-\frac{{t}_{1}{b}_{2}}{{b}_{2}})+\frac{1}{{b}_{1}}+\frac{1}{{b}_{2}}=\frac{1}{{b}_{1}}+\frac{1}{{b}_{2}}>0.}\end{array}$$

This leads to the following

#### Conclusion 4.1

*Any orthogonal block design is not optimal w*.*r*.*t*. *invariant quadratic estimation*. *Moreover*, *for any* **t** = (*t*_{1}, …, *t*_{v})^{T} *and* **b** = (*b*_{1}, …, *b*_{k})^{T} *there is no optimal design in the class* 𝓓 = 𝓓(**t**; **b**).

By the way we have demonstrated that, with reference to the orthogonal block design, the meaning of the optimality w.r.t. linear estimation may be strengthened in the sense that the words “at least as good” may be replaced by “better”.

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