OCDs in Randomized Block Design Set-Up

Abstract

We have already mentioned in Chap. 2 that the question of optimal estimation of the covariate parameters under the covariate model was first studied by Troya Lopes, J Stat Plan Inference 6:373–419, (1982a); 7:49–75, (1982b) in Completely Randomized Design (CRD) set-up. Later, Das et al., J Stat Plan Inference 115:273–285, (2003) extended her work to the construction of optimum covariate designs in the set-up of Randomized Block Designs (RBD). Also, Rao et al., Electron Notes Discret Math 15:157–160, (2003) established a relationship between mixed orthogonal arrays (MOAs) and OCDs in CRD and RBD set-ups. In this chapter we discuss the construction procedures of OCDs in RBD set-up as considered by Das et al., J Stat Plan Inference 115:273–285, (2003) and Rao et al., Electron Notes Discret Math 15:157–160, (2003).

Appendix

Proof of Theorem 3.4.3 It is observed that X is a symmetric matrix. Now $(i,i)\mathrm{th}$ block matrix of X X $^\prime$ is

$\begin{array}{rll} (\mathbf{X}\mathbf{X}^\prime )_{ii} &{}=\displaystyle \sum _{k=1,~k\ne i}^p(a_{ik}\mathbf{A}-\mathbf{I}_p)(a_{ki}\mathbf{A}-\mathbf{I}_p)+(a_{ii}\mathbf{A}+\mathbf{J}_p-\mathbf{I}_p)(a_{ii}\mathbf{A}+\mathbf{J}_p-\mathbf{I}_p)&{} \\ &{}=\displaystyle \sum _{k=1~k\ne i}^p(a_{ik}^2\mathbf{A}^2-2a_{ik}\mathbf{A}+\mathbf{I})+(\mathbf{J}_p-\mathbf{I}_p)(\mathbf{J}_p-\mathbf{I}_p)\,\mathrm{since}~a_{ij}=a_{ji}\,\text {and}\,a_{ii}=0~\forall i \\ &{} =(p\mathbf{I}_p-\mathbf{J}_p)\displaystyle \sum _{k=1}^pa_{ik}^2-2\mathbf{A}\displaystyle \sum _{k=1}^pa_{ik}+(p-1)\mathbf{I}_p+p\mathbf{J}_p-2\mathbf{J}_p+\mathbf{I}_p\\ &{}=(p\mathbf{I}_p-\mathbf{J}_p)(p-1) -2\mathbf{A}.0+(p-1)\mathbf{I}_p+(p-2)\mathbf{J}_p+\mathbf{I}_p&{}\\ &{}=(p(p-1)+p)\mathbf{I}_p+(p-2)\mathbf{J}_p-(p-1)\mathbf{J}_p&{} \\ &{}=p^2\mathbf{I}_p-\mathbf{J}_p&{} \end{array}$

$(i,j)\mathrm{th}$ block matrix of X X $^\prime$ is

$\begin{array}{rl} (\mathbf{X}\mathbf{X}^\prime )_{ij} &{}=\displaystyle \sum _{k=1,~k\ne i,j}^p(a_{ik}\mathbf{A}-\mathbf{I}_p)(a_{kj}\mathbf{A}-\mathbf{I}_p)+(a_{ii}\mathbf{A}+\mathbf{J}_p-\mathbf{I}_p)(a_{ji}\mathbf{A}-\mathbf{I}_p)\\ &{}~~~~~~~~~~~~~~+(a_{ij}\mathbf{A}-\mathbf{I}_p)(a_{jj}\mathbf{A}+\mathbf{J}_p-\mathbf{I}_p)\\ &{}=\displaystyle \sum _{k=1,~k\ne i,j}^p(a_{ik}a_{jk}\mathbf{A}^2-a_{ik}\mathbf{A}-a_{jk}\mathbf{A}+\mathbf{I}_p+(\mathbf{J}_p-\mathbf{I}_p)(a_{ji}\mathbf{A}-\mathbf{I}_p)\\ {} &{}\quad +(a_{ij}\mathbf{A}-\mathbf{I}_p)(\mathbf{J}_p-\mathbf{I}_p)\\ &{}=-(p\mathbf{I}_p-\mathbf{J}_p)+a_{ij}\mathbf{A}+a_{ij}\mathbf{A}+(p-2)\mathbf{I}_p-\mathbf{J}_p-a_{ij}\mathbf{A}-\mathbf{I}_p-\mathbf{J}_p-a_{ij}\mathbf{A}+\mathbf{I}_p\\ &{}=-\mathbf{J}_p\\ \mathrm{{Thus}}\,\mathbf{X}\mathbf{X}^\prime &{}=p^2\mathbf{I}_{p^2}-\mathbf{J}_{p^2}. \end{array}$

$\square$

Theorem 3.4.5

The columns of the matrix D are orthogonal.

Proof

The cross product of $i\mathrm{th}$ and $j\mathrm{th}$ elements of B,

$\begin{array}{rl} \displaystyle \sum _{k=1}^pb_{ki}b_{kj} &{}= \displaystyle \sum _{k=1,~k\ne i,j}^pb_{ki}b_{kj}+b_{ii}b_{ij}+b_{ji}b_{jj}+b_{(p+1),i}b_{(p+1),j} \\ &{}=\displaystyle \sum _{k=1,~k\ne i,j}^{p}(a_{ki}\alpha )(a_{kj}\alpha )+(-\beta )(a_{ij}\alpha )+(a_{ji}\alpha )(-\beta )+\beta .\beta \\ &{}=\displaystyle \sum _{k=1,~k\ne i,j}^{p}a_{ki}a_{kj}(\alpha .\alpha )+a_{ij}((-\beta ).\alpha )+a_{ji}(\alpha .(-\beta ))+c \\ &{}=c\displaystyle \sum _{k=1}^{p}a_{ki}a_{kj}+0+0+c \\ &{} =-c+c=0. \end{array}$

Similarly, it can be shown that the columns of C are also orthogonal. Now we want to show that any column of B is orthogonal to any column of C. For this, we consider the cross product of $i\mathrm{th}$ column of B and $j\mathrm{th}$ column of C: $\begin{array}{rl} \displaystyle \sum _{k=1}^pb_{ki}c_{kj} &{}= \displaystyle \sum _{k=1,~k\ne i,j}^pb_{ki}c_{kj}+b_{ii}c_{ij}+b_{ji}c_{jj}+b_{(p+1),i}c_{(p+1),j} \\ &{}=\displaystyle \sum _{k=1,~k\ne i,j}^{p}(a_{ki}\alpha )(a_{kj}\beta )+(-\beta )(a_{ij}\beta )+(a_{ji}\alpha )(\alpha )+\beta .(-\alpha ) \\ &{}=\displaystyle \sum _{k=1,~k\ne i,j}^{p}a_{ki}b_{kj}(\alpha .\beta )+a_{ij}((-\beta ).\beta )+a_{ji}(\alpha .(-\alpha ))+0=0-c+c+0=0 \end{array}$

$\square$