Block Design, Incidence, and Concurrence Matrix (2018)

Block Design Properties

The necessary conditions that the parameters of a Balanced Incomplete Block Design (BIB design) must satisfy are

  • $bk = vr$, where $r=\frac{bk}{v}$ each treatment has $r$ replications
  • no treatment appears more than once in any block
  • all unordered pairs of treatments appear exactly in $\lambda$ blocks (equi-concurrence)
    where $\lambda=\frac{r(k-1)}{v-1}=\frac{bk(k-1}{v(v-1)}$ is often referred to as the concurrence parameter of a BIB design.

A design say $d$ with parameters $(v, b, r, k, \lambda)$ can be represented as a $v \times b$ treatment block incidence matrix (having $v$ rows and $b$ columns). Let denote it by $N=n_{ij}$ whose elements $n_{ij}$ signify the number of units in block $j$ allocated to treatment $i$. The rows of the incidence matrix are labeled with varieties (treatments) of the design and the columns with the blocks.

We have to put 1 in the ($i$, $j$)th cell of the matrix if variety $i$ is contained in block $j$ and 0 otherwise. Each row of the incidence matrix has $r$ 1’s, each column has $k$ 1’s, and each pair of distinct rows has $\lambda$ column 1’s, leading to a useful identity matrix.
The matrix $NN’$ has $v$ rows and $v$ columns, referred to as the concurrence matrix of design $d$, and its entries, the concurrence parameters are denoted by $\lambda_{dij}$. For a BIBD, $n_{ij}$ is either one or zero, and $n_{ij}^2= n_{ij}$.

Theorem: If $N$ is the incidence matrix of a $(v, b, r, k, \lambda)$-design then $NN’=(r-\lambda)I+\lambda J$ where $I$ is $v\times v$ identity matrix and $J$ is the $v\times v$ matrix of all 1’s.

Example: For Block Design {1,2,3}, {2,3,4}, {3,4,1}, {4,1,2} construct incidence matrix

Block Design: incidence matrix
Incidence and Concurrence matrix

Denoting the elements of $NN’$ by $q_{ih}$, we see that $q_{ii}=\sum_j n_{ij}^2$ and $q_{ih}=\sum_j n_{ij} n_{hj}, (i \ne h)$. For any block design $NN’$, the treatment concurrence with diagonal elements equal to $q_{ii}=r$ and off-diagonal elements are $q_{ih}=\lambda, (i\ne h)$ equal to the number of times any pairs of treatment occur together within the block. In a balanced design, the off-diagonal entries in $NN’$ are all equal to a constant $\lambda$ i.e., the common replication for a BIBD is $r$, and the common pairwise treatment concurrence is $\lambda$.

$N$ is a matrix of $v$ rows and $b$ columns that $r(N)\le min(b, c)$. Hence, $t\le min(b, v)$. If design is symmetric $b=v$ and $N$ is square the $|NN’|=|N|^2$, so $(r-\lambda)^{v-1}r^2$ is a perfect square.

Using R Packages

MCQs General Knowledge

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