2. Successive difference replication

Stephen Ash

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2.1  Definition of successive difference replication

F&T present a method called successive difference replication (SDR) that estimates the variance from a sample selected with $sys$ by mimicking ${\widehat{v}}_{\text{SD}2}\left(\widehat{Y}\right),$ i.e., SDR is equivalent or nearly equivalent to ${\widehat{v}}_{\text{SD}2}\left(\widehat{Y}\right)$ . We show how SDR can be used to produce replicate factors and weights for a general replicate variance estimator that is equivalent to the SD2 estimator. Before we define the SDR estimator in the first theorem, we first establish some terms and provide a lemma that is used by the theorem.

A row assignment scheme, or more simply RA, is an assignment of two rows of a matrix to each unit in the sample. We usually denote the pair of rows as $\left(a_i,b_i\right)$ for unit $i.$ A connected loop is an RA that does not repeat any of the rows, i.e., $a_i\ne a_j$ and $b_i\ne b_j$ for all $i$ and $j$ in the connected loop, and is circular, i.e., $b_i=a_{i+1}$ for all $i<n$ and $b_n=a_1.$ For example, one possible connected loop for three observations is (1,2), (2,3), (3,1).

A shift matrix $S$ can be used to move either the rows or columns of a matrix. We will explain how to move rows, which is similar to columns. A shift matrix is a square matrix that has all 0s, except a single 1 in each column. If we wanted to move row $p$ to row $q,$ we would put a 1 in the $q^{\text{th}}$ row of the $p^{\text{th}}$ column and 0s elsewhere. We emphasize that order is important in applying a shift matrix to another matrix. The application of $S$ to another square matrix $A$ as $AS$ shifts the columns of $A$ and $SA$ shifts the rows of $A.$

Lemma: Let $S_1,S_2,\dots ,S_c$ be shift matrices, then $\text{block}\left({S^{\prime }}_1{S}_1,{S^{\prime }}_2{S}_2,\dots ,{S^{\prime }}_C{S}_C\right)=I.$

Proof. We first define a general block diagonal matrix $A$ that is formed by the square matrices $A_1,A_2,\dots ,A_C$ as

$$A=\text{block}\left(A_1,A_2,\dots ,A_C\right)=\left[\begin{array}{cccc}{A}_1& 0& \mathrm{...}& 0\\ 0& A_2& \mathrm{...}& ⋮\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \mathrm{...}& A_C\end{array}\right].$$

It can be shown that if both $A$ and $B$ are block diagonal matrices and the square matrices $$A_1,A_2,\dots ,A_C$$ have the same dimensions as $B_1,B_2,\dots ,B_C,$ respectively, then $AB=\text{block}\left(A_1{B}_1,A_2{B}_2,\dots ,A_C{B}_C\right).$ For a given shift matrix, we also know that $S^{\prime }S=I,$ since a one row down shift of a shift matrix is $I.$ With the two previous items, the lemma follows.

We also define a one row shift matrix as a shift matrix that either shifts all the rows of another matrix down one row and the last row moved to the first or shifts all the rows of another matrix up one row and the last row moved to the last. If $S_D$ is a one row shift matrix that moves rows down then it has 1s along the upper superdiagonal and a 1 in the bottom left entry of the matrix, for example $S_1.$ Similarly, if $S_U$ is a one row shift matrix that moves rows up, then it has 1s along the lower superdiagonal and a 1 in the top entry of the matrix, for example the subsequently defined $S_2.$ Note the property that $S_D={S^{\prime }}_U$ and $S_U={S^{\prime }}_D;$ therefore $S_U+{S^{\prime }}_U=S_D+{S^{\prime }}_D.$ We now present the main theorem of the paper that establishes the conditions under which SDR is equivalent to SD2.

Theorem 1: Let $n$ be the sample size of a given $sys$ sample and ${\stackrel{⌣}{y}}^{\prime }=\left[{\stackrel{⌣}{y}}_1{\stackrel{⌣}{y}}_2\dots {\stackrel{⌣}{y}}_n\right]$ be defined as the $n\times 1$ weighted observation vector, where the order of the observations reflects the sort order of $sys.$

• (a) Choose a Hadamard matrix of order $k$ $\left(H{H}^{\prime }=kI\right),$ where $n\le k.$
• (b) Choose a RA that assigns two rows $\left(a_i,b_i\right)$ to each unit $i$ in the sample. Let the RA define $C$ connected loops of $m_c$ units in each connected loop $c.$
• (c) Choose the $m=n$ rows of $H$ corresponding to the RA to make the $m\times k$ matrix $M.$ The order of the rows of $M$ should correspond to the first row of the RA. For example, the first row of $M$ should be row $a_{i=1}$ of $H,$ the second row should be row $a_{i=2}$ of $H,$ etc. Next define the $m\times m$ shift matrix as $S=\text{block}\left(S_1,S_2,\dots ,S_C\right)$ where the $m_c\times m_c$ one row shift matrices $S_c$ are defined to identify the position of the second row $b_i$ of the RA in $M.$ In general, each shift matrix $S_c$ will be a shift-up, shift-down, or a $2\times 2$ shift matrix (see the subsequently defined $S_4$ ).

Define the estimator for each replicate total $r$ as ${\widehat{Y}}_r={\displaystyle {\sum }_{i=1}^n{f}_{i,r}{\stackrel{⌣}{y}}_i,}$ where the matrix of replicate factors is $F=1_m{1^{\prime }}_k+\left(2^{-3/2}{I}_m-2^{-3/2}S\right)M$ and individual values within the matrix are defined for each unit $i$ (rows of $F$ ) of replicate $r$ (columns of $F$ ) as $f_{i,r}=1+2^{-3/2}{h}_{a_i,r}-2^{-3/2}{h}_{b_i,r}.$ $I_m$ is a $m\times m$ identity matrix and $1_m$ is a $m\times 1$ vector of 1s. Then the SDR variance estimator ${\widehat{v}}_{\text{SDR}}\left(\widehat{Y}\right)=\left(1-f\right)4/k{\displaystyle {\sum }_{r=1}^m{\left({\widehat{Y}}_r-\widehat{Y}\right)}^2}$ is equivalent to the sum of $C$ different SD2 estimators.

Proof. The SDR estimator can be written in matrix notation as

$$\begin{array}{l}\left(1-f\right)\frac{4}{k}\left({\stackrel{⌣}{y}}^{\prime }\left(1_m{1^{\prime }}_k+\left(2^{-3/2}{I}_m-2^{-3/2}S\right)M\right)-{\stackrel{⌣}{y}}^{\prime }{1}_m{1^{\prime }}_k\right){\left({\stackrel{⌣}{y}}^{\prime }\left(1_m{1^{\prime }}_k+\left(2^{-3/2}{I}_m-2^{-3/2}S\right)M\right)-{\stackrel{⌣}{y}}^{\prime }{1}_m{1^{\prime }}_k\right)}^{\prime }\\ =\left(1-f\right)\frac{4}{k}{\left(2^{-3/2}\right)}^2{\stackrel{⌣}{y}}^{\prime }\left(I_m-S\right)M{M}^{\prime }{\left(I_m-S\right)}^{\prime }\stackrel{⌣}{y}\end{array}$$

Because $\left\{rowsofM\right\}\subseteq \left\{rowsofH\right\}$ , it can be shown that $M{M}^{\prime }=kI.$ With this result, the variance becomes

$$\begin{array}{cc}\hfill \left(1-f\right)\frac{1}{2k}{\stackrel{⌣}{y}}^{\prime }\left(I_m-S\right)\left(k{I}_m\right){\left(I_m-S\right)}^{\prime }\stackrel{⌣}{y}& =\frac{1}{2}\left(1-f\right){\stackrel{⌣}{y}}^{\prime }\left(I_m-S\right){\left(I_m-S\right)}^{\prime }\stackrel{⌣}{y}\hfill \\ & =\frac{1}{2}\left(1-f\right){\stackrel{⌣}{y}}^{\prime }\left(2I_m-S-S^{\prime }\right)\stackrel{⌣}{y}\hfill \end{array}$$

The last line follows from the lemma and has a constant value for any choice of $H.$ By noting the block diagonal structure of $S,$ we can write the estimator as

$$\frac{1}{2}\left(1-f\right){\displaystyle \sum _{c=1}^C{{\stackrel{⌣}{y}}^{\prime }}_c\left(2I_m-S_c-{S^{\prime }}_c\right){\stackrel{⌣}{y}}_c},$$

where ${\stackrel{⌣}{y}}_c$ corresponds to the vector of the weighted observations in connected loop $c,$ which is a result of partitioning the weighted observation vector as ${\stackrel{⌣}{y}}^{\prime }=\left[{\stackrel{⌣}{y}}_{c=1}{\stackrel{⌣}{y}}_{c=2}\dots {\stackrel{⌣}{y}}_{c=C}\right].$ The choice of the RA does not change the result since we know that $2I_m-S_c-{S^{\prime }}_c$ is constant for either an up or down one row shift matrix $S_c.$

Note 1: Theorem 1 defines the SDR estimator in terms of replicate factors, but we can alternatively express the estimator in terms of replicate weights as

$$\left(1-f\right)\frac{4}{k}{y}^{\prime }\left(W-1_m{1^{\prime }}_k\right){\left(W-1_m{1^{\prime }}_k\right)}^{\prime }y.$$

Here, $W$ is the $m\times k$ matrix of replicate weights defined as $W=w*F,$ where $w=\left(w_1,w_2,\dots ,w_n\right)$ is the vector of design weights for the $n$ units of the sample and the operator $*$ multiplies element-wise the vector $w$ by each of the columns of $F,$ i.e., if $W_{i,r}$ and $w_i$ are entries of $W$ and $w,$ respectively, then the entries of $W$ are defined as $W_{i,r}=w_i\times f_{i,r}.$

Note 2: Huang and Bell (2009) similarly defined SDR as a quadratic form and used it to establish some general properties of the estimator when $y_k$ is $\text{i}\text{.i}\text{.d}\text{.}\left(\mu ,{\sigma }^2\right).$ Our interest lies with the interpretation of how and how well SDR works. Defining the quadratic form with shift matrices and connected loops leads to insights into the row assignments and the efficiency of the estimator.

For a large sample size, it is not usually practical to use $H,$ where $n<k.$ The second theorem shows one way that we can use $H$ with $k<n$ to produce a larger Hadamard matrix $\tilde{H}$ with $k\ge n$ that will result in the SDR estimator being equivalent to the SD2 estimator. The second theorem also builds upon and clarifies the instructions F&T give for the case of $n>k.$ In F&T's instructions, they use the term cycle to denote every $m_d\le k$ units of the sample. Theorem 2 does not make conditions on the RA, but otherwise it does follow the setup of F&T.

Theorem 2: Let $n$ be the sample size of a given $sys$ sample.

• (a) Choose a Hadamard matrix $H_A$ of order $k_A,$ where $n>k_A.$
• (b) Choose a RA that assigns rows to $H_A$ to the sample. Retaining their original order, split the $n$ sample units into $D$ cycles. Each cycle $d$ has $m_d\le k_A$ units. Within each cycle, the RA defines one or more connected loops.
• (c) Choose a seminormal Hadamard matrix $H_B$ of order $k_B$ and use it to define a larger Hadamard matrix $\tilde{H}$ of order $\tilde{k}$ generated from the original $H_A.$ This can be done by applying a Welsch construction to $H_A,$ i.e., $\tilde{H}=H_B\otimes H_A.$
• (d) Choose the $m={\displaystyle {\sum }_{d=1}^D{m}_d}$ rows of $\tilde{H}$ that correspond to the RA to make the $m\times \tilde{k}$ matrix $\tilde{M}.$ The order of the rows of $\tilde{M}$ should correspond to the first row of the RA. Next define the $m\times m$ shift matrix as $S=\text{block}\left(S_1,S_2,\dots ,S_D\right)$ where the $m_d\times m_d$ shift matrices $S_d$ identify the position of the second row $b_i$ of the RA in $\tilde{M}.$

With this prescription, the SDR estimator is defined as

$${\widehat{v}}_{\text{SDR}}\left(\widehat{Y}\right)=\left(1-f\right)\frac{4}{\tilde{k}}{\displaystyle \sum _{r=1}^{\tilde{k}}{\left({\widehat{Y}}_r-\widehat{Y}\right)}^2}$$

and is equivalent to the sum of at least $D$ SD2 estimators.

Proof. The result follows by applying Theorem 1. The specific value of $D$ follows from the fact that each of the $D$ cycles can have one or more connected loops, so there will be a total of at least $D$ connected loops.

Example 1: Let $n=14$ and choose the nonnormal Hadamard $H_A=H_{4b}$ of order $k_A=4.$ The number of cycles will be $D=4$ and the RA within each cycle is given in the second column of Table 2.1 for each unit. We define $\tilde{H}$ of $\tilde{k}=$ 16 using a Welsh construction of the original normal Hadamard matrix as

$$H_{16}=H_{4a}\otimes H_{4b}=\left[\begin{array}{rrrr}\hfill H_{4b}& \hfill H_{4b}& \hfill H_{4b}& \hfill H_{4b}\\ \hfill H_{4b}& \hfill -H_{4b}& \hfill H_{4b}& \hfill -H_{4b}\\ \hfill H_{4b}& \hfill H_{4b}& \hfill -H_{4b}& \hfill -H_{4b}\\ \hfill H_{4b}& \hfill -H_{4b}& \hfill -H_{4b}& \hfill H_{4b}\end{array}\right]$$

where

$$H_{4a}=\left[\begin{array}{rrrr}\hfill 1& \hfill 1& \hfill 1& \hfill 1\\ \hfill 1& \hfill -1& \hfill 1& \hfill -1\\ \hfill 1& \hfill 1& \hfill -1& \hfill -1\\ \hfill 1& \hfill -1& \hfill -1& \hfill 1\end{array}\right]\text{ and }{H}_{4b}=\left[\begin{array}{rrrr}\hfill 1& \hfill -1& \hfill 1& \hfill 1\\ \hfill -1& \hfill -1& \hfill -1& \hfill 1\\ \hfill 1& \hfill -1& \hfill -1& \hfill -1\\ \hfill 1& \hfill 1& \hfill -1& \hfill 1\end{array}\right].$$

Using $H_{16},$ we can calculate the replicate factors for 16 replicates as Table 2.1. In matrix notation, $\tilde{M}$ includes all the rows of $\tilde{H}=H_{16}$ except rows 13 and 16. The rows of $\tilde{M}$ are ordered by $a_i,$ the first row assigned in the RA. The shift matrix is defined as $$S=\text{block}\left(S_1,S_2,S_3,S_4\right),$$ where the shift matrices corresponding to each cycle are

$$S_1=\left[\begin{array}{cccc}0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\\ 1& 0& 0& 0\end{array}\right]S_2=\left[\begin{array}{cccc}0& 1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 1& 0\end{array}\right],S_3=\left[\begin{array}{cccc}0& 0& 0& 1\\ 1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\end{array}\right],S_4=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right].$$

Table 2.1
Matrix of replicate factors $\left(f_{i,r}\right)$ for example 1
Table summary
This table displays the Matrix of replicate factors . The information is grouped by Unit # 1 (appearing as row headers), $H_A=H_{4b}$ , $\tilde{H}=H_{16}$ , Cycle, Replicate (appearing as column headers).

Unit #	RA $H_A=H_{4b}$	RA $\tilde{H}=H_{16}$	Cycle	Replicate
				1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16
1	(1,2)	(1,2)	1	1.7	1.0	1.7	1.0	1.7	1.0	1.7	1.0	1.7	1.0	1.7	1.0	1.7	1.0	1.7	1.0
2	(2,3)	(2,3)		0.3	1.0	1.0	1.7	0.3	1.0	1.0	1.7	0.3	1.0	1.0	1.7	0.3	1.0	1.0	1.7
3	(3,4)	(3,4)		1.0	0.3	1.0	0.3	1.0	0.3	1.0	0.3	1.0	0.3	1.0	0.3	1.0	0.3	1.0	0.3
4	(4,1)	(4,1)		1.0	1.7	0.3	1.0	1.0	1.7	0.3	1.0	1.0	1.7	0.3	1.0	1.0	1.7	0.3	1.0
5	(1,3)	(5,7)		1.0	1.0	1.7	1.7	1.0	1.0	0.3	0.3	1.0	1.0	1.7	1.7	1.0	1.0	0.3	0.3
6	(3,1)	(7,5)	2	1.0	1.0	0.3	0.3	1.0	1.0	1.7	1.7	1.0	1.0	0.3	0.3	1.0	1.0	1.7	1.7
7	(2,4)	(6,8)		0.3	0.3	1.0	1.0	1.7	1.7	1.0	1.0	0.3	0.3	1.0	1.0	1.7	1.7	1.0	1.0
8	(4,2)	(8,6)		1.7	1.7	1.0	1.0	0.3	0.3	1.0	1.0	1.7	1.7	1.0	1.0	0.3	0.3	1.0	1.0
9	(1,4)	(9,12)		1.0	0.3	1.7	1.0	1.0	0.3	1.7	1.0	1.0	1.7	0.3	1.0	1.0	1.7	0.3	1.0
10	(4,3)	(12,11)	3	1.0	1.7	1.0	1.7	1.0	1.7	1.0	1.7	1.0	0.3	1.0	0.3	1.0	0.3	1.0	0.3
11	(3,2)	(11,10)		1.7	1.0	1.0	0.3	1.7	1.0	1.0	0.3	0.3	1.0	1.0	1.7	0.3	1.0	1.0	1.7
12	(2,1)	(10,9)		0.3	1.0	0.3	1.0	0.3	1.0	0.3	1.0	1.7	1.0	1.7	1.0	1.7	1.0	1.7	1.0
13	(2,3)	(14,15)	4	0.3	1.0	1.0	1.7	1.7	1.0	1.0	0.3	1.7	1.0	1.0	0.3	0.3	1.0	1.0	1.7
14	(3,2)	(15,14)		1.7	1.0	1.0	0.3	0.3	1.0	1.0	1.7	0.3	1.0	1.0	1.7	1.7	1.0	1.0	0.3

Given the replicate factors in Table 2.1, the SDR estimator is equivalent to the sum of five different SD2 estimators, one for each connected loop of the RA, i.e.,

             $\left(1-f\right)\frac{4}{\tilde{k}}{\displaystyle \sum _{r=1}^{\tilde{k}}{\left({\widehat{Y}}_r-\widehat{Y}\right)}^2}=\frac{1}{2}\left(1-f\right)\left[\begin{array}{l}{\displaystyle \sum _{i=2}^4{\left(y_i-y_{i-1}\right)}^2}+{\left(y_4-y_1\right)}^2+2{\left(y_6-y_5\right)}^2\\ +2{\left(y_8-y_7\right)}^2+{\displaystyle \sum _{i=10}^{12}{\left(y_i-y_{i-1}\right)}^2}+{\left(y_{12}-y_9\right)}^2\\ +2{\left(y_{13}-y_{13}\right)}^2\end{array}\right].\text{ (2}\text{.1)}$

There are a few items to note with Example 1. First, the number of replicates needed is greater than the sample size. This happens when $m_d$ is not constant across all cycles. The fourth cycle had only two sample units, but we had to use four replicates from each $H_{4b}$ because at least one of the cycles used four rows.

To make the example more interesting, we chose a nonnormal Hadamard matrix $H_{4b}$ for $H_A.$ This nonnormal Hadamard was generated by starting with the normal Hadamard $H_{4a}$ and reversing the procedure for finding a normal Hadamard as described by Hedayat and Wallis (1978).Here we simply changed the sign of all units in the second row and then changed all the signs for the second column.

If we would have used the normal Hadamard matrix $H_{4a}$ for both $H_A$ and $H_B,$ the replicate factors for replicates 1, 5, 9, and 13 would have all been 1.0. We call a replicate a dead replicate when every unit gets a value of 1.0 and thereby the replicate estimate is equal to the original estimate.In SDR, there is nothing wrong with dead replicates, it is just the way the replicate factors are distributed by the Hadamard matrix. With a dead replicate, many of the values of 1.0 are in the dead replicate, and the other replicates are more mixed with values of 1.7 and 0.3. However, all the replicates, even the dead replicates, are needed in estimation.

The real value of Theorem 2 is in understanding F&T's original prescription for SDR when $n>k.$ In F&T, the RA is applied repeatedly to the $m=k-1$ rows of $H_A$ (skipping the first row of $H_A$ ), where $H_A$ is chosen as a normal Hadamard matrix. Replicates are then formed using the $k_A$ columns of $H_A.$ If we apply the larger framework of Theorem 2, we would say that they implicitly used a normal $H_B,$ which results in $\tilde{H}=H_B\otimes H_A$ and only includes the first $k_A$ replicates in the variance estimator. Since a subset of the replicates needed for SDR to be equivalent to SD2 is used, we say that the resultant estimator is an approximation of the SD2 estimator.

Example 1 (continued): If we only used the first four replicates of Table 2.1, the SDR estimator would be equivalent to (2.1) plus the remainder term $R$ that is defined as

$$R=\left[\begin{array}{l}\left(y_1-y_2\right)\left(y_8-y_7\right)+\left(y_1-y_2\right)\left(y_{11}-y_{12}\right)+\left(y_1-y_2\right)\left(-y_{13}+y_{14}\right)\\ +\left(y_8-y_7\right)\left(y_{11}-y_{12}\right)+\left(y_8-y_7\right)\left(y_{14}-y_{13}\right)+\left(y_{11}-y_{12}\right)\left(y_{14}-y_{13}\right)\\ +\left(y_4-y_3\right)\left(y_8-y_7\right)+\left(y_4-y_3\right)\left(y_{10}-y_9\right)+\left(y_8-y_7\right)\left(y_{10}-y_9\right)\\ +\left(y_1-y_4\right)\left(y_5-y_6\right)+\left(y_1-y_4\right)\left(y_9-y_{12}\right)+\left(y_5-y_6\right)\left(y_9-y_{12}\right)\\ +\left(y_2-y_3\right)\left(y_5-y_6\right)+\left(y_2-y_3\right)\left(y_{10}-y_{11}\right)+\left(y_2-y_3\right)\left(y_{13}-y_{14}\right)\\ +\left(y_5-y_6\right)\left(y_{10}-y_{11}\right)+\left(y_5-y_6\right)\left(y_{13}-y_{14}\right)+\left(y_{10}-y_{11}\right)\left(y_{13}-y_{14}\right)\end{array}\right]$$

Note that $R$ includes the same number of positive and negative terms, which do not cancel exactly, but has the result that $R$ is usually close to zero.Similarly, using replicates 1 to $q\times k_A,$ where $q=1,2,\dots ,k_B,$ will result in a $R$ that has an equal number of positive and negative terms. Only with all the replicates of $\tilde{H}$ will the remainder term $R$ equal 0.

Example 2: The Current Population Survey (CPS) has a monthly sample size of $n=$ 72,000 households per month (U.S. Census Bureau 2006). CPS has a two-stage sample design, where a first-stage sample of Primary Sample Units (PSUs), which are generally counties or groups of counties, are selected and then in the second-stage households are selected within the sample PSUs.Some PSUs, generally the metropolitan areas, are selected with certainty, i.e., their first-stage probability of selection is 1.0. With the certainty PSUs, the $sys$ sample can be treated as the first-stage sample design in variance estimation, i.e., SDR is applied to produce replicates. In the noncertainty PSUs, Balanced Repeated Replication (BRR)[McCarthy 1966] is applied to produce replicates. Roughly 75% of the sample or 54,000 units are in SR PSUs, where SDR is applied.

The CPS application of SDR uses a Hadamard matrix with $k=$ 160 and excludes two rows, i.e., $m=$ 158. Replicate weights are produced for 160 replicates. Although it may seem like a logical conclusion of the paper, we do not suggest that CPS should use a Hadamard matrix of order $k=$54,000 or produce 54,000 sets of replicate weights.That would result in an unreasonable number of replicates. Instead, we suggest that the subset of 160 replicates used by CPS is large and therefore provides a reasonable approximation to SD2. Later in the empirical examples, we examine the impact of using a reduced set of replicates.

2.2  Row assignment when $n>k$

Until this point we have assumed a given RA and have not discussed how to generate the RA for a given sample, where $n>k.$ In this section, we review two RAs and discuss some considerations about RAs in general. The first RA is similar to the RA described by Sukasih and Jang (2003) and is intended for use with $k<n$ and Theorem 2.

RA1: The RA assigns a pair of rows $a_i$ and $b_i$ to every $m_d$ units of the sample, which we call cycle $d,$ where $m_d\le k.$ After $m_d-1$ cycles, the RA is repeated until all units of the sample have been assigned a pair of rows.

Step 1: Sort the sample in the order in which it was sorted prior to sample selection.

Step 2: Initialize the cycle number as $d=1$ and the number of connected loops as $c=1.$

Step 3: Start the RA at the beginning of a cycle or a connected loop as $a_1=c.$

Step 4: Repeat the following RA: $b_i=\mathrm{mod}\left(a_i+d,k\right)$ and $a_i=b_i$ until all $m_d$ rows of the cycle have been used or the RA becomes a connected loop. Here, the modulo function or $\mathrm{mod}\left(a,b\right)$ is defined as the remainder of the division of $a$ by $b.$ If all $m_d$ rows of the cycle have been used, start a new cycle: let $d=d+1$ and go back to step 3.Otherwise, (end of a connected loop, but not the end of a cycle) start a new connected loop: let $c=c+1$ and go back to step 3.

Step 5: At the end of $d=m_d-1$ cycles, start over with the first cycle $–$ go back to step 2.

RA1 has the following characteristics:

• - Each of the cycles $d=1,2,\dots ,m_d-1$ of the RA, assigns $m_d$ pairs of rows. This generates a total of $m_d\left(m_d-1\right)$ pairs of rows.
• - The RA repeats itself after $m_d-1$ cycles. F&T suggest that after 10 cycles, the RA be restarted. We suggest that all $m_d-1$ cycles be used before restarting the RA.
• - The values of $a_i$ and $b_i$ are always $c$ units apart.
• - Halfway through the sequence, the pattern repeats itself in reverse order. If $m$ is even, the cycles before and after the $\left(m_d+1\right)/2^{\text{th}}$ cycle repeat themselves in reverse order.

RA1 differs from the RA of Sukasih and Jang (2003) in that we do not suggest that row 1 be skipped, nor that the RA be repeated after 10 cycles, or require that $k-1$ be prime. First, a row of all 1s may seem odd, but it is not a problem. Similar to a column of all 1s in $M$ which made a dead replicate, a row of all 1s will only effect the distribution of the replicate factors.The replicate factors for a unit $i$ that are assigned row 1 (either $a_i=1$ or $b_i=1$) will have more replicate factors of 1.0 than otherwise. This is not wrong; it is just how the replicate factors are distributed by $H_A.$ The second difference is that we suggest repeating the assignment after $m$ cycles, which is when the pattern repeat, instead of a fixed number of 10 cycles. Lastly, we do not require that $k-1$ be prime but note that if $m_d=k-1$ and $k-1$ is prime, then every cycle is guaranteed to have only one connected loop.

We also provide a second simpler-to-implement RA called RA2 that will be compared with RA1 in the empirical examples.

RA2: No mixing of row assignments. Repeat the same simple RA for every $m_d$ units, i.e., $\left(1,2\right),\left(2,3\right),\dots ,\left(m_d,1\right).$

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