7. Discussion

Takis Merkouris

Previous

The proposed estimation method for matrix sampling involves a single-step calibration of the weights of the combined sample. Estimates of totals for all variables can be obtained by using only the units of sample $S_3$  and their calibrated weights which incorporate all the available information from all three samples. These weights could be used to calculate other weighted statistics, including means, ratios, quantiles and regression coefficients. When the second-order inclusion probabilities are known, including cross-sample inclusion probabilities in the nested case, the calibration procedure of Section 2 can produce composite optimal regression estimators and their variances, but with great computational difficulty. For general sampling settings, the much simpler calibration scheme of Section 3 generates readily composite generalized regression estimators, which for certain sampling strategies are optimal regression estimators.

Estimation of the variance of a CGR estimator may, in principle, be based on the method of Taylor linearization of the generalized regression estimator (see, e.g., Särndal et al. 1992, pages 235, 237). This approach requires calculations that may not be practical, or even feasible for complex sampling designs because the second-order inclusion probabilities are rarely known. Replication methods for variance estimation, such as the jackknife method or the bootstrap method (see, for example, Rust and Rao 1996), can be applied to the CGR estimators of the previous sections. For example, the jackknife method, customarily used in surveys with stratified multistage sampling design, could be used to replicate the calibration procedures that give rise to the CGR estimators. For the non-nested design, this requires applying the jackknife method to the combined sample, with the three independent samples treated as sample superstrata containing the sample strata. The replication procedure would involve then the combined sample sorted by sample and by strata within each sample, to produce replicates of the calibrated weights defined in the previous sections. The total number of strata used in the jackknife replication procedure is the total number of strata in the three samples, with each replicate involving all strata. Public-use microfiles may include the replicate calibrated weights for easy variance estimation by users. For this purpose too, replicate weights for $S_3$  only need to be included, bringing about substantial economy of data storage in such microfiles. The case of nested design is more complicated. Further investigation in this direction will be a topic of separate study.

The described estimation method may be readily adapted to matrix sampling designs with more than two subquestionnaires or more than three subsamples, making more evident the operational power of the calibration procedure. In each case, the crucial step is to determine the design matrix $\mathcal{X}.$  In such designs there may be more complex patterns with respect to the number of subquestionnaires administered to the various subsamples. All composite estimates can then be obtained using the weighted variable values only from the minimum number of subsamples that in combination contain all items.

Acknowledgements

The author thanks the Editor, Associate Editor and two referees for their comments and suggestions, which have substantially improved the paper.

Appendix

Proof of Lemma 1

For the partitioned matrix $\mathcal{X}=\left(X\mathrm{,}\Psi \right),$  the vector $c=w+R\mathcal{X}{\left({\mathcal{X}}^{\prime }R\mathcal{X}\right)}^{-1}\left(t_{\mathcal{X}}-{\mathcal{X}}^{\prime }w\right)$  takes the form

$$\begin{array}{lll}c\hfill & =\hfill & w+\left(RX\mathrm{,}R\Psi \right){\left(\begin{array}{cc}{X}^{\prime }RX& X^{\prime }R\Psi \\ {\Psi }^{\prime }RX& {\Psi }^{\prime }R\Psi \end{array}\right)}^{-1}\left(\begin{array}{c}{t}_X-X^{\prime }w\\ t_{\Psi }-{\Psi }^{\prime }w\end{array}\right)\hfill \\ \hfill & =\hfill & w+\left(RX{A}_{11}+R\Psi A_{21}\right)\left(t_X-X^{\prime }w\right)+\left(RX{A}_{12}+R\Psi A_{22}\right)\left(t_{\Psi }-{\Psi }^{\prime }w\right)\mathrm{,}\hfill \end{array}$$

where, from algebra of partitioned matrices, $A_{11}={\left[X^{\prime }RX-X^{\prime }R\Psi {\left({\Psi }^{\prime }R\Psi \right)}^{-1}{\Psi }^{\prime }RX\right]}^{-1}={\left[X^{\prime }R\left(I-P_{\Psi }\right)X\right]}^{-1}$  with $P_{\Psi }=\Psi {\left({\Psi }^{\prime }R\Psi \right)}^{-1}{\Psi }^{\prime }R,$   $A_{22}={\left[{\Psi }^{\prime }R\left(I-P_X\right){\Psi }^{\prime }\right]}^{-1}$  with $P_X=X{\left(X^{\prime }RX\right)}^{-1}{X}^{\prime }R,$   $A_{12}=-{\left(X^{\prime }RX\right)}^{-1}\left(X^{\prime }R\Psi \right)A_{22}$  and $A_{21}=-{\left({\Psi }^{\prime }R\Psi \right)}^{-1}\left({\Psi }^{\prime }RX\right)A_{11}.$  Then, equation (2.9) follows without difficulty. To prove equation (2.10), we set $c_{\Psi }=w+R\Psi {\left({\Psi }^{\prime }R\Psi \right)}^{-1}\left(t_{\Psi }-{\Psi }^{\prime }w\right),$  so that $(X^{\prime }R\Psi ){({\Psi }^{\prime }R\Psi )}^{-1}(t_{\Psi }-{\Psi }^{\prime }w)=X^{\prime }{c}_{\Psi }-X^{\prime }w,$  and use the alternative form $A_{22}={\left({\Psi }^{\prime }R\Psi \right)}^{-1}+{\left({\Psi }^{\prime }R\Psi \right)}^{-1}\left({\Psi }^{\prime }RX\right)A_{11}\left(X^{\prime }R\Psi \right){\left({\Psi }^{\prime }R\Psi \right)}^{-1}$  to write $c$  above without the second term as

$$\begin{array}{lll}w\hfill & +\hfill & R\Psi A_{22}\left(t_{\Psi }-{\Psi }^{\prime }w\right)-RX{\left(X^{\prime }RX\right)}^{-1}\left(X^{\prime }R\Psi \right)A_{22}\left(t_{\Psi }-{\Psi }^{\prime }w\right)\hfill \\ \hfill & =\hfill & w+\left[R\Psi {\left({\Psi }^{\prime }R\Psi \right)}^{-1}+R\Psi {\left({\Psi }^{\prime }R\Psi \right)}^{-1}\left({\Psi }^{\prime }RX\right)A_{11}\left(X^{\prime }R\Psi \right){\left({\Psi }^{\prime }R\Psi \right)}^{-1}\right]\left(t_{\Psi }-{\Psi }^{\prime }w\right)\hfill \\ \hfill & -\hfill & RX{\left(X^{\prime }RX\right)}^{-1}\left[I+\left(X^{\prime }R\Psi \right){\left({\Psi }^{\prime }R\Psi \right)}^{-1}\left({\Psi }^{\prime }RX\right)A_{11}\right]\left(X^{\prime }R\Psi \right){\left({\Psi }^{\prime }R\Psi \right)}^{-1}\left(t_{\Psi }-{\Psi }^{\prime }w\right)\hfill \\ \hfill & =\hfill & c_{\Psi }+R\Psi {\left({\Psi }^{\prime }R\Psi \right)}^{-1}\left({\Psi }^{\prime }RX\right)A_{11}\left(X^{\prime }{c}_{\Psi }-X^{\prime }w\right)\hfill \\ \hfill & -\hfill & RX{\left(X^{\prime }RX\right)}^{-1}\left[I+\left(X^{\prime }R\Psi \right){\left({\Psi }^{\prime }R\Psi \right)}^{-1}\left({\Psi }^{\prime }RX\right)A_{11}\right]\left(X^{\prime }{c}_{\Psi }-X^{\prime }w\right)\hfill \\ \hfill & =\hfill & c_{\Psi }+R\Psi {\left({\Psi }^{\prime }R\Psi \right)}^{-1}\left({\Psi }^{\prime }RX\right)A_{11}\left(X^{\prime }{c}_{\Psi }-X^{\prime }w\right)\hfill \\ \hfill & -\hfill & RX{\left(X^{\prime }RX\right)}^{-1}\left[I+\left(X^{\prime }RX-A_{11}^{-1}\right)A_{11}\right]\left(X^{\prime }{c}_{\Psi }-X^{\prime }w\right)\hfill \\ \hfill & =\hfill & c_{\Psi }+\left[R\Psi {\left({\Psi }^{\prime }R\Psi \right)}^{-1}\left({\Psi }^{\prime }RX\right)-RX\right]A_{11}\left(X^{\prime }{c}_{\Psi }-X^{\prime }w\right)\hfill \\ \hfill & =\hfill & c_{\Psi }-R\left(I-P_{\Psi }\right)X{\left[X^{\prime }R\left(I-P_{\Psi }\right)X\right]}^{-1}\left(X^{\prime }{c}_{\Psi }-X^{\prime }w\right)\mathrm{.}\hfill \end{array}$$

Adding to this the second term of $c$  from (2.9) gives (2.10), in the explicit form

$$c_{\Psi }+R\left(I-P_{\Psi }\right)X{\left[X^{\prime }R\left(I-P_{\Psi }\right)X\right]}^{-1}\left(t_X-X^{\prime }{c}_{\Psi }\right)\mathrm{.}$$

Proof of Theorem 1

• $\left(a\right)$   Calibration with design matrix $\mathcal{Z}=\left(\mathcal{X}\mathrm{,}D\right)$  and vector of totals $t_{\mathcal{Z}}={\left(0^{\prime }\mathrm{,}{N}^{\prime }\right)}^{\prime },$  with $0={\left(0^{\prime }\mathrm{,}{0}^{\prime }\right)}^{\prime },$   $N={\left({N^{\prime }}_1\mathrm{,}{N^{\prime }}_2\mathrm{,}{N^{\prime }}_3\right)}^{\prime },$  gives the vector of calibrated weights $c=w+\Lambda \mathcal{Z}{\left({\mathcal{Z}}^{\prime }\Lambda \mathcal{Z}\right)}^{-1}\left(t_{\mathcal{Z}}-{\mathcal{Z}}^{\prime }w\right),$  which by Lemma 1 is written as $c=c_D+L_D\mathcal{X}{\left({\mathcal{X}}^{\prime }{L}_D\mathcal{X}\right)}^{-1}\left(0-{\mathcal{X}}^{\prime }{c}_D\right),$  where $c_D=w+\Lambda D{\left(D^{\prime }\Lambda D\right)}^{-1}\left(N-D^{\prime }w\right)$  and $L_D=\Lambda \left(I-P_D\right),$  with $P_D=D{\left(D^{\prime }\Lambda D\right)}^{-1}{D}^{\prime }\Lambda .$  For STRSRS with $f_{ih}=n_{ih}/N_{ih},$   $D^{\prime }w=\widehat{N}=N,$  and thus $c=w+L_D\mathcal{X}{\left({\mathcal{X}}^{\prime }{L}_D\mathcal{X}\right)}^{-1}\left(0-{\mathcal{X}}^{\prime }w\right).$  Then, in view of (2.8), in order to show that $\widehat{ℬ}={\widehat{ℬ}}^o$  it suffices to show that $L_D={\Lambda }^0.$  For STRSRS it is easy to show that ${\Lambda }^0=\text{diag}\left\{{\lambda }_{ih}\left(I-P_{1ih}\right)\right\},$  where ${\lambda }_{ih}=N_{ih}^2\left(1-f_{ih}\right)/\left[n_{ih}\left(n_{ih}-1\right)\right]$  and $P_{1ih}=1_{ih}{\left({1^{\prime }}_{ih}{1}_{ih}\right)}^{-1}{1^{\prime }}_{ih}.$  Next, observe that the matrix $P_D$  is diagonal with $i{h}^{\text{th}}$  entry $1_{ih}{\left({1^{\prime }}_{ih}{\Lambda }_{ih}{1}_{ih}\right)}^{-1}{1^{\prime }}_{ih}{\Lambda }_{ih}=P_{1ih},$  because the elements of ${\Lambda }_{ih}$  are constant. Since this constant element is $w_{ik}/q_{ik}=\left(N_{ih}/n_{ih}\right)\left[N_{ih}\left(1-f_{ih}\right)/\left(n_{ih}-1\right)\right]={\lambda }_{ih},$  we get $L_D=\text{diag}\left\{{\Lambda }_{ih}\left(I-P_{1ih}\right)\right\}={\Lambda }^0,$  o.e.d.

• $\left(b\right)$   For Poisson sampling, ${\Lambda }_i^0=\text{diag}\left\{\left(1-{\pi }_{ihk}\right)/{\pi }_{ihk}^2\right\},h=1,\dots \mathrm{,}{H}_i.$  The proof follows immediately upon observing that with the specified constants $q_{ik}$  in the entries of ${\Lambda }_i$  we have ${\Lambda }_i={\Lambda }_i^0.$

• $\left(a’\right)$ For simplicity drop the stratum subscript. Simple random subsampling is done sequentially with fixed sizes $n_1,n_2$  and $n_3.$  It can be shown that the first-and-second order marginal inclusion probabilities for $S_i$  are ${\pi }_{ik}=n_i/N$  and ${\pi }_{ikl}=n_i\left(n_i-1\right)/\left[N\left(N-1\right)\right],$  as if $S_i$  was drawn directly from $U.$  A combinatorial argument shows that the conditional (given $S)$  second-order inclusion probability for $S_i$  and $S_j$  is ${\pi }_{ikjl|S}=n_i{n}_j/\left[n\left(n-1\right)\right]$  and thus the marginal inclusion probability is ${\pi }_{ikjl}=n_i{n}_j/\left[N\left(N-1\right)\right].$  For $k=l,{\pi }_{ikjk}=0.$  Then ${\Delta }_{kl}={\pi }_{ikjl}-{\pi }_{ik}{\pi }_{jl}=n_i{n}_j/\left[N^2\left(N-1\right)\right]$  and ${\Delta }_{kk}=-n_i{n}_j/N^2.$  Thus ${\Delta }_{kl}\approx 0,$  for $k\mathrm{,}l\in U$  when the sampling fractions are small, and then ${\Lambda }^0\approx \text{diag}\left\{{\Lambda }_i^0\right\}.$  Optimality of the CGR then follows from Theorem 1 (a).

• $\left(b’\right)$ Randomly assigning the units of $S$  to three subsamples, with fixed expected subsample size, implies that inclusion of the units is done independently within and between the subsamples. Since in Poisson sampling the units of $U$  are also included in $S$  independently, ${\Delta }_{kl}={\pi }_{ikjl}-{\pi }_{ik}{\pi }_{jl}=0$  and ${\Delta }_{kk}=-{\pi }_{ik}{\pi }_{jl}.$   ${\Delta }_{kk}$  is approximately zero for small sampling fractions, and then ${\Lambda }^0\approx \text{diag}\left\{{\Lambda }_i^0\right\}.$  Optimality of the CGR follows then from Theorem $1\left(b\right).$

Proof of Theorem 2

We start with the expression of the CGR estimator. By Lemma 1, with partitioned design matrix $\left(\mathcal{X}\mathrm{,}Z\right)$  and $R=\Lambda ,$  the calibrated weight vector $c$  can be written as $c=c_Z+L_Z\mathcal{X}{\left({\mathcal{X}}^{\prime }{L}_Z\mathcal{X}\right)}^{-1}\left(0-{\mathcal{X}}^{\prime }{c}_Z\right),$  where $c_Z=w+\Lambda Z{\left(Z^{\prime }\Lambda Z\right)}^{-1}\left(t_{(z)}-Z^{\prime }w\right)$  and $L_Z=\Lambda \left(I-P_Z\right).$  Then ${\widehat{\mathcal{X}}}_3^{\text{GR}}={{\mathcal{X}}^{\prime }}_3{c}_Z={\widehat{\mathcal{X}}}_3+{{\mathcal{X}}^{\prime }}_3\Lambda Z{\left(Z^{\prime }\Lambda Z\right)}^{-1}\left(t_{(z)}-\widehat{Z}\right)$  and ${\widehat{\mathcal{X}}}^{\text{GR}}=\widehat{\mathcal{X}}+{\mathcal{X}}^{\prime }\Lambda Z{\left(Z^{\prime }\Lambda Z\right)}^{-1}\left(t_{(z)}-\widehat{Z}\right).$  It follows that the CGR estimator is given by ${{\mathcal{X}}^{\prime }}_{3}c={\widehat{\mathcal{X}}}_3^{\text{GR}}-\widehat{ℬ}{\widehat{\mathcal{X}}}^{\text{GR}},$  where $\widehat{ℬ}=\left[{{\mathcal{X}}^{\prime }}_3\Lambda \left(I-P_Z\right)\mathcal{X}\right]{\left[{\mathcal{X}}^{\prime }\Lambda \left(I-P_Z\right)\mathcal{X}\right]}^{-1}.$

• (a)  Since $P_Z=\text{diag}\left\{P_{Z_i}\right\}$  and, for SRS, ${\Lambda }^0=\text{diag}\left\{{\lambda }_i\left(I-P_{1i}\right)\right\},$  where ${\lambda }_i=N^2\left(1-f_i\right)/\left[n_i\left(n_i-1\right)\right]$  and $P_{1i}=1_i{\left({1^{\prime }}_i{1}_i\right)}^{-1}{1^{\prime }}_i,$  we have ${\Lambda }^0\left(I-P_Z\right)=\text{diag}\left\{{\lambda }_i\left(I-P_{1i}\right)\left(I-P_{Z_i}\right)\right\}.$  Now, by assumption $1=Z_i{h}_i,$  so that $1^{\prime }{P}_{Z_i}=1^{\prime }$  and hence $P_{1i}\left(I-P_{Z_i}\right)=0.$  It follows that ${\Lambda }^0\left(I-P_Z\right)=\text{diag}\left\{{\lambda }_i\left(I-P_{Z_i}\right)\right\}$  and, since the matrices $I-P_{Z_i}$  are idempotent, ${\left(I-P_Z\right)}^{\prime }{\Lambda }^0\left(I-P_Z\right)=\text{diag}\left\{{\lambda }_i\left(I-P_{Z_i}\right)\right\}.$  But ${\lambda }_i=w_{ik}/q_{ik},$  where $w_{ik}=N/n_i$  and $q_{ik}$  are the specified constants in the entries of ${\Lambda }_i.$  It follows that ${\left(I-P_Z\right)}^{\prime }{\Lambda }^0\left(I-P_Z\right)=\text{diag}\left\{{\Lambda }_i\left(I-P_{Z_i}\right)\right\}=\Lambda \left(I-P_Z\right)$  and thus $\widehat{ℬ}={\widehat{ℬ}}^{wo},$  so that ${\widehat{\mathcal{X}}}_3^{\text{GR}}-\widehat{ℬ}{\widehat{\mathcal{X}}}^{\text{GR}}={\widehat{\mathcal{X}}}_3^{\text{GR}}-{\widehat{ℬ}}^{wo}{\widehat{\mathcal{X}}}^{\text{GR}}.$

• (b)  By Lemma 1, with the partitioned design matrix $\mathcal{Z}=\left(\mathcal{X}\mathrm{,}Z\mathrm{,}D\right)$  and vector of totals $t_{\mathcal{Z}}={\left(0^{\prime }\mathrm{,}{t^{\prime }}_{(z)}\mathrm{,}{N}^{\prime }\right)}^{\prime },$  the vector of calibrated weights $c=w+\Lambda \mathcal{Z}{\left({\mathcal{Z}}^{\prime }\Lambda \mathcal{Z}\right)}^{-1}\left(t_{\mathcal{Z}}-{\mathcal{Z}}^{\prime }w\right)$  can be written as $c=c_D+L_D\left(\mathcal{X}\mathrm{,}Z\right){\left[{\left(\mathcal{X}\mathrm{,}Z\right)}^{\prime }{L}_D\left(\mathcal{X}\mathrm{,}Z\right)\right]}^{-1}\left[{\left(0^{\prime }\mathrm{,}{t^{\prime }}_{(z)}\right)}^{\prime }-{\left(\mathcal{X}\mathrm{,}Z\right)}^{\prime }{c}_D\right],$  where $c_D=w+\Lambda D{\left(D^{\prime }\Lambda D\right)}^{-1}\left(N-D^{\prime }w\right)$  and $L_D=\Lambda \left(I-P_D\right),$  with $P_D=D{\left(D^{\prime }\Lambda D\right)}^{-1}{D}^{\prime }\Lambda .$  But, as shown in the proof of Theorem 1(a), $c_D=w$  and $L_D={\Lambda }^0.$  Thus, $c=w+{\Lambda }^0\left(\mathcal{X}\mathrm{,}Z\right){\left[{\left(\mathcal{X}\mathrm{,}Z\right)}^{\prime }{\Lambda }^0\left(\mathcal{X}\mathrm{,}Z\right)\right]}^{-1}\left[{\left(0^{\prime }\mathrm{,}{t^{\prime }}_{(z)}\right)}^{\prime }-{\left(\mathcal{X}\mathrm{,}Z\right)}^{\prime }w\right].$  Next, by applying again Lemma 1, now with $R={\Lambda }^0$  and design matrix $\left(\mathcal{X}\mathrm{,}Z\right),$  we get $c=c_Z+L_Z^0\mathcal{X}{\left(\mathcal{X}{L}_Z^0\mathcal{X}\right)}^{-1}\left(0-{\mathcal{X}}^{\prime }{c}_Z\right),$  where $c_Z=w+{\Lambda }^{0}Z{\left(Z^{\prime }{\Lambda }^{0}Z\right)}^{-1}\left(t_{(z)}-Z^{\prime }w\right)$  and $L_Z^0={\Lambda }^0\left(I-P_Z^0\right).$  Then it follows that the CGR estimator is ${{\mathcal{X}}^{\prime }}_{3}c={{\mathcal{X}}^{\prime }}_3{c}_Z-{{\mathcal{X}}^{\prime }}_3{L}_Z^0\mathcal{X}{\left({\mathcal{X}}^{\prime }{L}_Z^0\mathcal{X}\right)}^{-1}{\mathcal{X}}^{\prime }{c}_Z={\widehat{\mathcal{X}}}_3^{\text{OR}}-{\widehat{ℬ}}^o{\widehat{\mathcal{X}}}^{\text{OR}},$  in obvious expressions for ${\widehat{\mathcal{X}}}_3^{\text{OR}},{\widehat{\mathcal{X}}}^{\text{OR}}$  and ${\widehat{ℬ}}^o.$

• (c)  It was shown in the proof of Theorem 1 that $\Lambda ={\Lambda }^0.$  Clearly then it holds that ${\widehat{\mathcal{X}}}_3^{\text{GR}}={\widehat{\mathcal{X}}}_3^{\text{OR}},$   ${\widehat{\mathcal{X}}}^{\text{GR}}={\widehat{\mathcal{X}}}^{\text{OR}}$  and $\widehat{ℬ}={\widehat{ℬ}}^o,$  and thus ${\widehat{\mathcal{X}}}_3^{\text{GR}}-\widehat{ℬ}{\widehat{\mathcal{X}}}^{\text{GR}}={\widehat{\mathcal{X}}}_3^{\text{OR}}-{\widehat{ℬ}}^o{\widehat{\mathcal{X}}}^{\text{OR}}.$

Proof of Proposition 1

All matrices appearing in this proof are defined at the population level. Partitioning the matrix $\mathcal{X}$  in (4.4) as $\left(Z\mathrm{,}\Psi \right),$  where $Z$  consists of the second and fourth columns, and $\Psi$  of the rest, and applying Lemma 1 with $R={\Lambda }^0=\left\{({\pi }_{kl}-{\pi }_k{\pi }_l)/{\pi }_k{\pi }_l\right\},$  we obtain the vector of calibrated weights decomposed as

$$c=w+L_{\Psi }^{0}Z{\left(Z^{\prime }{L}_{\Psi }^{0}Z\right)}^{-1}\left[0-Z^{\prime }w\right]+L_Z^0\Psi {\left({\Psi }^{\prime }{L}_Z^0\Psi \right)}^{-1}\left[0-{\Psi }^{\prime }w\right]\mathrm{,}$$

where $L_Z^0={\Lambda }^0\left(I-P_Z^0\right)$  with $P_Z^0=Z{\left(Z^{\prime }{\Lambda }^{0}Z\right)}^{-1}{Z}^{\prime }{\Lambda }^0.$  The estimator ${\widehat{Z}}^B$  in (4.2) is obtained as ${Z^{\prime }}_{3-}c,$  where $Z_{3-}={\left(0^{\prime }\mathrm{,}{0}^{\prime }\mathrm{,}{Z^{\prime }}_3\right)}^{\prime }.$  The last two terms of (4.2) are consolidated in the term ${Z^{\prime }}_{3-}{L}_Z^0\Psi {\left({\Psi }^{\prime }{L}_Z^0\Psi \right)}^{-1}\left[0-{\Psi }^{\prime }w\right].$  These two terms vanish only if ${Z^{\prime }}_{3-}{L}_Z^0\Psi (={Z^{\prime }}_{3-}{\Lambda }^0\Psi -{Z^{\prime }}_{3-}{\Lambda }^{0}Z{(Z^{\prime }{\Lambda }^{0}Z)}^{-1}{Z}^{\prime }{\Lambda }^0\Psi )=0.$  First, we easily get ${Z^{\prime }}_{3-}{\Lambda }^0\Psi =\left({Z^{\prime }}_3{\Lambda }_3^0{X}_3\mathrm{,}{Z^{\prime }}_3{\Lambda }_3^0{Y}_3\right)$  and ${Z^{\prime }}_{3-}{\Lambda }^{0}Z={Z^{\prime }}_3{\Lambda }_3^0{Z}_3\left(I\mathrm{,}I\right),$  as well as

$$Z^{\prime }{\Lambda }^0\Psi =\left(\begin{array}{cc}{Z^{\prime }}_1{\Lambda }_1^0{X}_1+{Z^{\prime }}_3{\Lambda }_3^0{X}_3& {Z^{\prime }}_3{\Lambda }_3^0{Y}_3\\ {Z^{\prime }}_3{\Lambda }_3^0{X}_3& {Z^{\prime }}_2{\Lambda }_2^0{Y}_2+{Z^{\prime }}_3{\Lambda }_3^0{Y}_3\end{array}\right)\mathrm{,}$$

and

$$Z^{\prime }{\Lambda }^{0}Z=\left(\begin{array}{cc}{Z^{\prime }}_1{\Lambda }_1^0{Z}_1+{Z^{\prime }}_3{\Lambda }_3^0{Z}_3& {Z^{\prime }}_3{\Lambda }_3^0{Z}_3\\ {Z^{\prime }}_3{\Lambda }_3^0{Z}_3& {Z^{\prime }}_2{\Lambda }_2^0{Z}_2+{Z^{\prime }}_3{\Lambda }_3^0{Z}_3\end{array}\right)\mathrm{.}$$

Next we write

$${\left(Z^{\prime }{\Lambda }^{0}Z\right)}^{-1}={\left(\begin{array}{cc}A& B\\ B^{\prime }& D\end{array}\right)}^{-1}=\left(\begin{array}{cc}{A}^{-1}+F{E}^{-1}{F}^{\prime }& -F{E}^{-1}\\ -E^{-1}{F}^{\prime }& E^{-1}\end{array}\right)\mathrm{,}$$

where $E=D-B^{\prime }{A}^{-1}B$  and $F=A^{-1}B.$  It follows then that ${Z^{\prime }}_{3-}{\Lambda }^{0}Z{(Z^{\prime }{\Lambda }^{0}Z)}^{-1}=(B{A}^{-1}+BF{E}^{-1}{F}^{\prime }-B{E}^{-1}{F}^{\prime }\mathrm{,}B\left(I-F\right)E^{-1})=((D-B)E^{-1}{F}^{\prime }\mathrm{,}B(I-F)E^{-1}).$  Using the analytic expressions $B={Z^{\prime }}_3{\Lambda }_3^0{Z}_3,$   $D={Z^{\prime }}_2{\Lambda }_2^0{Z}_2+{Z^{\prime }}_3{\Lambda }_3^0{Z}_3,$   $F={\left({Z^{\prime }}_1{\Lambda }_1^0{Z}_1+{Z^{\prime }}_3{\Lambda }_3^0{Z}_3\right)}^{-1}{Z^{\prime }}_3{\Lambda }_3^0{Z}_3$  and $E={Z^{\prime }}_2{\Lambda }_2^0{Z}_2+{Z^{\prime }}_1{\Lambda }_1^0{Z}_{1}F,$  we obtain after some algebra

$${Z^{\prime }}_{3-}{\Lambda }^{0}Z{\left(Z^{\prime }{\Lambda }^{0}Z\right)}^{-1}=K^{-1}\left[{\left({Z^{\prime }}_1{\Lambda }_1^0{Z}_1\right)}^{-1}\mathrm{,}{\left({Z^{\prime }}_2{\Lambda }_2^0{Z}_2\right)}^{-1}\right]\mathrm{,}$$

where $K={\left({Z^{\prime }}_1{\Lambda }_1^0{Z}_1\right)}^{-1}+{\left({Z^{\prime }}_2{\Lambda }_2^0{Z}_2\right)}^{-1}+{\left({Z^{\prime }}_3{\Lambda }_3^0{Z}_3\right)}^{-1}.$  We can now obtain without much difficulty

$$\begin{array}{lll}{Z^{\prime }}_{3-}{L}_Z^0\Psi \hfill & =\hfill & {Z^{\prime }}_{3-}{\Lambda }^0\Psi -{Z^{\prime }}_{3-}{\Lambda }^{0}Z{\left(Z^{\prime }{\Lambda }^{0}Z\right)}^{-1}{Z}^{\prime }{\Lambda }^0\Psi \hfill \\ \hfill & =\hfill & K^{-1}[{\left({Z^{\prime }}_3{\Lambda }_3^0{Z}_3\right)}^{-1}{Z^{\prime }}_3{\Lambda }_3^0{X}_3-{\left({Z^{\prime }}_1{\Lambda }_1^0{Z}_1\right)}^{-1}{Z^{\prime }}_1{\Lambda }_1^0{X}_1\mathrm{,}\hfill \\ \hfill & \hfill & {\left({Z^{\prime }}_3{\Lambda }_3^0{Z}_3\right)}^{-1}{Z^{\prime }}_3{\Lambda }_3^0{Y}_3-{\left({Z^{\prime }}_2{\Lambda }_2^0{Z}_2\right)}^{-1}{Z^{\prime }}_2{\Lambda }_2^0{Y}_2]\mathrm{.}\hfill \end{array}$$

It follows that ${Z^{\prime }}_{3-}{L}_Z^0\Psi =\left(0\mathrm{,}0\right)$  only if ${\left({Z^{\prime }}_3{\Lambda }_3^0{Z}_3\right)}^{-1}{Z^{\prime }}_3{\Lambda }_3^0{X}_3={\left({Z^{\prime }}_1{\Lambda }_1^0{Z}_1\right)}^{-1}{Z^{\prime }}_1{\Lambda }_1^0{X}_1$  and ${\left({Z^{\prime }}_3{\Lambda }_3^0{Z}_3\right)}^{-1}{Z^{\prime }}_3{\Lambda }_3^0{Y}_3={\left({Z^{\prime }}_2{\Lambda }_2^0{Z}_2\right)}^{-1}{Z^{\prime }}_2{\Lambda }_2^0{Y}_2.$  But these two equations are identical to the equations in (4.6). Since all the matrices in ${\left({Z^{\prime }}_i{\Lambda }_i^0{Z}_i\right)}^{-1}{Z^{\prime }}_i{\Lambda }_i^0{X}_i$  are defined at the population level, with the subscript $i=1,3$  indicating survey, this quantity is constant across surveys only if the design-specific matrix ${\Lambda }_i^0$  is constant, or if ${\Lambda }_i^0$  differs among surveys by a constant multiple (depending on the sample size). This holds true also for ${\left({Z^{\prime }}_i{\Lambda }_i^0{Z}_i\right)}^{-1}{Z^{\prime }}_i{\Lambda }_i^0{Y}_i,$   $i=2,3.$  This completes the proof.

Proof of Proposition 2

Under the sampling scheme (a) of Theorem 1, composite calibration at population level with design matrix $\mathcal{Z}=\left(\mathcal{X}\mathrm{,}D\right)$  and vector of totals $t_{\mathcal{Z}}={\left(0^{\prime }\mathrm{,}{N}^{\prime }\right)}^{\prime }$  produces the joint CGR domain estimator of ${\left({t^{\prime }}_{xd}\mathrm{,}{t^{\prime }}_{yd}\right)}^{\prime }$  based on the weights of $S_3$  and written in the form ${\widehat{\mathcal{X}}}_{3d}^{\text{CGR}}={\widehat{\mathcal{X}}}_{3d}+{\widehat{ℬ}}_d\left(t_{\mathcal{Z}}-\widehat{\mathcal{Z}}\right),$  where ${\widehat{ℬ}}_d={{\mathcal{X}}^{\prime }}_{3d}\Lambda \mathcal{Z}{\left({\mathcal{Z}}^{\prime }\Lambda \mathcal{Z}\right)}^{-1}.$  The associated matrix of regression residuals is ${\mathcal{X}}_{3d}-\mathcal{Z}{{\widehat{ℬ}}^{\prime }}_d,$  alternatively written as $\left(I-P_{\mathcal{Z}}\right){\mathcal{X}}_{3d},$  with $P_{\mathcal{Z}}=\mathcal{Z}{({\mathcal{Z}}^{\prime }\Lambda \mathcal{Z})}^{-1}{\mathcal{Z}}^{\prime }\Lambda .$  Then $\widehat{\text{AV}}\left({\widehat{\mathcal{X}}}_{3d}^{\text{CGR}}\right)={{\mathcal{X}}^{\prime }}_{3d}{\left(I-P_{\mathcal{Z}}\right)}^{\prime }{\Lambda }^0\left(I-P_{\mathcal{Z}}\right){\mathcal{X}}_{3d}.$  Next recall from the proof of Theorem 1 that ${\Lambda }^0=\Lambda \left(I-P_D\right),$  with $P_D=D{\left(D^{\prime }\Lambda D\right)}^{-1}{D}^{\prime }\Lambda ,$  and notice that $D=\mathcal{Z}H$  for a suitable constant matrix $H.$  It is easy to verify that $P_D{P}_{\mathcal{Z}}=P_D.$  It follows then that ${\Lambda }^0\left(I-P_{\mathcal{Z}}\right)=\Lambda \left(I-P_{\mathcal{Z}}\right)$  and ${\left(I-P_{\mathcal{Z}}\right)}^{\prime }{\Lambda }^0\left(I-P_{\mathcal{Z}}\right)=\Lambda \left(I-P_{\mathcal{Z}}\right).$  Thus $\widehat{\text{AV}}\left({\widehat{\mathcal{X}}}_{3d}^{\text{CGR}}\right)={{\mathcal{X}}^{\prime }}_{3d}\Lambda \left(I-P_{\mathcal{Z}}\right){\mathcal{X}}_{3d}.$  Now, composite calibration at domain level involves the design matrix ${\mathcal{Z}}_d=\left({\mathcal{X}}_d\mathrm{,}D\right);$  no need to restrict $D$  to the domain $U_d.$  The resulting CGR estimator is ${\stackrel{⌣}{\mathcal{X}}}_{3d}^{\text{CGR}}={\widehat{\mathcal{X}}}_{3d}+{\stackrel{⌣}{ℬ}}_d\left(t_{{\mathcal{Z}}_d}-{\widehat{\mathcal{Z}}}_d\right)$  where ${\stackrel{⌣}{ℬ}}_d={{\mathcal{X}}^{\prime }}_{3d}\Lambda {\mathcal{Z}}_d{\left({{\mathcal{Z}}^{\prime }}_d\Lambda {\mathcal{Z}}_d\right)}^{-1}.$  As with ${\widehat{\mathcal{X}}}_{3d}^{\text{CGR}}$  above, it can be shown that $\widehat{\text{AV}}\left({\stackrel{⌣}{\mathcal{X}}}_{3d}^{\text{CGR}}\right)={{\mathcal{X}}^{\prime }}_{3d}\Lambda \left(I-P_{{\mathcal{Z}}_d}\right){\mathcal{X}}_{3d},$  where $P_{{\mathcal{Z}}_d}={\mathcal{Z}}_d{\left({{\mathcal{Z}}^{\prime }}_d\Lambda \mathcal{Z}\right)}_d^{-1}{{\mathcal{Z}}^{\prime }}_d\Lambda .$  Then $\widehat{\text{AV}}\left({\widehat{\mathcal{X}}}_{3d}^{\text{CGR}}\right)-\widehat{\text{AV}}\left({\stackrel{⌣}{\mathcal{X}}}_{3d}^{\text{CGR}}\right)={{\mathcal{X}}^{\prime }}_{3d}\Lambda \left(P_{{\mathcal{Z}}_d}-P_{\mathcal{Z}}\right){\mathcal{X}}_{3d}.$  Noticing that ${{\mathcal{X}}^{\prime }}_{3d}\Lambda \mathcal{Z}={{\mathcal{X}}^{\prime }}_{3d}\Lambda {\mathcal{Z}}_d,$  we can write $P_{\mathcal{Z}}={\mathcal{Z}}_d{\left({\mathcal{Z}}^{\prime }\Lambda \mathcal{Z}\right)}^{-1}{{\mathcal{Z}}^{\prime }}_d\Lambda .$  It is trivial then to show that $\left(P_{{\mathcal{Z}}_d}-P_{\mathcal{Z}}\right)={\left(P_{{\mathcal{Z}}_d}-P_{\mathcal{Z}}\right)}^2,$  and since the matrix $\Lambda$  is diagonal with positive entries, it follows that ${{\mathcal{X}}^{\prime }}_{3d}\Lambda \left(P_{{\mathcal{Z}}_d}-P_{\mathcal{Z}}\right){\mathcal{X}}_{3d}\text{>}0$  and hence $\widehat{\text{AV}}\left({\stackrel{⌣}{\mathcal{X}}}_{3d}^{\text{CGR}}\right)\text{<}\widehat{\text{AV}}\left({\widehat{\mathcal{X}}}_{3d}^{\text{CGR}}\right).$

Under the conditions of part $\left(b\right),$   $\Lambda ={\Lambda }^0$  and the CGR domain estimator is identical to the COR domain estimator ${\widehat{\mathcal{X}}}_{3d}^{\text{COR}}={\widehat{\mathcal{X}}}_{3d}-{\widehat{ℬ}}_d^0\widehat{\mathcal{X}},$  where ${\widehat{ℬ}}_d^0={{\mathcal{X}}^{\prime }}_{3d}{\Lambda }^0\mathcal{X}{\left({\mathcal{X}}^{\prime }{\Lambda }^0\mathcal{X}\right)}^{-1}.$  The associated matrix of regression residuals is $\left(I-P_{\mathcal{X}}\right){\mathcal{X}}_{3d},$  with $P_{\mathcal{X}}=\mathcal{X}{\left({\mathcal{X}}^{\prime }{\Lambda }^0\mathcal{X}\right)}^{-1}{\mathcal{X}}^{\prime }{\Lambda }^0.$  Then $\widehat{\text{AV}}\left({\widehat{\mathcal{X}}}_{3d}^{\text{COR}}\right)={{\mathcal{X}}^{\prime }}_{3d}{\left(I-P_{\mathcal{X}}\right)}^{\prime }{\Lambda }^0\left(I-P_{\mathcal{X}}\right){\mathcal{X}}_{3d}={{\mathcal{X}}^{\prime }}_{3d}{\Lambda }^0\left(I-P_{\mathcal{X}}\right){\mathcal{X}}_{3d}.$  On the other hand, for the estimator ${\stackrel{⌣}{\mathcal{X}}}_{3d}^{\text{COR}}={\stackrel{⌣}{\mathcal{X}}}_{3d}-{\stackrel{⌣}{ℬ}}_d^0\widehat{\mathcal{X}},$  where ${\widehat{ℬ}}_d^0={{\mathcal{X}}^{\prime }}_{3d}{\Lambda }^0{\mathcal{X}}_d{\left({{\mathcal{X}}^{\prime }}_d{\Lambda }^0{\mathcal{X}}_d\right)}^{-1}$  we have $\widehat{\text{AV}}\left({\stackrel{⌣}{\mathcal{X}}}_{3d}^{\text{COR}}\right)={{\mathcal{X}}^{\prime }}_{3d}{\Lambda }^0\left(I-P_{{\mathcal{X}}_d}\right){\mathcal{X}}_{3d},$  with $P_{{\mathcal{X}}_d}={\mathcal{X}}_d{\left({{\mathcal{X}}^{\prime }}_d{\Lambda }^0{\mathcal{X}}_d\right)}^{-1}{{\mathcal{X}}^{\prime }}_d{\Lambda }^0.$  Then $\widehat{\text{AV}}({\widehat{\mathcal{X}}}_{3d}^{\text{COR}})-\widehat{\text{AV}}({\stackrel{⌣}{\mathcal{X}}}_{3d}^{\text{COR}})={{\mathcal{X}}^{\prime }}_{3d}{\Lambda }^0(P_{{\mathcal{X}}_d}-P_{\mathcal{X}}){\mathcal{X}}_{3d}.$  Notice that ${{\mathcal{X}}^{\prime }}_{3d}{\Lambda }^0{\mathcal{X}}_d={{\mathcal{X}}^{\prime }}_{3d}{\Lambda }^0{\mathcal{X}}_{3d}$  and since ${\Lambda }^0$  is diagonal ${{\mathcal{X}}^{\prime }}_{3d}{\Lambda }^0\mathcal{X}={{\mathcal{X}}^{\prime }}_{3d}{\Lambda }^0{\mathcal{X}}_{3d}.$  It follows that ${{\mathcal{X}}^{\prime }}_{3d}{\Lambda }^0\left(P_{{\mathcal{X}}_d}-P_{\mathcal{X}}\right){\mathcal{X}}_{3d}={{\mathcal{X}}^{\prime }}_{3d}{\Lambda }^0{\left(P_{{\mathcal{X}}_d}-P_{\mathcal{X}}\right)}^2{\mathcal{X}}_{3d}$  and hence $\widehat{\text{AV}}\left({\stackrel{⌣}{\mathcal{X}}}_{3d}^{\text{COR}}\right)\text{<}\widehat{\text{AV}}\left({\widehat{\mathcal{X}}}_{3d}^{\text{COR}}\right).$

For parts $\left(a’\right)$  and $\left(b’\right),$  the proof is the same as in $\left(a\right)$  and $\left(b\right),$  in view of the proof of Theorem 1.

References

Andersson, P.G., and Thorburn, D. (2005). An optimal calibration distance leading to the optimal regression estimator. Survey Methodology, 31, 1, 95-99.

Australian Bureau of Statistics (2011). Household Expenditure Survey and Survey of Income and Housing, User Guide, Australia, 2009-10 (cat. no. 6503.0).

Chipperfield, J.O., and Steel, D.G. (2009). Design and estimation for split questionnaire surveys. Journal of Official Statistics, 25, 227-244.

Chipperfield, J.O., and Steel, D.G. (2011). Efficiency of split questionnaire surveys. Journal of Statistical Planning and Inference, 141, 1925-1932.

Deville, J.-C., and Särndal, C.-E. (1992). Calibration estimators in survey sampling. Journal of the American Statistical Association, 87, 376-382.

Fuller, W.A. (1990). Analysis of repeated surveys. Survey Methodology, 16, 2, 167-180.

Gonzalez, J.M., and Eltinge, J.L. (2007). Multiple matrix sampling: A review. Proceedings of the Survey Research Methods Section, American Statistical Association, 3069-3075.

Gonzalez, J.M., and Eltinge, J.L. (2008). Adaptive matrix sampling for the consumer expenditure quarterly interview survey. Proceedings of the Survey Research Methods Section, American Statistical Association, 3069-3075.

Hidiroglou, M.A. (2001). Double sampling. Survey Methodology, 27, 2, 143-154.

Houbiers, M. (2004). Towards a social statistical database on unified estimates at Statistics Netherlands. Journal of Official Statistics, 20, 55-75.

Jones, R.G. (1980). Best linear unbiased estimators for repeated surveys. Journal of the Royal Statistical Society, Serie B, 42, 221-226.

Kim, J.K., and Rao, J.N.K. (2012). Combining data from two independent surveys: A model-assisted approach. Biometrika, 99, 1, 85-100.

Merkouris, T. (2004). Combining independent regression estimators from multiple surveys. Journal of the American Statistical Association, 99, 1131-1139.

Merkouris, T. (2010). Combining information from multiple surveys by using regression for more efficient small domain estimation. Journal of the Royal Statistical Society, Serie B, 72, 27-48.

Montanari, G.E. (1987). Post-sampling efficient QR-prediction in large-scale surveys. International Statistics Review, 55, 191-202.

Montanari, G.E. (1998). On regression estimation of finite population means. Survey Methodology, 24, 1, 69-77.

Raghunathan, T.E., and Grizzle, J.E. (1995). A split questionnaire survey design. Journal of the American Statistical Association, 90, 54-63.

Rao, J.N.K. (1994). Estimating totals and distribution functions using auxiliary information at the estimation stage. Journal of Official Statistics, 10, 153-165.

Renssen, R.H. (1998). Use of statistical matching techniques in calibration estimation. Survey Methodology, 24, 2, 171-183.

Renssen, R.H., and Nieuwenbroek, N.J. (1997). Aligning estimates for common variables in two or more sample surveys. Journal of the American Statistical Association, 92, 368-375.

Rust, K.F., and Rao, J.N.K. (1996). Variance estimation for complex surveys using replication techniques. Statistical Methods in Medical Research, 5, 283-310.

Särndal, C.-E., Swensson, B. and Wretman, J.H. (1992). Model-Assisted Survey Sampling, New York: Springer.

Smith, P. (2009). Survey harmonization in official household surveys in the United Kingdom. Proceedings of the ISI World Statistical Congresses, Dublin.

Thomas, N., Raghunathan, T.E., Schenker, N., Katzoff, M.J. and Johnson, C.L. (2006). An evaluation of matrix sampling methods using data from the National Health and Nutrition Examination Survey. Survey Methodology, 32, 2, 217-231.

Wolter, K.M. (1979). Composite estimation in finite populations. Journal of the American Statistical Association, 74, 604-613.

Wu, C. (2004). Combining information from multiple surveys through the empirical likelihood method. Canadian Journal of Statistics, 32, 15-26.

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