Optimal linear estimation in two-phase sampling
Section 8. Discussion

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The described method of optimal and regression estimation for two-phase sampling involves a single-step calibration of the weights of the combined first-and-second phase samples. Thus, using a single set of calibrated weights that incorporate all the available information from the two phases, a substantially improved estimate of the total of a target variable can be obtained, as shown by the simulation study. These weights could be used to calculate other weighted statistics, including means, ratios, quantiles and regression coefficients. The framework of the method is general enough to encompass complex designs with multiple stages and different stratification at the two phases, as well as various types of auxiliary variables known at the population or sample level ‒ ten different cases of auxiliary information are identified in Estevao and Särndal (2002). Furthermore, the method may be extended to multi-phase sampling designs through the appropriate calibration setup.

Estimation of a total for any domain (subpopulation) of interest, $U_d\subset U,$ can be carried out readily using the calibrated weights and summing the weighted sample values of the variable of interest over $U_d.$ For the resulting domain estimator to be optimal linear estimator, the domain estimates of $t_y,$ $t_{x_1}$ and $t_{x_2}$ need to be combined linearly, by carrying out optimal calibration at the domain level with domain calibration totals and with the appropriate modification of the matrix $\text{X}.$ A number of calibration options, regarding the use of the available auxiliary information at the population, domain and two-phase sample levels, could be considered for the most efficient estimation of domain totals in any particular application. Related work in Merkouris (2010) would be helpful in this context.

The estimated approximate variances of the two-phase optimal estimator and the two-phase regression estimator, based on Taylor linearization, were given in Sections 4.1 and Section 5, respectively. For the two-phase regression estimator, replication methods of variance estimation, such as the jackknife method or the bootstrap method, could be alternatively applied, or would be the only option when first-phase or second-phase joint inclusion probabilities are not known. There is extensive literature on such replication methods for existing regression estimators in two-phase sampling. The single-step calibration feature of the proposed regression estimation method may be helpful in this direction; detailed study of this is beyond the scope of this paper.

Appendix

Proof of Lemma 1

The symmetric matrix $\text{Var}\mathrm{<mo\; stretchy="false">(</mo>}{w}_U^{*}\mathrm{<mo\; stretchy="false">)</mo>}$ has the form of (3.8) but with $\text{Cov}\mathrm{<mo\; stretchy="false">(</mo>}{w}_{1U}\mathrm{,}{w}_U\mathrm{<mo\; stretchy="false">)</mo>}$ as off-diagonal block. The $k{l}^{\text{th}}$ element of the matrix $\text{Var}\mathrm{<mo\; stretchy="false">(</mo>}{w}_{1U}\mathrm{<mo\; stretchy="false">)</mo>}$ is

$$\text{Cov}\mathrm{<mo\; stretchy="false">(</mo>}{w}_{1U_k}\mathrm{,}{w}_{1U_l}\mathrm{<mo\; stretchy="false">)</mo>}=\mathrm{<mo\; stretchy="false">[</mo>}E\mathrm{<mo\; stretchy="false">(</mo>}{I}_{1k}{I}_{1l}\mathrm{<mo\; stretchy="false">)</mo>}-E\mathrm{<mo\; stretchy="false">(</mo>}{I}_{1k}\mathrm{<mo\; stretchy="false">)</mo>}E\mathrm{<mo\; stretchy="false">(</mo>}{I}_{1l}\mathrm{<mo\; stretchy="false">)</mo><mo\; stretchy="false">]</mo>}/{\pi }_{1k}{\pi }_{1l}=\mathrm{<mo\; stretchy="false">(</mo>}{\pi }_{1kl}-{\pi }_{1k}{\pi }_{1l}\mathrm{<mo\; stretchy="false">)</mo>}/{\pi }_{1k}{\pi }_{1l}.$$

The $k{l}^{\text{th}}$ element of the matrix $\text{Var}\mathrm{<mo\; stretchy="false">(</mo>}{w}_U\mathrm{<mo\; stretchy="false">)</mo>}$ is

$$\begin{array}{ll}\text{Cov}\mathrm{<mo\; stretchy="false">(</mo>}{w}_{U_k}\mathrm{,}{w}_{U_l}\mathrm{<mo\; stretchy="false">)</mo>}\hfill & \mathrm{<mo>=</mo>}\mathrm{<mo\; stretchy="false">[</mo>}E\mathrm{<mo\; stretchy="false">(</mo>}{I}_{1k}{I}_{2k}{I}_{1l}{I}_{2l}\mathrm{<mo\; stretchy="false">)</mo>}-E\mathrm{<mo\; stretchy="false">(</mo>}{I}_{1k}{I}_{2k}\mathrm{<mo\; stretchy="false">)</mo>}E\mathrm{<mo\; stretchy="false">(</mo>}{I}_{1l}{I}_{2l}\mathrm{<mo\; stretchy="false">)</mo><mo\; stretchy="false">]</mo>}/{\pi }_{1k}{\pi }_{2k}{\pi }_{1l}{\pi }_{2l}\hfill \\ \hfill & \mathrm{<mo>=</mo>}\mathrm{<mo\; stretchy="false">[</mo>}{E}_1\mathrm{<mo\; stretchy="false">(</mo>}{I}_{1k}{I}_{1l}{E}_2\mathrm{<mo\; stretchy="false">(</mo>}{I}_{2k}{I}_{2l}\mathrm{<mo\; stretchy="false">)</mo><mo\; stretchy="false">)</mo>}-E_1\mathrm{<mo\; stretchy="false">(</mo>}{I}_{1k}{E}_2\mathrm{<mo\; stretchy="false">(</mo>}{I}_{2k}\mathrm{<mo\; stretchy="false">)</mo><mo\; stretchy="false">)</mo>}{E}_1\mathrm{<mo\; stretchy="false">(</mo>}{I}_{1l}{E}_2\mathrm{<mo\; stretchy="false">(</mo>}{I}_{2l}\mathrm{<mo\; stretchy="false">)</mo><mo\; stretchy="false">)</mo><mo\; stretchy="false">]</mo>}/{\pi }_{1k}{\pi }_{2k}{\pi }_{1l}{\pi }_{2l}\hfill \\ \hfill & \mathrm{<mo>=</mo>}\mathrm{<mo\; stretchy="false">[</mo>}{\pi }_{1kl}{\pi }_{2kl}-{\pi }_{1k}{\pi }_{2k}{\pi }_{1l}{\pi }_{2l}\mathrm{<mo\; stretchy="false">]</mo>}/{\pi }_{1k}{\pi }_{2k}{\pi }_{1l}{\pi }_{2l}\mathrm{,}\hfill \end{array}$$

where $E_1$ and $E_2$ denote expectation under first and second phase of sampling, respectively. Using similar arguments it follows that the $k{l}^{\text{th}}$ element of the matrix $\text{Cov}\mathrm{<mo\; stretchy="false">(</mo>}{w}_{1U}\mathrm{,}{w}_U\mathrm{<mo\; stretchy="false">)</mo>}$ is

$$\text{Cov}\mathrm{<mo\; stretchy="false">(</mo>}{w}_{1U_k}\mathrm{,}{w}_{U_l}\mathrm{<mo\; stretchy="false">)</mo>}\mathrm{<mo>=</mo>}[E\mathrm{<mo\; stretchy="false">(</mo>}{I}_{1k}{I}_{1l}{I}_{2l}\mathrm{<mo\; stretchy="false">)</mo>}-E\mathrm{<mo\; stretchy="false">(</mo>}{I}_{1k}\mathrm{<mo\; stretchy="false">)</mo>}E\mathrm{<mo\; stretchy="false">(</mo>}{I}_{1l}{I}_{2l}\mathrm{<mo\; stretchy="false">)</mo>}]/{\pi }_{1k}{\pi }_{1l}{\pi }_{2l}\mathrm{<mo>=</mo>}\mathrm{<mo\; stretchy="false">(</mo>}{\pi }_{1kl}-{\pi }_{1k}{\pi }_{1l}\mathrm{<mo\; stretchy="false">)</mo>}/{\pi }_{1k}{\pi }_{1l}.$$

This shows that $\text{Cov}\mathrm{<mo\; stretchy="false">(</mo>}{w}_{1U_k}\mathrm{,}{w}_{U_l}\mathrm{<mo\; stretchy="false">)</mo>}\mathrm{<mo>=</mo>}\text{Cov}\mathrm{<mo\; stretchy="false">(</mo>}{w}_{1U_k}\mathrm{,}{w}_{1U_l}\mathrm{<mo\; stretchy="false">)</mo>}$ and thus $\text{Cov}\mathrm{<mo\; stretchy="false">(</mo>}{w}_{1U}\mathrm{,}{w}_U\mathrm{<mo\; stretchy="false">)</mo>}\mathrm{<mo>=</mo>}\text{Var}\mathrm{<mo\; stretchy="false">(</mo>}{w}_{1U}\mathrm{<mo\; stretchy="false">)</mo>},$ which completes the proof.

Proof of Theorem 1

Matrix $\Delta =\text{Var}\mathrm{<mo\; stretchy="false">(</mo>}{w}_U^{*}\mathrm{<mo\; stretchy="false">)</mo>}$ is nonsingular if and only if $\text{Var}\mathrm{<mo\; stretchy="false">(</mo>}{w}_U\mathrm{<mo\; stretchy="false">)</mo>}-\text{Var}\mathrm{<mo\; stretchy="false">(</mo>}{w}_{1U}\mathrm{<mo\; stretchy="false">)</mo>}$ is nonsingular. This follows from a general result on inverses of partitioned matrices (see Harville, 2008, page 98). But $\text{Var}\mathrm{<mo\; stretchy="false">(</mo>}{w}_U\mathrm{<mo\; stretchy="false">)</mo>}-\text{Var}\mathrm{<mo\; stretchy="false">(</mo>}{w}_{1U}\mathrm{<mo\; stretchy="false">)</mo>}\mathrm{<mo>=</mo>}\text{Var}\mathrm{<mo\; stretchy="false">(</mo>}{w}_{1U}-w_U\mathrm{<mo\; stretchy="false">)</mo>},$ because $\text{Cov}\mathrm{<mo\; stretchy="false">(</mo>}{w}_{1U}\mathrm{,}{w}_U\mathrm{<mo\; stretchy="false">)</mo>}\mathrm{<mo>=</mo>}\text{Var}\mathrm{<mo\; stretchy="false">(</mo>}{w}_{1U}\mathrm{<mo\; stretchy="false">)</mo>},$ and therefore $\text{Var}\mathrm{<mo\; stretchy="false">(</mo>}{w}_U\mathrm{<mo\; stretchy="false">)</mo>}-\text{Var}\mathrm{<mo\; stretchy="false">(</mo>}{w}_{1U}\mathrm{<mo\; stretchy="false">)</mo>}$ is nonsingular, being a variance-covariance matrix. Next, to find the vector $c_U^{*}$ that minimizes $\mathrm{<mo\; stretchy="false">(</mo>}{c}_U^{*}-w_U^{*}{)}^{\prime }{\Delta }^{-1}\mathrm{<mo\; stretchy="false">(</mo>}{c}_U^{*}-w_U^{*}\mathrm{<mo\; stretchy="false">)</mo>}$ subject to the constraints ${\text{X\hspace{0.17em}}}_U^{\prime }{c}_U^{*}=t_{\text{X}},$  consider the function $F={\mathrm{<mo\; stretchy="false">(</mo>}{c}_U^{*}-w_U^{*}\mathrm{<mo\; stretchy="false">)</mo>}}^{\prime }{\Delta }^{-1}\mathrm{<mo\; stretchy="false">(</mo>}{c}_U^{*}-w_U^{*}\mathrm{<mo\; stretchy="false">)</mo>}-{\lambda }^{\prime }{\text{X\hspace{0.17em}}}_U^{\prime }{c}_U^{*}$  where $\lambda$ is a vector of Langrange multipliers. We then get the system of equations

$$\begin{array}{rr}\hfill \frac{\partial F}{\partial c_U^{*}}\mathrm{<mo>=</mo>}2{\Delta }^{-1}\mathrm{<mo\; stretchy="false">(</mo>}{c}_U^{*}-w_U^{*}\mathrm{<mo\; stretchy="false">)</mo>}-{\text{X}}_U\lambda & \hfill \mathrm{<mo>=</mo>}0\\ \hfill {\text{X\hspace{0.17em}}}_U^{\prime }{c}_U^{*}-t_{\text{X}}& \hfill =0.\end{array}$$

Multiplying the first equation by ${\text{X\hspace{0.17em}}}_U^{\prime }\Delta ,$  using ${\text{X\hspace{0.17em}}}_U^{\prime }{c}_U^{*}=t_{\text{X}}$  and solving for $\lambda$ gives $\lambda \mathrm{<mo>=</mo>}2{\mathrm{<mo\; stretchy="false">(</mo>}{\text{X\hspace{0.17em}}}_U^{\prime }\Delta {\text{X}}_U\mathrm{<mo\; stretchy="false">)</mo>}}^{-1}\left(t_{\text{X}}-{\text{X\hspace{0.17em}}}_U^{\prime }{w}_U^{*}\right).$  Inserting this into the first equation and solving for $c_U^{*}$ gives $c_U^{*}=w_U^{*}+\Delta {\text{X}}_U{\mathrm{<mo\; stretchy="false">(</mo>}{\text{X\hspace{0.17em}}}_U^{\prime }\Delta {\text{X}}_U\mathrm{<mo\; stretchy="false">)</mo>}}^{-1}\left(t_{\text{X}}-{\text{X\hspace{0.17em}}}_U^{\prime }{w}_U^{*}\right).$

Proof of Proposition 1

Clearly, the coefficients of ${\widehat{t}}_x-{\tilde{t}}_x$ in (3.14) and (6.2) are identical if ${\Delta }_1=\delta {\Delta }_2.$ Next, using the partition $X_U=\mathrm{<mo\; stretchy="false">(</mo>}{X}_{1U},X_{2U}),$ the coefficient of $t_{x_1}-{\widehat{t}}_{x_1}$ in (6.2) is expressed as follows. First we obtain

$$\begin{array}{ll}{Y}_U^{\prime }{\Delta }_2{X}_U{\left(X_U^{\prime }{\Delta }_2{X}_U\right)}^{-1}{X}_U^{\prime }{\Delta }_1{X}_{1U}\hfill & =Y_U^{\prime }{\Delta }_2\mathrm{<mo\; stretchy="false">(</mo>}{X}_{1U}\mathrm{,}{X}_{2U}\mathrm{<mo\; stretchy="false">)</mo>}{\left(\begin{array}{cc}{X}_{1U}^{\prime }{\Delta }_2{X}_{1U}& X_{1U}^{\prime }{\Delta }_2{X}_{2U}\\ X_{2U}^{\prime }{\Delta }_2{X}_{1U}& X_{2U}^{\prime }{\Delta }_2{X}_{2U}\end{array}\right)}^{-1}\left(\begin{array}{l}{X}_{1U}^{\prime }{\Delta }_1{X}_{1U}\hfill \\ X_{2U}^{\prime }{\Delta }_1{X}_{1U}\hfill \end{array}\right)\hfill \\ \hfill & =Y_U^{\prime }{\Delta }_2{X}_{1U}\left[A_{11}\mathrm{<mo\; stretchy="false">(</mo>}{X}_{1U}^{\prime }{\Delta }_1{X}_{1U}\mathrm{<mo\; stretchy="false">)</mo>}+A_{12}\mathrm{<mo\; stretchy="false">(</mo>}{X}_{2U}^{\prime }{\Delta }_1{X}_{1U}\mathrm{<mo\; stretchy="false">)</mo>}\right]\hfill \\ \hfill & +Y_U^{\prime }{\Delta }_2{X}_{2U}\left[A_{21}\mathrm{<mo\; stretchy="false">(</mo>}{X}_{1U}^{\prime }{\Delta }_1{X}_{1U}\mathrm{<mo\; stretchy="false">)</mo>}+A_{22}\mathrm{<mo\; stretchy="false">(</mo>}{X}_{2U}^{\prime }{\Delta }_1{X}_{1U}\mathrm{<mo\; stretchy="false">)</mo>}\right]\mathrm{,}\hfill \end{array}$$

where $A_{11},$ $A_{12},$ $A_{21},$ $A_{22}$ are derived by algebra of inverses of partitioned matrices. In particular,

$A_{11}\mathrm{<mo>=</mo>}{\mathrm{<mo\; stretchy="false">(</mo>}{X}_{1U}^{\prime }{\Delta }_2{X}_{1U}\mathrm{<mo\; stretchy="false">)</mo>}}^{-1}-A_{12}\mathrm{<mo\; stretchy="false">(</mo>}{X}_{2U}^{\prime }{\Delta }_2{X}_{1U}\mathrm{<mo\; stretchy="false">)</mo>}{\mathrm{<mo\; stretchy="false">(</mo>}{X}_{1U}^{\prime }{\Delta }_2{X}_{1U}\mathrm{<mo\; stretchy="false">)</mo>}}^{-1}$

and $A_{21}=-A_{22}\mathrm{<mo\; stretchy="false">(</mo>}{X}_{2U}^{\prime }{\Delta }_2{X}_{1U}\mathrm{<mo\; stretchy="false">)</mo>}\times {\mathrm{<mo\; stretchy="false">(</mo>}{X}_{1U}^{\prime }{\Delta }_2{X}_{1U}\mathrm{<mo\; stretchy="false">)</mo>}}^{-1}.$  Then,

$$\begin{array}{ll}{A}_{11}\mathrm{<mo\; stretchy="false">(</mo>}{X}_{1U}^{\prime }{\Delta }_1{X}_{1U}\mathrm{<mo\; stretchy="false">)</mo>}+A_{12}\mathrm{<mo\; stretchy="false">(</mo>}{X}_{2U}^{\prime }{\Delta }_1{X}_{1U}\mathrm{<mo\; stretchy="false">)</mo>}=\hfill & {\mathrm{<mo\; stretchy="false">(</mo>}{X}_{1U}^{\prime }{\Delta }_2{X}_{1U}\mathrm{<mo\; stretchy="false">)</mo>}}^{-1}{X}_{1U}^{\prime }{\Delta }_1{X}_{1U}+A_{12}B\hfill \\ A_{21}\mathrm{<mo\; stretchy="false">(</mo>}{X}_{1U}^{\prime }{\Delta }_1{X}_{1U}\mathrm{<mo\; stretchy="false">)</mo>}+A_{22}\mathrm{<mo\; stretchy="false">(</mo>}{X}_{2U}^{\prime }{\Delta }_1{X}_{1U}\mathrm{<mo\; stretchy="false">)</mo>}\mathrm{<mo>=</mo>}\hfill & A_{22}B\mathrm{,}\hfill \end{array}$$

where

$$B=X_{2U}^{\prime }{\Delta }_1{X}_{1U}-X_{2U}^{\prime }{\Delta }_2{X}_{1U}{\mathrm{<mo\; stretchy="false">(</mo>}{X}_{1U}^{\prime }{\Delta }_2{X}_{1U}\mathrm{<mo\; stretchy="false">)</mo>}}^{-1}\mathrm{<mo\; stretchy="false">(</mo>}{X}_{1U}^{\prime }{\Delta }_1{X}_{1U}\mathrm{<mo\; stretchy="false">)</mo>.}$$

It is then easy to verify that if ${\Delta }_1=\delta {\Delta }_2,$ we have ${\mathrm{<mo\; stretchy="false">(</mo>}{X}_{1U}^{\prime }{\Delta }_2{X}_{1U}\mathrm{<mo\; stretchy="false">)</mo>}}^{-1}{X}_{1U}^{\prime }{\Delta }_1{X}_{1U}=\delta I$  and $B=0.$ It follows that $Y_U^{\prime }{\Delta }_2{X}_U{\left(X_U^{\prime }{\Delta }_2{X}_U\right)}^{-1}{X}_U^{\prime }{\Delta }_1{X}_{1U}=Y_U^{\prime }{\Delta }_1{X}_{1U},$  and thus the coefficients of $t_{x_1}-{\widehat{t}}_{x_1}$ in (3.14) and (6.2) are also identical.

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