Statistical matching using fractional imputation 4. Split questionnaire survey designStatistical matching using fractional imputation 4. Split questionnaire survey design

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In Section 3, we consider the situation where Sample A and Sample B are two independent samples from the same target population. We now consider another situation of a split questionnaire design where the original sample $S$ is selected from a target population and then Sample A and Sample B are randomly chosen such that $A\cup B\mathrm{<mo>=</mo>}S$ and $A\cap B\mathrm{<mo>=</mo>}\varphi .$ We observe $\left(x\mathrm{,}{y}_1\right)$ from Sample A and observe $\left(x\mathrm{,}{y}_2\right)$ from Sample B. We are interested in creating fully augmented data with observation $\left(x\mathrm{,}{y}_1\mathrm{,}{y}_2\right)$ in $S.$

Such split questionnaire survey designs are gaining popularity because they reduce response burden (Raghunathan and Grizzle 1995; Chipperfield and Steel 2009). Split questionnaire designs have been investigated, for example, for the Consumer Expenditure survey (Gonzalez and Eltinge 2008) and the National Assessment of Educational Progress (NAEP) survey in the US. In applications of split-questionnaire designs, analysts may be interested in multiple parameters such as the mean of $y_1$ and the mean of $y_2,$ in addition to the coefficient in the regression of $y_2$ on $y_1.$

We consider a design where the original Sample $S$ is partitioned into two subsamples: $A$ and $B.$ We assume that $x_i$ is observed for $i\in S,$ $y_{1i}$ is collected for $i\in A$ and $y_{2i}$ is collected for $i\in B.$ The probability of selection into $A$ or $B$ may depend on $x_i$ but does not depend on $y_{1i}$ or $y_{2i}.$ As a consequence, the design used to select subsample $A$ or $B$ is non-informative for the specified model (Fuller 2009, Chapter 6). We let $w_i$ denote the sampling weight associated with the full sample $S.$ We assume a procedure is available for estimating the variance of an estimator of the form $\widehat{Y}\mathrm{<mo>=</mo>}{\displaystyle {\sum }_{i\in S}}{w}_i{y}_i,$ and we denote the variance estimator by ${\widehat{V}}_s\left({\displaystyle {\sum }_{i\in S}}{w}_i{y}_i\right).$

A procedure for obtaining a fully imputed data set is as follows. First, use the procedure of Section 3 to obtain imputed values $\left\{y_{1i}^{\mathrm{<mo>*</mo>}\left(j\right)}\mathrm{<mo>:</mo>}i\in B\mathrm{,}j\mathrm{<mo>=</mo>\; 1,}\dots \mathrm{,}m\right\}$ and an estimate, $\widehat{\theta },$ of the parameter in the distribution $f\left(y_2|y_1\mathrm{,}x\mathrm{<mo>;</mo>}\theta \right).$ The estimate $\widehat{\theta }$ is obtained by solving

$${\displaystyle \sum _{i\in B}}{w}_i{\displaystyle \sum _{j\mathrm{<mo>=</mo>\; 1}}^m}{w}_{ij}^{*}{S}_2\left(\theta \mathrm{<mo>;</mo>}{x}_i\mathrm{,}{y}_{1i}^{\mathrm{<mo>*</mo>}\left(j\right)}\mathrm{,}{y}_{2i}\right)\mathrm{<mo>=</mo>\; 0,}(4.1)$$

where $S_2\left(\theta \mathrm{<mo>;</mo>}x\mathrm{,}{y}_1\mathrm{,}{y}_2\right)\mathrm{<mo>=</mo>}\partial \log f\left(y_2|y_1\mathrm{,}x\mathrm{<mo>;</mo>}\theta \right)/\partial \theta .$ Given $\widehat{\theta },$ generate imputed values $y_{2i}^{\mathrm{<mo>*</mo>}\left(j\right)}\sim f\left(y_2|y_{1i}\mathrm{,}{x}_i\mathrm{<mo>;</mo>}\widehat{\theta }\right),$ for $i\in A$ and $j\mathrm{<mo>=</mo>\; 1,}\dots \mathrm{,}m.$

Under the assumption that the model is identified, the parameter estimator $\widehat{\theta }$ generated by solving (4.1) is fully efficient in the sense that the imputed value of $y_{2i}$ for Sample A leads to no efficiency gain. To see this, note that the score equation using the imputed value of $y_{2i}$ is computed by

$${\displaystyle \sum _{i\in A}}{w}_i{m}^{-1}{\displaystyle \sum _{j\mathrm{<mo>=</mo>\; 1}}^m}{S}_2\left(\theta \mathrm{<mo>;</mo>}{x}_i\mathrm{,}{y}_{1i}\mathrm{,}{y}_{2i}^{\mathrm{<mo>*</mo>}\left(j\right)}\right)+{\displaystyle \sum _{i\in B}}{w}_i{\displaystyle \sum _{j\mathrm{<mo>=</mo>\; 1}}^m}{w}_{ij}^{*}{S}_2\left(\theta \mathrm{<mo>;</mo>}{x}_i\mathrm{,}{y}_{1i}^{\mathrm{<mo>*</mo>}\left(j\right)}\mathrm{,}{y}_{2i}\right)\mathrm{<mo>=</mo>\; 0.}(4.2)$$

Because $y_{2i}^{\mathrm{<mo>*</mo>}\left(1\right)}\mathrm{,}\dots \mathrm{,}{y}_{2i}^{\mathrm{<mo>*</mo>}\left(m\right)}$ are generated from $f\left(y_2|y_{1i}\mathrm{,}{x}_i\mathrm{<mo>;</mo>}\widehat{\theta }\right),$

$$p\underset{m\to \infty }{{\displaystyle \lim }}{\displaystyle \sum _{i\in A}}{w}_i{m}^{-1}{\displaystyle \sum _{j\mathrm{<mo>=</mo>\; 1}}^m}{S}_2\left(\theta \mathrm{<mo>;</mo>}{x}_i\mathrm{,}{y}_{1i}\mathrm{,}{y}_{2i}^{\mathrm{<mo>*</mo>}\left(j\right)}\right)\mathrm{<mo>=</mo>}{\displaystyle \sum _{i\in A}}{w}_{i}E\left\{S_2\left(\theta \mathrm{<mo>;</mo>}{x}_i\mathrm{,}{y}_{1i}\mathrm{,}{Y}_2\right)|y_{1i}\mathrm{,}{x}_i\mathrm{<mo>;</mo>}\widehat{\theta }\right\}\mathrm{.}$$

Thus, by the property of score function, the first term of (4.2) evaluated at $\theta \mathrm{<mo>=</mo>}\widehat{\theta }$ is close to zero and the solution to (4.2) is essentially the same as the solution to (4.1). That is, there is no efficiency gain in using the imputed value of $y_{2i}$ in computing the MLE for $\theta$ in $f\left(y_2|y_1\mathrm{,}x\mathrm{<mo>;</mo>}\theta \right).$

However, the imputed values of $y_{2i}$ can improve the efficiency of inferences for parameters in the joint distribution of $\left(y_{1i}\mathrm{,}{y}_{2i}\right).$ As a simple example, consider estimation of ${\mu }_2,$ the marginal mean of $y_{2i}.$ Under simple random sampling, the imputed estimator of $\mu \mathrm{<mo>=</mo>}E\left(Y_2\right)$ is

$${\widehat{\mu }}_{I\mathrm{,}m}\mathrm{<mo>=</mo>}\frac{1}{n}\left\{{\displaystyle \sum _{i\in A}}\left(m^{-1}{\displaystyle \sum _{j\mathrm{<mo>=</mo>\; 1}}^m}{y}_{2i}^{\mathrm{<mo>*</mo>}\left(j\right)}\right)+{\displaystyle \sum _{i\in B}}{y}_{2i}\right\},(4.3)$$

where $y_{2i}^{\mathrm{<mo>*</mo>}\left(1\right)}\mathrm{,}\dots \mathrm{,}{y}_{2i}^{\mathrm{<mo>*</mo>}\left(m\right)}$ are generated from $f\left(y_2|y_{1i}\mathrm{,}{x}_i\mathrm{<mo>;</mo>}\widehat{\theta }\right).$ For sufficiently large $m,$ we can write

$$\begin{array}{ll}{\widehat{\mu }}_{I\mathrm{,}\infty }\hfill & \mathrm{<mo>=</mo>}\frac{1}{n}\left\{{\displaystyle \sum _{i\in A}}{\widehat{y}}_{2i}+{\displaystyle \sum _{i\in B}}{y}_{2i}\right\}\hfill \\ \hfill & \mathrm{<mo>=</mo>}\frac{1}{n}\left\{{\displaystyle \sum _{i\in A}}E\left(y_2|y_{1i}\mathrm{,}{x}_i\mathrm{<mo>;</mo>}\widehat{\theta }\right)+{\displaystyle \sum _{i\in B}}{y}_{2i}\right\}\mathrm{.}\hfill \end{array}$$

Under the setup of Example 2.1, we can express ${\widehat{y}}_{2i}\mathrm{<mo>=</mo>}{\widehat{\beta }}_0+{\widehat{\beta }}_1{y}_{1i}+{\widehat{\beta }}_2{x}_{2i}$ where $\left({\widehat{\beta }}_0\mathrm{,}{\widehat{\beta }}_1\mathrm{,}{\widehat{\beta }}_2\right)$ satisfies

$${\displaystyle \sum _{i\in B}}\left(y_{2i}-{\widehat{\beta }}_0-{\widehat{\beta }}_{1i}{\widehat{y}}_{1i}-{\widehat{\beta }}_2{x}_{2i}\right)\mathrm{<mo>=</mo>\; 0}$$

and ${\widehat{y}}_{1i}\mathrm{<mo>=</mo>}{\widehat{\alpha }}_0+{\widehat{\alpha }}_1{x}_{1i}+{\widehat{\alpha }}_2{x}_{2i}$ with $\left({\widehat{\alpha }}_0\mathrm{,}{\widehat{\alpha }}_1\mathrm{,}{\widehat{\alpha }}_2\right)$ satisfying ${\displaystyle {\sum }_{i\in A}}\left(y_{1i}-{\widehat{\alpha }}_0-{\widehat{\alpha }}_1{x}_{1i}-{\widehat{\alpha }}_2{x}_{2i}\right)\mathrm{<mo>=</mo>\; 0.}$ Thus, ignoring the smaller order terms, we have

$$V\left({\widehat{\mu }}_{I\mathrm{,}\infty }\right)\mathrm{<mo>=</mo>}\frac{1}{n}V\left(y_2\right)+\left(\frac{1}{n_b}-\frac{1}{n}\right)V\left(y_2-{\widehat{y}}_2\right)$$

which is smaller than the variance of the direct estimator ${\widehat{\mu }}_b\mathrm{<mo>=</mo>}{n}_b^{-1}{\displaystyle {\sum }_{i\in B}}{y}_{2i}.$

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