Reducing the response imbalance: Is the accuracy of the survey estimates improved? Section 2. Background and notationReducing the response imbalance: Is the accuracy of the survey estimates improved? Section 2. Background and notation

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A probability sample $s$ is drawn from the finite population $U=\left\{1,2,\dots ,k,\dots ,N\right\}.$ Unit $k$ has the known inclusion probability ${\pi }_k=\mathrm{Pr}\left(k\in s\right)$ and the known design weight $d_k=1/{\pi }_k.$ Nonresponse occurs. The response set, denoted $r,$ is that subset of $s$ for which the study variable is observed. We do not know how $r$ was generated from $s;$ the response probabilities are unknown (if assumed to “exist”, they are not needed in this article). The (design weighted) response rate is

$$P={\displaystyle {\sum }_r{d}_k}/{\displaystyle {\sum }_s{d}_k}.(2.1)$$

If $A$ is a set of units, $A\subseteq U,$ a sum ${\displaystyle {\sum }_{k\in A}\text{}}$ will be written as ${\displaystyle {\sum }_A\text{}}.$ The survey may have many study variables. A typical one, denoted $y$ (continuous or categorical), has value $y_k$ recorded for $k\in r$ but missing for $k\in s-r.$ Our objective is to estimate the population $y‑$total, $Y={\displaystyle {\sum }_U{y}_k}.$ The response indicator $I$ has value $I_k=1$ for $k\in r,$ $I_k=0$ for $k\in s-r.$ A goal for practice is to get a response $r$ that is well balanced, in the sense specified later. We are led to consider the different $r$ that may arise from a given $s.$

The auxiliary vector $x$ of dimension $J\ge 1$ has value $x_k$ known at least for all units $k\in s.$ Auxiliary information can be used in the data collection (for monitoring the data inflow to achieve improved balance) and/or in the estimation (for calibrated weight computation). The auxiliary vector need not be the same for the two purposes, but this article assumes that they agree, and that the $x‑$information used is for $k\in s.$ This includes the important case of paradata, that is, data about the data collection process.

An important type of auxiliary vector is a group vector. It identifies membership of every unit $k$ in one of $J$ mutually exclusive and exhaustive sample groups, so that $x_k={\left(0,\dots ,1,\dots 0\right)}^{\prime },$ where the only “1” indicates the unique group (out of $J$ possible) to which $k$ belongs.

A group vector occurs when several categorical auxiliary variables are completely crossed. To illustrate, if $x=\left(sex\times education\times age\right)$ represents a crossing of 2 sexes, 3 exhaustive education categories and 4 exhaustive age categories, then $x$ is a group vector with dimension $J=2\times 3\times 4=24$ and equally many possible values $x_k.$ When several categorical variables are used although not in completely crossed manner $–$ important for practice in statistical agencies $–$ then the dimension $J$ of $x_k$ can be kept relatively modest (say less than 15) while still coding a much larger number (say more than one hundred) of possible properties $x_k$ of the units $k.$ For a study of the Swedish Living Conditions Survey, Särndal and Lundquist (2014a) used an $x‑$vector of dimension 14 with 256 possible values. The group vector case and the non-group vector case give important differences in the results that follow.

All auxiliary vectors used here satisfy a requirement that grants mathematical convenience without severely restricting the choice of vector: There exists a constant vector $\mu$ such that

$${\mu }^{\prime }{x}_k=1\text{for}\text{all}k.(2.2)$$

For example, when $J=2$ and $x_k={\left(1,x_k\right)}^{\prime },$ then $\mu ={\left(1,0\right)}^{\prime }$ satisfies the requirement. In the group vector case where $x_k={\left(0,\dots ,1,\dots ,0\right)}^{\prime },$ then $\mu ={\left(1,\dots ,1,\dots ,1\right)}^{\prime }$ satisfies the requirement. If $x$ is not a group vector, say, one used to code “education” with three mutually exclusive and exhaustive categories and “gender” as a univariate variable equal to 1 or 0, then $J=3+1=4$ (education and gender not crossed), and $\mu ={\left(1,1,1,0\right)}^{\prime }$ satisfies the requirement.

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