6. Discussion

David G. Steel and Robert Graham Clark

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Incorporating unequal unit costs can improve the efficiency of sample designs. For the gains to be appreciable, the unit costs need to vary considerably. Even with no estimation error, a coefficient of variation of 50% may lead to a gain of only 6% in the anticipated variance. When this coefficient of variation is 75%, as can happen in a mixed mode survey, the reduction in the anticipated variance (or in the sample size for fixed precision) can be over 12%. Costs will be estimated with some error and this reduces the gain by a factor determined by the relative variation of the relative errors in estimating the costs at the individual level.

Appendix

A.1 Detailed derivations

Lemma 1: Let $u_i$ be defined for $i\in U.$ Let $u_i=\overline{u}+\theta e_i,$ where ${\displaystyle {\sum }_{i\in U}{e}_i}=0$ and $\theta$ is small. Then:

1. $\overline{\sqrt{u}}=\sqrt{\overline{u}}-\frac{1}{8}{\theta }^2{\overline{u}}^{-3/2}{S}_e^2+o\left({\theta }^2\right).$
2. $S_{\sqrt{u}}^2=\frac{1}{4}{\theta }^2{\overline{u}}^{-1}{S}_e^2+o\left({\theta }^2\right)=\frac{1}{4}{\overline{u}}^{-1}{S}_u^2+o\left({\theta }^2\right).$
3. $N^{-2}\left({\displaystyle {\sum }_{i\in U}{u}_i^{1/2}}\right)\left({\displaystyle {\sum }_{i\in U}{u}_i^{-1/2}}\right)=1+\frac{1}{4}{\theta }^2{\overline{u}}^{-2}{S}_e^2+o\left({\theta }^2\right)=1+\frac{1}{4}{C}_u^2+o\left({\theta }^2\right).$
4. $C_{\sqrt{u}}^2=\frac{1}{4}{\theta }^2{\overline{u}}^{-2}{S}_e^2+o\left({\theta }^2\right)=\frac{1}{4}{C}_u^2+o\left({\theta }^2\right).$

The notation $o\left(C_u^2\right)$ can be used in place of $o\left({\theta }^2\right),$ since $C_u^2={\theta }^2{C}_e^2.$ This will be done in the remainder of the Appendix.

Proof:

We start by writing $\overline{\sqrt{u}}$ as a function of $\theta :$

$$\overline{\sqrt{u}}=N^{-1}{\displaystyle \sum _{i\in U}}\sqrt{u_i}=N^{-1}{\displaystyle \sum _{i\in U}}\sqrt{\overline{u}+\theta e_i}.$$

Call this $g\left(\theta \right),$ then differentiating about $\theta =0$ gives $g(0)=\sqrt{\overline{u}},$ $g^{\prime }(0)=0$ and

$$g″(0)=-\frac{1}{4}{N}^{-1}{\overline{u}}^{-3/2}{\displaystyle \sum _{i\in U}}{e}_i^2=-\frac{1}{4}{\overline{u}}^{-3/2}{S}_e^2\mathrm{.}$$

Hence

$$\overline{\sqrt{u}}=g\left(\theta \right)=g(0)+g\prime (0)\theta +\frac{1}{2}g″(0){\theta }^2+o\left({\theta }^2\right)=\sqrt{\overline{u}}-\frac{1}{8}{\theta }^2{\overline{u}}^{-3/2}{S}_e^2+o\left({\theta }^2\right)$$

which is result a.

Result b is proven using result a:

$$\begin{array}{cc}{S}_{\sqrt{u}}^2& =N^{-1}{\displaystyle \sum _{i\in U}}{\left(\sqrt{u_i}\right)}^2-{\left(N^{-1}{\displaystyle \sum _{i\in U}}\sqrt{u_i}\right)}^2\hfill \\ & =\overline{u}-{\left(\overline{\sqrt{u}}\right)}^2\hfill \\ & =\overline{u}-{\left(\sqrt{\overline{u}}-\frac{1}{8}{\theta }^2{\overline{u}}^{-3/2}{S}_e^2+o\left({\theta }^2\right)\right)}^2\hfill \\ & =\overline{u}-\left(\overline{u}+\frac{1}{64}{\theta }^4{\overline{u}}^{-3}{S}_e^4-\frac{1}{4}{\theta }^2{\overline{u}}^{-1}{S}_e^2+o\left({\theta }^2\right)\right)\hfill \\ & =\frac{1}{4}{\theta }^2{\overline{u}}^{-1}{S}_e^2+o\left({\theta }^2\right)\hfill \\ & =\frac{1}{4}{\overline{u}}^{-1}{S}_u^2+o\left({\theta }^2\right).\hfill \end{array}$$

To derive c, we firstly write $N^{-1}{\displaystyle {\sum }_{i\in U}{u}_i^{-1/2}}$ as a function $g()$ of $\theta$ and take a Taylor Series expansion:

$$\begin{array}{cc}{N}^{-1}{\displaystyle \sum _{i\in U}}{u}_i^{-1/2}& =N^{-1}{\displaystyle \sum _{i\in U}}{\left(\overline{u}+\theta e_i\right)}^{-1/2}\hfill \\ & =g\left(\theta \right)=g(0)+g^{\prime }(0)\theta +\frac{1}{2}g″(0){\theta }^2+o\left({\theta }^2\right)(\text{A}.1)\hfill \\ & ={\overline{u}}^{-1/2}+0\theta +\frac{1}{2}\frac{3}{4}{\overline{u}}^{-5/2}{N}^{-1}{\displaystyle \sum _{i\in U}}{e}_i^2{\theta }^2+o\left({\theta }^2\right)\hfill \\ & ={\overline{u}}^{-1/2}+\frac{3}{8}{\overline{u}}^{-5/2}{S}_e^2{\theta }^2+o\left({\theta }^2\right)\hfill \end{array}$$

Note that $N^{-1}{\displaystyle {\sum }_{i\in U}{u}_i^{1/2}}=\overline{\sqrt{u}}.$ Multiplying the expression for $\overline{\sqrt{u}}$ in result a and (A.1) gives

$$\begin{array}{cc}{N}^{-2}\left({\displaystyle \sum _{i\in U}}{u}_i^{1/2}\right)\left({\displaystyle \sum _{i\in U}}{u}_i^{-1/2}\right)& =\left\{\sqrt{\overline{u}}-\frac{1}{8}{\theta }^2{\overline{u}}^{-3/2}{S}_e^2+o\left({\theta }^2\right)\right\}\left\{{\overline{u}}^{-1/2}+\frac{3}{8}{\overline{u}}^{-5/2}{S}_e^2{\theta }^2+o\left({\theta }^2\right)\right\}\hfill \\ & =1+\frac{1}{4}{\overline{u}}^{-2}{S}_e^2{\theta }^2+o\left({\theta }^2\right)\hfill \\ & =1+\frac{1}{4}{C}_u^2+o\left({\theta }^2\right)\hfill \end{array}$$

which is result c.

For result d, firstly note that $\overline{\sqrt{u}}=\sqrt{\overline{u}}+o\left(\theta \right)$ from result a, and so, from a first order Taylor Series,

$${\left(\overline{\sqrt{u}}\right)}^{-2}={\left(\sqrt{\overline{u}}\right)}^{-2}+o\left(\theta \right)={\overline{u}}^{-1}+o\left(\theta \right)\mathrm{.}$$

Combining this with result b, we obtain

$$\begin{array}{cc}{C}_{\sqrt{u}}^2& =S_{\sqrt{u}}^2{\left(\overline{\sqrt{u}}\right)}^{-2}\hfill \\ & =\left\{\frac{1}{4}{\theta }^2{\overline{u}}^{-1}{S}_e^2+o\left({\theta }^2\right)\right\}\left\{{\overline{u}}^{-1}+o\left(\theta \right)\right\}\hfill \\ & =\frac{1}{4}{\theta }^2{\overline{u}}^{-2}{S}_e^2+o\left({\theta }^2\right)\hfill \\ & =\frac{1}{4}{C}_u^2+o\left({\theta }^2\right)\hfill \end{array}$$

giving result d.

Derivation of (3.3)

For the special case where $u_i=v_i,$ (2.5) becomes

$${\displaystyle \sum _{i\in U}}{u}_i^2=N{\overline{u}}^2\left(1+C_u^2\right)\mathrm{.}(\text{A}.2)$$

Applying (2.5),

$${\displaystyle \sum _{i\in U}}{c}_i^{1/2}{z}_i^{1/2}=N\overline{\sqrt{c}}\overline{\sqrt{z}}\left(1+C_{\sqrt{c}\mathrm{,}\sqrt{z}}\right)(\text{A}.3)$$

where $\overline{\sqrt{c}}=N^{-1}{\displaystyle {\sum }_{i\in U}\sqrt{c_i}}$ and $\overline{\sqrt{z}}=N^{-1}{\displaystyle {\sum }_{i\in U}}\sqrt{z_i}.$ Using (A.2), we can express $\overline{\sqrt{c}}$ in terms of $\overline{c}:$

$$\overline{c}=N^{-1}{\displaystyle \sum _{i\in U}}{c}_i=N^{-1}{\displaystyle \sum _{i\in U}}{\left(\sqrt{c_i}\right)}^2={\left(\overline{\sqrt{c}}\right)}^2\left(1+C_{\sqrt{c}}^2\right)\mathrm{.}(\text{A}.4)$$

Similarly,

$$\overline{z}={\left(\overline{\sqrt{z}}\right)}^2\left(1+C_{\sqrt{z}}^2\right)\mathrm{.}(\text{A}.5)$$

Assuming the last term of (3.2) is negligible, applying (A.3), (A.4) and (A.5) gives (3.3).

Derivation of (3.4)

Lemma 1d implies that $C_{\sqrt{c}}^2=\left(1/4\right)C_c^2+o\left(C_c^2\right)\approx \left(1/4\right)C_c^2$ and $C_{\sqrt{z}}^2=\left(1/4\right)C_z^2+o\left(C_z^2\right)\approx \left(1/4\right)C_z^2.$ Result (3.4) follows from (3.3) by using these approximations, as well as assuming that $C_{\sqrt{c}\mathrm{,}\sqrt{z}}=0.$

Derivation of (3.7)

Firstly, ${\displaystyle {\sum }_{i\in U}{c}_i{z}_i^{1/2}}=N\overline{c}\overline{\sqrt{z}}\left(1+C_{c\mathrm{,}\sqrt{z}}\right),$ from (2.5), where $C_{c\mathrm{,}\sqrt{z}}$ is the population relative covariance between the values of $z_i^{1/2}$ and $c_i.$ It is assumed that the values of $c_i$ and $z_i$ are unrelated, so that $C_{c\mathrm{,}\sqrt{z}}=0.$ It is also assumed that the second term of (3.6) is negligible, corresponding to small sampling fraction. Hence (3.6) becomes:

$$A{V}_{nocosts}={\sigma }^2{N}^2{C}_f^{-1}\overline{c}{\left(\overline{\sqrt{z}}\right)}^2\mathrm{.}(\text{A}.6)$$

From (A.5), and Lemma 1d, we have

$${\left(\overline{\sqrt{z}}\right)}^2=\frac{\overline{z}}{1+C_{\sqrt{z}}^2}\approx \frac{\overline{z}}{1+\left(1/4\right)C_z^2}.$$

Substituting into (A.6) gives (3.7).

Derivation of (4.2)

Two terms in (4.1) will be simplified using (2.5). Firstly,

$$\begin{array}{cc}{\displaystyle \sum _{i\in U}}{\widehat{c}}_i^{1/2}{z}_i^{1/2}& ={\displaystyle \sum _{i\in U}}{b}_i^{1/2}{c}_i^{1/2}{z}_i^{1/2}\hfill \\ & =N\left(N^{-1}{\displaystyle \sum _{i\in U}}{b}_i^{1/2}\right)\left(N^{-1}{\displaystyle \sum _{i\in U}}{c}_i^{1/2}{z}_i^{1/2}\right)+C_{\sqrt{b}\mathrm{,}\sqrt{cz}}(\text{A}.7)\hfill \end{array}$$

where $C_{\sqrt{b}\mathrm{,}\sqrt{cz}}$ is the covariance between the population values of $b_i^{1/2}$ and $c_i^{1/2}{z}_i^{1/2}.$ Secondly,

$$\begin{array}{cc}{\displaystyle \sum _{i\in U}}{z}_i^{1/2}{\widehat{c}}_i^{-1/2}{c}_i& ={\displaystyle \sum _{i\in U}}{b}_i^{-1/2}{c}_i^{1/2}{z}_i^{1/2}\hfill \\ & =N\left(N^{-1}{\displaystyle \sum _{i\in U}}{b}_i^{-1/2}\right)\left(N^{-1}{\displaystyle \sum _{i\in U}}{c}_i^{1/2}{z}_i^{1/2}\right)+C_{1/\sqrt{b}\mathrm{,}\sqrt{cz}}(\text{A}.8)\hfill \end{array}$$

where $C_{1/\sqrt{b}\mathrm{,}\sqrt{cz}}$ is the covariance between the population values of $b_i^{-1/2}$ and $c_i^{1/2}{z}_i^{1/2}.$

If we assume that the population values of $b_i$ are unrelated to the values of $c_i$ and $z_i,$ so that $C_{\sqrt{b}\mathrm{,}\sqrt{cz}}=C_{1/\sqrt{b}\mathrm{,}\sqrt{cz}}=0,$ and subsitute (A.7) and (A.8) into (4.1), then we obtain (4.2).

Derivation of (4.3)

We can express (4.2) in terms of $A{V}_{opt}$ which is defined in (3.2), assuming the last term of (3.2) is negligible, corresponding to small sampling fraction:

$$A{V}_{ests}\approx A{V}_{opt}{N}^{-2}{\displaystyle \sum _{i\in U}}{b}_i^{-1/2}{\displaystyle \sum _{i\in U}}{b}_i^{1/2}(\text{A}.9)$$

Lemma 1c implies that

$$N^{-2}{\displaystyle \sum _{i\in U}}{b}_i^{-1/2}{\displaystyle \sum }{b}_i^{1/2}=1+\frac{1}{4}{C}_b^2+o\left(C_b^2\right)\approx 1+\frac{1}{4}{C}_b^2\mathrm{.}$$

Substituting this, and (3.3), into (A.9) gives (4.3).

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