Improved Horvitz-Thompson estimator in survey sampling

Section 6. Concluding remarks

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In this paper, we have proposed a novel and simple method to improve the Horvitz-Thompson estimator in survey sampling. Compared with the HT estimator, the proposed IHT estimator improves the estimation accuracy at the expense of introducing a small bias. Empirical studies show that the improvement is substantial. This new idea has also been used to construct an improved ratio estimator. Naturally, applying it to other estimators, such as the regression estimator and the treatment effect estimator, is of interest as well, and this warrants further study.

The choice of the threshold $K$ is important in our method. Although we have suggested an easy algorithm for the choice and have numerically showed that our choice is very close to the optimal one in terms of MSE, it may not be optimal in terms of MSE. How to choose an optimal threshold is a meaningful topic for future research.

Acknowledgements

The author are grateful to the referees, the associate editor and the editor for their meticulous reading of the manuscript and their invaluable comments. Zhu’s work was supported by the National Natural Science Foundation of China (grant nos. 11871459, 71532013 and 71771208). Zou’s work was partially supported by the Ministry of Science and Technology of China (Grant no. 2016YFB0502301) and the National Natural Science Foundation of China (Grant nos. 11529101 and 11331011).

Appendix

A.1  Proof of Theorem 1

To obtain the MSE of the IHT estimator, we first define $I_k\mathrm{<mo>=</mo>\; 1}$ or 0, $k\mathrm{<mo>=</mo>\; 1,}\dots \mathrm{,}N,$ if the $k^{\text{th}}$ unit is drawn or not, then

$$E\left(I_k\right)\mathrm{<mo>=</mo>}{\pi }_k\mathrm{,}\text{Var}\left(I_k\right)\mathrm{<mo>=</mo>}{\Delta }_{kk},\text{Cov}\left(I_k\mathrm{,}{I}_l\right)\mathrm{<mo>=</mo>}{\Delta }_{kl}\text{for}k\ne l,$$

where ${\Delta }_{kk}\mathrm{<mo>=</mo>}{\pi }_k\left(1-{\pi }_k\right)\mathrm{,}{\Delta }_{kl}\mathrm{<mo>=</mo>}{\pi }_{kl}-{\pi }_k{\pi }_l\mathrm{.}$ So the bias of the IHT estimator is

$$\text{Bias}\left({\widehat{t}}_{\text{IHT}}\right)=E\left({{\displaystyle \sum }}_U\frac{y_k}{{\pi }_k^{*}}{I}_k\right)-{{\displaystyle \sum }}_U{y}_k\mathrm{<mo>=</mo>}{{\displaystyle \sum }}_{U_2}\left(\frac{{\pi }_k}{{\pi }_{\left(K\right)}}-1\right)y_k\mathrm{.}(\mathrm{A}.1)$$

The variance of the IHT estimator is given by

$$\begin{array}{ll}\text{Var}\left({\widehat{t}}_{\text{IHT}}\right)\hfill & \mathrm{<mo>=</mo>}\text{Var}\left({\displaystyle {\sum }_s\frac{y_k}{{\pi }_k^{*}}}\right)\mathrm{<mo>=</mo>}\text{Var}\left({{\displaystyle \sum }}_U\frac{y_k}{{\pi }_k^{*}}{I}_k\right)\hfill \\ \hfill & \mathrm{<mo>=</mo>}{{\displaystyle \sum }}_U\left[{\left(\frac{y_k}{{\pi }_k^{*}}\right)}^2\text{Var}\left(I_k\right)\right]+\underset{k\ne l}{{\displaystyle \sum }{{\displaystyle \sum }}_U}\left(\frac{y_k}{{\pi }_k^{*}}\frac{y_l}{{\pi }_l^{*}}\text{Cov}\left(I_k\mathrm{,}{I}_l\right)\right)\hfill \\ \hfill & ={{\displaystyle \sum }}_{U_1}\frac{{\Delta }_{kk}}{{\pi }_k^2}{y}_k^2+{{\displaystyle \sum }}_{U_2}\frac{{\Delta }_{kk}}{{\pi }_{\left(K\right)}^2}{y}_k^2+\underset{k\ne l}{{\displaystyle \sum }{{\displaystyle \sum }}_U}\frac{{\Delta }_{kl}}{{\pi }_k^{*}{\pi }_l^{*}}{y}_k{y}_l\mathrm{.}(\text{A}\text{.2})\hfill \end{array}$$

Combining (A.1) and (A.2), we obtain

$$\begin{array}{ll}\text{MSE}\left({\widehat{t}}_{\text{IHT}}\right)\hfill & ={\text{Bias}}^2\left({\widehat{t}}_{\text{IHT}}\right)+\text{Var}\left({\widehat{t}}_{\text{IHT}}\right)\hfill \\ \hfill & ={\left[{{\displaystyle \sum }}_{U_2}\left(\frac{{\pi }_k}{{\pi }_{\left(K\right)}}-1\right)y_k\right]}^2+{{\displaystyle \sum }}_U\frac{{\Delta }_{kk}}{{\pi }_k^{*2}}{y}_k^2+\underset{k\ne l}{{\displaystyle \sum }{{\displaystyle \sum }}_U}\frac{{\Delta }_{kl}}{{\pi }_k^{*}{\pi }_l^{*}}{y}_k{y}_l\hfill \\ \hfill & =\left\{{{\displaystyle \sum }}_U\frac{{\Delta }_{kk}}{{\pi }_k^{*2}}{y}_k^2+{\left[{{\displaystyle \sum }}_{U_2}\left(\frac{{\pi }_k}{{\pi }_{\left(K\right)}}-1\right)y_k\right]}^2\right\}+\underset{k\ne l}{{\displaystyle \sum }{{\displaystyle \sum }}_U}\frac{{\Delta }_{kl}}{{\pi }_k^{*}{\pi }_l^{*}}{y}_k{y}_l\hfill \\ \hfill & \triangleq F_1+F_2\mathrm{.}(\text{A}\text{.3})\hfill \end{array}$$

It is directly verified that $E\left(\widehat{\text{MSE}}\left({\widehat{t}}_{\text{IHT}}\right)\right)\mathrm{<mo>=</mo>}\text{MSE}\left({\widehat{t}}_{\text{IHT}}\right).$ Therefore, Theorem 1 is proved.

A.2  Proof of Theorem 2

Using Conditions C.1 and C.2, we see that $\lambda \le {\pi }_k\le {\pi }_{\left(K\right)}\le 1$ for each $k\in U_2,$ and ${{\displaystyle \max }}_{k\ne l\in U_2}\left|{\pi }_{kl}-{\pi }_k{\pi }_l\right|\mathrm{<mo>=</mo>}O\left(n^{-1}\right).$ Then, from equation (2.1), we have

$$\begin{array}{ll}\left|E{\left({\widehat{\overline{t}}}_{\text{HT}}-\overline{t}\right)}^2\right|\hfill & \mathrm{<mo>=</mo>}\left|\frac{1}{N^2}{\displaystyle \sum _U}\frac{{\Delta }_{kk}}{{\pi }_k^2}{y}_k^2+\frac{1}{N^2}\underset{k\ne l}{{\displaystyle \sum }{{\displaystyle \sum }}_U}\frac{{\Delta }_{kl}}{{\pi }_k{\pi }_l}{y}_l{y}_k\right|\hfill \\ \hfill & \le \frac{1}{N^2}{\displaystyle \sum _U}\frac{1-{\pi }_k}{{\pi }_k}{y}_k^2+\frac{1}{N^2}\underset{k\ne l}{{\displaystyle \sum }{{\displaystyle \sum }}_U}\left|\frac{{\pi }_{kl}-{\pi }_k{\pi }_l}{{\pi }_k{\pi }_l}\right|\left|y_l{y}_k\right|\hfill \\ \hfill & \mathrm{<mo>=</mo>}O\left(n^{-1}\right)\mathrm{.}\hfill \end{array}$$

Similarly, by the MSE of the IHT estimator given in (3.1), we observe

$$\begin{array}{ll}\left|E{\left({\widehat{\overline{t}}}_{\text{IHT}}-\overline{t}\right)}^2\right|\hfill & \mathrm{<mo>=</mo>}\left|{\left[\frac{1}{N}{{\displaystyle \sum }}_{U_2}\left(\frac{{\pi }_k}{{\pi }_{\left(K\right)}}-1\right)y_k\right]}^2+\frac{1}{N^2}{{\displaystyle \sum }}_U\frac{{\Delta }_{kk}}{{\pi }_k^{*2}}{y}_k^2+\frac{1}{N^2}\underset{k\ne l}{{\displaystyle \sum }{{\displaystyle \sum }}_U}\frac{{\Delta }_{kl}}{{\pi }_k^{*}{\pi }_l^{*}}{y}_k{y}_l\right|\hfill \\ \hfill & \le {\left[\frac{K}{N}\frac{1}{K}{{\displaystyle \sum }}_{U_2}\left(\frac{{\pi }_k}{{\pi }_{\left(K\right)}}-1\right)y_k\right]}^2+\frac{1}{N^2}{{\displaystyle \sum }}_U\left|\frac{{\pi }_k\left(1-{\pi }_k\right)}{{\pi }_k^{*2}}\right|y_k^2\hfill \\ \hfill & +\frac{1}{N^2}\underset{k\ne l}{{\displaystyle \sum }{{\displaystyle \sum }}_U}\left|\frac{{\pi }_{kl}-{\pi }_k{\pi }_l}{{\pi }_k^{*}{\pi }_l^{*}}\right|\left|y_k{y}_l\right|\hfill \\ \hfill & \mathrm{<mo>=</mo>}O\left(n^{-1}\right).\hfill \end{array}$$

From Conditions C.1 and C.2, it is readily seen that

$$\text{Bias}\left({\widehat{\overline{t}}}_{\text{IHT}}\right)\mathrm{<mo>=</mo>}\left|\frac{1}{N}{{\displaystyle \sum }}_{U_2}\left(\frac{{\pi }_k}{{\pi }_{\left(K\right)}}-1\right)y_k\right|\le \frac{K}{N}\frac{1}{K}{{\displaystyle \sum }}_{U_2}\left|\frac{{\pi }_k}{{\pi }_{\left(K\right)}}-1\right|\left|y_k\right|\le \frac{1}{N}{{\displaystyle \sum }}_{U_2}\left|y_k\right|\mathrm{<mo>=</mo>}O\left(n^{-1}\right)\mathrm{,}$$

where the third and fourth steps are valid due to $\lambda \le {\pi }_k\le {\pi }_{\left(K\right)}\le 1$ for each $k\in U_2$ and $K/N\mathrm{<mo>=</mo>}O\left(n^{-1}\right),$ respectively.

A.3  Proof of Theorem 3

From equation (2.1), since the HT estimator is unbiased, we have

$$\text{MSE}\left({\widehat{Y}}_{\text{HT}}\right)\mathrm{<mo>=</mo>}\left\{{{\displaystyle \sum }}_{U_1}\frac{{\Delta }_{kk}}{{\pi }_k^2}{y}_k^2+{{\displaystyle \sum }}_{U_2}\frac{{\Delta }_{kk}}{{\pi }_k^2}{y}_k^2\right\}+\underset{k\ne l}{{\displaystyle \sum }{{\displaystyle \sum }}_U}\frac{{\Delta }_{kl}}{{\pi }_k{\pi }_l}{y}_k{y}_l\triangleq F_3+F_4\mathrm{.}(\text{A}\text{.4})$$

To illustrate the effectiveness of the new estimator, we compare equation (A.3) and equation (A.4). We prove $F_3\ge F_1$ at first. It is clear that

$$\begin{array}{ll}{F}_3-F_1\hfill & \mathrm{<mo>=</mo>}{{\displaystyle \sum }}_U\frac{{\Delta }_{kk}}{{\pi }_k^2}{y}_k^2-\left\{{{\displaystyle \sum }}_{U_1}\frac{{\Delta }_{kk}}{{\pi }_k^2}{y}_k^2+{{\displaystyle \sum }}_{U_2}\frac{{\Delta }_{kk}}{{\pi }_{\left(K\right)}^2}{y}_k^2+{\left[{{\displaystyle \sum }}_{U_2}\left(\frac{{\pi }_k}{{\pi }_{\left(K\right)}}-1\right)y_k\right]}^2\right\}\hfill \\ \hfill & \mathrm{<mo>=</mo>}{{\displaystyle \sum }}_{U_2}\frac{{\Delta }_{kk}}{{\pi }_k^2}{y}_k^2-{{\displaystyle \sum }}_{U_2}\frac{{\Delta }_{kk}}{{\pi }_{\left(K\right)}^2}{y}_k^2-{\left[{{\displaystyle \sum }}_{U_2}\left(\frac{{\pi }_k}{{\pi }_{\left(K\right)}}-1\right)y_k\right]}^2\hfill \\ \hfill & \mathrm{<mo>=</mo>}{{\displaystyle \sum }}_{U_2}\frac{\left({\pi }_{\left(K\right)}^2-{\pi }_k^2\right)\left(1-{\pi }_k\right)}{{\pi }_{\left(K\right)}^2{\pi }_k}{y}_k^2-{\left[{{\displaystyle \sum }}_{U_2}\left(\frac{{\pi }_k}{{\pi }_{\left(K\right)}}-1\right)y_k\right]}^2\hfill \\ \hfill & \triangleq D-C.\hfill \end{array}$$

Using the Cauchy-Schwarz inequality, we have

$$C\mathrm{<mo>=</mo>}{\left({{\displaystyle \sum }}_{U_2}\frac{{\pi }_k-{\pi }_{\left(K\right)}}{{\pi }_{\left(K\right)}}{y}_k\right)}^2\le K{{\displaystyle \sum }}_{U_2}\frac{{\left({\pi }_k-{\pi }_{\left(K\right)}\right)}^2}{{\pi }_{\left(K\right)}^2}{y}_k^2\triangleq E\mathrm{,}$$

where the strict inequality holds if there exist $k\ne l\in U_2$ such that $\left({\pi }_k-{\pi }_{\left(K\right)}\right)y_k\ne \left({\pi }_l-{\pi }_{\left(K\right)}\right)y_l.$ Further,

$$\begin{array}{ll}{F}_3-F_1\ge D-E\hfill & \mathrm{<mo>=</mo>}{{\displaystyle \sum }}_{U_2}\frac{\left({\pi }_{\left(K\right)}^2-{\pi }_k^2\right)\left(1-{\pi }_k\right)}{{\pi }_{\left(K\right)}^2{\pi }_k}{y}_k^2-K{{\displaystyle \sum }}_{U_2}\frac{{\left({\pi }_k-{\pi }_{\left(K\right)}\right)}^2}{{\pi }_{\left(K\right)}^2}{y}_k^2\hfill \\ \hfill & \mathrm{<mo>=</mo>}{{\displaystyle \sum }}_{U_2}\frac{\left({\pi }_{\left(K\right)}-{\pi }_k\right)\left[\left(1-{\pi }_k-K{\pi }_k\right){\pi }_{\left(K\right)}+\left({\pi }_k-{\pi }_k^2+K{\pi }_k^2\right)\right]}{{\pi }_{\left(K\right)}^2{\pi }_k}{y}_k^2\mathrm{.}(\text{A}\text{.5})\hfill \end{array}$$

From Definition 1, we have ${\pi }_k\le {\pi }_{\left(K\right)}\le {\left(K+1\right)}^{-1}$ for each $k\in U_2,$ thus $D-E\ge 0.$ So $F_3-F_1\mathrm{<mo>=</mo>}D-C\ge D-E\ge 0$ holds.

For the terms $F_2$ and $F_4,$ we note that

$$\begin{array}{ll}{F}_2-F_4\hfill & \mathrm{<mo>=</mo>}\underset{k\ne l}{{\displaystyle \sum }{{\displaystyle \sum }}_U}\frac{{\Delta }_{kl}}{{\pi }_k^{*}{\pi }_l^{*}}{y}_k{y}_l-\underset{k\ne l}{{\displaystyle \sum }{{\displaystyle \sum }}_U}\frac{{\Delta }_{kl}}{{\pi }_k{\pi }_l}{y}_k{y}_l\hfill \\ \hfill & \mathrm{<mo>=</mo>}\underset{k\ne l}{{\displaystyle \sum }{{\displaystyle \sum }}_{U_2}}\left(\frac{{\Delta }_{kl}}{{\pi }_{\left(K\right)}^2}-\frac{{\Delta }_{kl}}{{\pi }_k{\pi }_l}\right)y_k{y}_l+{\displaystyle \sum _{k\in U_1}}{\displaystyle \sum _{l\in U_2}}\left(\frac{{\Delta }_{kl}}{{\pi }_{\left(K\right)}{\pi }_k}-\frac{{\Delta }_{kl}}{{\pi }_k{\pi }_l}\right)y_k{y}_l\hfill \\ \hfill & +{\displaystyle \sum _{k\in U_2}}{\displaystyle \sum _{l\in U_1}}\left(\frac{{\Delta }_{kl}}{{\pi }_{\left(K\right)}{\pi }_l}-\frac{{\Delta }_{kl}}{{\pi }_k{\pi }_l}\right)y_k{y}_l\hfill \\ \hfill & \triangleq {\Delta }_1+{\Delta }_2+{\Delta }_3\mathrm{.}\hfill \end{array}$$

Using Conditions C.1 and C.2, it is seen that

$$\begin{array}{ll}\frac{\left|{\Delta }_1\right|}{N^2}\hfill & =\frac{1}{N^2}\left|\underset{k\ne l}{{\displaystyle \sum }{{\displaystyle \sum }}_{U_2}}\frac{{\pi }_{kl}-{\pi }_k{\pi }_l}{{\pi }_k{\pi }_l}\left(\frac{{\pi }_k{\pi }_l}{{\pi }_{\left(K\right)}^2}-1\right)y_k{y}_l\right|\hfill \\ \hfill & \le \frac{1}{N^2}\underset{k\ne l}{{\displaystyle \sum }{{\displaystyle \sum }}_{U_2}}\left|\frac{{\pi }_{kl}-{\pi }_k{\pi }_l}{{\pi }_k{\pi }_l}\right|\left|\frac{{\pi }_k{\pi }_l}{{\pi }_{\left(K\right)}^2}-1\right|\left|y_k{y}_l\right|\hfill \\ \hfill & \le \frac{K^2}{N^2}\frac{1}{K^2}\underset{k\ne l}{{\displaystyle \sum }{{\displaystyle \sum }}_{U_2}}\left|\frac{{\pi }_{kl}-{\pi }_k{\pi }_l}{{\pi }_k{\pi }_l}\right|\left|y_k{y}_l\right|\mathrm{<mo>=</mo>}O\left(n^{-3}\right)\mathrm{,}\hfill \end{array}$$

where the third and fourth steps are valid due to $\lambda \le {\pi }_k\le {\pi }_{\left(K\right)}\le 1$ for each $k\in U_2,$ $K/N\mathrm{<mo>=</mo>}O\left(n^{-1}\right),$ and ${{\displaystyle \max }}_{k\ne l\in U_2}\left|{\pi }_{kl}-{\pi }_k{\pi }_l\right|\mathrm{<mo>=</mo>}O\left(n^{-1}\right).$ Similarly, we obtain

$$\begin{array}{ll}\frac{\left|{\Delta }_2\right|}{N^2}\hfill & \mathrm{<mo>=</mo>}\frac{1}{N^2}\left|{\displaystyle \sum _{k\in U_1}}{\displaystyle \sum _{l\in U_2}}\frac{{\pi }_{kl}-{\pi }_k{\pi }_l}{{\pi }_k{\pi }_l}\left(\frac{{\pi }_l}{{\pi }_{\left(K\right)}}-1\right)y_k{y}_l\right|\hfill \\ \hfill & \le \frac{1}{N^2}{\displaystyle \sum _{k\in U_1}}{\displaystyle \sum _{l\in U_2}}\left|\frac{{\pi }_{kl}-{\pi }_k{\pi }_l}{{\pi }_k{\pi }_l}\right|\left|\frac{{\pi }_l}{{\pi }_{\left(K\right)}}-1\right|\left|y_k{y}_l\right|\hfill \\ \hfill & \le \frac{1}{N^2}{\displaystyle \sum _{k\in U_1}}{\displaystyle \sum _{l\in U_2}}\left|\frac{{\pi }_{kl}-{\pi }_k{\pi }_l}{{\pi }_k{\pi }_l}\right|\left|y_k{y}_l\right|\mathrm{<mo>=</mo>}O\left(n^{-2}\right)\mathrm{,}\hfill \end{array}$$

and $\frac{\left|{\Delta }_3\right|}{N^2}\mathrm{<mo>=</mo>}O\left(n^{-2}\right).$

Thus, together with $F_3\ge F_1,$ we have

$$\text{MSE}\left(N^{-1}{\widehat{t}}_{\text{IHT}}\right)\le \text{MSE}\left(N^{-1}{\widehat{t}}_{\text{HT}}\right)+o\left(n^{-1}\right)\mathrm{.}$$

For the Poisson sampling case, we have $F_4\mathrm{<mo>=</mo>}{F}_2\mathrm{<mo>=</mo>\; 0.}$ Hence, for Poisson sampling, we obtain

$$\text{MSE}\left(N^{-1}{\widehat{t}}_{\text{IHT}}\right)\le \text{MSE}\left(N^{-1}{\widehat{t}}_{\text{HT}}\right)\mathrm{.}$$

A.4  Proof of Theorem 4

First note that

$${\left(\widehat{R}-R\right)}^2\mathrm{<mo>=</mo>}{\left(\frac{{\widehat{t}}_{y\pi }-R{\widehat{t}}_{z\pi }}{{\widehat{t}}_{z\pi }}\right)}^2\mathrm{<mo>=</mo>}\frac{{\left({\widehat{t}}_{y\pi }-R{\widehat{t}}_{z\pi }\right)}^2}{t_z^2}-\frac{\left({\widehat{t}}_{z\pi }^2-t_z^2\right){\left({\widehat{t}}_{y\pi }-R{\widehat{t}}_{z\pi }\right)}^2}{t_z^2{\widehat{t}}_{z\pi }^2}\triangleq \text{I}+\text{III},$$

and

$${\left({\widehat{R}}^{*}-R\right)}^2\mathrm{<mo>=</mo>}{\left(\frac{{\widehat{t}}_{y\pi }^{*}-R{\widehat{t}}_{z\pi }}{{\widehat{t}}_{z\pi }^{*}}\right)}^2\mathrm{<mo>=</mo>}\frac{{\left({\widehat{t}}_{y\pi }^{*}-R{\widehat{t}}_{z\pi }^{*}\right)}^2}{t_z^2}-\frac{\left({\widehat{t}}_{z\pi }^{*2}-t_z^2\right){\left({\widehat{t}}_{y\pi }^{*}-R{\widehat{t}}_{z\pi }^{*}\right)}^2}{t_z^2{\widehat{t}}_{z\pi }^{*2}}\triangleq \text{II}+\text{IV}\text{.}$$

Let $u_k\mathrm{<mo>=</mo>}{y}_k-R{z}_k\mathrm{.}$ By Theorem 3, we have

$$N^{-2}E{\left({\widehat{t}}_u^{*}-t_u\right)}^2\le N^{-2}E{\left({\widehat{t}}_u-t_u\right)}^2+o\left(n^{-1}\right)\mathrm{.}$$

Thus, for the terms I and II, we get

$$E\left(\text{I}\right)\le E\left(\text{II}\right)+o\left(n^{-1}\right)\mathrm{.}(\text{A}\text{.6})$$

Now, we need to prove that the expectations of III and IV are negligible. Observe that,

$$\begin{array}{ll}\left|E\left(\text{III}\right)\right|\hfill & \mathrm{<mo>=</mo>}\left|E\frac{\left({\widehat{t}}_{z\pi }+t_z\right)\left({\widehat{t}}_{z\pi }-t_z\right){\left({\widehat{t}}_{y\pi }-R{\widehat{t}}_{z\pi }\right)}^2}{t_z^2{\widehat{t}}_{z\pi }^2}\right|\hfill \\ \hfill & \le E\frac{\left|{\widehat{t}}_{z\pi }+t_z\right|\left|{\widehat{t}}_{z\pi }-t_z\right|{\left({\widehat{t}}_{y\pi }-R{\widehat{t}}_{z\pi }\right)}^2}{t_z^2{\widehat{t}}_{z\pi }^2}\hfill \\ \hfill & \le \frac{Z^{*}+\left|t_z\right|}{t_z^2{Z}_{*}^2}E\left(\left|{\widehat{t}}_{z\pi }-t_z\right|{\left({\widehat{t}}_{y\pi }-R{\widehat{t}}_{z\pi }\right)}^2\right)\hfill \\ \hfill & \le \frac{Z^{*}+\left|t_z\right|}{t_z^2{Z}_{*}^2}\sqrt{E{\left({\widehat{t}}_{z\pi }-t_z\right)}^{2}E{\left({\widehat{t}}_{y\pi }-R{\widehat{t}}_{z\pi }\right)}^4}\mathrm{,}\hfill \end{array}$$

where $Z^{*}\mathrm{<mo>=</mo>}\frac{n}{N}{{\displaystyle \max }}_{k\in U}\left(\frac{z_k}{{\pi }_k}\right)\mathrm{,}{Z}_{*}\mathrm{<mo>=</mo>}\frac{n}{N}{{\displaystyle \min }}_{k\in U}\left(\frac{z_k}{{\pi }_k}\right).$ Similarly,

$$\left|E\left(\text{IV}\right)\right|\le \frac{{\tilde{Z}}^{*}+\left|t_z\right|}{t_z^2{\tilde{Z}}_{*}^2}\sqrt{E{\left({\widehat{t}}_{z\pi }^{*}-t_z\right)}^{2}E{\left({\widehat{t}}_{y\pi }^{*}-R{\widehat{t}}_{z\pi }^{*}\right)}^4}\mathrm{,}$$

where ${\tilde{Z}}^{*}\mathrm{<mo>=</mo>}\frac{n}{N}{{\displaystyle \max }}_{k\in U}\left(\frac{z_k}{{\pi }_k^{*}}\right)\mathrm{,}{\tilde{Z}}_{\mathrm{<mo>*</mo>}}\mathrm{<mo>=</mo>}\frac{n}{N}{{\displaystyle \min }}_{k\in U}\left(\frac{z_k}{{\pi }_k^{*}}\right).$

Using Theorem 2 and Lemma 1, we see that $\left|E\mathrm{<mo\; stretchy="false">(</mo>}\text{III}\mathrm{<mo\; stretchy="false">)</mo>}\right|\mathrm{<mo>=</mo>}O\left(n^{-3/2}\right)$ and $\left|E\left(\text{IV}\right)\right|\mathrm{<mo>=</mo>}O\left(n^{-3/2}\right)\mathrm{.}$ Combining these and equation (A.6), we get

$$\text{MSE}\left({\widehat{R}}^{*}\right)\le \text{MSE}\left(\widehat{R}\right)+o\left(n^{-1}\right)\mathrm{.}$$

It implies that $\text{MSE}\left(N^{-1}{\widehat{Y}}_R^{*}\right)\le \text{MSE}\left(N^{-1}{\widehat{Y}}_R\right)+o\left(n^{-1}\right).$

A.5  Discussion on Condition C.4

Case 1: Simple random sampling without replacement

Under the simple random sampling without replacement, we have that ${\pi }_i\mathrm{<mo>=</mo>}\frac{n}{N}$ for $i\in U,$ ${\pi }_{ij}\mathrm{<mo>=</mo>}\frac{n\left(n-1\right)}{N\left(N-1\right)}$ for $i\ne j\in U,$ ${\pi }_{ijk}\mathrm{<mo>=</mo>}\frac{n\left(n-1\right)\left(n-2\right)}{N\left(N-1\right)\left(N-2\right)}$ for $i\ne j\ne k\in U,$ and ${\pi }_{ijkl}\mathrm{<mo>=</mo>}\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{N\left(N-1\right)\left(N-2\right)\left(N-3\right)}$ for $i\ne j\ne k\ne l\in U.$ It follows that

$${\pi }_{ijk}-{\pi }_{ij}{\pi }_k\mathrm{<mo>=</mo>}-\frac{2n\left(n-1\right)\left(N-n\right)}{N^2\left(N-1\right)\left(N-2\right)}\mathrm{<mo>=</mo>}O\left(n^{-1}\right)\mathrm{,}$$

where the last equality is from Condition C.3. We also obtain

$$\begin{array}{l}{\pi }_{ijkl}-4{\pi }_{ijk}{\pi }_l+6{\pi }_{ij}{\pi }_k{\pi }_l-3{\pi }_i{\pi }_j{\pi }_k{\pi }_l\\ \begin{array}{ll}\hfill & \mathrm{<mo>=</mo>}\left({\pi }_{ijkl}-{\pi }_{ijk}{\pi }_l\right)-3\left({\pi }_{ijk}{\pi }_l-{\pi }_{ij}{\pi }_k{\pi }_l\right)+3\left({\pi }_{ij}{\pi }_k{\pi }_l-{\pi }_i{\pi }_j{\pi }_k{\pi }_l\right)\hfill \\ \hfill & \mathrm{<mo>=</mo>\; 3}\frac{n\left(n-1\right)\left(n-2\right)\left(n-N\right)}{N^2\left(N-1\right)\left(N-2\right)\left(N-3\right)}-6\frac{n^2\left(n-1\right)\left(n-N\right)}{N^3\left(N-1\right)\left(N-2\right)}+3\frac{n^3\left(n-N\right)}{N^4\left(N-1\right)}\hfill \\ \hfill & \mathrm{<mo>=</mo>}O\left(n^{-2}\right)\mathrm{,}\hfill \end{array}\end{array}$$

where the last equality is from Condition C.3. Thus, Condition C.4 holds under the simple random sampling without replacement.

Case 2: Poisson sampling

From the independence of Poisson sampling, ${\pi }_{ij}\mathrm{<mo>=</mo>}{\pi }_i{\pi }_j$ for $i\ne j\in U,$ ${\pi }_{ijk}\mathrm{<mo>=</mo>}{\pi }_i{\pi }_j{\pi }_k$ for $i\ne j\ne k\in U,$ and ${\pi }_{ijkl}\mathrm{<mo>=</mo>}{\pi }_i{\pi }_j{\pi }_k{\pi }_l$ for $i\ne j\ne k\ne l\in U.$ Hence, ${\pi }_{ijk}-{\pi }_{ij}{\pi }_k\mathrm{<mo>=</mo>\; 0,}$ and ${\pi }_{ijkl}-4{\pi }_{ijk}{\pi }_l+6{\pi }_{ij}{\pi }_k{\pi }_l-3{\pi }_i{\pi }_j{\pi }_k{\pi }_l\mathrm{<mo>=</mo>\; 0.}$ It follows that Poisson sampling satisfies Condition C.4.

A.6  A lemma for proving Theorem 4

Lemma 1. For the HT estimator ${\widehat{\overline{t}}}_{HT}$  and the IHT estimator ${\widehat{\overline{t}}}_{IHT},$  under the Conditions C.1-C.4, we have

$$E{\left({\widehat{\overline{t}}}_{HT}-\overline{t}\right)}^4\mathrm{<mo>=</mo>}O\left(n^{-2}\right)\mathrm{,}andE{\left({\widehat{\overline{t}}}_{IHT}-\overline{t}\right)}^4\mathrm{<mo>=</mo>}O\left(n^{-2}\right)\mathrm{.}$$

Proof. Noting that

$${\widehat{\overline{t}}}_{\text{HT}}-\overline{t}\mathrm{<mo>=</mo>}\frac{1}{N}{{\displaystyle \sum }}_U\frac{I_k-{\pi }_k}{{\pi }_k}{y}_k\triangleq \frac{1}{N}{{\displaystyle \sum }}_U{J}_k{y}_k\mathrm{,}$$

we have

$$\begin{array}{ll}{\left({\widehat{\overline{t}}}_{\text{HT}}-\overline{t}\right)}^4\hfill & \mathrm{<mo>=</mo>}\frac{1}{N^4}{\displaystyle \sum _k}{\displaystyle \sum _l}{\displaystyle \sum _i}{\displaystyle \sum _j}\left(J_k{y}_k\right)\left(J_l{y}_l\right)\left(J_i{y}_i\right)\left(J_j{y}_j\right)\hfill \\ \hfill & \mathrm{<mo>=</mo>}\frac{1}{N^4}{\displaystyle \sum _U}{\left(J_k{y}_k\right)}^4+\frac{4}{N^4}{\displaystyle \sum _{k\ne l}}{\left(J_k{y}_k\right)}^3\left(J_l{y}_l\right)+\frac{3}{N^4}{\displaystyle \sum _{k\ne l}}{\left(J_k{y}_k\right)}^2{\left(J_l{y}_l\right)}^2\hfill \\ \hfill & +\frac{6}{N^4}{\displaystyle \sum _{i\ne k\ne l}}{\left(J_i{y}_i\right)}^2\left(J_k{y}_k\right)\left(J_l{y}_l\right)+\frac{1}{N^4}{\displaystyle \sum _{i\ne j\ne k\ne l}}\left(J_i{y}_i\right)\left(J_j{y}_j\right)\left(J_k{y}_k\right)\left(J_l{y}_l\right)\hfill \\ \hfill & \triangleq \text{I}+\text{II}+\text{III}+\text{IV}+\text{V}.\hfill \end{array}$$

For the first term I, using $\lambda \le {\pi }_k\le 1$ and $\left|I_k-{\pi }_k\right|\le 1$ for any $k\in U,$ we get

$$\left|E\left(I\right)\right|\mathrm{<mo>=</mo>}E\left(\frac{1}{N^4}{\displaystyle \sum _U}{\left(J_k{y}_k\right)}^4\right)\mathrm{<mo>=</mo>}\frac{1}{N^4}{\displaystyle \sum _U}{\left(\frac{y_k}{{\pi }_k}\right)}^{4}E{\left(I_k-{\pi }_k\right)}^4\le \frac{1}{N^4}{\displaystyle \sum _U}{\left(\frac{y_k}{{\pi }_k}\right)}^4\mathrm{<mo>=</mo>}O\mathrm{<mo\; stretchy="false">(</mo>}{n}^{-2}\mathrm{<mo\; stretchy="false">)</mo>\; .}$$

For the terms II and III, we have

$$\left|E\mathrm{<mo\; stretchy="false">(</mo>}{J}_k^3{J}_l\mathrm{<mo\; stretchy="false">)</mo>}\right|\mathrm{<mo>=</mo>}\left|\frac{1}{{\pi }_k^3{\pi }_l}E\left[{\left(I_k-{\pi }_k\right)}^3\left(I_l-{\pi }_l\right)\right]\right|\le \frac{1}{{\pi }_k^3{\pi }_l}E\left[{\left|I_k-{\pi }_k\right|}^3\left|I_l-{\pi }_l\right|\right]\le \frac{1}{{\pi }_k^3{\pi }_l}\le \frac{1}{{\lambda }^4}\mathrm{,}$$

and

$$E\left(J_k^2{J}_l^2\right)\mathrm{<mo>=</mo>}\frac{1}{{\pi }_k^2{\pi }_l^2}E\left[{\left(I_k-{\pi }_k\right)}^2{\left(I_l-{\pi }_l\right)}^2\right]\le \frac{1}{{\pi }_k^2{\pi }_l^2}\le \frac{1}{{\lambda }^4}\mathrm{.}$$

Thus, $\left|E\left(\text{II}\right)\right|\mathrm{<mo>=</mo>}O\left(n^{-2}\right)$ and $\left|E\mathrm{<mo\; stretchy="false">(</mo>}\text{III}\mathrm{<mo\; stretchy="false">)</mo>}\right|\mathrm{<mo>=</mo>}O\left(n^{-2}\right).$ For the fourth term IV, we have that

$$\begin{array}{ll}\left|E\left(J_i^2{J}_k{J}_l\right)\right|\hfill & \mathrm{<mo>=</mo>}\frac{1}{{\pi }_i^2{\pi }_k{\pi }_l}\left|E\left[{\left(I_i-{\pi }_i\right)}^2\left(I_k-{\pi }_k\right)\left(I_l-{\pi }_l\right)\right]\right|\hfill \\ \hfill & \mathrm{<mo>=</mo>}\frac{1}{{\pi }_i^2{\pi }_k{\pi }_l}\left|\left(1-2{\pi }_i\right)\left[\left({\pi }_{ikl}-{\pi }_{ik}{\pi }_l\right)-{\pi }_k\left({\pi }_{il}-{\pi }_i{\pi }_l\right)\right]+{\pi }_i^2\left({\pi }_{kl}-{\pi }_k{\pi }_l\right)\right|\hfill \\ \hfill & \mathrm{<mo>=</mo>}O\left(n^{-1}\right)\mathrm{,}\hfill \end{array}$$

where the last step is from Conditions C.1 and C.4. It implies that $\left|E\left(\text{IV}\right)\right|\mathrm{<mo>=</mo>}O\left(n^{-2}\right).$ For the last term V, we have that

$$\begin{array}{l}\frac{1}{N^4}E\left({\displaystyle \sum _{i\ne j\ne k\ne l}}\left(J_i{y}_i\right)\left(J_j{y}_j\right)\left(J_k{y}_k\right)\left(J_l{y}_l\right)\right)\\ \begin{array}{ll}\hfill & \mathrm{<mo>=</mo>}\frac{1}{N^4}{\displaystyle \sum _{i\ne j\ne k\ne l}}\frac{{\pi }_{ijkl}-4{\pi }_{ijk}{\pi }_l+6{\pi }_{ij}{\pi }_k{\pi }_l-3{\pi }_i{\pi }_j{\pi }_k{\pi }_l}{{\pi }_i{\pi }_j{\pi }_k{\pi }_l}{y}_i{y}_j{y}_k{y}_l\hfill \\ \hfill & \mathrm{<mo>=</mo>}O\left(n^{-2}\right)\mathrm{,}\hfill \end{array}\end{array}$$

where the last step is from Conditions C.1 and C.4. Thus, $E{\left({\widehat{\overline{t}}}_{\text{HT}}-\overline{t}\right)}^4\mathrm{<mo>=</mo>}O\left(n^{-2}\right)$ holds.

Next we shall prove $E{\left({\widehat{\overline{t}}}_{\text{IHT}}-\overline{t}\right)}^4\mathrm{<mo>=</mo>}O\left(n^{-2}\right).$ Noting that

$${\widehat{\overline{t}}}_{\text{IHT}}-\overline{t}\mathrm{<mo>=</mo>}\frac{1}{N}{\displaystyle \sum _U}\frac{I_k-{\pi }_k^{*}}{{\pi }_k^{*}}{y}_k\mathrm{<mo>=</mo>}\frac{1}{N}{\displaystyle \sum _U}\frac{I_k-{\pi }_k}{{\pi }_k^{*}}{y}_k+\frac{1}{N}{\displaystyle \sum _U}\frac{{\pi }_k-{\pi }_k^{*}}{{\pi }_k^{*}}{y}_k\triangleq A+\Delta \mathrm{,}$$

we have

$$E{\left({\widehat{\overline{t}}}_{\text{IHT}}-\overline{t}\right)}^4\mathrm{<mo>=</mo>}E{\left(A+\Delta \right)}^4\mathrm{<mo>=</mo>}E\left(A^4\right)+4\Delta E\left(A^3\right)+6{\Delta }^{2}E\left(A^2\right)+4{\Delta }^{3}E\left(A\right)+{\Delta }^4\mathrm{.}(\text{A}\text{.7})$$

Similar as the proofs of the result $E{\left({\widehat{\overline{t}}}_{\text{HT}}-\overline{t}\right)}^4\mathrm{<mo>=</mo>}O\left(n^{-2}\right),$ using $\lambda \le {\pi }_k^{*}\le 1,$ it is easy to obtain

$$E\left(A^4\right)\mathrm{<mo>=</mo>}E{\left(\frac{1}{N}{\displaystyle \sum _U}\frac{I_k-{\pi }_k}{{\pi }_k^{*}}{y}_k\right)}^4\mathrm{<mo>=</mo>}O\left(n^{-2}\right)\mathrm{.}(\text{A}\text{.8})$$

From equation (A.8), we have that $E\left(A^2\right)\mathrm{<mo>=</mo>}O\left(n^{-1}\right)$ and $E\left(A^3\right)\mathrm{<mo>=</mo>}O\left(n^{-3/2}\right).$ Meanwhile $E\left(A\right)\mathrm{<mo>=</mo>\; 0}$ and

$$\Delta \mathrm{<mo>=</mo>}\frac{1}{N}{\displaystyle \sum _U}\frac{{\pi }_k-{\pi }_k^{*}}{{\pi }_k^{*}}{y}_k\mathrm{<mo>=</mo>}\frac{K}{N}\left(\frac{1}{K}{\displaystyle \sum _{U_2}}\frac{{\pi }_k-{\pi }_k^{*}}{{\pi }_k^{*}}{y}_k\right)\mathrm{<mo>=</mo>}O\left(n^{-1}\right)\mathrm{.}$$

Therefore, from equation (A.7), we prove that $E{\left({\widehat{\overline{t}}}_{\text{IHT}}-\overline{t}\right)}^4\mathrm{<mo>=</mo>}O\left(n^{-2}\right).$

References

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