Simultaneous confidence bands for the distribution function of a finite population in stratified sampling


Stratified sampling is one of the most important survey sampling approaches and is widely used in practice. In this paper, we consider the estimation of the distribution function of a finite population in stratified sampling by the empirical distribution function (EDF) and kernel distribution estimator (KDE), respectively. Under general conditions, the rescaled estimation error processes are shown to converge to a weighted sum of transformed Brownian bridges. Moreover, simultaneous confidence bands (SCBs) are constructed for the population distribution function based on EDF and KDE. Simulation experiments and illustrative data example show that the coverage frequencies of the proposed SCBs under the optimal and proportional allocations are close to the nominal confidence levels.

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Corresponding author

Correspondence to Lijian Yang.

Additional information

This research was supported in part by Jiangsu Specially-Appointed Professor Program SR10700111, Jiangsu Province Key-Discipline Program ZY107992, National Natural Science Foundation of China Awards NSFC 11371272, 11771240, 11701403, Research Fund for the Doctoral Program of Higher Education of China Award 20133201110002, 2017 Jiangsu Overseas Visiting Scholar Program for University Prominent Young and Middle-aged Teachers and Presidents, and the Simons Foundation Mathematics and Physical Sciences Program Award #499650. Helpful comments from a reviewer are greatly appreciated.



In this Appendix, we use \(a_{n}=o\left( b_{n}\right) \) to denote that \( \lim _{n\rightarrow \infty }a_{n}/b_{n}=0\), and \(a_{n}=O\left( b_{n}\right) \) to denote that \(\limsup _{n\rightarrow \infty }a_{n}/b_{n}=c\), where c is a constant. In addition, we denote by \(o_{p}\)\(\left( O_{p}\right) \) and \( o_{a.s.}\) a sequence of random variables of order o\(\left( O\right) \) in probability and almost surely, respectively, while \(u_{a.s.}\) means \( o_{a.s.} \) uniformly in the domain.

In the following we will prove Lemma 1 and Theorems 24.

A.1 Proof of Lemma 1

Our framework given in Sect. 2 and Condition (C2) ensure that, for any \(s\in \left\{ 1,\ldots ,S\right\} \),

$$\begin{aligned} \lim _{k\rightarrow \infty }\left( N_{sk}/N_{k}\right) =W_{s},\ \ \lim _{k\rightarrow \infty }\left( n_{sk}/n_{k}\right) =w_{s},\ \ \lim _{k\rightarrow \infty }\left( n_{k}/N_{k}\right) =C. \end{aligned}$$


$$\begin{aligned} CW_{s}^{-1}w_{s}={\normalsize \lim _{k\rightarrow \infty }}\left( n_{k}/N_{k}\right) \left( N_{sk}/N_{k}\right) ^{-1}\left( n_{sk}/n_{k}\right) ={\normalsize \lim _{k\rightarrow \infty }}\left( n_{sk}/N_{sk}\right) \le 1. \end{aligned}$$

Making use of the simple inequality

$$\begin{aligned} n_{k}/N_{k}=\frac{\sum \nolimits _{s=1}^{S}n_{sk}}{\sum \nolimits _{s=1}^{S}N_{sk}}\ge \min _{1\le s\le S}\left( n_{sk}/N_{sk}\right) \end{aligned}$$

and letting \(k\rightarrow \infty \), one obtains that

$$\begin{aligned} C= & {} \lim _{k\rightarrow \infty }n_{k}/N_{k}\ge \lim _{k\rightarrow \infty }\min _{1\le s\le S}\left( n_{sk}/N_{sk}\right) =\min _{1\le s\le S} \lim _{k\rightarrow \infty }\left( n_{sk}/N_{sk}\right) \\= & {} \min _{1\le s\le S}CW_{s}^{-1}w_{s}, \end{aligned}$$

since \(C<1\) according to Condition (C2), one obtains that \(\min _{1\le s\le S}CW_{s}^{-1}w_{s}<1\). The Lemma 1 is proved. \(\square \)

A.2 Proof of Theorem 2

For \(s=1,\ldots ,S\), combining (8) in Theorem 1 with Skorohod’s Representation Theorem shown in Theorem 6.7 of Billingsley (1999), there exits a version \(\tilde{B}_{sk}\left( \cdot \right) \) of Brownian bridge \(B_{s}\left( \cdot \right) \) that satisfies \(\tilde{B} _{sk}\left( F_{s}( x) \right) \overset{d}{\rightarrow } B_{s}\left( F_{s}( x) \right) \) as \(k\rightarrow \infty \) such that

$$\begin{aligned} \sup \nolimits _{x\in \mathbb {R}}\left| \lambda _{sk}\left\{ F_{n_{sk}}( x) -F_{N_{sk}}(x) \right\} -\tilde{B} _{sk}\left( F_{s}( x) \right) \right| \rightarrow 0,\,a.s., \end{aligned}$$

which implies that

$$\begin{aligned} F_{n_{sk}}( x) -F_{N_{sk}}(x) =\lambda _{sk}^{-1} \tilde{B}_{sk}\left( F_{s}( x) \right) +u_{a.s.}\left( \lambda _{sk}^{-1}\right) \text {.} \end{aligned}$$

Recalling the definitions of \(F_{N_{k}}( x) \) and \( F_{n_{k}}( x) \) given in (1) and (4), one has

$$\begin{aligned} \lambda _{k}\left\{ F_{n_{k}}( x) -F_{N_{k}}( x) \right\}= & {} \lambda _{k}\left\{ \sum \limits _{s=1}^{S}W_{sk}F_{n_{sk}}( x) -\sum \limits _{s=1}^{S}W_{sk}F_{N_{sk}}( x) \right\} \\= & {} \lambda _{k}\sum \limits _{s=1}^{S}W_{sk}\left\{ F_{n_{sk}}( x) -F_{N_{sk}}( x) \right\} \\= & {} \lambda _{k}\sum \limits _{s=1}^{S}W_{sk}\left\{ \lambda _{sk}^{-1}\tilde{ B}_{sk}\left( F_{s}( x) \right) +u_{a.s.}\left( \lambda _{sk}^{-1}\right) \right\} . \end{aligned}$$

According to Condition (C2), as \(k\rightarrow \infty \),

$$\begin{aligned} \frac{n_{sk}}{N_{sk}}=\frac{n_{sk}}{n_{k}}\cdot \frac{n_{k}}{N_{k}}\cdot \frac{N_{k}}{N_{sk}}\rightarrow _{p} w_{s}CW_{s}^{-1}, \end{aligned}$$


$$\begin{aligned} \lambda _{k}W_{sk}\lambda _{sk}^{-1}=\frac{N_{sk}}{N_{k}}\sqrt{\frac{n_{k}}{ n_{sk}}\cdot \frac{1-n_{sk}/N_{sk}}{1-n_{k}/N_{k}}}\rightarrow _{p} W_{s}\sqrt{ w_{s}^{-1}\frac{1-w_{s}CW_{s}^{-1}}{1-C}}. \end{aligned}$$


$$\begin{aligned} \lambda _{k}\left\{ F_{n_{k}}( x) -F_{N_{k}}( x) \right\} \overset{d}{\rightarrow }\sum \limits _{s=1}^{S}W_{s}\sqrt{\left( w_{s}^{-1}-CW_{s}^{-1}\right) /\left( 1-C\right) }B_{s}\left\{ F_{s}( x) \right\} .\ \ \end{aligned}$$

The proof of Theorem 2 is completed. \(\square \)

A.3 Proof of Theorem 3

Note that \(\lambda _{k}N_{k}^{-1/2}=\left( n_{k}^{-1}-N_{k}^{-1}\right) ^{-1/2}N_{k}^{-1/2}=\left( n_{k}/N_{k}\right) ^{1/2}\left( 1-n_{k}/N_{k}\right) ^{-1/2}\)\(\rightarrow 0\) when \(n_{k}/N_{k}\rightarrow C\equiv 0\) as \(k\rightarrow \infty \). Because of a sequence of populations \( \left\{ \pi _{k}\right\} _{k=1}^{\infty }\) as i.i.d. random samples generated from F(x) , Donsker’s Theorem entails that \( N_{k}^{1/2}\left\{ F_{N_{k}}( x) -F(x) \right\} \overset{d}{\rightarrow }B\left\{ F( x) \right\} \). Hence, as \( k\rightarrow \infty \),

$$\begin{aligned} \lambda _{k}M( F_{N_{k}},F) =\lambda _{k}O_{p}\left( N_{k}^{-1/2}\right) =o_{p}\left( 1\right) . \end{aligned}$$

Then Theorem 3 follows by Theorem 2 and Slutsky’s Theorem. \(\square \)

A.4 Proof of Theorem 4

According to the definitions of \(F_{n_{k}}( x) \) and \(\hat{F} _{k}( x) \) given in (4) and (6), one has

$$\begin{aligned} \lambda _{k}\left\{ F_{n_{k}}( x) -\hat{F}_{k}( x) \right\} =\lambda _{k}\left\{ \sum \limits _{s=1}^{S}W_{sk}F_{n_{sk}}( x) -\sum \limits _{s=1}^{S}W_{sk}\hat{F}_{sk}( x) \right\} . \end{aligned}$$

Then (9) and (A.1) imply that

$$\begin{aligned} \lambda _{k}M( \hat{F}_{k},F_{N_{k}})= & {} \lambda _{k}\sup \nolimits _{x\in \mathbb {R}}\left| \hat{F}_{k}( x) -F_{n_{k}}( x) \right| \\\le & {} \lambda _{k}\sum \limits _{s=1}^{S}W_{sk}\sup \nolimits _{x\in \mathbb {R }}\left| \hat{F}_{sk}( x) -F_{n_{sk}}( x) \right| \\= & {} \lambda _{k}\sum \limits _{s=1}^{S}W_{sk}\lambda ^{-1} _{sk}\times o_{p}\left( 1\right) =o_{p}\left( 1\right) . \end{aligned}$$

Applying Theorems 2 and 3 and Slutsky’s Theorem, Theorem 4 is proved. \(\square \)

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Gu, L., Wang, S. & Yang, L. Simultaneous confidence bands for the distribution function of a finite population in stratified sampling. Ann Inst Stat Math 71, 983–1005 (2019).

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  • Confidence band
  • Stratified population distribution
  • Allocation
  • Brownian bridge
  • Kernel
  • Superpopulation