Nonparametric confidence intervals for ranked set samples

Abstract

In this work, we propose several different confidence interval methods based on ranked-set samples. First, we develop bootstrap bias-corrected and accelerated method for constructing confidence intervals based on ranked-set samples. Usually, for this method, the accelerated constant is computed by employing jackknife method. Here, we derive an analytical expression for the accelerated constant, which results in reducing the computational burden of this bias-corrected and accelerated bootstrap method. The other proposed confidence interval approaches are based on a monotone transformation along with normal approximation. We also study the asymptotic properties of the proposed methods. The performances of these methods are then compared with those of the conventional methods. Through this empirical study, it is shown that the proposed confidence intervals can be successfully applied in practice. The usefulness of the proposed methods is further illustrated by analyzing a real-life data on shrubs.

Appendix A: Proofs

Proof of Theorem 2.1:

Let us define $Y_{(r),i}=\frac{X_{(r),i}-\mu _{r}}{\sigma _r}$, for $r=1,2,\ldots ,k$. Then, $T_{RSS}$ can be expressed as

$$\begin{aligned} T_{RSS}=\frac{n\sum _{r=1}^{k} \sigma _{r}\bar{Y}_{r}}{\sqrt{\sum _{r=1}^{k}\sigma ^{2}_{r}\frac{s^{2}_{r,Y}}{\lambda _{r,n}}}}, \end{aligned}$$

where $s^{2}_{r,Y}=m^{-1}_{r}\sum _{i=1}^{m_{r}}(Y_{(r),i}-\bar{Y}_{r})^2.$ We can express

$$\begin{aligned} \sum _{r=1}^{k}\sigma ^{2}_{r}\frac{s^{2}_{r,Y}}{\lambda _{r,n}}= a_{n}\Biggl [1+\sum _{r=1}^{k}a_{r,n} m^{-1}_{r}\sum _{i=1}^{m_{r}}(Y^{2}_{(r),i}-1)-\sum _{r=1}^{k}a_{r,n}\bar{Y}^{2}_{r}\Biggr ], \end{aligned}$$

(7.1)

where $a_{r,n}=\frac{\lambda ^{-1}_{r,n}\sigma ^{2}_{r}}{a_{n}}$ and $a_{n}=\sum _{r=1}^{k}\lambda ^{-1}_{r,n}\sigma ^{2}_{r}$. Using (7.1), $T_{RSS}$ can be expressed as

$$\begin{aligned} T_{RSS}= & {} a^{-1/2}_{n} T_{1}, \end{aligned}$$

(7.2)

where

$$\begin{aligned} T_1= & {} \sqrt{n}\sum _{r=1}^{k} \sigma _{r}\bar{Y}_{r}\Biggl [1-\frac{1}{2}\sum _{r=1}^{k}a_{r,n} m^{-1}_{r}\sum _{i=1}^{m_{r}}(Y^{2}_{(r),i}-1)+\frac{1}{2}\sum _{r=1}^{k}a_{r,n}\bar{Y}^{2}_{r}\nonumber \\&+ \frac{3}{8}\left\{ \sum _{r=1}^{k}a_{r,n} m^{-1}_{r}\sum _{i=1}^{m_{r}}(Y^{2}_{(r),i}-1)\right\} ^{2}\nonumber \\&+\frac{3}{4}\Biggl \{\sum _{r=1}^{k} a_{r,n} m^{-1}_{r}\sum _{i=1}^{m_{r}}(Y^{2}_{(r),i}-1)\Biggr \} \sum _{r=1}^{k}a_{r,n}\bar{Y}^{2}_{r}\Biggr ]+O_{p}(n^{-2}). \end{aligned}$$

(7.3)

In order to obtain the Edgeworth expansion of $T_{RSS}$ given in Theorem 2.1, we first need to derive asymptotic expansions for the first three cumulants of $T_{RSS}$, which are given in the following lemma. $\square$

Lemma 7.1

Under the assumptions of Theorem 2.1, we have:

• (1) $E(T_{RSS})=-\frac{1}{2}n^{-1/2}\biggl (\sum _{r=1}^{k} \frac{\sigma ^{2}_{r}}{\lambda _{r}}\biggr )^{-3/2}\sum _{r=1}^{k}\frac{\gamma _{r}}{\lambda ^{2}_{r,n}}+O(n^{-3/2}),$

• (2) $E(T^{2}_{RSS})=1+2\biggl (\sum _{r=1}^{k} \frac{\sigma ^{2}_{r}}{\lambda _{r}}\biggr )^{-3}\biggl [\sum _{r=1}^{k}\frac{\gamma _{r}}{\lambda ^{2}_{r,n}}\biggr ]^{2}+ \biggl (\sum _{r=1}^{k} \frac{\sigma ^{2}_{r}}{\lambda _{r}}\biggr )^{-2}\biggl [\sum _{r=1}^{k}\frac{3\sigma ^{4}_{r}}{\lambda ^{3}_{r,n}}+\sum _{r=1}^{k}\sum _{r^{'}>r} \frac{\sigma ^{2}_{r} \sigma ^{2}_{r^{'}}}{\lambda ^{2}_{r}\lambda ^{2}_{r^{'}}}(\lambda _{r}+\lambda _{r^{'}})\biggr ] +O(n^{-2}),$

• (3) $E(T^{3}_{RSS})=-\frac{7}{2}n^{-1/2}\biggl (\sum _{r=1}^{k} \frac{\sigma ^{2}_{r}}{\lambda _{r}}\biggr )^{-3/2}\sum _{r=1}^{k}\frac{\gamma _{r}}{\lambda ^{2}_{r,n}}+O(n^{-3/2})$.

Proof of (1)

Note that $\lambda _{r,n}\rightarrow {\lambda _{r}}$, and so $\lambda ^{-1}_{r,n}=O(1)$ and $a_{r,n}=O(1)$. Now, $a_{r,n} m^{-1}_{r}\sum _{i=1}^{m_{r}}(Y^{2}_{(r),i}-1)=a_{r,n}m^{-1/2}_{r}O_{p}(1)=a_{r,n}(\lambda ^{-1}_{r,n})^{1/2} O_{p}(n^{-1/2})=O_{p}(n^{-1/2})$, and similarly we have $\bar{Y}^{2}_{r}=O_{p}(n^{-1})$. These facts imply that

$$\begin{aligned} \Biggl \{\sum _{r=1}^{k} a_{r,n} m^{-1}_{r}\sum _{i=1}^{m_{r}}(Y^{2}_{(r),i}-1)\Biggr \} \sum _{r=1}^{k}a_{r,n}\bar{Y}^{2}_{r}=O_{p}(n^{-3/2}). \end{aligned}$$

From (7.3), we have

$$\begin{aligned} E(T_1)= & {} -\frac{1}{2} E\Biggl [\sqrt{n}\sum _{r=1}^{k}\sigma _{r}\bar{Y}_{r}\Biggl \{\sum _{r=1}^{k}a_{r,n} m^{-1}_{r}\sum _{i=1}^{m_{r}}(Y^{2}_{(r),i}-1)\Biggr \}\Biggr ]\nonumber \\&+\frac{1}{2}E\Biggl [\sqrt{n}\sum _{r=1}^{k}\sigma _{r}\bar{Y}_{r}\sum _{r=1}^{k} a_{r,n}\bar{Y}^{2}_{r}\Biggr ]\nonumber \\&+\frac{3}{8}E\Biggl [\sqrt{n}\sum _{r=1}^{k}\sigma _{r}\bar{Y}_{r}\Biggl \{\sum _{r=1}^{k}a_{r,n} m^{-1}_{r}\sum _{i=1}^{m_{r}}(Y^{2}_{(r),i}-1)\Biggr \}^{2}\Biggr ]+O(n^{-3/2}).\qquad \quad \end{aligned}$$

(7.4)

Since $E(Y_{(r^{'}),i} (Y^{2}_{(r),i}-1))=0$ for $r\ne r^{'}$, we have

$$\begin{aligned} E\Biggl [\sqrt{n}\sum _{r=1}^{k}\sigma _{r}\bar{Y}_{r}\Biggl \{\sum _{r=1}^{k}a_{r,n} m^{-1}_{r}\sum _{i=1}^{m_{r}}(Y^{2}_{(r),i}-1)\Biggr \}\Biggr ]= & {} \sqrt{n}\sum _{r=1}^{k} \frac{\sigma _{r}a_{r,n}}{m^{2}_{r}}\sum _{i=1}^{m_{r}}E(Y^{3}_{i,r})\nonumber \\= & {} n^{-1/2}a^{-1}_{n}\sum _{r=1}^{k}\frac{\gamma _{r}}{\lambda ^{2}_{r,n}}, \end{aligned}$$

(7.5)

where the last part follows from the fact that $a^{-1}_{n}=(\frac{\sigma ^{2}_{r}}{\lambda _{r,n}})^{-1} a_{r,n}$ and $E\{Y^{3}_{(r),i}\}=\sigma ^{-3}_{r}\gamma _{r}$. Using arguments similar to those above, we have

$$\begin{aligned} E\Biggl [\sqrt{n}\sum _{r=1}^{k}\sigma _{r}\bar{Y}_{r}\sum _{r=1}^{k} a_{r,n}\bar{Y}^{2}_{r}\Biggr ]=\sqrt{n}\sum _{r=1}^{k} a_{r,n}\sigma _{r} E(\bar{Y}^{3}_{r})= & {} n^{-3/2}\sum _{r=1}^{k}\frac{a_{r,n}\sigma _{r}}{\lambda _{r,n}} E(Y^{3}_{r})\nonumber \\= & {} O(n^{-3/2}), \end{aligned}$$

(7.6)

since $\lambda ^{-1}_{r,n}=O(1)$ and $a_{r,n}=O(1)$. For notational simplification, set $U_{(r),i}=Y^{2}_{(r),i}-1$. Then,

$$\begin{aligned} E\Biggl [\sqrt{n}\sum _{r=1}^{k}\sigma _{r}\bar{Y}_{r}\Biggl \{\sum _{r=1}^{k}a_{r,n} m^{-1}_{r}\sum _{i=1}^{m_{r}}(Y^{2}_{(r),i}-1)\Biggr \}^{2}\Biggr ]= & {} \sqrt{n}E\Biggl [\sum _{i,j,l=1}^{k}\sigma _{i}a_{j,n} a_{l,n}\bar{Y}_{i}\bar{U}_{j}\bar{U}_{l}\Biggr ]\nonumber \\= & {} \sqrt{n}E\Biggl [\sum _{r=1}^{k}\sigma _{r}a^{2}_{r,n} \bar{Y}_{r}\bar{U}^{2}_{r}\Biggr ], \end{aligned}$$

(7.7)

where the last part follows from the fact that for other choices of (i, j, l), $E(\bar{Y}_{i}\bar{U}_{j}\bar{U}_{l})=0$. Further, we have

$$\begin{aligned} E\Biggl [\sum _{r=1}^{k}\sigma _{r}a^{2}_{r,n} \bar{Y}_{r}\bar{U}^{2}_{r}\Biggr ]=\sum _{r=1}^{k}\sigma _{r}a^{2}_{r,n} \frac{1}{m^{2}_{r}} E\biggl (Y_{(r)}(Y_{(r)}-1)^{2}\biggr )=O(n^{-2}). \end{aligned}$$

Now, from (7.7), we have

$$\begin{aligned} \ E\Biggl [\sqrt{n}\sum _{r=1}^{k}\sigma _{r}\bar{Y}_{r}\Biggl \{\sum _{r=1}^{k}a_{r,n} m^{-1}_{r}\sum _{i=1}^{m_{r}}(Y^{2}_{(r),i}-1)\Biggr \}^{2}\Biggr ]=O(n^{-3/2}). \end{aligned}$$

(7.8)

Hence, Eqs. (7.4)–(7.6) and (7.8) imply that

$$\begin{aligned} E(T_{RSS})=a^{-1/2}_{n}E(T_{1})=-\frac{n^{-1/2}}{2}\biggl (\sum _{r=1}^{k} \frac{\sigma ^{2}_{r}}{\lambda _{r,n}}\biggr )^{-3/2} \sum _{r=1}^{k}\frac{\gamma _{r}}{\lambda ^{2}_{r,n}}+O(n^{-3/2}). \end{aligned}$$

$\square$

Proof of (3):

From (7.3), we have

$$\begin{aligned} \begin{aligned} E(T^{3}_1)&=n^{3/2}E\Biggl [\biggl (\sum _{r=1}^{k} \sigma _{r}\bar{Y}_{r}\biggr )^{3}\Biggl [1-\frac{3}{2}\sum _{r=1}^{k}a_{r,n} m^{-1}_{r}\sum _{i=1}^{m_{r}}(Y^{2}_{(r),i}-1)-\frac{3}{2}\sum _{r=1}^{k}a_{r,n}\bar{Y}^{2}_{r}\\&\quad -\frac{9}{8}\biggl \{\sum _{r=1}^{k}a_{r,n} m^{-1}_{r}\sum _{i=1}^{m_{r}}(Y^{2}_{(r),i}-1)\biggr \}^{2}\Biggr ]+O(n^{-3/2})\\&=n^{3/2}\biggl [A-\frac{3}{2}B-\frac{3}{2}C-\frac{9}{8}D\biggr ]+O(n^{-3/2}), \end{aligned} \end{aligned}$$

(7.9)

where

$$\begin{aligned} A= & {} E\biggl (\sum _{r=1}^{k} \sigma _{r}\bar{Y}_{r}\biggr )^{3} \;,\;B=E\biggl \{\biggl (\sum _{r=1}^{k} \sigma _{r}\bar{Y}_{r}\biggr )^{3}\sum _{r=1}^{k}a_{r,n} m^{-1}_{r}\sum _{i=1}^{m_{r}}(Y^{2}_{(r),i}-1)\biggr \}\;,\; C\\= & {} E\biggl \{\biggl (\sum _{r=1}^{k} \sigma _{r}\bar{Y}_{r}\biggr )^{3}\sum _{r=1}^{k}a_{r,n}\bar{Y}^{2}_{r}\biggr \} \end{aligned}$$

and

$$\begin{aligned} D=E\Biggl [\biggl (\sum _{r=1}^{k} \sigma _{r}\bar{Y}_{r}\biggr )^{3}\biggl \{\sum _{r=1}^{k}a_{r,n} m^{-1}_{r}\sum _{i=1}^{m_{r}}(Y^{2}_{(r),i}-1)\biggr \}^{2}\Biggr ]. \end{aligned}$$

Now,

$$\begin{aligned} A= & {} E\biggl (\sum _{r=1}^{k}\sigma ^{3}_{r}\bar{Y}^{3}_{r}+3\sum _{r=1}^{k}\sigma _{r}\bar{Y}_{r}\sum _{r^{'}\ne r=1}^{k} \sigma ^{2}_{r^{'}}\bar{Y}^{2}_{r^{'}}+\sum _{r\ne r^{'}\ne r^{''}=1}^{k}\sigma _{r}\sigma _{r^{'}}\sigma _{r^{''}}\bar{Y}_{r^{}}\bar{Y}_{r^{'}}\bar{Y}_{r^{''}}\biggr )\nonumber \\= & {} \sum _{r=1}^{k}\sigma ^{3}_{r}m^{-2}_{r}E(Y^{3}_{(r)})= n^{-2}\sum _{r=1}^{k}\frac{\gamma _{r}}{\lambda ^{2}_{r,n}}. \end{aligned}$$

(7.10)

Let us recall $U_{(r),i}=Y^{2}_{(r),i}-1$, and so $\bar{U}_{r}=m^{-1}_{r}\sum _{i=1}^{m_{r}}(Y^{2}_{(r),i}-1)$. Now,

$$\begin{aligned} B= & {} E\biggl \{\biggl (\sum _{r=1}^{k} \sigma _{r}\bar{Y}_{r}\biggr )^{3}\sum _{r=1}^{k}a_{r,n}\bar{U}_{r}\biggr \}\nonumber \\= & {} E\Biggl (\sum _{r=1}^{k} \sigma ^{3}_{r}a_{r,n}\bar{Y}^{3}_{r}\bar{U}_{r}\Biggl )+ 3\sum _{r=1}^{k}\sigma _{r} a_{r,n}E\biggl (\bar{Y}_{r} \bar{U}_{r}\biggr )\sum _{r^{'}\ne r=1}^{k} \frac{\sigma ^{2}_{r^{'}}}{m_{r^{'}}}. \end{aligned}$$

(7.11)

Again, after some algebraic manipulations, we obtain

$$\begin{aligned} E\biggl (\bar{Y}^{3}_{r}\bar{U}_{r}\biggl )= & {} m^{-4}_{r}\sum _{i=j=l=q=1}E\biggl \{Y_{(r),i}Y_{(r),j}Y_{(r),l}(Y^{2}_{(r),q}-1)\biggr \}\nonumber \\= & {} m^{-3}_{r}E(Y^{5}_{(r)})+3\frac{(m_{r}-1)}{m^{3}_{r}}E(Y^{3}_{(r)})\nonumber \\= & {} 3n^{-2}\frac{\sigma ^{-3}_{r}\gamma _{r}}{\lambda ^{2}_{r,n}}+O(n^{-3}) \end{aligned}$$

(7.12)

and

$$\begin{aligned} \sum _{r=1}^{k}\sigma _{r} a_{r,n}E\biggl (\bar{Y}_{r} \bar{U}_{r}\biggr )\sum _{r^{'}\ne r=1}^{k} \frac{\sigma ^{2}_{r^{'}}}{m_{r^{'}}}= & {} \sum _{r=1}^{k}\sigma _{r} a_{r,n}\frac{\sigma ^{-3}_{r}\gamma _{r}}{m_{r}}\sum _{r^{'}\ne r=1}^{k} \frac{\sigma ^{2}_{r^{'}}}{m_{r^{'}}}\nonumber \\= & {} n^{-2}\sum _{r=1}^{k} a_{r,n}\frac{\sigma ^{-2}_{r}\gamma _{r}}{\lambda _{r,n}}\sum _{r^{'}\ne r=1}^{k} \frac{\sigma ^{2}_{r^{'}}}{\lambda _{r^{'},n}}. \end{aligned}$$

(7.13)

From Eqs. (7.11)–(7.13), we obtain

$$\begin{aligned} B= & {} 3n^{-2}\biggl (\sum _{r=1}^{k}a_{r,n}\frac{\gamma _{r}}{\lambda ^{2}_{r,n}}+\sum _{r=1}^{k} a_{r,n}\frac{\sigma ^{-2}_{r}\gamma _{r}}{\lambda ^{2}_{r,n}}\sum _{r^{'}\ne r=1}^{k} \frac{\sigma ^{2}_{r^{'}}}{\lambda _{r^{'}}}\biggr )+O(n^{-3})=3n^{-2}\sum _{r=1}^{k}\frac{\gamma _{r}}{\lambda ^{2}_{r,n}}\nonumber \\&+O(n^{-3}), \end{aligned}$$

(7.14)

where the last part follows from the fact that $\sum _{r=1}^{k}a_{r,n}=1$. Similarly, it can be shown that $C=O(n^{-3})$ and $D=O(n^{-3})$. Hence, Eqs. (7.9), (7.10) and (7.14) imply

$$\begin{aligned} E(T^{3}_{RSS})=a^{-3/2}_{n} E(T^{3}_{1}) =-n^{-1/2}\frac{7}{2}\biggl (\sum _{r=1}^{k} \frac{\sigma ^{2}_{r}}{\lambda _{r,n}}\biggr )^{-3/2}\sum _{r=1}^{k}\frac{\gamma _{r}}{\lambda ^{2}_{r,n}}+O(n^{-3/2}). \end{aligned}$$

$\square$

Proof of (2):

Now, from (7.3), we have

$$\begin{aligned} E(T^{2}_{1})= & {} E\Biggl [n\biggl (\sum _{r=1}^{k}\sigma _{r}\bar{Y}_{r}\biggr )^{2}\Biggl \{1-\sum _{r=1}^{k}a_{r,n} m^{-1}_{r}\sum _{i=1}^{m_{r}}(Y^{2}_{(r),i}-1)\biggr \}\Biggr ] +O(n^{-1})\nonumber \\= & {} nE\Biggl [\Biggl \{\sum _{r=1}^{k}\sigma ^{2}_{r}\bar{Y}^{2}_{r}+2\sum _{r^{'}>r=1}^{k}\sigma _{r}\sigma _{r^{'}}\bar{Y}_{r}\bar{Y}_{r^{'}}\Biggr \}\Biggl \{1-\sum _{r=1}^{k}a_{r,n} m^{-1}_{r}\sum _{i=1}^{m_{r}}(Y^{2}_{(r),i}-1)\Biggr \}\Biggr ]\nonumber \\&+O(n^{-1}). \end{aligned}$$

(7.15)

It is easy to prove that

$$\begin{aligned}&E\biggl (\sum _{r^{'}>r=1}^{k}\sigma _{r}\sigma _{r^{'}}\bar{Y}_{r}\bar{Y}_{r^{'}}\biggr )=0\quad \hbox {and}\quad E\Biggl [\Biggl \{\sum _{r^{'}>r=1}^{k}\sigma _{r}\sigma _{r^{'}}\bar{Y}_{r}\bar{Y}_{r^{'}}\Biggr \}\\&\quad \sum _{r=1}^{k}a_{r,n} m^{-1}_{r}\sum _{i=1}^{m_{r}}(Y^{2}_{(r),i}-1)\Biggr ]=0. \end{aligned}$$

From (7.15) and after performing some tedious algebraic manipulations, we obtain

$$\begin{aligned} E(T^{2}_{1})= & {} \Biggl [nE\biggl (\sum _{r=1}^{k}\sigma ^{2}_{r}\bar{Y}^{2}_{r}\biggr )-nE\biggl \{\sum _{r=1}^{k}\sigma ^{2}_{r}\bar{Y}^{2}_{r}\biggr \}\biggl \{\sum _{r=1}^{k}a_{r,n} m^{-1}_{r}\sum _{i=1}^{m_{r}}(Y^{2}_{(r),i}-1)\biggr \}\Biggr ]\nonumber \\&+O(n^{-1})\nonumber \\= & {} a_{n}-n^{-1}\sum _{r=1}^{k}\frac{\sigma ^{2}_{r}a_{r,n}}{\lambda ^{2}_{r}}(\kappa _{r,Y}+2)+O(n^{-1})=a_{n}+O(n^{-1}), \end{aligned}$$

(7.16)

where $\kappa _{r,Y}=E(Y^{4}_{r})$. Equations (7.2) and (7.16) imply that $E(T^{2}_{RSS})=1+O(n^{-1})$. Now, Lemma 7.1 together with some algebraic calculations, give the following asymptotic expansions for the first three cumulants, $\kappa _{1}(T_{RSS})$, $\kappa _{2}(T_{RSS})$ and $\kappa _{3}(T_{RSS})$, of $T_{RSS}$:

$$\begin{aligned} \kappa _{1}(T_{RSS})=-\frac{1}{2}n^{-1/2}\biggl (\sum _{r=1}^{k} \frac{\sigma ^{2}_{r}}{\lambda _{r}}\biggr )^{-3/2}\sum _{r=1}^{k}\frac{\gamma _{r}}{\lambda ^{2}_{r,n}}+O(n^{-3/2}),\\ \kappa _{2}(T_{RSS})=E(T^{2})-E(T)^{2}=1+O(n^{-1}) \end{aligned}$$

and

$$\begin{aligned} \kappa _{3}(T_{RSS})=-\frac{7}{2}n^{-1/2}\biggl (\sum _{r=1}^{k} \frac{\sigma ^{2}_{r}}{\lambda _{r}}\biggr )^{-3/2}\sum _{r=1}^{k}\frac{\gamma _{r}}{\lambda ^{2}_{r,n}}+O(n^{-3/2}). \end{aligned}$$

The characteristic function of $T_{RSS}$ is

$$\begin{aligned} E(e^{iT_{RSS}})=\exp \biggl \{it\kappa _{1}(T)+\frac{1}{2}(it)^{2}\kappa _{2}(T)+\frac{1}{6}(it)^{3}\kappa _{3}(T)\biggr \} \end{aligned}$$

(7.17)

and so substituting the above asymptotic formulae for the cumulants in (7.17) and expanding the right-hand side as

$$\begin{aligned} e^{-\frac{{t}^{2}}{2}}\biggl \{1+n^{-1/2}p_{1}(it)+O(n^{-1})\biggr \} \end{aligned}$$

for polynomial $p_{1}$ and inverting the Fourier transformation (for details, see Section 2.3 of Hall 1992b), we get the Edgeworth expansion of $P(T_{RSS}\le x)$ as given in Theorem 2.1.

The derivation of the Edgeworth expansion of $P(S_{RSS}\le x)$ is straightforward, hence omitted. $\square$

Proof of Theorem 3.1:

To facilitate the proof of Theorem 3.1, let $S^{*}_{RSS}=({\bar{X}}^{*}_{RSS}-{\bar{X}}_{RSS})/{\hat{\tau }}$ be the bootstrap version of $S_{RSS}$. Analogous to Theorem 2.1, we have

$$\begin{aligned} P\biggl \{S^{*}_{RSS}\le x|\mathcal {X}_{RSS}\biggr \}= & {} \Phi (x)+n^{-1/2}\hat{p}_{1}(x)\phi (x)+ O_{p}(n^{-1}),\nonumber \\ P\biggl \{T^{*}_{RSS}\le x|\mathcal {X}_{RSS}\biggr \}= & {} \Phi (x)+n^{-1/2}\hat{q}_{1}(x)\phi (x)+ O_{p}(n^{-1}), \end{aligned}$$

(7.18)

respectively, where $\hat{p}_{1}(x)$ and $\hat{q}_{1}(x)$ are as given in Theorem 2.1 except that population moments are now replaced by sample moments. Let $t_{\xi }$, $s_{\xi }$, $\hat{t}_{\xi }$ and $\hat{s}_{\xi }$ be the $\xi$th quantiles of $P(T_{RSS}\le x)$, $P(S_{RSS}\le x)$, $P(T^{*}_{RSS}\le x|\mathcal {X}_{RSS})$ and $P(S^{*}_{RSS}\le x|\mathcal {X}_{RSS})$, respectively. Then, the Cornish–Fisher expansions of these quantiles are given by

$$\begin{aligned} t_{\xi }= & {} z_{\xi }+n^{-1/2}q_{11}(z_{\xi })+ O(n^{-1}),\nonumber \\ s_{\xi }= & {} z_{\xi }+n^{-1/2}p_{11}(z_{\xi })+ O(n^{-1}),\nonumber \\ \hat{t}_{\xi }= & {} z_{\xi }+n^{-1/2}\hat{q}_{11}(z_{\xi })+ O_{p}(n^{-1}),\nonumber \\ \hat{s}_{\xi }= & {} z_{\xi }+n^{-1/2}\hat{p}_{11}(z_{\xi })+ O_{p}(n^{-1}); \end{aligned}$$

(7.19)

see the review of Cornish–Fisher expansion in Hall (1992b). The quantities $\hat{q}_{11}(x)$ and $\hat{p}_{11}(x)$ are the sample versions of $q_{11}(x)=-q_{1}(x)$ and $p_{11}(x)=-p_{1}(x)$, respectively. We now provide the proof for $P(\mu \in I_{1,{BC_{a}}})$. Equation (7.18) implies that

$$\begin{aligned} \hat{G}_{RSS}({\bar{X}}_{RSS})= & {} P\biggl \{S^{*}_{RSS}\le 0|\mathcal {X}_{RSS} \biggr \}\nonumber \\= & {} \Phi (0)+n^{-1/2}\hat{p}_{1}(0)\phi (0)+O_{p}(n^{-1}). \end{aligned}$$

(7.20)

Therefore, from Eqs. (3.1) and (7.20), we have

$$\begin{aligned} \hat{d}= & {} \Phi ^{-1}\biggl \{\Phi (0)+n^{-1/2}\hat{p}_{1}(0)\phi (0)+O_{p}(n^{-1})\biggr \}\nonumber \\= & {} n^{-1/2}\hat{p}_{1}(0)+O_{p}(n^{-1}), \end{aligned}$$

(7.21)

with the last line following from Taylor series expansion. Equation (7.19) implies that

$$\begin{aligned} \hat{u}_{l_{\hat{a}}}(\alpha )= & {} {\bar{X}}_{RSS}+ \hat{s}_{l_{\hat{a}}(\alpha )}{\hat{\tau }}={\bar{X}}_{RSS}+{\hat{\tau }}\biggl \{z_{l_{\hat{a}}(\alpha )}+n^{-1/2}\hat{p}_{11}(z_{l_{\hat{a}}(\alpha )})+ n^{-1}\hat{p}_{21}(z_{l_{\hat{a}}(\alpha )})\biggr \}\nonumber \\&+O_{p}(n^{-3/2}). \end{aligned}$$

(7.22)

Equations (3.2) and (7.20) imply that

$$\begin{aligned} z_{l_{\hat{a}}(\alpha )} 2\hat{d}+z_{\alpha }+\hat{a}(\hat{d}^{2}+2\hat{d}z_{\alpha }+z^{2}_{\alpha })+O_{p}(n^{-1})= & {} z_{\alpha }+n^{-1/2}(2\hat{p}_{1}(0)+\hat{c}z^{2}_{\alpha })\nonumber \\&+O_{p}(n^{-1}), \end{aligned}$$

(7.23)

where $\hat{a}=n^{-1/2}\hat{c}$ and $\hat{c}=\hat{\eta }^{-3/2}_{1}\hat{\eta }_2$. From (7.21) and (7.22), upon using Taylor series expansion, we get

$$\begin{aligned} \hat{u}_{l_{\hat{a}}}(\alpha )= & {} {\bar{X}}_{RSS}+{\hat{\tau }}\biggl [z_{\alpha }+n^{-1/2}\{2\hat{p}_{1}(0)+\hat{c}z^{2}_{\alpha }+\hat{p}_{11}(z_{\alpha })\}\biggr ]+O_{p}(n^{-1})\nonumber \\= & {} {\bar{X}}_{RSS}+{\hat{\tau }}[z_{\alpha }+n^{-1/2}\hat{q}_{1}(z_{\alpha })]+O_{p}(n^{-1}), \end{aligned}$$

(7.24)

since $2\hat{p}_{1}(0)+\hat{c}z^{2}_{\alpha }+\hat{p}_{11}(z_{\alpha })=\hat{q}_{1}(z_{\alpha })$. So, from (7.24), we have

$$\begin{aligned} P(\mu \in I^{*}_{1,{BC_{a}}})= & {} P\{ {\hat{\tau }}^{-1} ({\bar{X}}_{RSS}-\mu )+O_{p}(n^{-1})\ge -z_{\alpha }-n^{-1/2}q_{1}(z_{\alpha })\}\\ {}= & {} P\{ {\hat{\tau }}^{-1} ({\bar{X}}_{RSS}-\mu )\ge z_{1-\alpha }-n^{-1/2}q_{1}(z_{1-\alpha })\}+O(n^{-1}), \end{aligned}$$

where the last line is obtained by the delta method (see Section 2.7, Hall 1992b) and by using the fact that $-z_{\alpha }=z_{1-\alpha }$. From Theorem 2.1, we then get

$$\begin{aligned} P(\mu \in I^{*}_{1,{BC_{a}}})= & {} 1- \biggl [ P\biggl \{ {\hat{\tau }}^{-1} ({\bar{X}}_{RSS}-\mu )\le z_{1-\alpha }-n^{-1/2}q_{1}(z_{1-\alpha })\biggr \}+O(n^{-1})\biggr ]\\= & {} 1- 1+\alpha +O(n^{-1})\\= & {} \alpha +O(n^{-1}). \end{aligned}$$

In a similar manner, we can show that $P(\mu \in I^{*}_{0,{BC_{a}}})=P(\mu \in I^{*}_{2,{BC_{a}}})=\alpha +O(n^{-1})$. $\square$