Bahadur representations for the bootstrap median absolute deviation and the application to projection depth weighted mean

Abstract

Median absolute deviation (hereafter MAD) is known as a robust alternative to the ordinary variance. It has been widely utilized to induce robust statistical inferential procedures. In this paper, we investigate the strong and weak Bahadur representations of its bootstrap counterpart. As a useful application, we utilize the results to derive the weak Bahadur representation of the bootstrap sample projection depth weighted mean—a quite important location estimator depending on MAD.

Appendix: Proofs of the lemmas

In this appendix, we provide the detailed proofs for all lemmas given above.

Proof of Lemma 1

By Hoeffding’s inequality in Serfling (1980), we have

$$\begin{aligned}{} & {} \textsf{P}^{*}\left( \hat{v}^{*}_{n,l}> v+\frac{\epsilon }{2}\right) \\{} & {} \quad = \textsf{P}^{*}\left( nF^{*}_{n}\left( v+\frac{\epsilon }{2}\right) \le \left\lfloor \frac{n+l}{2}\right\rfloor -1\right) \\{} & {} \quad = \textsf{P}^{*}\left( \sum _{i=1}^{n} I\left( X^{*}_{i}>v+\frac{\epsilon }{2}\right) \ge n - \left( \left\lfloor \frac{n+l}{2}\right\rfloor -1\right) \right) \\{} & {} \quad = \textsf{P}^{*}\left( \sum _{i=1}^{n}I\left( X^{*}_{i}>v+\frac{\epsilon }{2}\right) - \sum _{i=1}^{n} \left( 1 - F_{n}\left( v+\frac{\epsilon }{2}\right) \right) \right. \\{} & {} \quad \left. \ge nF_{n} \left( v+\frac{\epsilon }{2}\right) - \left( \left\lfloor \frac{n+l}{2}\right\rfloor -1\right) \right) \\{} & {} \quad \le \exp \left\{ -2n\left( F_{n}(v+\frac{\epsilon }{2}) - \frac{\left\lfloor \frac{n+l}{2}\right\rfloor -1}{n}\right) ^{2}\right\} . \end{aligned}$$

By the Glivenko–Cantelli theorem, $F_{n}(v+\frac{\epsilon }{2})\rightarrow F(v+\frac{\epsilon }{2}), a.s.$ Thus for sufficiently large n,

$$\begin{aligned} F_{n}\left( v+\frac{\epsilon }{2}\right) -\frac{\left\lfloor \frac{n+l}{2}\right\rfloor -1}{n} ~> ~\left( F\left( v+\frac{\epsilon }{2}\right) -\frac{\left\lfloor \frac{n+l}{2}\right\rfloor -1}{n}\right) \Big /\root 4 \of {2} ~ > ~0, \quad a.s. \end{aligned}$$

by noting that $F(v+\frac{\epsilon }{2})>\frac{1}{2}$ for any $\epsilon >0$. Hence

$$\begin{aligned} \textsf{P}\left( \hat{v}^{*}_{n,l}> v+\frac{\epsilon }{2}\right) = \textsf{E}\,\left[ \textsf{P}^{*}\left( \hat{v}^{*}_{n,l} > v+\frac{\epsilon }{2}\right) \right] \le e^{-\sqrt{2}n a_{0}^2(\epsilon ,l)}. \end{aligned}$$

Similar discussion leads to

$$\begin{aligned} \textsf{P}\left( \hat{v}^{*}_{n,l}< v-\frac{\epsilon }{2}\right) = \textsf{E}\,\left[ \textsf{P}^{*}\left( \hat{v}^{*}_{n,l} < v - \frac{\epsilon }{2}\right) \right] \le e^{-\sqrt{2}n b_{0}^2(\epsilon ,l)}. \end{aligned}$$

Then the conclusion follows. $\square$

Proof of Lemma 2

Denote by $G^{*}_{n}(y)$ the empirical distribution function of $W^{*}_{1,l},W^{*}_{2,l},\ldots , W^{*}_{n,l}$. Set $\alpha _{n}=(\left\lfloor \frac{n+m}{2}\right\rfloor -1)/n$, then we have

$$\begin{aligned}{} & {} \textsf{P}^{*}\left( \hat{\xi }^{*}_{n,m,l}>\xi +\epsilon \right) = \textsf{P}^{*}\left( W^{*}_{\left\lfloor \frac{n+m}{2}\right\rfloor l:n}>\xi +\epsilon \right) \\{} & {} \quad = \textsf{P}^{*}\left( G^{*}_{n}(\xi +\epsilon )\le \alpha _{n}\right) \\{} & {} \quad \le \textsf{P}^{*}\left( G^{*}_{n}(\xi +\epsilon )\le \alpha _{n},~ |\hat{v}^{*}_{n,l} - v|\le \frac{\epsilon }{2}\right) + \textsf{P}^{*}\left( |\hat{v}^{*}_{n,l}-v|>\frac{\epsilon }{2}\right) \\{} & {} \quad \le \textsf{P}^{*}\left( \sum _{i=1}^nI\left( v-\xi -\frac{\epsilon }{2} \le X_i^* \le v+\xi +\frac{\epsilon }{2}\right) \le n \alpha _{n} \right) \\{} & {} \quad + \textsf{P}^{*}\left( |\hat{v}^{*}_{n,l}-v|>\frac{\epsilon }{2}\right) , \end{aligned}$$

where the last inequality follows from

$$\begin{aligned}{} & {} \left\{ \sum _{i=1}^nI(\hat{v}^{*}_{n,l}-\xi -\epsilon \le X_i^*\le \hat{v}^{*}_{n,l}+\xi +\epsilon )\le n\alpha _n,~v-\frac{\epsilon }{2}\le \hat{v}^{*}_{n,l}\le v+\frac{\epsilon }{2}\right\} \\{} & {} \subset \left\{ \sum _{i=1}^nI\left( v-\xi -\frac{\epsilon }{2} \le X_i^* \le v+\xi +\frac{\epsilon }{2}\right) \le n \alpha _{n},~v-\frac{\epsilon }{2}\le \hat{v}^{*}_{n,l}\le v+\frac{\epsilon }{2}\right\} . \end{aligned}$$

For the first part, by Hoeffding’s inequality, we have

$$\begin{aligned}{} & {} { \textsf{P}^{*}\left( \sum _{i=1}^nI\left( v-\xi -\frac{\epsilon }{2} \le X_i^*\le v+\xi +\frac{\epsilon }{2}\right) \le n\alpha _{n}\right) }\\{} & {} \quad = \textsf{P}^{*}\left( \sum _{i=1}^{n}I\left( v-\xi -\frac{\epsilon }{2} \le X_i^*\le v+\xi +\frac{\epsilon }{2}\right) - np_{n} \le n(\alpha _{n}-p_{n1})\right) \\{} & {} \quad = \textsf{P}^{*}\left( \sum _{i=1}^{n}(Y_{i}- \textsf{E}\,^* Y_{i})\ge n(p_{n1}-\alpha _{n})\right) \\{} & {} \quad \le \exp \{-2n(p_{n1}-\alpha _{n})^{2}\}, \end{aligned}$$

where $Y_{i}=1-I\left( v-\xi -\frac{\epsilon }{2} \le X^{*}_{i}\le v+\xi +\frac{\epsilon }{2}\right)$, $i=1,\ldots ,n$ and $p_{n1}=F_{n}\left( v+\xi +\frac{\epsilon }{2}\right) -F_{n}\left( v-\xi -\frac{\epsilon }{2}-\right)$. Note that $G(\xi +\frac{\epsilon }{2})=F\left( v+\xi +\frac{\epsilon }{2}\right) -F\left( v-\xi -\frac{\epsilon }{2}-\right)$, the Glivenko–Cantelli theorem implies that

$$\begin{aligned} p_{n1}\rightarrow F\left( v+\xi +\frac{\epsilon }{2}\right) -F\left( v-\xi -\frac{\epsilon }{2}-\right) > \frac{1}{2}, \quad a.s. \end{aligned}$$

Then for sufficiently large n

$$\begin{aligned} p_{n1}-\alpha _n>\frac{F(v+\xi +\frac{\epsilon }{2}) -F(v-\xi -\frac{\epsilon }{2})-(\left\lfloor \frac{n+m}{2}\right\rfloor -1)/n}{\root 4 \of {2}}>0,\quad a.s. \end{aligned}$$

It follows from the discussion above and Lemma 1 that

$$\begin{aligned} \textsf{P}\left( \hat{\xi }^{*}_{n,m,l}>\xi +\epsilon \right)= & {} \textsf{E}\,\left[ \textsf{P}^{*}\left( \hat{\xi }^{*}_{n,m,l}>\xi +\epsilon \right) \right] \\\le & {} 2e^{-\sqrt{2}n\delta ^{2}_{\epsilon ,n}} + \textsf{E}\,\left[ e^{-2n(p_{n}-\alpha _{n})^{2}}\right] \\\le & {} 2e^{-\sqrt{2}n\delta ^{2}_{\epsilon ,n}} + e^{-\sqrt{2}nc_{0}^2(\epsilon ,m)}\\\le & {} 3e^{-\sqrt{2}n\Delta ^{2}_{\epsilon ,n}}. \end{aligned}$$

Set $\beta _{n}=\left\lfloor \frac{n+m}{2}\right\rfloor /n$, a similar argument leads to

$$\begin{aligned} \textsf{P}^{*}(\hat{\xi }^{*}_{n,m,l} < \xi - \epsilon )\le & {} \textsf{P}^{*}\left( \sum _{i=1}^nI\left( v-\xi +\frac{\epsilon }{2} \le X_i^*\le v+\xi -\frac{\epsilon }{2}\right) \ge n\beta _{n}\right) \\{} & {} + \textsf{P}^{*}\left( |\hat{v}^{*}_{n,l}-v|>\frac{\epsilon }{2} \right) \\\le & {} \exp \{-2n(\beta _{n}-p_{n2})^{2}\}+2e^{-\sqrt{2}n\delta ^{2}_{\epsilon ,n}}, \end{aligned}$$

where $p_{n2}=F_{n}(v+\xi -\frac{\epsilon }{2}) -F_{n}(v-\xi +\frac{\epsilon }{2}-)$, and $\beta _{n}-p_{n2}>0$ for sufficiently large n. Then we have

$$\begin{aligned} \textsf{P}(\hat{\xi }^{*}_{n,m,l}< \xi -\epsilon ) = \textsf{E}\,\left[ \textsf{P}^{*}(\hat{\xi }^{*}_{n,m,l} < \xi -\epsilon )\right] \le 3e^{-\sqrt{2}n\Delta ^{2}_{\epsilon ,n}}. \end{aligned}$$

The conclusion has been proved. $\square$

Proof of Lemma 3

Put $\epsilon _n=D \frac{(\log n)^{1/2}}{n^{1/2}}$. It follows from Lemmas 1 and 2 that, for any fixed $l \ge 1$ and $m \ge 1$,

$$\begin{aligned} \textsf{P}(|(\hat{v}^{*}_{n,l} + \hat{\xi }^{*}_{n,m,l})-(v+\xi )| > \epsilon _n) \le 8\exp \{-\sqrt{2}n\Delta ^2_{\epsilon _n/2, n}\}. \end{aligned}$$

Since $F(v)=1/2$, we have

$$\begin{aligned} a_0\left( \frac{\epsilon _n}{2}, l\right)= & {} F\left( v+\frac{\epsilon _n}{4}\right) -\frac{\left\lfloor \frac{n+l}{2}\right\rfloor -1}{n}\\= & {} F\left( v+\frac{\epsilon _n}{4}\right) - \frac{1}{2}+O\left( \frac{1}{n}\right) \\= & {} F\left( v+\frac{\epsilon _n}{4}\right) - F(v)+O\left( \frac{1}{n}\right) \\= & {} \frac{F'(v)}{4} \epsilon _n +o\left( \epsilon _n\right) +O\left( \frac{1}{n}\right) \\> & {} \frac{(\log n)^{1/2}}{n^{1/2}}, \quad \text {for sufficiently large } n. \end{aligned}$$

Similarly

$$\begin{aligned} b_0\left( \frac{\epsilon _n}{2}, l\right) > \frac{(\log n)^{1/2}}{n^{1/2}}, \quad \text {for sufficiently large } n. \end{aligned}$$

By similar arguments using $F(v+\xi )-F(v-\xi )=1/2$, we also obtain

$$\begin{aligned} c_0\left( \frac{\epsilon _n}{2}, m\right) > \frac{(\log n)^{1/2}}{n^{1/2}}, \quad \text {for sufficiently large } n \end{aligned}$$

and

$$\begin{aligned} d_0\left( \frac{\epsilon _n}{2}, m\right) > \frac{(\log n)^{1/2}}{n^{1/2}}, \quad \text {for sufficiently large }n. \end{aligned}$$

The conclusion follows from the inequalities above and the Borel-Cantelli lemma. $\square$

Proof of Lemma 4

Denote by $\theta _p$ the p-th quantile of F for $p\in (0, 1)$. Let $a_n =\frac{c\log n}{n^{1/2}}$ for some positive constant c, and define

$$\begin{aligned} H_{pn}(x):=[F^*_n(x)-F^*_n(\theta _p)]- [F(x)-F(\theta _p)]. \end{aligned}$$

It follows from Lemma 3.7 of Zuo (2015) that

$$\begin{aligned} \sup _{|x-\theta _p|<a_n}|H_{pn}(x)|=O\left( n^{-3/4}\log n\right) , \text {as}\, n\rightarrow \infty , \quad a.s. \end{aligned}$$

Let we express $v-\xi$ as the p-th quantile of F: $v-\xi =F^{-1}(p)=\theta _p$, and put $x_n=\hat{v}^{*}_{n,l} - \hat{\xi }^{*}_{n,m,l}$ for any fixed $l,m \ge 1$, then Lemma 3 implies

$$\begin{aligned} |x_n-\theta _p| \le D \frac{(\log n)^{1/2}}{n^{1/2}} < a_n, \,\text {for sufficiently large }n. \end{aligned}$$

Now we have

$$\begin{aligned} H_{n1} \le \sup _{|x-\theta _p|<a_n}|H_{pn}(x)|=O\left( n^{-3/4}\log n\right) , \quad a.s. \end{aligned}$$

Similarly, we can obtain

$$\begin{aligned} H_{n2} =O\left( n^{-3/4}\log n\right) , \quad a.s. \end{aligned}$$

The proof has been completed. $\square$

Proof of Lemma 5

For convenience, we set $l=m=1$. Recall that $G^{*}_{n}(y)$ is the empirical distribution function of $W^{*}_{1,l},W^{*}_{2,l},\ldots , W^{*}_{n,l}$. Since $\hat{\xi }^{*}_{n,1,1} = W^{*}_{\left\lfloor \frac{n+1}{2}\right\rfloor :n, 1}$, we have $G_{n}^{*}(\hat{\xi }^{*}_{n,1,1})=\left\lfloor \frac{n+1}{2}\right\rfloor /n$ unless there is a tie. If such a tie exists, we have some $X_{i}^{*}=\hat{v}^{*}_{n,1} \pm \hat{\xi }^{*}_{n,1,1}$. It follows from Zuo (2015) that for large n

$$\begin{aligned} \sum _{i=1}^{n}I\left( X_{i}^{*} = \hat{v}^{*}_{n,1}\pm \hat{\xi }^{*}_{n,1,1} \right) < 2\log n, \quad a.s. \end{aligned}$$

That is, we have for large n

$$\begin{aligned} nG_{n}^{*}(\hat{\xi }^{*}_{n,1,1})\le \left\lfloor \frac{n+1}{2}\right\rfloor + 2\log n,\quad a.s. \end{aligned}$$

Then almost surely

$$\begin{aligned} G_{n}^{*}(\hat{\xi }^{*}_{n,1,1})=\frac{1}{2}+ O\left( \frac{\log n}{n} \right) , \quad n\rightarrow \infty . \end{aligned}$$

This completes the proof of this lemma. $\square$

Proof of Lemma 6

For any $\epsilon >0$, let $M > \sqrt{\log (1/\epsilon )}/\root 4 \of {2}$. Put $\epsilon _n= D\frac{M}{n^{1/2}}$, where the constant D is defined in Lemma 3. It can be seen from Lemmas 1–2 that

$$\begin{aligned} \textsf{P}(|(\hat{v}^{*}_{n,l}+\hat{\xi }^{*}_{n,m,l})-(v+\xi )|>\epsilon _n)\le 8\exp \{-\sqrt{2} n\Delta ^2_{\epsilon _n/2, n}\}. \end{aligned}$$

Similar to Lemma 3, we have

$$\begin{aligned} a_0\left( \frac{\epsilon _n}{2}, l\right)= & {} F\left( v+\frac{\epsilon _n}{4}\right) -\frac{\left\lfloor \frac{n+l}{2}\right\rfloor -1}{n}\\= & {} F\left( v+\frac{\epsilon _n}{4}\right) - F(v)+O\left( \frac{1}{n}\right) \\= & {} \frac{F'(v)}{4} \epsilon _n +o(\epsilon _n)+O\left( \frac{1}{n}\right) \\> & {} \frac{M}{n^{1/2}}, \quad \text {for all sufficiently large } n. \end{aligned}$$

The same results hold for $b_0(\frac{\epsilon _n}{2}, l),c_0(\frac{\epsilon _n}{2}, m)$ and $d_0(\frac{\epsilon _n}{2},m)$. Now we have

$$\begin{aligned} \sqrt{2} n\Delta ^2_{\epsilon _n/2, n} \ge \sqrt{2} M^2\quad \text {for all sufficiently large } n, \end{aligned}$$

whence for n large enough

$$\begin{aligned} \textsf{P}(n^{1/2}|(\hat{v}^{*}_{n, l}+\hat{\xi }^{*}_{n,m,l})-(v+\xi )|>DM) \le e^{-\sqrt{2} M^2}<\epsilon , \end{aligned}$$

which implies

$$\begin{aligned} |(\hat{v}^{*}_{n, l}+\hat{\xi }^{*}_{n, m,l})-(v+\xi )|=O_p(n^{-1/2}). \end{aligned}$$

The rest parts can be proved by using the same steps. $\square$

Proof of Lemma 7

Let

$$\begin{aligned} U_n= & {} n^{1/2}[F_n^*(\hat{v}^{*}_{n,l}+\hat{\xi }^{*}_{n,m,l})-F_n^*(v+\xi )]\\ V_n= & {} n^{1/2}[F(\hat{v}^{*}_{n, l}+\hat{\xi }^{*}_{n,m,l})-F(v+\xi )]. \end{aligned}$$

By Taylor expansion and Lemma 6,

$$\begin{aligned} F(\hat{v}^{*}_{n, l}+\hat{\xi }^{*}_{n, m,l})-F(v+\xi )=O(|\hat{v}^{*}_{n, l}+\hat{\xi }^{*}_{n, m,l}-v-\xi |)=O_p(n^{-1/2}),\, n\rightarrow \infty . \end{aligned}$$

Thus $V_n$ satisfies (a) of Lemma 13 of Serfling and Mazumder (2009).

Consider the case $t>0$. Define the right limit as

$$\begin{aligned} \beta :=\lim _{t\rightarrow 0^+}F^{-1}(F(v+\xi )+t/\sqrt{n}). \end{aligned}$$

Since $F^{-1}$ maybe not continuous at $F(v+\xi )$, there are two cases to consider. When $\beta =v+\xi$, using $F(x) < p$ if and only if $x<F^{-1}(p)$, we have

$$\begin{aligned} \begin{aligned} \{V_n\le t\} =&\left\{ F(\hat{v}^{*}_{n, l}+\hat{\xi }^{*}_{n, m,l})-F(v+\xi )\le \frac{t}{\sqrt{n}}\right\} \\ \subset&\left\{ F(\hat{v}^{*}_{n, l}+\hat{\xi }^{*}_{n, m,l})<F(v+\xi )+ \frac{t+\epsilon /2}{\sqrt{n}}\right\} \\ =&\left\{ \hat{v}^{*}_{n, l}+\hat{\xi }^{*}_{n, m,l} < F^{-1}\left( F(v+\xi )+\frac{t+\epsilon /2}{\sqrt{n}}\right) \right\} \\ \subset&\left\{ F_n^*(\hat{v}^{*}_{n, l}+\hat{\xi }^{*}_{n, m,l})\le F_n^*(\eta _n(t))\right\} \end{aligned} \end{aligned}$$

(21)

where

$$\begin{aligned} \eta _n(t)=F^{-1}\left( F(v+\xi )+\frac{t+\epsilon /2}{\sqrt{n}}\right) . \end{aligned}$$

By (21) and the expressions of $U_n$ and $V_n$, we have

$$\begin{aligned} \textsf{P}(U_n\ge t+\epsilon , V_n \le t)\le \textsf{P}\left( F_n^*(\eta _n(t))-F_n^*(v+\xi )\ge \frac{t+\epsilon }{\sqrt{n}}\right) . \end{aligned}$$

(22)

Since F is continuous at $v+\xi$, which implies that $F(\eta _n(t))-F(v+\xi )=\frac{t+\epsilon /2}{\sqrt{n}} >0$. Then for all n sufficiently large

$$\begin{aligned} p_n:=F_n(\eta _n(t))-F_n(v+\xi )>0,\quad a.s. \end{aligned}$$

Then for a sufficiently large n, given $X_1,X_2,\ldots ,X_n$, we have

$$\begin{aligned} Z_n^*=:n\left( F_n^*(\eta _n(t))-F_n^*(v+\xi )\right) \sim \text {Binomial}(n, p_n). \end{aligned}$$

By using the Chebyshev inequality, and noting that $E(p_n)=\frac{t+\epsilon /2}{\sqrt{n}}$, we have

$$\begin{aligned}{} & {} \textsf{P}\left( F_n^*(\eta _n(t))-F_n^*(v+\xi )\ge \frac{t+\epsilon }{\sqrt{n}}\right) \\{} & {} = \textsf{E}\,\left[ \textsf{P}^*\left( F_n^*(\eta _n(t))-F_n^*(v+\xi )\ge \frac{t+\epsilon }{\sqrt{n}}\right) \right] \\{} & {} = \textsf{E}\,\left[ \textsf{P}^*\left( Z_n^*-np_n\ge \sqrt{n}(t+\epsilon )-np_n\right) \right] \\{} & {} \le \textsf{E}\,\left[ \textsf{P}^*\left( |Z_n^*-np_n|\ge \frac{\epsilon }{3}\sqrt{n}\right) \right] \\{} & {} \le \textsf{E}\,\left[ \frac{9p_n(1-p_n)}{\epsilon ^2}\right] \le \frac{9(t+\epsilon /2)}{\sqrt{n} \epsilon ^2}\rightarrow 0,\quad n\rightarrow \infty . \end{aligned}$$

Returning to (22), the first condition in (b) of Lemma 13 in Mazumder and Serfling (2009) is established for $t>0$ and $\beta =v+\xi$.

When $t>0$ and $\beta >v+\xi$, let $\theta$ be any point in the open interval $(v+\xi , \beta )$. As has been proved in Sect. 2 that $\hat{v}^{*}_{n, l}+\hat{\xi }^{*}_{n, m,l}\rightarrow v+\xi ,\, a.s.$ which implies $\textsf{P}(\hat{v}^{*}_{n, l}+\hat{\xi }^{*}_{n, m,l}>\theta )\rightarrow 0$ and

$$\begin{aligned} \textsf{P}(U_n\ge t+\epsilon , V_n\le t)= \textsf{P}(U_n\ge t+\epsilon , V_n\le t, \hat{v}^{*}_{n, l}+\hat{\xi }^{*}_{n, m,l}\le \theta )+o(1), n\rightarrow \infty . \end{aligned}$$

Since $\eta _n(t)\rightarrow \beta >\theta$, then for sufficiently large n

$$\begin{aligned}{} & {} \{V_n\le t, \hat{v}^{*}_{n, l}+\hat{\xi }^{*}_{n, m,l}\le \theta \}\\{} & {} \quad \subset \left\{ F(\hat{v}^{*}_{n, l}+\hat{\xi }^{*}_{n, m,l})<F(v+\xi )+ \frac{t+\epsilon /2}{\sqrt{n}},~ \hat{v}^{*}_{n, l}+\hat{\xi }^{*}_{n, m,l}\le \theta \right\} \\{} & {} \quad \subset \left\{ \hat{v}^{*}_{n, l}+\hat{\xi }^{*}_{n, m,l}<F^{-1}\left( F(v+\xi )+\frac{t+\epsilon /2}{\sqrt{n}}\right) , ~ \hat{v}^{*}_{n,l}+\hat{\xi }^{*}_{n, m,l}\le \theta \right\} \\{} & {} \quad \subset \{F_n^*(\hat{v}^{*}_{n,l}+\hat{\xi }^{*}_{n, m,l})\le F_n^*(\theta )\}. \end{aligned}$$

Then similar to (22), we have

$$\begin{aligned} \textsf{P}(U_n\ge t+\epsilon , V_n\le t, \hat{v}^{*}_{n,l}+\hat{\xi }^{*}_{n, m,l}\le \theta ) \le \textsf{P}\left( F_n^*(\theta )-F_n^*(v+\xi )\ge \frac{t+\epsilon }{\sqrt{n}}\right) .\nonumber \\ \end{aligned}$$

(23)

Note that by the definition of $\beta$ and $\theta$, almost surely there are no sample in the interval $[v+\xi , \theta ]$, hence no bootstrap sample in the same interval. So $F_n^*(\theta )-F_n^*(v+\xi )=0,\, a.s.$ Hence

$$\begin{aligned} \textsf{P}\left( F_n^*(\theta )-F_n^*(v+\xi )\ge \frac{t+\epsilon }{\sqrt{n}}\right) =0. \end{aligned}$$

Thus we establish the first condition in (b) of Lemma 13 in Mazumder and Serfling (2009) for $t>0$. The case $t\le 0$ and the second condition of (b) can be proved similarly. That is, we obtain $H_{2n}=o_p(n^{-1/2})$.

The proof of $H_{1n}=o_p(n^{-1/2})$ follows a similar fashion. We omit the details. $\square$