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Simultaneous confidence bands for comparing variance functions of two samples based on deterministic designs

Abstract

Asymptotically correct simultaneous confidence bands (SCBs) are proposed in both multiplicative and additive form to compare variance functions of two samples in the nonparametric regression model based on deterministic designs. The multiplicative SCB is based on two-step estimation of ratio of the variance functions, which is as efficient, up to order \(n^{-1/2}\), as an infeasible estimator if the two mean functions are known a priori. The additive SCB, which is the log transform of the multiplicative SCB, is location and scale invariant in the sense that the width of SCB is free of the unknown mean and variance functions of both samples. Simulation experiments provide strong evidence that corroborates the asymptotic theory. The proposed SCBs are used to analyze several strata pressure data sets from the Bullianta Coal Mine in Erdos City, Inner Mongolia, China.

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Acknowledgements

This research is part of the first author’s dissertation under the supervision of the second author, and has been supported exclusively by National Natural Science Foundation of China award 11771240. The authors are grateful to Professor Yaodong Jiang for providing the strata pressure data, and two Reviewers for thoughtful comments.

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Correspondence to Lijian Yang.

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Appendix

Appendix

The following is a reformulation of Theorems 11.1.5 and 12.3.5 in Leadbetter et al. (1983).

Lemma 1

If a Gaussian process \(\varsigma \left( s\right) ,0\le s\le T\) is stationary with mean zero and variance one, and covariance function statisfying

$$\begin{aligned} r\left( t\right) =\text{ E }\varsigma \left( s\right) \varsigma \left( t+s\right) =1-C\left| t\right| ^{\alpha }+o\left( \left| t\right| ^{\alpha }\right) ,\text { as }t\rightarrow 0 \end{aligned}$$

for some constant \(C>0,0<\alpha \le 2.\) Then as \(T\rightarrow \infty \),

$$\begin{aligned} \text{ P }\left[ a_{T}\left\{ \sup _{t\in \left[ 0,T\right] }\left| \varsigma \left( t\right) \right| -b_{T}\right\} \le z\right] \rightarrow e^{-2e^{-z}},\forall z\in {\mathbb {R}}, \end{aligned}$$

where \(a_{T}=\left( 2\log T\right) ^{1/2}\) and

$$\begin{aligned} b_{T}=a_{T}+a_{T}^{-1}\times \left\{ \left( \frac{1}{\alpha }-\frac{1}{2} \right) \log \left( a_{T}^{2}/2\right) +\log \left( 2\pi \right) ^{-1/2}\left( C^{\frac{1}{\alpha }}H_{\alpha }2^{\frac{2-\alpha }{2\alpha } }\right) \right\} \end{aligned}$$

with \(H_{1}=1,H_{2}=\pi ^{-1/2}\).

Lemmas 24 are from Cai et al. (2019).

Lemma 2

Under Assumption (A6), for \(s=1,2\), as \(n\rightarrow \infty \),

$$\begin{aligned} \sup _{x\in {\mathcal {I}}_{n}}\left| {\hat{f}}_{s}\left( x\right) -1\right| ={\mathcal {O}}\left( n_{s}^{-1}h^{-2}\right) . \end{aligned}$$

Lemma 3

Under Assumptions (A2), (A6) and (A7), for \(s=1,2\), as \( n\rightarrow \infty \),

$$\begin{aligned} \sup _{x\in {\mathcal {I}}_{n}}\left| A_{s,n_{s}}\left( x\right) \right| ={\mathcal {O}}\left( h^{\theta _{0}+p_{0}-1}+n_{s}^{-1}h^{-1}\right) . \end{aligned}$$

Lemma 4

Under Assumptions (A2)–(A4), (A6), (A7), for \(s=1,2\), as \(n\rightarrow \infty ,\)

$$\begin{aligned}&(a) \sup _{x\in \left[ 0,1\right] }\left| B_{s,n_{s}}\left( x\right) -B_{s,n_{s},1}\left( x\right) \right| ={\mathcal {O}}_{p}\left( n_{s}^{\beta _{s}-1}h^{-1}\right) , \\&\quad (b) \sup _{x\in \left[ 0,1\right] }\left| B_{s,n_{s},1}\left( x\right) -B_{s,n_{s},2}\left( x\right) \right| ={\mathcal {O}}_{p}\left( n_{s}^{-1/2}h^{1/2}\log ^{1/2}n_{s}\right) , \\&\quad (c) \sup _{x\in {\mathcal {I}}_{n}}\left| B_{s,n_{s},2}\left( x\right) -B_{s,n_{s},3}\left( x\right) \right| ={\mathcal {O}}_{p}\left( n_{s}^{-3/2}h^{-2}\log ^{1/2}n_{s}\right) , \\&\quad (d) \sup _{x\in \left[ 0,1\right] }\left| B_{s,n_{s},3}\left( x\right) \right| ={\mathcal {O}}_{p}\left( n_{s}^{-1/2}h^{-1/2}\log ^{1/2}n_{s}\right) . \end{aligned}$$

Denote

$$\begin{aligned} B_{n_{1},n_{2}}\left( x\right) =\sigma _{1}^{-2}\left( x\right) B_{1,n_{1}}\left( x\right) -\sigma _{2}^{-2}\left( x\right) B_{2,n_{2}}\left( x\right) \\ B_{n_{1},n_{2,}3}\left( x\right) =\sigma _{1}^{-2}\left( x\right) B_{1,n_{1},3}\left( x\right) -\sigma _{2}^{-2}\left( x\right) B_{2,n_{2},3}\left( x\right) . \end{aligned}$$

Lemma 5

Under Assumptions (A2)–(A4), (A6), (A7), as \( n\rightarrow \infty ,\)

$$\begin{aligned}&\sup _{x\in {\mathcal {I}}_{n}}\left| \ln \frac{{\tilde{\sigma }} _{1}^{2}\left( x\right) }{{\tilde{\sigma }}_{2}^{2}\left( x\right) }-\ln \frac{ \sigma _{1}^{2}\left( x\right) }{\sigma _{2}^{2}\left( x\right) } -B_{n_{1},n_{2,}3}\left( x\right) \right| \\&\quad ={\mathcal {O}}_{p}\left( n_{1}^{-1/2}h^{-1/2}\log ^{1/2}n_{1}+n_{2}^{-1/2}h^{-1/2}\log ^{1/2}n_{2}\right) \\&\qquad +{\mathcal {O}}_{p}\left( h^{\theta _{0}+p_{0}-1}+n_{1}^{\beta _{1}-1}h^{-1}+n_{2}^{\beta _{2}-1}h^{-1}\right) +o_{p}\left( 1\right) . \end{aligned}$$

Consequently,

$$\begin{aligned} a_{h}\left\{ v_{n}^{-1}\sup _{x\in {\mathcal {I}}_{n}}\left| \ln \frac{ {\tilde{\sigma }}_{1}^{2}\left( x\right) }{{\tilde{\sigma }}_{2}^{2}\left( x\right) }-\ln \frac{\sigma _{1}^{2}\left( x\right) }{\sigma _{2}^{2}\left( x\right) }\right| \right\} =a_{h}\left\{ v_{n}^{-1}\sup _{x\in {\mathcal {I}} _{n}}\left| B_{n_{1},n_{2,}3}\left( x\right) \right| \right\} +o_{p}\left( 1\right) , \end{aligned}$$

where \(a_{h}\) and \(v_{n}\) are given in (5).

Proof According to Lemmas 24, one has

$$\begin{aligned}&\sup _{x\in {\mathcal {I}}_{n}}\left| {\hat{f}}_{s}^{-1}\left( x\right) \left\{ A_{s,n_{s}}\left( x\right) +B_{s,n_{s}}\left( x\right) \right\} \right| \\&\quad \le \sup _{x\in {\mathcal {I}}_{n}}\left| {\hat{f}}_{s}^{-1}\left( x\right) \right| \sup _{x\in {\mathcal {I}}_{n}}\left| A_{s,n_{s}}\left( x\right) +B_{s,n_{s}}\left( x\right) \right| \\&\quad \le \left\{ 1+{\mathcal {O}}\left( n_{s}^{-1}h^{-2}\right) \right\} \left\{ \sup _{x\in {\mathcal {I}}_{n}}\left| A_{s,n_{s}}\left( x\right) \right| +\sup _{x\in {\mathcal {I}}_{n}}\left| B_{s,n_{s}}\left( x\right) \right| \right\} \\&\quad \le {\mathcal {O}}\left( h^{\theta _{0}+p_{0}-1}+n_{s}^{-1}h^{-1}\right) +\sup _{x\in {\mathcal {I}}_{n}}\left| B_{s,n_{s}}\left( x\right) \right| \\&\le {\mathcal {O}}\left( h^{\theta _{0}+p_{0}-1}+n_{s}^{-1}h^{-1}\right) +\sup _{x\in \left[ 0,1\right] }\left| B_{s,n_{s}}\left( x\right) -B_{s,n_{s},1}\left( x\right) \right| \\&\qquad +\sup _{x\in \left[ 0,1\right] }\left| B_{s,n_{s},1}\left( x\right) -B_{s,n_{s},2}\left( x\right) \right| \\&\qquad +\sup _{x\in {\mathcal {I}}_{n}}\left| B_{s,n_{s},2}\left( x\right) -B_{s,n_{s},3}\left( x\right) \right| +\sup _{x\in \left[ 0,1\right] }\left| B_{s,n_{s},3}\left( x\right) \right| \\&\quad \le {\mathcal {O}}_{p}\left( h^{\theta _{0}+p_{0}-1}+n_{s}^{\beta _{s}-1}h^{-1}+n_{s}^{-1/2}h^{-1/2}\log ^{1/2}n_{s}\right) \end{aligned}$$

Now applying Taylor series expansions to \(\ln {\tilde{\sigma }}_{s}^{2}\left( x\right) -\ln \sigma _{s}^{2}\left( x\right) \), for \(s=1,2\)

$$\begin{aligned}&\sup _{x\in {\mathcal {I}}_{n}}\left| \ln {\tilde{\sigma }}_{s}^{2}\left( x\right) -\ln \sigma _{s}^{2}\left( x\right) \right| \\= & {} \sup _{x\in {\mathcal {I}}_{n}}\left| \ln \left[ \sigma _{s}^{2}\left( x\right) +{\hat{f}}_{s}^{-1}\left( x\right) \left\{ A_{s,n_{s}}\left( x\right) +B_{s,n_{s}}\left( x\right) \right\} \right] -\ln \sigma _{s}^{2}\left( x\right) \right| \\\le & {} \sup _{x\in {\mathcal {I}}_{n}}\left| \sigma _{s}^{-2}\left( x\right) {\hat{f}}_{s}^{-1}\left( x\right) \left\{ A_{s,n_{s}}\left( x\right) +B_{s,n_{s}}\left( x\right) \right\} \right| +o_{p}\left( 1\right) . \end{aligned}$$

Then one obtains

$$\begin{aligned}&\ln \frac{{\tilde{\sigma }}_{1}^{2}\left( x\right) }{{\tilde{\sigma }} _{2}^{2}\left( x\right) }-\ln \frac{\sigma _{1}^{2}\left( x\right) }{\sigma _{2}^{2}\left( x\right) }-B_{n_{1},n_{2,},3}\left( x\right) \\&\quad =\ln {\tilde{\sigma }}_{1}^{2}\left( x\right) -\ln \sigma _{1}^{2}\left( x\right) -\left\{ \ln {\tilde{\sigma }}_{2}^{2}\left( x\right) -\ln \sigma _{2}^{2}\left( x\right) \right\} -B_{n_{1},n_{2,},3}\left( x\right) \\&\quad =\sigma _{1}^{-2}\left( x\right) {\hat{f}}_{1}^{-1}\left( x\right) \left\{ A_{1,n_{1}}\left( x\right) +B_{1,n_{1}}\left( x\right) \right\} \\&\qquad -\sigma _{2}^{-2}\left( x\right) {\hat{f}}_{2}^{-1}\left( x\right) \left\{ A_{2,n_{2}}\left( x\right) +B_{2,n_{2}}\left( x\right) \right\} -B_{n_{1},n_{2,},3}\left( x\right) +u_{p}\left( 1\right) \\&\quad =\sigma _{1}^{-2}\left( x\right) {\hat{f}}_{1}^{-1}\left( x\right) A_{1,n_{1}}\left( x\right) -\sigma _{2}^{-2}\left( x\right) {\hat{f}} _{2}^{-1}\left( x\right) A_{2,n_{2}}\left( x\right) \\&\qquad +\sigma _{1}^{-2}\left( x\right) \left\{ {\hat{f}}_{1}^{-1}\left( x\right) -1\right\} B_{1,n_{1}}\left( x\right) -\sigma _{2}^{-2}\left( x\right) \left\{ {\hat{f}}_{2}^{-1}\left( x\right) -1\right\} B_{2,n_{2}}\left( x\right) \\&\qquad +B_{n_{1},n_{2}}\left( x\right) -B_{n_{1},n_{2,},3}\left( x\right) +u_{p}\left( 1\right) . \end{aligned}$$

Since one has

$$\begin{aligned}&B_{n_{1},n_{2}}\left( x\right) -B_{n_{1},n_{2,},3}\left( x\right) =\sigma _{1}^{-2}\left( x\right) B_{1,n_{1}}\left( x\right) -\sigma _{2}^{-2}\left( x\right) B_{2,n_{2}}\left( x\right) \nonumber \\&\quad =\sigma _{1}^{-2}\left( x\right) \left\{ B_{1,n_{1}}\left( x\right) -B_{1,n_{1},1}\left( x\right) \right\} +\sigma _{1}^{-2}\left( x\right) \left\{ B_{1,n_{1},1}\left( x\right) -B_{1,n_{1},2}\left( x\right) \right\} \nonumber \\&\qquad +\sigma _{1}^{-2}\left( x\right) \left\{ B_{1,n_{1},2}\left( x\right) -B_{1,n_{1},3}\left( x\right) \right\} -\sigma _{2}^{-2}\left( x\right) \left\{ B_{2,n_{2}}\left( x\right) -B_{2,n_{2},1}\left( x\right) \right\} \nonumber \\&\qquad -\sigma _{2}^{-2}\left( x\right) \left\{ B_{2,n_{2},1}\left( x\right) -B_{2,n_{2},2}\left( x\right) \right\} -\sigma _{2}^{-2}\left( x\right) \left\{ B_{2,n_{2},2}\left( x\right) -B_{2,n_{2},3}\left( x\right) \right\} , \nonumber \\ \end{aligned}$$
(14)

and according to Lemmas 24, one has

$$\begin{aligned}&\sup _{x\in {\mathcal {I}}_{n}}\left| \sigma _{s}^{-2}\left( x\right) {\hat{f}} _{s}^{-1}\left( x\right) A_{s,n_{s}}\left( x\right) \right| ={\mathcal {O}} \left( h^{\theta _{0}+p_{0}-1}+n_{s}^{-1}h^{-1}\right) , \end{aligned}$$
(15)
$$\begin{aligned}&\sup _{x\in {\mathcal {I}}_{n}}\left| \sigma _{s}^{-2}\left( x\right) \left\{ {\hat{f}}_{s}^{-1}\left( x\right) -1\right\} B_{1,n_{s}}\left( x\right) \right| ={\mathcal {O}}_{p}\left( n_{s}^{\beta _{s}-1}h^{-1}+n_{s}^{-1/2}h^{-1/2}\log ^{1/2}n_{s}\right) , \nonumber \\ \end{aligned}$$
(16)

Hence combining (14), (15) and (16), the proof is completed.

Denote the following processes

$$\begin{aligned}&Y_{1,n_{1},1}\left( x\right) =h^{-1}n_{1}^{-1/2}\left( \mu _{1,4}-1\right) ^{1/2}\int K\left( x-u/h\right) dW_{1,n_{1}}(u),x\in \left[ 1,h^{-1}-1\right] , \\&\quad Y_{2,n_{2,}1}\left( x\right) =h^{-1}n_{2}^{-1/2}\left( \mu _{2,4}-1\right) ^{1/2}\int K\left( x-u/h\right) dW_{2,n_{2}}(u),x\in \left[ 1,h^{-1}-1\right] , \\&\quad Y_{1,n_{1},2}\left( x\right) =h^{-1/2}n_{1}^{-1/2}\left( \mu _{1,4}-1\right) ^{1/2}\int K\left( x-r\right) dW_{1,n_{1}}(r),x\in \left[ 1,h^{-1}-1\right] , \\&\quad Y_{2,n_{2},2}\left( x\right) =h^{-1/2}n_{2}^{-1/2}\left( \mu _{2,4}-1\right) ^{1/2}\int K\left( x-r\right) dW_{2,n_{2}}(r),x\in \left[ 1,h^{-1}-1\right] . \end{aligned}$$

As \(\text{ E }\left\{ B_{n_{1},n_{2,}3}^{2}\left( x\right) \right\} =h^{-1}\left\{ n_{1}^{-1}\left( \mu _{1,4}-1\right) +n_{2}^{-1}\left( \mu _{2,4}-1\right) \right\} \int _{-1}^{1}K^{2}\left( u\right) du\), one obtains the following standard Gaussian processes,

$$\begin{aligned}&\bigtriangleup _{1}\left( x\right) =\frac{B_{n_{1},n_{2,}3}\left( x\right) }{ h^{-1/2}\left[ \left\{ n_{1}^{-1}\nu _{1,4}+n_{2}^{-1}\nu _{2,4}\right\} \int _{-1}^{1}K^{2}\left( u\right) du\right] ^{1/2}},x\in \left[ h,1-h\right] , \end{aligned}$$
(17)
$$\begin{aligned}&\bigtriangleup _{2}\left( x\right) =\frac{Y_{1,n_{1},1}\left( x\right) -Y_{2,n_{2,}1}\left( x\right) }{h^{-1/2}\left[ \left\{ n_{1}^{-1}\nu _{1,4}+n_{2}^{-1}\nu _{2,4}\right\} \int _{-1}^{1}K^{2}\left( u\right) du \right] ^{1/2}},x\in \left[ 1,h^{-1}-1\right] , \nonumber \\ \end{aligned}$$
(18)

where \(\nu _{1,4}=\mu _{1,4}-1\) and \(\nu _{2,4}=\mu _{2,4}-1\).

Another standard Gaussian process is

$$\begin{aligned} \frac{Y_{1,n_{1},2}\left( x\right) -Y_{2,n_{2,}2}\left( x\right) }{h^{-1/2} \left[ \left\{ n_{1}^{-1}\nu _{1,4}+n_{2}^{-1}\nu _{2,4}\right\} \int _{-1}^{1}K^{2}\left( u\right) du\right] ^{1/2}},x\in \left[ 1,h^{-1}-1 \right] , \end{aligned}$$

which is \(\zeta \left( x\right) \) defined in (10).

Lemma 6

The absolute maximum of the process \(\bigtriangleup _{1}\left( x\right) \) follows the same as that of \(\bigtriangleup _{2}\left( x\right) \), and the absolute maximum of the process \(\bigtriangleup _{2}\left( x\right) \) follows the same as that of \(\zeta \left( x\right) \), that is

$$\begin{aligned} \sup _{x\in \left[ h,1-h\right] }\left| \bigtriangleup _{1}\left( x\right) \right| \overset{d}{=}\sup _{x\in \left[ 1,h^{-1}-1\right] }\left| \bigtriangleup _{2}\left( x\right) \right| \overset{d}{=} \sup _{x\in \left[ 1,h^{-1}-1\right] }\left| \zeta \left( x\right) \right| . \end{aligned}$$

Proof This lemma can be easily obtained by noting the fact that for \(s=1,2\), the process \(B_{n_{1},n_{2,}3}\left( x\right) ,x\in \left[ h,1-h\right] \) has the same probability law as \(Y_{1,n_{1},1}\left( x\right) -Y_{2,n_{2,}1}\left( x\right) ,x\in \left[ 1,h^{-1}-1\right] \), and the process \(Y_{s,n_{s},1}\left( x\right) ,x\in \left[ h,1-h\right] \) has the same probability law as \(Y_{s,n_{s},2}\left( x\right) ,x\in \left[ 1,h^{-1}-1\right] \).

Proof of Proposition 1

Proposition 1 is a direct corollary of Lemma 5, Lemma 6 and Proposition 3.

Proof of Proposition 2

According to Theorem 2 in Cai et al. (2019) and applying Taylor expansion, one has

$$\begin{aligned}&\sup _{x\in {\mathcal {I}}_{n}}\left| \ln \frac{{\hat{\sigma }}_{1}^{2}\left( x\right) }{{\hat{\sigma }}_{2}^{2}\left( x\right) }-\ln \frac{{\tilde{\sigma }} _{1}^{2}\left( x\right) }{{\tilde{\sigma }}_{2}^{2}\left( x\right) }\right| =\sup _{x\in {\mathcal {I}}_{n}}\left| \ln {\hat{\sigma }}_{1}^{2}\left( x\right) -\ln {\tilde{\sigma }}_{1}^{2}\left( x\right) -\left\{ \ln {\hat{\sigma }} _{2}^{2}\left( x\right) -\ln {\tilde{\sigma }}_{2}^{2}\left( x\right) \right\} \right| \\&\quad \le \sup _{x\in {\mathcal {I}}_{n}}\left| \ln {\hat{\sigma }}_{1}^{2}\left( x\right) -\ln {\tilde{\sigma }}_{1}^{2}\left( x\right) \right| +\sup _{x\in {\mathcal {I}}_{n}}\left| \ln {\hat{\sigma }}_{2}^{2}\left( x\right) -\ln {\tilde{\sigma }}_{2}^{2}\left( x\right) \right| \\&\quad =\sup _{x\in {\mathcal {I}}_{n}}\left| {\tilde{\sigma }}_{1}^{-2}\left( x\right) \left\{ {\hat{\sigma }}_{1}^{2}\left( x\right) -{\tilde{\sigma }} _{1}^{2}\left( x\right) \right\} \right| +\sup _{x\in {\mathcal {I}} _{n}}\left| {\tilde{\sigma }}_{2}^{-2}\left( x\right) \left\{ {\hat{\sigma }} _{2}^{2}\left( x\right) -{\tilde{\sigma }}_{2}^{2}\left( x\right) \right\} \right| +{\mathcal {O}}_{p}(n_{1}^{-1}+n_{2}^{-1}) \\&\quad \le c_{\sigma }^{-2}\sup _{x\in {\mathcal {I}}_{n}}\left| {\hat{\sigma }} _{1}^{2}\left( x\right) -{\tilde{\sigma }}_{1}^{2}\left( x\right) \right| +c_{\sigma }^{-2}\sup _{x\in {\mathcal {I}}_{n}}\left| {\hat{\sigma }} _{2}^{2}\left( x\right) -{\tilde{\sigma }}_{2}^{2}\left( x\right) \right| + {\mathcal {O}}_{p}(n_{1}^{-1}+n_{2}^{-1})=o_{p}(n^{-1/2}), \end{aligned}$$

which completes the proof.

Proof of Proposition 3

For Gaussian process \(\zeta \left( x\right) \), its correlation function is

$$\begin{aligned}&r\left( x-y\right) =\text{ corr }\left( \zeta \left( x\right) ,\zeta \left( y\right) \right) =\frac{\text{ E }\left\{ \zeta \left( x\right) \zeta \left( y\right) \right\} }{\text{ var}^{1/2}\left\{ \zeta \left( x\right) \right\} \text{ var}^{1/2}\left\{ \zeta \left( y\right) \right\} } \\&\quad =\frac{\left( n_{1}^{-1}\nu _{1,4}+n_{2}^{-1}\nu _{2,4}\right) \left( K*K\right) \left( x-y\right) }{\left( n_{1}^{-1}\nu _{1,4}+n_{2}^{-1}\nu _{2,4}\right) \int _{-1}^{1}K^{2}\left( u\right) du} \\&\quad =\frac{\left( K*K\right) \left( x-y\right) }{\int _{-1}^{1}K^{2}\left( u\right) du}, \end{aligned}$$

which implies that

$$\begin{aligned} r\left( t\right) =\frac{\int K\left( u\right) K\left( u-t\right) du}{ \int _{-1}^{1}K^{2}\left( u\right) du}. \end{aligned}$$

Define next a Gaussian process \(\varsigma \left( t\right) ,0\le t\le T=T_{n}=h^{-1}-2\),

$$\begin{aligned} \varsigma \left( t\right) =\zeta \left( t+1\right) \left\{ \int _{-1}^{1}K^{2}\left( u\right) du\right\} ^{-1/2}, \end{aligned}$$

which is stationary with mean zero and variance one, and covariance function

$$\begin{aligned} r\left( t\right) =\text{ E }\varsigma \left( s\right) \varsigma \left( t+s\right) =1-Ct^{2}+o\left( t^{2}\right) ,t\rightarrow 0, \end{aligned}$$

with \(C=\int _{-1}^{1}K^{\left( 1\right) }\left( u\right) ^{2}du/2\int _{-1}^{1}K^{2}\left( u\right) du\). Hence applying Lemmas 16, one has for \( h\rightarrow 0\) or \(T\rightarrow \infty ,\)

$$\begin{aligned} \text{ P }\left[ a_{T}\left\{ \sup _{t\in \left[ 0,T\right] }\left| \varsigma \left( t\right) \right| -b_{T}\right\} \le z\right] \rightarrow e^{-2e^{-z}},\forall z\in {\mathbb {R}}, \end{aligned}$$

where \(a_{T}=\left( 2\log T\right) ^{1/2}\) and \(b_{T}=a_{T}+a_{T}^{-1}\left\{ 2^{-1}\mathrm {log}\left( C_{K}/\left( 4\pi ^{2}\right) \right) \right\} \) . Note that

$$\begin{aligned} a_{h}a_{T}^{-1}\rightarrow 1,a_{T}\left( b_{T}-b_{h}\right) \rightarrow 0. \end{aligned}$$

Hence, applying Slutsky’s Theorem twice, one obtains that

$$\begin{aligned} a_{h}\left\{ \sup _{t\in \left[ 0,T\right] }\left| \varsigma \left( t\right) \right| -b_{h}\right\}= & {} a_{h}a_{T}^{-1}\left[ a_{T}\left\{ \sup _{t\in \left[ 0,T\right] }\left| \varsigma \left( t\right) \right| -b_{T}\right\} \right] \\&+a_{h}\left( b_{T}-b_{h}\right) , \end{aligned}$$

which converges in distribution to the same limit as \(a_{T}\left\{ \sup _{t\in \left[ 0,T\right] }\left| \varsigma \left( t\right) \right| -b_{T}\right\} \). Thus

$$\begin{aligned} \text{ P }\left[ a_{h}\left\{ \sup _{s\in \left[ 1,h^{-1}-1\right] }\left| \zeta \left( s\right) \right| -b_{h}\right\} <z\right] \rightarrow \exp \left\{ -2\exp \left( -z\right) \right\} ,z\in {\mathbb {R}}. \end{aligned}$$

Hence the proof is completed.

Proof of Theorem 1

According to Proposition 1, as \(n\rightarrow \infty ,\)

$$\begin{aligned} {\mathbb {P}}\left[ a_{h}\left\{ v_{n}^{-1}\sup _{x\in {\mathcal {I}} _{n}}\left| \ln \frac{{\tilde{\sigma }}_{1}^{2}\left( x\right) }{{\tilde{\sigma }}_{2}^{2}\left( x\right) }-\ln \frac{\sigma _{1}^{2}\left( x\right) }{ \sigma _{2}^{2}\left( x\right) }\right| -b_{h}\right\} \le z\right] \rightarrow \exp \left\{ -2\exp \left( -z\right) \right\} ,z\in {\mathbb {R}} , \nonumber \\ \end{aligned}$$
(19)

where \(a_{h},b_{h}\) and \(v_{n}\) are given in (5). Finally applying Proposition 2, one obtains

$$\begin{aligned} a_{h}\left\{ v_{n}^{-1}\sup _{x\in {\mathcal {I}}_{n}}\left| \ln \frac{{\hat{\sigma }}_{1}^{2}\left( x\right) }{{\hat{\sigma }}_{2}^{2}\left( x\right) }-\ln \frac{{\tilde{\sigma }}_{1}^{2}\left( x\right) }{{\tilde{\sigma }}_{2}^{2}\left( x\right) }\right| \right\} =o_{p}\left( \left\{ \log \left( h^{-1}\right) \right\} ^{1/2}h^{1/2}\right) =o_{p}\left( 1\right) . \end{aligned}$$

Using Slutsky’s Theorem one can substitute \(\ln \frac{{\hat{\sigma }} _{1}^{2}\left( x\right) }{{\hat{\sigma }}_{2}^{2}\left( x\right) }\) for \(\ln \frac{{\tilde{\sigma }}_{1}^{2}\left( x\right) }{{\tilde{\sigma }}_{2}^{2}\left( x\right) }\) in (19). Hence the proof of Theorem 1 is completed.

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Zhong, C., Yang, L. Simultaneous confidence bands for comparing variance functions of two samples based on deterministic designs. Comput Stat 36, 1197–1218 (2021). https://doi.org/10.1007/s00180-020-01043-6

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Keywords

  • Brownian motion
  • B-spline
  • Kernel
  • Oracle efficiency
  • Strata pressure
  • Variance ratio