The proof of Theorem 2.1 will use the famous Berman inequality, which was first presented by Slepian [17] and Berman [18] and then polished up by Li and Shao [19]. For the latest results related to Berman’s inequality, we refer the reader to Hashorva and Weng [20] and Lu and Wang [21]. The upper bound of Berman’s inequality gives an estimate of the difference between two standardized n-dimensional distribution functions by a convenient function of their covariances. According to Hashorva and Weng [20], some results for normal sequences may be extended to nonnormal cases. The proof of Theorem 2.1 also depends on the following lemma of Zhang [7].
Lemma 3.1
Suppose that
\(\{X_{i}, i\geq1\}\)
is a standardized normal sequence with correlation coefficient
\(r_{ij}\)
satisfying (2.1). Define
\(u_{n}=u_{n}(x)=x/a_{n}+b_{n}\)
and
\(\rho_{n}=\gamma/\log n\). Then
-
(i)
\(r_{ij}\rightarrow0\)
as
\(j-i\rightarrow+\infty\),
-
(ii)
\(\sum_{1\leq i< j\leq nb}|r_{ij}-\rho_{n}|\exp (-\frac{u_{n}^{2}}{1+w_{ij}} )\rightarrow 0\)
as
\(n\rightarrow+\infty\),
where
\(0< b<+\infty\)
and
\(w_{ij}=\max\{|r_{ij}|,\rho_{n}\}\).
Proof of Theorem 2.1
It is sufficient to show that, as n goes to ∞,
-
(a)
\(E(N_{n}(B))\rightarrow E(N(B))\) for all sets B of the form \((c,d]\times(r,\delta]\), \(r<\delta\), \(0< c< d\), where \(d\leq1\), and \(E(\cdot)\) is the expectation, and
-
(b)
\(P(N_{n}(B)=0)\rightarrow P(N(B)=0)\) for all sets B that are finite unions of disjoint sets of this form.
First, consider (a). If \(B=(c,d]\times(r,\delta]\) intersects any of the lines, suppose that these are \(L_{s},L_{s+1},\ldots,L_{t}\) (\(1\leq s< t\leq r\)). Then
$$N_{n}(B)=\sum^{t}_{k=s}N_{n}^{(k)}\bigl((c,d]\bigr), \qquad N(B)=\sum^{t}_{k=s}N^{(k)}\bigl((c,d]\bigr), $$
and the number of points \(j/n\) in \((c,d]\) is \(([nd]-[nc])\). As in the proof Theorem 5.5.1 on p.113 in Leadbetter et al. [1], we have \(E(N_{n}(B))=([nd]-[nc])\sum^{t}_{k=s}(1-F(u_{n}^{(k)}))\) and
$$1-F\bigl(u_{n}^{(k)}\bigr)=1-\Phi\bigl(u_{n}^{(k)}-m_{j} \bigr),\quad 1\leq j\leq n. $$
Using conditions (2.2) and (2.3) yields
$$\begin{aligned} n\bigl(1-\Phi\bigl(u_{n}^{(k)}-m_{j} \bigr)\bigr)=n\bigl(1-\Phi (x_{k}/a_{n}+b_{n}+m_{n}-m_{j}) \bigr)\sim e^{-x_{k}}\quad \mbox{as } n\rightarrow\infty, \end{aligned}$$
(3.1)
where the last ‘∼’ attributes to the well-known fact that \(n(1-\Phi(x_{k}/a_{n}+b_{n}))\sim e^{-x}\) implies \(n(1-\Phi(x_{k}/\alpha_{n}+\beta_{n}))\sim e^{-x}\) if \(\alpha_{n}/a_{n}\rightarrow1\) and \((\beta_{n}-b_{n})/a_{n}\rightarrow0\). Thus, we have \(E(N_{n}(B))\sim n(d-c)\sum^{t}_{k=s}(\frac{e^{-x_{k}}}{n}+o(\frac{1}{n}))\rightarrow (d-c)\sum^{t}_{k=s}e^{-x_{k}}\). So, since
$$\begin{aligned} E\bigl(N(B)\bigr) =&\sum^{t}_{k=s}E \bigl((d-c)\exp(-x_{k}-\gamma+\sqrt{2\gamma}\zeta )\bigr) \\ =&\sum^{t}_{k=s}(d-c)e^{-x_{k}-\gamma} \cdot e^{\frac{(\sqrt{2\gamma})^{2}}{2}}=\sum^{t}_{k=s}(d-c)e^{-x_{k}}, \end{aligned}$$
the first result follows. In order to prove (b), we must prove that \(P(N_{n}(B)=0)\rightarrow P(N(B)=0)\), where \(B=\bigcup^{m}_{1}C_{k}\) with disjoint \(C_{k}=(c_{k}, d_{k}]\times(r_{k}, s_{k}]\). It is convenient to neglect any set \(C_{k}\) that does not intersect any of the lines \(L_{1}, L_{2}, \ldots, L_{r}\). Because there are intersections and differences of the intervals \((c_{k}, d_{k}]\), we may write B in the form \(\bigcup^{s}_{k=1}(c_{k}, d_{k}]\times E_{k}\), where \((c_{k}, d_{k}]\) are disjoint, and \(E_{k}\) is a finite union of semiclosed intervals. So we have
$$\begin{aligned} \bigl\{ N_{n}(B)=0\bigr\} =\bigcap ^{s}_{k=1}\bigl\{ N_{n}(F_{k})=0 \bigr\} , \end{aligned}$$
(3.2)
where \(F_{k}=(c_{k},d_{k}]\times E_{k}\). \(L_{l_{k}}\) stands for the lowest \(L_{j}\) intersecting \(F_{k}\). The aforementioned thinning property induces
$$\begin{aligned} \begin{aligned}[b] \bigl\{ N_{n}(F_{k})=0\bigr\} &=\bigl\{ N_{n}^{(l_{k})}\bigl((c_{k},d_{k}]\bigr)=0\bigr\} \\ &= \bigl\{ M_{n}(c_{k},d_{k})\leq u_{n}^{(l_{k})}\bigr\} , \end{aligned} \end{aligned}$$
(3.3)
where \(M_{n}(c_{k},d_{k})\) stands for the maximum of \(\{X_{k}\}\) with index k (\([cn]< k\leq[dn]\)). Calculating the probabilities of (3.2) and (3.3), we obtain
$$\begin{aligned} P\bigl(N_{n}(B)=0\bigr)=P \Biggl(\bigcap ^{s}_{k=1}\bigl\{ M_{n}(c_{k},d_{k}) \leq u_{n}^{(l_{k})}\bigr\} \Biggr). \end{aligned}$$
(3.4)
In order to get the limit of the right-hand side of (3.4), we first prove the following result. Define a sequence \(\{\bar{X}_{i}=\bar{Y}_{i}+m_{i}, i\geq1\}\), where \(\{\bar{Y}_{i}, i\geq1\}\) is a standardized normal sequence with correlation coefficient ρ, and \(\{m_{i}, i\geq1\}\) is the same as that in \(\{X_{i}, i\geq1\}\). \(M_{n}(c,d;\rho)\) stands for the maximum of \(\{\bar{X}_{k}\}\) with index k (\([cn]< k\leq[dn]\)). It is well known that \(M_{n}(c_{1}, d_{1}; \rho),\ldots,M_{n}(c_{k}, d_{k}; \rho)\) have the same distribution as \((1-\rho)^{1/2}M_{n}(c_{1}, d_{1}; 0)+\rho^{1/2}\zeta,\ldots, (1-\rho)^{1/2}M_{n}(c_{k}, d_{k}; 0)+\rho^{1/2}\zeta\), where \(c=c_{1}< d_{1}<\cdots<c_{k}<d_{k}=d\), and ζ is a standard normal variable. In the following, we estimate the bound of
$$\begin{aligned} \Biggl\vert P \Biggl(\bigcap^{s}_{k=1} \bigl\{ M_{n}(c_{k},d_{k})\leq u_{n}^{(l_{k})} \bigr\} \Biggr)-P \Biggl(\bigcap^{s}_{k=1} \bigl\{ M_{n}(c_{k},d_{k},\rho _{n})\leq u_{n}^{(l_{k})}\bigr\} \Biggr)\Biggr\vert , \end{aligned}$$
(3.5)
where \(\rho_{n}=\gamma/\log n\).
Using Berman’s inequality, the bound of (3.5) does not exceed
$$\begin{aligned} \begin{aligned}[b] & \frac{1}{2\pi}\sum \vert r_{ij}- \rho_{n}\vert \bigl(1-\rho_{n}^{2} \bigr)^{-1/2} \exp \biggl(-\frac{\frac{1}{2}((u_{n}^{i})^{2}+(u_{n}^{j})^{2})}{1+\omega _{ij}} \biggr) \\ & \quad \leq C \sum_{1\leq i< j\leq n}\vert r_{ij}-\rho_{n}\vert \exp \biggl(-\frac{\frac {1}{2}((x_{i}/a_{n}+b_{n})^{2}+(x_{j}/a_{n}+b_{n})^{2})}{1+\omega _{ij}} \biggr) \\ &\qquad {}\cdot \exp \bigl(\bigl( (x_{i}/a_{n}+b_{n})\bigl(m_{i}-m_{n}^{*}\bigr)+(x_{i}/a_{n}+b_{n})\bigl(m_{i}-m_{n}^{*}\bigr)\\ &\qquad {} -\frac{1}{2}\bigl(\bigl(m_{n}^{*}-m_{i}\bigr)^{2}+\bigl(m_{n}^{*}-m_{j}\bigr)^{2}\bigr)\bigr)/(1+\omega _{ij}) \bigr), \end{aligned} \end{aligned}$$
(3.6)
where the first sum is carried out over \(i,j\in\bigcup^{s}_{k=1}([c_{k}n],[d_{k}n]]\), \(i< j\), \(u_{n}^{i}\) or \(u_{n}^{j}\) stands for \(u_{n}^{(l_{k})}-m_{i}\) or \(u_{n}^{(l_{k})}-m_{j}\) when \(i\mbox{ or }j \in([c_{k}n],[d_{k}n]]\), and \(\omega_{ij}=\max\{|r_{ij}|,\rho_{n}\}\). Using the proof of Theorem 6.2.1 on p.129 in Leadbetter et al. [1], (2.3) implies that
$$\begin{aligned} \frac{1}{n}\sum^{n}_{i=1} \exp\biggl((x_{i}/a_{n}+b_{n}) \bigl(m_{i}-m_{n}^{*}\bigr)-\frac {1}{2} \bigl(m_{i}-m_{n}^{*}\bigr)^{2}\biggr) \rightarrow1. \end{aligned}$$
Since \(\omega_{ij}\) is bounded, we further get
$$\sup_{1 \leq i< j\leq n}\exp \biggl(\frac{(x_{i}/a_{n}+b_{n})(m_{i}-m_{n}^{*})-\frac {1}{2}(m_{i}-m_{n}^{*})^{2}}{1+\omega_{ij}} \biggr)< C. $$
So, (3.6) does not exceed
$$\begin{aligned} & C \sum_{1\leq i< j\leq n}\vert r_{ij}- \rho_{n}\vert \exp \biggl(-\frac{\frac {1}{2}((x_{i}/a_{n}+b_{n})^{2}+(x_{j}/a_{n}+b_{n})^{2})}{1+\omega _{ij}} \biggr) \\ & \quad < C\sum_{1\leq i< j\leq n}\vert r_{ij}- \rho_{n}\vert \exp \biggl(-\frac{((\min_{1\leq i\leq n}x_{i})/a_{n}+b_{n})^{2}}{1+\omega_{ij}} \biggr) \\ & \quad \rightarrow0. \end{aligned}$$
The last ‘→’ attributes to Lemma 3.1. In order to get the desired limit of (3.4), we only need to prove
$$P \Biggl(\bigcap^{s}_{k=1} \bigl\{ M_{n}(c_{k},d_{k},\rho_{n})\leq u_{n}^{(l_{k})}\bigr\} \Biggr)\rightarrow P\bigl(N(B)=0\bigr). $$
By the definition of \(M_{n}(c_{k}, d_{k}, \rho_{n})\) it clearly follows that
$$\begin{aligned} &P \Biggl(\bigcap^{s}_{k=1} \bigl\{ M_{n}(c_{k},d_{k},\rho_{n})\leq u_{n}^{(l_{k})}\bigr\} \Biggr)\\ &\quad =P \Biggl(\bigcap ^{s}_{k=1}\bigl\{ (1-\rho_{n})^{\frac {1}{2}}M_{n}(c_{k},d_{k},0) +\rho_{n}^{\frac{1}{2}}\zeta\leq u_{n}^{(l_{k})}\bigr\} \Biggr) \\ &\quad = \int^{+\infty}_{-\infty}P \Biggl(\bigcap ^{s}_{k=1}\bigl\{ M_{n}(c_{k},d_{k},0) \leq (1-\rho_{n})^{-\frac{1}{2}}\bigl(u_{n}^{(l_{k})}- \rho_{n}^{\frac{1}{2}}z\bigr)\bigr\} \Biggr)\phi(z)\,dz, \end{aligned}$$
where the proof of the last ‘=’ is the same as the argument on the first line from the bottom on p.136 in Leadbetter et al. [1]. Since \(a_{n}=(2\log n)^{\frac{1}{2}}\), \(b_{n}=a_{n}+O(a_{n}^{-1}\log\log n)\), and \(\rho_{n}=\gamma/\log n\), it is easy to show that
$$\begin{aligned} (1-\rho_{n})^{-\frac{1}{2}}\bigl(u_{n}^{(l_{k})}- \rho_{n}^{\frac{1}{2}}z\bigr) =\frac{x_{l_{k}}+\gamma-\sqrt{2\gamma}z}{a_{n}}+b_{n}+o \bigl(a_{n}^{-1}\bigr). \end{aligned}$$
Furthermore, we have
$$\begin{aligned} & P \Biggl(\bigcap^{s}_{k=1} \bigl\{ M_{n}(c_{k},d_{k},0) \leq (1- \rho_{n})^{-\frac{1}{2}}\bigl(u_{n}^{(l_{k})}- \rho_{n}^{\frac{1}{2}}z\bigr)\bigr\} \Biggr) \\ & \quad =P \Biggl(\bigcap^{s}_{k=1}\bigl\{ \tilde{\xi}_{[c_{k}n]+1}\leq (1-\rho_{n})^{-\frac{1}{2}} \bigl(u_{n}^{(l_{k})}-\rho_{n}^{\frac {1}{2}}z \bigr)-m_{[c_{k}n]+1}, \ldots, \tilde{\xi}_{[d_{k}n]}\\ & \qquad\leq(1- \rho_{n})^{-\frac {1}{2}}\bigl(u_{n}^{(l_{k})}- \rho_{n}^{\frac{1}{2}}z\bigr)-m_{[d_{k}n]}\bigr\} \Biggr) \\ & \quad \rightarrow\prod^{s}_{k=1}\exp \bigl(-(d_{k}-c_{k})e^{-x_{l_{k}}-\gamma+\sqrt {2\gamma}z}\bigr), \end{aligned}$$
where \(\tilde{\xi}_{k}\) stands for independent standard normal variables, and the proof of the last ‘→’ is the same as that of (3.1). Using the dominated convergence theorem yields that
$$\begin{aligned} & \int^{+\infty}_{-\infty}P \Biggl(\bigcap ^{s}_{k=1}\bigl\{ M_{n}(c_{k},d_{k},0) \leq (1-\rho_{n})^{-\frac{1}{2}}\bigl(u_{n}^{(l_{k})}- \rho_{n}^{\frac{1}{2}}z\bigr)\bigr\} \Biggr)\phi(z)\,dz \\ & \quad \rightarrow \int^{+\infty}_{-\infty}\prod^{s}_{k=1} \exp \bigl(-(d_{k}-c_{k})e^{-x_{l_{k}}-\gamma+\sqrt{2\gamma}z}\bigr)\phi(z)\,dz \\ & \quad =P\bigl(N(B)=0\bigr). \end{aligned}$$
The proof of (b) is completed. □
Proof of Corollary 2.1
Using Theorem 2.1, the proof is similar to that of Corollary 5.5.2 in Leadbetter et al. [1]. So we omit it. □
Proof of Theorem 2.2
By Corollary 2.1 the left-hand side of (2.6) converges to
$$\begin{aligned} P\bigl(S^{(1)}=k_{1},S^{(2)}=k_{1}+k_{2}, \ldots,S^{(r)}=k_{1}+k_{2}+\cdots+k_{r} \bigr), \end{aligned}$$
(3.7)
where \(S^{(i)}=N^{(i)}([0, 1])\) is the ith component of N. In our paper, the structure of the Cox process is similar to that of the Poisson process in plane in Leadbetter et al. [1]. So we can refer to the proof of Theorem 5.6.1 in Leadbetter et al. [1], and hence (3.7) equals
$$\begin{aligned} &\frac{(k_{1}+k_{2}+\cdots+k_{r})!}{k_{1}!k_{2}!\cdots k_{r}!} \biggl(\frac{\tau_{1}}{\tau_{r}} \biggr)^{k_{1}} \biggl(\frac{\tau _{2}-\tau_{1}}{\tau_{r}} \biggr)^{k_{2}}\biggl(\frac{\tau_{r}-\tau_{r-1}}{\tau_{r}} \biggr)^{k_{r}} \\ &\quad {}\cdot P\bigl(N^{(r)}\bigl((0,1]\bigr)=k_{1}+k_{2}+ \cdots+k_{r}\bigr). \end{aligned}$$
The proof is completed since
$$\begin{aligned} & P\bigl(N^{(r)}\bigl((0,1]\bigr)=k_{1}+k_{2}+ \cdots+k_{r}\bigr) \\ & \quad = \int^{+\infty}_{-\infty}\frac{(\exp(-x_{r}-\gamma+\sqrt{2\gamma }z))^{k_{1}+k_{2}+\cdots+k_{r}}}{ (k_{1}+k_{2}+\cdots+k_{r})!}\cdot\exp \bigl(-e^{-x_{r}-\gamma+\sqrt{2\gamma }z}\bigr)\phi(z)\,dz \\ & \quad =\frac{(\exp(-x_{r}))^{k_{1}+k_{2}+\cdots+k_{r}}}{ (k_{1}+k_{2}+\cdots+k_{r})!} \int^{+\infty}_{-\infty}\bigl(\exp(-\gamma+\sqrt {2\gamma}z) \bigr)^{k_{1}+k_{2}+\cdots+k_{r}} \\ & \qquad{}\times\exp\bigl(-e^{-x_{r}-\gamma+\sqrt{2\gamma}z}\bigr)\phi(z)\,dz. \end{aligned}$$
□
Proof of Theorem 2.3
Clearly, the left-hand side of (2.7) is equal to
$$\begin{aligned} & P\bigl(a_{n}\bigl(M_{n}^{(1)}-b_{n}-m_{n}^{*} \bigr)\leq x_{1},a_{n}\bigl(M_{n}^{(2)}-b_{n}-m_{n}^{*} \bigr)\leq x_{2}\bigr) \\ & \quad =P\bigl(S_{n}^{(2)}=0\bigr)+ P\bigl(S_{n}^{(1)}=0, S_{n}^{(2)}=1\bigr), \end{aligned}$$
where \(S^{(i)}_{n}\) is the number of exceedances of \(u^{(i)}_{n}\) by \(X_{1}, X_{2}, \ldots, X_{n}\). Using Theorem 2.2 can complete the proof. In order to prove (2.8), write I, J for intervals \(\{1, 2,\ldots, [nt]\}\), \(\{[nt] + 1,\ldots, n\}\), respectively, and \(M^{(1)}(I)\), \(M^{(2)}(I)\), \(M^{(1)}(J)\), \(M^{(2)}(J)\) for the maxima and the second largest of \(\{X_{i},1\leq i\leq n\}\) in the intervals I, J. Let \(K_{n}(x_{1}, x_{2}, x_{3}, x_{4})\) be the joint d.f. of the normalized r.v.s
$$\begin{aligned} & X_{n}^{(1)}=a_{n}\bigl(M_{n}^{(1)}(I)-b_{n}-m_{n}^{*} \bigr),\qquad X_{n}^{(2)}=a_{n}\bigl(M_{n}^{(2)}(I)-b_{n}-m_{n}^{*} \bigr), \\ & Y_{n}^{(1)}=a_{n}\bigl(M_{n}^{(1)}(J)-b_{n}-m_{n}^{*} \bigr),\qquad Y_{n}^{(2)}=a_{n}\bigl(M_{n}^{(2)}(J)-b_{n}-m_{n}^{*} \bigr). \end{aligned}$$
Consider an interesting case of \(x_{1} > x_{2}\) and \(x_{3} > x_{4}\), that is,
$$\begin{aligned} &H_{n}(x_{1},x_{2},x_{3},x_{4})\\ &\quad = P\bigl(M_{n}^{(1)}(I)\leq u_{n}^{(1)},M_{n}^{(2)}(I) \leq u_{n}^{(2)},M_{n}^{(1)}(J)\leq u_{n}^{(3)},M_{n}^{(2)}(J)\leq u_{n}^{(4)}\bigr) \\ &\quad = P\bigl(N_{n}^{(1)}\bigl(I' \bigr)=0,N_{n}^{(2)}\bigl(I'\bigr) \leq1,N_{n}^{(3)}\bigl(J'\bigr)=0, N_{n}^{(4)}\bigl(J'\bigr)\leq1\bigr), \end{aligned}$$
where \(I'=(0,t]\) and \(J'=(t,1]\). By Corollary 2.1 with \(B_{1}=I'\) and \(B_{2} = J'\) we have
$$\begin{aligned} & \lim_{n\rightarrow\infty}H_{n}(x_{1},x_{2},x_{3},x_{4}) \\ & \quad =\lim_{n\rightarrow\infty}P\bigl(N_{n}^{(1)} \bigl(I'\bigr)=0,N_{n}^{(2)}\bigl(I' \bigr)\leq1\bigr)\cdot P\bigl(N_{n}^{(3)}\bigl(J' \bigr)=0, N_{n}^{(4)}\bigl(J'\bigr)\leq1\bigr) \\ & \quad = \int^{+\infty}_{-\infty}\bigl(\bigl(e^{-x_{2}}-e^{-x_{1}} \bigr)t\exp(\sqrt{2\gamma }-\gamma)+1\bigr) \exp\bigl(-te^{-x_{2}-\gamma+\sqrt{2\gamma}z}\bigr) \phi(z)\,dz \\ & \qquad{}\times \int^{+\infty}_{-\infty}\bigl(\bigl(e^{-x_{4}}-e^{-x_{3}} \bigr) (1-t)\exp(\sqrt{2\gamma }-\gamma)+1\bigr) \exp\bigl(-(1-t)e^{-x_{4}-\gamma+\sqrt{2\gamma}z} \bigr)\phi(z)\,dz \\ & \quad =H_{t}(x_{1},x_{2})H_{1-t}(x_{3},x_{4})=H(x_{1},x_{2},x_{3},x_{4}). \end{aligned}$$
Now the left-hand side of (2.8) is equal to
$$\begin{aligned} \begin{aligned}[b] & P\bigl(M^{(2)}_{n}(I)\leq u^{(2)}_{n},M^{(2)}_{n}(I)\geq M^{(1)}_{n}(J)\bigr) \\ & \quad {} +P\bigl(M^{(1)}_{n}(I)\leq u^{(2)}_{n}, M^{(1)}_{n}(J)>M^{(1)}_{n}(I)\geq M^{(2)}_{n}(J)\bigr). \end{aligned} \end{aligned}$$
(3.8)
Obviously, H is absolutely continuous, and the boundaries of sets in \(R^{4}\) such as \(\{(w_{1},w_{2},w_{3}, w_{4}): w_{2}\leq x_{2},w_{2}>w_{3}\}\) and \(\{(w_{1},w_{2},w_{3},w_{4}) : w_{1}\leq x_{2},w_{3} > w_{1}\geq w_{4}\}\) have zero Lebesgue measure. Thus, by Corollary 2.1, (3.8) converges to
$$P(X_{2}\leq x_{2},X_{2}\geq Y_{1})+P(X_{1}\leq x_{2},Y_{1}>X_{1} \geq Y_{2}). $$
Using the joint distribution \(H(x_{1}, x_{2}, x_{3}, x_{4})\) and a simple evaluation complete the proof. □