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. □