It is sufficient to show that when
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,
(b)
\(P(N_{n}(B)=0)\rightarrow P(N(B)=0)\) for all sets B which are finite unions of disjoint sets of this form.
Focus on (a) firstly. If
\(B=(c,d]\times(r,\delta]\) intersects any of the lines, suppose these are
\(L_{s},L_{s+1},\ldots,L_{t}\) (
\(1\leq s\leq t\leq r\)). 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 of Theorem 5.5.1 on p.113 in Leadbetter
et al. [
1], we have
\(E(N_{n}(B))=([nd]-[nc])\sum^{t}_{k=s}(1-F(u_{n}^{(k)}))\), where
$$1-F\bigl(u_{n}^{(k)}\bigr)=1-\Phi\bigl(u_{n}^{(k)} \bigr),\quad 1\leq j\leq n. $$
Obviously
$$ n\bigl(1-\Phi\bigl(u_{n}^{(k)}\bigr)\bigr)=n \bigl(1-\Phi(x_{k}/a_{n}+b_{n})\bigr)\sim e^{-x_{k}}\quad \mbox{as } n\rightarrow\infty. $$
(2.3)
Thus, we have
\(E(N_{n}(B))\sim n(d-c)\sum^{t}_{k=s}(\frac{e^{-x_{k}}}{n}+o(\frac{1}{n}))\rightarrow (d-c)\sum^{t}_{k=s}e^{-x_{k}}\). Since
$$\begin{aligned} E\bigl(N(B)\bigr) =&\sum^{t}_{k=s}E \bigl((d-c)\exp(-x_{k}-\gamma+\sqrt{2\gamma}\zeta )\bigr) \\ =&\sum^{t}_{k=s}(d-c)e^{-x_{k}-\gamma} \cdot e^{\frac{(\sqrt{2\gamma})^{2}}{2}} \\ =&\sum^{t}_{k=s}(d-c)e^{-x_{k}}, \end{aligned}$$
(a) follows. In order to prove (b), we must show 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 discard any set
\(C_{k}\) which 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 semi-closed intervals. It therefore follows that
$$ \bigl\{ N_{n}(B)=0\bigr\} =\bigcap ^{s}_{k=1}\bigl\{ N_{n}(F_{k})=0 \bigr\} , $$
(2.4)
where
\(F_{k}=(c_{k},d_{k}]\times E_{k}\). Denote the lowest
\(L_{j}\) intersecting
\(F_{k}\) by
\(L_{l_{k}}\). By the above thinning property, obviously
$$ \bigl\{ N_{n}(F_{k})=0\bigr\} =\bigl\{ N_{n}^{(l_{k})}\bigl((c_{k},d_{k}]\bigr)=0\bigr\} = \bigl\{ M_{n}(c_{k},d_{k})\leq u_{n}^{(l_{k})}\bigr\} , $$
(2.5)
where
\(M_{n}(c_{k},d_{k})\) stands for the maximum of
\(\{\eta_{i}, i\geq1\}\) with index
k (
\([cn]< k\leq[dn]\)). Consider the probabilities of (
2.4) and (
2.5) and obtain
$$ P\bigl(N_{n}(B)=0\bigr)=P \Biggl(\bigcap ^{s}_{k=1}\bigl\{ M_{n}(c_{k},d_{k}) \leq u_{n}^{(l_{k})}\bigr\} \Biggr). $$
(2.6)
It is convenient to firstly prove the following result. Let
\(\{\bar{\xi}_{i}, i\geq1\}\) be a standardized normal sequence with the correlation coefficient
ρ.
\(M_{n}(c,d;\rho)\) stands for the maximum of
\(\{\bar{\xi}_{k}\}\) 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; see Leadbetter
et al. [
1]. Next we must estimate the bound of
$$ \Biggl\vert P \Biggl(\bigcap^{s}_{k=1} \bigl\{ M_{n}(c_{k},d_{k})\leq u_{n}^{(l_{k})} \bigr\} \Biggr)-P \Biggl(\bigcap^{s}_{k=1} \bigl\{ M_{n}(c_{k},d_{k},\rho _{n})\leq u_{n}^{(l_{k})}\bigr\} \Biggr)\Biggr\vert , $$
(2.7)
where
\(\rho_{n}=\gamma/\log n\).
Using Berman’s inequality, the bound of (
2.7) does not exceed
$$ \frac{1}{2\pi}\sum|r_{ij}- \rho_{n}|\bigl(1-\rho_{n}^{2}\bigr)^{-1/2} \exp \biggl(-\frac{\frac {1}{2}((u_{n}^{(i)})^{2}+(u_{n}^{(j)})^{2})}{1+\omega_{ij}} \biggr), $$
(2.8)
where the sum is carried out over
\(i< j\) and
\(i,j\in\bigcup^{s}_{k=1}([c_{k}n],[d_{k}n]]\),
\(u_{n}^{(i)}\) or
\(u_{n}^{(j)}\) stands for
\(x_{i}/a_{n}+b_{n}\) or
\(x_{j}/a_{n}+b_{n}\), and
\(\omega_{ij}=\max\{|r_{ij}|,\rho_{n}\}\). Furthermore, (
2.8) does not exceed
$$\begin{aligned}& C \sum_{1\leq i< j\leq n}|r_{ij}- \rho_{n}| \exp \biggl(-\frac{\frac {1}{2}((x_{i}/a_{n}+b_{n})^{2}+(x_{j}/a_{n}+b_{n})^{2})}{1+\omega _{ij}} \biggr) \\& \quad < C n\sum_{k=1}^{n}|r_{k}- \rho_{n}|\exp \biggl(-\frac{((\min_{1\leq i\leq n}x_{i})/a_{n}+b_{n})^{2}}{1+\omega_{k}} \biggr) \\ & \quad \rightarrow0. \end{aligned}$$
Noting
\(n(1-\Phi((\min_{1\leq i\leq n}x_{i})/a_{n}+b_{n}))\) is bounded, the last ‘→’ attributes to Lemma
2.1. So it suffices to prove
$$ P \Biggl(\bigcap^{s}_{k=1}\bigl\{ M_{n}(c_{k},d_{k},\rho_{n})\leq u_{n}^{(l_{k})}\bigr\} \Biggr)\rightarrow P \Biggl(\bigcap ^{s}_{k=1}\bigl\{ N(B)=0\bigr\} \Biggr). $$
Noting the definition of
\(M_{n}(c_{k}, d_{k}, \rho_{n})\), clearly, it 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 ‘=’ can be completed by using 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
$$ (1-\rho_{n})^{-\frac{1}{2}}\bigl(u_{n}^{(l_{k})}- \rho_{n}^{\frac{1}{2}}z\bigr) =\frac{x_{l_{k}}+\gamma-\sqrt{2\gamma}z}{a_{n}}+b_{n}+o \bigl(a_{n}^{-1}\bigr), $$
see also the proof of Theorem 6.5.1 on p.137 in Leadbetter
et al. [
1]. Furthermore, we may obtain the following result:
$$\begin{aligned}& P \Biggl(\bigcap^{s}_{k=1}\bigl\{ M_{n}(c_{k},d_{k},0) \leq (1- \rho_{n})^{-\frac{1}{2}}\bigl(u_{n}^{(l_{k})}- \rho_{n}^{\frac{1}{2}}z\bigr)\bigr\} \Biggr) \\& \quad =P \Biggl(\bigcap^{s}_{k=1}\bigl\{ \tilde{\zeta}_{[c_{k}n]+1}\leq (1-\rho_{n})^{-\frac{1}{2}} \bigl(u_{n}^{(l_{k})}-\rho_{n}^{\frac{1}{2}}z\bigr), \ldots, \\& \qquad \tilde{\zeta}_{[d_{k}n]}\leq(1-\rho_{n})^{-\frac {1}{2}} \bigl(u_{n}^{(l_{k})}-\rho_{n}^{\frac{1}{2}}z\bigr)\bigr\} \Biggr) \\& \quad \rightarrow\prod^{s}_{k=1}\exp \bigl(-(d_{k}-c_{k})e^{-x_{l_{k}}-\gamma+\sqrt {2\gamma}z}\bigr), \end{aligned}$$
where
\(\tilde{\zeta}_{k}\) is a sequence of independent standard normal variables and we used the same arguments as (
2.3) for the last step. Using the dominated convergence theorem, it follows 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. □