text:
\begin{document}

\title[Gaussian Approximation of spacings]{Gaussian Approximations and Related Questions for the Spacings process}
\author{Gane Samb LO}
\address{Research address. L.S.T.A., Universit\'{e} Paris VI. T.45/55, E.3. 4, Place
Jussieu, F-75230, Paris C\'{e}dex 05. France.}

\keywords{Key words. Spacings, empirical process, oscillation modulus, strong and weak
approximation, order statistics, gamma distribution and function, law of the
iterated logarithm.}

\begin{abstract}
\large
All the available results on the approximation of the k-spacings
process to Gaussian processes have only used one approach, that is the Shorack and Pyke's
one. Here, it is shown that this approach cannot yield a rate better than $
\left( N/\log \log N\right) ^{-\frac{1}{4}}\left( \log N\right) ^{\frac{1}{2}
}$. Strong and weak bounds for that rate are specified both where k is fixed
and where $k\rightarrow +\infty $. A Glivenko-Cantelli Theorem is given
while Stute's result for the increments of the empirical process based on
independent and indentically distributed random variables is extended to the spacings process. One of the
Mason-Wellner-Shorack cases is also obtained.

\end{abstract}

\maketitle

\noindent \textbf{Nota Bene.} This paper was part of the PhD thesis, Cheikh Anta Diop University, 1991, not yet published in a peer-reviewed journal by August 2014.

\bigskip

\section{Introduction}

\Large

\noindent The non-overlapping uniform k-spacings are defined by
\begin{equation*}
D_{i,n}^{k}=U_{ik,n}-U_{\left( i-1\right) k,n},\text{ \ }1\leq i\leq \left[
\frac{n+1}{k}\right] =N,
\end{equation*}

\noindent where $0\equiv U_{0,n}\leq U_{1,n}\leq ...\leq U_{n,n}\leq U_{n+1,n}\equiv 1$
are the order statistics of a sequence $U_{1},...,U_{n}$ of independent
random variables (r.v.'s) uniformly distributed on $\left( 0,1\right) $ and $[x]$ denotes the integer part of $x$. The study of these r.v.'s have received a
great amount of attention in recent years (see \cite{2}, \cite{5}, \cite{10}
and \cite{13}). Particularly the related empirical process
\begin{equation*}
\beta _{N}\left( x\right) =N^{\frac{1}{2}}\left\{ F_{N}\left( x\right)
-H_{k}\left( x\right) \right\} ,0\leq x\leq +\infty ,
\end{equation*}

\noindent  where
$$
F_{N}\left( x\right) =\#\left\{ i,1\leq i\leq N,\text{ }
NkD_{i,n}^{k}\leq x\right\} /N
$$

\noindent and
$$
H_{k}\left( x\right) =\int_{0}^{x}\frac{
t^{k}e^{-t}}{\left( k-1\right) !} dt, \ \ \ x\geq 0.
$$

\noindent plays a fundamental role in many areas in
statistics (see \cite{5}). All its aspects are have described by various
authors.

\bigskip

\noindent (i) For the convergence of statistics based on spacings, it is helpful to
have a Glivenko-Cantelli Theorem for $F_{N}\left( .\right) $. Such results
for the overlapping case are available in \cite{3}.\\

\bigskip

\noindent (ii) The limiting law of the spacings statistics may follow from suitable
approximations of $\beta _{N}$ to Gaussian processes. It is clear that the
better the rates of those approximations are the less restrictive the
conditions on the underlying random variables ($r.v.$). Such approximations also yield
Kolmogorov-Smirov's tests.\\

\bigskip

\noindent (iii) Finally, the oscillation modulus of $\beta _{N}$ has been studied in
\cite{7}, where is established the weak behaviour of the oscillation moduli of $\beta
_{N}$ is equivalent to that of the empirical process based on a sequence of
independent and indentically distributed ($i.i.d$) random variables.\\

\noindent Our aim is to give strong versions of weak characterizations of the
oscillation moduli that we have already given in \cite{7}. As to the
approximation of $\beta _{n}$ to Gaussian processes, we will show that the
rate given in \cite{7} is, in fact, a strong one. Our best achievement is
that this rate is the best attainable for the approach used until now and
we provide the corresponding bounds. With respect to \cite{1} and \cite{2}, we
do not let k fixed. We allow it to go to infinity. Finally we give the
Glivenko-Cantelli Theorem for $F_{N}$ with almost the same condition as in
\cite{3} for the overlapping case.

\bigskip

\section{The Gaussian approximation.}

\bigskip

\noindent Approximations of $\beta _{N}$ to Gaussian processes are available since
\cite{12}. The best rates among those already given are due to \cite{1} and
to \cite{2}. Among other results, \cite{2} proved the following theorem and
corollary.

\bigskip

\begin{theorem} \label{t1}. There exists a probability space carrying a sequence $U_{1},U_{2},...$ of
independent r.v.'s uniformly distributed on $\left( 0,1\right) $ and a
sequences of Gaussian processes $\left\{ W_{N}\left( x\right) ,0\leq x\leq
+\infty \right\} ,$ $N=1,2,...$ satisfying
\begin{equation*}
\forall N>1,\text{ }\mathbb{E}\left( W_{N}\left( x\right) W_{N}\left( y\right)
\right)
\end{equation*}

\begin{equation}
=\min \left( H_{k}\left( x\right) ,\text{ }H_{k}\left( y\right)
\right) -H_{k}\left( x\right) H_{k}\left( y\right) -k^{-1}xyH_{k}^{\prime
}\left( x\right) H_{k}^{\prime }\left( y\right)   \label{S}
\end{equation}

\noindent such that

\begin{equation*}
\lim_{N\rightarrow +\infty }\sup \left( \log N\right) ^{-\frac{3}{4}}N^{
\frac{1}{4}}\sup_{0\leq x\leq +\infty }\left| \beta _{N}\left( x\right)
-W_{N}\left( x\right) \right| <+\infty ,a.s.
\end{equation*}

\noindent whenever k is fixed. Here $H_{k}^{\prime }\left( x\right) =dH_{k}\left(
x\right) /dx.$
\end{theorem}

\bigskip

\begin{remark} \label {r1} From now on, we will say \underline{according to the wording of Theorem \ref{t1}} at the
place of \underline{There exist a probability space ... such that}.
\end{remark}

\bigskip

\begin{definition} A Gaussian process whose covariance function is given by (\ref{S}) will be
called a Shorack process of parameter k or a k-Shorack process.
\end{definition}

\bigskip

\begin{corollary} \label{c1} According to wording of Theorem \ref{t1}, we have
\begin{equation*}
N^{\frac{1}{4}}\left( \log N\right) ^{-\frac{1}{2}}\left( \log \log N\right)
^{-\frac{1}{4}}\sup_{0\leq x<+\infty }\left| \beta _{N}\left( x\right)
-W_{N}\left( x\right) \right| =0_{p}\left( 1\right) ,\text{ as }N\rightarrow
+\infty .
\end{equation*}

\noindent This means that $a_{N}^{o}=\left( \log N\right) ^{\frac{3}{4}}N^{-\frac{1}{4}
}$ is a strong rate of convergence while $a_{N}=\left( \log N\right) ^{\frac{
1}{2}}\left( 2\log \log N\right) ^{\frac{1}{4}}N^{-\frac{1}{4}}$ is a weak
one. In fact \cite{1} has showed
\end{corollary}

\bigskip

\begin{theorem} \label{t2}. There exist another sequence of processes $\beta _{N}^{1},$ $N=1,2,...$
and a sequence of k-Shorack processes $W_{N}^{1},$ $N=1,2,...$ such that,
for k fixed, the two following assertions hold :

\begin{equation*}
\beta _{N}^{1}\text{ } =^{d}\beta _{N},\text{ }\forall N\geq 1 \tag{i}
\end{equation*}

\begin{equation*}
\sup_{0\leq x<+\infty }\left| \beta _{N}^{1}\left( x\right) -W_{N}^{1}\left(
x\right) \right| a.s.=0\left( a_{N}\right)N\rightarrow +\infty, \text{ a.s. }. \tag{ii}
\end{equation*}

\noindent All these results are based on representations of spacings by exponential
r.v.'s. Namely, when $n+1=kN,$

\begin{equation*}
\left\{ D_{i,n}^{k},1\leq i\leq N\right\}=^{d} \left\{ \frac{\left(
\sum_{j=\left( i-1\right) k}^{j=ik}E_{j}\right) }{S_{n+1}},1\leq i\leq
N\right\}
\end{equation*}

\begin{equation}
=:\left\{ Y_{i}/S_{n+1},1\leq i\leq N\right\}, \label{11}
\end{equation}

\noindent where $E_{1},$ $E_{2},...$ is a sequence of independent exponential rv's
with mean one and whose partial sums are $S_{n},n\geq 1.$ If $\mu
_{N}=\delta _{n}=S_{n+1}/Nk$, it follows that

\begin{equation*}
\left\{ \beta _{N}\left( x\right) ,0\leq x<+\infty \right\} =^{d}\left\{
N^{\frac{1}{2}}\left( \xi _{N}\mu _{N}\left( x\right) -H_{k}\left( x\right)
\right) +0\left( N^{\frac{1}{2}}\right) \right\}
\end{equation*}

\begin{equation}
= \left\{ \Lambda_{N}\left( x\right) +R_{N}\left( x\right) ,0\leq
x<+\infty \right\} =:\left\{ \beta _{N}^{\ast }\left( x\right) ,0\leq
x<+\infty \right\}, \label{12}
\end{equation}

\noindent where $\xi _{N}\left( .\right) $ (resp. $\Lambda _{N}\left( .\right) $) is
the empirical distribution function (resp. empirical process) based on $
Y_{1},...,Y_{N}$. The cited results are derived from simultaneous
approximations of $\Lambda _{N}$ and $R_{N}$.\\

\noindent First, we establish that the best rate attainable through this approach is that
of \cite{1} even when $k\rightarrow +\infty $.
\end{theorem}

\begin{theorem} \label{t3} According to the wording of Theorem \ref{t2}, for any k satisfying
\begin{equation}
\exists \delta _{0}<0,\text{ }\forall 0<\delta <\delta _{0},kN^{-\delta
}\rightarrow 0\text{ as }N\rightarrow +\infty ,  \tag{L}
\end{equation}

\noindent we have

\begin{equation*}
\lim_{N\rightarrow +\infty }\sup a_{N}^{-1}\sup_{0\leq x<+\infty }\left|
\beta _{N}^{\ast }\left( x\right) -W_{N}^{\ast }\left( x\right) \right|
a.s.=\left\{
\begin{array}{c}
K\left( k\right) =\left( k^{k+\frac{1}{2}}e^{-k}/k!\right) ^{\frac{1}{2}},
\text{ (k fixed)} \\
K_{0}=\left( 2\pi \right) ^{-\frac{1}{4}},\text{ }\left( k\rightarrow
+\infty \right).
\end{array}
\right.
\end{equation*}

\bigskip

\noindent Our second result is an improvement of Theorem 1 of \cite{2}.
\end{theorem}

\bigskip

\begin{theorem} \label{t4}. According to the wording of Theorem \ref{t1}, we have for any $k$ such that for
some $\delta _{0},0<\delta _{0}<\frac{1}{4},kN^{-\frac{1}{4}+\delta
_{0}}\rightarrow 0$ as $N\rightarrow +\infty ,$
\begin{equation*}
\lim_{N\rightarrow +\infty }\sup a_{N}^{-1}\sup_{0\leq x<+\infty }\left|
\beta _{N}\left( x\right) -W_{N}\left( x\right) \right| \leq \left\{
\begin{array}{c}
K\left( k\right) ,\text{ (k fixed)} \\
K_{0}\text{ \ (k}\rightarrow +\infty \text{)}
\end{array}
a.s.,\right.
\end{equation*}
\end{theorem}

\bigskip

\noindent \textbf{Proof of Theorem \ref{t4}}. From (\ref{12}), we have $\beta _{N}$ $=^{d}$ $\beta _{N}^{\ast }$ for all $
N\geq 1$. Furthermore,
\begin{eqnarray*}
\beta _{N}^{\ast }\left( x\right)  &=&\Lambda _{N}\left( x\right) +N^{\frac{1
}{2}}\left( H_{k}\left( \mu _{N}x\right) -H_{k}\left( x\right) \right)
-\left\{ \Lambda _{N}\left( \mu _{N}x\right) -\Lambda _{N}\left( x\right)
\right\} +0\left( N^{-\frac{1}{2}}\right)  \\
&=&:\Lambda _{N}\left( x\right) +R_{N1}\left( x\right) +R_{N2}\left(
x\right) +R_{N3}\left( x\right) .
\end{eqnarray*}

\noindent We shall proceed by steps, approximating each of the $R_{Ni}$'s$.$

\bigskip

\begin{lemma} \label{l1} Let $N_{p}=\left[ \left( 1+\rho \right) ^{p}\right] ,$ $p>0,$ $
p=1,2,...,\varepsilon >0$ and
\begin{equation*}
C_{N_{p}}=\bigcup_{N=N_{p}}^{N=N_{p+1^{-1}}}\left\{ \sup_{0\leq x<+\infty
}\left| R_{N1}\left( x\right) -N^{\frac{1}{2}}\text{ }xH_{n}^{\prime }\left(
x\right) \left( \mu _{N}-1\right) \right| >\varepsilon a_{N}K\left( k\right)
/4\right\} .
\end{equation*}

\noindent Then if $k/N\rightarrow 0$ as $N\rightarrow +\infty $, $\sum_{p}\mathbb {P}\left(
C_{N_{p}}\right) <+\infty $.
\end{lemma}

\bigskip

\noindent \textbf{Proof of Lemma \ref{l1}} Apply the mean value theorem twice and get
\begin{equation}
A_{N1}=R_{N1}\left( x\right) -N^{\frac{1}{2}}\left( \mu _{N}-1\right)
xH_{k}^{\prime }\left( x\right) =N^{\frac{1}{2}}\left( \mu _{N}-1\right)
^{2}x^{2}H_{k}^{\prime \prime }\left( x_{N}\right) ,  \label{13}
\end{equation}

\noindent Where $0<\left| x_{N}/x\right| <\max \left( 1,\mu _{N}\right) $. First, it
may be easily seen that
\begin{equation}
\sup_{0\leq x<+\infty }\frac{xH_{k}^{\prime }\left( x\right) }{k^{\frac{1}{2}
}}=\frac{k^{\frac{1}{2}+k}e^{-k}}{k!}=K\left( k\right) ^{2},  \label{14}
\end{equation}

\begin{equation}
\lim_{k\rightarrow +\infty }\sup_{0\leq x<+\infty }\left| x\text{ }
H_{k}^{\prime }\left( x\right) /k^{\frac{1}{2}}\right| =K_{0}^{2},
\label{15}
\end{equation}

\noindent and

\begin{equation}
0<M=\sup_{k\geq 1}\sup_{0\leq x<+\infty }\left| x^{2}H_{k}^{\prime \prime
}\left( x\right) /k\right| <+\infty .  \label{16}
\end{equation}

\noindent Recall that for all $\varepsilon >0$,
\begin{equation}
\sum_{p}\mathbb {P}\left( \max \left( 1,\mu _{N}\right) >1+\varepsilon \right) \leq
\sum_{N}\mathbb {P}\left( \left| \mu _{N}\right| >1+\varepsilon \right) <+\infty,
\label{17}
\end{equation}

\noindent by the strong law of large numbers (SLLN) and
\begin{equation}
\sum_{p}\mathbb {P}\left( \bigcup_{N=N_{p}}^{N=N_{p+1^{-1}}}\left( \frac{Nk}{2\log
\log nk}\right) \left| \mu _{N}-1\right| >1+\varepsilon \right) <+\infty
\label{18}
\end{equation}

\noindent by the law of the iterated logarithm (loglog-law). We show in the Appendix
how to adapt the classical SLLN and loglog-law to these cases.\\

\noindent Now by (\ref{13}), (\ref{14}) and (\ref{15})

\begin{equation*}
\mathbb {P}\left( C_{N_{p}}\right) \leq \sum_{N=N_{p}}^{N=N_{p+1^{-1}}}\mathbb {P}\left( \max
\left( 1,\mu _{N}\right) ^{2}>1+\varepsilon \right)
\end{equation*}

\begin{equation}
+\mathbb {P}\left(
\bigcup_{N=N_{p}}^{N=N_{p+1^{-1}}}\left| \mu _{N}-1\right| ^{2}\left( \frac{
Nk}{2\log \log Nk}\right) >ce_{N}\right) ,  \label{19}
\end{equation}

\noindent with $c=\varepsilon K\left( k\right) ^{2}/4M\left( 1+\varepsilon \right)
,e_{N}=\left( \log \log N\right) ^{\frac{1}{4}}N^{\frac{1}{4}}\left( \log
N\right) ^{\frac{1}{2}}\left( 2\log \log Nk\right) ^{-\frac{1}{2}}$. But $
\log \log Nk=\left( \log N\right) \left( 1+o\left( 1\right) \right) ,$ $
K\left( k\right) $ is bounded and thus $ce_{N}>\left( 1+\varepsilon \right)
^{2}$ for large N. Thus we can apply (\ref{17}) and (\ref{18}) to (\ref{19})
and this completes the proof.

\begin{lemma} \label{l2} Let $\varepsilon >0$ and
\begin{equation*}
D_{N_{p}}=\left\{ \bigcup_{N=N_{p}}^{N=N_{p+1^{-1}}}\left( \sup_{0\leq
x<+\infty }\left| R_{N2}\left( x\right) \right| >\left( 1+\varepsilon
/4\right) a_{N}\text{ }K\left( k\right) \right) \right\} ,\text{ }p=1,2,...
\end{equation*}

\noindent Then for any $k=k\left( N\right) $ such that $k/N\rightarrow 0$ as $
N\rightarrow +\infty ,$ $\sum_{p}p\left( D_{N_{p}}\right) <+\infty .$
\end{lemma}

\bigskip

\noindent \textbf{Proof of Lemma \ref{l2}} The mean value theorem implies
\begin{equation}
\left| H_{k}\left( \mu _{N}x\right) -H_{k}\left( x\right) \right| \leq
\left| \mu _{N}-1\right| \text{ }K\left( k\right) ^{2}\max \left( 1,\text{ }
\mu _{N}\right) k^{\frac{1}{2}}.  \label{120}
\end{equation}

\noindent By proceeding similarly to (\ref{19}), we get
\begin{equation*}
\mathbb {P}\left( D_{N_{p}}\right)  \leq \sum_{N=N_{p}}^{N=N_{p+1^{-1}}}\mathbb {P}\left( \max
\left( 1,\mu _{N}\right) >\left( 1+\varepsilon /4\right) ^{1/3}\right)
\end{equation*}

\begin{equation*}
+\mathbb {P}\left( \bigcup_{N=N_{p}}^{N=N_{p+1^{-1}}}\left\{ \sup_{\left|
H_{k}\left( x\right) -H_{k}\left( y\right) \right| <c_{N}}\left| \Lambda
_{N}\left( x\right) -\Lambda _{N}\left( y\right) \right| >\left(
1+\varepsilon /4\right) a_{N}K\left( k\right) \right\} \right)
\end{equation*}

\begin{equation}
=R_{N21}+R_{N22}, \label{121}
\end{equation}

\noindent with $c_{N}=K\left( k\right) ^{2}k^{\frac{1}{2}}\left| \mu _{N}-1\right|
\left( 1+\varepsilon /4\right) ^{1/3}$. Now,

\begin{equation}
R_{N22} \leq \mathbb{P}\left( \bigcup_{N=N_{p}}^{N=N_{p+1^{-1}}}\left\{ \left| \mu
_{N}-1\right| >\left( 1+\varepsilon /4\right) ^{1/3}\left( \frac{2\log \log N
}{Nk}\right) ^{\frac{1}{2}}\right\} \right) \label{122}
\end{equation}

\begin{equation*}
+\mathbb{P}\left( \bigcup_{N=N_{p}}^{N=N_{p+1^{-1}}}\left\{ \sup_{\left|
H_{k}\left( x\right) -H_{k}\left( y\right) \right| \leq b_{N}}\left| \Lambda
_{N}\left( x\right) -\Lambda _{N}\left( y\right) \right| >\left(
1+\varepsilon /4\right) a_{N}K\left( k\right) \right\} \right),
\end{equation*}

\noindent where $b_{N}=\left( \frac{2\log \log \text{ }N}{N}\right) ^{\frac{1}{2}
}K\left( k\right) ^{2}\left( 1+\varepsilon /4\right) ^{2/3}$. Let $\gamma
_{N}\left( .\right) $ be the empirical process based on $U_{1},...,U_{N}$
and $P_{N_{p}}$ be the second term of the right member of the inequality (
\ref{122}). Thus (\ref{11}) implies

\begin{equation}
P_{N_{p}}\leq \mathbb {P}\left( \bigcup_{N=N_{p}}^{N=N_{p+1^{-1}}}\left\{ \sup_{0\leq
u\leq 1-b_{N}}\frac{\gamma _{N}\left( u\right) -\gamma _{N}\left(
u+b_{N}\right) }{\left( 2b_{N}\log b_{N}^{-1}\right) ^{\frac{1}{2}}}
>1+\varepsilon _{1}\right\} \right),  \label{123}
\end{equation}

\noindent $1+\varepsilon _{1}<\left(1+\varepsilon \right) ^{2/3}$, where we have used the fact that $\left( 2b_{N}\log b_{N}^{-1}\right) ^{\frac{1}{2}}/a_{N}k\left( k\right) \rightarrow \left( 1+\varepsilon \right)
^{1/3}$ as $k/N\rightarrow 0$, as $N\rightarrow +\infty $. Finally, from
line 14, p.95 and line 23, p.98 in \cite{13}, we get $\sum_{p}P_{N_{p}}<+
\infty $. This and (\ref{120}), (\ref{121}), (\ref{122}) and (\ref{123})
together imply Lemma \ref{l2}.

\bigskip

\begin{lemma} \label{l3} (Koml\'os, M\'ajor,Tusn\'ady, 1975). There exist a probability space carrying a
sequence $Y_{1},Y_{2},...$ as defined in (\ref{11}) and a sequence of
Brownian bridges
\begin{equation*}
B_{N}^{1}\left( s\right) ,0\leq s\leq 1,\text{ }N=1,2,...
\end{equation*}

\noindent such that
\begin{equation*}
\forall N\geq N_{1},\mathbb {P}\left( \sup_{0\leq x<+\infty }\left| \Lambda _{N}\left(
x\right) -B_{N}^{1}\left( H_{k}\left( x\right) \right) \right| >\frac{A\log
\text{ }N+x}{N^{\frac{1}{2}}}\right) \leq Be^{-\lambda x},
\end{equation*}

\noindent for all sequence $\left( k=k\left( N\right) \right) _{N\geq 1}$ and for all
x, where $N_{1},A,B$ and $\lambda $ are absolute positive constants.
\end{lemma}

\bigskip

\noindent \textbf{Proof of Lemma \ref{l3}} This doesn't need to be proved. It is directly derived from \cite{6} and
Corollary 4.4.4 of \cite{4}.

\bigskip

\noindent \textbf{Proof of Theorem \ref{t3} continued.} On the probability space of Lemma \ref{l3}, Lemmas \ref{l1} and \ref{l2} combined with the fact $
R_{N3}\leq N^{-\frac{1}{2}}$ imply that

\begin{equation}
\sum_{p}\mathbb {P}\left( \bigcup_{N=N_{p}}^{N=N_{p+1^{-1}}}\sup_{0\leq x\leq +\infty
}\left| \beta _{N}^{\ast }\left( x\right) -\beta _{N}^{\ast \ast }\left(
x\right) \right| >\left( 1+3\varepsilon /4\right) a_{N}K\left( k\right)
\right) <+\infty ,  \label{124}
\end{equation}

\noindent where $\beta _{N}^{\ast \ast }\left( x\right) =\Lambda _{N}\left( x\right)
-N^{\frac{1}{2}}$ x $H_{k}^{\prime }\left( x\right) \left( \mu _{N}-1\right)
,0\leq x<+\infty $. Hence, the proof will be complete if we approximate $
\beta _{N}^{\ast \ast }$ in the right way. But by Lemma \ref{l3}, for any $
\varepsilon >0$, for large N
\begin{equation}
\mathbb {P}\left( \sup_{0\leq x<+\infty }\left| \Lambda _{N}\left( x\right)
-B_{N}^{1}\left( H_{k}\left( x\right) \right) \right| >A_{1}\left( \log
N\right) ^{2}N^{-\frac{1}{2}}\right) \leq N^{-1-\varepsilon },  \label{125}
\end{equation}

\noindent where $A_{1}$ is some absolute constant. From Lemma 3.1 of \cite{2}

\begin{equation*}
N^{\frac{1}{2}}\left( \mu _{N}-1\right) =N^{\frac{1}{2}}k\frac{S_{n+1}-Nk}{Nk
}+k^{-1}\int_{0}^{+\infty }\left\{ \Lambda _{N}\left( x\right)
-B_{N}^{1}\left( H_{k}\left( x\right) \right) \right\} dx
\end{equation*}

\begin{equation}
+k^{-1}\int_{0}^{+\infty }B_{N}^{1}\left( H_{k}\left( x\right) \right) dx. \label{126}
\end{equation}

\noindent Let $t_{N}=N^{\frac{1}{4}-\delta },0\leq \delta \leq \delta _{0}$. On the
one hand, one has for large N.

\begin{equation*}
\mathbb {P}\left( \left| \int_{0}^{t_{N}}\left\{ \Lambda _{N}\left( x\right)
-B_{N}^{1}\left( x\right) \right\} dx\right| >\varepsilon a_{N}/12\right)
\end{equation*}

\begin{equation*}
\leq \mathbb {P}\left( \sup_{0\leq x<+\infty }\left| \Lambda _{N}\left( x\right)
-B_{N}^{1}\left( H_{k}\left( x\right) \right) \right| >\frac{\varepsilon
\left( 2\log \log N\right) ^{\frac{1}{4}}\left( \log N\right) ^{\frac{1}{2}}
}{12N^{\frac{1}{4}-\delta }}\right)
\end{equation*}

\begin{equation*}
\leq \mathbb {P}\left( \sup_{0\leq x<+\infty }\left| \Lambda _{N}\left( x\right)
-B_{N}^{1}\left( H_{k}\left( x\right) \right) \right| >A_{1}\log N/N^{\frac{1
}{2}}\right).
\end{equation*}

\noindent This and (\ref{15}) together imply

\begin{equation}
\mathbb {P}\left( \sup_{0\leq x<+\infty }\left| xH_{k}^{\prime }\left( x\right)
k^{-1}\int_{0}^{t_{N}}\left\{ \Lambda _{N}\left( t\right) -B_{N}^{1}\left(
H_{k}\left( t\right) \right) \right\} dt\right| >\varepsilon a_{N}K\left(
k\right) /12\right) \leq N^{-1-\varepsilon },  \label{127}
\end{equation}

\noindent for N large enough. On the other hand, as $N\rightarrow +\infty ,$

\begin{equation}
\mathbb {P}\left( \sup_{0\leq x<+\infty }\left| \int_{t_{N}}^{+\infty }\left\{ \Lambda
_{N}\left( t\right) -B_{N}^{1}\left( H_{k}\left( t\right) \right) \right\}
dt\right| >N^{-\frac{1}{2}}\right) \leq N^{\frac{1}{2}}\exp \left( -N^{\frac{
1}{4}-\delta }/4\right) .  \label{128}
\end{equation}

\noindent To see that, apply Markov's inequality with
\begin{equation*}
\mathbb {E}\int_{t_{N}}^{+\infty }\left| \Lambda _{N}\left( x\right) -B_{N}^{1}\left(
H_{k}\left( x\right) \right) \right| dx\leq \int_{t_{N}}^{+\infty
}4k^{-1}e^{-x/2}\frac{x^{\left( k-1\right) /2}}{\left( k-1\right) !}dx\leq
4k^{-1}t_{N}^{k}\exp \left( -t_{N}/2\right) .
\end{equation*}

\noindent Since $k=o\left( N^{\frac{1}{4}-\delta }o\right) $, as $N\rightarrow +\infty
$, (\ref{128}) follows. Finally for large N,

\begin{equation*}
\mathbb {P}\left( \sup_{0\leq x<+\infty }\left| xH_{k}^{\prime }\left( x\right) /k^{
\frac{1}{2}}\frac{S_{Nk}-S_{n+1}}{\left( Nk\right) ^{\frac{1}{2}}}\right|
>\varepsilon a_{N}K\left( k\right) /16\right)
\end{equation*}

\begin{equation*}
\leq \mathbb {P}\left( S_{k}>N^{\frac{1}{
2}}k^{\frac{1}{2}}\right) =1-H_{k}\left( N^{\frac{1}{2}}k^{\frac{1}{2}
}\right).
\end{equation*}

\noindent Integrating by parts we have : $k/x\leq \frac{1}{2}\Rightarrow 1-H_{k}\left(
x\right) \leq 2x^{k-1}e^{-x}/\left( k-1\right) !$. Then if $k/N\leq \frac{1}{
2}$ for large N, we get by Sterling's formula,
\begin{equation}
1-H_{k}\left( k^{\frac{1}{2}}N^{\frac{1}{2}}\right) \leq const.\text{ }\exp
\left( -k^{\frac{1}{2}}N^{\frac{1}{2}}\left( 1+\left( k/N\right) ^{\frac{1}{2
}}\log \left( k/N\right) \right) \right) .  \label{129}
\end{equation}

\noindent Thus,

\begin{equation}
\mathbb {P}\left( \sup_{0\leq x<+\infty } xH_{k}^{\prime }\left( x/k\right)
\left( \left( S_{Nk}-S_{n+1}\right) /N^{\frac{1}{2}}\right) >\varepsilon
a_{N}K\left( k\right) /12\right)
\end{equation}

\begin{equation}
\leq const.\text{ }\exp \left( -\frac{1}{4}
k^{\frac{1}{2}}N^{\frac{1}{2}}\right),  \label{130}
\end{equation}

\noindent ultimately as $N\rightarrow +\infty $ whenever $k/N\rightarrow 0$ as $
N\rightarrow +\infty $. Put together (\ref{125}), (\ref{126}), (\ref{127}), (
\ref{128}) and (\ref{130}) to get
\begin{equation}
\sum_{N}\mathbb {P}\left( \sup_{0\leq x<+\infty }\left| \beta _{N}^{\ast \ast }\left(
x\right) -W_{N}^{\ast \ast }\left( x\right) \right| >\varepsilon
a_{N}K\left( k\right) /4\right) <+\infty ,  \label{131}
\end{equation}

\noindent where $W_{N}^{\ast \ast }\left( x\right) =B_{N}^{1}\left( H_{k}\left(
x\right) \right) -xk^{-1}H_{k}^{\prime }\left( x\right) \int_{0}^{+\infty }t$
$dB_{N}^{1}\left( H_{k}\left( t\right) \right) ,$ $x\geq 0$. And combine (
\ref{124}) with (\ref{131}) to have
\begin{equation}
\sum_{p}\mathbb {P}\left( \bigcup_{N=N_{p}}^{N=N_{p+1^{-1}}}\left\{ \sup_{0\leq
x<+\infty }\left| \beta _{N}^{\ast }\left( x\right) -W_{N}^{\ast \ast
}\left( x\right) \right| >\left( 1+\varepsilon \right) a_{N}K\left( k\right)
\right\} <+\infty \right) .  \label{132}
\end{equation}

\noindent This together with Lemma 4.4.4. of \cite{4} completes the proof.

\bigskip

\begin{proof} of Theorem \ref{t3}. As in the proof of Theorem \ref{t4}, the spacings are always defined on the
probability space of Lemma \ref{l3}. We shall study each of the $R_{Ni}$'s once
again. First we put together (\ref{13}), (\ref{14}), (\ref{15}) and (\ref{16}
) to get

\begin{equation}
\sup_{0\leq x<+\infty }\left| R_{N1}\left( x\right) -N^{\frac{1}{2}}\left(
\delta _{n}-1\right) xH_{k}^{\prime }\left( x\right) \right| =0\left( N^{-
\frac{1}{2}}\log \log N\right) ,a.s.,\text{ as }N\rightarrow +\infty .
\label{133}
\end{equation}

\noindent Now Lemma \ref{l2} says nothing else but
\begin{equation}
\lim_{N\rightarrow +\infty }\sup \sup_{0\leq x<+\infty }\left| R_{N2}\left(
x\right) /a_{N}\right| \leq K\left( k\right) \text{ or }K_{0},a.s.,
\label{134}
\end{equation}

\noindent whenever k is fixed or $k\rightarrow +\infty $ while $k/N\rightarrow 0$ as $
N\rightarrow +\infty $. And the proof will be completed through our
fundamental Lemma which is the following.
\end{proof}

\bigskip

\begin{lemma} \label{l4} Under the assumptions of Theorem \ref{t3}, we have
\begin{equation*}
\lim_{N\rightarrow +\infty }\sup \sup_{0\leq x<+\infty }\left|
a_{N}^{-1}R_{N2}\left( x\right) \right| \geq K\left( k\right) \text{ or }
K_{0},a.s.,
\end{equation*}

\noindent according whether k is fixed or $k\rightarrow +\infty $ and satisfies $(L)$.
\end{lemma}

\bigskip

\begin{proof} of Lemma \ref{l4}.

\noindent Let $\psi \left( x\right) =\left( \left( k-1\right) !\right)
^{-1}x^{k}e^{-x},$ $x\geq 0$. By the mean value theorem,
\begin{equation*}
\left| \psi \left( x\right) -\psi \left( k\right) \right| \leq
he^{h}k^{k-1}\left( 1-1/k\right) ^{k-1}\left( \left( k-1\right) !\right)
^{-1},\text{ if }\left| x-k\right| \leq h\leq 1\text{.}
\end{equation*}

\noindent By Sterling's formula we can find a constant $\tau >0$ such that
\begin{equation}
\sup_{\left| x-k\right| \leq h\leq 1}k^{\frac{1}{2}}\left| \psi \left(
x\right) -\psi \left( k\right) \right| \leq \tau hk^{-1},\text{ for all }
k\geq 1.  \label{135}
\end{equation}

\noindent Now,
\begin{equation}
A_{N}\left( x\right) =H_{k}\left( \delta _{n}x\right) -H_{k}\left( x\right)
=\left( \delta _{n}-1\right) \psi \left( x_{n}\right) \left( x_{n}/x\right)
,0\leq x_{n}/x\leq \max \left( 1,\delta _{n}\right) .  \label{136}
\end{equation}

\noindent If $\left| x-k\right| \leq h\leq 1,\left| x_{n}-k\right| \leq k+\left(
k+h\right) \left| 1-\delta _{n}\right| ,$ and thus by (\ref{135}),

\begin{equation*}
\left| x-k\right| \leq h\leq 1\Rightarrow A_{N}\left( x\right) =\left(
1+o\left( 1\right) \right) k^{\frac{1}{2}}\left( \delta _{n}-1\right)
\end{equation*}

\begin{equation*}
\times \left\{ K\left( k\right) +0\left( \left\{ h+\left( h+k\right) \left|
1-\delta _{n}\right| \right\} /k\right) \right\} ,a.s.
\end{equation*}

\noindent Let $h=h\left( N\right) \rightarrow 0$ as $N\rightarrow +\infty $. Then by
the loglog-law, there exists $\Omega ^{1}\subset \Omega $ and a sequence $
\left( N_{j\left( \omega \right) }\right) $ extracted from $\left( N\right) $
(let $n_{j}$ and $k_{j}$ be the corresponding subsequences) satisfying

\begin{equation*}
\mathbb {P}\left( \Omega ^{1}\right) =1,\text{ }\forall \omega \in \Omega ^{1},\text{ }
A_{N_{j}}\left( x\right) =\left( \left( 2\log \log n_{j}\right)
/N_{j}\right) ^{\frac{1}{2}}K\left( k_{j}\right) ^{\frac{1}{2}}\left(
1+o\left( 1\right) \right)
\end{equation*}

\begin{equation}
=:\left( 1+o\left( 1\right) \right) d_{N_{j}}, \label{137}
\end{equation}

\noindent uniformly in $x,$ $k_{j}-h_{j}\leq x\leq k_{j}-h_{j}$, where $h_{j}=h\left(
N_{j}\right) $ as $N\rightarrow +\infty $. Thus we have uniformly un $x$ $ \in
\left[ k_{j}-h_{j},\text{ }k_{j}+h_{j}\right] =I_{k},$
\begin{equation}
\left| R_{N_{j^{2}}}\left( x\right) \right| d_{=}\left| \gamma
_{N_{j}}\left( H_{k_{j}}\left( x\right) +d_{N_{j}}\left( 1+o\left( 1\right)
\right) -\gamma _{N_{j}}\left( H_{k_{j}}\left( x\right) \right) \right)
\right| =:\left| R_{N_{j^{2}}}^{\ast }\left( x\right) \right| .  \label{138}
\end{equation}

\noindent We now prove that
\begin{equation}
\exists \Omega \subset \Omega ^{1},\text{ }\mathbb {P}\left( \Omega _{0}\right) =1,
\text{ }\forall \omega \in \Omega _{0},\text{ }\lim_{j\rightarrow +\infty
}\inf \sup_{x\in I_{k_{j}}}\left\{ \left| R_{N_{j^{2}}}^{\ast }\left(
x\right) /b\left( d_{N_{j}}\right) \right| \right\} \geq 1,  \label{139}
\end{equation}

\noindent where $b\left( s\right) =\left( 2s\log \log s^{-1}\right) ^{\frac{1}{2}},$ $
0<s<1$.\\

\noindent \textbf{Proof of (\ref{139})}.

\noindent Let
\begin{equation*}
C_{N_{1}}\left( p\right) =\sup_{0\leq v\leq d_{N}/p}\sup_{0\leq s\leq
1-v}\left| \gamma _{N}\left( s\right) -\gamma _{N}\left( s+v\right) \right|
/b\left( d_{N}\right) ,\text{ }p\geq 1.
\end{equation*}

\noindent By Theorem 0.2 of \cite{13},

\begin{equation}
\forall p\geq 1,\text{ }\exists \Omega _{p}\subset \Omega \text{, }P\left(
\Omega _{p}\right) =1,\text{ }\forall \omega \in \Omega _{p},\text{ }
\lim_{N\rightarrow +\infty }\sup C_{N_{1}}\left( p\right) \left( \omega
\right) <p^{-\frac{1}{2}}.  \label{140}
\end{equation}

\noindent Let
\begin{equation*}
\Omega =\Omega ^{1} \bigcap \bigcup_{p=1}^{p=+\infty }\Omega _{p}.
\end{equation*}

\noindent Obviously $\mathbb {P}\left( \Omega ^{2}\right) =1$. And for any $\omega \in \Omega ^{2}$, $C_{N_{j^{2}}}\left( \omega \right)=$

\begin{equation*}
\sup_{0\leq x<+\infty }\gamma _{N_{j}}\left( H_{k_{j}}\left( x\right)
+d_{N_{j}}\left( 1+o\left( 1\right) \right)-\gamma _{N_{j}}\left(
H_{k_{j}}\left( x\right) +d_{N_{j}}\right)\right) =o\left( b\left( d_{N_{j}}\right)
\right),    \label{141}
\end{equation*}

\noindent This, together with the following, as $\text{ }j \rightarrow +\infty$,

\begin{equation*}
\forall x \in I_{k_{j}},\text{ }R_{N_{j^{2}}}^{\ast }\left( x\right)
=\gamma _{N_{j}}\left( H_{k_{j}}\left( x\right) +d_{N_{j}}\right) -\gamma
_{N_{j}}\left( H_{k}\left( x\right) \right)
\end{equation*}

\begin{equation}
+\gamma _{N_{j}}\left( H_{k_{j}}\left( x\right) +d_{N_{j}}\left( 1+o\left(
1\right) \right) \right) -\gamma _{N_{j}}\left( H_{k_{j}}\left( x\right)
+d_{N_{j}}\right), \label{142}
\end{equation}

\noindent implies that

\begin{equation*}
\sup_{x\in I_{k_{j}}}R_{N_{j^{2}}}^{\ast }\left( x\right)  \geq \sup_{x\in
I_{k_{j}}}\gamma _{N_{j}}\left( H_{k_{j}}\left( x\right) +d_{N_{j}}\right)
-\gamma _{H_{j}}\left( H_{k_{j}}\left( x\right) \right) +o\left( b\left(
d_{N_{j}}\right) \right)
\end{equation*}

\begin{equation}
\geq :C_{N_{j^{3}}}\left( h\left( N_{j}\right) \right) +o\left( b\left(
d_{N_{j}}\right) \right). \label{143}
\end{equation}

\noindent Now put $J_{k}=H_{k}\left( I_{k}\right) $ and remark that the lenght of $
J_{k}$ is $\rho \left( J_{k}\right) =2K\left( k\right) ^{2}nk^{-\frac{1}{2}
}\left( 1+o\left( 1\right) \right).$\\

\noindent For any $p\geq 1,$ choose $h=h\left( N,\text{ }p\right) =h_{p}$ (with $
h_{j,p}=h\left( N_{j},p\right) $ such that $2K\left( k\right) ^{2}h_{p}k^{-
\frac{1}{2}}d_{N}^{-1/4p}=1+o\left( 1\right) $, as $N\rightarrow +\infty $.
Thus, $h\rightarrow 0$ as $N\rightarrow +\infty $ when $(L)$ holds. Also
$m_{N}=\max \left\{ i,\text{ }i\geq 0,\text{ }H_{k}\left( k-h_{p}\right)
+id_{N}\in J_{k}\right\} \rightarrow +\infty $ as $N\rightarrow +\infty $.
Therefore we may use the lines of the proof of Lemma 2.9 of \cite{13} to
conclude that for any $p\geq 1$,

\begin{equation*}
\mathbb {P}\left( D_{N}\right) =\mathbb {P}\left( \max_{1\leq i\leq m_{N}}\left\{ \gamma
_{N}\left( C_{i+1}^{N}\right) -\gamma _{N}\left( C_{i}^{N}\right) \right\}
/b\left( d_{N}\right) \leq \left( 1-1/p\right) ^{\frac{1}{2}}\right)
\end{equation*}

\begin{equation*}
=0\left( N^{\frac{1}{2}}\exp \left( -m_{N}d_{N}^{1-1/2p}\right) \right) ,
\end{equation*}

\noindent as $N\rightarrow +\infty $, where $C_{i}^{N}=H_{k}\left( k-h_{p}\right)
+id_{N},$ $i=1,...,m_{N}$. But $m_{N}d_{N}=\left( 2K\left( k\right)
^{2}h_{p}k^{-\frac{1}{2}}xd_{N}^{-1/4p}\right)
d_{N}^{1/4p}=d_{N}^{1/4p}\left( 1+o\left( 1\right) \right) $. Hence $\mathbb {P}\left(
D_{N}\right) =0\left( d_{N}^{-1/8p}\right) $ for large N. Thus $
\sum_{N}\mathbb {P}\left( D_{N}\right) <+\infty $, that is
\begin{equation}
\forall p\geq 1,\text{ }\exists \Omega _{p}^{\prime },\text{ }\mathbb {P}\left( \Omega
_{p}^{\prime }\right) =1,\text{ }\forall \omega \in \Omega _{p}^{\prime },
\text{ }\lim_{N\rightarrow +\infty }\inf C_{N3}\left( h_{p}\right) /b\left(
d_{N}\right) \geq \left( 1-1/p\right) ^{\frac{1}{2}}.
\end{equation}

\noindent Letting
\begin{equation*}
\Omega _{0}^{\prime }=\Omega ^{2}\bigcup \bigcup_{p=1}^{p=+\infty },
\end{equation*}

\noindent we get $\mathbb {P}\left( \Omega _{0}^{\prime }\right) =1$ and for all $\omega \in
\Omega _{0}^{\prime },$
\begin{equation}
\lim_{j\rightarrow +\infty }\inf \sup_{x\in I_{j_{k}}}\left|
R_{N_{j^{2}}}^{\ast }\left( x\right) \right| /b\left( d_{N}\right) \geq 1.
\label{145}
\end{equation}

\noindent We have used in (\ref{138}) that representation for commodity reasons as it
has appeared in the proof. The same may be done, step by step, following
Stute's results (see \cite{13}) to get the version of (\ref{145}) for $
R_{N_{j^{2}}\text{ }}$itself. This remark completes the proof of (\ref{139}).
\end{proof}

\bigskip

\begin{proof} of Lemma \ref{l4} (Continued). Remark that

\begin{equation}
\lim_{N\rightarrow +\infty }\sup \sup_{0\leq x<+\infty }\left| R_{N2}\left(
x\right) \right| /b\left( d_{N}\right)  \geq \lim_{j\rightarrow +\infty
}\sup \sup_{0\leq x<+\infty }\left\{ \left| R_{N_{j^{2}}}\left( x\right)
\right| /b\left( d_{N_{j}}\right) \right\}
\end{equation}

\begin{equation*}
\geq \lim_{j\rightarrow +\infty }\inf \sup_{0\leq x<+\infty
}R_{N_{j^{2}}}\left( x\right) /b\left( d_{N_{j}}\right) \geq
\lim_{j\rightarrow +\infty }\inf \sup_{x\in I_{k}}R_{N_{j^{2}}}\left(
x\right) /b\left( d_{N_{j}}\right) .
\end{equation*}

\noindent This combined with (\ref{139}) and with the fact that $b\left( d_{N}\right)
=K\left( k\right) a_{N}\left( 1+o\left( 1\right) \right) $ as $N\rightarrow
+\infty $ proves the Lemma \ref{l4}.
\end{proof}

\bigskip

\begin{conclusion}
It is clear by Theorem \ref{t3}. that the approach used until now cannot yield a
rate better than $a_{N}$. The problem is now : what new approach would be
used to reach, if possible, the very best rate, that of \cite{6} which is $
N^{-\frac{1}{2}}\log N$.
\end{conclusion}

\bigskip

\section{The Glivenko-Cantelli Theorem}

\noindent For the overlapping case, \cite{3} obtained a Glivenko-Cantelli theorem when
the step satisfies $kN^{-1+a}\rightarrow 0$ as $N\rightarrow +\infty $ for
some $0<a<1$. As to the overlapping case only fixed steps have been handled
in \cite{2}. We give the general result in

\bigskip

\begin{theorem} \label{t5}. Let $k\geq 1$ be fixed or $k\rightarrow +\infty $ while $k/N\rightarrow 0$
as $N\rightarrow +\infty $. Then
\begin{equation*}
\lim_{N\rightarrow +\infty }\sup_{0\leq x<+\infty }\left| F_{N}\left(
x\right) -H_{k}\left( x\right) \right| =0,a.s.
\end{equation*}

\noindent on the probability space where the spacings are defined.
\end{theorem}

\bigskip

\begin{proof} of Theorem \ref{t5}. We have

\begin{equation*}
\forall N\geq 1,\left\{ F_{N}\left( x\right) -H_{k}\left( x\right) ,0\leq
x<+\infty \right\}
\end{equation*}

\begin{equation}
=^{d}\left\{ \xi _{N}\left( x\right) -H_{k}\left(
x\right) +R_{N4}\left( x\right) +N^{-\frac{1}{2}}R_{N2}\left( x\right)
+0\left( N^{-\frac{1}{2}}\right) ,0\leq x<+\infty \right\} .  \label{21}
\end{equation}

\noindent First, it follows from Lemma \ref{l2} that for all $\varepsilon >0$,
\begin{equation*}
\sum_{p}\mathbb {P}\left( \bigcup_{N=N_{p}}^{N=N_{p+1^{-1}}}N^{-\frac{1}{2}
}\sup_{0\leq x<+\infty }\left| R_{N2}\left( x\right) \right| >\varepsilon
/4\right) <+\infty .
\end{equation*}

\noindent Next,
\begin{equation*}
\mathbb {P}\left( \sup_{0\leq x<+\infty }\left| R_{N4}\left( x\right) \right|
>\varepsilon /4\right) \leq \mathbb {P}\left( \left| 1-\mu _{N}\right| k^{\frac{1}{2}
}K\left( k\right) ^{2}>\varepsilon /4\right) \text{. }
\end{equation*}

\noindent And direct calculations imply that for all $\lambda >1$, we have
\begin{equation*}
\mathbb {P}\left( \left| 1-\mu _{N}\right| k^{\frac{1}{2}}K\left( k\right)
^{2}>\varepsilon /4\right) \leq \mathbb {P}\left( \left| 1-\mu _{N}\right| \left(
\frac{Nk}{2\log \log Nk}\right) ^{\frac{1}{2}}>\lambda \right)
\end{equation*}

\noindent for large N. Thus by (\ref{18})
\begin{equation*}
\sum_{p}\mathbb {P}\left( \bigcup_{N=N_{p}}^{N=N_{p+1^{-1}}}\sup_{0\leq x<+\infty
}\left| R_{N4}\left( x\right) \right| >\varepsilon /4\right) <+\infty
\end{equation*}

\noindent whenever $k/N\rightarrow 0$ as $N\rightarrow +\infty $. Finally,

\begin{equation*}
\mathbb {P}\left( \sup_{0\leq x<+\infty }\left| \xi _{N}\left( x\right) -H_{k}\left(
x\right) \right| >N^{-\frac{1}{4}}\right)  =\mathbb {P}\left( \sup_{0\leq s<1}N^{-
\frac{1}{2}}\left| \gamma _{N}\left( s\right) \right| >N^{-\frac{1}{4}
}\right)
\end{equation*}

\begin{equation*}
\leq 2N\text{ }\max_{0\leq i\leq N}\text{ }\mathbb {P}\left( U_{i,N}-\frac{i}{N}>N^{-
\frac{1}{4}}-N^{-1}\right) =J_{N},
\end{equation*}

\noindent by the fact that $\gamma _{N}\left( .\right) $ has stationary increments.
Using now a representation of $\gamma _{N}$ by a Poisson process and an
approximation of a Poisson distribution by a Gaussian one
(see Lemmas 2.7 and 2.9 in \cite{13}) to get for large N that
\begin{equation*}
J_{N}\leq const.\text{ }N^{3/2}\mathbb {P}\left( N\left( 0,1\right) >N^{-\frac{1}{4}
}const.\right) \leq const.\text{ }N^{5/4}\exp \left( -N^{1/8}\right) .
\end{equation*}

\noindent Thus $\sum_{N}J_{N}<+\infty $. And the proof of Theorem \ref{t5} is now complete.
\end{proof}

\bigskip

\section{The oscillation moduli}
\noindent The oscillation modulus of a function $R\left( s\right) $, $0\leq s<1$, is
defined by
\begin{equation*}
\kappa \left( d,R\right) =\sup_{0\leq h\leq d}\sup_{0\leq s<1-h}\left|
R\left( s+h\right) -R\left( s\right) \right| ,0<d<1\text{.}
\end{equation*}

\noindent That of the empirical process pertaining to $iid$ $rv$'s has been studied
for several choices of $d$ in \cite{9} and \cite{13}. It is remarkable that
the weak versions of all those results are inherited by the reduced spacings
process $\alpha _{N}\left( s\right) =\beta _{N}\left( H_{k}^{-1}\left(
s\right) \right) ,0\leq s<1$, (see \cite{7}). For the strong case, we obtain
these two results.

\bigskip

\begin{theorem} \label{t6}

\noindent I. The Stute's case.\\

\noindent If $\left( d_{N}\right) _{N\geq 1}$ is a sequence of non-increasing positive
reals such that
\begin{equation}
Nd_{N}\rightarrow +\infty ,  \tag{S1}
\end{equation}

\begin{equation}
\left( \log d_{N}^{-1}\right) /\left( Nd_{N}\right) \rightarrow 0,
\tag{S2}
\end{equation}

\begin{equation}
\left( \log d_{n}^{-1}\right) /\log \log N\rightarrow +\infty ,  \tag{S3}
\end{equation}

\begin{equation}
\left( 2d_{N}\log d_{N}^{-1}\right) ^{\frac{1}{2}}/a_{N}=:q_{N}/a_{N}
\rightarrow +\infty ,\text{ as }N\rightarrow +\infty ,  \tag{S4}
\end{equation}

\noindent then for $k\geq 1$ fixed or $k=k\left( N\right) \rightarrow +\infty $ as $
N\rightarrow +\infty $ and satisfying
\begin{equation}
\exists N_{o},\delta >2,\text{ }\forall N\geq N_{o},0<d_{N}<k^{k\left(
\delta -2\right) }\exp \left( -\frac{1}{2}k^{\delta }\right) .
\end{equation}

\noindent we have $\lim_{N\rightarrow +\infty }\sup \kappa \left( d_{N},\alpha
_{N}\right) /q_{N}=1$ $\ a.s.$\\

\noindent II. A Mason-Wellner-Shorack case.\\

\noindent Let $a_{N}=\alpha \left( \log N\right) ^{-c},\alpha >0,c>0$. Then under the
same assumptions on k used in Part I, we have $\lim_{N\rightarrow +\infty
}\sup \kappa \left( d_{N},\alpha _{N}\right) /q_{N}\leq \left( 1+c\right) ^{
\frac{1}{2}},a.s.$
\end{theorem}

\bigskip

\noindent \textbf{Proof  of Part I of Theorem \ref{t6}}. We have by Lemmas \ref{l1} and \ref{l2},

\begin{equation*}
\forall N\geq 1,\left\{ \alpha _{N}\left( s\right) ,0\leq s<1\right\}
d_{=}\left\{ \Lambda _{N}\left( H_{k}^{-1}\left( s\right) \right)
+R_{N5}\left( s\right) +R_{N6}\left( s\right) ,0\leq s<1\right\}
\end{equation*}

\begin{equation}
=:\left\{
\bar{\alpha}_{N}\left( s\right) ,0\leq s<1\right\} ,  \label{30}
\end{equation}

\noindent with
\begin{equation*}
R_{N5}\left( s\right) =N^{\frac{1}{2}}\left( \mu _{N}-1\right)
H_{k}^{-1}\left( s\right) H_{k}^{\prime }\left( H_{k}^{-1}\left( s\right)
\right) =:N^{\frac{1}{2}}\left( \mu _{N}-1\right) \phi \left( s\right)
,0\leq s<1,
\end{equation*}

\noindent and

\begin{equation}
\sum_{p}\mathbb {P}\left( \bigcup_{N=N_{p}}^{N=N_{p+1^{-1}}}\sup_{0\leq s<1}\left|
R_{N6}\left( s\right) \right| >\left( 1+\varepsilon \right) a_{N}K\left(
k\right) \right) <+\infty,  \label{31}
\end{equation}

\noindent by (\ref{31}) and $(S4)$, we have

\begin{equation}
\sum_{p}\mathbb {P}\left( \bigcup_{N=N_{p}}^{N=N_{p+1^{-1}}}\kappa \left(
d_{N},R_{N6}\right) >\varepsilon q_{N}/3\right) <+\infty .  \label{32}
\end{equation}

\noindent By Lemma A4 in \cite{7}, $\kappa \left( d_{N},\phi \right) =\left( 1+o\left(
1\right) \right) q_{N}^{2}$ as $N\rightarrow +\infty $ for all k satisfying $(S5)$. Thus, by the loglog-law,

\begin{equation*}
\sum_{p}\mathbb {P}\left( \bigcup_{N=N_{p}}^{N=N_{p+1^{-1}}}\kappa \left(
d_{N},R_{N5}\right) >\varepsilon q_{N}/3\right) <+\infty,
\end{equation*}

\noindent whenever

\begin{equation}
\lim_{N\rightarrow +\infty }k^{-1}d_{N}\log \log \left( 1/d_{N}\right) \log
\log \text{ }Nk=0  \label{33}
\end{equation}

\noindent is satisfied. This obviously follows from $(S1)$, $(S2)$, $(S3)$, $(S4)$ and $(S5)$. By the results of \cite{13} as recalled in (\ref{123}), for $
\varepsilon >0$,

\begin{equation}
\sum_{p}\mathbb {P}\left( \bigcup_{N=N_{p}}^{N=N_{p+1^{-1}}}\kappa \left(
d_{N},\Lambda _{N}\left( H_{k}^{-1}\right) \right) >\left( 1+\varepsilon
/3\right) q_{N}\right) <+\infty,   \label{34}
\end{equation}

\noindent when $(S1)$, $(S2)$ and $(S3)$ hold. Since $\varepsilon $ is arbitrary
and since $(S1)$ and $(S3)$ imply (\ref{33}), we get
\begin{equation}
\lim_{N\rightarrow +\infty }\sup q_{N}^{-1}\kappa \left( d_{N},\alpha
_{N}\right) \leq 1,a.s.  \label{35}
\end{equation}

\noindent To get the other inequality, define for $0<c_{1}<c_{2}<+\infty ,0<d<1$, for
any function $R\left( s\right) ,0\leq s<1,$

\begin{equation}
\kappa ^{\prime }\left( d,\text{ }R\right) =\sup_{c_{1}d<u-t<c_{2}d}\left|
R\left( u\right) -R\left( t\right) \right| /\sqrt{u-t},0\leq u,\text{ }t\leq1. \label{36}
\end{equation}

\noindent Let $R_{N}\left( .\right) =R_{N5}\left( .\right) +R_{N6}\left( .\right) $
and $r_{N}\left( .\right) =\Lambda _{N}\left( H_{k}^{-1}\left( .\right)
\right).$ Now remark that for all $\varepsilon _{1}>0,$ there exists $
\varepsilon _{2}>0$ such that for

$$a=\left( \left( 1-\varepsilon _{1}\right)
\log d_{N}^{-1}\right) ^{\frac{1}{2}}
$$

\noindent and

$$
b=\left( \varepsilon _{2}\log
d_{N}^{-1}\right) ^{\frac{1}{2}},
$$

$$
a+b=\left( \left( 1-\varepsilon
_{1}+\varepsilon _{2}+2\left( \varepsilon _{2}\left( 1-\varepsilon
_{1}\right) \right) ^{\frac{1}{2}}\right) \log d_{N}^{-1}\right) ^{\frac{1}{2
}}=\left( \left( 1-\varepsilon _{3}\right) \log d_{N}^{-1}\right) ^{\frac{1}{
2}}
$$

\noindent with $\varepsilon _{3}>0$, $\varepsilon _{3},\varepsilon_{2}\rightarrow 0$ as $\varepsilon _{1}\rightarrow 0$. Thus,

\begin{equation*}
\mathbb {P}\left( \kappa ^{\prime }\left( d_{N},\alpha _{N}\right) \leq a\right)
\leq \mathbb {P}\left( \left\{ \kappa ^{\prime }\left(d_{N},\bar{\alpha}_{N}\right)\leq a\right\} \bigcup \left\{ \kappa ^{\prime }\left( d_{N},R_{N}\right)>b\right\} \right)
\end{equation*}

\begin{equation*}
+\mathbb {P}\left( \left\{ \kappa ^{\prime }\left( d_{N},\bar{\alpha}_{n}\right)
\leq a\right\} \prod \left\{ \kappa ^{\prime }\left( d_{N},R_{N}\right) \leq
b\right\} \right)
\end{equation*}

\begin{equation}
\leq \mathbb {P}\left( \kappa ^{\prime }\left( d_{N},R_{N}\right) >b\right) +\mathbb {P}\left(
\kappa ^{\prime }\left( d_{N},r_{N}\right) \leq a+b\right), \label{37}
\end{equation}

\noindent By (\ref{31}) and (\ref{32})

\begin{equation}
\sum_{p}\mathbb {P}\left( \bigcup_{N=N_{p}}^{N=N_{p+1^{-1}}}\kappa ^{\prime }\left(
d_{N},R_{N}\right) >b\right) <+\infty ,  \label{38}
\end{equation}

\noindent for all $\varepsilon _{2}>0$. Thus by (\ref{30}), (\ref{36}), (\ref{37}) and
(\ref{38}) and Lemma 2.9 of \cite{13} and some straightforward
considerations, we get $\lim_{N\rightarrow +\infty }\inf \kappa ^{\prime
}\left( d_{N},\alpha _{N}\right) \geq 1,a.s.,$ under $(S1)$, $(S2)$, $(S3)$ and $(S4)$. Letting $c_{1}=c_{2}=1$,

\begin{equation}
\liminf_{N \rightarrow +\infty} \kappa \left( d_{N},\alpha _{N}\right) \geq
\lim_{N\rightarrow +\infty }\inf \kappa ^{\prime }\left( d_{N},\alpha
_{N}\right) \geq 1,a.s. \label{39}
\end{equation}

\noindent (\ref{35}) and (\ref{39}) together complete the proof of Part I of Theorem \ref{t6}.

\bigskip

\noindent \textbf{Proof of Part II of Theorem \ref{t6}}.

\bigskip

\noindent Here $(S3)$ and $(S4)$ are satisfied. It suffices thus to write
again the proof of the part one where one should use the probability
inequality (\ref{13}) of \cite{9}. It must be noticied that Part III of Theorem
1 in \cite{9} holds for the general case where $a_{N}=\alpha \left( \log
_{N}\right) ^{-c},0<\alpha ,0<c$.

\newpage

\textbf{APPENDIX. PROOFS OF STATEMENTS (\ref{17}) AND (\ref{18})}

\bigskip
\noindent a) \textbf{Proof of Statement (\ref{17})}.\\

\noindent Tchebychev's inequality yields $\alpha >1$ and $\beta >1$ such that $\mathbb {P}\left(
S_{n}2/n^{2}>1+\varepsilon \right) \leq A_{2}n^{-\alpha }$ and $\mathbb {P}\left(
\left| S_{n}-S_{m\left( n\right) }\right| >n\varepsilon /2\right) \leq
A_{3}n^{-\beta }$ as $n\rightarrow +\infty ,$ where $m\left( n\right) =\max
\left\{ j^{2},j^{2}\leq n,\text{ }j=1,2,...\right\}$. Thus

\begin{equation*}
\mathbb {P}\left( \left| \mu _{N}-S_{n+1}/\left( n+1\right) \right| >\varepsilon
/2\right) +\mathbb {P}\left( S_{n+1}\geq 1+\varepsilon /2\right)  \\
\end{equation*}

\begin{equation}
\leq \mathbb {P}\left( \left| \mu _{N}-S_{n+1}/\left( n+1\right) \right|
>\varepsilon /2\right) +\left( A_{2}+o\left( 1\right) \right) k^{-\alpha
}N^{-\alpha }+\left( A_{3}+o\left( 1\right) \right) k^{-\beta }N^{-\beta }, \label{a1}
\end{equation}

\noindent since $\left( n+1\right) \sim Nk$ as $N\rightarrow +\infty $. Furthermore,
by Tchebychev's inequality,

\begin{equation*}
\mathbb {P}\left( S_{Nk}/Nk-\left( Nk\right) ^{2}>\varepsilon /8\right) \leq
64N^{-3}k^{-3}/\varepsilon ^{2}
\end{equation*}

\begin{equation*}\mathbb {P}\left( S_{k}/\left( Nk\right)
^{2}-\left( Nk\right) ^{2}>\varepsilon /8\right) \leq
64N^{-4}k^{-3}/\varepsilon ^{2}
\end{equation*}

\noindent and

\begin{equation*}
\mathbb {P}\left( \left| \mu _{N}-S_{n+1}/\left( n+1\right) \right| \geq \varepsilon
/2\right) \leq \mathbb {P}\left( S_{Nk}/Nk>\varepsilon /4\right) +\mathbb {P}\left( S_{k}/\left(
Nk\right) ^{2}>\varepsilon /4\right) .
\end{equation*}

\noindent Hence since $Nk\rightarrow +\infty ,$ $N^{2}k\rightarrow +\infty $ as $
N\rightarrow +\infty ,$

\begin{equation}
\sum_{N}\mathbb {P}\left( \left| \mu _{N}-S_{n+1}/n+1\right| >\varepsilon /2\right)
<+\infty .  \label{a2}
\end{equation}

\noindent Thus (\ref{a1}) and (\ref{a2}) together imply (\ref{17}).\\

\bigskip

\noindent \textbf{Proof of (\ref{18})}.\\

\noindent We have
\begin{equation*}
\frac{S_{n+1}-Nk}{\left( 2Nk\log \log Nk\right) ^{\frac{1}{2}}}=\frac{
S_{n+1}-S_{Nk}}{\left( 2Nk\log \log Nk\right) ^{\frac{1}{2}}}+\frac{
S_{n+1}-Nk}{\left( 2Nk\log \log Nk\right) ^{\frac{1}{2}}}=:S_{N}^{\prime
}+S_{N}^{\prime \prime }\text{.}
\end{equation*}

\noindent First, since $0\leq \left( n+1\right) -Nk\leq k,$

\begin{equation*}
\mathbb {P}\left( S_{N}^{\prime }>\varepsilon /2\right) \leq \mathbb {P}\left( S_{k}>\varepsilon
\left( 2Nk\log \log Nk\right) ^{\frac{1}{2}}/2\right)
\end{equation*}

\begin{equation*}
 \leq 1-H_{k}\left( k^{
\frac{1}{2}}N^{\frac{1}{2}}\right) \leq const.\exp \left( -\frac{1}{4}k^{
\frac{1}{2}}N^{\frac{1}{2}}\right)
\end{equation*}

\noindent as $k/N\rightarrow 0$, $N\rightarrow +\infty $ (see Statement (\ref{129})). Thus
\begin{equation}
\sum_{N}\mathbb {P}\left( S_{N}^{\prime }>\varepsilon /2\right) <+\infty .  \label{a3}
\end{equation}

\noindent Now, let
\begin{equation*}
p=p\left( N\right) =\inf \left\{ j,N>N_{j}\right\}
\end{equation*}

\noindent and
\begin{equation*}
q\left( N\right) =\inf \left\{ j,k\left( N\right) >N_{j}=\left[ \left(
1+\rho \right) ^{j}\right] ,\text{ }j=1,2,...\right\}
\end{equation*}

\noindent Then $N_{p-1}\leq N\leq N_{p},N_{p-1}N_{q-1}\leq NK\leq N_{p}N_{q},\log \log
N_{p}N_{q}=\left( \log \log N_{p}\right) \left( 1+o\left( 1\right) \right) ,$
as $N/k\rightarrow +\infty ,N\rightarrow +\infty ,N_{p+1}/N_{p}\rightarrow
1+\rho $, as $N\rightarrow +\infty $. Thus (see \cite{8}, p.259-262).

\begin{equation*}
\mathbb {P} \left(\bigcup_{N=N_{p}}^{N=N_{p+1^{-1}}}\left\{ S_{N}^{\prime \prime }\geq
1+\varepsilon /2\right\} \right) \leq A_{4}
\mathbb {P}\left(S_{N_{p}N_{q}}>1+\delta\left( \varepsilon ,\rho \right)\left( 2N_{p}loglogN_{p} \right) ^{ \frac{1}{2} } \right)
\end{equation*}

\begin{equation*}
\leq A_{5}p^{-\left( 1+\delta \left( \varepsilon ,\rho \right)
\right) }
\end{equation*}

\noindent as $p\rightarrow +\infty $, for $\rho $ small enough, $\delta \left(
\varepsilon ,\rho \right) >0$. The same holds for $-S_{N}^{\prime \prime }$.
\noindent Thus

\begin{equation}
\sum_{p}\mathbb {P}\left( \bigcup_{N=N_{p}}^{N=N_{p+1^{-1}}}\left( \left|
S_{N}^{\prime \prime }\right| >1+\varepsilon /2\right) \right) <+\infty. \label{a4}
\end{equation}

\noindent Finally (\ref{a3}) and (\ref{a4}) together imply (\ref{18}).

\newpage

config: arxiv


TITLE: Weak convergence in the space of Lipschitz Functions
QUESTION [3 upvotes]: Consider the Banach space $X$ of Lipschitz functions $g:[0,1] \to \mathbb{R}$ such that $g(0)=0$,  with the norm of $g$ given by its Lipschitz constant, i.e. $||g||_L = \sup_{x \neq y } \frac{g(x)-g(y)}{|x-y|}$.
Suppose that for some constant $K$ we have a bounded sequence of functions $\{f_n\} \subseteq X$ such that $||f_n||_L \leq K$ for all $n$. By the Arzelà–Ascoli theorem, there is a subsequence  which converges (in the $C^0$ norm) to a continuous function $f:[0,1] \to \mathbb{R}$. However, this convergence is not necessarily in the $||\cdot ||_L$ norm. Nevertheless, the function $f$ is Lipschitz and $||f||_L \leq K$.
When could one expect the subsequence of $f_n$ to converge weakly to the function $f$?

REPLY [2 votes]: Since my comments above didn't evince any response, I have decided to elaborate them to an answer, especially as  it is part of a longer and interesting story which
 goes back to the distinguished mathematicians Arens and Eells in (1955) (Pac. J. Math, vol. 78). They showed that a metric space can be embedded into a Banach space in such a way that a suitable universal property is satisfied
(the special case of the unit interval is not particularly relevant here---the natural setting is that of a general metric space).  In the language of category theory (which they did not use), this means that you can reflect a suitable category of metric spaces into one of Banach spaces. In order to simplify the notation, I shall assume that the metric space has a base point and its radius is at most $1$---then every Lipschitz function which vanishes at the base point is automatically bounded.  It turns out that this space is precisely that predual of the space of Lipschitz functions in question which I mentioned in my comment (more or less a direct consequence of the universal property).  Then the remarks in my comments apply.  The sequence in your question will converge weak $\ast$ but not weakly, in general.
Your question of when it will converge weakly is a bit vague for me to answer but the following considerations suggest that it won't be easy to give a simple necessary and sufficient condition.  One can embed your space in a natural way into the bounded, continuous functions on the square of the metric space minus the diagonal set.    So your question can be reinterpreted as one concerning weak convergence in a $C^b(S)$.  But the latter is the same as $C(\beta S)$ (the Stone-Čech compactification) and so this convergence is equivalent to uniform boundedness (no problem here) and pointwise convergence on $\beta S$.  Interpreting this in your original setting might be a problem.


TITLE: Equivariant homology of Hilb and torus stable curves
QUESTION [1 upvotes]: The torus equivariant homology of a compact homogeneous variety $G/B$ may be described by using the "skeleton" consisting of torus fixed points and the torus invariant curves that connect them.  Combinatorially, this produces a copy of the Hasse diagram of Bruhat order inside $G/B$, with edges labeled by appropriate roots.  The torus invariant curves can be given explicitly as the images of certain $SL_2$'s inside $G/B$.  Is there a similarly explicit construction of the torus invariant curves inside the Hilbert scheme of $n$ points in $\mathbb{C}^2$?  If so, does one obtain the same description of equivariant homology via localization (with dominance order replacing Bruhat order)?
Edit: Thanks to David Speyer for the explicit construction of a torus invariant curve here: Reference for combinatorics of cell decomposition of the Hilbert scheme of points in the plane

REPLY [3 votes]: It doesn't work as simply as it does for $G/B$, because there aren't finitely many $T$-invariant curves.  Instead they will come in families whose unions are things like products of projective spaces.  Although there will be a larger torus acting on each such family which has finitely many $T$-invariant curves, it won't extend to an action on the whole variety.
If you can work out the $T$-equivariant cohomology of the components of the fixed point sets for all codimension $1$ subtori of $T$, then you can take the intersection of all those things inside $H_T(X^T)$ just as you would if there were finitely many $T$-invariant curves.  This has been worked out for $Hilb_n(C^2)$, and more generally the Hilbert scheme of a smooth toric surface, in Evain, "The Chow ring of punctual Hilbert schemes of toric surfaces".


TITLE: About maximal invariant statistics on a group family
QUESTION [2 upvotes]: I searched and I didn't find any answer (positive or negative),
Suppose I have a group G, and a group family of probability measures .
(i.e, there is a probability measure P on G, and we define P_g(A)=P(gA) as g in G is a parameter). Also this family is assumed to be complete.
The maximal invariant statistic is always ancillary. Is it always maximal ancillary? Is it always the greatest ancillary?
Are there any reference to a proof or a counter example?

REPLY [1 votes]: the way your question is posed, it suggests that the sample space is $G$. in that case, there is only one orbit and the maximal invariant is a constant [function on $G$], so it is also [trivially] ancillary.
let $G$ be a compact lie group - the orthogonal group on $R^n$, for example. let P be normalized haar measure on $G$. the "group family of probability measures" $G$ generates from P is just P itself.
then any function on $G$ is ancillary - including the identity map of $G$ into itself. the latter is then obviously a maximal ancillary - and is not the same as the maximal invariant.
so in general the maximal invariant is not maximal ancillary.
btw - the above example even works for n = 1, when $G$ contains just 2 elements: the identity map on $R$ and its negative: $x \to -x$.


TITLE: Nullity of the linear transformation
QUESTION [0 upvotes]: Find the nullity of the linear transformation $\mathbb{T:M_{22}
\rightarrow M_{22}}$ given that it has rank $2$.

T can be a matrix transformation, for instance, it can be $I_2$. Now it's given that $I_2$ has rank 2 so by dimension theorem (rank + nullity = number of columns) it's nullity should be $2-2 = 0$. This is what I think is correct.
However, another approach gives a different answer.
 $\text{dim(ker T) + rank(T) = dim(M$_{22}$)} \implies \text{nullity} = 2\times 2 - 2 = 2$
Which approach is correct and what's wrong with the other one?

REPLY [2 votes]: The rank nullity theorem gives that $M_{22}$ is of dimension $4$, so the answer is $4-2 = 2$.
$I_2$ is of rank $2$ as a linear transformation from $\Bbb R^2 \to \Bbb R^2$.  But if you take the identity transformation from $\Bbb M_{22} \to \Bbb M_{22}$, that has matrix representation $I_{\mathbf{4}}$, so has rank $4$.
A matrix is used to represent a transformation from $\Bbb R^n \to \Bbb R^m$. But when $\Bbb R^n$ and $\Bbb R^m$ themselves are hidden as spaces of matrices, secretly $\mathbb M_{22}$, then we are technically looking at a map from $\mathbb R^\color{blue}4 \to \mathbb R^{\color{blue}4}$, not $\mathbb R^\color{red}2$.
This confusion should be promptly removed at this stage.


TITLE: What are some good explanations of the epsilon-delta definition of limits?
QUESTION [1 upvotes]: I am reading Stewart Calculus 7e and have encountered the epsilon-delta definition of a limit again. There was a time when I understood it, but it required a great amount of thought. Does anyone know of any exceptionally clear explanations of the definition that include media such as pictures, videos, etc.? I have seen some really artistic answers on this site and was hoping somebody could post one about the epsilon-delta definition of the limit.

REPLY [1 votes]: $$
\lim_{x\to a} f(x) = L
$$
means $f(x)$ can be made as close to $L$ as desired by making $x$ close enough, but not equal, to $0$.
$\varepsilon$ is how close we want to make $f(x)$ to $L$, i.e. we want to make the distance between $f(x)$ and $L$ less than $\varepsilon$.
$\delta$ is how close we will make $x$ to $a$, i.e. we will make the distance between $x$ and $a$ less than $\delta$, but not $0$.
So $\lim\limits_{x\to a}f(x)=L$ means no matter how tiny $\varepsilon$ is made there is some value of $\delta$ small enough to guarantee that if the distance from $x$ to $a$ is less than $\delta$, but not $0$, then the distance from $f(x)$ to $L$ is less than $\varepsilon$.
I've done some graphics related to this in the past, and I'll have to look for that stuff.