\begin{document}  
\maketitle{}
{\small \textit{Laboratoire de probabilit\'es et mod\`eles al\'eatoires, UMR7599, Universit\'e Pierre et Marie Curie,}}

 {\small \textit{Case 188, 4 place Jussieu, 75252 Cedex 05, Paris, France. }}
\vskip 1.5cm
\begin{abstract}
Let $P$ be a probability distribution on $\mathbb{R}^d$ (equipped  with an Euclidean norm). Let $ r,s > 0 $ and assume  $(\alpha_n)_{n \geq 1}$ is an (asymptotically) $L^r(P)$-optimal sequence of $n$-quantizers.  In this paper we investigate the asymptotic behavior of the  maximal radius sequence  induced by the sequence $(\alpha_n)_{n \geq 1}$ and defined to be for every $n \geq 1$, $\ \rho(\alpha_n) = \max \{\vert a \vert, \ a \in \alpha_n \}$. We show that if ${\rm card(supp}(P))$ is  infinite, the maximal radius sequence goes to  $\sup \{ \vert x \vert, x \in {\rm supp}(P) \}$ as $n$ goes to infinity. We then give the rate of convergence for  two classes of  distributions with unbounded support : distributions with exponential tails and distributions with polynomial tails. 
\end{abstract}


\section{ Introduction}

 Quantization has become an important field of information theory since the early $1940$'s. Nowadays, it plays an important rule in Digital Signal Processing (DSP), the basis of many areas of  technology, from mobile phones to modems and multimedia PCs. In  DSP, quantization is the process of approximating a continuous range of values or a very large set of discrete values by a relatively small set of discrete values. A common use of quantization is the conversion of a continuous signal  into a digital signal. This is  performed in analog-to-digital converters with a given quantization level. Beside these fields, quantization has recently  become a domain of  interest in Numerical Probability specially in numerical pricing of financial derivatives when their  prices read as an expectation (or involve conditional expectations) of some random processes (see e.g. $\cite{BPP}$). 

   From  a mathematical point of view, the $L^r$-optimal quantization  problem at level $n$ for  a  $\mathbb{R}^d$-valued random vector $X$  lying in   $L^r(\Omega,\mathcal{A},\mathbb{P})$ consists in finding  the best approximation of  $X$  by $q(X)$, where $q$ is a  Borel function taking  at most  $n$ values. This reads as the following minimization problem:
   $$ e_{n,r}(X) = \inf{\{ \Vert X - q(X) \Vert_r, q: \mathbb{R}^d \stackrel{\textrm{Borel}}{\longrightarrow} \mathbb{R}^d, \ \textrm{card}(q(\mathbb{R}^d)) \leq n \}}.$$  
   Note that in fact $e_{n,r}(X)$ only depends on the distribution $P=\mathbb{P}_X$ of $X$  so that  we will also use the notation $e_{n,r}(P)$ . However, for any Borel function $q: \mathbb{R}^d \rightarrow \alpha$ we have 
$$ \vert X -q(X) \vert \geq \min_{a \in \alpha} d(X,a) = d(X,\alpha)=\vert X - \widehat{X}^{\alpha} \vert  \quad \mathbb{P} \textrm{  a.s}  $$
so that the quantization problem reduces to 
\begin{eqnarray}
 e_{n,r}(X)  & = & \inf{\{ \Vert X - \widehat{X}^{\alpha} \Vert_r, \alpha \subset \mathbb{R}^d,  \textrm{ card}(\alpha) \leq n \}}  \nonumber \\
  & = &  \inf_{ \substack{\alpha \subset \mathbb{R}^d \\ \textrm{card}(\alpha) \leq n}} \left(\int_{\mathbb{R}^d} d(x,\alpha)^r dP(x) \right)^{1/r}. \label{er.quant}
 \end{eqnarray}
where $ \widehat{X}^{\alpha} = \sum_{a \in \alpha} a \mbox{\bf{1}}_{\{X \in C_a(\alpha)\}}$  and $(C_a(\alpha))_{a \in \alpha}$  corresponds to a Voronoi partition of  $\mathbb{R}^d$ (with respect to a norm $\vert \cdot \vert$ on $\mathbb{R}^d$), that is, a Borel partition of $\mathbb{R}^d$ satisfying for every $a \in \alpha$, $$ C_a(\alpha) \subset \{ x \in \mathbb{R}^d : \vert x-a \vert = \min_{b \in \alpha} \vert x-b \vert \}.$$

For every $n \geq 1$, the infimum in $(\ref{er.quant})$  holds as a finite minimum reached (at least) at  one   grid   $\alpha^{\star}$. In this case $\alpha^{\star}$ is called an  \textbf{$L^r(P)$-optimal} (or $L^r$-optimal for $X$)   and a sequence of $n$-quantizers $ (\alpha_n)_{n \geq 1} $ is $L^r(P)$-optimal if for every  $n \geq 1$,  $\alpha_n$ is $L^r(P)$-optimal. A sequence $ (\alpha_n)_{n \geq 1} $  is  said  \textbf{asymptotically $L^r(P)$-optimal} if
$$  \int_{\mathbb{R}^d}d(x,\alpha_n)^r P(dx) \ = \  e_{n,r}^r(X) + o(e_{n,r}^r(X)) \qquad  \textrm{as } n \rightarrow \infty.$$ 

  Moreover the  $L^r$-quantization error  $e_{n,r}(X)$ decreases to $0$ as $n$ goes to infinity and if there  is  an  $(r+\eta)$-moment of  $X$, for $\eta >0$,  the so-called Zador's theorem recalled below rules its rate of convergence  to $0$.
 \vskip 0.3cm
\noindent  \textbf{Zador's Theorem }(see $\cite{GL}$).  Let $P=P_a + P_s$ be the Lebesgue decomposition of $P$ with respect to the Lebesgue measure $\lambda_d$, where $P_a$ denotes the absolutely  continuous  part and $P_s$ the singular part of $P$.  Suppose  $\mathbb{E}\vert X \vert^{r+ \eta} < + \infty \textrm{ for some } \eta >0 $. Then $$  \lim_{n \rightarrow +\infty} n^{r/d} (e_{n,r}(P))^r = Q_r(P).$$
with $$ Q_r(P) = J_{r,d} \left( \int_{\mathbb{R}^d} f^{\frac{d}{d+r}} d\lambda_d \right)^{\frac{d+r}{d}} = J_{r,d} \ \Vert f \Vert_{\frac{d}{d+r}} \ \in [0,+\infty), $$
 $$ J_{r,d} = \inf_{n \geq 1} n^{r/d} e_{n,r}^r(U([0,1]^d)) \in (0,+\infty), $$
 where $ U([0,1]^d) $ denotes the uniform distribution on the set  $[0,1]^d$ and  $f = \frac{dP_{a}}{d\lambda_d}$. Note that the moment assumption : $\mathbb{E}\vert X \vert^{r+ \eta} < + \infty $ ensure that  $\Vert f \Vert_{\frac{d}{d+r}} $ is finite. 
  
Very little is known about the geometric properties of optimal quantizers. In this paper we address a first problem in this direction: we study the asymptotic behavior of the radii of a sequence $(\alpha_n)_{n \geq 1}$ of $L^r$-optimal quantizers. The maximal radius (or simply radius) $\rho(\alpha)$ of a quantizer $\alpha \subset \mathbb{R}^{d}$ is defined by 
$$ \rho(\alpha) = \max \{ \vert a \vert, \ a \in \alpha  \} .$$
In our framework, $\vert \cdot \vert$ will be  an Euclidean norm on $\mathbb{R}^d$. For the sake of simplicity, we will denote from now on  by $(\rho_n)_{n \geq 1}$  the sequence $(\rho(\alpha_n))_{n \geq 1}$  of radii of a sequence  $(\alpha_n)_{n \geq 1}$ of optimal quantizers  (although it may be not unique).  

We will show that, as soon as ${\rm supp}(P)$ is unbounded, $ \underset{n \rightarrow +\infty}{\lim} \rho_n = +\infty.$  Besides, our key inequalities  to get the upper and lower estimates of the maximal radius sequence are provided  in  Theorem $\ref{prop_princip_limsup}$ and Theorem $\ref{prop_gen_liminf}$. The first theorem yields  amount others the maximal rate of convergence of $\bar{F}_r({\frac{\rho_n}{c_{r,d}+\varepsilon}})$ (when $n \rightarrow +\infty$) to $0$, for every $\varepsilon>0$, with $c_{r,d}=1$ if $d=1; r \geq 1$ and $c_{r,d} = 2$ otherwise. It claims that  this rate is at most equals to $n^{-(1+r/d)}$. 

 Theorem $\ref{prop_gen_liminf}$  maintains in particular that for every $u>1$, $\bar{F}_r(\rho_n u)$ goes to $0$, as $n$ goes to infinity, at a rate at less equals to $n^{-\frac{r+\nu}{d}}$, where $\nu$ is such that the random vector $X$  has an $(r,r+\nu)$-distribution (see Definition $\ref{def_(r,s)_dist}$). We will see later on that the index  $\nu^{\star}_{X}$  ensuring that  $X$ has an $(r,r+\nu)$-distribution for every $\nu \in (0,\nu^{\star})$ will play a crucial role in the lower limit estimates of the maximal radius sequence.  
 
  Then  we will emphasize how knowing  the asymptotic behavior  of  the function $-\log \bar{F}_r$  allow  to derive the asymptotic estimates of $\rho_n$ (or $\log \rho_n$ ).  
  
As an important example  we can already mention distributions with density function $f$ satisfying 
\begin{equation*}  
 f(x) \ \varpropto \  \frac{(\log \vert x \vert)^{\beta}}{\vert x \vert^c} \ \mbox{\bf{1}}_{\{ \vert x \vert >1 \}}  \qquad  x \in \mathbb{R}^d, \ \beta \in \mathbb{R}, \ c> r+d
 \end{equation*}
for which  the  optimal rate of convergence of    $\log \rho_n$ is computed and given by 
\begin{equation*}
\lim_{n \rightarrow +\infty} \frac{\log \rho_n}{\log(n)} = \frac{1}{c-r-d} \frac{r+d}{d}.
\end{equation*}
Of course, this result is less accurate as giving the rate of convergence of the sequence $(\rho_n)$ itself  for which the exact limit can not be computed with our approach because the upper and lower limits  make appear no identified constants. Another example concerns distributions with exponential tail for which the upper and lower rates of convergence of  the sequence $(\rho_n)$ are provided. This is the case for 
 the normally distributed random vector on $\mathbb{R}^d$ for which we have
 \begin{equation*}
\sqrt{ \frac{2(r+d)}{d} } \leq  \liminf_{n \rightarrow +\infty}  \frac{\rho_n}{ \sqrt{  \log(n)} }  \leq   \limsup_{n \rightarrow +\infty} \frac{\rho_n}{ \sqrt{ \log(n)}}  \leq 2  \sqrt{ \frac{2(r+d)}{d}} .
 \end{equation*} 


Our general conjecture for such  distributions, which  is proved  when $d=1$ and $r \geq 1$, is that the liminf bound is sharp, that is, 
$$ \lim_{n \rightarrow +\infty}  \frac{\rho_n}{\sqrt{\log(n)}} = \sqrt{  \frac{2(r+d)}{d}}.$$
Moreover, an  alternative approach is given for the lower limit estimates. This approach  is based on random  quantization and relies $\rho_n$ to the expectation of  an  $i.i.d$ sequence of  random variables distributed as $X$.
 \vskip 0.3cm
The paper  is  organized as follows.  In Section 2, upper and lower estimates of the maximal radius sequence are given  and  the exact limit  is provided when the  cardinal of the support  of $P$ is infinite. This limit corresponds to $\sup\{ \vert x \vert, \ x \in {\rm supp}(P) \}$ and  Section 3 is entirely devoted to the convergence rate  of the sequence of radii  to this limit value.
\vskip 0.4cm
\noindent  { \small { \textbf{ Notations :}  Throughout the paper $X$ will denote an $\mathbb{R}^d$-valued random vector defined in the probability space $(\Omega,\mathcal{A},\mathbb{P})$ with distribution $P$ having a moment of order $r>0\ i.e. \ \mathbb{E} \vert X \vert^r<+\infty.$  We define  $$L^{r+}(\mathbb{P}) = \underset{\varepsilon>0}{\bigcup}L^{r+\varepsilon}(\mathbb{P}).$$

We will denote  by $\lambda_d$   the Lebesgue  measure on $\big(\mathbb{R}^d, \mathcal{B}(\mathbb{R}^d) \big)$. We will also denote  by $\bar{F}$ the survival function of $X$, that is, the $(0,1]$-valued function defined on $\mathbb{R}_{+}$ by 
 $$ \bar{F}: x \mapsto \bar{F}(x) = \mathbb{P} (\{ \vert X \vert > x \}) $$
 and for every $r>0$, we  define  the generalized  survival function of $X$  by
  $$\bar{F}_r: x \mapsto \bar{F}_r(x) = \mathbb{E}  \big(\vert X \vert^r \mbox{\bf{1}}_{\{ \vert X \vert > x \}} \big).$$
  Note that this last function is defined on $\mathbb{R}_{+}$ and takes values on the set  $(0,\mathbb{E} \vert X \vert^r]$ .
  
  For a given set $A$, $\overline{A}$ will stand for its closure, $\partial A$ its boundary,  ${\rm Conv}(A)$ its convex hull and $\mathring{A}$ or ${\rm  Int}(A)$ its interior. The cardinal of $A$ is denoted by ${\rm card}(A)$. For every $x \geq 0$, $[x]$ will denote the integral part of $x$. 
 } } 


\begin{comment}
\section{The main result}
 Let $\bar{F}_r: x \mapsto \bar{F}_r(x) = \mathbb{E}  \big(\vert X \vert^r \mbox{\bf{1}}_{\{ \vert X \vert > x \}} \big)$  and $\bar{F}: x \mapsto \bar{F}(x) = \mathbb{P} (\{ \vert X \vert > x \})$. It is clear that these two functions, defined on $\mathbb{R}_{+}^{\star}$, are monotone, decreasing to $0$ ($\bar{F}_r$ tends to $0$ provided $\mathbb{E} \vert X \vert^{r}<+\infty$) and take their values respectively in $(0,m_r]$ and $(0,1]$ where  $m_r= \mathbb{E} \vert X \vert^r$. Consequently, both functions $-\log \bar{F}_r(x)$ and  $-\log \bar{F}(x)$ are  nondecreasing and goes to $+\infty$ as $x$ goes to $+\infty$.
 
  We know that if  a function $f$ defined  on $(0,+\infty)$ is  continuous  increasing to $+\infty$ (at $+\infty)$,  its  generalized inverse function $f^{\leftarrow}$  defined by $\forall x>0$,
 $$  f^{\leftarrow}(x)=\inf\{t>0, \ f(t) \geq x \}.$$
  is a continuous function, monotone increasing to $+ \infty$. In the other hand, the following result holds (see \cite{BGT}):  If $f$ is regularly varying  (at $+\infty$) with index $1/ \delta, \delta>0$ then  there exists a function $\psi$, regularly varying  with index $\delta$ and satisfying
 \begin{equation}  \label{Asymp_equiv}
\psi(f(x)) \sim f(\psi(x)) \sim x    \quad \textrm{ as }\  x \rightarrow +\infty.
\end{equation}

 The function $\psi$ is an asymptotic inverse of  $f$ and it  is not necessarily increasing neither continuous. Moreover,  $\psi$ is unique up to asymptotic equivalence and $f^{\leftarrow}$ is one version of $\psi$.

Suppose  $-\log \bar{F}_r$ is  continuous and varies regularly with index $1/ \delta, \delta>0$ and  let  $\phi_r$  be  its  generalized inverse function. For a given probability distribution $P$ and an $L^r(P)$-optimal sequence of quantizers $(\alpha_n)_{n \geq 1}$  we show (under some technical hypotheses which will be specified later on) that  the generalized inverse  function $\phi_r$, or more generally  any continuous increasing function $\psi_r$, regularly varying with index $\delta$ (at $+\infty$) such that  $\psi_r(-\log \bar{F}_r(x)) = x + o(x)$,  satisfies 
\begin{equation*}
\left( 1+ \frac{r}{d} \right)^{\delta} \leq  \liminf_{n \rightarrow +\infty}  \frac{\rho_n}{ \psi_r(\log(n)) }  \leq   \limsup_{n \rightarrow +\infty} \frac{\rho_n}{ \psi_r(\log(n)) }  \leq c_{r,d}\left( 1+\frac{r}{d} \right)^{\delta}.
 \end{equation*} 
 where $c_{2,1}=1$ and $c_{r,d}=2$ otherwise.  This leads to the estimates  of  the maximal radius sequence for distributions  with exponential tails like the Gaussian distribution, the Gamma  and the  Weibull distributions. For example, if $P=\mathcal{N}(0,I_d)$ we have 
 \begin{equation*}
1 \leq  \liminf_{n \rightarrow +\infty}  \frac{\rho_n}{ \sqrt{  \frac{2(r+d)}{d} \log(n)} }  \leq   \limsup_{n \rightarrow +\infty} \frac{\rho_n}{ \sqrt{ \frac{2(r+d)}{d}  \log(n)}}  \leq 2 
 \end{equation*} 
 and in case $r=2$ and $d=1$ one has an exact limit, that is, $\underset{n \rightarrow +\infty}{\lim} \frac{\rho_n}{\sqrt{6 \log(n)}}=1.$
 
 However,  when the distribution has a polynomial tail like the  Pareto distribution, taking the logarithm of $\bar{F}_r$  make appear (asymptotically) the term $\log(x)$ in  $(\ref{Asymp_equiv})$ instead of  $x$. In this case the result is as follows: if $\psi_r$ is a continuous increasing function, regularly varying with index $\delta$ and satisfying $\psi_r(-\log \bar{F}_r(x)) = \log(x) + o(\log(x)), \ $ then
 $$ \lim_{n \rightarrow +\infty} \frac{\log \rho_n}{\psi_r(\log(n))} = \left(1+\frac{r}{d}\right)^{\delta}$$ 
 which, for the Pareto distribution with parameter $\gamma>r$ reads
$$  \lim_{n \rightarrow +\infty} \frac{\log \rho_n}{ \frac{r+1}{\gamma -r} \log(n)} = 1. $$ 
\end{comment}
 
\section{Asymptotics of the the maximal radius sequence}
  In this section we give an asymptotic  upper bound and a  lower bound of  the sequence  of radii. For  distributions supported by a  infinite  set, the  exact limit is provided.
 \vskip 0.2cm
  \begin{prop}  \label{prop_liminf_raymax}
  Let  $X \in L^r_{\mathbb{R}^d}(\mathbb{P})$. Let  $(\alpha_n)_{n\geq 1}$ be  a sequence of $n$-quantizers such that $e_{n,r}(X) \rightarrow 0 $  as $n \rightarrow +\infty$. Then, 
 \begin{equation} \label{hyp6.1}
  \liminf_{n \rightarrow +\infty} \rho_n \geq  \sup \{ \vert x \vert, \ x \in {\rm supp}(P) \}.
 \end{equation}
 \end{prop}
 Remark that this result  also holds  for any norm on $\mathbb{R}^d$.
\vskip 0.2cm
\begin{proof}[\textbf{Proof}] \ Let $ x \in \textrm{ supp}(P)$. Suppose that there exists $\varepsilon_0 >0 $ and  a subsequence $(\rho_{n_k})_{k \geq 1}$   such that  
\begin{equation} \label{hyp_contrap}
 \forall k \geq 1, \qquad \rho_{n_k} <  \vert x \vert -2 \varepsilon_0.
\end{equation}
Thus
$$ \exists \ \eta >0 \textrm{ such that } \forall k, \ d(B(0,\rho_{n_k}),B(x,\varepsilon_0))> \eta > 0.$$
Then one has for every  $k \geq 1$,
\begin{eqnarray*}
e_{n_k,r}(X) & = & \Vert d(X,\alpha_{n_k}) \Vert_r \\
& \geq & \Vert d(X,B(0,\rho_{n_k})) \Vert_r   \hspace{3.5cm} ( \textrm{since } \ \alpha_{n_k} \subset B(0, \rho_{n_k}))  \\ 
& \geq & \Vert d(X,B(0,\rho_{n_k})) \mbox{\bf{1}}_{\{X \in B(x,\varepsilon_0)\}} \Vert_r \\
& \geq & \Vert d(B(x,\varepsilon_0),B(0,\rho_{n_k})) \mbox{\bf{1}}_{\{X \in B(x,\varepsilon_0)\}} \Vert_r \\
& = & d(B(x,\varepsilon_0),B(0,\rho_{n_k})) \mathbb{P}(X \in B(x,\varepsilon_0))^{1/r} \\
& > & \eta \  \mathbb{P}(X \in B(x,\varepsilon_0))^{1/r} >0.
\end{eqnarray*}
This  is not possible since  $e_{n,r}(X) \rightarrow 0 $. 
Then, we have  shown that  
\begin{equation*}
 \forall x \in \textrm{ supp}(P), \quad  \liminf_{n} \rho_n \geq \vert x \vert. 
\end{equation*}
Hence   $ \underset{n}{\lim\inf} \rho_n \geq \sup \{ \vert x \vert, \ x \in {\rm supp}(P) \}.$
\end{proof}
\vskip 0.2cm
Among other results, the next proposition provides the limit of the sequence  $(\rho_n)_{n \geq 1}$  when the support of $P$ is infinite.
\vskip 0.3cm 
\begin{prop} \label{proplimitRM}
$\textbf{(a)}$\  Let  $\alpha$ be an $L^r$-optimal quantizer at level $n$. If  ${\rm card(supp}(P)) \geq n$  then  
 \begin{equation}
  \alpha \subset \overline{{\rm Conv(supp}(P))} \quad  \textrm{ and } \quad  \rho_n  \leq  \sup \{ \vert x \vert, \ x \in {\rm supp}(P) \}. 
  \end{equation}
\vskip 0.2cm
 \noindent $\textbf{(b)}$ If ${\rm card(supp}(P)) = +\infty$  then
\begin{equation}
 \lim_{n \rightarrow +\infty} \rho_n   = \sup_{n \geq 1} \rho_n =  \sup \{ \vert x \vert, \ x \in {\rm supp}(P) \}.
\end{equation}
for any  $L^r(P)$-optimal sequence of quantizers $(\alpha_n)_{n \geq 1}$.
 \end{prop}
\vskip 0.2cm
\begin{proof}[\textbf{Proof}] $\textbf{(a)} \ $ If $\alpha$ is $L^r$-optimal at level $n$  then $\textrm{card}(\alpha) = n$ since  ${\rm card(supp}(P)) \geq n$  (see $\cite{P}$).

Now, suppose that $\alpha \not \subset \overline{{\rm Conv(supp}(P))}$. Then let  $\ a \in \alpha \cap \left(\overline{{\rm Conv(supp}(P))} \right)^{c}$ and set  
$$ \alpha' = (\alpha \backslash \{ a \}) \cup \{ \Pi(a) \} $$
where $\Pi$  denotes the projection on the  non empty  closed convex set $\overline{{\rm Conv(supp}(P))}$. The projection is $1$-Lipschitz and $X$ is $\mathbb{P}$-$a.s \ {\rm supp}(P)$-valued, hence 
  $$ d(X,a) \geq d(\Pi(X),\Pi(a))  \stackrel{\mathbb{P}\textrm{-}a.s}{=}  d(X,\Pi(a)).$$
It follows that $$ d(X,\alpha) \geq d(X,\alpha')  \qquad \mathbb{P}\textrm{-}a.s . $$
 Since $\alpha$ is $L^r(P)$-optimal at level $n$ and ${\rm card}(\alpha') \leq {\rm card}(\alpha)=n$,
$$  \mathbb{E}(d(X,\alpha')^r) = \mathbb{E}(d(X,\alpha)^r) $$
so that the three statements hold:
 \begin{itemize}
\item[-] $d(X,\alpha') = d(X,\alpha) \quad \mathbb{P-}a.s$
\item[-] $\Pi(a) \not \in \alpha \backslash \{ a \} $ since $\alpha'$ is $L^r(P)$-optimal (which implies that ${\rm card} (\alpha') = n$),
\item[-] $\mathbb{P}(X \in C_{\Pi(a)}(\alpha'))>0$ (otherwise $\alpha' \backslash \{ \Pi(a) \}$ would be optimal).
\end{itemize} 
On the other hand, $X \in {\rm Conv(supp}(P))\ \mathbb{P}$-$a.s$ so that  $$\big(a-\Pi(a) \vert X-\Pi(a) \big) \leq 0  \quad \mathbb{P} \textrm{-}a.s.$$
Consequently
\begin{eqnarray*}
 \vert X-a \vert^2 - \vert X - \Pi(a) \vert^2 & = &  2(\Pi(a)-a \vert X-\Pi(a)) + \vert a - \Pi(a) \vert^2 \\
& \geq & \vert a-\Pi(a) \vert^2 >0   \qquad  \textrm{since } a \not \in \overline{{\rm Conv(supp}(P))} .
\end{eqnarray*}
As a consequence 
$$ d(X,\alpha') < d(X,\alpha) \qquad \mathbb{P}\textrm{-a.s \  on } \  \{ X \in \mathring{C}_{\Pi\textbf{(a)}}(\alpha') \} $$
where  $ \mathring{C}_{\Pi(a)}(\alpha') = \{\xi \in \mathbb{R}^d,\ d(\xi,\Pi(a)) < d(\xi,\alpha \backslash \{ a \})  \}  $  since the norm is Euclidean. 

This implies that $\mathbb{P}(X \in \mathring{C}_{\Pi\textbf{(a)}}(\alpha')) = 0$ and then $\mathbb{P}(X \in \partial C_{\Pi\textbf{(a)}}(\alpha'))>0$; this is impossible since $\alpha'$ is $L^r$-optimal (see $\cite{GL}$). Hence  $ \alpha \subset \overline{{\rm Conv(supp}(P))} $.

Now, let us prove that  $ \rho_n  \leq  \sup \{ \vert x \vert, \ x \in {\rm supp}(P) \}.$  Note first that this assertion is obvious if  ${\rm supp}(P)$ is unbounded.

On the other hand if  ${\rm supp}(P)$ is bounded then it is compact and so is ${\rm Conv(supp}(P))$. Let $x_0 \in  {\rm Conv(supp}(P))$  be such that $\vert x_0 \vert = \sup\{ \vert x \vert, x \in {\rm Conv(supp}(P)) \}.$  Thus  
$$ x_0 = \lambda_0 \xi_1 + (1-\lambda_0) \xi_2, \qquad \xi_1,\xi_2  \in  {\rm supp}(P) $$ and $\lambda \longmapsto \vert \lambda \xi_1 + (1-\lambda) \xi_2 \vert $ is convex so  that it  reaches its maximum at $\lambda=0$ or $\lambda=1$. Consequently $x_0 \in {\rm supp}(P).$
\vskip 0.2cm
\noindent $\textbf{(b)} $  This follows from the assertion about $\rho(\alpha_n)$ in the  item $(a)$ and from Proposition $\ref{prop_liminf_raymax}.$
\end{proof}

\begin{rem}
If the norm on $\mathbb{R}^d$ is an arbitrary norm, the assertion  $(a)$ of the  proposition may fail. An example  is given  with the $l_{\infty}$-norm in $\cite{GL}$, p. $ 25$.
\end{rem}
\vskip 0.2cm

 


\section{Convergence rate of the  maximal radius sequence}
We first start  by giving  two examples of distributions for which the sharp convergence  rate of the maximal radius sequence can be computed rather easily. In fact the semi-closed forms  established in $\cite{FP}$  for the $L^r$-optimal quantizers of the exponential and the Pareto distributions  and summed up in the following proposition allow to derive some sharp asymptotics for the maximal radius sequence  $(\rho_n)_{n \geq 1}$ induced by the unique  sequence of $L^r$-optimal quantizers at level $n$. These rates will be very useful  to validate the asymptotic rates  obtained by others approaches.
 \vskip 0.2cm
  \begin{prop}  \label{prop_semi_closed}  (see  \cite{FP} )
   $\textbf{(a)}$ Let $r>0$ and let $X$ be an  exponentially distributed random variable with scale parameter $\lambda>0$. Then, for every $n \geq 1$, the $L^r$-optimal quantizer $\alpha_n = (\alpha_{n,1}, \cdots, \alpha_{n,n})$ is unique and given by 
\begin{equation} \label{eq_comp_grid}
 \alpha_{n,k} = \frac{1}{\lambda}\left(\frac{a_n}{2}  +  \sum_{i=n+1-k}^{n-1} a_i \right), \quad 1 \leq k \leq n,
\end{equation}   
where  $(a_k)_{k \geq 1}$ is an $\mathbb{R}_{+}$-valued sequence recursively  defined by the following implicit  equation: 
\begin{equation*}
a_0 := +\infty, \quad \phi_r(-a_{k+1}) := \phi_r(a_k), \ k \geq 0 
\end{equation*}
with  $\phi_r(x) := \int_0^{x/2} \vert u \vert^{r-1} {\rm sign}(u) e^{-u} du $ (convention : $ 0^0 = 1$).

Furthermore, the sequence  $(a_k)_{k \geq 1}$ decreases to zero and  for every $k \geq 1$, $$a_k = \frac{r+1}{k}\left(1+ \frac{c_r}{k} + \mbox{O}(\frac{1}{k^2}) \right) $$
for some positive real constant $c_r$. \\
 $\textbf{(b)}$  Let  $r>0$ and let  $X$  be a random variable having a  Pareto distribution with index $\gamma>r$. Let $f$ be the density function : $ f(x) = \gamma x^{-(\gamma+1)} \mbox{\bf{1}}_{\{x >1\}}.$  Then for every  $n \geq 1$, the $L^r$-optimal quantizer $\alpha_n = (\alpha_{n,1}, \cdots, \alpha_{n,n})$ is  unique and given by  
\begin{equation} \label{form_Pareto}
\alpha_{n,k} = \frac{1}{1+a_n} \prod_{i=n-k+1}^{n-1} (1+a_i), \quad 1 \leq k \leq n,
\end{equation}
 where  $(a_k)_{k \geq 1}$  is an $\mathbb{R}_{+}$-valued sequence recursively defined by the following implicit equation: 
\begin{equation*}
a_0 = +\infty, \quad \phi_{\gamma}\left(- \frac{a_{k+1}}{1+a_{k+1}} \right) := \phi_{\gamma}(a_k), \ k \geq 1, 
\end{equation*}
with  $\phi_{\gamma}(x):= \int_0^{x/2} \gamma \vert u \vert^{r-1} {\rm sign}(u) (1+u)^{-(\gamma+1)}du.$
 The sequence $(a_k)_{k \geq 1}$ decreases to zero and there is some positive real constant $c$ such that for every $k \geq 1$,
$$ a_k = \frac{r+1}{(\gamma -r)k} \left(1+\frac{c}{k} + O(\frac{1}{k^2}) \right).$$
\end{prop}
Let us give now the sharp asymptotic derived from  these semi-closed forms. 
\vskip 0.4cm
\begin{prop} \label{prop_loi_expon}
 $\textbf{(a)}$ \ Let $r>0$ and let  $X$ be an exponentially distributed  random variable with parameter $\lambda >0$. 
 Then
\begin{equation} \label{asymp_exp}
\rho_n  \quad  = \quad  \frac{r+1}{\lambda} \log(n)+ \frac{C_{r}}{\lambda} + \mbox{O}\big(\frac{1}{n} \big),
\end{equation}
where  $C_{r}$ is a real constant depending only on  $r$. 
\vskip 0.2cm
\noindent   $\textbf{(b)}$ \  Let $r>0$ and let $X$  be a random variable with Pareto distribution of index $\gamma$  such that $ \gamma>r $.     Then,
\begin{equation} \label{asymp_Par}
\log(\rho_n)    =  \frac{r+1}{\gamma-r}  \log(n) + C_r + \mbox{O} \big(\frac{1}{n} \big),
\end{equation}
where  $C_r$ is a real constant depending only on $r$.
\end{prop}
\vskip 0.2cm
\begin{proof}[\textbf{Proof}] 
$\textbf{(a)}$ \ It follows from $(\ref{eq_comp_grid})$ that 
 $$\lambda \ \rho_n = \frac{a_n}{2} + \sum_{i=1}^{n-1} a_i  \  $$ 
where the sequence  $(a_n)_{n\geq 1}$  decreases  to zero and  satisfies  for  every  $ \ n \geq 1, \ a_n = \big((r+1)/ n \big) \big( 1 + c_r/n + \mbox{O}(1/n^2) \big) $, for some real constant  $c_r$. Thus, 
\begin{eqnarray*} 
\lambda \ \rho_n  

& = &   \frac{a_n}{2} + (r+1) \sum_{i=1}^{n-1} \frac{1}{i}  +  c_r \sum_{i=1}^{n-1} \frac{1}{i^2} + \sum_{i=1}^{n-1} \mbox{O}(1/i^3) \\
& =  &  (r+1) \log(n) + C_r + \mbox{O}\big(\frac{1}{n} \big).
\end{eqnarray*}
$\textbf{(b)}$ \ It follows from  $(\ref{form_Pareto})$ that  $$ \rho_n = \frac{1}{1+a_n} \prod_{i=1}^{n-1} (1+a_i)$$ 
 where $(a_n)_{n \geq 1}$ is an $\mathbb{R}_{+}$-valued sequence, decreasing to zero and satisfying  $: \ \forall n \geq 1$,\\
$ a_n = \frac{r+1}{(\gamma - r)n}  \big( 1 + c_r /n + \mbox{O}(1/n^2) \big)$, for some real constant  $ c_r $.

 Then,
\begin{eqnarray*}
\log(\rho_n) 

& = &  -\log(1+a_n) + \sum_{i=1}^{n-1} \big( a_i - \frac{a_i^2}{2} + \mbox{O}(a_i^3)   \big) \\
& = &   \frac{r+1}{\gamma-r}  \log(n) + C_r + \mbox{O} \big(\frac{1}{n} \big)
\end{eqnarray*}
 where we used that   $ \ \sum_{i=1}^{\infty} a_i^2 < \infty $ and   $\sum_{i=1}^{\infty} \mbox{O}(a_i^3)  < \infty $.
\end{proof}


\subsection{Upper estimate}
We investigate in this section the rate of convergence of $(\rho_n)$ to infinity. Let us give first  some definitions and  some hypotheses which will be useful later on. 

Let  $(\alpha_n)_{n \geq 1}$ be an $L^r(P)$-optimal sequence of quantizers at level $n$. For $n \geq 1$, we define $M(\alpha_n)$ to be  $$M(\alpha_n) = \{a \in \alpha_n \textrm{ such that }  \vert a \vert = \max_{b \in \alpha_n}  \vert b \vert  \}.$$
We will need  the following  assumption on $P$ 
\begin{equation*}
  \textbf{(H)} \ \equiv \hspace{1cm} P(dx) \geq \varepsilon_0 \ \mbox{\bf{1}}_{\{x \in \bar{B}(x_0,r_0)\}}  \lambda_d(dx), \quad \varepsilon_0,\  r_0>0, \ x_0 \in \mathbb{R}^d.
\end{equation*}


In the one dimensional setting, we will need the following specific assumption depending on $r \in [1,+\infty)$ :

\vskip 0.2cm
\noindent $\mathbf{(G_r)}$ \  $P=f \cdot \lambda_d$  where $f$ is  non-increasing to $0$ on $[A,+\infty)$ for some real constant $A$ and 

\begin{equation} \label{Assump_exact_limit}
 \lim_{y \rightarrow +\infty} \int_1^{+\infty} (u-1)^{r-1} \frac{f(uy)}{f(y)} du =0.  
 \end{equation}


\vskip 0.3cm
Let us make some brief comments on theses assumptions as well as some simple criterions.

 $\bullet$ Note that Assumption $\textbf{(H)}$ holds as soon as  $X$ has a density $f$ which is  bounded away from $0$  on a closed ball $\bar{B}(x_0,r_0), r_0>0, x_0 \in \mathbb{R}^d$, i.e.  $\varepsilon_0 := \min_{x \in \bar{B}(x_0,r_0)} f(x) >0$. This is  a very light assumption satisfied  by all usual distributions (Gaussian distribution, the exponential distribution,  the Pareto distribution, etc).


 
 $\bullet$  Assumption  $\mathbf{(G_r)}$ holds  for distributions with  density functions of the form 
 \begin{equation*}
 f(x) \ \varpropto \  \vert x \vert^{c} \ e^{- \vartheta  \vert x \vert^{\kappa}} \qquad  x \in \mathbb{R}; \ \vartheta, \kappa>0;\ c>-1.
 \end{equation*}
 Indeed,  we have for large enough $y$, $f$ is non-increasing and 
 \begin{eqnarray*}
  \int_1^{+\infty} (u-1)^{r-1} \frac{f(uy)}{f(y)} du 
  & = & \int_1^{+\infty} (u-1)^{r-1}u^c e^{-\vartheta y^{\kappa}(u^{\kappa}-1)} du \\
  & \leq &  \int_1^{+\infty} (u-1)^{r-1}u^c e^{-\vartheta A^{\kappa}(u^{\kappa}-1)} du < +\infty.
\end{eqnarray*}
 The existence of the last integral follows from the  existence of the moment of every order. 
 
 It follows from the Lebesgue convergence theorem that $(\ref{Assump_exact_limit})$  holds.  Then Assumption $\mathbf{(G_r)}$ holds in particular for the Gaussian distribution, for the Weibull distribution and for the Gamma distribution. However it fails for example for  the Pareto distribution. But, we will see  later that we do not need this assumption for distributions with polynomial tails to estimate the sequence $(\log \rho_n)_{n \geq 1}$.

\vskip 0.2cm 
 
 Let us recall the $L^r$-stationary property  which will be  also useful. 
 


Assume  $P= f \cdot \lambda_d$. The so-called $L^r$-distorsion  function  $ D_{n,r}^X : (\mathbb{R}^{d})^n \longrightarrow \mathbb{R}_{+}$ is   defined by :
$$ \alpha = (\alpha_1,\cdots,\alpha_n) \longmapsto  \mathbb{E}\left( \min_{i=1,\cdots,n} \vert X - \alpha_i \vert^r \right). $$ 
Then, for every $r \geq 1$, $ D_{n,r}^X $ is  differentiable at any codebook  having pairwise  distinct components and (see $\cite{GLP1}$ for details)  
\begin{equation} \label{station.}
\nabla D_{n,r}^X(\alpha) = r \left( \int_{C_i(\alpha)} (\alpha_i - u) \vert u - \alpha_i\vert^{r-2} f(u) du \right)_{1 \leq i \leq n} .
\end{equation}


An optimal $L^r$-quantizer at level $n$ $\ \alpha=\{\alpha_1, \cdots,\alpha_n \}$ for $P$  has full size $n$, so that,  
$$ \nabla D_{n,r}^X(\alpha) =0 .$$
$\alpha$ is said to satisfy an $L^r$-stationary property.

When $d=1$ then for any (ordered) quantizer $\alpha_n = \{x_{1}^{(n)},\cdots,x_{n}^{(n)}\}$, $x_{1}^{(n)} <\cdots<x_{n}^{(n})$ at level $n$, its Voronoi partition is given by
$$C_{1}(\alpha_n)=(-\infty,x_{\frac{1}{2}}^{(n)}], \  C_n(\alpha_n)=(x_{n-\frac{1}{2}}^{(n)},+\infty),\ C_i(\alpha_n) = (x_{i-\frac{1}{2}}^{(n)},x_{i+\frac{1}{2}}^{(n)}],\ i=2,\cdots,n-1, $$
with $\ x_{i-\frac{1}{2}}^{(n)} = \frac{x_{i}^{(n)}+x_{i-1}^{(n)}}{2} $ and $\ x_{i+\frac{1}{2}}^{(n)} = \frac{x_{i}^{(n)}+x_{i+1}^{(n)}}{2} $. 


 \vskip 0.2cm
 The main result of this section is  the following.

\begin{thm} \label{prop_princip_limsup}
 Suppose  that  $X$  has an  unbounded support and  that $\textbf{(H)}$ holds.  Let  $(\alpha_n)_{n \geq 1}$ be an $L^r(P)$-optimal  sequence of  quantizers. Then,

\vskip 0.2cm
\noindent $\textbf{(a)}$  
\begin{equation} \label{eqthm2.6}
 \lim_{\varepsilon \downarrow 0} \  \liminf_{n \rightarrow +\infty}  \left(  n^{1+ \frac{r}{d}}  \bar{F}_r \big(\frac{\rho_n}{2+\varepsilon} \big) \right) \geq C_{r,d,U}  .
\end{equation} 
\vskip 0.2cm
\noindent  $\textbf{(b)}$   If  $d=1,\ r\geq 1$ and if furthermore $\mathbf{(G_r)}$  holds then,
\begin{equation} \label{eqthm2.6dim1}
\lim_{\varepsilon \downarrow 0} \ \liminf_{n \rightarrow +\infty }   \left( n^{r+1} \bar{F}_r \big(\frac{\rho_n}{1+\varepsilon} \big) \right) \geq C_{r,1,U}  .
\end{equation} 

$C_{r,d,U}$ is a positive real constant depending on $r,d$ and the uniform distribution $U$ on $[0,1]^d$.
\end{thm}
The Lemmas below are used to prove this result. 
\vskip 0.2cm
\begin{lem} \label{lem1_princip_limsup}
 Let $X$ be an $\mathbb{R}^d$ valued random variable with unbounded support and   probability distribution $P$ and let $(\alpha_n)_{n \geq 1}$ be an  $L^r(P)$-optimal sequence of $n$-quantizers, $r>0$. Let  $(\rho_n)_{n \geq 1}$ be the maximal radius sequence induced by $(\alpha_n)_{n \geq 1}$. Then,
 
 \vskip 0.2cm
\noindent  $\textbf{(a)}$   $\ \forall \ \varepsilon >0, \ \exists \ n_{\varepsilon} $ such that $\forall n\geq n_{\varepsilon}$,
\begin{equation}  \label{equaconjecture}
\forall \ a \in M(\alpha_n), \ \forall \ \xi \in C_a(\alpha_n), \qquad \vert \xi \vert \geq \frac{\rho_n}{2+\varepsilon}.
\end{equation}
\vskip 0.2cm
 \noindent $\textbf{(b)}$  If  $d=1,\ r \geq 1$ and if furthermore $\mathbf{(G_r)}$ holds then, for large enough  $n$,
\begin{equation}  \label{equaconjecturedim1}
\forall \ a \in M(\alpha_n), \ \forall \ \xi \in C_a(\alpha_n), \qquad \vert \xi \vert \geq \frac{\rho_n}{1+\varepsilon}.
\end{equation}

\end{lem} 
\vskip 0.2cm
\begin{proof}[\textbf{Proof}] $\textbf{(a)}$  Since $(\alpha_n)$ is $L^r(P)$-optimal, $e_{n,r}(X) \rightarrow 0$ as $n \rightarrow +\infty$. Then,  the following asymptotic density  property  of $(\alpha_n)$ in the support of $P$ holds:
\begin{equation}
\forall \varepsilon>0, \ \forall x \in {\rm supp}(P) \qquad B(x,\varepsilon) \cap \alpha_n \not= \emptyset.
\end{equation}
Otherwise, if  there exists  $x \in {\rm supp}(P), \varepsilon>0$ and a subsequence $(\alpha_{n_k})_{k \geq 1}$ so that $ \forall k \geq 1, \  B(x,\varepsilon) \cap \alpha_{n_k} = \emptyset , $ then, for every $k \geq 1$, 
$$ e_{n_k,r}(X) \geq \Vert d(X,\alpha_{n_k}) \mbox{\bf{1}}_{X \in B(x, \varepsilon/2)} \Vert_r  \geq \frac{\varepsilon}{2}  P(B(x,\varepsilon/2))^{1/r} >0.$$
 Which contradicts the fact that $e_{n,r}(X) \rightarrow 0$ as $n \rightarrow +\infty$.
 
Now, to prove the result assume first $0 \in {\rm supp}(P)$.  Let $ \varepsilon >0,\ a \in M(\alpha_n)$. Then,  $\exists N_1 \in \mathbb{N}$ such that $B(0,\varepsilon) \cap \alpha_n \not= \emptyset, \ \forall n\geq N_1$. Now $\rho_n \rightarrow +\infty$ implies that $B(0,\varepsilon) \cap (\alpha_n \backslash M(\alpha_n)) \not= \emptyset $ for $n \geq N'_1$.

Let $b\in B(x,\varepsilon) \cap(\alpha_n \backslash M(\alpha_n)) $. We have for every $\xi \in C_a(\alpha_n)$,
$$ \vert \xi -b \vert^2  \geq  \vert \xi -a \vert^2, $$
namely
\begin{eqnarray*}
2(\xi|a-b) & \geq & \vert a \vert^2 -\vert b \vert^2 \qquad (\geq 0) \\
& \geq & \rho_n^2 - \vert b \vert^2.
\end{eqnarray*}
Now, $\ \vert \xi \vert \vert a-b \vert \geq  (\xi|a-b), $ then,
$$ \vert \xi \vert \vert a-b \vert \geq \frac{(\rho_n+\vert b \vert)(\rho_n-\vert b \vert)}{2}. $$
Moreover, $ \vert a-b \vert \leq \vert a \vert + \vert b \vert  \leq \rho_n+\vert b \vert$. One finally gets
$$ \vert \xi \vert \geq \frac{\rho_n-\vert b \vert}{2} \geq \frac{\rho_n-\varepsilon}{2}.  $$
Since $\rho_n \rightarrow +\infty$ as $n \rightarrow +\infty$, $\ \vert \xi \vert \geq \frac{\rho_n}{2+\varepsilon}$, for large enough $n$.

If $0 \not\in {\rm supp}(P)$ we show likewise that  $\vert \xi \vert \geq \frac{\rho_n - \vert x_0 \vert - \varepsilon}{2}, \ \forall x_0 \in  {\rm supp}(P)$ which implies the announced result since $\rho_n \rightarrow +\infty.$ 
\vskip 0.2cm
\noindent $\textbf{(b)}$  We will make an abuse of notation by considering that 
$$ \rho_n = \rho_n^{+} := \max \{x,\ x \in \alpha \} .$$
In what follows all results on  $ \rho_{-}(\alpha) := \max \{-x,\ x \in \alpha \}$ can be derived by using $-X$ instead of $X$.
 Let $\alpha_n=\{x_1^{(n)},\cdots,x_n^{(n)} \}$ and  suppose that (up to a subsequence) $$ \frac{x_{n-1}^{(n)}}{x_{n}^{(n)}} \rightarrow   \rho <1 .$$
 Let $\varepsilon >0$ such that $\rho + \varepsilon<1$. We have for large enough $n$,
$$ \frac{x_{n-1}^{(n)}}{x_{n}^{(n)}} < \rho + \varepsilon<1$$
or equivalently,
\begin{equation}  \label{equa_ref1}
 \frac{x_{n-1}^{(n)}+x_n^{(n)}}{2} < x_n^{(n)} \left( \frac{1+\rho+\varepsilon}{2}\right) .
 \end{equation}
Let $\rho'$ be such that  $0<\rho'< \frac{1-(\rho+\varepsilon)}{2}$, that is,
\begin{equation}  \label{equa_ref2}
\left( \frac{1+\rho+\varepsilon}{2}\right) < 1-\rho'<1.
\end{equation}


It follows from $(\ref{equa_ref1})$  and  $(\ref{equa_ref1})$  that 
\begin{eqnarray*}
 \int_{\frac{x_{n-1}^{(n)} +x_{n}^{(n)}}{2}}^{x_n^{(n)}} \left( 1-\frac{u}{x_n^{(n)}}\right)^{r-1} f(u)du 
 & \geq & 
 \int_{\frac{x_n^{(n)}(1+\rho+\varepsilon)}{2}}^{x_n^{(n)}(1-\rho')} \left( 1-\frac{u}{x_n^{(n)}}\right)^{r-1} f(u)du  \\
 & \geq & (\rho')^{r-1} \int_{\frac{x_n^{(n)}(1+\rho+\varepsilon)}{2}}^{x_n^{(n)}(1-\rho')}  f(u)du   \\
 & \geq &  \rho''  x_n^{(n)} f(c_n) 
 \end{eqnarray*}
 with $\rho'' = (\rho')^{r-1}(\frac{1}{2}-\rho' - \frac{\rho+\varepsilon}{2}) >0$  and $c_n \in (x_n^{(n)}(1+\rho+\varepsilon)/2,x_n^{(n)}(1-\rho'))$.

On the other hand, we have 
$$  \frac{1}{x_n^{(n)} f(x_n^{(n)})} \int_{x_n^{(n)}}^{+\infty} \left(  \frac{u}{x_n^{(n)}} -1 \right)^{r-1}f(u) du = \int_{1}^{+\infty} (u-1)^{r-1} \frac{f(u x_n^{(n)})}{f(x_n^{(n)})} du.$$


 It follows from Assumption $\mathbf{(G_r)}$ that 
$$ \limsup_{n \rightarrow +\infty} \ \frac{1}{x_n^{(n)} f(x_n^{(n)})} \int_{x_n^{(n)}}^{+\infty} \left(  \frac{u}{x_n^{(n)}} -1 \right)^{r-1}f(u) du =0. $$
Consequently we have for large enough $n$,
 $$  \frac{1}{x_n^{(n)} f(x_n^{(n)})} \int_{x_n^{(n)}}^{+\infty} \left(  \frac{u}{x_n^{(n)}} -1 \right)^{r-1}f(u) du < \rho'' $$
 so that ($A<c_n < x_n^{(n)}$ for large enough $n$ and $f$ is non-increasing in $[A,+\infty)$) 
 \begin{equation*}
  \int_{x_n^{(n)}}^{+\infty} \left(  \frac{u}{x_n^{(n)}} -1 \right)^{r-1}f(u) du  <  \rho'' x_n^{(n)} f(x_n^{(n)})   \leq  \rho'' x_n^{(n)} f(c_n)  \leq   \int_{\frac{x_{n-1}^{(n)} +x_{n}^{(n)}}{2}}^{x_n^{(n)}} \left( 1-\frac{u}{x_n^{(n)}}\right)^{r-1} f(u)du
  \end{equation*}
  which is not possible since the $L^r$-stationary equation  implies
$$ \int_{\frac{x_{n-1}^{(n)} +x_{n}^{(n)}}{2}}^{x_n^{(n)}} \left( 1-\frac{u}{x_n^{(n)}}\right)^{r-1} f(u)du =  \int_{x_n^{(n)}}^{+\infty} \left( \frac{u}{x_n^{(n)}} -1\right)^{r-1} f(u)du. $$

\begin{comment}
Set $\ z_n = \frac{x_{n,n} +x_{n,n-1}}{2}$.  It follows from the stationnary equation (given  by  $(\ref{eq_station})$ ) that, $\forall n \geq 1$,
$$ x_{n,n} = \frac{1}{P([z_n,l^{+}])} \int_{[z_n,l^{+}]} u P(du). $$
Then
$$ \frac{x_{n,n}}{z_n} = \frac{2}{1+\frac{x_{n,n}}{x_{n,n-1}}} = \frac{1 }{z_nP([z_n,l^{+}])} \int_{[z_n,l^{+}]} u P(du).$$
On the other hand one has
\begin{equation*}
 \frac{1}{z_nP([z_n,l^{+}[) } \int_{z_n}^{l^{+}} u P(du) 
  =  \frac{1}{z_n G(z_n) } \int_{z_n}^{l^{+}} u d (-G(u))   =   \frac{z_n G(z_n) + \int_{z_n}^{l^{+}} G(u)du}{z_n G(z_n)} .
 \end{equation*}
 where $G$  is the  survival function of $X$. Hence 
  $$ \frac{2}{1+\frac{x_{n,n}}{x_{n,n-1}}} = 1 + \frac{\int_{z_n}^{l^{+}}G(u) du }{z_n G(z_n)} .$$
If $l^{+} <\infty$ then $z_n \rightarrow l^{+}$ as $n \rightarrow +\infty$ by $\textbf{(G2)}, \ a)$;  otherwise if $l^{+}= +\infty$ then $z_n \rightarrow +\infty $ as $n \rightarrow +\infty$ because the quantization error goes to $0$ as $n$ goes to $+\infty$ (see Proposition \ref{prop_liminf_raymax}). Consequently, it follows from $\mathbf{(G_r)}$ that  $$  \lim_{n \rightarrow +\infty} \frac{2}{1+\frac{x_{n,n}}{x_{n,n-1}}} = 1 $$ 
 $i.e$
 \end{comment}

We have then shown that
 $$ \lim_{n \rightarrow +\infty} \frac{x_{n}^{(n)}}{x_{n-1}^{(n)}} =1. $$
 It follows that $\ \forall \varepsilon >, \exists \ n_{\varepsilon}$ such that $\forall n \geq n_{\varepsilon}$,
$\ x_{n}^{(n)} < (1+\varepsilon) x_{n-1}^{(n)}$.  Thus, $$ \rho_n = x_{n}^{(n)} < (1+\varepsilon) x_{n-1}^{(n)} < (1+\varepsilon) \xi \quad \forall \ \xi \in C_a(\alpha_n), a \in M(\alpha_n).$$
This completes the proof.
\end{proof}
\vskip 0.2cm
\begin{lem} \label{lem_minor_diff} Let $(\alpha_n)_{n \geq 1}$ be a sequence of  $L^r$-optimal  $n$-quantizers of the  distribution $P$. Suppose  that $\textbf{(H)}$ holds.
 Then for large enough $n$,
\begin{equation}
e^r_{n,r}(X) - e^r_{n+1,r}(X)  \geq  C_{r,d,U} \ n^{-\frac{r+d}{d}}, 
\end{equation}
where  
\begin{equation} \label{def_Const}
\ C_{r,d,U} = 2^{-(r+d)} \left(\frac{r}{d+r} \right)^{1/r} \left( \frac{d}{d+r}\right)^{d/r} \frac{\varepsilon_0}{1+\varepsilon_0} \ Q_{d+r}(U(\bar{B}(x_0,\frac{r_0}{2}))).
\end{equation}
\end{lem}
\vskip 0.2cm
\begin{proof}[\textbf{Proof}] \ \textbf{Step $1$}. Let  $y \in \mathbb{R}^d$. Without lost of generality we temporarily set $ \delta_n=d(y,\alpha_n)$. Following the lines of the proof of Theorem $2$ in $\cite{GLP}$  we have for every $x \in B(y,\delta_n/2)$ and $a \in \alpha_n$, 
$$  \vert x -a \vert \geq \vert y-a \vert-\vert x-a \vert \geq \delta_n/2$$
and hence 
$$ d(x,\alpha_n) \geq \delta_n/2 \geq \vert x-y \vert, \quad x \in B(y,\delta_n/2). $$
  It follows, by setting $\beta_n = \alpha_n \cup \{ y\}$, that
  $$ d(x,\beta_n) = \vert x -y \vert, \quad x \in B(y,\delta_n/2). $$
  Consequently  for every $b \in (0,1/2)$,
  \begin{eqnarray*}
  e^r_{n,r}(X) -e^r_{n+1,r}(X) & \geq  & \int d(x,\alpha_n)^rdP(x) - \int d(x,\beta_n)^rdP(x) \\
  &  \geq & \int_{B(y,\delta_n  b)}(d(x,\alpha_n)^r-d(x,\beta_n)^r) dP(x) \\
  & = &  \int_{B(y,\delta_n  b)} (d(x,\alpha_n)^r- \vert x-y \vert^r) dP(x) \\
  & \geq & \int_{B(y,\delta_n  b)} ((\delta_n/2)^r -( \delta_n b)^r)dP(x) \\
  & = & (2^{-r} -b^r) \ \delta_n^r \ P(B(y,\delta_n b)) .
  \end{eqnarray*}
  


\noindent \textbf{Step $2$}. Now, coming back  to the core of our proof let  $x_0$  and $r_0$ be as in $\textbf{(H)}$. For every $y \in \bar{B}(x_0,\frac{r_0}{2})  $  we have
\begin{eqnarray*}
 e^r_{n,r}(X) - e^r_{n+1,r}(X)  & \geq & (2^{-r} - b^{r}) \ \delta_n^r \ P(B(y,(b \ \delta_n ) \wedge  \frac{r_0}{2} )) \\
& \geq & (2^{-r} - b^{r}) \ \delta_n^r \varepsilon_0 \left((b\ \delta_n)^d \wedge  (\frac{r_0}{2})^d \right) \mbox{\bf{1}}_{\{y \in \bar{B}(x_0,\frac{r_0}{2})\}}. 
 \end{eqnarray*}
One checks that $$ \sup_{y \in \bar{B}(x_0,\frac{r_0}{2})} d(y,\alpha_n) \rightarrow 0.$$
Otherwise  $\exists  y_{\infty} \in \bar{B}(x_0,\frac{r_0}{2}),\ \eta>0 $ and a subsequence $(\alpha_{\varphi(n)})_{n \geq 1}$ of $(\alpha_n)_{n \geq 1}$ such that for every  $ n \geq 1$, $\ d(y_{\infty},\alpha_{\varphi(n)}) > \frac{\eta}{2}.$  Then 
$$ \int d(\alpha_{\varphi(n)},\xi)^rP(d\xi) \geq \int_{\bar{B}(y_{\infty},\frac{\eta}{4})} d(\alpha_{\varphi(n)},\xi)^rP(d\xi) . $$
Moreover for every $ \xi \in B(y_{\infty,\frac{\eta}{4}})$  $$d(\alpha_{\varphi(n)},\xi) \geq d(y_{\infty},\alpha_{\varphi(n)}) - d(y_{\infty},\xi) \geq \frac{\eta}{2} - \frac{\eta}{4} $$  so that
$$ \int d(\alpha_{\varphi(n)},\xi)^rP(d\xi) \geq (\frac{\eta}{2})^r  P(B(y_{\infty,\frac{\eta}{4}})).$$
This contradicts the fact that  $e_{n,r}(X) \rightarrow 0$ as $n$ goes to infinity. Consequently, for  large enough $n$, $$\ \sup_{y \in \bar{B}(x_0,\frac{r_0}{2})} d(y,\alpha_n) \leq \frac{r_0}{2}$$ so that 
$$e^r_{n,r}(X) - e^r_{n+1,r}(X)  \geq (2^{-r} - b^{r})\ b^d d(y,\alpha_n)^{d+r} \varepsilon_0 \  \mbox{\bf{1}}_{\{y \in \bar{B}(x_0,\frac{r_0}{2})\}}.$$


It follows that 
\begin{eqnarray*}
 \int_{\bar{B}(x_0,\frac{r_0}{2})}(e^r_{n,r}(X) - e^r_{n+1,r}(X)) \lambda_d(dy) 
& \geq & (2^{-r} - b^{r})\ \varepsilon_0\ b^d \int_{\bar{B}(x_0,\frac{r_0}{2})} d(y,\alpha_n)^{d+r}\lambda_d(dy) \\
 & \geq & (2^{-r} - b^{r}) \ \varepsilon_0\ \ b^d \ \lambda_d(\bar{B}(x_0,\frac{r_0}{2})) \ e_{n,r+d}^{r+d}(U(\bar{B}(x_0,\frac{r_0}{2}))).
\end{eqnarray*}
Then,
$$ e^r_{n,r}(X) - e^r_{n+1,r}(X) \geq  (2^{-r} - b^{r}) \ \varepsilon_0\ b^d \ e_{n,r+d}^{r+d}(U(\bar{B}(x_0,\frac{r_0}{2}))). $$
Consequently, for large enough $n$, 
$$ e^r_{n,r}(X) - e^r_{n+1,r}(X) \geq (2^{-r} - b^{r})\ \frac{\varepsilon_0}{1+\varepsilon_0} \ b^d \ Q_{d+r}(U(\bar{B}(x_0,\frac{r_0}{2})))\ n^{-\frac{d+r}{d}}. $$
As a function of $b$, the right hand side of this last inequality reaches its maximum at $b^{\star} = \frac{1}{2} \left( \frac{d}{d+r}\right)^{1/r}$. Which completes the proof.
\end{proof}
\vskip 0.2cm

\begin{proof}[\textbf{Proof of Theorem  $\ref{prop_princip_limsup}$}]  Let  $a \in M(\alpha_n)$ and $\varepsilon>0$. We have,
 $$ \mathbb{E} \vert X - \widehat{X}^{\alpha_{n-1 }} \vert ^r  \leq \mathbb{E} \vert X - \widehat{X}^{\alpha_{n} \backslash \{a\} } \vert ^r    $$ and
 \begin{eqnarray*}
\mathbb{E} \vert X - \widehat{X}^{\alpha_{n} \backslash a } \vert ^r  & = & 
\mathbb{E} \big(  \vert X - \widehat{X}^{\alpha_{n} } \vert ^r  \mbox{\bf{1}}_{ \{ X \in C_a^{c}(\alpha_{n}) \}} \big) + 
\mathbb{E} \big( \min_{b \in \alpha_{n} \backslash \{a \}}  \vert X - b  \vert ^r  \mbox{\bf{1}}_{ \{ X \in C_a(\alpha_{n}) \}} \big) \\ 
& \leq  & \mathbb{E}  \vert X - \widehat{X}^{\alpha_{n} } \vert ^r  + \mathbb{E} \big( \min_{b \in \alpha_{n} \backslash \{ a \} } ( \vert X \vert + \vert  b  \vert )^r  \mbox{\bf{1}}_{ \{ X \in C_a(\alpha_{n}) \}} \big).
\end{eqnarray*}
It follows from Lemma $\ref{lem1_princip_limsup}\ (a)$ that $ \exists \  n_{\varepsilon} \in \mathbb{N} \textrm{ such that  } \forall n \geq n_{\varepsilon}, \ \vert X \vert > \frac{\rho_n}{2+\varepsilon}$ on $ \{ X \in C_a(\alpha_{n}) \}$. Consequently, for all  $ b \in \alpha_{n} \backslash \{ a \},  \  \vert b \vert \leq \vert a \vert = \rho_n < (2+\varepsilon) \  \vert X \vert $. 

Hence,
 \begin{equation*} 
 \mathbb{E} \vert X - \widehat{X}^{\alpha_{n-1 }} \vert ^r -  \mathbb{E}  \vert X - \widehat{X}^{\alpha_{n} } \vert ^r  \leq (3+\varepsilon)^r \mathbb{E} \big(\vert X \vert^r \mbox{\bf{1}}_{\{ \vert X \vert > \frac{\rho_n}{2+\varepsilon} \}} \big).
\end{equation*}
 Lemma $\ref{lem_minor_diff}$  yields (since $(n-1)^{-\frac{r+d}{d}} \sim n^{-\frac{r+d}{d}}$ as $n \rightarrow +\infty$)
$$ (1+\varepsilon)^{-1} C_{r,d,U} \ n^{-\frac{r+d}{d}}  \leq (3+\varepsilon)^r \mathbb{E} \big(\vert X \vert^r \mbox{\bf{1}}_{\{ \vert X \vert > \frac{\rho_n}{2+\varepsilon} \}} \big) $$
for large enough $n$, so that  for every $\varepsilon>0$,
$$ \liminf_{n }  \left( n^{\frac{r+d}{d}}  \bar{F}_r \big(\frac{\rho_n}{2+\varepsilon} \big) \right) \geq \frac{C_{r,d,U}}{ (3+\varepsilon)^r (1+\varepsilon) } . $$
Taking the limit as $\varepsilon \rightarrow 0$ gives  the statement  $(\ref{eqthm2.6})$.
Assertion $(\ref{eqthm2.6dim1})$ is proved as above by using Lemma $\ref{lem1_princip_limsup}\ (b)$ instead of Lemma $\ref{lem1_princip_limsup}\ (a)$.
\end{proof}
\vskip 0.2cm
Recall that $\bar{F}_r(x) = \mathbb{E}  \big(\vert X \vert^r \mbox{\bf{1}}_{\{ \vert X \vert > x \}} \big)$.  It is clear that this function is non-increasing and goes to $0$ as $x \rightarrow +\infty$ (provided $\mathbb{E} \vert X \vert^{r}<+\infty$). Consequently, $-\log \bar{F}_r(x)$  is monotone nondecreasing and goes to $+\infty$ as $x$ goes to $+\infty$. Moreover, we know that if  a function $f$ defined  on $(0,+\infty)$ is   increasing to $+\infty$ (at $+\infty)$,  its  generalized inverse function $f^{\leftarrow}$  defined by $\forall x>0$,
 \begin{equation} 
   f^{\leftarrow}(x)=\inf\{t>0, \ f(t) \geq x \}
 \end{equation}
  is  monotone increasing to $+ \infty$. On the other hand, the following result holds (see \cite{BGT}):  If furthermore $f$ is regularly varying  (at $+\infty$) with index $1/ \delta, \delta>0$, then  there exists a function $\psi$, regularly varying  with index $\delta$ and satisfying
 \begin{equation}  \label{Asymp_equiv}
\psi(f(x)) \sim f(\psi(x)) \sim x    \quad \textrm{ as }\  x \rightarrow +\infty.
\end{equation}

 The function $\psi$ is an asymptotic inverse of  $f$ and it  is not necessarily increasing neither continuous. Moreover,  $\psi$ is unique up to asymptotic equivalence and $f^{\leftarrow}$ is one version of $\psi$.

We next show that for distributions with exponential tails, specifying either the asymptotic  inverse $\phi_r$ (if any) of the function $-\log \bar{F}_r$ or finding some asymptotic upper bound $\psi_r$ of  $\phi_r$ (having some " nice" properties)  leads to an upper estimate of the maximal radius sequence. This estimate is connected to the  chosen function $\psi_r$.

When the distribution has a polynomial tail,  we will look for  the  asymptotic inverse function of  $-\log \bar{F}_r(e^{x})$ or  some asymptotic upper bound $\psi_r$  of it to   provide  an upper estimate of $(\log \rho_n )_{n \geq 1}$.  
\vskip 0.3cm
 \begin{prop} \label{thm_princip_limsup}
 Assume  that the distribution $P$ of $X$ has an unbounded support and satisfies $(\textbf{H})$. Let $(\alpha_n)_{n \geq 1}$ be an $L^r(P)$-optimal sequence of $n$-quantizers. 
\vskip 0.2cm
\noindent $\textbf{(a)}$ If   $\psi_r$ is a measurable nondecreasing function, regularly varying  with index $\delta$ and  
\begin{equation}  \label{hyp_surv1}
\psi_r(-\log \bar{F}_r(x)) \geq x + o(x) \quad \textrm{ as } x \rightarrow +\infty,
\end{equation}
 then
\begin{equation} \label{result1_thm_result_gen}
\limsup_{n \rightarrow +\infty} \frac{\rho_n}{ \psi_r(\log(n))} \leq  2 \left( 1+\frac{r}{d} \right)^{\delta}.
\end{equation}
If $d=1$ and $r \geq 1$ and if $\mathbf{(G_r)}$  holds then, one has 
 \begin{equation}  \label{result1_thm_result_gen_part}
 \limsup_{n \rightarrow +\infty} \frac{\rho_n}{ \psi_r(\log(n))} \leq   (r+1)^{\delta}. 
 \end{equation}
 In particular if $-\log \bar{F}_r$ has regular variation with index $ 1/ \delta$ then $(\ref{result1_thm_result_gen})$ holds with $\psi_r = (-\log \bar{F}_r)^{\leftarrow}$.
\vskip 0.2cm
\noindent $\textbf{(b)}$  If   $\psi_r$ is a measurable nondecreasing function,  regularly varying  with index $\delta$ and  
\begin{equation}   \label{hyp_surv2}
\psi_r(-\log \bar{F}_r(e^x)) \geq  x + o(x) \quad \textrm{ as } x \rightarrow +\infty,
\end{equation}
 then
\begin{equation} \label{result2_thm_result_gen}
\limsup_{n \rightarrow +\infty} \frac{ \log \rho_n}{ \psi_r(\log(n))} \leq  \left( 1+\frac{r}{d} \right)^{\delta}.
\end{equation}
If $-\log \bar{F}_r(e^{\centerdot})$ has regular variation of index $ 1/ \delta$ then $(\ref{result2_thm_result_gen})$ holds with $\psi_r = (-\log \bar{F}_r(e^{\centerdot}))^{\leftarrow}$.
\end{prop}
\vskip 0.2cm
Prior to the proof, let us make some comment on the proposition. First  note that  the measurability of $\psi_r$ is necessary to define the regular varying property. On the other hand we have  for every $r > 0$ and for every $x>0, $ $$ \bar{F}_r(x)  \geq  x^r \bar{F}(x) .$$ 
Then
 $$ -\log \bar{F}_r(x)  \leq -\log \bar{F}(x) -r \log(x).$$
According to the nondecreasing hypothesis on $\psi$ we have for every  $x>1$, 
\begin{equation}  \label{eq_com_limsup}
  \psi_r(-\log \bar{F}_r(x)) \leq \psi_r( -\log \bar{F}(x) -r \log(x) ) \leq \psi_r(-\log \bar{F}(x))  
  \end{equation}
 since $log(x)>0$. Hence if Assumption $(\ref{hyp_surv1})$ holds then 
  $$ \psi_r(-\log \bar{F}(x)) \geq x + o(x) . $$
 We will see  further on  that for distributions with exponential tails, the function  $\psi_r$ in the statement $(a)$ of the proposition  does not depend on $r$.  However  in the situation of the item $(b)$ of the proposition, Assumption $(\ref{hyp_surv2})$ implies that
\begin{equation*}
\psi_r(-\log \bar{F}(e^x)) \geq (r+1) x + o(x).
\end{equation*}
 Consequently,  taking $\bar{F}$ instead of $\bar{F}_r$ in  Assumption $(\ref{hyp_surv2})$   will  induce a  loss of  precision in the upper estimate of $\log \rho_n$.

Also remark that if  $-\log \bar{F}_r$ (resp. $-\log \bar{F}_r(e^{\centerdot})$  ) is  measurable, locally bounded and  regularly varying  with index $1/ \delta, \delta>0$ then  its generalized inverse function  $\phi_r$  (resp. $\Phi_r$ ) is  measurable increasing to $+\infty$,  regularly varying  with index $\delta$ and, $\phi_r(-\log \bar{F}_r(x)) = x + o(x)$ (resp. $\Phi_r(-\log \bar{F}_r(e^x)) = x + o(x)$). Consequently,  both inequalities $(\ref{result1_thm_result_gen})$ and $(\ref{result1_thm_result_gen_part})$ (resp. claim $(\ref{result2_thm_result_gen})$) hold with  $\phi_r$ (resp. $\Phi_r$) in place of $\psi_r$. However,  $\phi_r$ (resp. $\Phi_r$ ) is in general not easy to compute and the examples below show that it is often easier to exhibit directly a  function $\psi_r$ satisfying the announced hypotheses without inducing any asymptotic loss of accuracy.    

We prove now  the proposition. 
\begin{proof} [\textbf{Proof}] $\textbf{(a)} \ $ It follows from $(\ref{eqthm2.6})$ and $(\ref{eqthm2.6dim1})$ that  for every $\varepsilon>0$, there is a positive real constant $C_{r,d,U,\varepsilon}$ depending on the indexing parameters such that 
$$ n^{-\frac{d+r}{d}}  C_{r,d,U,\varepsilon} \leq \bar{F}_r \big(\frac{\rho_n}{c_{r,d} + \varepsilon} \big)$$
where (from now on) $c_{r,1}=1$ if $r>1$; $c_{r,d}=2$ otherwise. Therefore, one has
$$ \frac{r+d}{d} \log(n)-\log(C_{r,d,U,\varepsilon}) \geq -\log \bar{F}_r \big(\frac{\rho_n}{c_{r,d} + \varepsilon} \big).$$
Combining the fact that  $\psi_r$ is nondecreasing  and Assumption $(\ref{hyp_surv1})$  yield
\begin{eqnarray*}
\psi_r \big( \frac{r+d}{d} \log(n)-\log(C_{r,d,U,\varepsilon}) \big) & \geq &  \psi_r \big(-\log \bar{F}_r \big(\frac{\rho_n}{c_{r,d} + \varepsilon} \big) \big) \\
& \geq & \frac{\rho_n}{c_{r,d} + \varepsilon} + o(\rho_n ).
\end{eqnarray*}
Moreover, dividing by $\psi_r(\log(n))$ (which is  positive for large enough $n$) yields
$$\frac{\rho_n}{\psi_r(\log(n))} \leq (c_{r,d}+\varepsilon) \big(1+ \frac{o(\rho_n)}{\rho_n} \big)^{-1} \frac{\psi_r \big( \frac{r+d}{d} \log(n)-\log(C_{r,d,U,\varepsilon}) \big)}{\psi_r(\log(n))}.$$
 It follows from the regular varying hypothesis on $\psi_r$ and $\underset{n}{\lim} \ \rho_n =+\infty$ that 
$$ \limsup_{n \rightarrow +\infty} \frac{\rho_n}{\psi_r(\log(n))} \leq  (c_{r,d}+\varepsilon) \left(\frac{r+d}{d} \right)^{\delta}, \quad \forall \varepsilon>0.  $$
The result follows  by letting $\varepsilon \rightarrow 0.$ \\
$\textbf{(b)}\ $ As previously,  one derives from $(\ref{eqthm2.6})$ and  from Assumption $(\ref{hyp_surv2})$  and  the nondecreasing  hypothesis  on $\psi_r$    that
\begin{eqnarray*}
\psi_r \big( \frac{r+d}{d} \log(n)-\log(C_{r,d,U,\varepsilon}) \big) & \geq &  \psi_r \big(-\log \bar{F}_r \big(\frac{\rho_n}{c_{r,d} + \varepsilon} \big) \big) \\
& \geq &  \log \rho_n - \log(c_{r,d} + \varepsilon)+ o(\log \rho_n ).
\end{eqnarray*}
It follows that
$$ \frac{ \log \rho_n}{\psi_r(\log(n))}  \leq \left( 1-\frac{\log(c_{r,d}+\varepsilon)}{ \log \rho_n} + \frac{o(\log \rho_n)}{\log \rho_n} \right)^{-1} \frac{\psi_r \big( \frac{r+d}{d} \log(n)-\log(C_{r,d,U,\varepsilon}) \big)}{\psi_r(\log(n))}. $$
Owing to  the regular varying hypothesis on $\psi_r$ and the fact that  $\underset{n}{\lim} \ \rho_n = +\infty$, we have
$$ \limsup_{n \rightarrow +\infty} \frac{ \log \rho_n}{ \psi_r(\log(n))} \leq  \left( \frac{r+d}{d} \right)^{\delta}.  $$
\end{proof}

We  next give an explicit  asymptotic upper bound for the convergence rate of the maximal radius sequence in the sense  that the function $\psi_r$ is made explicit. These  bounds are derived  on the rate of decay of the generalized survival function $\bar{F}_r$. 

\begin{crit} \label{cor_princip_limsup}  $\textbf{(a)} $ \ Let  $X $  be a random variable with unbounded support. Let $r>0$ and  let $(\alpha_n)_{n \geq 1}$ be an $L^r$-optimal  sequence of $n$-quantizers for $X$. Let  $\kappa>0$  such that $e^{\vert X \vert^{\kappa}} \in L^{0+}(\mathbb{P})$. Set
\begin{equation}  \label{Assum_Cor_Exp_tail1}
 {\theta}^{\star} =  \sup \big\{ \theta>0, \  \limsup_{x \rightarrow +\infty} \ e^{\theta  x^{\kappa}} \bar{F}_r(x) < +\infty \big\} = \sup \big\{ \theta>0, \  \mathbb{E}\ e^{\theta \vert X \vert^{\kappa}} < +\infty \big\}.
 \end{equation}
Then $\theta^{\star} \in (0,+\infty]$ and 
\begin{equation} \label{Ineq_Cor_Exp_tail1}
\limsup_{n \rightarrow +\infty} \ \frac{\rho_n}{\big(\log(n) \big)^{1/ \kappa}}  \leq 2  \left( \frac{r+d}{  d \ \theta^{\star} } \right)^{1/ \kappa}.
\end{equation}
When  $d=1$ and  $r \geq 1$, if $\mathbf{(G_r)}$ holds then $$ \underset{n \rightarrow +\infty}{\limsup} \ \frac{\rho_n}{\big(\log(n) \big)^{1/ \kappa}}  \leq \left(\frac{r+1}{\theta^{\star}} \right)^{1/ \kappa} .$$
\vskip 0.2cm
\noindent \textbf{(b)} \ Let  $X \in L^{r+}(\mathbb{P})$ be a random variable with unbounded support. Set 
\begin{equation}  \label{Assum_Cor_pol_tail1}
\zeta^{\star}=\sup \big\{ \zeta>0, \  \limsup_{x \rightarrow +\infty} \ x^{\zeta -r}  \bar{F}_{r}(x) <+\infty  \big\} = \sup \big\{ \zeta>r, \ \mathbb{E} \vert X \vert^{\zeta} <+\infty  \big\}.
\end{equation}
Then $\zeta^{\star} \in (r,+\infty]$ and 
\begin{equation}  \label{Ineq_Cor_pol_tail1}
\limsup_{n \rightarrow +\infty} \  \frac{\log \rho_n}{\log(n)} \leq \frac{1}{\zeta^{\star}-r} \frac{r+d}{d}.
\end{equation}
\end{crit}
Prior to the proof we can make the following remark.
\begin{rem}  { \rm If $X \in \underset{r>0}{\bigcap} L^r(\mathbb{P})$ then $\zeta^{\star}=+\infty$ and consequently $\ \underset{n \rightarrow +\infty}{\lim} \  \frac{\log \rho_n}{\log(n)} =0.$ }
\end{rem}
\vskip 0.3cm
\begin{proof} [$\textbf{Proof }$]
$\textbf{(a)}$ \ The equalities in $(\ref{Assum_Cor_Exp_tail1})$ and  $(\ref{Assum_Cor_pol_tail1})$ are obvious. 

 Let $\theta \in (0,\theta^{\star})$. We have 
\begin{equation*} \label{Ineq_tail}
 \mathbb{E} \big(\vert X \vert^r \mbox{\bf{1}}_{\{ \vert X \vert > x \}} \big) = \mathbb{E} \big(\vert X \vert^r \mbox{\bf{1}}_{\big\{ e^{\theta \vert X \vert^{\kappa}} > e^{\theta x^{\kappa}}  \big\} } \big) \leq e^{-\theta x^{\kappa}} \mathbb{E} \big(\vert X \vert^r e^{\theta \vert X \vert^{\kappa}}\big).  
 \end{equation*}
Now, the right hand side of this last inequality is finite because if  $\theta' \in (\theta,\theta^{\star})$,
$$ \vert x \vert^r e^{\theta \vert  x \vert^{\kappa}} \leq 1 + C_{\theta, \theta'} e^{\theta' \vert  x \vert^{\kappa}} .$$

As a consequence,
$$ -\log \bar{F}_r(x) \geq \theta x^{\kappa} + C_{X}, \qquad C_{X} \in \mathbb{R}.$$
Let $\psi_{\theta} (y)= \left( \frac{y}{\theta} \right)^{1/ \kappa}$. As a function of $y$, $\psi_{\theta}$ is continuous (then measurable) increasing to $+\infty$, regularly varying with index $\delta=\frac{1}{\kappa}$ and we have 
$$\psi_{\theta}(-\log \bar{F}_r(x))  \geq \left(x^{\kappa} + \frac{C_X}{\theta}\right)^{1/ \kappa} = x + o(x), \quad \textrm{as } x \rightarrow +\infty. $$
It follows  from Proposition $\ref{thm_princip_limsup}\ (a) $ that 
$$ \limsup_{n \rightarrow +\infty} \frac{\rho_n}{\big(\log(n) \big)^{1/ \kappa}}  \leq c_{r,d}  \left(\frac{d+r}{d \ \theta} \right)^{1/ \kappa} \qquad \forall \theta \in (0,\theta^{\star}).$$ Letting $\theta \rightarrow \theta^{\star}$ completes the proof.  


\vskip 0.2cm
\noindent $\textbf{(b)}$   Let $\zeta \in (r,\zeta^{\star})$. We have 
\begin{eqnarray*} 
\mathbb{E} \big(\vert X \vert^r \mbox{\bf{1}}_{\{ \vert X \vert > x \}} \big) 
&  =  & \mathbb{E} \big(\vert X \vert^r \mbox{\bf{1}}_{\{ 1< x^{-\zeta+r} \vert X \vert^{\zeta-r} \}} \big)  \\ 
& \leq & x^{-\zeta+r} \mathbb{E} \vert X \vert^{\zeta} .
\end{eqnarray*}
Then 
$$ -\log \bar{F}_r(x)  \geq (\zeta-r) \log(x) + C $$
so that by setting  $\psi_r(x) = \frac{x}{\zeta - r}$,   it follows from Proposition $\ref{thm_princip_limsup}\ (b) $  that 
\begin{equation*}
\limsup_{n \rightarrow +\infty} \  \frac{\log \rho_n}{\log(n)} \leq \frac{1}{\zeta -r} \frac{r+d}{d}.
\end{equation*}
Letting $\zeta$ go to $\zeta^{\star}$  yields the assertion  $(\ref{Ineq_Cor_pol_tail1})$ .
\end{proof}
\vskip 0.2cm
\begin{rem}
 { \rm  Note that the choice of the function  $\psi_r$ in the statement $(a)$ of Proposition $\ref{thm_princip_limsup}$ does not depend on $r$ as approved in the proof of the item $(a)$ of the  previous criterion. But  for distributions with polynomial tails the choice of $\psi_r$ clearly depends on $r$.}
\end{rem}
 
 We now give  more explicit results  for  specified density functions.
 
 \begin{cor} \label{cor_gen_density}
  \textbf{(a)} \  Suppose  that  the  density $f$ of $X$ satisfies  
\begin{equation}  \label{gen_density_exp_limsup}
f(x) \ \varpropto \  \vert x \vert^{c} \ e^{- \vartheta  \vert x \vert^{\kappa}} \qquad  x \in \mathbb{R}^d; \ \vartheta, \kappa>0;\ c>-d.
\end{equation}
Then $\theta^{\star}= \vartheta$ and
\begin{equation}  \label{Ineq_Cor_Exp_tail_density1}
 \limsup_{n \rightarrow +\infty} \frac{\rho_n}{\big(\log(n) \big)^{1/ \kappa}}  \leq \frac{c_{r,d}}{\vartheta^{1/ \kappa}}  \left(1+ \frac{r}{d} \right)^{1/ \kappa}.
\end{equation}
 \vskip 0.2cm 
 \noindent $\textbf{(b)}$ \ If   the density of $X$ reads 
 \begin{equation}   \label{gen_density_pol_limsup}
 f(x) \ \varpropto \  \frac{(\log \vert x \vert)^{\beta}}{\vert x \vert^c} \ \mbox{\bf{1}}_{\{ \vert x \vert >1 \}}  \qquad  x \in \mathbb{R}^d, \ \beta \in \mathbb{R}, \ c> r+d
 \end{equation}
 then $ \zeta^{\star}  = c-d $ and 
 \begin{equation}  \label{Ineq_Cor_pol_tail_density2}
\limsup_{n \rightarrow +\infty} \  \frac{\log \rho_n}{\log(n)} \leq \frac{1}{c-d-r} \frac{r+d}{d}.
\end{equation}
\vskip 0.2cm
\noindent $\textbf{(c)}$ \ The statement $(\ref{Ineq_Cor_Exp_tail_density1})$ (resp $(\ref{Ineq_Cor_pol_tail_density2})$) holds if the density of $X$ is simply equivalent to the specified density in $(\ref{gen_density_exp_limsup})$ (resp in $(\ref{gen_density_pol_limsup})$).
 \end{cor}
 \vskip 0.2cm
 Notice that the restriction $c> r+d$ in  $(\ref{gen_density_pol_limsup})$ ensures that $\mathbb{E} \vert X \vert^r < +\infty.$
 \begin{proof} [$\textbf{Proof}$]  $\textbf{(a)}$   


We have 
 \begin{eqnarray*}
\bar{F}_r(x) = \mathbb{E}\big(\vert X \vert^r \mbox{\bf1{}}_{\{\vert X \vert >x \}}\big) & = & K \int_{\{ \vert u \vert >x \}} \vert  u \vert^{r+c}\ e ^{-\vartheta \vert u \vert^{\kappa}} d\lambda_d(u) \\
&  = & K \ V_d  \int_{x}^{+\infty} \rho^{r+c+d-1}  \ e ^{-\vartheta \rho^{\kappa}} d\rho
\end{eqnarray*}
where $K= \int \vert x \vert^c e^{-\vartheta \vert x \vert^{\kappa}} d\lambda_d(x)$ is the normalizing positive real constant in  $(\ref{gen_density_exp_limsup})$  and $V_d$  denotes  the volume of the Euclidean unit ball. 
Integrating by parts and using  usual  integral comparison rules  yields
\begin{equation*} 
\bar{F}_r(x)   = C_{d,\vartheta,\kappa} \  x^{r+c+d-\kappa} \ e^{-\vartheta x^{\kappa}}\big(1 + o(1) \big)
\end{equation*}
for a positive real constant $C_{d,\vartheta,\kappa}$.  It follows that  if $\theta<\vartheta$ then 
$$ \limsup_{x \rightarrow +\infty} \ e^{\theta x^{\kappa}} \bar{F}_r(x)  <+\infty $$
and if  $\theta>\vartheta$ then 
$$ \limsup_{x \rightarrow +\infty} \ e^{\theta x^{\kappa}} \bar{F}_r(x)  = +\infty.$$
Which means that $\theta^{\star} = \vartheta$  and the statement $(\ref{Ineq_Cor_Exp_tail_density1})$ follows from  Criterion  $\ref{cor_princip_limsup} \ (a)$.


\begin{comment}
 On the other hand we have 
 \begin{eqnarray*}
\bar{F}_r(x) = \mathbb{E}\big(\vert X \vert^r \mbox{\bf1{}}_{\{\vert X \vert >x \}}\big) & = & K \int_{\{ \vert u \vert >x \}} \vert  u \vert^{r+c}\ e ^{-\vartheta \vert u \vert^{\kappa}} d\lambda_d(u) \\
&  = & K C_d  \int_{x}^{+\infty} \rho^{r+c+d-1}  \ e ^{-\vartheta \rho^{\kappa}} d\rho
\end{eqnarray*}
where $K$ is the normalizing positive real constant in  $(\ref{gen_density_exp_limsup})$ and $C_d=   \frac{\pi^{d/2}}{\Gamma(1+d/2)}$.  In fact define the hyperspherical coordinate by setting
$$ u_1 =  \prod_{j=1}^{n-1} \sin(\phi_j);\  u_i =  \rho \prod_{j=1}^{n-i} \sin(\phi_j) \cos(\phi_{n-i+1}), \ i=2,\cdots,d-1  \textrm{ and }   u_d = \rho \cos(\phi_1),$$
with   $$ \rho >0; \ \phi_i \in (0,\pi), \ i=1,\cdots,d-2  \textrm{ and  }  \phi_{d-1} \in (0,2\pi). $$
 We show by induction on $d$ that the Jacobian  is equals to  $\rho^{d-1} \prod_{j=1}^{d-2} \sin^{d-j-1}(\phi_j).$ Hence
 \begin{equation*}
   \int_{\{ \vert u \vert >x \}} \vert  u \vert^{r+c}\ e ^{-\vartheta \vert u \vert^{\kappa}} d\lambda_d(u) = C_d \int_{x}^{+\infty} \rho^{r+c+d-1}  \ e ^{-\vartheta \rho^{\kappa}} d\rho
 \end{equation*}
 with $$ C_d =  \prod_{j=1}^{n-2} \int_0^{\pi}\sin^{n-j-1}(\phi_j) d\phi_j \int_0^{2 \pi} d \phi_{d-1}   = \frac{\pi^{d/2}}{\Gamma(1+d/2)}.$$
Now, making the change of variable $u=\vartheta \rho^{\kappa}$ yields
$$ \bar{F}_r(x) = C_{d,\vartheta,\kappa} \int_{\vartheta x^{\kappa}}^{+\infty} u^{\frac{r+c+d}{\kappa} -1} \ e^{-u} du$$
for a positive real constant $C_{d,\vartheta,\kappa}$. One gets by a integration by parts,
$$ \bar{F}_r(x)  = C_{d,\vartheta,\kappa} \  x^{r+c+d-\kappa} \ e^{-\vartheta x^{\kappa}}\big(1 + o(1) \big).$$
Then for every $\theta>0$,
$$ e^{\theta x^{\kappa}} \bar{F}_r(x)  = C_{\vartheta,\kappa} \  x^{r+c+d-\kappa} \ e^{(\theta-\vartheta) x^{\kappa}}\big(1 + o(1) \big)  $$
so that if $\theta<\vartheta$ then 
$$ \limsup_{x \rightarrow +\infty} \ e^{\theta x^{\kappa}} \bar{F}_r(x)  <+\infty $$
and if  $\theta>\vartheta$ then 
$$ \limsup_{x \rightarrow +\infty} \ e^{\theta x^{\kappa}} \bar{F}_r(x)  = +\infty.$$
Which means that $\theta^{\star} = \vartheta.$ 
\end{comment}

\vskip 0.2cm
\noindent \textbf{(b)}  


We have  for every $r>0$, for every $x>1$,
$$  \bar{F}_{r}(x) = \int_{\{ \vert u \vert >x\}} \frac{(\log \vert u \vert)^{\beta}}{\vert u \vert^{c-r}} d\lambda_d(u) = V_d  \int_x^{+\infty} \frac{(\log \rho)^{\beta}}{ \rho ^{c'}} d\rho.$$
 with $c'= c-r-d+1$ ($c'>1$) and $V_d$ defined as previously.  Integrating  by parts    and multiplying by $x^{\zeta-r} $  yields

 $$ x^{\zeta-r} \bar{F}_{r}(x) =  \ \frac{K\ V_d}{c-(r+d)} \log(x) ^{\beta} x^{\zeta-c+d}(1+o(1)). $$
 Consequently,  $\zeta^{\star} = c-d.$ The statement   $(\ref{Ineq_Cor_pol_tail_density2})$ follows from Criterion $\ref{cor_princip_limsup}\ (b)$.
\vskip 0.2cm
\noindent $\textbf{(c)}$ Obvious from the forgoing. 

\begin{comment}
Now let us show that $\zeta^{\star}=c-d$. We have  for every $r>0$, for every $x>1$,
$$  \bar{F}_{r}(x) = \int_{\{ \vert u \vert >x\}} \frac{(\log \vert u \vert)^{\beta}}{\vert u \vert^{c+r}} d\lambda_d(u) .$$
Making a  change of variable with  hyperspherical coordinate yields 
 $$  \bar{F}_{r}(x) =  C_d  \int_x^{+\infty} \frac{(\log \rho)^{\beta}}{ \rho ^{c'}} d\rho $$
 with $c':= c-r-d+1$ ($c'>1$) and $C_d$ defined as previously. We make  the change of variable $u=(c'-1) \log(x)$  to have
 \begin{equation*}
 \int_x^{+\infty} \frac{(\log \rho)^{\beta}}{ \rho ^{c'}} d\rho  =   (c'-1)^{-\beta-1}\int_{(c'-1)\log(x)}^{+\infty} u^{\beta} \ e^{-u} d u.
 \end{equation*}
 Integrating  by parts    and multiplying by $x^{\zeta-r} $  yields

 $$ x^{\zeta-r} \bar{F}_{r}(x) = K \big[(c'-1) \log(x) \big]^{\beta+1} x^{\zeta-c+d}(1+o(1)) $$
  where $K_{\beta}$ is a positive real constant. Therefore, if $\zeta>c-d$ then $$ \limsup_{x \rightarrow +\infty} \ x^{\zeta-r} \bar{F}_{r}(x) = +\infty $$
and $\zeta<c-d$ then $$ \limsup_{x \rightarrow +\infty} \ x^{\zeta-r} \bar{F}_{r}(x) < +\infty .$$
It follows that $\zeta^{\star} = c-d.$
\end{comment}

 \end{proof}


 

 We now give some examples for usual distributions.
\vskip 0.2cm
\begin{exam}  \ {\rm  $\textbf{(a)}$  $\bullet$ If $X \sim \mathcal{N}(0;I_d)$,  we have $$f(x) = (2 \pi)^{-d/2} e^{-\frac{1}{2} \vert x \vert^2}.$$ It follows from the item $(a)$ of  the previous corollary (with $\vartheta=1/2, \kappa=2, c=0$) that  for every $r>0$, for every $d \geq 1$,
$$\limsup_{n \rightarrow +\infty}  \  \frac{\rho_n}{\sqrt{\log(n)}} \leq 2 \sqrt{2 \left(1+\frac{r}{d}\right)}.$$
In particular, when $r \geq1,d=1$ we have  $ \underset{n \rightarrow +\infty}{\limsup} \  \frac{\rho_n}{\sqrt{\log(n)}} \leq \sqrt{2(r+1)}.$
\vskip 0.2cm
\noindent  $\bullet$ For a double Gamma distribution in the real line where  
 $$f(x)=\frac{\lambda^a}{2 \Gamma(a)} \vert x \vert^{a-1}e^{-\lambda \vert x \vert}, \quad x \in \mathbb{R}; \  \lambda, a>0 $$
 or a Gamma distribution for which 
 $$f(x)=\frac{\lambda^a}{ \Gamma(a)} x^{a-1}e^{-\lambda x} \mbox{\bf{1}}_{\{x \geq 0 \}}, \quad  \lambda, a>0 $$
 we have  (from Corollary $\ref{cor_gen_density} \ (a) $ with $c=a-1, \vartheta = \lambda, \kappa=1$) for every $r \in (0,1)$,
$$ \limsup_{n \rightarrow +\infty} \frac{\rho_n}{ \log(n)}  \leq  \frac{2(r+1)}{\lambda}$$
and in case $r \geq 1$ we have $$ \underset{n \rightarrow +\infty}{\limsup} \ \frac{\rho_n}{\log(n)}  \leq \frac{r+1}{\lambda};$$  which coincides with the sharp rate given in $(\ref{asymp_exp}).$  
\vskip 0.2cm
\noindent  $\bullet$  When $X$ has a logistic distribution with density $f(x) = \frac{e^{-x}}{(1+e^{-x})^2}$ we have $$ f(x)  \ \sim \ e^{-x} \qquad \textrm{as} \ x \rightarrow +\infty.$$
Then, it follows from Corollary $\ref{cor_gen_density} \ (c) $  that $(\rho_n)_{n \geq 1}$ has the same upper asymptotic as the exponential distribution with parameter $\lambda=1.$
\vskip 0.2cm
\noindent  $\bullet$ As concern  the Weibull distribution with shape parameter $\kappa>0$ with  density function  $$ f(x)=\kappa x^{\kappa-1} e^{-x^{\kappa}}  \mbox{\bf{1}}_{\{x \geq 0 \}}$$ 
 it follows from Corollary $\ref{cor_gen_density}, (a)$ (with $\vartheta =1$) that for  $r \in (0,1)$,
$$   \limsup_{n \rightarrow +\infty} \ \frac{ \rho_n}{ \big(\log(n) \big)^{1/ \kappa}} \leq 2 (r+1)^{1/ \kappa}$$
and  if $r\geq 1$
$$   \limsup_{n \rightarrow +\infty} \ \frac{\rho_n}{ \big(\log(n) \big)^{1/ \kappa}} \leq (r+1)^{1/ \kappa} .$$
\vskip 0.2cm
\noindent    $\textbf{(b)}$ Suppose $X$ is a random variable having a Pareto distribution with index $\gamma>r$. The density function reads  $$ f(x) = \gamma x^{-(\gamma+1)} \mbox{\bf{1}}_{\{ x>1\}}. $$
Then we deduce from Corollary $\ref{cor_gen_density} \ (b)$ (with $d=1, c=\gamma+1, \beta=0$) that
 \begin{equation*} 
   \limsup_{n \rightarrow +\infty} \ \frac{\log \rho_n}{\log(n)} \leq \frac{r+1}{\gamma -r}.  
  \end{equation*}  
  }
  \end{exam}
\vskip 0.3cm 

\subsection{Lower estimate }
 In this section we investigate the asymptotic lower estimate of the maximal radius sequence  $(\rho_n)_{n \geq 1}$ induced by an $L^{r}$-optimal sequence of $n$-quantizers. First we introduce  the family of  the $(r,s)$-distributions  which will play  a crucial role  to obtain the optimal  lower estimate  for the rate of the maximal radius sequence. 
 
 Let $r>0, s>r$. Since the $L^r$-norm is increasing, it is clear that  for every $s \leq r$  any   $L^r$-optimal  sequence of quantizers  $(\alpha_n)_{n \geq 1}$   is \textbf{$L^s$-rate optimal} $i.e.$
\begin{equation}  \label{eq_rate_optim}
  \limsup_{n \rightarrow +\infty} \ n^{1/d}  \Vert X -  \widehat{X}^{\alpha_n} \Vert_s  < +\infty. \end{equation}
But if $s>r$ (and $X \in L^s(\mathbb{P})$) this asymptotic  rate optimality  usually fails . So is always the case when  $s>r+d$ and $X$ has a probability distribution $f$ satisfying $\lambda_d(f>0) = +\infty$, see $\cite{GLP}$. But it is established in $\cite{SAG}$ that some linear transformation of the $L^r$-optimal quantizers $(\alpha_n)$ makes possible to overcome the critical exponent $r+d$, that is, one can always construct an $L^s$-rate-optimal sequence of quantizers by a linear transformation of  the $L^r$-optimal sequence of quantizers $(\alpha_n)$.  However there is some distributions for which $(\ref{eq_rate_optim})$  holds for every $s \in [ r,r+d)$.  This leads to  the following definition:
\begin{defin} \label{def_(r,s)_dist}
Let $s,r>0,s>r$. A random vector  $X \in L^{s}(\mathbb{P})$  has an \textbf{$(r,s)$-distribution} if  any $L^r$-optimal sequence $(\alpha_n)_{n \geq 1}$ is $L^s$-rate optimal.
\end{defin}  
 For $\nu \in (0,d)$, sufficient conditions  such that  $X$ has an $(r,r+\nu)$-distribution  are provided in $\cite{GLP}$.  Let us  mention two criterions ensuring that a random vector $X$ has an $(r,r+\nu)$-distribution. The first one  deals with distributions with radial tails. 
 
\begin{crit}  \label{crit_radial}
$\textbf{(a)}$ Let $d \geq 1.$  If  $f = h(\vert  \cdot  \vert)$  on $B_{\vert \cdot \vert}(0,N)^{c}$ with $h : (R,+\infty) \rightarrow  \mathbb{R}_{+},\ R \in \mathbb{R}_{+}$, a decreasing function  and $\vert \cdot \vert$ any norm on $\mathbb{R}^{d}$.  If 
\begin{equation}  \label{asscor1}
 \int f(cx)^{-\frac{r+\nu}{r+d}} dP(x)  < +\infty  
 \end{equation}
for some  $c>1$.  Then  $X$ has an $(r,r+\nu)$-distribution. \\
$\textbf{(b)}$ If $d=1$ and if $ {\rm supp}(P) \subset [R_0, +\infty)$  for some $ \ R_0 \in \mathbb{R}$ and $f_{|(R_0^{'}, +\infty)}$ is  decreasing  for $R_0^{'} \geq R_0$. If further Assumption  $(\ref{asscor1})$  holds  for some  $c>1$.  Then  $X$ has an $(r,r+\nu)$-distribution.
 \end{crit} 
  The following criterion  works for distributions with non radial tails.
  \begin{crit}  \ Let $r>0, \ \nu \in (0,d), \ P=f\cdot \lambda_d$ and  $\int \vert x \vert^{r+ \eta} P(dx) < +\infty$ for some $\eta >0$. Assume  that  ${\rm supp}(P)$ is convex
  
and that $f$ satisfies   the local growth control assumption : there exists real numbers $\varepsilon \geq 0, \ \eta \in (0,1/2), \ M, C >0 $ such that
$$ \forall x,y \in \textrm{supp}(P), \ \vert x \vert \geq M, \ \vert y-x \vert \leq 2 \eta \vert x \vert \  \Longrightarrow \ f(y) \geq C f(x)^{1+\varepsilon}. $$
If   
\begin{equation} \label{Hypoth2}
\int f(x)^{-\frac{(r+\nu)(1+\varepsilon)}{r+d}} dP(x) < +\infty, 
\end{equation}
  then  $X$ has an $(r,r+\nu)$-distribution. 
  


\end{crit}
 Furthermore  a necessary condition for $X$ (with density $f$) to have  an  $(r,r+\nu)$-distribution is (see  $\cite{GLP}$) :
\begin{equation} \label{necess_suffic_rate_opt}
X \textrm{ has an $(r,r+\nu)$-distribution } \Longrightarrow \int f(x)^{-\frac{(r+\nu)}{d+r}} dP(x) < +\infty.
\end{equation}

 It  follows from $(\ref{asscor1})$ and  $(\ref{necess_suffic_rate_opt})$ that  the Gaussian distribution, the Weibull and the Gamma distributions have an $(r,r+\nu)$-distribution if and only if $\nu \in (0,d)$. The Pareto distribution with index $\gamma>r$ has an $(r,r+\nu)$-distribution if and only if  $ \nu \in (0,\frac{\gamma-r}{\gamma+1})$. 
 
 Now, suppose  that $X$ has an $(r,r+\nu)$-distribution for some $\nu \in (0,d)$ and set 
 $$ \nu^{\star}_{X}:= \sup\{\nu>0  \textrm{ s.t.  $X$ has an $(r,r+\nu)$-distribution}\} .$$ 
Note that 
$$  \{\nu>0  \textrm{ s.t.  $X$ has an $(r,r+\nu)$-distribution}\} = (0,\nu^{\star}_{X}) \textrm{ or } (0,\nu^{\star}_{X}]$$ and that $X \in L^{r+\nu}(\mathbb{P}), \ \forall \nu \in (0,\nu^{\star}_{X}).$  When  $$  \{\nu>0  \textrm{ s.t.  $X$ has an $(r,r+\nu)$-distribution}\} = \emptyset$$
we set $\nu^{\star}_{X}=0$. 

This index $\nu^{\star}_{X}$ will play a crucial role to determine the lower bound of the maximal radius sequence. Recall that if $X$ has a density $f$ satisfying $\lambda_d(f>0)=+\infty$ then a necessary condition for $X$ to have  $(r,r+\nu)$-distribution is that $\nu<d$. Which means that  $\nu^{\star}_{X} \leq d$. However, this inequality  may stand strictly as approved by the Pareto distribution with index $\gamma$ for which $\nu^{\star}_{X}=\frac{\gamma-r}{\gamma+1}<1.$ 
\vskip 0.3cm
 We  present below two different  approaches to get the  lower bound for the maximal radius sequence. The first one  involves the  generalized survival functions $\bar{F}_r$ like for upper bounds and  is  based on tail estimates. The second one is probably more original. It is based on  random quantization: its specificity is to provide a close connection between the sequence  $(\rho_n)_{n \geq 1}$  and the maximum of the norm of  an $i.i.d$ sequence of random variables with distributions $P$. 


\subsubsection{Distribution tail approach}
The main result of this section is the following theorem.

\begin{thm} \label{prop_gen_liminf}
  Let $r>0$ and let   $X$ be a  $\mathbb{R}^d$-valued  random variable with probability distribution  $P$.   Let  $(\alpha_n)_{n \geq 1}$ be an  $L^r(P)$-optimal sequence of $n$-quantizers. For every  $\nu  \in (0,\nu^{\star}_{X})$, the following statements hold:  
\begin{itemize}
\item[$\textbf{(a)}$] 
\begin{equation} \label{eqthm1.6}
   \limsup_{n \rightarrow +\infty} \  \sup_{c>0} \left( c^{r+\nu} \ n^{\frac{r+\nu}{d}} \bar{F}\big( \rho_n + c\big) \right)  < +\infty.
\end{equation}
\item[$\textbf{(b)}$]  
\begin{equation} \label{eqthm1.6.1}
   \limsup_{n \rightarrow +\infty}  \  \sup_{u >1} \left(  \big( 1- 1/u \big)^{r+\nu} \ n^{\frac{r+\nu}{d}} \bar{F}_r \big( u  \rho_n  \big)   \right)   < +\infty.
\end{equation}

\end{itemize}
\end{thm} 
\vskip 0.2cm
\begin{proof}[\textbf{Proof}] \ $\textbf{(a)}$
Let  $c>0$ and let $\nu \in (0,\nu^{\star}_{X})$. Then
\begin{equation*} \label{eq.bad_estim}
 \mathbb{E} \vert X - \widehat{X}^{\alpha_n} \vert ^{r+\nu}   \geq 
\mathbb{E} \left( \min_{a \in \alpha_n}\vert X - a \vert^{r+\nu}   \mbox{\bf{1}}_{ \{ \vert X \vert >  \rho_n + c \}}  \right).
\end{equation*}
On the events $\{\vert X \vert >  \rho_n + c \}$, we have: $\vert X \vert > \rho_n +c > \rho_n \geq  \vert a \vert$, $ \ \forall a \in \alpha_n$. Then 
\begin{eqnarray}  
\mathbb{E} \vert X - \widehat{X}^{\alpha_n} \vert ^{r+\nu}  & \geq &
\mathbb{E} \left( \min_{a \in \alpha_n}\vert X - a \vert^{r+\nu}   \mbox{\bf{1}}_{ \{ \vert X \vert >  \rho_n + c \}}  \right)  \label{estim_error} \nonumber  \\
& \geq & \mathbb{E} \left( \min_{a \in \alpha_n} \big(\vert X \vert - \vert a \vert \big)^{r+\nu}   \mbox{\bf{1}}_{ \{ \vert X \vert >  \rho_n +c \}}  \right) \nonumber  \\
& \geq & \mathbb{E} \left( \big(\vert X \vert - \rho_n \big)^{r+\nu}   \mbox{\bf{1}}_{ \{ \vert X \vert >  \rho_n + c \}}  \right)  \label{result_inter} \\
&  \geq & c^{r+\nu} \ \mathbb{P} \left( \{ \vert X \vert >  \rho_n + c \}  \right) \nonumber .
\end{eqnarray}
It follows that 
$$  \mathbb{E} \vert X - \widehat{X}^{\alpha_n} \vert ^{r+\nu}  \geq  \sup_{c>0} \left( c^{r+\nu} \ \mathbb{P} \big( \{ \vert X \vert >  \rho_n + c \} \big) \right).$$
Since $X$ has an $(r,r+\nu)$-distribution  we have   
$$\limsup_{n} \ n^{\frac{r+\nu}{d}} \Vert X - \widehat{X}^{\alpha_n} \Vert_{r+\nu}^{r+\nu} < +\infty.$$
Which completes the proof.
\vskip 0.2cm 


\noindent $\textbf{(b)}$ is proved  like  $(a)$. Inequality $(\ref{result_inter})$ becomes: for  every $u>1$, 
$$ \mathbb{E} \vert X - \widehat{X}^{\alpha_n} \vert ^{r+\nu}  \geq  \mathbb{E} \left( \big(\vert X \vert - \rho_n \big)^{r+\nu}   \mbox{\bf{1}}_{ \{ \vert X \vert >  u \rho_n  \}}  \right) \geq \mathbb{E} \left(\vert  X \vert^{r+\nu} \left(1- 1/u \right)^{r+\nu}  \mbox{\bf{1}}_{ \{ \vert X \vert > u  \rho_n \}}  \right)  . $$  
Then,
$$ \mathbb{E} \vert X - \widehat{X}^{\alpha_n} \vert ^{r+\nu}  \geq  \sup_{u>1}  \left[ \left(1- 1/u \right)^{r+\nu}  \mathbb{E} \left(\vert  X \vert^{r+\nu}   \mbox{\bf{1}}_{ \{ \vert X \vert > u  \rho_n \}}  \right)  \right].$$


Inequality $(\ref{eqthm1.6.1})$ follows by noticing that   $\  \underset{n}{\limsup} \ n^{\frac{r+\nu}{d}} \Vert X - \widehat{X}^{\alpha_n} \Vert_{r+\nu}^{r+\nu} < +\infty$.
\end{proof}

Like for the upper estimate, given the asymptotic inverse function $\phi$ of $-\log \bar{F}$  or  given  an asymptotic lower bound $\psi$  of $\phi$ satisfying some standard  hypotheses specified below, we provide the asymptotic lower estimate for the maximal radius sequence for distributions with exponential tails. 

For distributions with polynomial tails, we will rather look  for the asymptotic inverse function $\Phi_{r,\nu}$, $\nu \in (0,\nu^{\star}_{X})$ (if any) of   $ -\log \bar{F}_{r+\nu}(e^x)$ or some asymptotic lower bound of it  to provide a lower estimate of  $\log \rho_n$.  

\begin{prop} \label{thm_princip_liminf}
  Let $r>0$ and let   $X$ be an  $\mathbb{R}^d$-valued  random variable with distribution  $P$.  Suppose that $X$ has an unbounded support. Let  $(\alpha_n)_{n \geq 1}$ be an  $L^r(P)$-optimal sequence of $n$-quantizers.
 \vskip 0.2cm
\noindent $\textbf{(a)}$ If $\psi$ is a measurable nondecreasing function going to $+\infty$ as $x \rightarrow +\infty$, regularly varying  with index $\delta$ and satisfying 
\begin{equation}  \label{Hyp_liminf1}
 \psi(-\log \bar{F}(x)) \leq x + o(x),
 \end{equation}
 then
 \begin{equation}  \label{eq_thm_gen_liminf1}
  \liminf_{n \rightarrow +\infty}  \ \frac{\rho_n}{ \psi(\log(n)) }  \geq  \left( \frac{r +\nu^{\star}_{X}}{d} \right)^{\delta}.
 \end{equation} 
 If $-\log \bar{F}$  is regularly varying of index $1/ \delta$ then $(\ref{eq_thm_gen_liminf1})$  holds with $\psi= (-\log \bar{F})^{\leftarrow}$.
 \vskip 0.2cm
\noindent $\textbf{(b)}$ Let  $\nu \in (0,\nu^{\star}_{X})$. If  there is  a measurable nondecreasing function $\psi_{r,\nu}(x)$  going to $+\infty$ as $x \rightarrow +\infty$, regularly varying  with index $\delta$ and satisfying
\begin{equation}  \label{Hyp_liminf2}
\psi_{r,\nu}(-\log \bar{F}_{r+\nu}(e^x)) \leq  x + o(x),
\end{equation}
  then
\begin{equation}  \label{eq_thm_gen_liminf2}
  \liminf_{n \rightarrow +\infty}  \ \frac{\log \rho_n}{\psi_{r,\nu}(\log(n))} \geq  \left( \frac{r + \nu}{d} \right)^{\delta}.
\end{equation}
In particular if $-\log \bar{F}_{r+\nu}(e^{x})$  has regular variation with index  $1/ \delta$ then $(\ref{eq_thm_gen_liminf2})$  holds with $\psi_{r,\nu}(x)= (-\log \bar{F}_{r+\nu}(e^{x}))^{\leftarrow}$. 
\end{prop}
\vskip 0.2cm
Let us provide a few comments on this proposition. We have for every $r > 0$ and for every $x>0, \ \bar{F}_r(x)  > x^r \bar{F}(x)$. Then
 $$ -\log \bar{F}_r(x)  \leq -\log \bar{F}(x) -r \log(x).$$
According to the nondecreasing hypothesis on $\psi$ we have for every $x>1$  
$$  \psi(-\log \bar{F}_r(x)) \leq \psi( -\log \bar{F}(x) -r \log(x) ) \leq \psi(-\log \bar{F}(x))  $$
so that if  $(\ref{Hyp_liminf1})$ holds then  for every $r>0$, for every $\nu \in (0,\nu^{\star}_{X})$,
 \begin{equation*} 
   \psi(-\log \bar{F}_{r+\nu}(x)) \leq x + o(x) . 
   \end{equation*}
Reproducing  the proof  (given below) of Proposition $\ref{thm_princip_liminf}\ (a)$ by using $(\ref{eqthm1.6.1})$ instead of  $(\ref{eqthm1.6})$ shows that  $(\ref{eq_thm_gen_liminf1})$  still holds true. This means that  for distribution with exponential tails,  the function $\psi$ do not depend on $r$ and $\nu$ even if in Assumption $(\ref{Hyp_liminf1})$ we take the generalized survival function $\bar{F}_{r+\nu}$  in place of the regular survival function $\bar{F}$. However for distributions with polynomial tails like Pareto distribution  the function $\psi_{r,\nu}$  in $(\ref{Hyp_liminf2})$   may depend on $r$ and taking  the regular survival function $\bar{F}$ in place of  the generalized survival function $\bar{F}_{r+\nu}$  would make lose the dependance upon $r$ and consequently lead to a less accurate result. 

We next prove the proposition.

\vskip 0.2cm 
\begin{proof}[\textbf{Proof}] \ $\textbf{(a)}$  Assume $\nu^{\star}_{X}>0$ and let 
$ \nu \in (0,\nu^{\star}_{X})$.  It follows from  $(\ref{eqthm1.6.1})$  that for  large enough $n$,
$$- \log \bar{F}( \rho_n +c) \geq - \log(C_{\nu,c}) + \frac{r+\nu}{d} \log(n)$$ where $C_{\nu,c}$ is a positive real constant depending on the indexing parameters. It follows  from the fact that $\psi$ is nondecreasing and goes to $+\infty$ and from Assumption $(\ref{Hyp_liminf1})$  that
$$ \frac{\rho_n}{\psi(\log(n))} \geq \left( 1+ \frac{c}{\rho_n} + \frac{o(\rho_n)}{ \rho_n} \right)^{-1} \frac{\psi( \frac{r+\nu}{d} \log(n) - \log(C_{\nu,c})) }{\psi(\log(n))}.$$
Since $\psi$ is regularly varying with index $\delta$ we have 
$$ \liminf_{n \rightarrow +\infty} \frac{\rho_n}{\psi(\log(n))} \geq  \left( \frac{r+\nu}{d} \right)^{\delta} , \qquad \ \forall \nu \in (0,\nu^{\star}_{X}). $$
Letting  $\nu \rightarrow \nu^{\star}_{X}$  give the announced result. If $\nu^{\star}_{X}=0$, one follows the same proof with  $\nu = 0.$
\vskip 0.2cm
\noindent $\textbf{(b)}$  This is proved like the statement $\textbf{(b)}$ in Proposition $\ref{thm_princip_limsup}$ by considering $\bar{F}_{r+\nu}$ instead of $\bar{F}_r$, for $\nu \in (0,\nu^{\star}_{X})$.
\end{proof}


\vskip 0.2cm


The next criterion is the lower limit counterpart  of Criterion $\ref{cor_princip_limsup}$.
\begin{crit}  \label{cor_princip_liminf}
$\textbf{(a)}$ \ Let $X$ be an  $\mathbb{R}^d$-valued  random variable with unbounded support  and suppose that
\begin{equation} \label{def_expon_tail}
\theta_{\star} = \inf \big\{\theta>0, \  \liminf_{x \rightarrow +\infty} \ e^{\theta x^{\kappa}} \mathbb{P}(\vert X \vert >x) >0 \big\}  \in (0, +\infty]. 
\end{equation}
Then 
\begin{equation}
\liminf_{n \rightarrow +\infty} \ \frac{\rho_n}{\big(\log(n) \big)^{1/ \kappa}}  \geq \left(\frac{r + \nu^{\star}_{X}}{d \ \theta_{\star}} \right)^{1/ \kappa}.
\end{equation}
\vskip 0.2cm
\noindent $\textbf{(b)}$  Let  $X$ be a random variable with unbounded support such that  $\nu^{\star}_{X}>0$. Set 
\begin{equation}
\zeta_{\star} = \inf \big\{ \zeta>0, \ \forall \nu \in (0,\nu^{\star}_{X}), \  \liminf_{x \rightarrow +\infty} \  x^{\zeta -r -\nu} \bar{F}_{r+\nu}(x) >0 \big\}   \in [r+\nu^{\star}_{X}, +\infty].
\end{equation}
Then 
\begin{equation}
\liminf_{n \rightarrow +\infty} \ \frac{\log \rho_n}{\log(n)} \geq \frac{1}{\zeta_{\star}-r-\nu^{\star}_{X}}\ \frac{r+\nu^{\star}_{X}}{d} .
\end{equation}
\end{crit}
\vskip 0.2cm
\begin{proof}[\textbf{Proof }] \ $\textbf{(a)}$\ Let  $\theta \in  (\theta_{\star},+\infty)$.  Then 
$$ \bar{F}(x)  \geq C \ e^{-\theta x^{\kappa}} $$
for large enough $x$ and for a  positive real  constant $C$.  Therefore 
$$ -\log \bar{F}(x) \leq \theta x^{\kappa} (1 - \frac{\log(C)}{x^{\kappa}}) $$
so that by setting $\psi_{\theta}(x) = (x/\theta)^{1/ \kappa}$  we have 
$$  \psi_{\theta}(-\log \bar{F}(x)) \leq x + o(x). $$
It follows from Proposition $\ref{thm_princip_liminf}\ (a)$ that 
$$ \liminf_{n \rightarrow +\infty} \ \frac{\rho_n}{\big(\log(n) \big)^{1/ \kappa}}  \geq \frac{1}{\theta^{1/ \kappa}}  \left(\frac{r + \nu^{\star}_{X}}{d} \right)^{1/ \kappa}.$$
We let $\theta$ go to $\theta_{\star}$ to get the announced result.
\vskip 0.2cm
\noindent $\textbf{(b)}$  Let $\zeta \in (0, \zeta_{\star})$. We have, for every $\nu \in (0,\nu^{\star}_{X}),$ 
 $$ \bar{F}_{r+\nu}(x) \geq C_{\nu} \ x^{-\zeta+r+\nu}$$
 for large enough $x$ and for a  positive real  constant $C_{\nu}$. Then, by setting $\psi_{r,\nu}(x) = \frac{x}{\zeta-r-\nu}$ we get $\psi_{r,\nu}(-\log \bar{F}_{r+\nu}(x))  \leq \log(x) +o(\log(x))$. It follows from Proposition $\ref{thm_princip_liminf}\ (b)$ that 
 $$ \liminf_{n \rightarrow +\infty} \ \frac{\log \rho_n}{\log(n)}   \geq \frac{1}{\zeta-r-\nu} \frac{r+\nu}{d}.$$
 The right hand side of this last inequality is increasing on $(0,\nu^{\star}_{X})$ (as a function of $\nu$) and on $(0,\zeta_{\star})$ (as a function of $\zeta$) so that by letting $\nu$ and $\zeta$ go respectively to $\nu^{\star}_{X}$ and $\zeta_{\star}$ we get 
 $$ \liminf_{n \rightarrow +\infty} \ \frac{\log \rho_n}{\log(n)}   \geq \frac{1}{\zeta_{\star}-r-\nu^{\star}_{X}} \frac{r+\nu^{\star}_{X}}{d}.$$
\end{proof}
\vskip 0.3cm
\begin{cor}  \label{cor_explic_dens_liminf}
$\textbf{(a)}$ \ If  the density function of $X$ reads 
\begin{equation}  \label{gen_density_exp_liminf}
f(x) \ \varpropto \  \vert x \vert^{c} \ e^{- \vartheta  \vert x \vert^{\kappa}} \ x \in \mathbb{R}^d; \ \vartheta, \kappa>0;\ c>-d
\end{equation}
  then $  \nu^{\star}_{X} = d  \  \textrm{ and } \  \theta^{\star} = \theta_{\star} = \vartheta.$ In this case  we have for every $r>0$, for every $d \geq 1$,
\begin{equation}  \label{eq_liminf_limsup_exp}
\frac{1}{\vartheta^{1/ \kappa}}  \left(1+\frac{r }{d} \right)^{1/ \kappa} \leq \liminf_{n \rightarrow +\infty} \ \frac{\rho_n}{\big(\log(n) \big)^{1/ \kappa}}  \leq \limsup_{n \rightarrow +\infty} \ \frac{\rho_n}{\big(\log(n) \big)^{1/ \kappa}} \leq \frac{2}{\vartheta^{1/ \kappa}}  \left(1+\frac{r}{d} \right)^{1/ \kappa} .
\end{equation}
  When $d=1$ and $ r \geq 1$  we have  
 \begin{equation}
\lim_{n \rightarrow +\infty} \ \frac{\rho_n}{\big(\log(n) \big)^{1/ \kappa}} = \left(\frac{r+1}{\vartheta} \right)^{1/ \kappa}.
\end{equation}
\vskip 0.2cm
\noindent  $\textbf{(b)} \ $If  $X$ has a density $f$ satisfying
 \begin{equation}   \label{gen_density_pol_liminf}
 f(x) \ \varpropto \  \frac{(\log \vert x \vert)^{\beta}}{\vert x \vert^c} \ \mbox{\bf{1}}_{\{ \vert x \vert  >1 \}}  \qquad  x \in \mathbb{R}^d,   \ \beta \in \mathbb{R},\ c> r+d
 \end{equation}
then  $ \nu^{\star}_{X} = d\left(1 -\frac{r+d}{c} \right) \in  (0,d) \  \textrm{ and }  \ \zeta_{\star}=\zeta^{\star}=c-d$. Furthermore  we have for every $r>0$ and  every $d \geq 1$,
\begin{equation}\label{eq_liminf_limsup_pol}
 \lim_{n \rightarrow +\infty} \ \frac{\log \rho_n}{\log(n)}  = \frac{1}{c-r-d} \frac{r+d}{d}.
\end{equation}
\vskip 0.2cm
\noindent  $\textbf{(c)}$ \  The claim $(\ref{eq_liminf_limsup_exp})$ (resp $(\ref{eq_liminf_limsup_pol})$) holds if the density of $X$ is simply  equivalent to the specified density in $(\ref{gen_density_exp_liminf})$ (resp in $(\ref{gen_density_pol_liminf})$).
\end{cor}
\vskip 0.2cm
\begin{proof}[\textbf{Proof }]  $\textbf{(a)}$ \ It is obvious  from  Criterion $\ref{crit_radial}$ and $(\ref{necess_suffic_rate_opt})$  that $\nu^{\star}_{X} = d.$  Let  $K$ be  the normalizing positive real constant in  $(\ref{gen_density_exp_liminf})$. We have for every $x>0$,
\begin{eqnarray*}
\mathbb{P}(\vert X \vert >x) & = & K \int_{\{ \vert u \vert >x \}} \vert  u \vert^c\ e ^{-\vartheta \vert u \vert^{\kappa}} d\lambda_d(u) \\
&  = & K_d  \int_{x}^{+\infty} \rho^{c+d-1}  \ e ^{-\vartheta \rho^{\kappa}} d\rho \\
& = & K_d \  \frac{x^{c+d-\kappa}}{\vartheta} e^{-\vartheta x^{\kappa}} (1+o(1))
\end{eqnarray*}
where we used an integration by parts and usual integral comparison criterions. Consequently,


 


 if $ \theta > \vartheta $  then  $$ \liminf_{x \rightarrow +\infty}\ e^{\theta x^{\kappa}} \mathbb{P}(\vert X \vert >x) = +\infty$$ and if  $ \theta < \vartheta $  then  $$ \liminf_{x \rightarrow +\infty}\ e^{\theta x^{\kappa}} \mathbb{P}(\vert X \vert >x)  = 0 .$$ 
Which means that  $\theta_{\star}=\vartheta$ and the statement  $(\ref{eq_liminf_limsup_exp})$ follows from Criterion $\ref{cor_princip_limsup}\ (a)$ and Criterion $\ref{cor_princip_liminf}\ (a)$.
\vskip 0.2cm
\noindent $\textbf{(b)}$ \ Let $\bar{c}>1$.  We  have 
$$ \int_{\{f>0 \}} f(\bar{c}x)^{-\frac{r+\nu}{r+d}} f(x) d\lambda_d(x)  =( \bar{c})^{-\frac{r+\nu}{r+d}} \int_{\{ \vert x \vert >1 \}}  \frac{(\log \vert u \vert)^{\beta'}}{\vert u \vert^{c'}} d\lambda_d(u) = K_{d,\bar{c}}  \int_{1}^{+\infty}  \frac{(\log \rho )^{\beta'}}{\rho^{c'-d+1}} d\rho$$ 
with $c'=c(1-\frac{r+\nu}{r+d})$. $K_{d,\bar{c}}$ is some positive real constant and $\beta' \in \mathbb{R}$. We deduce that if  $$c'>d \ \Longleftrightarrow \  \nu <d \left( 1- \frac{r+d}{c}\right) \ \textrm{ then } \ \int_{\{f>0 \}} f(\bar{c}x)^{-\frac{r+\nu}{r+d}} f(x) d\lambda_d(x)   < +\infty $$ 
and if $$c'<d \ \Longleftrightarrow \  \nu >d \left( 1- \frac{r+d}{c}\right) \ \textrm{ then } \ \int_{\{f>0 \}} f(x)^{-\frac{r+\nu}{r+d}} f(x) d\lambda_d(x)   = +\infty $$
 so that (from Criterion $\ref{crit_radial}$ and Statement $(\ref{necess_suffic_rate_opt})$) \ $\nu^{\star}_{X}=d \left( 1- \frac{r+d}{c}\right) .$ 

Let us show that $\zeta_{\star}=c-d$. For every $r>0$, for every $\nu \in (0, \nu^{\star}_{X})$, for every $x>1$, integrating by parts and using integral comparison criterions yield 
\begin{eqnarray*} 
 \bar{F}_{r+\nu}(x) = \mathbb{E} \left( \vert X \vert^{r+\nu} \mbox{\bf{1}}_{\{ \vert X \vert >x\}} \right) &  =  & \int_{\{ \vert u \vert >x\}} \frac{(\log \vert u \vert)^{\beta}}{\vert u \vert^{c-r-\nu}} d\lambda_d(u) \\
 & = & V_d  \int_x^{+\infty} \frac{(\log \rho)^{\beta}}{ \rho^{c'}} d\rho \\
 & = & V_d \ \frac{x^{-c'+1}}{c'-1} \log(x)^{\beta}(1+o(1))
 \end{eqnarray*}


 with $c':= c-r-\nu-d+1 >1$. It follows that 
 
 


 
 $$ x^{\zeta-r-\nu} \bar{F}_{r+\nu}(x) = \frac{V_d}{c'-1}  \log(x)^{\beta} x^{\zeta-c+d}(1+o(1)) $$
so that  for every $r>0$ and for every $\nu \in (0,\nu^{\star}_{X})$, if  $\zeta >c-d$ then
$$\liminf_{x \rightarrow +\infty} x^{\zeta-r-\nu} \bar{F}_{r+\nu}(x) =+\infty  $$ 
and if   $\zeta <c-d$, 
$$\liminf_{x \rightarrow +\infty} x^{\zeta-r-\nu} \bar{F}_{r+\nu}(x) =0 . $$ 
Hence  $\zeta_{\star}=c-d$. It follows from Criterion $\ref{cor_princip_limsup}\ (b)$ and Criterion $\ref{cor_princip_liminf}\ (b)$ that 
$$ \frac{1}{\zeta_{\star}-r-\nu^{\star}_{X}} \frac{r+\nu^{\star}_{X}}{d} \leq \liminf_{n \rightarrow +\infty}  \ \frac{\log \rho_n}{ \log(n)} \leq \limsup_{n \rightarrow +\infty}  \ \frac{\log \rho_n}{ \log(n)}  \leq \frac{1}{c-r-d} \frac{r+d}{d}.$$
Now \big(recall that $\zeta_{\star} = c-d$ and $\nu^{\star}_{X} = d\left(1 -\frac{r+d}{c} \right)$ \big),
$$ \frac{1}{\zeta_{\star}-r-\nu^{\star}_{X}} \frac{r+\nu^{\star}_{X}}{d} = \frac{1}{c-r-d} \frac{r+d}{d}$$
so that 
$$ \lim_{n \rightarrow +\infty} \ \frac{\log \rho_n}{\log(n)}  = \frac{1}{c-r-d} \frac{r+d}{d}. $$
\vskip 0.2cm 
\noindent  $\textbf{(c)}$ Obvious from what forgoes. 
\end{proof}


We deal now with examples.
\vskip 0.2cm
\begin{exam} \ { \rm  $\textbf{(1)}$  It follows from Corollary $\ref{cor_explic_dens_liminf} \ (a)$ that  
\vskip 0.2cm
\noindent  $\bullet$   When $X \sim \mathcal{N}(0,I_d)$,  for every $r>0$, for every $d \geq 1$,
\begin{equation*}
\sqrt{\frac{2(r+d)}{d}} \leq \liminf_{n \rightarrow +\infty}  \frac{\rho_n}{\sqrt{\log(n)} } \leq \limsup_{n \rightarrow +\infty}  \frac{\rho_n}{\sqrt{ \log(n)} } \leq 2 \sqrt{\frac{2(r+d)}{d}}. 
 \end{equation*}
 In  case  $d=1$ and $r\geq 1$ we have  $$ \underset{n \rightarrow +\infty}{\lim} \frac{\rho_n}{\sqrt{\log(n)}} = \sqrt{2(r+1)} .$$

 \vskip 0.2cm
\noindent   $\bullet$ If $X \sim \Gamma(a,\lambda), a>0, \lambda>0$  or if $X$ has a double gamma distribution we have for every  $r\geq 1$, 
$$\  \underset{n \rightarrow +\infty}{\lim} \frac{\rho_n}{\log(n)} = \frac{r+1}{\lambda} $$ 
(which coincides to the exact rate given in $(\ref{asymp_exp})$ for the exponential distribution) and for every $r \in (0,1)$,
 \begin{equation*}
\frac{r+1}{\lambda}  \leq  \liminf_{n \rightarrow +\infty} \ \frac{\rho_n}{\log(n)} \leq \limsup_{n \rightarrow +\infty} \ \frac{\rho_n}{\log(n)} \leq  \frac{2(r+1)}{\lambda}  .
 \end{equation*}
 
 \vskip 0.2cm
\noindent   $\bullet$ As concern the logistic distribution, the maximal radius sequence has the same asymptotic as the exponential distribution with parameter $\lambda=1$ following  
 Corollary $\ref{cor_explic_dens_liminf} \ (c)$.
\vskip 0.2cm
\noindent   $\bullet$  For a  Weibull distribution with shape parameter $\kappa>0$ we have   for every $r\geq 1$, $$ \ \underset{n \rightarrow +\infty}{\lim} \frac{\rho_n}{\big( \log(n) \big)^{1/ \kappa}} =(r+1)^{1/ \kappa}.$$
For $r \in (0,1)$, one has
\begin{equation*}
  (r+1)^{1/ \kappa} \leq \liminf_{n \rightarrow +\infty} \frac{\rho_n}{\big( \log(n) \big)^{1/ \kappa}} \leq \limsup_{n \rightarrow +\infty} \frac{\rho_n}{\big( \log(n) \big)^{1/ \kappa}}  \leq 2 (r+1)^{1/ \kappa}.
  \end{equation*}
\vskip 0.2cm
\noindent    $\textbf{(2)}$   Suppose $X$ is a random variable having a Pareto distribution with index $\gamma>r$ where the density reads $f(x) = \gamma x^{-(\gamma+1)} \mbox{\bf{1}}_{\{ x>1\}}$. It follows from Corollary $\ref{cor_explic_dens_liminf}\ (b)$ (with $c=\gamma+1$, $d=1$)
\begin{equation}
   \lim_{n \rightarrow +\infty} \ \frac{\log \rho_n}{\log(n)} =  \frac{r+1}{\gamma -r} .  
  \end{equation}
 We retrieve  of course the sharp rate given in $(\ref{asymp_Par})$. }
 \end{exam} 


\subsubsection{An alternative approach by random quantization approach}
Let $X \sim P$. Random quantization  is another tool to compute the lower estimate of the maximal radius sequence. It makes  a connection between $\rho_n$ and the maximum of an $i.i.d$ sequence of random variables with distributions $P$.
\begin{thm} \label{theoreme1}   Let $r>0$ and let $ X $ be a random variable taking values in $\mathbb{R}^d$ with probability distribution $P $ with $P_a \not=0$. Assume  $(\alpha_n)_{n \geq 1}$ is an  $L^r(P)$-optimal sequence of $n$-quantizers.  Let  $( X_k)_{k \geq 1}$ be an  $\ i.i.d$ sequence of  $\mathbb{R}^d$-valued  random variables  with probability distribution  $P$. For every  $\nu \in (0,\nu^{\star}_{X})$,
\begin{equation}  \label{first_low_estim}
 \liminf_{n \rightarrow +\infty} \big( \rho_n  -  \mathbb{E} \big( \max_{k \leq [n^{(r+\nu)/d}]} \vert X_k \vert \big) \big) \geq  -  \ \textrm{C}_{\nu}
\end{equation}
where $\textrm{C}_{\nu}$ is a positive real constant. 
\end{thm}
\vskip 0.2cm
\begin{proof}[\textbf{Proof}] \ Let  $\nu \in (0,\nu^{\star}_{X})$  and let  $\widehat{X}_{k}^{\alpha_n} = \sum_{a \in \alpha_n} a \mbox{\bf{1}}_{\{ X_k \in C_a (\alpha_n) \}}$.  We have,
\begin{eqnarray*}
\rho_n  & \geq  &   \max_{k \leq m} \vert \widehat{X}_{k}^{\alpha_n}  \vert  \\
 & \geq & \sum_{k=1}^{m} \max_{l \leq m} \vert \widehat{X}_{l}^{\alpha_n} \vert \mbox{\bf{1}}_{\{ \vert X_k \vert> \max_{i \not= k} \vert X_i \vert\}} \\ 
& \geq &  \sum_{k=1}^{m}  \vert \widehat{X}_{k}^{\alpha_n} \vert \mbox{\bf{1}}_{\{ \vert X_k \vert> \max_{i \not= k} \vert X_i \vert\}} \\
& \geq & \sum_{k=1}^{m}  \big(\vert X_k \vert - \vert X_k - \widehat{X}_{k}^{\alpha_n} \vert \big) \mbox{\bf{1}}_{\{ \vert X_k \vert> \max_{i \not= k} \vert X_i \vert\}} \\
& \geq & \mathbb{E} \max_{k \leq m} \vert X_k \vert - \sum_{k=1}^{m} \mathbb{E} \left(\vert X_k - \widehat{X}_{k}^{\alpha_n} \vert \mbox{\bf{1}}_{\{ \vert X_k \vert> \max_{i \not= k} \vert X_i \vert\}} \right).
\end{eqnarray*}
Furthermore, $$ \forall k \geq 1,  \qquad  \vert X_k - \widehat{X}_k^{\alpha_n} \vert \mbox{\bf{1}}_{\{ \vert X_k \vert> \max_{i \not= k} \vert X_i \vert\}} \stackrel{\mathscr{L}}{=}  \vert X_1 - \widehat{X}_1^{\alpha_n} \vert \mbox{\bf{1}}_{\{ \vert X_1 \vert> \max_{i \not= 1} \vert X_i \vert\}}.$$
Hence, 
\begin{eqnarray*}
 \rho_n & \geq  &  \mathbb{E} \max_{k \leq m} \vert X_k \vert - m \ \mathbb{E} \left(\vert X_1 - \widehat{X}_{1}^{\alpha_n} \vert \mbox{\bf{1}}_{\{ \vert X_1 \vert > \max_{i \not= 1} \vert X_i \vert\}}\right) \\
 & \geq &   \mathbb{E} \max_{k \leq m} \vert X_k \vert - m \ \Vert X_1 - \widehat{X}^{\alpha_n}_1  \Vert_{r+\nu} \left(\mathbb{P} \big(\vert X_1 \vert > \max_{i \not= 1}\vert X_i \vert \big) \right)^{1-1/(r+\nu)}. 
 \end{eqnarray*}
 Since the events
$$ \{ \vert X_k \vert> \max_{i \not= k} \vert X_i \vert \},\  k=1,\cdots,m  $$
are  pairwise disjoint  with the same probability   we   have 
$$\mathbb{P} \big(\vert X_1 \vert > \max_{i \not= 1}\vert X_i \vert \big)\leq \frac{1}{m}.$$
Finally,
$$ \rho_n \geq  \mathbb{E} \max_{k \leq m} \vert X_k \vert - m^{\frac{1}{r+\nu}} \ \Vert X_1 - \widehat{X}_1^{\alpha_n}  \Vert_{r+\nu}. $$
It follows, by setting $\ m = [n^{(r+\nu)/d}]  $,  that 
$$ \liminf_{n \rightarrow +\infty} \big( \rho_n  -  \mathbb{E} \big( \max_{k \leq [n^{(r+\nu)/d}]} \vert X_k \vert \big) \big) \geq - \limsup_{n \rightarrow +\infty} n^{\frac{1}{d}}\Vert X_1 - \widehat{X}_1^{\alpha_n}  \Vert_{r+\nu}. $$
However, since $X$ has an $(r,r+\nu)$-distribution, the upper limit on the right hand side of the equation is finite.
\end{proof}

\begin{exam} { \rm (Exponential distribution)  Let $r>0$ and  let  $X$ be an exponentially distributed random variable with parameter $\lambda>0$. If $(\alpha_n)_{n \geq 1}$ is  an  $L^r$-optimal sequence of $n$-quantizers for $X$ then Theorem $\ref{theoreme1}$ implies 
 \begin{equation} \label{eq_exp_comment}
 \liminf_{ n \rightarrow +\infty} \ \frac{\rho_n}{ \log(n)}  \geq \frac{r+1}{\lambda}.
 \end{equation}
which is the sharp rate given by   $(\ref{asymp_exp})$.}
\end{exam}

Indeed, let $\nu \in (0,\nu^{\star}_{X})$ and let $(X_i)_{\{i=1,\cdots,[n^{r+\nu}] \}}$,  be an $i.i.d$ exponentially distributed  sequence of  random variables with parameter $\lambda$. We have for every   $u \geq 0$,
$$ \mathbb{P}(\max_{i \leq [n^{r+\nu}]} X_i \geq u) = 1 - \mathbb{P}(X_1 \leq u)^{[n^{r+\nu}]} = 1 - F(u)^{[n^{r+\nu}]}, $$
where  $F$  is the  distribution function of $X$ (we will denote by  $f$ its density function). Then
\begin{eqnarray*}
\mathbb{E} \big( \max_{i \leq [n^{r+\nu}]} X_i \big) 
& = & \int_0^{+ \infty} \mathbb{P}(\max_{i \leq [n^{r+\nu}]} X_i \geq u) du = \int_0^{+\infty} (1 - (1-e^{- \lambda u})^{[n^{r+\nu}]}) du \\
& = &  \int_0^{+ \infty} \big(\sum_{i=0}^{[n^{r+\nu}]-1} F(u)^i \big) e^{- \lambda u} du \\
& = & \int_0^{+\infty} \left(1 + F(u) + \cdots + F(u)^{[n^{r+\nu}]-1} \right) \frac{f(u)}{\lambda} du \\
& = & \frac{1}{\lambda} (1 + \frac{1}{2} + \cdots + \frac{1}{[n^{r+\nu}]}) \\
& \geq  & \frac{1}{\lambda}  \log(1+ [n^{r+\nu}]) \ \geq \  \frac{r+\nu}{\lambda} \log(n). 
\end{eqnarray*}
Consequently, it follows from the super-additivity of the liminf that  for every $\nu \in (0,1)$,
\begin{eqnarray*}
\liminf_{ n \rightarrow +\infty} \frac{\rho_n}{\log(n)} & \geq & \liminf_{ n \rightarrow +\infty} \frac{ \rho_n  -  \mathbb{E} \big( \max_{i \leq [n^{r+\nu}] } \big)}{\log(n)} +  \liminf_{ n \rightarrow +\infty} \frac{\mathbb{E} \big( \max_{i \leq [n^{r+\nu}]} \big)}{ \log(n)} \\
& \geq &  \frac{r+ \nu}{\lambda}. 
\end{eqnarray*}
The result follows  by letting $\nu$ go to $\nu^{\star}_{X}=1$.
\vskip 0.4cm
\begin{exam} {\rm (Pareto distribution)  Let $X$ be a random variable having a  Pareto  distribution with  index $\gamma>0 $. If $(\alpha_n)_{n \geq 1}$ is an  asymptotically $L^r$-optimal sequence of $n$-quantizers for $X$, $r$ is such that  $ \gamma >  r $, then   Theorem $\ref{theoreme1}$ yields
$$\liminf_{n \rightarrow +\infty} \ \frac{\log \rho_n}{ \log(n)} \geq \frac{r+1}{\gamma +1}.$$
Which is not the sharp rate given by   $(\ref{asymp_Par})$. }
\end{exam} 

Notice  that if $\gamma>r$ then $X \in L^{r+\eta}( \mathbb{P})$ for $ \eta \in \big(0,\gamma - r \big)$. Now, to prove this result,  let  $\nu \in (0, \nu^{\star}_{X})$ and let $(X_i)_{\{i=1,\cdots,[n^{r+\nu}] \}}$ be  an $i.i.d$  sequence of random variables with Pareto distribution with index $\gamma$. We have 
$$  \forall m \geq 1, \forall u \geq 1, \quad  \mathbb{P} (\max_{i \leq m} X_i \leq u ) = (1-u^{- \gamma})^{m} .$$ 
Then, the density function of $\max_{1 \leq i \leq m} X_i$  is $ \  m  \gamma u^{-(\gamma +1)}(1 - u^{-\gamma})^{m-1}.$

Hence  
\begin{eqnarray*}
  \mathbb{E} \big( \max_{1 \leq i \leq m } X_i \big) & = &  m \gamma \int_1^{+ \infty} x^{- \gamma } (1-x^{-\gamma})^{m-1} dx  \\
& = &  m \int_0^1 u^{-1/  \gamma} (1-u)^{m-1} du  \qquad \ (u = x^{-\gamma}) \\
  & = & m \  B(1- \frac{1}{ \gamma},m)  \\
  & = &  \frac{\Gamma(1- \frac{1}{ \gamma})\Gamma(m+1)}{ \Gamma(m+1- \frac{1}{\gamma})}\\
 & \sim & \Gamma(1- \frac{1}{ \gamma}) \  m^{\frac{1}{\gamma}}  \qquad \textrm{as }   m \rightarrow +\infty 
  \end{eqnarray*}
where we used Stirling's formula for the last statement.
 We finally  set $m = [n^{r+\nu}] $ to get  
 $$ \mathbb{E} \big( \max_{1 \leq i \leq [n^{r+\nu}] } X_i \big)  \sim  \Gamma(1- \frac{1}{ \gamma}) \ n^{\frac{r+\nu}{\gamma}} . $$
 It follows from  $(\ref{first_low_estim})$ that for every  $\varepsilon \in (0,1)$,
 $$  \rho_n - (1-\varepsilon)  \Gamma(1- \frac{1}{ \gamma}) \ n^{\frac{r+\nu}{\gamma}}  \geq -\varepsilon - \textrm{C}_{\nu}.  $$
 Dividing both side of the inequality  by $n^{\frac{r+\nu}{\gamma}}$  and taking the logarithm yields
 $$ \log \rho_n - \frac{r+\nu}{\gamma} \log(n)  \geq \log \left((1-\varepsilon)  \Gamma(1- \frac{1}{ \gamma}) - (\varepsilon + \textrm{C}_{\nu}) n^{-\frac{r+\nu}{\gamma}} \right).$$
 Consequently 
 $$ \liminf_{n \rightarrow +\infty} \ \frac{\log \rho_n}{\log(n)} \geq \frac{r+\nu}{\gamma}  $$ for every $\nu \in (0,\nu^{\star}_{X})$.  The announced result follows  by letting $\nu$ go to $\nu^{\star}_{X}= \frac{\gamma-r}{\gamma+1}$.
 \vskip 0.2cm
\noindent  \textbf{Comment.} \  Let $\phi$ be the inverse (if any) function of $-\log \bar{F}$.  It can be noticed that in both previous examples we have 
   \begin{equation} \label{eq_par_comment}
    \mathbb{E} \big(\max_{k \leq [n^{r+\nu^{\star}_{X}}]} \vert X_k \vert \big) \sim \phi ((r+\nu^{\star}_{X}) \log(n)) \qquad \textrm{ as } n \rightarrow +\infty 
    \end{equation}
 which, for distributions  with exponential tail leads  to the same asymptotic lower bound for the sequence $(\rho_n)_{n\geq 1}$  as in $(\ref{eq_thm_gen_liminf1})$.  For Pareto distribution, using the approximation $(\ref{eq_par_comment})$ to compute the asymptotic lower estimate of the maximal radius sequence  make us loose  the $"-r"$ term in the exact asymptotic. To recover this reminding term we have simply to consider  the inverse function of $-\log \bar{F}_{r+\nu^{\star}_{X}}$ (as done in the previous section) instead of $-\log \bar{F}$, and, the random quantization approach clearly  does not allow us to do so. 
 \vskip 0.2cm
 \subsubsection{A conjecture about the sharp rate}
 The previous  results related to  distributions with exponential tails  strongly suggest the following conjecture: suppose $X$ is a distribution with exponential tail in the sense of claim $(\ref{def_expon_tail})$. Then  for every $r>0$, for every $d \geq 1$,
$$ \lim_{n \rightarrow +\infty} \ \frac{\rho_n}{\big(\log(n) \big)^{1/ \kappa}} =  \left(\frac{r+d}{d \ \theta^{\star}} \right)^{1/ \kappa} .$$

This conjecture is proved for $d=1$ and $r\geq 1$. To be satisfied for hight dimension we need to proof that the geometric statement  $(\ref{equaconjecture})$ of Lemma $\ref{lem1_princip_limsup}\ (a)$ holds true with $"1+\varepsilon"$ instead of $"2+\varepsilon"$ like in $1$-dimension. Although this inequality looks quite intuitive in any  dimension its proof seems out of reach when $d \geq 2$. 
 \vskip 0.2cm
 \subsubsection{Numerical experiments }
We now attempt to focus on numerical experiment of the maximal radius sequence $(\rho_n)_{n \geq 1}$ for the quadratic optimal quantizers of the Gaussian, the Weibull  and the exponential distributions. A whole package of quadratic  optimal $n$-quantizers of the $\mathcal{N}(0,I_d)$ distributions are available in the website
\begin{center}
  {\tt  www.quantize.maths-fi.com} 
 \end{center}
\noindent for $d \in \{1, \cdots,10 \}$ and $n \in \{1, \cdots, 5000 \}.$  When $d=1,$ these $L^2$-optimal grids are obtained by the Newton method, see  e.g. $\cite{PP}$ for details. For the exponential distribution the quadratic optimal quantizers are computed  by using   the semi-closed formulae given in Proposition $\ref{prop_semi_closed}$. 

As concern the Weibull distribution with shape parameter $\kappa=2$, we compute the quadratic optimal quantizers  up to $3000$ using the Lloyd's I algorithm  described in $\cite{PP}$ (see  $\cite{GG}$ for a more itemized  description of the algorithm).

In these three cases we depicted the ratio between $\rho_n$ and the expected  asymptotic optimal rate. For  the exponential distribution we  represent  the graph of $ \frac{\rho_n}{ 3 \log(n)}$ as a function of the grid sizes (see Figure $\ref{asymp_gauss_exp}$). One remarks that the convergence of  $ \frac{\rho_n}{ 3 \log(n)}$  to $1$ as $n$ goes to infinity is  almost instantaneous. 

 However, the cases of the Gaussian and the Weibull distributions  are more delicate.  Indeed, for the Gaussian distribution the ratio $\frac{\rho_n}{\sqrt{6 \log(n)}}$  seems increasing  but has not  reached yet the value $9$ even for a grid size equals $100000$, as emphasized by Figure  $\ref{asymp_gauss_exp}$ (right hand side graph).  For the Weibull distribution,  $\frac{\rho_n}{\sqrt{3 \log(n)}}$ also seems increasing but takes values  around $0.927$ for a grid size equal to $3000$ (see Figure $2$). Then for both cases, the convergence to $1$  of  the ratio between  the maximal radius and  the expected asymptotic optimal rate  seems increasing  but very low.
 
\begin{figure}[htpb] 
  \includegraphics[width=5cm,angle=270]{raymax_exp.ps} 
  \hfill \includegraphics[width=5cm,angle=270]{raymax.ps}
 \caption{\small{Left: $\frac{\rho_n}{ 3 \log(n)}$ as a  function of the grid size $n$ for the exponential distribution. Right: $\frac{\rho_n}{\sqrt{6 \log(n)}}$ as a function of the grid size $n$  for the normal distribution.}}
\label{asymp_gauss_exp}
  \end{figure}
 

  \begin{figure}[htpb] 
   \begin{center} 
  \includegraphics[width=6cm,height=10cm,angle=270]{Weibullraymax.ps} 
  \caption{\small{ $\frac{\rho_n}{\sqrt{3 \log(n)}}$  as a function of the grid  size for the Weibull distribution with  shape  parameter  $\kappa =2$. } }
  \end{center}
  \label{asymptotic_weibull}
\end{figure}

 

\begin{comment}
\section{Lloyd's I algorithm}
One of the important issues from the computational point of view is the search of the optimal quantizers. For this, different algorithms like the Competitive Learning Vector Quantization (CLVQ) algorithm and Lloyd's algorithms (I and II) are used.  In practice, Lloyd's I algorithm is widely used to compute optimal (or stationary) quantizers because it can be easily implemented. It starts with an initial points and consists of repeatedly executing one relaxation step: the Voronoi tesselation of all the points is computed at first, then each cell of the Voronoi tesselation is integrated and the centroid is computed, at last each point is moved at the centroid of its Voronoi cell. 

 As a matter of fact, starting with an initial quantizer $\alpha^{(0)}$ of size $n$, one defines recursivily a sequence $(\alpha^{(k)})_{k=1,\cdots,p}$ of $n$-quantizers by setting 
\begin{equation}
\widehat{X}^{\alpha^{(k)}} = \mathbb{E}(X |\widehat{X}^{\alpha^{(k)}} )
\end{equation}
$i.e$ by forcing stationary. In each step $k \geq 1$, $\alpha_i^{(k)}$ is then given by 

\begin{equation} \label{eq_station1}
\alpha_i^{(k)} = \frac{\mathbb{E}(X \mbox{\bf{1}}_{\{X \in C_i(\alpha^{(k)}) \}})}{\mathbb{P}(X \in C_i(\alpha^{(k)})) }, \qquad i=1,\cdots,n.
\end{equation}
 For every  $i \in \{1,\cdots,n \}$ and in each iteration $ k \in \{1,\cdots,p \} $,  $\alpha_i^{(k)}$ is approximated by  Monte Carlo method.  In  fact, by the low of large numbers, 
$$  \frac{1}{m} \sum_{j=1}^m  X_j  \mbox{\bf{1}}_{\{x_j \in C_i(\alpha^{(k)}) \}}   \stackrel{a.s} { \longrightarrow }\mathbb{E}(X \mbox{\bf{1}}_{\{X \in C_i(\alpha^{(k)}) \}}) $$
$$  \frac{1}{m} \sum_{j=1}^m   \mbox{\bf{1}}_{\{X_j \in C_i(\alpha^{(k))} \}}  \stackrel{a.s} { \longrightarrow }  \mathbb{P}(X \in C_i(\alpha^{(k)}))$$
where $(X_j)_{j=1,\cdots,m}$ is an $i.i.d$ sample of the random variable $X$. According to  Equation $(\ref{equivalence})$, if the algorithm converges to a quantizer $\alpha_n$ of size $n$ in the sense that further iterations no longer make any changes in the set of previously reproduced values, then $\alpha_n$ is a local minimum (or stationary) point. In one dimension $\alpha_n$ will be the unique optimal $n$-quantizer if $X$ has log-concave $p.d.f$.

  However, one of the difficulties for this algorithm lies in the choice of the initial point. In the case of the Gaussian distribution there is different initialization methods like random initialization (see $\cite{GG}$) and the Competitive Learning Vector Quantization (CLVQ) algorithm initialization used in $\cite{PP}$  which consists on processing the CLVQ algorithm  before starting Lloyd's algorithm. We propose in this section an initialization method called a \textit{Hyperspherical Initialization Method} (HIM) which consists on putting the initial grid of the Lloyd procedure "uniformly" in a hypersphere of radius $r$. We shall see that the radius $r$ can be chosen around the expected value of the maximal radii of the grid, for a fixed $n \geq 1$. The most important advantage  of this method is that we can carry out the Lloyd's  algorithm for every dimension $d$ and for every grid size without using  the previously computed grids as done in the initialization of the CLVQ algorithm (by the so called split method).

 \subsection{Numerical results for the multivariate  Gaussian distribution }
 In this section we focus on the numerical computation of  stationary quantizers (which are local minima as mentioned above) for the multivariate Gaussian distribution. To this aim we only use Lloyd's I algorithm described previously by  choosing  the initial point in an  hypersphere of radius $r>0$. We consider only the bivariate Gaussian distribution case. The case $d \geq 3$ works as well as previously (at least up to $d=12$).

According to the previous paragraph where we have  conjectured  the  maximal  radius sequence of  $n$-quantizers  $(\rho_n)_{n \geq 1}$ to be  such that  
 $$\ \rho_n  \sim  \sqrt{\frac{2(2+d)}{d} \log(n)} ,$$ 
 set  $Raymax =  \sqrt{\frac{2+d}{d} \log(n)} $.

In the next graphs, we depicted the distortion as a fonction of the  grid's size  $n  \in \{10,\cdots,99 \}$. In  Figure $\ref{figure1}\ $ and Figure $\ref{figure2}\ $ we compare distortions obtained  (after  $p=10$ and $p=80$ Lloyd's iterations) by putting  the initial grid on  an  hypersphere of radius  $1.3 * Raymax, \ 2*Raymax$ and $0.5*Raymax$, namely, almost at, more than  and less  than the  conjectured (asymptotic) maximal radii. The distortions we get are identical no matter how we choose the initial grid. However, after $80$ Lloyd's iterations, the distortions tend to be more precise that the ones got for $p=10$ as shown in figure $\ref{figure3} \ $ where we compare the distorsions obtained after $p=10$ and $p=80$ iterations of the Lloyd's procedure. This is in fact known since
for every $k \geq 1$,
$$ \Vert X - \widehat{X}^{\alpha^{(k+1)}} \Vert_2^2 =  \Vert X - \mathbb{E}(X | \widehat{X}^{\alpha^{(k)}}) \Vert_2^2 \leq \Vert X - \widehat{X}^{\alpha^{(k)}} \Vert_2^2 $$ 
since $\widehat{X}^{\alpha^{(k)}}$ is $\sigma(\widehat{X}^{\alpha^{(k)}})$-mesurable. It then suggests to increase the number of Lloyd's iterations to have more precise estimations of the computed grids.  From $10$ to $80$ Lloyd's iterations, the average gain is about $4.22 \%$. It follows from these computations that the choice of the radius $r$ does not really depends on the maximal radii $\rho_n$  since the distorsions are almost identical apart from the choice of $r$. Anyway, one gets the same distortion by initializing the grid on a hypersphere of radius $15*Raymax$ or $0.1*Raymax$, means, far from the maximal radii. We finally compare the distortion we obtain with the one obtained by the CLVQ algorithm (with a split initialization method)  followed to $100$ Lloyd's iterations. These grids can be found in the following URL's:

 {\small{www.proba.jussieu.fr/pagesperso/pages.html}} or {\small{perso-math.univ-mlv.fr/users/printems.jacques/n01/}}

One observes that we obtain almost the same distortions  by using only the Lloyd's I algorithm and after $80$ iterations.
 \end{comment}
 
\newpage
{"config": "arxiv", "file": "0806.0918/Asymp_MaxRadius.tex"}