# Conditional constrained and unconstrained quantization for probability
distributions

1Megha Pandey  and  2Mrinal K. Roychowdhury  1Department of Mathematical
Sciences
Indian Institute of Technology (Banaras Hindu University)
Varanasi, 221005, India.  2School of Mathematical and Statistical Sciences
University of Texas Rio Grande Valley
1201 West University Drive
Edinburg, TX 78539-2999, USA<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

In this paper, we present the idea of conditional quantization for a Borel
probability measure $P$ on a normed space $\mathbb{R}^{k}$. We introduce the
concept of conditional quantization in both constrained and unconstrained
scenarios, along with defining the conditional quantization errors,
dimensions, and coefficients in each case. We then calculate these values for
specific probability distributions. Additionally, we demonstrate that for a
Borel probability measure, the lower and upper quantization dimensions and
coefficients do not depend on the conditional set of the conditional
quantization in both constrained and unconstrained quantization.

###### Key words and phrases:

Probability measure, conditional constrained and unconstrained quantization,
optimal sets of $n$-points, conditional quantization dimension, conditional
quantization coefficient

###### 2010 Mathematics Subject Classification:

60Exx, 94A34.

##  1\. Introduction

Quantization involves the process of discretizing signals. In the context of
probability distributions, quantization refers to finding the best
approximation of a $k$-dimensional probability distribution $P$ using a
discrete probability distribution with a specified number $n$ of supporting
points (referred to as the optimal set of $n$-points). In other words, it aims
to find the best approximation of a $k$-dimensional random vector $X$ with
distribution $P$ using a random vector $Y$ that has nearly $n$ values in its
range.

A plethora of research is given on quantization for probability distributions
without using any constraint. The concept of constrained quantization was
recently introduced by Pandey and Roychowdhury (see [PR1, PR2, PR3]). This new
approach allows us to categorize quantization into two types: unconstrained
quantization and constrained quantization. Unconstrained quantization is
traditionally known as quantization. For some recent work in the direction of
unconstrained quantization, one can see [GL, DFG, DR, GL2, GL3, KNZ, PRRSS,
P1, R1, R2, R3]. Quantization theory has broad applications in communications,
information theory, signal processing, and data compression (see [GG, GL1, GN,
P, Z1, Z2]). This paper deals with conditional quantization in both
constrained and unconstrained cases. Conditional quantization also has
significant interdisciplinary applications: for example, in radiation therapy
of cancer treatment to find the optimal locations of $n$ centers of radiation,
where $k$ centers for some $k<n$ of radiation are preselected, the conditional
quantization technique can be useful.

Let $P$ be a Borel probability measure on $\mathbb{R}^{k}$ equipped with a
metric $d$ induced by a norm $\|\cdot\|$ on $\mathbb{R}^{k}$, and
$r\in(0,\infty)$. Let $\mathbb{N}:=\\{1,2,3,\cdots\\}$ be the set of natural
numbers. For a finite set $\gamma\subset\mathbb{R}^{k}$ and $a\in\gamma$, by
$M(a|\gamma)$ we denote the set of all elements in $\mathbb{R}^{k}$ which are
nearest to $a$ among all the elements in $\gamma$, i.e.,
$M(a|\gamma)=\\{x\in\mathbb{R}^{k}:d(x,a)=\mathop{\min}\limits_{b\in\gamma}d(x,b)\\}.$
$M(a|\gamma)$ is called the Voronoi region in $\mathbb{R}^{k}$ generated by
$a\in\gamma$.

######  Definition 1.1.

Let $\\{S_{j}\subseteq\mathbb{R}^{k}:j\in\mathbb{N}\\}$ be a family of closed
sets with $S_{1}$ nonempty. Let $\beta\subset\mathbb{R}^{k}$ be given with
$\text{card}(\beta)=\ell$ for some $\ell\in\mathbb{N}$. Then, for
$n\in\mathbb{N}$ with $n\geq\ell$, the $n$th conditional constrained
quantization error for $P$, of order $r$, with respect to the family of
constraints $\\{S_{j}\subseteq\mathbb{R}^{k}:j\in\mathbb{N}\\}$ and the set
$\beta$, is defined as

$V_{n,r}:=V_{n,r}(P)=\inf_{\alpha}\Big{\\{}\int\mathop{\min}\limits_{a\in\alpha\cup\beta}d(x,a)^{r}dP(x):\alpha\subseteq\bigcup_{j=1}^{n}S_{j},~{}0\leq\text{card}(\alpha)\leq
n-\ell\Big{\\}},$ (1)

where $\text{card}(A)$ represents the cardinality of the set $A$.

Notice the error $V_{n,r}$ explicitly depends on $\beta$. For any
$\alpha\subseteq\mathbb{R}^{k}$, the number

$V_{r}(P;\alpha):=\int\mathop{\min}\limits_{a\in\alpha}d(x,a)^{r}dP(x)$ (2)

is called the distortion error for $P$, of order $r$, with respect to the set
$\alpha$. We assume that $\int d(x,0)^{r}dP(x)<\infty$ to make sure that the
infimum in (1) exists (see [PR1]).

######  Definition 1.2.

A set $\alpha\cup\beta$, where
$\alpha\subseteq\mathop{\bigcup}_{j=1}^{n}S_{j}$ and
$P(M(b|\alpha\cup\beta))>0$ for $b\in\beta$, for which the infimum in (1)
exists and contains no less than $\ell$ elements, and no more than $n$
elements is called an optimal set of $n$-points for $P$, or more specifically
a conditional optimal set of $n$-points for $P$. Elements of an optimal set
are called optimal elements.

######  Definition 1.3.

Instead of the family of constraints
$\\{S_{j}\subseteq\mathbb{R}^{k}:j\in\mathbb{N}\\}$ if there is a single
constraint $S$, i.e., if $S_{j}=S$ for all $j\in\mathbb{N}$, then Definition
1.1 reduces to

$V_{n,r}:=V_{n,r}(P)=\inf_{\alpha}\Big{\\{}\int\mathop{\min}\limits_{a\in\alpha\cup\beta}d(x,a)^{r}dP(x):\alpha\subseteq
S,~{}0\leq\text{card}(\alpha)\leq n-\ell\Big{\\}},$

which is called the $n$th conditional constrained quantization error for $P$,
of order $r$, with respect to the single constraint $S$ and the set $\beta$.

Write $V_{\infty,r}(P):=\mathop{\lim}\limits_{n\to\infty}V_{n,r}(P)$. The
numbers

$\underline{D}_{r}(P):=\liminf_{n\to\infty}\frac{r\log
n}{-\log(V_{n,r}(P)-V_{\infty,r}(P))}\text{ and
}\overline{D}_{r}(P):=\limsup_{n\to\infty}\frac{r\log
n}{-\log(V_{n,r}(P)-V_{\infty,r}(P))},$ (3)

are called the conditional lower and the conditional upper constrained
quantization dimensions of the probability measure $P$ of order $r$,
respectively. If $\underline{D}_{r}(P)=\overline{D}_{r}(P)$, the common value
is called the conditional constrained quantization dimension of $P$ of order
$r$ and is denoted by $D_{r}(P)$. For any $\kappa>0$, the two numbers
$\liminf_{n}n^{\frac{r}{\kappa}}(V_{n,r}(P)-V_{\infty,r}(P))$ and
$\limsup_{n}n^{\frac{r}{\kappa}}(V_{n,r}(P)-V_{\infty,r}(P))$ are,
respectively, called the $\kappa$-dimensional conditional lower and
conditional upper constrained quantization coefficients for $P$ of order $r$.
If both of them are equal, then it is called the $\kappa$-dimensional
conditional constrained quantization coefficient for $P$ of order $r$.

######  Definition 1.4.

In Definition 1.1 if $S_{j}=\mathbb{R}^{k}$ for all $j\in\mathbb{N}$, then for
$n\in\mathbb{N}$ with $n\geq\ell$, the $n$th conditional unconstrained
quantization error for $P$, of order $r$, with respect to the set $\beta$, is
defined as

$V_{n,r}:=V_{n,r}(P)=\inf_{\alpha}\Big{\\{}\int\mathop{\min}\limits_{a\in\alpha\cup\beta}d(x,a)^{r}dP(x):\alpha\subseteq\mathbb{R}^{k},~{}0\leq\text{card}(\alpha)\leq
n-\ell\Big{\\}}.$

The corresponding quantization dimension and the $\kappa$-dimensional
quantization coefficient, if they exist, are called the conditional
unconstrained quantization dimension and the $\kappa$-dimensional conditional
unconstrained quantization coefficient for $P$, respectively.

######  Remark 1.5.

The set $\beta$ that occurs in any of the above definitions is referred to as
the conditional set. Notice that in unconstrained quantization, if
$\alpha\cup\beta$ is a conditional optimal set of $n$-points with $\alpha$
nonempty, then all the elements in $\alpha$ are the means, i.e., the
conditional expectations in their own Voronoi regions. The elements of the
conditional set $\beta$ are not necessarily the means of their own Voronoi
regions.

This paper deals with $r=2$ and $k=2$, and the metric on $\mathbb{R}^{2}$ as
the Euclidean metric induced by the Euclidean norm $\|\cdot\|$. Thus, instead
of writing $V_{r}(P;\alpha)$ and $V_{n,r}:=V_{n,r}(P)$ we will write them as
$V(P;\alpha)$ and $V_{n}:=V_{n}(P)$.

######  Remark 1.6.

Although all the work done in the sequel is for uniform distributions,
interested researchers can explore them for any probability distribution.

######  Delineation.

In this paper, first we have given the preliminaries. Then, in Section 3, we
have investigated the conditional constrained quantization for a uniform
distribution on the boundary of a semicircular disc; in Section 4, we have
investigated the conditional unconstrained quantization for a uniform
distribution on an equilateral triangle; in Section 5, with some examples, we
mentioned whether a conditional optimal set of $n$-points always exist or not.
In the last section, Section 6, we have proved two theorems: first theorem
shows that if $\alpha_{n}$ are the optimal sets of $n$-means for $P$ for all
$n\in\mathbb{N}$, then $\mathop{\cup}\limits_{n=1}^{\infty}\alpha_{n}$ is
dense in the support of $P$; and in the 2nd theorem we have shown that in both
constrained and unconstrained quantization, the lower and upper quantization
dimensions, and the lower and upper quantization coefficients for a Borel
probability measure do not depend on the conditional set.

##  2\. Preliminaries

Let $\mathbb{N}$ be the set of natural numbers and $\mathbb{R}$ be the
collection of all real numbers. Recall that the boundary of the Voronoi
regions of any two elements $p$ and $q$ in an optimal set is the perpendicular
bisector of the line segment joining the elements. Hence, for any element $e$
on the boundary of the Voronoi regions of $p$ and $q$, we have

$\rho(p,e)-\rho(q,e)=0,$ (4)

where for any two elements $(a,b)$ and $(c,d)$ in $\mathbb{R}^{2}$,
$\rho((a,b),(c,d))$ represents the squared Euclidean distance between the two
elements. Equation (4) is known as a canonical equation. Let us now give the
following two propositions.

######  Proposition 2.1.

Let $P$ be a uniform distribution on the closed interval $[a,b]$. Let $a\leq
c<d\leq b$. Let $\alpha_{n}$ be an optimal set of $n$-points for $P$ such that
$\alpha_{n}$ contains $m$ elements, for some $m\leq n$, from the closed
interval $[c,d]$ including the endpoints $c$ and $d$, then

$\alpha_{n}\cap[c,d]=\\{c+\frac{j-1}{m-1}(d-c):1\leq j\leq m\\}.$

Then, the distortion error contributed by these $m$ elements in the closed
interval $[c,d]$ is given by

$V(P,\alpha_{n}\cap[c,d]):=\frac{1}{12}\frac{(d-c)^{3}}{b-a}\frac{1}{(m-1)^{2}}.$

###### Proof.

$P$ being a uniform distribution on the closed interval $[a,b]$, its density
function is given by $f(x)=\frac{1}{b-a}$ if $x\in[a,b]$, and zero otherwise.
Also, notice that the closed interval $[a,b]$ can be represented by

$[a,b]:=\\{(t,0):a\leq t\leq b\\}.$

Let ${c_{1},c_{2},c_{3},\cdots,c_{m}}$ be the $m$ elements that $\alpha_{n}$
contains from the closed interval $[c,d]$ such that
$c=c_{1}<c_{2}<\cdots<c_{m}=d$. Since the closed interval $[c,d]$ is a line
segment and $P$ is a uniform distribution, we have

$c_{2}-c_{1}=c_{3}-c_{2}=\cdots=c_{m}-c_{m-1}=\frac{c_{m}-c_{1}}{m-1}=\frac{d-c}{m-1}$

implying

$\displaystyle c_{2}$ $\displaystyle=c_{1}+\frac{d-c}{m-1}=c+\frac{d-c}{m-1},$
$\displaystyle c_{3}$
$\displaystyle=c_{2}+\frac{d-c}{m-1}=c+\frac{2(d-c)}{m-1},$ $\displaystyle
c_{4}$ $\displaystyle=c_{3}+\frac{d-c}{m-1}=c+\frac{3(d-c)}{m-1},$ and so on.

Thus, we have $c_{j}=c+\frac{j-1}{m-1}(d-c)$ for $1\leq j\leq m$. The
distortion error contributed by the $m$ elements in the closed interval
$[c,d]$ is given by

$\displaystyle V(P;\alpha_{n}\cap[c,d])$
$\displaystyle=\int_{[c,d]}\min_{x\in\alpha_{n}\cap[c,d]}\rho((t,0),x)\,dP$
$\displaystyle=\frac{1}{b-a}\Big{(}2\int_{c_{1}}^{\frac{c_{1}+c_{2}}{2}}\rho((t,0),(c_{1},0))\,dt+(m-2)\int_{\frac{c_{1}+c_{2}}{2}}^{\frac{c_{2}+c_{3}}{2}}\rho((t,0),(c_{2},0))\,dt\Big{)}$
$\displaystyle=\frac{1}{12}\frac{(d-c)^{3}}{b-a}\frac{1}{(m-1)^{2}}.$

Thus, the proof of the proposition is complete. ∎

######  Proposition 2.2.

Let $P$ be a uniform distribution on the closed interval $[a,b]$. Let
$\alpha_{n}$ be an optimal set of $n$-points for $P$ such that $\alpha_{n}$
contains $n$ elements from the closed interval $[a,b]$ including the endpoint
$a$. Then,

$\alpha_{n}=\left\\{a+\frac{2(j-1)(b-a)}{2n-1}:1\leq j\leq n\right\\}$

with the conditional unconstrained quantization error

$V_{n}=\frac{(b-a)^{2}}{3(2n-1)^{2}}.$

###### Proof.

As mentioned in Proposition 2.1, here the probability density function is
given by $f(x)=\frac{1}{b-a}$ if $x\in[a,b]$, and zero otherwise, and

$[a,b]:=\\{(t,0):a\leq t\leq b\\}.$

Let ${c_{1},c_{2},c_{3},\cdots,c_{n}}$ be the $n$ elements that $\alpha_{n}$
contains from the closed interval $[a,b]$ including the endpoint $a$, i.e.,
$c_{1}=a$. Let $c_{n}=d$, where $d<b$. Since the closed interval $[a,d]$ is a
line segment and $P$ is a uniform distribution, we have

$c_{2}-c_{1}=c_{3}-c_{2}=\cdots=c_{n}-c_{n-1}=\frac{c_{n}-c_{1}}{n-1}=\frac{d-a}{n-1}$

implying

$\displaystyle c_{2}$ $\displaystyle=c_{1}+\frac{d-a}{n-1}=a+\frac{d-a}{n-1},$
$\displaystyle c_{3}$
$\displaystyle=c_{2}+\frac{d-a}{n-1}=a+\frac{2(d-a)}{n-1},$ $\displaystyle
c_{4}$ $\displaystyle=c_{3}+\frac{d-a}{n-1}=a+\frac{3(d-a)}{n-1},$ and so on.

Thus, we have $c_{j}=a+\frac{j-1}{n-1}(d-a)$ for $1\leq j\leq n$. The
distortion error contributed by the $n$ elements is given by

$\displaystyle V(P;\alpha_{n})$
$\displaystyle=\int\min_{x\in\alpha_{n}}\rho((t,0),x)\,dP$
$\displaystyle=\frac{1}{b-a}\Big{(}\int_{c_{1}}^{\frac{c_{1}+c_{2}}{2}}\rho((t,0),(c_{1},0))\,dt+(n-2)\int_{\frac{c_{1}+c_{2}}{2}}^{\frac{c_{2}+c_{3}}{2}}\rho((t,0),(c_{2},0))\,dt$
$\displaystyle\qquad+\int_{\frac{c_{n-1}+c_{n}}{2}}^{b}\rho((t,0),(c_{n},0))\,dt\Big{)}$
$\displaystyle=\frac{\frac{(a-d)^{3}}{(n-1)^{2}}-4(b-d)^{3}}{12(a-b)},$

the minimum value of which is $\frac{(b-a)^{2}}{3(2n-1)^{2}}$ and it occurs
when $d=b-\frac{b-a}{2n-1}$. Putting the values of $d$, we have

$c_{j}=a+\frac{2(j-1)(b-a)}{2n-1}\text{ for }1\leq j\leq n$

with the conditional unconstrained quantization error

$V_{n}=\frac{(b-a)^{2}}{3(2n-1)^{2}}.$

Thus, the proof of the proposition is complete. ∎

######  Proposition 2.3.

Let $P$ be a uniform distribution on the closed interval $[a,b]$. Let
$\alpha_{n}$ be an optimal set of $n$-points for $P$ such that $\alpha_{n}$
contains $n$ elements from the closed interval $[a,b]$ including the endpoint
$b$. Then,

$\alpha_{n}=\left\\{a+\frac{(2j-1)(b-a)}{2n-1}:1\leq j\leq n\right\\}$

with the conditional unconstrained quantization error

$V_{n}=\frac{(b-a)^{2}}{3(2n-1)^{2}}.$

###### Proof.

As mentioned in Proposition 2.1, here the probability density function is
given by $f(x)=\frac{1}{b-a}$ if $x\in[a,b]$, and zero otherwise, and

$[a,b]:=\\{(t,0):a\leq t\leq b\\}.$

Let ${c_{1},c_{2},c_{3},\cdots,c_{n}}$ be the $n$ elements that $\alpha_{n}$
contains from the closed interval $[a,b]$ including the endpoint $b$, i.e.,
$c_{n}=b$. Let $c_{1}=d$, where $a<d$. Since the closed interval $[c_{1},b]$
is a line segment and $P$ is a uniform distribution, we have

$c_{2}-c_{1}=c_{3}-c_{2}=\cdots=c_{n}-c_{n-1}=\frac{c_{n}-c_{1}}{n-1}=\frac{b-d}{n-1}$

implying

$\displaystyle c_{2}$ $\displaystyle=c_{1}+\frac{b-a}{n-1}=d+\frac{b-d}{n-1},$
$\displaystyle c_{3}$
$\displaystyle=c_{2}+\frac{b-a}{n-1}=d+\frac{2(b-d)}{n-1},$ $\displaystyle
c_{4}$ $\displaystyle=c_{3}+\frac{b-a}{n-1}=d+\frac{3(b-d)}{n-1},$ and so on.

Thus, we have $c_{j}=d+\frac{j-1}{n-1}(b-d)$ for $1\leq j\leq n$. The
distortion error contributed by the $n$ elements is given by

$\displaystyle V(P;\alpha_{n})$
$\displaystyle=\int\min_{x\in\alpha_{n}}\rho((t,0),x)\,dP$
$\displaystyle=\frac{1}{b-a}\Big{(}\int_{a}^{\frac{c_{1}+c_{2}}{2}}\rho((t,0),(c_{1},0))\,dt+(n-2)\int_{\frac{c_{1}+c_{2}}{2}}^{\frac{c_{2}+c_{3}}{2}}\rho((t,0),(c_{2},0))\,dt$
$\displaystyle\qquad+\int_{\frac{c_{n-1}+c_{n}}{2}}^{c_{n}}\rho((t,0),(c_{n},0))\,dt\Big{)}$
$\displaystyle=\frac{4(a-d)^{3}+\frac{(d-b)^{3}}{(n-1)^{2}}}{12(a-b)},$

the minimum value of which is $\frac{(b-a)^{2}}{3(2n-1)^{2}}$ and it occurs
when $d=a+\frac{b-a}{2n-1}$. Putting the values of $d$, we have

$c_{j}=a+\frac{(2j-1)(b-a)}{2n-1}\text{ for }1\leq j\leq n$

with the conditional unconstrained quantization error

$V_{n}=\frac{(b-a)^{2}}{3(2n-1)^{2}}.$

Thus, the proof of the proposition is complete. ∎

######  Remark 2.4.

Although a detailed proof is given, the elements in $\alpha_{n}$ in
Proposition 2.3 can be obtained from the elements in $\alpha_{n}$ given in
Proposition 2.2 by translating the elements to the right in the amount of
$\frac{b-a}{2n-1}$.

##  3\. Conditional constrained quantization for a uniform distribution on the
boundary of a semicircular disc

Let $L$ be the boundary of the semicircular disc $x_{1}^{2}+x_{2}^{2}=1$,
where $x_{2}\geq 0$. Let the base of the semicircular disc be $AOB$, where $A$
and $B$ have the coordinates $(-1,0)$ and $(1,0$), and $O$ is the origin
$(0,0)$. Let $s$ represent the distance of any point on $L$ from the origin
tracing along the boundary $L$ in the counterclockwise direction. Notice that
$L=L_{1}\cup L_{2}$, where

$\displaystyle L_{1}$ $\displaystyle=\\{(x_{1},x_{2}):x_{1}=t,\,x_{2}=0\text{
for }-1\leq t\leq 1\\},\text{ and }$ $\displaystyle L_{2}$
$\displaystyle=\\{(x_{1},x_{2}):x_{1}=\cos t,\,x_{2}=\sin t\text{ for }0\leq
t\leq\pi\\}.$

Let $P$ be the uniform distribution on the boundary of the semicircular disc.
Then, the probability density function for $P$ is given by

$\displaystyle
f(x_{1},x_{2})=\left\\{\begin{array}[]{cc}\frac{1}{2+\pi}&\text{ if
}(x_{1},x_{2})\in L,\\\ 0&\text{ otherwise,}\end{array}\right.$

On both $L_{1}$ and $L_{2}$, we have
$ds=\sqrt{(\frac{dx_{1}}{dt})^{2}+(\frac{dx_{2}}{dt})^{2}}\,dt=dt$ yielding
$dP(s)=P(ds)=f(x_{1},x_{2})ds=f(x_{1},x_{2})dt$. In the definition of
conditional constrained quantization error, let $\beta:=\\{(-1,0),(1,0)\\}$,
and let $L$ be the single constraint. Upon the given set $\beta$, the $n$th
conditional constrained quantization errors are defined for all $n\geq 2$.
Notice that the boundary $L$ has ‘maximum symmetry’ with respect to the
$y$-axis. By the maximum symmetry of $L$ with respect to the $y$-axis, it is
meant that if two regions of the same geometrical shapes are equidistant and
are on opposite sides from the line, then they have the same probability. In
finding the conditional optimal sets of $n$-points, we will use this property.

Notice that the conditional optimal set of two-points is the set $\beta$
itself. In the sequel of this section, we investigate the conditional optimal
sets of $n$-points for $n\geq 3$. Let us now give the following proposition,
which plays a vital role in this section.

######  Proposition 3.1.

Let $\alpha_{n}$ be a conditional optimal set of $n$-points for $P$ for some
$n\geq 3$. Let $\text{card}(\alpha_{n}\cap L_{1})=n_{1}$ and
$\text{card}(\alpha_{n}\cap L_{2})=n_{2}$ with corresponding conditional
quantization error $V_{n}:=V_{n_{1},n_{2}}(P)$ for some $n_{1},n_{2}\geq 2$.
Then,

$\alpha_{n}\cap L_{1}=\\{(-1+\frac{2(j-1)}{n_{1}-1},0):1\leq j\leq
n_{1}\\},\text{ and }\alpha_{n}\cap
L_{2}=\\{(\cos\frac{(j-1)\pi}{n_{2}-1},\sin\frac{(j-1)\pi}{n_{2}-1}):1\leq
j\leq n_{2}\\},$

with

$V_{n_{1},n_{2}}(P)=\frac{2}{3(2+\pi)}\Big{(}\frac{1}{(n_{1}-1){}^{2}}-6(n_{2}-1)\sin\Big{(}\frac{\pi}{2(n_{2}-1)}\Big{)}+3\pi\Big{)}.$

###### Proof.

By the hypothesis, $\text{card}(\alpha_{n}\cap L_{1})=n_{1}$ and
$\text{card}(\alpha_{n}\cap L_{2})=n_{2}$ for some $n_{1},n_{2}\geq 2$. The
proof of

$\alpha_{n}\cap L_{1}=\\{(-1+\frac{2(j-1)}{n_{1}-1},0):1\leq j\leq n_{1}\\}$

directly follows from Proposition 2.1. Moreover, as shown in Proposition 2.1,
the distortion error for these $n_{1}$ elements is obtained as

$V(P;\alpha_{n}\cap L_{1})=\frac{2}{3(2+\pi)}\frac{1}{(n_{1}-1)^{2}}.$

Let $\alpha_{n}\cap L_{2}=\\{(\cos\theta_{j},\sin\theta_{j}):1\leq j\leq
n_{2}\\}$, where $\theta_{1}=0$, and $\theta_{n_{2}}=\pi$. $L_{2}$ being a
circular arc and $P$ is a uniform distribution, we have

$\theta_{2}-\theta_{1}=\theta_{3}-\theta_{2}=\cdots=\theta_{n_{2}}-\theta_{n_{2}-1}=\frac{\theta_{n_{2}}-\theta_{1}}{n_{2}-1}=\frac{\pi}{n_{2}-1}.$

Thus, proceeding in a similar way as Proposition 2.1, we have
$\theta_{j}=\frac{(j-1)\pi}{n_{2}-1}$ for $1\leq j\leq n_{2}$ yielding

$\alpha_{n}\cap
L_{2}=\\{(\cos\frac{(j-1)\pi}{n_{2}-1},\sin\frac{(j-1)\pi}{n_{2}-1}):1\leq
j\leq n_{2}\\}.$

The distortion error due to these $n_{2}$ elements is obtained as

$\displaystyle V(P;\alpha_{n}\cap L_{2})$
$\displaystyle=\frac{1}{2+\pi}\Big{(}2\int_{0}^{\frac{\theta_{1}+\theta_{2}}{2}}\rho((\cos
t,\sin
t),(1,0))\,dt+(n_{2}-2)\int_{\frac{\theta_{1}+\theta_{2}}{2}}^{\frac{\theta_{2}+\theta_{3}}{2}}\rho((\cos
t,\sin t),(\cos\theta_{2},\sin\theta_{2}))\,dt\Big{)}$
$\displaystyle=\frac{2}{3(2+\pi)}\Big{(}-6(n_{2}-1)\sin\Big{(}\frac{\pi}{2(n_{2}-1)}\Big{)}+3\pi\Big{)}.$

Thus, we have

$\displaystyle V_{n}$ $\displaystyle=V_{n_{1},n_{2}}(P)=V(P;\alpha_{n}\cap
L_{1})+V(P;\alpha_{n}\cap L_{2})$
$\displaystyle=\frac{2}{3(2+\pi)}\Big{(}\frac{1}{(n_{1}-1){}^{2}}-6(n_{2}-1)\sin\Big{(}\frac{\pi}{2(n_{2}-1)}\Big{)}+3\pi\Big{)}.$

Hence, the proof of the proposition is complete. ∎

######  Remark 3.2.

Proposition 3.1 plays a significant role in calculating the optimal sets of
$n$-points for the probability distribution $P$ on the boundary of the
semicircular disc, as shown in the following two propositions.

######  Proposition 3.3.

The conditional optimal set of three-points is given by
$\\{(1,0),(0,1),(-1,0)\\}$ with conditional constrained quantization error
$V_{3}=\frac{2}{2+\pi}(-2\sqrt{2}+\frac{1}{3}+\pi)$.

###### Proof.

Let $\alpha$ be an optimal set of three-points. By the hypothesis
$(1,0),(-1,0)\in\alpha$. Due to maximum symmetry, we can assume that either
$(0,0)$ or $(0,1)\in\alpha$. If $(0,0)\in\alpha$, then as $n_{1}=3$ and
$n_{2}=2$, using Proposition 3.1, we have the distortion error as
$V_{3,2}(P)=0.476477.$ If $(0,1)\in\alpha$, then as $n_{1}=2$ and $n_{2}=3$,
we have $V_{2,3}(P)=0.251478$. Since $V_{2,3}(P)<V_{3,2}(P)$, the conditional
optimal set of three-points is given by $\\{(1,0),(0,1),(-1,0)\\}$ with
conditional constrained quantization error
$V_{3}=\frac{2}{2+\pi}(-2\sqrt{2}+\frac{1}{3}+\pi)$. Thus, the proof of the
proposition is complete (see Figure 1). ∎

Figure 1. Optimal configuration of $n$-points for $3\leq n\leq 11$.

######  Proposition 3.4.

The conditional optimal set of four-points is given by
$\\{(0,0),(1,0),(0,1),(-1,0)\\}$ with conditional constrained quantization
error $V_{4}=\frac{-24\sqrt{2}+12\pi+1}{12+6\pi}$.

###### Proof.

Let $\alpha$ be an optimal set of four-points. By the hypothesis
$(1,0),(-1,0)\in\alpha$. Due to maximum symmetry, we can assume that the other
two elements in $\alpha$ are on the axis of symmetry, or they are
symmetrically located on $L$. Thus, the following cases can occur:

Case 1. The two elements in $\alpha\setminus\\{(1,0),(-1,0)\\}$ are on the
axis of symmetry.

In this case, we can assume that $(0,0),(0,1)\in\alpha$. Then, as $n_{1}=3$
and $n_{2}=3$, the distortion error is $V_{3,3}(P)=0.154232.$

Case 2. The two elements in $\alpha\setminus\\{(1,0),(-1,0)\\}$ are
symmetrically located on $L$.

In this case, the two elements are either symmetrically located on $L_{1}$ or
on $L_{2}$. If they are symmetrically located on $L_{1}$, then the distortion
error is $V_{4,2}(P)=0.458469$. If they are symmetrically located on $L_{2}$,
then the distortion error is $V_{2,4}(P)=0.184739.$

Considering all the above possible distortion errors, we see that the
distortion error is minimal when $n_{1}=3$ and $n_{2}=3$. Thus,
$\\{(0,0),(1,0),(0,1),(-1,0)\\}$ forms the conditional optimal set of four-
points with conditional constrained quantization error
$V_{4}=\frac{-24\sqrt{2}+12\pi+1}{12+6\pi}$, which is the proposition (see
Figure 1). ∎

Let us now give the following theorem, which gives the main result in this
section.

######  Theorem 3.5.

Let $\alpha_{n}$ be a conditional optimal set of $n$-points for $P$ for some
$n\geq 3$. Let $\text{card}(\alpha_{n}\cap L_{1})=n_{1}$ be known. Then,

$\displaystyle\alpha_{n}=\\{(-1+\frac{2(j-1)}{n_{1}-1},0):1\leq j\leq
n_{1}\\}\bigcup\\{(\cos\frac{(j-1)\pi}{n-n_{1}+1},\sin\frac{(j-1)\pi}{n-n_{1}+1}):2\leq
j\leq n-n_{1}+1\\}$

with the conditional constrained quantization error

$V_{n}=\frac{2}{3(2+\pi)}\Big{(}\frac{1}{(n_{1}-1){}^{2}}-6(n-n_{1}+1)\sin(\frac{\pi}{2n-2n_{1}+2})+3\pi\Big{)}.$

###### Proof.

Let $\alpha_{n}$ be a conditional optimal set of $n$-points for $P$. Let
$\text{card}(\alpha_{n}\cap L_{1})=n_{1}$ and $\text{card}(\alpha_{n}\cap
L_{2})=n_{2}$. Notice that $\alpha_{n}\cap L_{1}\cap
L_{2}=\\{(-1,0),(0,1)\\}$, and hence $n_{1}+n_{2}=n+2$ yielding
$n_{2}=n-n_{1}+2$. Thus, if $n_{1}$ is known, one can easily calculate
$n_{2}$, and then the conditional optimal set $\alpha_{n}$ of $n$-points and
the corresponding conditional constrained quantization error can be deduced by
Proposition 3.1, and they are given by

$\displaystyle\alpha_{n}=\\{(-1+\frac{2(j-1)}{n_{1}-1},0):1\leq j\leq
n_{1}\\}\bigcup\\{(\cos\frac{(j-1)\pi}{n_{2}-1},\sin\frac{(j-1)\pi}{n_{2}-1}):2\leq
j\leq n-n_{1}+1\\}$

with
$V_{n}=V_{n_{1},n-n_{1}+2}=\frac{2}{3(2+\pi)}\Big{(}\frac{1}{(n_{1}-1){}^{2}}-6(n-n_{1}+1)\sin(\frac{\pi}{2n-2n_{1}+2})+3\pi\Big{)}$.
Thus, the proof of the theorem is complete. ∎

For a given positive integer $n\geq 3$, to determine the positive integer
$n_{1}$ as mentioned in Theorem 3.5, we proceed as follows:

######  Definition 3.6.

Define the sequence $\\{a(n)\\}$ such that
$a(n)=\lfloor\frac{5(n+4)}{13}\rfloor$ for $n\geq 1$, i.e.,

$\displaystyle\\{a(n)\\}_{n=1}^{\infty}=$
$\displaystyle\\{1,2,2,3,3,3,4,4,5,5,5,6,6,6,7,7,8,8,8,9,9,10,10,10,11,11,11,12,12,13,13,13,\cdots\\},$

where $\lfloor x\rfloor$ represents the greatest integer not exceeding $x$.

The following algorithm helps us to determine the exact value of $n_{1}$
mentioned in Theorem 3.5.

###  3.7. Algorithm

Let $n\geq 3$, and let $V(n,n_{1}):=V_{n_{1},n-n_{1}+2}$, as given by Theorem
3.5, denote the distortion error if an optimal set $\alpha_{n}$ contains
$n_{1}$ elements from the base of the semicircular disc. Let $\\{a(n)\\}$ be
the sequence defined by Definition 3.6. Then, the algorithm runs as follows:

$(i)$ Write $n_{1}:=a(n)$.

$(ii)$ If $n_{1}=2$ go to step $(v)$, else step $(iii)$.

$(iii)$ If $V(n,n_{1}-1)<V(n,n_{1})$ replace $n_{1}$ by $n_{1}-1$ and go to
step $(ii)$, else step $(iv)$.

$(iv)$ If $V(n,n_{1}+1)<V(n,n_{1})$ replace $n_{1}$ by $n_{1}+1$ and return,
else step $(v)$.

$(v)$ End.

When the algorithm ends, then the value of $n_{1}$, obtained, is the exact
value of $n_{1}$ that an optimal set $\alpha_{n}$ contains from the base of
the semicircular disc.

######  Remark 3.8.

If $n=50$, then $a(n)=20$, and by the algorithm we also obtain $n_{1}=20$; if
$n=80$, then $a(n)=32$, and by the algorithm we also obtain $n_{1}=32$. If
$n=1200$, then $a(n)=463$, and by the algorithm, we obtain $n_{1}=468$; if
$n=2000$, then $a(n)=770$, and by the algorithm, we obtain $n_{1}=779$; and if
$n=3000$, then $a(n)=1155$, and by the algorithm, we obtain $n_{1}=1168$.
Thus, we see that with the help of the sequence and the algorithm, one can
easily determine the exact value of $n_{1}$ for any positive integer $n\geq
3$.

##  4\. Conditional unconstrained quantization for a uniform distribution on
an equilateral triangle

Let $\triangle OAB$ be an equilateral triangle with vertices
$O(0,0),A(1,0),B(\frac{1}{2},\frac{\sqrt{3}}{2})$. Let $L_{1},L_{2},L_{3}$ be
the sides $OA,AB$ and $BO$, respectively. Let $P$ be the uniform distribution
defined on the equilateral triangle $\triangle$ formed by the sides
$L_{1},L_{2},L_{3}$. Let $s$ represent the distance of any point on
$\triangle$ from the origin tracing along the boundary of the triangle in the
counterclockwise direction. Then, the points $O,A,B$ are, respectively,
represented by $s=0,s=1,s=2$. The probability density function (pdf) $f$ of
the uniform distribution $P$ is given by $f(s):=f(x_{1},x_{2})=\frac{1}{3}$
for all $(x_{1},x_{2})\in L_{1}\cup L_{2}\cup L_{3}$, and zero otherwise. The
sides $L_{1},L_{2},L_{3}$ are represented by the parametric equations as
follows:

$\displaystyle L_{1}$ $\displaystyle=\\{(t,0)\\},\quad
L_{2}=\Big{\\{}\Big{(}1-\frac{t}{2},\frac{\sqrt{3}t}{2}\Big{)}\Big{\\}},\quad
L_{3}=\Big{\\{}\Big{(}\frac{1-t}{2},\frac{\sqrt{3}(1-t)}{2}\Big{)}\Big{\\}},$

where $0\leq t\leq 1$. Again, $dP(s)=P(ds)=f(x_{1},x_{2})ds$. On each $L_{j}$
for $1\leq j\leq 3$, we have $(ds)^{2}=(dx_{1})^{2}+(dx_{2})^{2}=(dt)^{2}$
yielding $ds=dt$. In the definition of conditional unconstrained quantization
error, let $\beta:=\\{(0,0),(1,0),(\frac{1}{2},\frac{\sqrt{3}}{2})\\}$. Let
$L:=L_{1}\cup L_{2}\cup L_{3}$ which is the support of $P$. Upon the given set
$\beta$, the $n$th conditional unconstrained quantization errors are defined
for all $n\geq 3$. Notice that the conditional optimal set of three-points is
the set $\beta$ itself. In the sequel of this section, we investigate the
conditional optimal sets of $n$-points for $n\geq 4$.

Consider the following two affine transformations:

$T_{1}(x,y)=(-\frac{1}{2}x+1,\frac{\sqrt{3}}{2}x)\text{ and
}T_{2}(x,y)=(-\frac{1}{2}x+\frac{1}{2},-\frac{\sqrt{3}}{2}x+\frac{\sqrt{3}}{2}).$
(5)

Let us now give the following proposition, which plays a vital role in this
section. In this regard, one can also see [RR].

######  Proposition 4.1.

Let $\alpha_{n}$ be a conditional optimal set of $n$-points for $P$ for some
$n\geq 4$. Let $\text{card}(\alpha_{n}\cap L_{1})=n_{1}$,
$\text{card}(\alpha_{n}\cap L_{2})=n_{2}$ and $\text{card}(\alpha_{n}\cap
L_{3})=n_{3}$ with corresponding conditional quantization error
$V_{n}:=V_{n_{1},n_{2},n_{3}}(P)$ for some $n_{1},n_{2},n_{3}\geq 2$. Then,

$\displaystyle\alpha_{n}\cap L_{1}$
$\displaystyle=\Big{\\{}(\frac{j-1}{n_{1}-1},0):1\leq j\leq n_{1}\Big{\\}},$
$\displaystyle\alpha_{n}\cap L_{2}$
$\displaystyle=\Big{\\{}T_{1}(\frac{j-1}{n_{2}-1},0):1\leq j\leq
n_{2}\Big{\\}},\text{ and }$ $\displaystyle\alpha_{n}\cap L_{3}$
$\displaystyle=\Big{\\{}T_{2}(\frac{j-1}{n_{3}-1},0):1\leq j\leq
n_{3}\Big{\\}},$

with

$V_{n_{1},n_{2},n_{3}}(P)=\frac{1}{36}\Big{(}\frac{1}{(n_{1}-1)^{2}}+\frac{1}{(n_{2}-1)^{2}}+\frac{1}{(n_{3}-1)^{2}}\Big{)}.$

###### Proof.

By the hypothesis, $\text{card}(\alpha_{n}\cap L_{1})=n_{1}$ for some
$n_{1}\geq 2$. Let

$\alpha_{n}\cap L_{1}=\\{(a_{j},0):1\leq j\leq n_{1}\\},$ (6)

By Proposition 2.1, we have $a_{j}=\frac{j-1}{n_{1}-1}$ for $1\leq j\leq
n_{1}$ implying

$\alpha_{n}\cap L_{1}=\\{(\frac{j-1}{n_{1}-1},0):1\leq j\leq n_{1}\\}.$

Given $\text{card}(\alpha_{n}\cap L_{2})=n_{2}$. If
$\text{card}(\alpha_{n}\cap L_{1})=n_{2}$, then as before we have

$\alpha_{n}\cap L_{1}=\\{(\frac{j-1}{n_{2}-1},0):1\leq j\leq n_{2}\\}.$

Hence, $T_{2}$ being an affine transformation such that $T_{2}(L_{1})=L_{2}$,
we have

$\alpha_{n}\cap L_{2}=\\{T_{1}(\frac{j-1}{n_{2}-1},0):1\leq j\leq n_{2}\\}.$

Similarly, we have

$\alpha_{n}\cap L_{3}=\\{T_{2}(\frac{j-1}{n_{3}-1},0):1\leq j\leq n_{3}\\}.$

To find the quantization error, we proceed as follows. Let $V(P;\alpha_{n}\cap
L_{j})$ denote the distortion error contributed by the elements in
$\alpha_{n}\cap L_{j}$ for $j=1,2,3$. Then,

$V_{n_{1},n_{2},n_{3}}(P)=V(P;\alpha_{n}\cap L_{1})+V(P;\alpha_{n}\cap
L_{2})+V(P;\alpha_{n}\cap L_{3}).$

By Proposition 2.1, we have

$\displaystyle V(P;\alpha_{n}\cap
L_{1})=\frac{1}{36\left(n_{1}-1\right){}^{2}}.$

Due to rotational symmetry, we have

$\displaystyle V(P;\alpha_{n}\cap
L_{2})=\frac{1}{36\left(n_{2}-1\right){}^{2}}\text{ and }V(P;\alpha_{n}\cap
L_{3})=\frac{1}{36\left(n_{3}-1\right){}^{2}}.$

Hence,

$V_{n_{1},n_{2},n_{3}}(P)=\frac{1}{36}\Big{(}\frac{1}{(n_{1}-1)^{2}}+\frac{1}{(n_{2}-1)^{2}}+\frac{1}{(n_{3}-1)^{2}}\Big{)}.$

Thus, the proof of the proposition is complete. ∎

######  Note 4.2.

Let $\alpha_{n}(L_{j})$ be the set consisting of all the elements in
$\alpha_{n}\cap L_{j}$ except the right endpoint for each $j=1,2,3$. Then,

$\displaystyle\alpha_{n}(L_{1})$
$\displaystyle=\Big{\\{}(\frac{j-1}{n_{1}-1},0):1\leq j\leq n_{1}-1\Big{\\}},$
$\displaystyle\alpha_{n}(L_{2})$
$\displaystyle=\Big{\\{}T_{1}(\frac{j-1}{n_{2}-1},0):1\leq j\leq
n_{2}-1\Big{\\}},\text{ and }$ $\displaystyle\alpha_{n}(L_{3})$
$\displaystyle=\Big{\\{}T_{2}(\frac{j-1}{n_{3}-1},0):1\leq j\leq
n_{3}-1\Big{\\}}$

with $\text{card}(\alpha_{n}(L_{1}))=n_{1}-1$,
$\text{card}(\alpha_{n}(L_{2}))=n_{2}-1$, and
$\text{card}(\alpha_{n}(L_{3}))=n_{3}-1$. Notice that the sets
$\alpha_{n}(L_{j})$ are disjoints and $(n_{1}-1)+(n_{2}-1)+(n_{3}-1)=n$.

Figure 2. Optimal configuration of $n$-points for $4\leq n\leq 12$.

######  Lemma 4.3.

Let $n_{j}\in\mathbb{N}$ for $j=1,2,3$ be the numbers as defined in
Proposition 4.1. Then, $|n_{i}-n_{j}|=0$, or $1$ for $1\leq i\neq j\leq 3$.

###### Proof.

We first show that $|n_{1}-n_{2}|=0,\text{ or }1$. Write
$m:=(n_{1}-1)+(n_{2}-1)$. Now, the distortion error contributed by the $m$
elements in $\alpha_{1}(L_{1})\cup\alpha_{n}(L_{2})$ is given by

$\frac{1}{36}\Big{(}\frac{1}{(n_{1}-1)^{2}}+\frac{1}{(n_{2}-1)^{2}}\Big{)}.$

The above expression is minimum if $n_{1}-1\approx\frac{m}{2}$ and
$n_{2}-1\approx\frac{m}{2}$. Thus, we see that if $m=2k$ for some positive
integer $k$, then $n_{1}-1=n_{2}-1=k$, and if $m=2k+1$ for some positive
integer $k$, then either $(n_{1}-1=k+1$ and $n_{2}-1=k)$ or $(n_{1}-1=k$ and
$n_{2}-1=k+1)$ which yields the fact that $|n_{1}-n_{2}|=0,\text{ or }1$.
Similarly, we can show that $|n_{i}-n_{j}|=0$, or $1$ for any $1\leq i\neq
j\leq 3$. Thus, the proof of the lemma is complete. ∎

Let us now give the following theorem, which gives the main result in this
section.

######  Theorem 4.4.

Let $\alpha_{n}$ be a conditional optimal set of $n$-points for $P$ for some
$n\geq 4$. Then,

$\displaystyle\alpha_{n}$
$\displaystyle=\Big{\\{}(\frac{j-1}{n_{1}-1},0):1\leq j\leq
n_{1}-1\Big{\\}}\bigcup\Big{\\{}T_{1}(\frac{j-1}{n_{2}-1},0):1\leq j\leq
n_{2}-1\Big{\\}}$
$\displaystyle\qquad\qquad\bigcup\Big{\\{}T_{2}(\frac{j-1}{n_{3}-1},0):1\leq
j\leq n_{3}-1\Big{\\}}$

with the conditional unconstrained quantization error

$V_{n}=\frac{1}{36}\Big{(}\frac{1}{(n_{1}-1)^{2}}+\frac{1}{(n_{2}-1)^{2}}+\frac{1}{(n_{3}-1)^{2}}\Big{)},$

where $n_{1},n_{2},n_{3}$ are given as follows: For some $k\in\mathbb{N}$, if
$n=3k$, then $n_{1}-1=n_{2}-1=n_{3}-1=k$; if $n=3k+1$, then $n_{1}-1=k+1$ and
$n_{2}-1=n_{3}-1=k$; and if $n=3k+2$, then $n_{1}-1=n_{2}-1=k+1$ and
$n_{3}-1=k$.

###### Proof.

Let $\alpha_{n}(L_{j})$ be the sets defined by Note 4.2 for $j=1,2,3$. Notice
that for $j=1,2,3,$ the sets $\alpha_{n}(L_{j})$ are nonempty, and so we can
find three positive integers $n_{j}\geq 2$ as defined in Proposition 4.1 such
that $\text{card}(\alpha_{n}(L_{j}))=n_{j}-1$. Since

$\alpha_{n}=\bigcup_{j=1}^{3}\alpha_{n}(L_{j}),$

the expression for $\alpha_{n}$ follows by Note 4.2. The expression for the
conditional constrained quantization error $V_{n}$ follows from Proposition
4.1. By Lemma 4.3, it follows that for some $k\in\mathbb{N}$, if $n=3k$, then
$n_{1}-1=n_{2}-1=n_{3}-1=k$; if $n=3k+1$, then $n_{1}-1=k+1$ and
$n_{2}-1=n_{3}-1=k$; and if $n=3k+2$, then $n_{1}-1=n_{2}-1=k+1$ and
$n_{3}-1=k$. Thus, the proof of the theorem is complete. ∎

Using Theorem 4.4, the following example can be obtained.

######  Example 4.5.

A conditional optimal set of four-points is given by
$\\{(0,0),(\frac{1}{2},1),(1,0),(\frac{1}{2},\frac{\sqrt{3}}{2})\\}$ with
conditional constrained quantization error $V_{4}=\frac{1}{16}$; a conditional
optimal set of five-points is given by
$\\{(0,0),(\frac{1}{2},1),(1,0),(\frac{1}{4},\frac{\sqrt{3}}{4}),(\frac{1}{2},\frac{\sqrt{3}}{2})\\}$
with conditional constrained quantization error $V_{5}=\frac{1}{24}$; and so
on (see Figure 2).

######  Theorem 4.6.

The conditional unconstrained quantization dimension $D(P)$ of the probability
measure $P$ exists, and $D(P)=1$, and the $D(P)$-dimensional unconstrained
quantization coefficient for $P$ exists as a finite positive number and equals
$\frac{3}{4}$.

###### Proof.

For $n\in\mathbb{N}$ with $n\geq 4$, let $\ell(n)$ be the unique natural
number such that $3{\ell(n)}\leq n<3({\ell(n)+1})$. Then,
$V_{3(\ell(n)+1)}\leq V_{n}\leq V_{3{\ell(n)}}$. We can take $n$ large enough
so that $V_{3{\ell(n)}}<1$. Then,

$0<-\log V_{3{\ell(n)}}\leq-\log V_{n}\leq-\log V_{3({\ell(n)+1})}$

yielding

$\frac{2\log 3\ell(n)}{-\log V_{3({\ell(n)+1})}}\leq\frac{2\log n}{-\log
V_{n}}\leq\frac{2\log 3(\ell(n)+1)}{-\log V_{3{\ell(n)}}}.$

Notice that

$\displaystyle\lim_{n\to\infty}\frac{2\log 3\ell(n)}{-\log
V_{3({\ell(n)+1})}}$ $\displaystyle=\lim_{n\to\infty}\frac{2\log
3\ell(n)}{-\log\frac{1}{12(\ell(n)+1)^{2}}}=1,\text{ and }$
$\displaystyle\lim_{n\to\infty}\frac{2\log 3(\ell(n)+1)}{-\log
V_{3{\ell(n)}}}$ $\displaystyle=\lim_{n\to\infty}\frac{2\log
3(\ell(n)+1)}{-\log\frac{1}{12(\ell(n))^{2}}}=1.$

Hence, $\lim_{n\to\infty}\frac{2\log n}{-\log V_{n}}=1$, i.e., the conditional
unconstrained quantization dimension $D(P)$ of the probability measure $P$
exists and $D(P)=1$. Since

$\displaystyle\lim_{n\to\infty}n^{2}V_{n}\geq\lim_{n\to\infty}(3\ell(n))^{2}V_{3(\ell(n)+1)}=\lim_{n\to\infty}(3\ell(n))^{2}\frac{1}{12(\ell(n)+1)^{2}}=\frac{3}{4},\text{
and }$
$\displaystyle\lim_{n\to\infty}n^{2}V_{n}\leq\lim_{n\to\infty}(3(\ell(n)+1))^{2}V_{3\ell(n)}=\lim_{n\to\infty}(3(\ell(n)+1))^{2}\frac{1}{12(\ell(n))^{2}}=\frac{3}{4},$

by the squeeze theorem, we have $\lim_{n\to\infty}n^{2}V_{n}=\frac{3}{4}$,
i.e., the $D(P)$-dimensional unconstrained quantization coefficient for $P$
exists as a finite positive number and equals $\frac{3}{4}$. Thus, the proof
of the theorem is complete. ∎

##  5\. Observations

To distinguish, let $V_{c,n}(P)$ denote the $n$th conditional quantization
error for $n$-points, and $V_{n}(P)$ denote the $n$th unconstrained
quantization error for $n$-means for a Borel probability measure $P$. Also, to
distinguish we denote the conditional quantization dimension by $D_{c}(P)$ and
the unconstrained quantization dimension by $D(P)$, respectively.

In the following remarks, we give some observations.

######  Remark 5.1.

Conditional optimal sets of $n$-points for a Borel probability measure can
exist with $V_{c,n}>V_{n}$ for all $n\in\mathbb{N}$ with $n\geq\ell$, where
$\ell$ is the number of elements in the conditional set $\beta$. In this
regard, we give the following example: Let $P$ be a Borel probability measure
on $\mathbb{R}$, which is uniform on its support $[0,1]$, and let
$\beta:=\\{0\\}$ be the conditional set. Then, by Proposition 2.2, we see that
the conditional optimal sets of $n$-points for all $n\geq 1$ are given by

$\Big{\\{}\frac{2(j-1)}{2n-1}:1\leq j\leq n\Big{\\}}\text{ with
}V_{c,n}(P)=\frac{1}{3(2n-1)^{2}}.$

On the other hand, $P$ is a uniform distribution, the optimal sets of
$n$-means are given by

$\Big{\\{}\frac{2j-1}{2n}:1\leq j\leq n\Big{\\}}\text{ with
}V_{n}(P)=\frac{1}{12n^{2}}.$

Thus, we see that for all $n\geq 1$, the conditional optimal sets of
$n$-points exist in this case and

$V_{c,n}=\frac{1}{3(2n-1)^{2}}=\frac{1}{3(n+n-1)^{2}}>\frac{1}{3(n+n)^{2}}=\frac{1}{12n^{2}}>V_{n}.$

Moreover, we see that $D_{c}(P)=D(P)=1$.

######  Remark 5.2.

In unconstrained quantization, for a Borel probability measure with infinite
support, an optimal set of $n$-means contains exactly $n$ elements. It is not
true in constrained quantization (see [PR1]). In constrained quantization for
a Borel probability measure with infinite support, an optimal set of
$n$-points for all $n$ may not contain exactly $n$ elements. Let $k$ be the
largest positive integer for which an optimal set of $k$-points contains
exactly $k$ elements. In this case, there is no conditional optimal set of
$n$-points for any $n\geq(k+1)$.

######  Remark 5.3.

Optimal sets of $n$-points for all positive integers can exist, but a
conditional optimal set may not exist for all $n$. In this regard, we give the
following example:

Let $P$ be a probability measure on $\mathbb{R}^{2}$ with its support the
closed interval $\\{(t,0):0\leq t\leq 1\\}$, and let $P$ be uniform on its
support. Let $\beta:=\\{(0,\frac{1}{100})\\}$ be the conditional set. Let
$\alpha:=\\{(0,\frac{1}{100})\\}\cup\\{(t_{j},0):1\leq j\leq n\\}$ be a
conditional optimal set of $(n+1)$-points such that
$t_{1}<t_{2}<\cdots<t_{n}$. Let the boundary of the Voronoi regions of
$(0,\frac{1}{100})$ and $(t_{1},0)$ intersect the support of $P$ at the point
$(d,0)$. Clearly $0\leq d<\frac{t_{1}}{2}$. Also, notice that
$t_{j}=d+\frac{(2j-1)(1-d)}{2n}$ for $j=1,2,\cdots n$. Then, the distortion
error is given by

$\displaystyle V(P;\alpha)$ $\displaystyle=\text{distortion error contributed
by }(0,\frac{1}{100})+\text{distortion error contributed by all }t_{j}$
$\displaystyle=\int_{0}^{d}\rho((t,0),(0,\frac{1}{100}))\,dt+\frac{(1-d)^{3}}{12n^{2}}$
$\displaystyle=\frac{d^{3}}{3}+\frac{d}{10000}+\frac{(1-d)^{3}}{12n^{2}}$

the minimum value of which is

$\frac{n\left(4n\left(n\sqrt{10001-4n^{2}}+249925\right)-10001\sqrt{10001-4n^{2}}\right)+250075}{750000\left(1-4n^{2}\right)^{2}}$

and it occurs when $d=\frac{50-n\sqrt{10001-4n^{2}}}{50-200n^{2}}$. Notice
that $d$ is a decreasing sequence of real numbers for $n=1,2,\cdots,50$, and
becomes imaginary if $n\geq 51$. In fact, we see that $d>0$ if $1\leq n<50$,
and $d=0$ if $n=50$. Hence, we can say that the conditional optimal sets of
$n$-points exist for $1\leq n\leq 49$, and it does not exist if $n\geq 50$.
But, notice that the optimal sets of $n$-points for all $n\geq 50$ still exist
which are given by $\\{(\frac{2j-1}{2n},0):1\leq j\leq n\\}$ with quantization
error $\frac{1}{12n^{2}}$.

######  Remark 5.4.

In Section 4, we have seen that for a uniform distribution defined on the
equilateral triangle, the conditional unconstrained quantization dimension is
one, and the conditional unconstrained quantization coefficient is
$\frac{3}{4}$. These values are the same as the unconstrained quantization
dimension and the unconstrained quantization coefficient for the uniform
distribution on the equilateral triangle (see [RR]). In the next section, we
give a general proof to show that the lower and upper quantization dimensions
and the lower and upper quantization coefficients for a Borel probability
measure do not depend on the underlying conditional set.

##  6\. Some important properties

In this section, in the first theorem, we show that in unconstrained
quantization, the union of the optimal sets of $n$-means is dense in the
support of $P$. In the next theorem, we show that in both constrained and
unconstrained quantization, the lower and upper quantization dimensions and
the lower and upper quantization coefficients for a Borel probability measure
do not depend on the conditional set of the conditional quantization.

######  Theorem 6.1.

Let $P$ be a Borel probability measure on $\mathbb{R}^{k}$. Let $\alpha_{n}$
be the optimal sets of $n$-means for $P$ for all $n\in\mathbb{N}$. Then,
$\mathop{\cup}\limits_{n=1}^{\infty}\alpha_{n}$ is dense in the support of
$P$.

###### Proof.

Let $x\in\text{Supp}(P)$. Our aim is to show that for each $\epsilon>0$,
$B(x,\epsilon)\cap\mathop{\bigcup}\limits_{n=1}^{\infty}\alpha_{n}\neq\emptyset.$
We prove it by contradiction. Let there exists an $\epsilon>0$ such that
$B(x,\epsilon)\cap\bigcup_{n=1}^{\infty}\alpha_{n}=\emptyset.$ Then,

$\displaystyle V_{n,r}(P)=\int
d(x,\alpha_{n})^{r}dP(x)\geq\epsilon^{r}P(B(x,\epsilon)\cap\text{Supp}(P)).$

We claim that $P(B(x,\epsilon)\cap\text{Supp}(P))>0$. Assume that
$P(B(x,\epsilon)\cap\text{Supp}(P))=0$. This implies that
$x\notin\text{Supp}(P)$, which is a contradiction. Thus
$P(B(x,\epsilon)\cap\text{Supp}(P))>0$. Therefore, we get

$\lim_{n\to\infty}V_{n,r}(P)\geq\epsilon^{r}P(B(x,\epsilon)\cap\text{Supp}(P))>0,$

which contradicts the fact that $\lim_{n\to\infty}V_{n,r}(P)=0$. This implies
that for each $\epsilon>0$, we have
$B(x,\epsilon)\cap\mathop{\bigcup}\limits_{n=1}^{\infty}\alpha_{n}\neq\emptyset.$
Thus, $\mathop{\bigcup}\limits_{n=1}^{\infty}\alpha_{n}$ is dense in the
support of $P$. This completes the proof. ∎

######  Theorem 6.2.

In both constrained and unconstrained quantization, the lower and upper
quantization dimensions and the lower and upper quantization coefficients for
a Borel probability measure do not depend on the conditional set.

###### Proof.

Since the unconstrained quantization is a special case of constrained
quantization, we give the proof of the theorem for constrained cases only. Let
$\beta\subset\mathbb{R}^{k}$ be a conditional set with
$\text{card}(\beta)=\ell$ for some $\ell\in\mathbb{N}$. Let $V_{n,r}(P)$ and
$V_{c,n,r}(P)$ denote the $n$th constrained and the $n$th conditional
constrained quantization errors, respectively. Take $n>\ell$. Then, one can
easily see that

$\displaystyle V_{n,r}(P)\leq V_{c,n,r}(P)\leq V_{n-\ell,r}(P).$ (7)

By the previous inequalities and application of the squeeze theorem, we have
$V_{\infty,r}(P)=V_{c,\infty,r}(P)$. Thus, by (7), we have

$\displaystyle V_{n,r}(P)-V_{\infty,r}(P)\leq
V_{c,n,r}(P)-V_{c,\infty,r}(P)\leq V_{n-\ell,r}(P)-V_{\infty,r}(P).$ (8)

Again, by the previous inequalities and application of the squeeze theorem, we
obtain that the lower and upper constrained quantization coefficients are the
same in both constrained and conditional constrained cases. Choose
$n\in\mathbb{N}$ large enough such that $V_{n-\ell,r}(P)-V_{\infty,r}(P)<1$.
Then, by Equation (8), we get

$\frac{r\log n}{-\log(V_{n,r}(P)-V_{\infty,r}(P))}\leq\frac{r\log
n}{-\log(V_{c,n,r}(P)-V_{c,\infty,r}(P))}\leq\frac{r\log
n}{-\log(V_{n-\ell,r}(P)-V_{\infty,r}(P))}.$

Thus, by the squeeze theorem, we see that the lower and upper constrained
quantization dimensions are the same in both constrained and conditional
constrained cases. This completes the proof of the theorem. ∎

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