Conditional Quantization for Some Discrete Distributions

Abstract

Quantization for a Borel probability measure refers to the idea of estimating a given probability by a discrete probability with support containing a finite number of elements. If in the quantization some of the elements in the finite support are preselected, then the quantization is called a conditional quantization. In this paper, we have determined the conditional quantization, first for two different finite discrete distributions with a same conditional set, and for a finite discrete distribution with two different conditional sets. Next, we have determined the conditional and unconditional quantization for an infinite discrete distribution with support $\{{\displaystyle \frac{1}{2^n}}:n\in \mathbb{N}\}$. We have also investigated the conditional quantization for an infinite discrete distribution with support $\{{\displaystyle \frac{1}{n}}:n\in \mathbb{N}\}$. At the end of the paper, we have given a conjecture and discussed about some open problems based on the conjecture.

1. Introduction

Quantization is a vast area of research with a huge application in engineering and technology (see [1,2,3]). For surveys on the subject and comprehensive lists of references to the literature, one can see [4,5,6,7]. For mathematical treatment of quantization, one can see [8]). Recently, Pandey and Roychowdhury introduced the concepts of constrained quantization and the conditional quantization (see [9,10,11]). A quantization without a constraint is known as an unconstrained quantization, which traditionally in the literature is known as quantization. On unconstrained quantization, there is a number of papers written by many authors, for example, one can see [1,2,3,6,7,8,12,13,14,15,16,17,18,19]. This paper mainly deals with conditional quantization in the unconstrained scenario.

Definition 1.

Let P be a Borel probability measure on ${\mathbb{R}}^k$ equipped with a Euclidean metric d induced by the Euclidean norm $\parallel ·\parallel$. Let $\beta \subset {\mathbb{R}}^k$ be given with $card\left(\beta \right)=\mathcal{l}$ for some $\mathcal{l}\in \mathbb{N}$. Then, for $n\in \mathbb{N}$ with $n\ge \mathcal{l}$, the nth conditional quantization error for P with respect to the conditional set β, is defined as

$$V_n:=V_n\left(P\right)=\underset{\alpha }{inf}\left\{\int \underset{a\in \alpha \cup \beta }{min}d{(x,a)}^{2}dP\left(x\right):\mathrm{card}\left(\alpha \right)\le n-\mathcal{l}\right\},$$

(1)

where $card\left(A\right)$ represents the cardinality of the set A.

We assume that $\int d{(x,0)}^{2}dP\left(x\right)<\infty$ to make sure that the infimum in (1) exists (see [9]). For a finite set $\gamma \subset {\mathbb{R}}^k$ and $a\in \gamma$, by $M\left(a\right|\gamma )$ we denote the set of all elements in ${\mathbb{R}}^k$ that are nearest to a among all the elements in $\gamma$, i.e., $M\left(a\right|\gamma )=\{x\in {\mathbb{R}}^k:d(x,a)=\underset{b\in \gamma }{min}d(x,b)\}.$ $M\left(a\right|\gamma )$ is called the Voronoi region in ${\mathbb{R}}^k$ generated by $a\in \gamma$.

Definition 2.

A set $\alpha \cup \beta$, where $P\left(M\right(b|\alpha \cup \beta ))>0$ for $b\in \beta$, for which the infimum in $V_n$ exists and contains no less than ℓ elements, and no more than n elements is called a conditional optimal set of n-points for P with respect to the conditional set β.

Remark 1.

By Definition 1, we see that the nth conditional quantization error $V_n\left(P\right)$ forms a decreasing sequence, i.e., $V_{n+1}\left(P\right)\le V_n\left(P\right)$ for all $n\in \mathbb{N}$. By Definition 2, we see that a conditional optimal set of n-points for $n\ge \mathcal{l}$ contains at least ℓ elements the Voronoi regions of which with respect to the set $\alpha \cup \beta$ have positive probability. Notice that if $N\ge \mathcal{l}$ is the maximum number of elements in an optimal set $\alpha \cup \beta$ of n-points the Voronoi regions of which with respect to the set $\alpha \cup \beta$ have positive probability, then the sequence ${\left\{V_k\left(P\right)\right\}}_{k=\mathcal{l}}^N$ is strictly decreasing. For more details in this regard, one can see Proposition 1.4 and its corollary in [9].

Let $V_n\left(P\right)$ be a strictly decreasing sequence and write $V_{\infty }\left(P\right):=\underset{n\to \infty }{lim}{V}_n\left(P\right)$. Then, the number

$$\begin{array}{c}\hfill D\left(P\right):=\underset{n\to \infty }{lim}{\displaystyle \frac{2logn}{-log(V_n\left(P\right)-V_{\infty }\left(P\right))}}\end{array}$$

(2)

if it exists, is called the conditional quantization dimension of P and is denoted by $D\left(P\right)$. When quantizing a probability measure with infinite support with a discrete set, there will be an error or difference between the two. The quantization dimension measures how fast this error decreases as the number of points in the discrete approximation increases. In other words, conditional quantization dimension quantifies the rate at which the error in quantizing a probability measure with infinite support by a conditional optimal set of n-points decreases as the number of discrete points in the conditional optimal set increases. A higher quantization dimension suggests a faster decrease in error. For any $\kappa >0$, the number

$$\begin{array}{c}\hfill \underset{n}{lim}{n}^{{\displaystyle \frac{2}{\kappa }}}(V_n\left(P\right)-V_{\infty }\left(P\right)),\end{array}$$

(3)

if it exists, is called the κ-dimensional conditional quantization coefficient for P. The quantization coefficient, if it exists, gives the asymptotic behavior how the quantization error converges. Notice that in unconditional quantization the term $V_{\infty }=0$. Hence, if in (2) and (3) the term $V_{\infty }=0$, then they are reduced to the corresponding definitions of the quantization dimension and the quantization coefficient in unconditional scenario. In this paper, we have mainly investigated the conditional quantization for different discrete distributions.

Delineation

The arrangement of this paper is as follows: In Section 2, we first give the basic preliminaries. In Section 3, we have determined the conditional quantization for four different discrete distributions with support $\{1,2,3,4,5,6\}$: in Section 3.1 and Section 3.2, it was determined for two different discrete distributions with the same conditional set $\beta :=\{4,5,6\}$; in Section 3.3 and Section 3.4 it was determined for the same discrete distribution with the conditional sets $\beta :=\left\{{\displaystyle \frac{1}{6}}\right\}$ and $\beta :=\left\{{\displaystyle \frac{1}{4}}\right\}$, respectively. In Section 3, we also give two important remarks, Remark 3 and Remark 4. In Section 4, we have determined the conditional and unconditional quantization for an infinite discrete distribution with support $\{{\displaystyle \frac{1}{2^n}}:n\in \mathbb{N}\}$. In Section 5, we have investigated the conditional quantization for an infinite discrete distribution with support $\{{\displaystyle \frac{1}{n}}:n\in \mathbb{N}\}$. At the end of the paper, we have given a conjecture and discussed about some open problems based on the conjecture.

2. Preliminaries

In unconditional quantization, i.e., if there is no conditional set, then the following proposition is well known (see [1,8]).

Proposition 1.

Let α be an optimal set of n-points for P, and $a\in \alpha$. Then,

$$\left(i\right) P\left(M\right(a\left|\alpha \right))>0 \left(ii\right) P(\partial M(a\left|\alpha \right))=0 \left(iii\right) a=E(X:X\in M(a\left|\alpha \right)),$$

where X is a random variable with distribution P.

Thus, by the above proposition, we see that in unconditional quantization, the elements in an optimal set of n-points are the means in their own Voronoi regions. This fact is not true in conditional quantization. In conditional quantization, if $\beta$ is the conditional set and if $\alpha \cup \beta$ is a conditional optimal set of n-points, then the elements in $\beta$ may not be the means of their own Voronoi regions, but the elements in $\alpha$, which are not in $\beta$, are the means of their own Voronoi regions; for details one can see [11].

Let $\alpha$ be a discrete set. Let P be a Borel probability measure on $\mathbb{R}$ equipped with a metric d induced by the Euclidean norm $\parallel ·\parallel$ with probability mass function f. Then, $V(P;\alpha )$ means the distortion error for P with respect to the set $\alpha$, i.e.,

$$V(P;\alpha ):=\sum _{x\in \mathbb{R}}\underset{a\in \alpha }{min}d{(x,a)}^2 f\left(x\right).$$

Let P be a Borel probability measure on $\mathbb{R}$ with support $\{a_1,a_2,\cdots ,a_n\}$ for some $n\in \mathbb{N}$ associated with a probability mass function (pmf) f. Let X be an associated random variable. For $k,\mathcal{l}\in \mathbb{N}$, where $k\le \mathcal{l}$, write

$$[k,\mathcal{l}]:=\{a_x:x\in \mathbb{N} \mathrm{and} k\le x\le \mathcal{l}\}.$$

(4)

Further, write

$$Av[k,\mathcal{l}]:=E\left(X:X\in [k,\mathcal{l}]\right)={\displaystyle \frac{{\sum }_{n=k}^{\mathcal{l}}{a}_{x}f\left(x\right)}{{\sum }_{n=k}^{\mathcal{l}}f\left(x\right)}} \mathrm{and} Er[k,\mathcal{l}]:=\sum _{x=k}^{\mathcal{l}}f\left(x\right){\left(a_x-Av[k,\mathcal{l}]\right)}^2.$$

(5)

Let P be a uniform distribution defined on the set $\{1,2,3,4,5,6\}$. Then, the random variable X associated with the probability distribution is a discrete random variable with probability mass function f given by

$$f\left(x\right)=P(X:X=x)={\displaystyle \frac{1}{6}}, \mathrm{where} x\in \{1,2,3,4,5,6\}.$$

Let us take the conditional set as $\beta :=\left\{1.5\right\}$. Then, the conditional optimal set of n-points for all $n\ge 1$ exists. If ${\alpha }_n$ is a conditional optimal set of n-points with conditional quantization error $V_n$ for P, then it can be shown that

$$\begin{array}{cc}\hfill {\alpha }_1& =\left\{1.5\right\} \mathrm{with} V_1={\displaystyle \frac{83}{12}}, {\alpha }_2=\{1.5,Av[3,6]\} \mathrm{with} V_2={\displaystyle \frac{11}{12}},\hfill \\ \hfill {\alpha }_3& =\{1.5,Av[3,4],Av[5,6]\} \mathrm{with} V_3={\displaystyle \frac{1}{4}},\hfill \\ \hfill {\alpha }_4& =\{1.5,3,4,Av[5,6]\} \mathrm{or} \{1.5,Av[3,4],5,6\} \mathrm{with} V_4={\displaystyle \frac{1}{6}},\hfill \\ \hfill {\alpha }_5& =\{1.5,3,4,5,6\} \mathrm{with} V_5={\displaystyle \frac{1}{12}}, \mathrm{and} \hfill \\ \hfill {\alpha }_6& =\{1,1.5,3,4,5,6\} \mathrm{or} \{1.5,2,3,4,5,6\} \mathrm{with} V_6={\displaystyle \frac{1}{24}}.\hfill \end{array}$$

Remark 2.

Conditional optimal sets are not unique. For example, in the above, the sets ${\alpha }_4$ and ${\alpha }_6$ can be chosen in two different ways.

In the following sections we give our main results.

3. Conditional Quantization for Nonuniform Finite Discrete Distributions

Recall the notations given by (4) and (5). This section has four subsections.

3.1. Conditional Quantization for the Probability Distribution P with pmf $f\left(j\right)={\displaystyle \frac{1}{2^j}}$ for $1\le j\le 5$ and $f\left(6\right)={\displaystyle \frac{1}{2^5}}$ Where the Conditional Set Is $\beta :=\{4,5,6\}$

Notice that $\beta$ is the conditional optimal set of three-points with conditional quantization error $V_3={\displaystyle \frac{45}{8}}$, and the conditional optimal set of six-points is the support of P with conditional quantization error $V_6=0$. The following two propositions give the main results in this subsection.

Proposition 2.

The conditional optimal set of four-points is given by $\left\{Av\right[1,2],4,5,6\}$ with conditional quantization error $V_4={\displaystyle \frac{7}{24}}$.

Proof. 

Notice that $Av[1,2]={\displaystyle \frac{4}{3}}$. Let us consider the set $\gamma :=\left\{Av\right[1,2],4,5,6\}$. Since $2<{\displaystyle \frac{1}{2}}(Av[1,2]+4)={\displaystyle \frac{8}{3}}<3$, the distortion error due to the set $\gamma$ is given by

$$\begin{array}{c}\hfill \sum _{j=1}^{6}f\left(j\right)\underset{a\in \gamma }{min}{(j-a)}^2=\sum _{j=1}^{2}f\left(j\right){(j-Av[1,2])}^2+f\left(3\right){(3-4)}^2={\displaystyle \frac{7}{24}}.\end{array}$$

Since $V_4$ is the conditional quantization error for four-points, we have $V_4\le {\displaystyle \frac{7}{24}}$. Let $\alpha :=\{a_1,4,5,6\}$ be a conditional optimal set of four-points. Without any loss of generality, we can assume that $1\le a_1<4$. Notice that the Voronoi region of $a_1$ must contain 1. Suppose that the Voronoi region of $a_1$ contains the set $\{1,2,3\}$. Then,

$$V_4\ge \sum _{j=1}^{3}f\left(j\right){(j-Av[1,3])}^2={\displaystyle \frac{13}{28}}>V_4,$$

which gives a contradiction. Hence, we can assume that the Voronoi region of $a_1$ does not contain $\{1,2,3\}$. Next, suppose that the Voronoi region of $a_1$ contains only the element 1. Then, the Voronoi region of 4 must contain 2 and 3, and so

$$V_4>\sum _{j=2}^{3}f\left(j\right){(j-4)}^2={\displaystyle \frac{9}{8}}>V_4,$$

which yields a contradiction. Hence, we can assume that the Voronoi region of $a_1$ contains only the elements 1 and 2. On the other hand, 3 is contained in the Voronoi region of 4. Hence,

$$a_1=Av[1,2],$$

i.e., the conditional optimal set of four-points is given by $\left\{Av\right[1,2],4,5,6\}$ with conditional quantization error $V_4={\displaystyle \frac{7}{24}}$. Thus, the proof of the proposition is complete. □

Proposition 3.

The conditional optimal set of five-points is given by $\{1,Av[2,3],4,5,6\}$ with conditional quantization error $V_5={\displaystyle \frac{1}{12}}$.

Proof. 

Let $\gamma :=\{1,Av[2,3],4,5,6\}$. The distortion error due to the set $\gamma$ is given by

$$\begin{array}{c}\hfill \sum _{j=1}^{6}f\left(j\right)\underset{a\in \gamma }{min}{(j-a)}^2=\sum _{j=2}^{3}f\left(j\right){(j-Av[2,3])}^2={\displaystyle \frac{1}{12}}.\end{array}$$

Since $V_5$ is the quantization error for five-points, we have $V_5\le {\displaystyle \frac{1}{12}}$. Let $\{a_1,a_2,4,5,6\}$ be a conditional optimal set of five-points such that $1\le a_1<a_2<4$. If $a_1=Av[1,2]$ and $a_2=3$, then

$$V_5\ge \sum _{j=1}^{2}f\left(j\right){(j-Av[1,2])}^2={\displaystyle \frac{1}{6}},$$

which is a contradiction. If $a_1=1$ and $a_2=2$, then the element 3 is contained in the Voronoi region of 4 yielding

$$V_5\ge f\left(3\right){(x-4)}^2={\displaystyle \frac{1}{8}}>V_5,$$

which leads to a contradiction. Hence, we can conclude that $a_1=1$ and $a_2=a[2,3]$. Thus, the conditional optimal set of five-points is given by $\{1,a[2,3],4,5,6\}$ with conditional quantization error $V_5={\displaystyle \frac{1}{12}}$, which is the proposition. □

3.2. Conditional Quantization for the Probability Distribution P with pmf $f\left(j\right)={\displaystyle \frac{1}{10}}$ for $j\in \{1,2,4,5,6\}$ and $f\left(3\right)={\displaystyle \frac{1}{2}}$ Where the Conditional Set Is $\beta :=\{4,5,6\}$

Notice that in this subsection $\beta$ is the conditional optimal set of three-points with conditional quantization error $V_3={\displaystyle \frac{9}{5}}$, and the conditional optimal set of six-points is the support of P with conditional quantization error $V_6=0$. The following two propositions give the main results in this subsection.

Proposition 4.

The conditional optimal set of four-points is given by $\left\{Av\right[1,3],4,5,6\}$ with conditional quantization error $V_4={\displaystyle \frac{13}{35}}$.

Proof. 

Notice that $Av[1,3]={\displaystyle \frac{18}{7}}$. Let us consider the set $\gamma :=\left\{Av\right[1,3],4,5,6\}$. Since $3<{\displaystyle \frac{1}{2}}(Av[1,3]+4)={\displaystyle \frac{23}{7}}<4$, the distortion error due to the set $\gamma$ is given by

$$\begin{array}{c}\hfill \sum _{j=1}^{6}f\left(j\right)\underset{a\in \gamma }{min}{(j-a)}^2=\sum _{j=1}^{3}f\left(j\right){(j-Av[1,3])}^2={\displaystyle \frac{13}{35}}.\end{array}$$

Since $V_4$ is the conditional quantization error for four-points, we have $V_4\le {\displaystyle \frac{13}{35}}$. Let $\alpha :=\{a_1,4,5,6\}$ be a conditional optimal set of four-points. Without any loss of generality, we can assume that $1\le a_1<4$. Notice that the Voronoi region of $a_1$ must contain 1. If the elements 2 and 3 are contained in the Voronoi region of 4, then

$$V_4=\sum _{j=2}^{3}f\left(j\right){(j-4)}^2={\displaystyle \frac{9}{10}}>V_4,$$

which leads to a contradiction. Hence, we can assume that the Voronoi region of $a_1$ also contains 2. Suppose that the Voronoi region of $a_1$ contains only the set $\{1,2\}$. Then,

$$V_4\ge \sum _{j=1}^{2}f\left(j\right){(j-Av[1,2])}^2+f\left(3\right){(3-4)}^2={\displaystyle \frac{11}{20}}>V_4,$$

which gives a contradiction. Hence, we can assume that the Voronoi region of $a_1$ must contain $\{1,2,3\}$. Hence,

$$a_1=Av[1,3],$$

i.e., the conditional optimal set of four-points is given by $\left\{Av\right[1,3],4,5,6\}$ with conditional quantization error $V_4={\displaystyle \frac{13}{35}}$. Thus, the proof of the proposition is complete. □

Proposition 5.

The conditional optimal set of five-points is given by $\left\{Av\right[1,2],3,4,5,6\}$ with conditional quantization error $V_5={\displaystyle \frac{1}{20}}$.

Proof. 

Let us consider the set $\gamma :=\left\{Av\right[1,2],3,4,5,6\}$. Since $2<{\displaystyle \frac{1}{2}}(Av[1,2]+3)={\displaystyle \frac{9}{4}}<3$, the distortion error due to the set $\gamma$ is given by

$$\begin{array}{c}\hfill \sum _{j=1}^{6}f\left(j\right)\underset{a\in \gamma }{min}{(j-a)}^2=\sum _{j=1}^{2}f\left(j\right){(j-Av[1,2])}^2={\displaystyle \frac{1}{20}}.\end{array}$$

Since $V_5$ is the conditional quantization error for five-points, we have $V_5\le {\displaystyle \frac{1}{20}}$. Let $\alpha :=\{a_1,a_2,4,5,6\}$ be a conditional optimal set of five-points such that $1\le a_1<a_2<4$. Notice that the Voronoi region of $a_1$ must contain 1, and the Voronoi region of $a_2$ must contain 3. Suppose that the Voronoi region of $a_2$ also contains 2. Then,

$$V_5\ge \sum _{j=2}^{3}f\left(j\right){(j-Av[2,3])}^2={\displaystyle \frac{1}{12}}>V_5,$$

which gives a contradiction. Hence, we can assume that the Voronoi region of $a_2$ contains only the element 3, which yields the fact that the Voronoi region of $a_1$ contains only the elements 1 and 2. Hence,

$$a_1=Av[1,2] \mathrm{and} a_2=2,$$

i.e., the conditional optimal set of five-points is given by $\left\{Av\right[1,2],3,4,5,6\}$ with conditional quantization error $V_5={\displaystyle \frac{1}{20}}$. Thus, the proof of the proposition is complete. □

Remark 3.

From Section 3.1 and Section 3.2, we see that though the underlying support and the conditional sets remain constant, the conditional optimal sets and the corresponding conditional quantization errors depend on the associated probability distribution.

3.3. Conditional Quantization for the Probability Distribution P with pmf $f\left(j\right)={\displaystyle \frac{1}{2^j}}$ for $1\le j\le 5$ and $f\left(6\right)={\displaystyle \frac{1}{2^5}}$ Where the Conditional Set Is $\beta :=\left\{6\right\}$

Notice that $\beta$ is the conditional optimal set of one-point with conditional quantization error $V_1={\displaystyle \frac{573}{32}}$, and the conditional optimal set of six-points is the support of P with conditional quantization error $V_6=0$. The following proposition gives the main results in this subsection. The proof is similar to the proofs of the propositions given in the previous two subsections.

Proposition 6.

Let ${\alpha }_n$ be a conditional optimal set of n-points with the conditional quantization error $V_n$. Then,

$$\begin{array}{cc}\hfill {\alpha }_2& =\{Av[1,3],6\} \mathrm{with} V_2={\displaystyle \frac{167}{224}},\hspace{1em}{\alpha }_3=\{1,Av[2,4],6\} \mathrm{with} V_3={\displaystyle \frac{59}{224}},\hfill \\ \hfill {\alpha }_4& =\{1,2,Av[3,4],6\} \mathrm{with} V_4={\displaystyle \frac{7}{96}},\hspace{1em}{\alpha }_5=\{1,2,3,Av[4,5],6\} \mathrm{with} V_5={\displaystyle \frac{1}{48}}.\hfill \end{array}$$

3.4. Conditional Quantization for the Probability Distribution P with pmf $f\left(j\right)={\displaystyle \frac{1}{2^j}}$ for $1\le j\le 5$ and $f\left(6\right)={\displaystyle \frac{1}{2^5}}$ Where the Conditional Set Is $\beta :=\left\{4\right\}$

Notice that $\beta$ is the conditional optimal set of one-point with conditional quantization error $V_1={\displaystyle \frac{185}{32}}$, and the conditional optimal set of six-points is the support of P with conditional quantization error $V_6=0$. The following proposition gives the main results in this subsection. The proof is similar to the proofs of the propositions given in the first two subsections.

Proposition 7.

Let ${\alpha }_n$ be a conditional optimal set of n-points with the conditional quantization error $V_n$. Then,

$$\begin{array}{cc}\hfill {\alpha }_2& =\{Av[1,2],4\} \mathrm{with} V_2={\displaystyle \frac{43}{96}},\hspace{1em}{\alpha }_3=\{1,Av[2,3],4\} \mathrm{with} V_3={\displaystyle \frac{23}{96}},\hfill \\ \hfill {\alpha }_4& =\{1,Av[2,3],4,Av[5,6]\} \mathrm{with} V_4={\displaystyle \frac{19}{192}},\hspace{1em}{\alpha }_5=\{1,2,3,4,Av[5,6]\} \mathrm{with} V_5={\displaystyle \frac{1}{64}}.\hfill \end{array}$$

Remark 4.

From Section 3.3 and Section 3.4, we see that though the probability distributions remain same, the conditional optimal sets and the corresponding conditional quantization errors depend on the associated conditional set.

4. Conditional and Unconditional Quantization for an Infinite Discrete Distribution with Support $\mathbf{\{}{\displaystyle \frac{\mathbf{1}}{{\mathbf{2}}^{\mathit{n}}}}:\mathit{n}\mathbf{\in }\mathbb{N}\mathbf{\}}$

Let P be an infinite discrete distribution with support $\{{\displaystyle \frac{1}{2^n}}:n\in \mathbb{N}\}$ such that the probability mass function f for P is given by

$$f\left(x\right)=\left\{\begin{array}{cc}{\displaystyle \frac{1}{2^j}}& \mathrm{if} x={\displaystyle \frac{1}{2^j}} \mathrm{for} j\in \mathbb{N},\\ 0& \mathrm{otherwise}.\end{array}\right.$$

In this section, there are two subsections. In the first subsection, for the probability distribution P, we investigate the conditional optimal sets of n-points and the nth conditional quantization errors for $n\in \mathbb{N}$ with respect to the conditional set $\beta :=\left\{0\right\}$. In the next subsection, for the probability distribution P, we investigate the unconditional optimal sets of n-means and the nth unconditional quantization errors for $n\in \mathbb{N}$. We also show that the quantization dimensions in both the scenarios exist and equal zero, but the quantization coefficients do not exist.

Let X be a random variable with probability distribution P. For $k,\mathcal{l}\in \mathbb{N}$, where $k\le \mathcal{l}$, write

$$[k,\mathcal{l}]:=\{{\displaystyle \frac{1}{2^n}}:n\in \mathbb{N} \mathrm{and} k\le n\le \mathcal{l}\}, \mathrm{and} [k,\infty ):=\{{\displaystyle \frac{1}{2^n}}:n\in \mathbb{N} \mathrm{and} n\ge k\}.$$

Further, write

$$Av[k,\mathcal{l}]:=E(X:X\in [k,\mathcal{l}])={\displaystyle \frac{{\sum }_{n=k}^{\mathcal{l}}\frac{1}{2^n}\frac{1}{2^n}}{{\sum }_{n=k}^{\mathcal{l}}\frac{1}{2^n}}} \mathrm{and} Er[k,\mathcal{l}]:=\sum _{n=k}^{\mathcal{l}}{\displaystyle \frac{1}{2^n}}{\left({\displaystyle \frac{1}{2^n}}-Av[k,\mathcal{l}]\right)}^2,$$

and

$$Er[k,\infty ):=\sum _{n=k}^{\infty }{\displaystyle \frac{1}{2^n}}{\left({\displaystyle \frac{1}{2^n}}-0\right)}^2.$$

In addition, write

$$\widehat{A}v[k,\infty )={\displaystyle \frac{{\sum }_{n=k}^{\infty }\frac{1}{2^n}\frac{1}{2^n}}{{\sum }_{n=k}^{\infty }\frac{1}{2^n}}} \mathrm{and} \widehat{E}r[k,\infty )=\sum _{n=k}^{\infty }{\displaystyle \frac{1}{2^n}}{\left({\displaystyle \frac{1}{2^n}}-\widehat{A}v[k,\infty )\right)}^2.$$

4.1. Conditional Quantization for the Probability Distribution P with Respect to the Conditional Set $\beta :=\left\{0\right\}$

For $n=1$, the set $\beta$ forms a conditional optimal set of one-point with nth conditional quantization error given by

$$V_1\left(P\right)=Er[1,\infty )=\sum _{n=1}^{\infty }{\displaystyle \frac{1}{2^n}}{\displaystyle \frac{1}{4^n}}={\displaystyle \frac{1}{7}}.$$

Proposition 8.

The set $\{0,Av[1,2\left]\right\}$ forms the conditional optimal set of two-points with conditional quantization error $V_2\left(P\right)=Er[3,\infty )+Er[1,2]={\displaystyle \frac{17}{1344}}$.

Proof. 

Consider the set $\gamma :=\{0,Av[1,2\left]\right\}$. Since ${\displaystyle \frac{1}{2^3}}<{\displaystyle \frac{1}{2}}(0+Av[1,2])<{\displaystyle \frac{1}{2^2}}$, the Voronoi region of $Av[1,2]$ contains the set $[1,2]$, and the Voronoi region of 0 contains the set $[3,\infty )$. Hence, the distortion error due to the set $\gamma$ is given by

$$V(P;\gamma )=Er[3,\infty )+Er[1,2]={\displaystyle \frac{17}{1344}}.$$

Since $V_2$ is the conditional quantization error for two-points, we have $V_2\le {\displaystyle \frac{17}{1344}}$$=0.0126488$. Let $\alpha :=\{0,a_1\}$ be a conditional optimal set of two-points. Since $a_1$ is the mean of its own Voronoi region, we can assume that $0<a_1\le {\displaystyle \frac{1}{2}}$. The Voronoi region of $a_1$ must contain ${\displaystyle \frac{1}{2}}$. Suppose that the Voronoi region of $a_1$ also contains ${\displaystyle \frac{1}{2^k}}$ for $k=2,3$. Then,

$$V_2\ge Er[1,3]={\displaystyle \frac{5}{256}}>V_2,$$

which leads to a contradiction. Thus, we can assume that the Voronoi region of $a_1$ does not contain ${\displaystyle \frac{1}{2^3}}$. Next, assume that the Voronoi region of $a_1$ contains only the element ${\displaystyle \frac{1}{2}}$. Then,

$$V_2>Er[2,\infty )={\displaystyle \frac{1}{56}}>V_2,$$

which is a contradiction. Hence, we can assume that the Voronoi region of $a_1$ contains only the elements ${\displaystyle \frac{1}{2}}$ and ${\displaystyle \frac{1}{2^2}}$, i.e., the set $\{0,Av[1,2\left]\right\}$ forms the conditional optimal set of two-points with conditional quantization error $V_2\left(P\right)=Er[3,\infty )+Er[1,2]={\displaystyle \frac{17}{1344}}$. Thus, the proof of the proposition is complete. □

The following theorem gives the conditional optimal sets of n-points and the corresponding nth conditional quantization errors for all $n\ge 3$.

Theorem 1.

For $n\in \mathbb{N}$ with $n\ge 3$, the set $\{0,Av[n-1,n],{\displaystyle \frac{1}{2^{n-2}}},\cdots ,{\displaystyle \frac{1}{2^2}},{\displaystyle \frac{1}{2}}\}$ forms the conditional optimal set of n-points with conditional quantization error $V_n\left(P\right)=Er[n+1,\infty )+Er[n-1,n]$$={\displaystyle \frac{17}{21}}{8}^{-n}$.

Proof. 

Consider the set $\gamma :=\{0,Av[n-1,n],{\displaystyle \frac{1}{2^{n-2}}},\cdots ,{\displaystyle \frac{1}{2^2}},{\displaystyle \frac{1}{2}}\}$. Since ${\displaystyle \frac{1}{2^{n+1}}}<{\displaystyle \frac{1}{2}}(0+Av[n-1,n])<{\displaystyle \frac{1}{2^n}}<{\displaystyle \frac{1}{2^{n-1}}}<{\displaystyle \frac{1}{2}}(Av[n-1,n]+{\displaystyle \frac{1}{2^{n-2}}})<{\displaystyle \frac{1}{2^{n-2}}}$, the distortion error contributed by the set $\gamma$ is given by

$$V(P;\gamma )=Er[n+1,\infty )+Er[n-1,n]={\displaystyle \frac{17}{21}}{8}^{-n}.$$

Since $V_n$ is the conditional quantization error for n-points, we must have $V_n\le {\displaystyle \frac{17}{21}}{8}^{-n}$. Let us now prove the following claim.

Claim. Let ${\alpha }_n:=\{a_n,a_{n-1},\cdots ,a_3,a_2,a_1\}$ be a conditional optimal set of n-points such that $0=a_n<a_{n-1}<\cdots <a_3<a_2<a_1\le {\displaystyle \frac{1}{2}}$. Then, $a_j={\displaystyle \frac{1}{2^j}}$ for all $1\le j\le n-2$ and $a_{n-1}=Av[n-1,n]$.

Let the set ${\alpha }_n$ given by

$${\alpha }_n:=\{a_n,a_{n-1},\cdots ,a_3,a_2,a_1\}$$

form a conditional optimal set of n-points such that $0=a_n<a_{n-1}<\cdots <a_3<a_2<a_1\le {\displaystyle \frac{1}{2}}$. Clearly, the Voronoi region of $a_1$ contains the element ${\displaystyle \frac{1}{2}}$. Suppose that it also contains ${\displaystyle \frac{1}{2^2}}$. Then,

$$V_n\ge Er[1,2]={\displaystyle \frac{1}{96}}>{\displaystyle \frac{17}{21}}{8}^{-n}\ge V_n,$$

which leads to a contradiction. Hence, we can assume that the Voronoi region of $a_1$ contains only the element ${\displaystyle \frac{1}{2}}$, i.e., $a_1={\displaystyle \frac{1}{2}}$. Similarly, we can show that $a_j={\displaystyle \frac{1}{2^j}}$ for all $1\le j\le n-2$. Since $a_j={\displaystyle \frac{1}{2^j}}$ for $1\le j\le n-2$, the Voronoi region of $a_{n-1}$ must contain ${\displaystyle \frac{1}{2^{n-1}}}$. Suppose that the Voronoi region of $a_{n-1}$ also contains the elements ${\displaystyle \frac{1}{2^n}}$ and ${\displaystyle \frac{1}{2^{n+1}}}$. Then,

$$V_n\ge Er[n-1,n+1]=5\times 2^{-3n-2}>V_n,$$

which is a contradiction. Suppose that the Voronoi region of $a_{n-1}$ contains only the element ${\displaystyle \frac{1}{2^{n-1}}}$. Then, the Voronoi region of $a_n$ will contain the set $\{{\displaystyle \frac{1}{2^j}}:j\ge n\}$ implying

$$V_n\ge Er[n,\infty )={\displaystyle \frac{8^{1-n}}{7}}>V_n,$$

which leads to a contradiction. Hence, we can assume that the Voronoi region of $a_{n-1}$ contains only the set $Er[n-1,n]$, i.e., $a_{n-1}=Av[n-1,n]$. Hence, the claim is true.

Thus, we have proved that $a_j={\displaystyle \frac{1}{2^j}}$ for $1\le j\le n-2$, and $a_{n-1}=Av[n-1,n]$, i.e., the set $\{0,Av[n-1,n],{\displaystyle \frac{1}{2^{n-2}}},\cdots ,{\displaystyle \frac{1}{2^2}},{\displaystyle \frac{1}{2}}\}$ forms the conditional optimal set of n-points with conditional quantization error $V_n\left(P\right)=Er[n+1,\infty )+Er[n-1,n]={\displaystyle \frac{17}{21}}{8}^{-n}$. Thus, the proof of the theorem is complete. □

Let us now give the following theorem.

Theorem 2.

The conditional quantization dimension $D\left(P\right)$ of the measure P exists and equals zero, and the $D\left(P\right)$-dimensional conditional quantization coefficient does not exist.

Proof. 

By Theorem 1, we have $V_n\left(P\right)={\displaystyle \frac{17}{21}}{8}^{-n}$. Then, $V_{\infty }={lim}_{n\to \infty }{V}_n=0$. Hence, the conditional quantization dimension $D\left(P\right)$ is given by

$$D\left(P\right)=\underset{n\to \infty }{lim}{\displaystyle \frac{2logn}{-log(V_n-V_{\infty })}}=0,$$

i.e., $D\left(P\right)$ exists and equals zero. Since $D\left(P\right)=0$, the $D\left(P\right)$-dimensional conditional quantization coefficient does not exist. Hence, the conditional quantization dimension $D\left(P\right)$ of the measure P exists and equals zero, and the $D\left(P\right)$-dimensional conditional quantization coefficient does not exist. □

4.2. Unconditional Quantization for the Probability Distribution P

Recall that in unconditional quantization the elements in an optimal set are the conditional expectations in their own Voronoi regions. Since

$$E(X:X\in \{{\displaystyle \frac{1}{2^n}}:n\in \mathbb{N}\})=Av[1,\infty )={\displaystyle \frac{1}{3}},$$

we can say that the optimal set of one-mean is the set $\left\{{\displaystyle \frac{1}{3}}\right\}$ with quantization error $V_1=Er[1,\infty )={\displaystyle \frac{2}{63}}.$

The following theorem gives the unconditional optimal sets of n-means and the nth unconditional quantization errors for all $n\in \mathbb{N}$ with $n\ge 2$

Theorem 3.

Let $n\in \mathbb{N}$ with $n\ge 2$. Then, the set $\{\widehat{A}v[n,\infty ),{\displaystyle \frac{1}{2^{n-1}}},{\displaystyle \frac{1}{2^{n-2}}},\cdots ,{\displaystyle \frac{1}{2^3}},{\displaystyle \frac{1}{2}}\}$ forms an unconditional optimal set of n-means with the nth unconditional quantization error $V_n={\displaystyle \frac{1}{63}}{2}^{4-3n}$.

Proof. 

Consider the set $\gamma :=\{\widehat{A}v[n,\infty ),{\displaystyle \frac{1}{2^{n-1}}},\cdots ,{\displaystyle \frac{1}{2^2}},{\displaystyle \frac{1}{2}}\}$. Since ${\displaystyle \frac{1}{2^n}}<{\displaystyle \frac{1}{2}}(\widehat{A}v[n,\infty )+{\displaystyle \frac{1}{2^{n-1}}})$$<{\displaystyle \frac{1}{2^{n-1}}}$ for all $n\ge 2$, the distortion error contributed by the set $\gamma$ is given by

$$V(P;\gamma )=\widehat{E}r[n,\infty )={\displaystyle \frac{1}{63}}{2}^{4-3n}.$$

Since $V_n$ is the conditional quantization error for n-means, we must have $V_n\le {\displaystyle \frac{1}{63}}{2}^{4-3n}$. Let the set ${\alpha }_n$ given by

$${\alpha }_n:=\{a_n,a_{n-1},\cdots ,a_3,a_2,a_1\}$$

be an unconditional optimal set of n-means such that $0\le a_n<a_{n-1}<\cdots <a_3<a_2$$<a_1\le {\displaystyle \frac{1}{2}}$. Clearly, the Voronoi region of $a_1$ contains the element ${\displaystyle \frac{1}{2}}$. Suppose that it also contains ${\displaystyle \frac{1}{2^2}}$. Then,

$$V_n\ge Er[1,2]={\displaystyle \frac{1}{96}}>{\displaystyle \frac{1}{63}}{2}^{4-3n}\ge V_n,$$

which leads to a contradiction. Hence, we can assume that the Voronoi region of $a_1$ contains only the element ${\displaystyle \frac{1}{2}}$, i.e., $a_1={\displaystyle \frac{1}{2}}$. Similarly, we can show that $a_j={\displaystyle \frac{1}{2^j}}$ for all $2\le j\le n-1$. Thus, we deduce that $a_j={\displaystyle \frac{1}{2^j}}$ for $1\le j\le n-1$, and hence, $a_n=\widehat{A}v[n,\infty )$, i.e., the set $\{\widehat{A}v[n,\infty ),{\displaystyle \frac{1}{2^{n-1}}},{\displaystyle \frac{1}{2^{n-2}}},\cdots ,{\displaystyle \frac{1}{2^3}},{\displaystyle \frac{1}{2}}\}$ forms an unconditional optimal set of n-means with unconditional quantization error $V_n=\widehat{E}r[n,\infty )={\displaystyle \frac{1}{63}}{2}^{4-3n}$, what proves the theorem. □

Let us now give the following theorem.

Theorem 4.

The unconditional quantization dimension $D\left(P\right)$ of the measure P exists and equals zero, and the $D\left(P\right)$-dimensional unconditional quantization coefficient does not exist.

Proof. 

By Theorem 3, we have $V_n\left(P\right)={\displaystyle \frac{1}{63}}{2}^{4-3n}$. Hence, the unconditional quantization dimension $D\left(P\right)$ is given by $D\left(P\right)=\underset{n\to \infty }{lim}{\displaystyle \frac{2logn}{-log{V}_n\left(P\right)}}=0$, i.e., $D\left(P\right)$ exists and equals zero. Since $D\left(P\right)=0$, the $D\left(P\right)$-dimensional unconditional quantization coefficient does not exist, which proves the theorem. □

Remark 5.

The results in Theorems 2 and 4 indicate that the quantization dimension of the discrete distribution P does not depend on the conditional set β. In fact, for a general proof in this direction one can see Theorem 6.2 in [11].

5. Conditional Quantization for an Infinite Discrete Distribution with Support $\mathbf{\{}{\displaystyle \frac{\mathbf{1}}{\mathit{n}}}:\mathit{n}\mathbf{\in }\mathbb{N}\mathbf{\}}$

Let P be a Borel probability measure on the set $\{{\displaystyle \frac{1}{n}}:n\in \mathbb{N}\}$ with probability mass function f given by

$$f\left(x\right)=\left\{\begin{array}{cc}{\displaystyle \frac{1}{2^k}}\hfill & \mathrm{if} x={\displaystyle \frac{1}{k}} \mathrm{for} k\in \mathbb{N},\hfill \\ 0\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

Then, P is a Borel probability measure on $\mathbb{R}$, and the support of P is given by $\mathrm{supp}\left(P\right)=\{{\displaystyle \frac{1}{n}}:n\in \mathbb{N}\}$. In this section, for the probability measure P, we investigate the conditional optimal sets of n-points and the nth conditional quantization errors for $n\in \mathbb{N}$ with respect to the conditional set $\beta :=\left\{0\right\}$. Let X be a random variable with probability distribution P. For $k,\mathcal{l}\in \mathbb{N}$, where $k\le \mathcal{l}$, write

$$[k,\mathcal{l}]:=\{{\displaystyle \frac{1}{n}}:n\in \mathbb{N} \mathrm{and} k\le n\le \mathcal{l}\}, \mathrm{and} [k,\infty ):=\{{\displaystyle \frac{1}{n}}:n\in \mathbb{N} \mathrm{and} n\ge k\}.$$

Further, write

$$Av[k,\mathcal{l}]:=E\left(X:X\in [k,\mathcal{l}]\right)={\displaystyle \frac{{\sum }_{n=k}^{\mathcal{l}}\frac{1}{2^n}\frac{1}{n}}{{\sum }_{n=k}^{\mathcal{l}}\frac{1}{2^n}}} \mathrm{and} Er[k,\mathcal{l}]:=\sum _{n=k}^{\mathcal{l}}{\displaystyle \frac{1}{2^n}}{\left({\displaystyle \frac{1}{n}}-Av[k,\mathcal{l}]\right)}^2,$$

and

$$\widehat{E}r[k,\mathcal{l}]:=\sum _{n=k}^{\mathcal{l}}{\displaystyle \frac{1}{2^n}}{\left({\displaystyle \frac{1}{n}}-0\right)}^2 \mathrm{and} Er[k,\infty ):=\sum _{n=k}^{\infty }{\displaystyle \frac{1}{2^n}}{\left({\displaystyle \frac{1}{n}}-0\right)}^2.$$

Notice that for $n=1$, the set $\beta$ forms a conditional optimal set of one-point with nth conditional quantization error given by

$$V_1\left(P\right)=Er[1,\infty )=\sum _{n=1}^{\infty }{\displaystyle \frac{1}{2^n}}{\displaystyle \frac{1}{n^2}}={\displaystyle \frac{{\pi }^2}{12}}-{\displaystyle \frac{{log}^2\left(2\right)}{2}}.$$

In the following, we investigate the conditional optimal sets of n-points for all $n\in \mathbb{N}$ with $n\ge 2$.

Proposition 9.

The set $\{0,Av[1,2\left]\right\}$ forms the conditional optimal set of two-points with conditional quantization error $V_2\left(P\right)=Er[3,\infty )+Er[1,2]={\displaystyle \frac{1}{48}}\left(4{\pi }^2-25-24{log}^2\left(2\right)\right)$.

Proof. 

Consider the set $\gamma :=\{0,Av[1,2\left]\right\}$. Since ${\displaystyle \frac{1}{3}}<{\displaystyle \frac{1}{2}}(0+Av[1,2])<{\displaystyle \frac{1}{2}}$, the Voronoi region of $Av[1,2]$ contains the set $[1,2]$, and the Voronoi region of 0 contains the set $[3,\infty )$. Hence, the distortion error due to the set $\gamma$ is given by

$$V(P;\gamma )=Er[3,\infty )+Er[1,2]={\displaystyle \frac{1}{48}}\left(4{\pi }^2-25-24{log}^2\left(2\right)\right)=0.0614072.$$

Since $V_2$ is the conditional quantization error for two-points, we have $V_2\le 0.0614072$. Let $\alpha :=\{0,a_1\}$ be a conditional optimal set of two-points. Since $a_1$ is the mean of its own Voronoi region, we can assume that $0<a_1\le 1$. The Voronoi region of $a_1$ must contain 1. Suppose that the Voronoi region of $a_1$ also contains ${\displaystyle \frac{1}{2}}$ and ${\displaystyle \frac{1}{3}}$. Then,

$$V_2\ge Er[1,3]={\displaystyle \frac{23}{336}}=0.0684524>V_2,$$

which leads to a contradiction. Thus, we can assume that the Voronoi region of $a_1$ does not contain ${\displaystyle \frac{1}{3}}$. Next, assume that the Voronoi region of $a_1$ contains only the element 1. Then,

$$V_2>Er[2,\infty )={\displaystyle \frac{1}{12}}\left({\pi }^2-6-6{log}^2\left(2\right)\right)=0.0822405>V_2,$$

which is a contradiction. Hence, we can assume that the Voronoi region of $a_1$ contains only the elements 1 and ${\displaystyle \frac{1}{2}}$, i.e., the set $\{0,Av[1,2\left]\right\}$ forms the conditional optimal set of two-points with conditional quantization error $V_2\left(P\right)=Er[3,\infty )+Er[1,2]={\displaystyle \frac{1}{48}}\left(4{\pi }^2-25-24{log}^2\left(2\right)\right)$. Thus, the proof of the proposition is complete. □

Proposition 10.

The set $\{0,Av[2,4],1\}$ forms the conditional optimal set of three-points with conditional quantization error $V_3\left(P\right)=Er[5,\infty )+Er[2,4]={\displaystyle \frac{{\pi }^2}{12}}-{\displaystyle \frac{1327}{2304}}-{\displaystyle \frac{1}{2}}{log}^2\left(2\right)$.

Proof. 

Consider the set $\gamma :=\{0,Av[2,4],1\}$. Since, ${\displaystyle \frac{1}{5}}<{\displaystyle \frac{1}{2}}(0+Av[2,4])<{\displaystyle \frac{1}{4}}$, and ${\displaystyle \frac{1}{2}}<{\displaystyle \frac{1}{2}}(Av[2,4]+1)<1$, the distortion error due to the set $\gamma$ is given by

$$V(P;\gamma )=Er[5,\infty )+Er[2,4]={\displaystyle \frac{{\pi }^2}{12}}-{\displaystyle \frac{1327}{2304}}-{\displaystyle \frac{1}{2}}{log}^2\left(2\right)=0.00628567.$$

Since $V_3$ is the conditional quantization error for three-points, we have $V_3\le 0.00628567$. Let $\alpha :=\{0,a_2,a_1\}$ be a conditional optimal set of three-points such that $0<a_2<a_1\le 1$. If the Voronoi region of $a_1$ contains 1 and ${\displaystyle \frac{1}{2}}$, then

$$V_3\ge Er[1,2]={\displaystyle \frac{1}{24}}=0.0416667>V_3,$$

which leads to a contradiction. Hence, we can assume that the Voronoi region of $a_1$ contains only the element 1, i.e., $a_1=1$. The Voronoi region of $a_2$ must contain ${\displaystyle \frac{1}{2}}$. Suppose that the Voronoi region of $a_2$ contains the set $\{{\displaystyle \frac{1}{6}},{\displaystyle \frac{1}{5}},{\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{2}}\}$. Then,

$$V_3\ge Er[2,6]={\displaystyle \frac{11693}{1785600}}=0.0065485>V_3,$$

which is a contradiction. Suppose that the Voronoi region of $a_2$ contains only the set $\{{\displaystyle \frac{1}{5}},{\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{2}}\}$. Then, the set $\{{\displaystyle \frac{1}{k}}:k\ge 6\}$ is contained in the Voronoi region of zero implying

$$V_3\ge Er[6,\infty )+Er[2,5]=0.00640488>V_3,$$

which leads to a contradiction. If the Voronoi region of $a_2$ contains only the set $\{{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{2}}\}$. Then,

$$V_3=Er[4,\infty )+Er[2,3]=0.00816645>V_3,$$

which gives a contradiction. Hence, we can assume that the Voronoi region of $a_2$ contains only the set $\{{\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{2}}\}$, i.e., $a_2=Av[2,4]$. Thus, the set $\{0,Av[2,4],1\}$ forms the conditional optimal set of three-points with conditional quantization error $V_3\left(P\right)=Er[5,\infty )+Er[2,4]={\displaystyle \frac{{\pi }^2}{12}}-{\displaystyle \frac{1327}{2304}}-{\displaystyle \frac{1}{2}}{log}^2\left(2\right)$. Hence, the proof of the proposition is complete. □

Using the similar technique as the proof of Proposition 10, the following proposition can be proved.

Proposition 11.

Let ${\alpha }_n$ be a conditional optimal set of n-points with conditional quantization error $V_n\left(P\right)$. Then,

$$\begin{array}{cc}\hfill {\alpha }_4& =\{0,Av[3,7],{\displaystyle \frac{1}{2}},1\} \mathrm{with} \hfill \\ & V_4\left(P\right)=Er[8,\infty )+Er[3,7]={\displaystyle \frac{{\pi }^2}{12}}-{\displaystyle \frac{3178127}{5468400}}-{\displaystyle \frac{1}{2}}{log}^2\left(2\right);\hfill \\ \hfill {\alpha }_5& =\{0,Av[4,9],{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{2}},1\} \mathrm{with} \hfill \\ \hfill & V_5\left(P\right)=Er[10,\infty )+Er[4,9]={\displaystyle \frac{{\pi }^2}{12}}-{\displaystyle \frac{29804538289}{51209625600}}-{\displaystyle \frac{1}{2}}{log}^2\left(2\right);\hfill \\ \hfill {\alpha }_6& =\{0,Av[5,11],{\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{2}},1\} \mathrm{with} \hfill \\ & V_6\left(P\right)=Er[12,\infty )+Er[5,11]={\displaystyle \frac{{\pi }^2}{12}}-{\displaystyle \frac{72720870121}{124910843904}}-{\displaystyle \frac{1}{2}}{log}^2\left(2\right);\hfill \\ \hfill {\alpha }_7& =\{0,Av[6,13],{\displaystyle \frac{1}{5}},{\displaystyle \frac{1}{4}},{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{2}},1\} \mathrm{with} \hfill \\ & V_7\left(P\right)=Er[14,\infty )+Er[6,13]={\displaystyle \frac{{\pi }^2}{12}}-{\displaystyle \frac{9871280467571329}{16954434072576000}}-{\displaystyle \frac{1}{2}}{log}^2\left(2\right).\hfill \end{array}$$

Let us now prove the following proposition.

Proposition 12.

Let ${\alpha }_n$ be a conditional optimal set of n-points with conditional quantization error $V_n\left(P\right)$. Then, for $n=200$, we have ${\alpha }_n=\{0,Av[n-1,2n-1],{\displaystyle \frac{1}{n-2}},\cdots ,{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{2}},1\}$ and

$$V_n=Er[2n,\infty )+Er[n-1,2n-1]=3.022245937265158212\times {10}^{-69}.$$

Proof. 

Fix $n=200$. Consider the set $\gamma :=\{0,Av[n-1,2n-1],{\displaystyle \frac{1}{n-2}},\cdots ,{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{2}},1\}$. Since,

$${\displaystyle \frac{1}{2n}}<{\displaystyle \frac{1}{2}}(Av[n-1,2n-1]+0)<{\displaystyle \frac{1}{2n-1}}<{\displaystyle \frac{1}{n-1}}<{\displaystyle \frac{1}{2}}(Av[n-1,2n-1]+{\displaystyle \frac{1}{n-2}})<{\displaystyle \frac{1}{n-2}}$$

the distortion error due to the set $\gamma$ is given by

$$V(P;\gamma )=Er[2n,\infty )+Er[n-1,2n-1]=3.022245937265158212\times {10}^{-69}.$$

Since $V_n$ is the conditional quantization error for n-points, we have $V_n\le 3.0222459372$$65158212\times {10}^{-69}$. We now prove the following claim.

Claim. Let ${\alpha }_n:=\{0,a_{n-1},a_{n-2},\cdots ,a_3,a_2,a_1\}$ be a conditional optimal set of n-points such that $0<a_{n-1}<a_{n-2}<\cdots <a_3<a_2<a_1\le 1$, where $n=200$. Then, $a_k={\displaystyle \frac{1}{k}}$ for $1\le k\le 198$, $a_{199}=Er[n-1,2n-1]$ and $a_{200}=Er[2n,\infty )$.

Clearly, the Voronoi region of $a_1$ contains 1. Let the Voronoi region of $a_1$ contains 1 and ${\displaystyle \frac{1}{2}}$. Then,

$$V_n\ge Er[1,2]={\displaystyle \frac{1}{24}}=0.0416667>V_n,$$

which leads to a contradiction. Hence, we can assume that the Voronoi region of $a_1$ contains only the element 1, i.e., $a_1=1$. Similarly, we can show that $a_k={\displaystyle \frac{1}{k}}$ for $1\le k\le 195$. We now show that $a_{196}={\displaystyle \frac{1}{196}}$.

Clearly, the Voronoi region of $a_{196}$ contains the element ${\displaystyle \frac{1}{196}}$. Suppose that the Voronoi region of $a_{196}$ contains ${\displaystyle \frac{1}{196}},{\displaystyle \frac{1}{197}}$, and ${\displaystyle \frac{1}{198}}$. Then,

$$V_n\ge Er[196,198]=6.1533869682152350215\times {10}^{-69}>V_n,$$

which gives a contradiction. Assume that the Voronoi region of $a_{196}$ contains only the elements ${\displaystyle \frac{1}{196}}$ and ${\displaystyle \frac{1}{197}}$. Then, the Voronoi region of $a_{197}$ must contain ${\displaystyle \frac{1}{198}}$. Suppose that the Voronoi region of $a_{197}$ also contains ${\displaystyle \frac{1}{199}}$ and ${\displaystyle \frac{1}{200}}$. Then,

$$V_n\ge Er[196,197]+Er[198,200]=3.7035461652740751683\times {10}^{-69}>V_n,$$

which is a contradiction. Suppose that the Voronoi region of $a_{197}$ contains only the elements ${\displaystyle \frac{1}{198}}$ and ${\displaystyle \frac{1}{199}}$. Then, the Voronoi region of $a_{198}$ must contain ${\displaystyle \frac{1}{200}}$. Suppose that the Voronoi region of $a_{198}$ also contains ${\displaystyle \frac{1}{201}},{\displaystyle \frac{1}{202}}$. Then,

$$V_n\ge Er[196,197]+Er[198,199]+Er[200,202]=3.1154552539006565704\times {10}^{-69}>V_n,$$

which leads to a contradiction. Suppose that the Voronoi region of $a_{198}$ contains only the set $\{{\displaystyle \frac{1}{200}},{\displaystyle \frac{1}{201}}\}$. Then, the Voronoi region of $a_{199}$ must contain ${\displaystyle \frac{1}{202}}$. Suppose that the Voronoi region of $a_{199}$ also contains the elements ${\displaystyle \frac{1}{k}}$ for $k=203,204,205$. Then,

$$\begin{array}{cc}\hfill V_n& \ge Er[196,197]+Er[198,199]+Er[200,201]+Er[202,205]\hfill \\ & =3.0364362494654583758\times {10}^{-69}>V_n,\hfill \end{array}$$

which leads to a contradiction. Suppose that the Voronoi region of $a_{199}$ contains only the elements ${\displaystyle \frac{1}{k}}$ for $k=202,203,204$. Then, the Voronoi region of $a_{200}$ contains the remaining elements in the support, yielding

$$\begin{array}{cc}\hfill V_n& \ge Er[196,197]+Er[198,199]+Er[200,201]+Er[202,204]+\widehat{E}r[205,205]\hfill \\ & =4.657207505681280412\times {10}^{-67}>V_n,\hfill \end{array}$$

which gives a contradiction. Suppose that the Voronoi region of $a_{199}$ contains only the elements ${\displaystyle \frac{1}{k}}$ for $k=202,203$. Then, the Voronoi region of $a_{200}$ contains the remaining elements in the support, yielding

$$\begin{array}{cc}\hfill V_n& \ge Er[196,197]+Er[198,199]+Er[200,201]+Er[202,203]+\widehat{E}r[204,205]\hfill \\ & =1.4002550754262693591\times {10}^{-66}>V_n,\hfill \end{array}$$

which gives a contradiction. Suppose that the Voronoi region of $a_{199}$ contains only the element ${\displaystyle \frac{1}{202}}$. Then, the Voronoi region of $a_{200}$ contains the remaining elements in the support, yielding

$$\begin{array}{cc}\hfill V_n& \ge Er[196,197]+Er[198,199]+Er[200,201]+\widehat{E}r[203,205]\hfill \\ & =3.2878626382684084966\times {10}^{-66}>V_n,\hfill \end{array}$$

which gives a contradiction. Because of these contradictions, we can assume that the Voronoi region of $a_{198}$ contains only the element ${\displaystyle \frac{1}{200}}$. Then, as before we can show that a contradiction arises. Hence, we can assume that the Voronoi region of $a_{197}$ contains only the element ${\displaystyle \frac{1}{198}}$, i.e., $a_{197}={\displaystyle \frac{1}{198}}$. Recall that for any $a\in {\alpha }_n$, $M\left(a\right|{\alpha }_n)$ represents the Voronoi region of a with respect to the set ${\alpha }_n$. Thus, we see that

$$M\left(a_{196}\right|{\alpha }_n)=\{{\displaystyle \frac{1}{196}},{\displaystyle \frac{1}{197}}\}, M\left(a_{197}\right|{\alpha }_n)=\{{\displaystyle \frac{1}{198}}\}.$$

Then, the Voronoi region of $a_{198}$ must contain the element ${\displaystyle \frac{1}{199}}$. Suppose that the Voronoi region of $a_{198}$ contains the elements ${\displaystyle \frac{1}{k}}$ for $199\le k\le 202$. Then,

$$\begin{array}{cc}\hfill V_n& \ge Er[196,197]+Er[199,202]=3.4783049085986551266\times {10}^{-69}>V_n,\hfill \end{array}$$

which gives a contradiction. Suppose that the Voronoi region of $a_{198}$ contains only the elements ${\displaystyle \frac{1}{k}}$ for $199\le k\le 201$. Then, the Voronoi region of $a_{199}$ must contain the element ${\displaystyle \frac{1}{202}}$. Suppose that the Voronoi region of $a_{199}$ also contains the elements ${\displaystyle \frac{1}{k}}$ for $k=203,204$. Then,

$$\begin{array}{cc}\hfill V_n& \ge Er[196,197]+Er[199,201]+Er[202,204]=3.0354447349842304616\times {10}^{-69}>V_n,\hfill \end{array}$$

which leads to a contradiction. Suppose that the Voronoi region of $a_{199}$ contains only the elements ${\displaystyle \frac{1}{k}}$ for $k=202,203$. Then, the Voronoi region of $a_{200}$ contains the remaining elements in the support, yielding

$$\begin{array}{cc}\hfill V_n& \ge Er[196,197]+Er[199,201]+Er[202,203]+\widehat{E}r[204,204]\hfill \\ & =9.3756976959312554841\times {10}^{-67}>V_n,\hfill \end{array}$$

which gives a contradiction. Suppose that the Voronoi region of $a_{199}$ contains only the element ${\displaystyle \frac{1}{202}}$. Then, the Voronoi region of $a_{200}$ contains the remaining elements in the support, yielding

$$\begin{array}{cc}\hfill V_n& \ge Er[196,197]+Er[199,201]+\widehat{E}r[203,204]=2.8251773324352646859\times {10}^{-66}>V_n,\hfill \end{array}$$

which is a contradiction. Because of these contradictions, we can assume that the Voronoi region of $a_{198}$ contains only the elements ${\displaystyle \frac{1}{199}}$ and ${\displaystyle \frac{1}{200}}$. Proceeding as before, we can show that a contradiction also arises in this case. Hence, we can assume that the Voronoi region of $a_{198}$ contains only the element ${\displaystyle \frac{1}{199}}$. Thus, we see that

$$M\left(a_{196}\right|{\alpha }_n)=\{{\displaystyle \frac{1}{196}},{\displaystyle \frac{1}{197}}\}, M\left(a_{197}\right|{\alpha }_n)=\{{\displaystyle \frac{1}{198}}\} \mathrm{and} M\left(a_{198}\right|{\alpha }_n)=\{{\displaystyle \frac{1}{199}}\}.$$

Then, using the similar arguments as before, we can show that $a_{199}={\displaystyle \frac{1}{200}}$. Then,

$$\begin{array}{cc}\hfill V_n& \ge Er[196,197]+\widehat{E}r[201,201]=7.7037871017668987188\times {10}^{-66}>V_n,\hfill \end{array}$$

which is a contradiction. Thus, we can deduce that the Voronoi region of $a_{196}$ contains only the element ${\displaystyle \frac{1}{196}}$, i.e., $a_{196}={\displaystyle \frac{1}{197}}$.

Using the similar arguments, we can also prove that

$$a_{197}={\displaystyle \frac{1}{197}} \mathrm{and} a_{198}={\displaystyle \frac{1}{198}}.$$

We now prove that $a_{199}=Er[n-1,2n-1]$ and $a_{200}=Er[2n,\infty )$, where $n=200$.

We have proved that $a_k={\displaystyle \frac{1}{k}}$ for $1\le k\le 198$. Then, the Voronoi region of $a_{199}$ must contain the element ${\displaystyle \frac{1}{199}}$. Suppose that the Voronoi region of $a_{199}$ contains the set $\{{\displaystyle \frac{1}{k}}:199\le k\le 406\}$. Then,

$$\begin{array}{cc}\hfill V_n& \ge Er[199,406]\hfill \\ \hfill & =3.02224593726515821199304444004276581088997570502723148484405\times {10}^{-69}\hfill \\ \hfill & >3.02224593726515821199304444004276581088997570502723148484404=V_n,\hfill \end{array}$$

which leads to a contradiction. Suppose that the Voronoi region of $a_{199}$ contains only the set $\{{\displaystyle \frac{1}{k}}:199\le k\le 405\}$. Then, the Voronoi region of $a_{200}$ contains the remaining elements in the support, yielding

$$\begin{array}{cc}\hfill V_n& \ge Er[199,405]+Er[406,\infty )>V_n,\hfill \end{array}$$

which is a contradiction. If the Voronoi region of $a_{199}$ contains only the set $\{{\displaystyle \frac{1}{k}}:199\le k\le 404\}$, $\{{\displaystyle \frac{1}{k}}:199\le k\le 403\}$, $\{{\displaystyle \frac{1}{k}}:199\le k\le 402\}$, $\{{\displaystyle \frac{1}{k}}:199\le k\le 401\}$, or $\{{\displaystyle \frac{1}{k}}:199\le k\le 400\}$, then also contradiction arises. Hence, we can assume that the Voronoi region of $a_{199}$ contains only the set $\{{\displaystyle \frac{1}{k}}:199\le k\le 399\}$, i.e.,

$$a_{199}=Av[n-1,2n-1] \mathrm{and} a_{200}=0.$$

Thus, for $n=200$, we have ${\alpha }_n=\{0,Av[n-1,2n-1],{\displaystyle \frac{1}{n-2}},\cdots ,{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{2}},1\}$ and

$$V_n=Er[2n,\infty )+Er[n-1,2n-1]=3.022245937265158212\times {10}^{-69},$$

which completes the proof of the proposition. □

Remark and Conjecture

Let P be the infinite discrete distribution as defined in this section. Let $n\in \mathbb{N}$ with $n\ge 4$, and let ${\alpha }_n\left(P\right)$ be a conditional optimal set of n-points with conditional quantization error $V_n\left(P\right)$. For $n\in \mathbb{N}$ with $n\ge 4$, set

$$\begin{array}{cc}\hfill {\alpha }_n\left(P\right):& =\{0,Av[n-1,2n-1],{\displaystyle \frac{1}{n-2}},{\displaystyle \frac{1}{n-1}},\cdots ,{\displaystyle \frac{1}{2}},1\} \mathrm{and} \hfill \\ \hfill V_n\left(P\right):& =Er[2n,\infty )+Er[n-1,2n-1].\hfill \end{array}$$

By Proposition 11, we see that the sets ${\alpha }_n\left(P\right)$ form the conditional optimal sets of n-points for P with conditional quantization errors $V_n\left(P\right)$ for all $4\le n\le 7$. By Proposition 12, this fact is also true for $n=200$. Indeed, proceeding in the similar way as Proposition 12, it can be proved that this fact is true for all $7<n<200$, and for many other values of $n>200$. We conjecture that the sets ${\alpha }_n\left(P\right)$ do not form the conditional optimal sets of n-points for P for all $n\in \mathbb{N}$ with $n\ge 4$. If the conjecture is true, then the following questions arise:

(i)

What is the least upper bound of $n\in \mathbb{N}$ for which the sets ${\alpha }_n\left(P\right)$ form the conditional optimal sets of n-points for P with conditional quantization errors $V_n\left(P\right)$?

(ii)

If k is the least upper bound of $n\in \mathbb{N}$ for which ${\alpha }_n\left(P\right)$ form the conditional optimal sets of n-points for P, then what are the forms of ${\alpha }_n\left(P\right)$ for all $n\in \mathbb{N}$ with $n>k$?