###### Abstract

The equation for canonical gravity produced by Wheeler and DeWitt in the late
1960s still presents difficulties both in terms of its mathematical solution
and its physical interpretation. One of these issues is, notoriously, the
absence of an explicit time. In this short note, we suggest one simple and
straightforward way to avoid this occurrence. We go back to the classical
equation that inspired Wheeler and DeWitt (namely, the
Hamilton–Jacobi–Einstein equation) and make explicit, before quantization, the
presence of a known, classically meaningful notion of time. We do this by
allowing Hamilton’s principal function to be explicitly dependent on this time
locally. This choice results in a Wheeler–DeWitt equation with time. A working
solution for the de Sitter minisuperspace is shown.

###### keywords:

Problem of Time; Wheeler–DeWitt Equation; WKB Time; Time in Quantum Cosmology

1 1 0 Academic Editor: Lorenzo Iorio 26 September 2022 26 October 2022
https://doi.org/ A Wheeler–DeWitt Equation with Time A Wheeler–DeWitt Equation
with Time Marcello Rotondo Marcello Rotondo Rotondo, M.

##  1 Introduction

One traditional avenue to the quantization of gravity is the geometrodynamical
one, represented by the infamous Wheeler–DeWitt (WDW) equation [1, 2]. The
equation is expected to describe the quantum evolution of the spatial
components of the metric tensor of General Relativity (GR), but its solution
and interpretation are long-standing problems [3]. In particular, a problem
with time occurs when we try to interpret the WDW equation as a Schrödinger-
type equation for gravity, because the state it describes appears to be
stationary.

To begin with, the absence of time from the WDW equation is a consequence of
the fact that the first-class Hamiltonian constraint of GR, of which the WDW
equation intends to be the quantization, specifically enforces time
diffeomorphism. In other words, it ensures its dynamical laws are valid
independently of our choice of time coordinate. When we consider the so-called
Hamilton–Jacobi–Einstein (HJE) equation developed by Peres [4], which
expresses the constraint on the 00 component of the Einstein field equations
in the Hamilton–Jacobi formalism, it is clear that in the classical case, time
is absent where it should appear, even though the theory is classical. We
know, however, that the HJE equation does not describe a timeless geometry.
The reason that the HJE equation is not problematic can be traced back to the
fact that a notion of time exists for the evolution of the spatial geometry,
as long as the classical notion of trajectory in superspace holds. In that
sense, it appears that the actual problem with time is not that it is absent
in the Schrödinger-type equation itself, but that we cannot introduce it as we
do in the classical case, since space no longer evolves along classical
trajectories. It is not clear what becomes of this time beyond the semi-
classical level, when gravity does not act as a stage for matter fields, but
rather partakes in the quantum dance.

The absence of an external time in the description of GR, which is inherited
by its quantization, is sometimes referred to as the “frozen formalism
problem”, and constitutes only one among other difficulties in the definition
of time in classical and quantum physics [5]. In the present work, we will
address only this particular aspect of the problem, ignoring its relations to
others (a strictly related one being the definition of time-evolving
observables for quantum gravity). Two classic reviews on this subject are
Isham’s [6] and Kuchař’s [7]. We invite the reader to read these reviews for
detailed references, and to gain a general idea of the large extent of
variable approaches. The study of this aspect of the problem of time has
certainly evolved significantly since the time of these reviews, with some
issues of each approach being successfully addressed, but it ultimately
remains open. Faced by the menace represented by the loss of a useful notion
of time, three alternative reactions have been adopted by researchers:
_flight_ , _fight_ , or _freeze_ (corresponding to Isham’s _tempus ante
quantum_ , _tempus post quantum_ , and _tempus nihil est_).

  1. 1.

_Flight_ : Time is recognized as a fundamental element of our description of
physical phenomena, and attempts are made to define it _before_ quantization,
as a functional of the canonical variables. This is a conservative approach
that tries to obtain an “external” notion of time as that appearing in
Schrödinger’s equation.

  2. 2.

_Fight_ : Time is recognized as a fundamental element of our description of
physical phenomena, but it is retrieved only _after_ the quantization. This
type of approach fights against the interpretative problems presented by the
quantum theory to obtain a novel definition of time.

  3. 3.

_Freeze_ : Timelessness is accepted, time is forsaken as a fundamental notion
for the description of quantum gravity, and attempts are made to provide a
complete quantum theory otherwise.

The present work adopts a definition of time resulting from the semi-classical
approximation of the WDW equation, an approach falling within the second
category above. However, we do not limit to the semi-classical regime the
definition of time identified by the semi-classical approximation. The point
of the present note is in fact to suggest that the “frozen formalism” could be
avoided by retaining the use of a classical notion of time suggested by the
semi-classical approximation of the theory, even though quantum space does not
evolve along classical trajectories. Therefore, our proposal belongs to the
first category, in that it carries over to the quantum regime the definition
of time justified by the semi-classical approximation.

The definition of time that we adopt as a starting point is that naturally
resulting from the semi-classical approximation of canonical quantum gravity
obtained by expanding the total wave functional in inverse squares of the
Planck mass [8, 9, 10]. Classical GR and the Schrödinger equation for non-
gravitational fields straightforwardly recovered this approximation. Time, for
the matter fields, is a multi-fingered (i.e., space dependent) functional
generated by the classical evolution of background geometry along its
trajectory in superspace. The operation has formal analogies with the
Wentzel–Kramers–Brillouin (WKB) approximations of quantum mechanics, and the
Born–Oppenheimer approximation from molecular physics [11, 12, 13]. In recent
years, special attention was drawn to the problem of re-establishing the
unitarity of time evolution in this approach (see, for example, [14, 15, 16,
17]). For a recent work by this author which is related to the present one,
see [18].

One key observation that motivated this approach was that this multi-fingered
(or “WKB”) functional time can be applied to the HJE equation itself, and we
can make the presence of that classical time explicit in the HJE equation,
which is the timeless form. This alone was the starting point of Wheeler and
DeWitt, as the following memoir by one of the authors recalls.

> One day in 1965, John Wheeler had a two-hour stopover between flights at the
> Raleigh–Durham airport in North Carolina. He called Bryce DeWitt, then at
> the University of North Carolina in Chapel Hill, proposing to meet at the
> airport during the wait. Bryce showed up with the Hamilton–Jacobi equation
> of general relativity, published by Asher Peres not long before […] Bryce
> mumbled the idea of repeating what Schröedinger did for the hydrogen atom:
> obtaining a wave equation by replacing the square of derivatives with (i
> times) a second derivative—a manner for undoing the optical approximation.
> […] Wheeler was tremendously excited (he was often enthusiastic) and
> declared on the spot that the equation for quantum gravity had been found.
> [19, 20]

What if DeWitt had presented Wheeler with the HJE equation together with the
notion of multi-fingered time? The resulting equation does present the
functional time variable just as time appears in the Schrödinger equation, and
preserves the correct classical limit for gravity.

In Section 2, we briefly review the definition of this multi-fingered time. In
Section 3, we rewrite the HJE equation, allowing an explicit local dependence
of Hamilton’s principal function on that time, and write the associated WDW
equation with time. Finally, in Section 4, we discuss a simple realization in
the de Sitter minisuperspace that is of special interest to quantum cosmology.

##  2 Classical Time Evolution from the Hamiltonian Constraint

For a straightforward introduction to the emergence of time in the semi-
classical approximation of canonical quantum gravity, we refer the reader to
the mentioned work by Isham [6], Section 5.4 and references therein. Here, we
follow, with some variation, the notation adopted by Kiefer [9]. Consider a
generic spacetime with line element

$\displaystyle ds^{2}=-{N^{2}}dt^{2}+N_{i}dtdx^{i}+h_{ij}dx^{i}dx^{j}\,.$ (1)

Here, $h_{ij}=g_{\mu\nu},\mu,\nu\in\\{1,2,3\\}$ is the spatial metric, with
Latin indices $i,j\in(1,2,3)$, $N_{i}=g_{0i}$ is the shift function, and
$N=(-g^{00})^{-1/2}$ is the lapse function. In the Hamiltonian formalism of
GR, time parametrization invariance is enforced by a first-class Hamiltonian
constraint (i.e., a constraint imposed on the Hamiltonian only after the
equations of motion are satisfied).

$\displaystyle\int
d^{3}x\left[(2M)^{-1}G_{AB}\pi^{A}\pi^{B}+\mathcal{V}(h_{A})+\mathcal{H}_{\phi}(\phi;h_{A})\right]=0\,.$
(2)

(This constraint can be obtained from the variation of the super-Hamiltonian
of GR with respect to the lapse function $N$ in the Arnowitt–Deser–Misner
(ADM) formalism [21]. Here, we intend to recall only the elements strictly
necessary to follow our discussion.)

In Equation (2), the capital indices, $A,B=\\{ij\\}$ represent pairs of Latin
indices, and $G_{AB}$ is the DeWitt metric

$\displaystyle G_{AB}\equiv
G_{ijkl}=\frac{1}{2\sqrt{h}}\left(h_{ik}h_{jl}+h_{il}h_{jk}-h_{ij}h_{kl}\right)\,,$
(3)

underlying superspace, i.e., the space of spatial metrics up to
differeomorphism invariance.

The physical scale (we have set $\hbar=c=1$) of quantum gravity is set by the
“geometrodynamical mass” $M$, which is proportional to the square of the
Planck mass $m_{P}$

$\displaystyle M=(m_{P}/2)^{2}\,,\quad m_{P}=\left(8\pi G\right)^{-1/2}\,.$
(4)

The geometrodynamical potential density $\mathcal{V}$ is

$\displaystyle\mathcal{V}=2M\sqrt{h}(2\Lambda-{}^{(3)}R)\,,$ (5)

where $h$ and ${}^{(3)}R$ are the determinant and the Ricci scalar of the
spatial metric, respectively. The Hamiltonian density operator
$\mathcal{H}_{\phi}$ is taken to describe bosonic matter.

The WDW equation results from an attempt to quantize the Hamiltonian
constraint (2) [-25]straightforwardly, applying it to the “wave functional of
the universe”, $\Psi[h_{A},\phi]$, thus obtaining

$\displaystyle\int
d^{3}x\left[(2M)^{-1}G_{AB}\partial^{A}\partial^{B}+\mathcal{V}(h_{A})+\mathcal{H}_{\phi}(h_{A},\phi)\right]\Psi[h_{A},\phi]=0\,,$
(6)

where all variables are promoted to the respective operators. For the sake of
simplicity, in this section, we have adopted the trivial ordering, and the
symbols $\partial_{A}$ are used to indicate functional derivatives with
respect to the metric component indicated by the double index.

The evolution of bosonic quantum fields in classical curved spacetime is
obtained by making the ansatz

$\displaystyle\Psi[h_{A},\phi]=\chi[h_{A}]\psi[\phi;h_{A}]$ (7)

for the total wave functional, and considering a WKB-like expansion in inverse
powers of $M$ [9, 10]. In doing so, one aims at wave functionals $\chi$ and
$\psi$ that describe the “heavy” (i.e., the spatial metric components) and
“light” degrees of freedom (i.e., matter), respectively. Notice that $\psi$
depends on the geometry only parametrically, which is indicated by the use of
the semicolon. The method consists of substituting the expansion

$\displaystyle\Psi[h_{A},\phi]=\exp\left(i\sum_{n=0}^{\infty}M^{1-n}S_{n}[h_{A},\phi]\right)$
(8)

in the WDW equation, and equating contributions to equal powers of $M$.

To order $M^{2}$, one obtains $S_{0}=S_{0}[h_{A}]$: the leading contribution
is purely geometrodynamical. To order $M^{1}$, one obtains the vacuum HJE
equation

$\displaystyle\int
d^{3}x\left[(2M)^{-1}G_{AB}\partial^{A}S_{G}\partial^{B}S_{G}+\mathcal{V}\right]=0\,,$
(9)

$S_{G}=MS_{0}$ being the leading contribution to the phase of the wave
functional (8). The HJE Equation (9) appears to be timeless due to the
vanishing of the RHS. However, the time evolution of space can still be
obtained from the Hamiltonian constraint (2) by expressing the canonical
momenta (defined by the Lagrangian as
$\pi^{A}=\partial\mathcal{L}/\partial\dot{h}_{A}$, and identified with
$\pi^{A}=\partial^{A}S_{G}$ in the Hamilton–Jacobi formalism) in terms of the
geometrodynamical velocities. From the Hamiltonian equations of motion in the
ADM formalism, these are given by

$\displaystyle\dot{h}_{A}=N\pi_{A}+2N_{(A)}\,,$ (10)

$N_{(A)}$ being a shorthand for $N_{(i;j)}$. The fact that time evolution is
retrieved by such substitution, as the Hamiltonian is constrained, is an
important point.

At this point, define
$\psi_{1}[\phi;h_{A}]=\exp\left(iS_{1}[\phi,h_{A}]\right)$ and require
conservation of the current associated with $\chi$. Then, to order $M^{0}$,
one gets the following functional equation for matter

$\displaystyle\int
d^{3}x\left[iM^{-1}G_{AB}\partial^{A}S_{G}\partial^{B}-\mathcal{H}_{\phi}\right]\psi_{1}[\phi;h_{A}]=0\,.$
(11)

By using as time the local “multi-fingered” time $\tau=\tau(x)$ of the
parametrization generated by the classical momenta along the classical
trajectory in superspace

$\displaystyle G_{AB}\pi^{A}\partial^{B}\Big{|}_{x}\tau(y)=\delta(y-x)\,,$
(12)

Equation (11) gives the Schrödinger-type functional equation

$\displaystyle\int
d^{3}x\left[i\overset{\circ}{\psi}_{1}[\phi;h_{A}]-\mathcal{H}_{\phi}\psi_{1}[\phi;h_{A}]\right]=0\,,$
(13)

where we employ a circle over the variable to indicate the functional
derivative with respect to $\tau$

$\displaystyle\dfrac{\delta}{\delta\tau}=M^{-1}G_{AB}\pi^{A}\partial^{B}\,.$
(14)

Notice that in normal coordinates ($N=1$, $N_{0i}=0$), the Equation (13)
reduces to the functional Schrödinger equation for bosonic fields. In this
case, the passage to the partial derivative with respect to time is granted by
the fact that the matter wave functional $\psi_{1}[\phi;h_{A}]$ depends
explicitly on time only through the background metric, which appears as a set
of local parameters, and employing the resulting relation between momenta and
velocities (10), (3) reduces to an application of the chain rule.

##  3 The Time Evolution of Quantum Space

The functional derivative with respect to $\tau$ of a metric component,
intended as the functional

$\displaystyle h_{A}(x)=\int d^{3}y\,h_{A}(y)\,\delta(y-x)\,,$ (15)

gives the relation between velocities and momenta,

$\displaystyle\overset{\circ}{h}_{A}=M^{-1}G_{AB}\pi^{A}\,.$ (16)

Putting the relation (16) back into the Hamiltonian equation gives the
equations of motion with respect to $\tau$. In other words, we can use $\tau$
not only to describe the evolution of quantum fields with respect to the
background metric, but also to describe the evolution of the background metric
itself. It is a favorable parametrization in that it makes the form of the
equations of motion simpler and not explicitly dependent on the coordinate
choice (1). Incidentally, notice that, in Peres’ HJE equation, the spatial
metric components are defined as functions of space alone. How they become a
function of time seems to be a problem that is not addressed in the
literature. In the present treatment, they become functions of time precisely
by defining their dependence on time according to (16).

The main objection in extending the use of multi-fingered time to the quantum
evolution is that classical trajectories are lost in that regime. While this
is true, we still know what the classical trajectory is, and we can use the
“natural” parametrization (i.e., the parametrization in which all equations of
motion appear simple as (16)) along it to describe evolution along non-
classical trajectories between wave fronts of constant multi-fingered time.

Working with the vacuum model, we can make multi-fingered time explicit in the
HJE Equation (9) by rewriting it as

$\displaystyle\int
d^{3}x\left[\partial_{\tau}{S_{G}}+(2M)^{-1}G_{AB}\partial^{A}S_{G}\partial^{B}S_{G}+\mathcal{V}\right]=0\,,$
(17)

and requiring that

$\displaystyle\int d^{3}x\partial_{\tau}{S_{G}}=0\,.$ (18)

Notice that the derivative with respect to $\tau$ is partial. As we previously
observed, WKB time (3) only takes into account the dependency on $\tau$
through the geometrodynamical degrees of freedom, but $S_{G}$ could, in
principle, also be explicitly dependent on it. What we are doing is simply
allowing for this possibility. Adding and subtracting the $\tau$-derivative of
$S_{G}$, and substituting the velocities (16), integration of the HJE Equation
(17) tells us that $S_{G}$ is indeed Hamilton’s principal function, defined as
a functional integral of the Lagrangian.

Notwithstanding the fact that the condition (18) ensures that the Hamiltonian
constraint still holds, it allows for the action to remain dependent
explicitly on $\tau$ locally. We may then try to quantize the HJE Equation
(17) à la Schrödinger, and require the global condition (18) to be true only
in the classical limit. What we obtain is a WDW equation with time

$\displaystyle\int
d^{3}x\left[-i\frac{\partial\Psi}{\partial\tau}+\left((2M)^{-1}G_{AB}\partial^{A}\partial^{B}+\mathcal{V}\right)\Psi\right]=0\,.$
(19)

Both the introduction of the coordinate independent functional time derivative
and the (classically redundant) condition (18) are necessary to obtain
Equation (19). The second condition is somehow reminiscent of another approach
[22], where time is recovered by weakening the classical Hamiltonian
constraint, required to hold only on average in the quantum regime. In that
work, the problem of which time to use to describe the evolution is not
addressed, and the condition on time dependence is stricter than ours (see (6)
in the referenced paper), resulting in a wave function whose phase does not
depend on time even locally.

##  4 The de Sitter Minisuperspace

The results of the previous section, i.e., the time evolution Equation (19)
combined with the global condition (18), are only formal. Equation (19) still
presents the same issues as the “timeless” WDW Equation (6). Besides that,
using a spatial-dependent $\tau$ to parametrize the spatial geometry is more
easily said than done. In the following, we will consider, by way of example,
the solution for the spatially flat de Sitter universe, described by the line
element

$\displaystyle ds^{2}=-dt^{2}+a(t)^{2}\delta^{ij}dx_{i}dx_{j}\,.$ (20)

Here, $a(t)$ is the scale factor. We will set $M=1/12$.

As a geometrodynamical variable, rather than the scale factor itself. It will
be convenient to adopt

$\displaystyle q=\frac{(2a)^{\frac{3}{2}}}{3}\left(\int
d^{3}x\right)^{\frac{1}{2}}\,,$ (21)

which is proportional to the square root of the co-moving spatial volume
considered.

The Ricci scalar of the spatial metric vanishes in this model, and the
Hilbert–Einstein action Lagrangian, reads simply

$\displaystyle L=-\frac{1}{2}\left(\dot{q}^{2}+\omega^{2}q^{2}\right)\,.$ (22)

Here, we have defined the constant $\omega=\sqrt{3\Lambda/4}$. The canonical
momentum is $\pi_{q}=-\dot{q}$, and the Hamiltonian is

$\displaystyle H={-}\frac{1}{2}\pi_{q}^{2}+\frac{1}{2}\omega^{2}\,q^{2}$ (23)

The Hamiltonian constraint then gives the Friedmann equation

$\displaystyle\left(\frac{\dot{q}}{q}\right)^{2}=\omega^{2}\,.$ (24)

This allows us to obtain the classical time evolution of spatial volume

$\displaystyle q(t)={q(0)}\exp\left(\omega t\right)\,.$ (25)

The HJE Equation (9) reduces to

$\displaystyle-\frac{1}{2}\left(\frac{\partial S}{\partial
q}\right)^{2}+\frac{1}{2}\omega^{2}\,q^{2}=0\,.$ (26)

Classically, Hamilton’s principal function depends on time only implicitly,
and is of the form

$\displaystyle S=\mp\frac{\omega}{2}\left(q^{2}-q_{i}^{2}\right)\,.$ (27)

Using Equation (25), one can check that the (one-fingered) time associated
with this action indeed coincides with forward coordinate time when we choose
the negative sign.

Moving onto the quantization, notice that with our choice of variable, the
DeWitt metric, that in terms of the scale factor reads $G_{aa}=-2a$, is now

$\displaystyle G_{qq}=-(3q)^{\frac{2}{3}}\,.$ (28)

This simplifies the measure

$\displaystyle\sqrt{-G_{(a)}}\,da\to dq$ (29)

and, adopting the Laplace–Beltrami operator ordering for the kinetic operator,
we have

$\displaystyle\sqrt{-G_{(a)}}^{-1}\partial_{a}\left(\sqrt{-G_{(a)}}G^{aa}\partial_{a}\right)\to\partial_{q}^{2}\,.$
(30)

Then, in this minisuperspace, the WDW Equation (19) reads

$\displaystyle
i\dot{\Psi}=\frac{1}{2}\partial_{q}^{2}\Psi+\frac{1}{2}\omega^{2}q^{2}\Psi\,,$
(31)

which is essentially the one-dimensional Schrödinger equation for the so-
called inverted harmonic oscillator. See [23] for a recent review, and [24]
for an application to the study of a scalar field in slow-roll inflation. The
difference in our case is that the variable is constrained to the positive
axis only, and time appears with the opposite sign. As in [24], we require the
Gaussian ansatz

$\displaystyle\psi(q,t)=A(t)\exp\left(-B(t)q^{2}\right)\,.$ (32)

The equality in (31) of the terms of null and second order in $q$ imposes

$\displaystyle-i\dot{A}=A\,B$ (33)

and

$\displaystyle-i\dot{B}=\frac{1}{2}\omega^{2}+2B^{2}\,.$ (34)

From the evolution Equation (34), we have

$\displaystyle B(t)=\frac{\omega}{2}\tan\left(\phi+i\omega t\right)\,.$ (35)

Here $\phi$ is the real part of the constant of integration. The imaginary
part of the constant is merged into the choice of initial time, instead, so
that the width of the state is minimized at $t=0$. The value of $\phi$
determines the greater ($<$$\pi$/4) or lesser ($>$$\pi$/4) peaking of the
distribution in $q$, rather than the conjugate momentum at time $t=0$.

Substituting $B(t)$ in Equation (33), for a normalized solution, we obtain

$\displaystyle
A(t)=\left(\frac{2}{\pi}\omega\sin(2\phi)\right)^{1/4}\left(\cos(\phi+i\omega
t)\right)^{-\frac{1}{2}}\,.$ (36)

The expectation value for $q$ is

$\displaystyle\langle{q}{\rangle}$
$\displaystyle=(2\pi\omega\sin(2\phi))^{-\frac{1}{2}}\frac{\cos(2\phi)+\cosh(2\omega
t)}{|\cos(\phi+i\omega t)|}\,.$ (37)

At late times, we correctly recover the classical inflationary expansion
$\langle{q}{\rangle}\propto\exp(\omega t)$ of Equation (25). In this limit,
the phase of $\psi$ approximates the classical action (27) (up to a global
phase, which fixes the initial value) and loses its explicit dependence on
time, thus reducing to the classical action (18). On the other hand, when we
approach the time $t=0$, where the state is maximally contracted for the given
value of $\phi$, the expectation value of the scale variable deviates from the
classical one, which is directed at asymptotic convergence to zero: we observe
instead the state of the de Sitter universe bouncing back from an earlier
phase of contraction (see Figure 1).

Figure 1: Time evolution of the expectation value of the scale variable _q_
for $\phi=\pi/4$. We set $\omega=1$.

##  5 Conclusions

In this short note, we have proposed to extend to the quantum regime the use
of the multi-fingered time originating from the semi-classical WKB
approximation of geometrodynamics. We have done so by rewriting the classical
HJE equation for vacuum space to include an explicit time dependency on
Hamilton’s principal function, and requiring this dependency to vanish
globally. The quantization provides a WDW equation that describes the
evolution of the state with respect to classical multi-fingered time. Quantum
matter fields can be included by appropriately augmenting the Hamiltonian
operator. The classical limit will still be granted.

The main purpose of this work was to provide a formal WDW equation, (19), with
a clear and working notion of time. This result, however, does not help with
the original mathematical difficulties of the WDW equation, such as the
indefiniteness of the superspace metric, or the divergence of functional
derivatives in the full theory. However, we showed by example of a
minisuperspace model of flat de Sitter universe that exact normalizable
solutions can be easily found and interpreted. In particular, for the de
Sitter universe, we found a well-behaved Gaussian solution, which shows a
state that bounces back at the time of maximal contraction, thus avoiding the
classical asymptotic regression to _nihil_.

The next step along this line of research could be the formalization of our
heuristic approach both in the Hamiltonian and the Lagrangian formalism, where
the wave functional could be constructed in terms of a path integral over
foliations of constant multi-fingered time. On the side of application, it
would be fundamental to work out exact solutions that include quantum matter
degrees of freedom. Expanding on the de Sitter model introduced here by adding
perturbations could be relevant for early inflationary cosmology. Furthermore,
beyond cosmology, an application to time evolution during the last stage of
gravitational collapse could be of interest.

This research received no external funding

The authors declare no conflict of interest.

-0cm References

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# Quantization for the mixtures of uniform distributions on connected and
disconnected line segments

1Asha Barua  ,  2Gustavo Fernandez  ,  3Ashley Gomez  ,  4Ogla Lopez  and
5Mrinal Kanti Roychowdhury

###### Abstract.

In this paper, we have considered different mixed distributions which are
generated by two uniform distributions: first supported by two connected line
segments, and then supported by two disconnected line segments. For these
mixed distributions, we have determined the optimal sets of $n$-means and the
$n$th quantization errors for all positive integers $n$. The technique of this
paper can be utilized to investigate the optimal quantization for any mixed
distributions $P:=pP_{1}+(1-p)P_{2}$, generated by any two probability
distributions $P_{1}$ and $P_{2}$ associated with any probability vector
$(p,1-p)$ for $0<p<1$ with support a connected or the union of disconnected
line segments.

###### Key words and phrases:

Mixed distribution, uniform distribution, optimal sets of $n$-means,
quantization error

###### 2010 Mathematics Subject Classification:

60Exx, 94A34.

##  1\. Introduction

Let $\mathbb{R}^{d}$ denote the $d$-dimensional Euclidean space equipped with
a metric $\|\cdot\|$ compatible with the Euclidean topology. Let $P$ be a
Borel probability measure on $\mathbb{R}^{d}$ and $\alpha$ be a finite subset
of $\mathbb{R}^{d}$. Then, $\int\min_{a\in\alpha}\|x-a\|^{2}dP(x)$ is often
referred to as the cost, or distortion error for $\alpha$ with respect to the
probability measure $P$, and is denoted by $V(P;\alpha)$. Write
$\mathcal{D}_{n}:=\\{\alpha\subset\mathbb{R}^{d}:1\leq\text{card}(\alpha)\leq
n\\}$. Then, $\inf\\{V(P;\alpha):\alpha\in\mathcal{D}_{n}\\}$ is called the
$n$th quantization error for the probability measure $P$, and is denoted by
$V_{n}:=V_{n}(P)$. A set $\alpha$ for which the infimum occurs and contains no
more than $n$ points is called an optimal set of $n$-means. It is known that
for a Borel probability measure $P$, if its support contains infinitely many
elements and $\int\|x\|^{2}dP(x)$ is finite, then an optimal set of $n$-means
always has exactly $n$-elements [AW, GKL, GL1, GL2]. For some recent work in
this direction of optimal sets of $n$-means, one can see [CR, DR1, DR2, GL3,
L1, R1, R2, R3, R4, R5, R6, RR1]. Optimal quantization has broad application
in engineering and technology (see [GG, GN, Z]).

Let us now state the following proposition (see [GG, GL1]):

######  Proposition 1.1.

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

$(i)$ $P(M(a|\alpha))>0$, $(ii)$ $P(\partial M(a|\alpha))=0$, $(iii)$
$a=E(X:X\in M(a|\alpha))$, where $M(a|\alpha)$ is the Voronoi region of
$a\in\alpha,$ i.e., $M(a|\alpha)$ is the set of all elements $x$ in
$\mathbb{R}^{d}$ which are closest to $a$ among all the elements in $\alpha$.

Proposition 1.1 says that if $\alpha$ is an optimal set and $a\in\alpha$, then
$a$ is the conditional expectation of the random variable $X$ given that $X$
takes values in the Voronoi region of $a$. Mixed distributions are an exciting
new area for optimal quantization. For any two Borel probability measures
$P_{1}$ and $P_{2}$, and $p\in(0,1)$, if $P:=pP_{1}+(1-p)P_{2}$, then the
probability measure $P$ is called the mixture or the mixed distribution
generated by the probability measures $(P_{1},P_{2})$ associated with the
probability vector $(p,1-p)$.

In this paper, first, assume that $P_{1}$ and $P_{2}$ are two uniform
distributions on the two connected line segments $J_{1}:=[0,1]$ and
$J_{2}:=[1,2]$, respectively. Then, for the mixed distribution
$P:=pP_{1}+(1-p)P_{2}$ with support the closed interval $[0,2]$, in Subsection
2.8 for $p=\frac{1}{5}$ and in Subsection 2.9 for $p=\frac{1}{3}$, we
determine the optimal sets of $n$-means and the $n$th quantization errors for
all $n\in\mathbb{N}$. Next, assume that $P_{1}$ and $P_{2}$ are two uniform
distributions on the two disconnected line segments $J_{1}:=[0,\frac{1}{3}]$
and $J_{2}:=[\frac{2}{3},1]$, respectively. Then, for the mixed distribution
$P:=pP_{1}+(1-p)P_{2}$ with support the union of the closed intervals
$[0,\frac{1}{3}]$ and $[\frac{2}{3},1]$, in Subsection 3.4 for
$p=\frac{1}{100}$ and in Subsection 3.5 for $p=\frac{2}{5}$ and
$p=\frac{1}{1000}$, we determine the optimal sets of $n$-means and the $n$th
quantization errors for all $n\in\mathbb{N}$. The main significance of this
paper is that using the similar technique, given in this paper, one can
investigate the optimal sets of $n$-means and the $n$th quantization errors
where $n\in\mathbb{N}$, for any mixed distribution $P$ generated by
$(P_{1},P_{2})$ associated with any probability vectors $(p,1-p)$ with support
a connected or the union of disconnected line segments. In addition, in the
paper, we give a conjecture Conjecture 3.7, and two open problems Open 3.8 and
Open 3.9. Under a conjecture in Subsection 3.11, we give a partial answer of
the open problem Open 3.9.

######  Remark 1.2.

In the sequel, there are some decimal numbers which are rational
approximations of some real numbers.

##  2\. Mixed distribution $P$ with support the closed interval $[0,2]$

###  2.1. Basic preliminaries

Let $P_{1}$ and $P_{2}$ be two uniform distributions, respectively, on the
intervals given by

$J_{1}:=[0,1],\text{ and }J_{2}:=[1,2].$

Let $f_{1}$ and $f_{2}$ be their respective density functions. Then,

$f_{1}(x)=\left\\{\begin{array}[]{cc}1&\text{ if }x\in[0,1],\\\
0&\text{otherwise};\end{array}\right.\text{ and
}f_{2}(x)=\left\\{\begin{array}[]{cc}1&\text{ if }x\in[1,2],\\\
0&\text{otherwise}.\end{array}\right.$

Let us consider the mixed distribution $P:=pP_{1}+(1-p)P_{2}$, where $0<p<1$.
Notice that $P$ has support the closed interval $[0,2]$. By $E(X)$, it is
meant the expectation of a random variable $X$ with probability distribution
$P$, and $V(X)$ represents the variance of $X$. By
$\alpha_{n}:=\alpha_{n}(P)$, we denote an optimal set of $n$-means with
respect to the probability distribution $P$, and $V_{n}:=V_{n}(P)$ represents
the corresponding quantization error for $n$-means. Since $f_{1}$ and $f_{2}$
are the density functions for the probability distributions $P_{1}$ and
$P_{2}$, respectively, we have

$dP_{1}(x)=P_{1}(dx)=f_{1}(x)dx=dx,\text{ and
}dP_{2}(x)=P_{2}(dx)=f_{2}(x)dx=dx.$

######  Lemma 2.2.

Let $P$ be the mixed distribution defined by $P=pP_{1}+(1-p)P_{2}$. Then,
$E(X)=\frac{1}{2}(3-2p)$, and $V(X)=-p^{2}+p+\frac{1}{12}$.

###### Proof.

We have

$E(X)=\int xdP=p\int xd(P_{1}(x))+(1-p)\int
xd(P_{2}(x))=p\int_{J_{1}}x\,dx+(1-p)\int_{J_{2}}x\,dx$

yielding $E(X)=\frac{1}{2}(3-2p)$, and

$V(X)=\int(x-E(X))^{2}dP=p\int(x-E(X))^{2}d(P_{1}(x))+(1-p)\int(x-E(X))^{2}d(P_{2}(x)),$

implying $V(X)=-p^{2}+p+\frac{1}{12}$, and thus, the proposition is yielded. ∎

######  Note 2.3.

Lemma 2.2 implies that the optimal set of one-mean is the set
$\\{\frac{1}{2}(3-2p)\\}$, and the corresponding quantization error is the
variance $V:=V(X)$ of a random variable with probability distribution $P$. For
a subset $J$ of $\mathbb{R}$ with $P(J)>0$, by $P(\cdot|_{J})$, we denote the
conditional probability given that $J$ is occurred, i.e.,
$P(\cdot|_{J})=P(\cdot\cap J)/P(J)$, in other words, for any Borel subset $B$
of $\mathbb{R}$ we have $P(B|_{J})=\frac{P(B\cap J)}{P(J)}$.

Let us now give the following proposition.

######  Proposition 2.4.

Let $P$ be a Borel probability measure on $\mathbb{R}$ such that $P$ is
uniformly distributed over a closed interval $[a,b]$ with a constant density
function $f$ such that $f(x)=t$ for all $x\in[a,b]$, where $t\in\mathbb{R}$.
Then, the optimal set $\alpha_{n}(P(\cdot|_{[a,b]}))$ of $n$-means and the
corresponding quantization error $V_{n}(P(\cdot|_{[a,b]}))$ of $n$-means for
the probability distribution $P(\cdot|_{[a,b]})$ are, respectively, given by

$\alpha_{n}(P(\cdot|_{[a,b]})):=\Big{\\{}a+\frac{(2j-1)(b-a)}{2n}:1\leq j\leq
n\Big{\\}},\text{ and }V_{n}(P(\cdot|_{[a,b]}))=\frac{(b-a)^{3}t}{12n^{2}}.$

###### Proof.

Let $\alpha_{n}:=\\{a_{1}<a_{2}<\cdots<a_{n}\\}$ be an optimal set of
$n$-means for the probability distribution $P$ with a constant density
function $f$ on $[a,b]$ such that $f(x)=t$ for all $x\in[a,b]$, where
$t\in\mathbb{R}$. Then, proceedings analogously as [RR2, Theorem 2.1.1], we
can show that $a_{j}=a+\frac{(2j-1)(b-a)}{2n}$ implying

$\alpha_{n}(P(\cdot|_{[a,b]}))=\Big{\\{}a+\frac{(2j-1)(b-a)}{2n}:1\leq j\leq
n\Big{\\}}.$

Notice that the probability density function is constant, and the Voronoi
regions of the points $a_{j}$ for $1\leq j\leq n$ are of equal lengths. This
yields the fact that the distortion errors due to each $a_{j}$ are equal.
Hence, the $n$th quantization error is given by

$\displaystyle V_{n}(P(\cdot|_{[a,b]}))$
$\displaystyle=\int\min_{a\in\alpha_{n}(P(\cdot|_{[a,b]})}(x-a)^{2}dP=nt\int_{a}^{\frac{1}{2}(a_{1}+a_{2})}(x-a_{1})^{2}dx$
$\displaystyle=nt\int_{a}^{a+\frac{b-a}{n}}(x-(a+\frac{b-a}{2n}))^{2}dx$

implying

$V_{n}(P(\cdot|_{[a,b]}))=\frac{(b-a)^{3}t}{12n^{2}}.$

Thus, the proof of the proposition is complete. ∎

For $n\geq 3$, let $\alpha_{n}$ be an optimal set of $n$-means such that
$\alpha_{n}$ contains points from both $[0,1]$ and $(1,2]$. Then, there exist
two positive integers $k:=k(n)$ and $m:=m(n)$ such that
$\text{card}(\alpha_{n}\cap[0,1])=k$, and
$\text{card}(\alpha_{n}\cap(1,2])=m$. Let
$\alpha_{n}\cap[0,1]:=\\{a_{1}<a_{2}<\cdots<a_{k}\\}$ and
$\alpha_{n}\cap(1,2]=\\{b_{1}<b_{2}<\cdots<b_{m}\\}$. Then, $k+m=n$. Moreover,
notice that the following two cases can happen: either
$\frac{1}{2}(a_{k}+b_{1})\leq 1$, or $1\leq\frac{1}{2}(a_{k}+b_{1})$. Let
$V1(k,m)$ be the $n$th quantization error when $\frac{a_{k}+b_{1}}{2}\leq 1$,
and $V2(k,m)$ be the $n$th quantization error when
$1\leq\frac{a_{k}+b_{1}}{2}$. The following propositions give the optimal sets
of $n$-means and the $n$th quantization errors for all $n\geq 3$:

######  Proposition 2.5.

Let $k\geq 2$ and $m=1$. Then, if $\frac{1}{2}(a_{k}+b_{1})\leq 1$, we have
$a_{j}=\frac{(2j-1)(a_{k}+b_{1})}{4k}$ for $1\leq j\leq k$, and
$b_{1}=E(X:X\in[\frac{1}{2}(a_{k}+b_{1}),2])$. On the other hand, if
$1\leq\frac{1}{2}(a_{k}+b_{1})$, we have
$a_{j}=\frac{(2j-1)(a_{k-1}+a_{k})}{4(k-1)}$ for $1\leq j\leq(k-1)$,
$a_{k}=E(X:X\in[\frac{1}{2}(a_{k-1}+a_{k}),\frac{1}{2}(a_{k}+b_{1})])$, and
$b_{1}=E(X:X\in[\frac{1}{2}(a_{k}+b_{1}),2])$. The quantization errors for
$n$-means are given by

$\displaystyle V1(k,1)$
$\displaystyle=p\frac{(a_{k}+b_{1}){}^{3}}{96k^{2}}+p\int_{\frac{1}{2}(a_{k}+b_{1})}^{1}(x-b_{1}){}^{2}\,dx+(1-p)\int_{1}^{2}(x-b_{1}){}^{2}\,dx,\text{
and }$ $\displaystyle V2(k,1)$
$\displaystyle=p\frac{(a_{k-1}+a_{k}){}^{3}}{96(k-1)^{2}}+p\int_{\frac{1}{2}(a_{k-1}+a_{k})}^{1}(x-a_{k}){}^{2}\,dx+(1-p)\int_{1}^{\frac{1}{2}(a_{k}+b_{1})}(x-a_{k}){}^{2}\,dx$
$\displaystyle\qquad\qquad+(1-p)\int_{\frac{1}{2}(a_{k}+b_{1})}^{2}(x-b_{1}){}^{2}\,dx.$

###### Proof.

If $\frac{a_{k}+b_{1}}{2}\leq 1$, then $a_{1},a_{2},\cdots,a_{k}$ are
uniformly distributed over the closed interval $[0,\frac{a_{k}+b_{1}}{2}]$; on
the other hand, if $1\leq\frac{a_{k}+b_{1}}{2}$, then
$a_{1},a_{2},\cdots,a_{k-1}$ are uniformly distributed over the closed
interval $[0,\frac{a_{k-1}+a_{k}}{2}]$. Thus, by Proposition 1.1 and
Proposition 2.4, the expressions for $a_{j}$ and $b_{j}$ can be obtained. With
the help of the formula given in Proposition 2.4, the quantization errors are
also obtained as routine. ∎

######  Proposition 2.6.

Let $k=1$ and $m\geq 2$. Then, if $\frac{1}{2}(a_{1}+b_{1})\leq 1$, we have
$a_{1}=E(X:X\in[0,\frac{1}{2}(a_{1}+b_{1})]$,
$b_{1}=E(X:X\in[\frac{1}{2}(a_{1}+b_{1}),\frac{1}{2}(b_{1}+b_{2})])$, and

$b_{1+j}=\frac{1}{2}(b_{1}+b_{2})+\frac{(2j-1)}{2(m-1)}(2-\frac{1}{2}(b_{1}+b_{2}))\text{
for }1\leq j\leq m-1.$

On the other hand, if $1\leq\frac{1}{2}(a_{1}+b_{1})$, we have
$a_{1}=E(X:X\in[0,\frac{1}{2}(a_{1}+b_{1})]$, and

$b_{j}=\frac{1}{2}(a_{1}+b_{1})+\frac{(2j-1)}{2m}(2-\frac{1}{2}(a_{1}+b_{1}))\text{
for }1\leq j\leq m.$

The quantization errors for $n$-means are given by

$\displaystyle V1(1,m)$
$\displaystyle=p\int_{0}^{\frac{1}{2}(a_{1}+b_{1})}(x-a_{1}){}^{2}\,dx+p\int_{\frac{1}{2}(a_{1}+b_{1})}^{1}(x-b_{1}){}^{2}\,dx+(1-p)\int_{1}^{\frac{1}{2}(b_{1}+b_{2})}(x-b_{1}){}^{2}\,dx$
$\displaystyle\qquad\qquad+(1-p)\frac{(4-(b_{1}+b_{2}))^{3}}{96(m-1)^{2}},\text{
and }$ $\displaystyle V2(1,m)$
$\displaystyle=p\int_{0}^{1}(x-a_{1}){}^{2}\,dx+(1-p)\int_{1}^{\frac{1}{2}(a_{1}+b_{1})}(x-a_{1}){}^{2}\,dx+(1-p)\frac{(4-(a_{1}+b_{1})){}^{3}}{96m^{2}}.$

###### Proof.

If $\frac{1}{2}(a_{1}+b_{1})\leq 1$, then $b_{2},b_{3},\cdots,b_{m}$ are
uniformly distributed over the closed interval $[\frac{1}{2}(b_{1}+b_{2}),2]$,
and so by Proposition 1.1 and Proposition 2.4 the expressions for $a_{j}$ and
$b_{j}$, and the corresponding quantization error are obtained. Likewise, if
$1\leq\frac{1}{2}(a_{1}+b_{1})$, by Proposition 1.1 and Proposition 2.4, the
expressions for $a_{j}$ and $b_{j}$, and the corresponding quantization error
are obtained. ∎

######  Proposition 2.7.

Let $k\geq 2$ and $m\geq 2$. Then, if $\frac{1}{2}(a_{k}+b_{1})\leq 1$, we
have $a_{j}=\frac{(2j-1)(a_{k}+b_{1})}{4k}$ for $1\leq j\leq k$,
$b_{1}=E(X:X\in[\frac{1}{2}(a_{k}+b_{1}),\frac{1}{2}(b_{1}+b_{2})])$, and

$b_{1+j}=\frac{1}{2}(b_{1}+b_{2})+\frac{(2j-1)}{2(m-1)}(2-\frac{1}{2}(b_{1}+b_{2}))\text{
for }1\leq j\leq m-1.$

On the other hand, if $1\leq\frac{1}{2}(a_{k}+b_{1})$, we have
$a_{j}=\frac{(2j-1)(a_{k-1}+a_{k})}{4(k-1)}$ for $1\leq j\leq(k-1)$,
$a_{k}=E(X:X\in[\frac{1}{2}(a_{k-1}+a_{k}),\frac{1}{2}(a_{k}+b_{1})])$, and

$b_{j}=\frac{1}{2}(a_{k}+b_{1})+\frac{(2j-1)}{2m}(2-\frac{1}{2}(a_{k}+b_{1}))\text{
for }1\leq j\leq m.$

The quantization errors for $n$-means are given by

$\displaystyle V1(k,m)$
$\displaystyle=p\frac{(a_{k}+b_{1}){}^{3}}{192k^{2}}+p\int_{\frac{1}{2}(a_{k}+b_{1})}^{1}(x-b_{1}){}^{2}\,dx+(1-p)\int_{2}^{\frac{1}{2}(b_{1}+b_{2})}(x-b_{1}){}^{2}\,dx$
$\displaystyle\qquad\qquad+(1-p)\frac{(4-(b_{1}+b_{2}))^{3}}{96(m-1)^{2}},\text{
and }$ $\displaystyle V2(k,m)$
$\displaystyle=p\frac{(a_{k-1}+a_{k}){}^{3}}{192(k-1)^{2}}+p\int_{\frac{1}{2}(a_{k-1}+a_{k})}^{1}(x-a_{k}){}^{2}\,dx+(1-p)\int_{1}^{\frac{1}{2}(a_{k}+b_{1})}(x-a_{k}){}^{2}\,dx$
$\displaystyle\qquad\qquad+(1-p)\frac{\left(4-(a_{k}+b_{1})\right){}^{3}}{96m^{2}}.$

###### Proof.

Notice that Proposition 2.7 is a mixture of Proposition 2.5 and Proposition
2.6, and thus the proof follows in the similar way. ∎

###  2.8. Quantization for the mixed distribution $P$ when $p=\frac{1}{5}$

Notice that if $p=\frac{1}{5}$, then $E(X)=\frac{13}{10}$ and
$V(X)=\frac{73}{300}$, i.e., the optimal set of one-mean for $p=\frac{1}{5}$
is $\\{\frac{13}{10}\\}$ and the corresponding quantization error is
$V(X)=\frac{73}{300}$.

######  Proposition 2.8.1.

The optimal set of two-means is $\\{\frac{11}{16},\frac{25}{16}\\}$ with
quantization error $V_{2}=\frac{317}{3840}$.

###### Proof.

Let $\alpha:=\\{a_{1},a_{2}\\}$ be an optimal set of two-means. Recall that
the mixed distribution $P$ has support $[0,2]$. Since the points in an optimal
set are the conditional expectations in their own Voronoi regions, without any
loss of generality, we can assume that $0<a_{1}<a_{2}<2$. The following cases
can arise:

Case 1. $0<a_{1}<a_{2}\leq 1$.

Since the boundary of the Voronoi region is $\frac{1}{2}(a_{1}+a_{2})$, we
have the distortion error as

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP=\frac{1}{5}\int_{0}^{\frac{1}{2}\left(a_{1}+a_{2}\right)}\left(x-a_{1}\right){}^{2}\,dx+\frac{1}{5}\int_{\frac{1}{2}\left(a_{1}+a_{2}\right)}^{1}\left(x-a_{2}\right){}^{2}\,dx+\frac{4}{5}\int_{1}^{2}\left(x-a_{2}\right){}^{2}\,dx$
$\displaystyle=\frac{1}{60}\left(3a_{1}^{3}+3a_{2}a_{1}^{2}-3a_{2}^{2}a_{1}-3a_{2}^{3}+60a_{2}^{2}-156a_{2}+116\right),$

the minimum value of which is $\frac{37}{135}=0.274074$ and it occurs when
$a_{1}=\frac{1}{3},$ and $a_{2}=1$.

Case 2. $0<a_{1}\leq 1<a_{2}<2$.

In this case, the boundary $\frac{1}{2}(a_{1}+a_{2})$ of the Voronoi regions
of $a_{1}$ and $a_{2}$ satisfies one of the following two conditions: either
$\frac{a_{1}+a_{2}}{2}\leq 1$, or $1\leq\frac{a_{1}+a_{2}}{2}$. Consider the
following two subcases:

Subcase 1. $0<a_{1}<\frac{a_{1}+a_{2}}{2}\leq 1<a_{2}<2$.

In this subcase, the distortion error is given by

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP=\frac{1}{5}\int_{0}^{\frac{1}{2}\left(a_{1}+a_{2}\right)}\left(x-a_{1}\right){}^{2}\,dx+\frac{1}{5}\int_{\frac{1}{2}\left(a_{1}+a_{2}\right)}^{1}\left(x-a_{2}\right){}^{2}\,dx+\frac{4}{5}\int_{1}^{2}\left(x-a_{2}\right){}^{2}\,dx$
$\displaystyle=\frac{1}{60}\left(3a_{1}^{3}+3a_{2}a_{1}^{2}-3a_{2}^{2}a_{1}-3a_{2}^{3}+60a_{2}^{2}-156a_{2}+116\right),$

the minimum value of which is $\frac{1}{12}=0.0833333$ and it occurs when
$a_{1}=\frac{1}{2},$ and $a_{2}=\frac{3}{2}$.

Subcase 2. $0<a_{1}\leq 1\leq\frac{a_{1}+a_{2}}{2}<a_{2}<2$.

In this subcase, the distortion error is given by

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP=\frac{1}{5}\int_{0}^{1}\left(x-a_{1}\right){}^{2}\,dx+\frac{4}{5}\int_{1}^{\frac{1}{2}\left(a_{1}+a_{2}\right)}\left(x-a_{1}\right){}^{2}\,dx+\frac{4}{5}\int_{\frac{1}{2}\left(a_{1}+a_{2}\right)}^{2}\left(x-a_{2}\right){}^{2}\,dx$
$\displaystyle=\frac{1}{15}\left(3a_{1}^{3}+3\left(a_{2}-3\right)a_{1}^{2}-3\left(a_{2}^{2}-3\right)a_{1}-3a_{2}^{3}+24a_{2}^{2}-48a_{2}+29\right).$

the minimum value of which is $\frac{317}{3840}=0.0825521$ and it occurs when
$a_{1}=\frac{11}{16},$ and $a_{2}=\frac{25}{16}$.

Case 3. $1\leq a_{1}<a_{2}<2$.

In this case the distortion error is given by

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP$
$\displaystyle=\frac{1}{5}\int_{0}^{1}\left(x-a_{1}\right){}^{2}\,dx+\frac{4}{5}\int_{1}^{\frac{1}{2}\left(a_{1}+a_{2}\right)}\left(x-a_{1}\right){}^{2}\,dx+\frac{4}{5}\int_{\frac{1}{2}\left(a_{1}+a_{2}\right)}^{2}\left(x-a_{2}\right){}^{2}\,dx$
$\displaystyle=\frac{1}{15}\left(3a_{1}^{3}+3\left(a_{2}-3\right)a_{1}^{2}-3\left(a_{2}^{2}-3\right)a_{1}-3a_{2}^{3}+24a_{2}^{2}-48a_{2}+29\right),$

the minimum value of which is $\frac{13}{135}=0.0962963$ and it occurs when
$a_{1}=1,$ and $a_{2}=\frac{5}{3}$.

Comparing the distortion errors obtained in all the above possible cases, we
see that the distortion error in Subcase 2 is the smallest. Thus, the optimal
set of two-means is $\\{\frac{11}{16},\frac{25}{16}\\}$ with quantization
error $V_{2}=\frac{317}{3840}$, which is the proposition. ∎

######  Proposition 2.8.2.

The set $\\{0.400679,1.202036,1.734012\\}$ forms an optimal set of three-means
with quantization error $V_{3}=0.0295695$.

###### Proof.

Let $\alpha:=\\{a_{1},a_{2},a_{3}\\}$ be an optimal set of three-means with
quantization error $V_{3}$. Without any loss of generality, we can assume that
$0<a_{1}<a_{2}<a_{3}<2$. Consider the set of three points
$\beta:=\\{0.400679,1.202036,1.734012\\}$. The distortion error due to the set
$\beta$ is given by

$\int\min_{a\in\beta}(x-a)^{2}dP=0.0295695.$

Since $V_{3}$ is the quantization error for three-means, we have $V_{3}\leq
0.0295695$. If $1\leq a_{1}$, then

$V_{3}\geq\frac{1}{5}\int_{0}^{1}(x-1)^{2}\,dx=\frac{1}{15}=0.0666667>V_{3},$

which is a contradiction. Hence, $a_{1}<1$. Similarly, we can show that
$a_{3}\leq 1$ leads to a contradiction. Hence, $1<a_{3}$, and recall
Proposition 2.5 to Proposition 2.7. We now consider the following possible
cases:

Case 1. $0<a_{1}<\frac{1}{2}(a_{1}+a_{2})<a_{2}<\frac{1}{2}(a_{2}+a_{3})\leq
1<a_{3}<2$.

In this case the distortion error is given by

$\displaystyle
V1(2,1)=\int\min_{a\in\alpha}(x-a)^{2}dP=\frac{51a_{2}^{2}a_{3}-45a_{2}a_{3}^{2}-15a_{2}^{3}-111a_{3}^{3}+1920a_{3}^{2}-4992a_{3}+3712}{1920},$

the minimum value of which is $0.0708333$ and it occurs when $a_{2}=0.5$ and
$a_{3}=1.5$.

Case 2. $0<a_{1}<\frac{1}{2}(a_{1}+a_{2})<a_{2}\leq
1\leq\frac{1}{2}(a_{2}+a_{3})<a_{3}<2$.

In this case the distortion error is given by

$\displaystyle V2(2,1)=\int\min_{a\in\alpha}(x-a)^{2}dP$
$\displaystyle=\frac{1}{480}(96a_{2}^{2}(a_{3}-3)-96a_{2}(a_{3}^{2}-3)-3a_{1}^{3}+15a_{2}a_{1}^{2}-9a_{2}^{2}a_{1}+69a_{2}^{3}-32(3a_{3}^{3}-24a_{3}^{2}+48a_{3}-29))$

the minimum value of which is $0.037037$ and it occurs when $a_{1}=0.333333$,
$a_{2}=1$, and $b_{1}=1.66667$.

Case 3. $0<a_{1}<\frac{1}{2}(a_{1}+a_{2})\leq
1<a_{2}<\frac{1}{2}(a_{2}+a_{3})<a_{3}<2$.

In this case the distortion error is given by

$\displaystyle V1(1,2)=\int\min_{a\in\alpha}(x-a)^{2}dP$
$\displaystyle=\frac{1}{120}(6a_{1}^{2}a_{2}-6a_{1}a_{2}^{2}+6a_{1}^{3}+21a_{2}^{3}+3a_{3}^{3}+12a_{3}^{2}-48a_{3}+a_{2}^{2}(9a_{3}-60)-3a_{2}(5a_{3}^{2}-8a_{3}-8)+40),$

the minimum value of which is $0.0295695$ and it occurs when
$a_{1}=0.400679,a_{2}=1.202036$, and $a_{3}=1.734012$.

Case 4.
$0<a_{1}<1\leq\frac{1}{2}\left(a_{1}+a_{2}\right)<a_{2}<\frac{1}{2}\left(a_{2}+a_{3}\right)<a_{3}<2$.

In this case the distortion error is given by

$\displaystyle V2(1,2)=\int\min_{a\in\alpha}(x-a)^{2}dP$
$\displaystyle=\frac{1}{480}(3a_{1}^{2}(15a_{2}-92)+a_{1}(-51a_{2}^{2}+24a_{2}+240)+111a_{1}^{3}+15a_{2}^{3}+12a_{2}^{2}-48a_{2}-32),$

the minimum value of which is $0.0333333$ and it occurs when
$a_{1}=0.5,a_{2}=1.5$.

Thus, considering all the possible cases, we see that
$\\{0.400679,1.202036,1.734012\\}$ forms an optimal set of three-means with
quantization error $V_{3}=0.0295695$, which is the proposition. ∎

######  Lemma 2.8.3.

Let $\alpha_{n}$ be an optimal set of $n$-means for $P$ for any $n\geq 2$.
Then, $\alpha_{n}$ must contain a point from $[0,1]$ and a point from $(1,2]$,
i.e., $\alpha_{n}\cap[0,1]\neq\emptyset$ and
$\alpha_{n}\cap(1,2]\neq\emptyset$.

###### Proof.

By Proposition 2.8.1 and Proposition 2.8.2, the lemma is true for $n=2$ and
$n=3$. We now prove the lemma for $n\geq 4$. Let
$\alpha_{n}:=\\{a_{1},a_{2},\cdots,a_{n}\\}$ be an optimal set of $n$-means
with $0<a_{1}<a_{2}<\cdots<a_{n}$ where $n\geq 4$. Since $V_{n}$ is the
quantization error for $n$-means with $n\geq 4$, we have $V_{n}<V_{3}$, i.e,
$V_{n}<0.0295695$. We prove the lemma by contradiction. Suppose that $1\leq
a_{1}$, then

$V_{n}\geq\int_{0}^{1}(x-1)^{2}dP=\frac{1}{15}>V_{3},$

which leads to a contradiction. Next, suppose that $a_{n}\leq 1$, then

$V_{n}\geq\int_{1}^{2}(x-1)^{2}dP=\frac{4}{15}>V_{n},$

which gives a contradiction. Thus, we can deduce that $a_{1}<1$ and $1<a_{n}$,
i.e., the lemma is also true for $n\geq 4$. Thus, the proof of the lemma is
complete. ∎

######  Lemma 2.8.4.

For $n\geq 2$, let $\alpha_{n}$ be an optimal set of $n$-means for $P$. Assume
that $\text{card}(\alpha_{n}\cap[0,1])=k$ and
$\text{card}(\alpha_{n}\cap(1,2])=m$ for some positive integers $k$ and $m$.
Then, either $\text{card}(\alpha_{n+1}\cap[0,1])=k+1$ and
$\text{card}(\alpha_{n+1}\cap(1,2])=m$, or
$\text{card}(\alpha_{n+1}\cap[0,1])=k$ and
$\text{card}(\alpha_{n+1}\cap(1,2])=m+1$.

###### Proof.

Assume that $\text{card}(\alpha_{n}\cap[0,1])=k$ and
$\text{card}(\alpha_{n}\cap(1,2])=m$ for some positive integers $k$ and $m$.
Notice that $k$ and $m$ depend on $n$, i.e., we can write $k:=k(n)$ and
$m:=m(n)$. Let $V(k(n),m(n))$ be the corresponding distortion error. By
Proposition 2.8.1 and Proposition 2.8.2, we know that
$\alpha_{2}=\\{\frac{11}{16},\frac{25}{16}\\}$ and
$\alpha_{3}=\\{0.400679,1.202036,1.734012\\}$. Notice that here
$\text{card}(\alpha_{2}\cap[0,1])=\text{card}(\alpha_{3}\cap[0,1])=1$, on the
other hand, $\text{card}(\alpha_{2}\cap(1,2])=1$ and
$\text{card}(\alpha_{3}\cap(1,2])=2$. Thus, the lemma is true for $n=2$.

Let the lemma be true for $n=N$ for some given positive integer $N\geq 2$.
Then, $\text{card}(\alpha_{N}\cap[0,1])=k(N)\text{ and
}\text{card}(\alpha_{N}\cap(1,2])=m(N)$ imply that either
$\text{card}(\alpha_{N+1}\cap[0,1])=k(N)+1$ and
$\text{card}(\alpha_{N+1}\cap(1,2])=m(N)$, or
$\text{card}(\alpha_{N+1}\cap[0,1])=k(N)$ and
$\text{card}(\alpha_{N+1}\cap(1,2])=m(N)+1$. Suppose that
$\text{card}(\alpha_{N+1}\cap[0,1])=k(N)+1$ and
$\text{card}(\alpha_{N+1}\cap(1,2])=m(N)$ hold. Now, for the given $N$, by
calculating the distortion errors $V(\ell,N+2-\ell)$ for all $1\leq\ell\leq
N+1$, we see that the distortion error is smallest if
$\text{card}(\alpha_{N+2}\cap[0,1])=k(N)+2\text{ and
}\text{card}(\alpha_{N+2}\cap(1,2])=m(N),\text{ or if
}\text{card}(\alpha_{N+2}\cap[0,1])=k(N)+1\text{ and
}\text{card}(\alpha_{N+2}\cap(1,2])=m(N)+1$, i.e., the lemma is true for
$n=N+1$ whenever it is true for $n=N$. Similarly, we can show that the lemma
is true for $n=N+1$ if $\text{card}(\alpha_{N+1}\cap[0,1])=k(N)\text{ and
}\text{card}(\alpha_{N+1}\cap(1,2])=m(N)+1$ hold. Thus, by the induction
principle, the proof of the lemma is complete. ∎

######  Remark 2.8.5.

By Lemma 2.8.3 and Lemma 2.8.4, we see that for $n\geq 2$, if $\alpha_{n}$ is
an optimal set of $n$-means for $P$, then there exist two positive integers
$k:=k(n)$ and $m:=m(n)$ such that $\text{card}(\alpha_{n}\cap[0,1])=k$ and
$\text{card}(\alpha_{n}\cap(1,2])=m$, and $k+m=n$. Let $F(k,m)$ be the
corresponding quantization error for $n$-means. Notice that the function
$F(k,m)$ satisfies

$F(k,m)=\min\\{V1(k,m),V2(k,m)\\}.$

Let us now give the following definition.

######  Definition 2.8.6.

Define the sequence $\\{a(n)\\}$ such that
$a(n)=\lfloor\frac{8(n+1)}{21}\rfloor$ for $n\in\mathbb{N}$, i.e.,

$\displaystyle\\{a(n)\\}_{n=1}^{\infty}=$
$\displaystyle\\{0,1,1,1,2,2,3,3,3,4,4,4,5,5,6,6,6,7,7,8,8,8,9,9,9,10,10,11,11,11,12,$
$\displaystyle\qquad 12,12,13,13,14,14,14,15,15,16,16,16,17,17,17,\cdots\\},$

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

The following algorithm helps us to determine the exact value of $k$, and so
the value of $m:=n-k$, as mentioned in Remark 2.8.5.

####  2.8.7. Algorithm.

For $n\geq 2$ let $k$ and $m:=n-k$ be the positive integers as defined in
Remark 2.8.5, and let $F(k,n-k)$ be the corresponding distortion error. Let
$\\{a(n)\\}$ be the sequence defined by Definition 2.8.6. Then, the algorithm
runs as follows:

$(i)$ Write $k:=a(n)$ and calculate $F(k,n-k)$.

$(ii)$ If $F(k-1,n-k+1)<F(k,n-k)$ replace $k$ by $k-1$ and return, else step
$(iii)$.

$(iii)$ If $F(k+1,n-k-1)<F(k,n-k)$ replace $k$ by $k+1$ and return, else step
$(iv)$.

$(iv)$ End.

When the algorithm ends, then the value of $k$, obtained, is the exact value
of $k$ that $\alpha_{n}$ contains from the closed interval $[0,1]$.

####  2.8.8. Optimal sets of $n$-means and the $n$th quantization errors for
all $n\geq 2$ when $p=\frac{1}{5}$.

If $n=30$, then $a(n)=11$, and by the algorithm, we also obtain $k=11$; if
$n=51$, then $a(n)=19$, and by the algorithm, we also obtain $k=19$. If
$n=1000$, then $a(n)=381$, and by the algorithm, we obtain $k=386$. Thus, we
see that with the help of the sequence and the algorithm, we can easily
determine the exact value of $k$ and $m:=n-k$ for any positive integer $n\geq
2$. Notice that $F(k,m)=\min\\{V1(k,m),V2(k,m)\\}$, i.e., either
$V_{n}=V1(k,m)$, or $V_{n}=V2(k,m)$. Thus, in either case, once $k$ and $m$
are known, by using Proposition 2.5 to Proposition 2.7, we can calculate the
optimal sets $\alpha_{n}$ of $n$-means and the corresponding quantization
errors $V_{n}:=F(k,m)$ for any $n\geq 2$.

###  2.9. Quantization for the mixed distribution $P$ when $p=\frac{1}{3}$

Notice that if $p=\frac{1}{3}$, then $E(X)=\frac{7}{6}$ and
$V(X)=\frac{11}{36}$, i.e., the optimal set of one-mean for $p=\frac{1}{3}$ is
$\\{\frac{7}{6}\\}$ and the corresponding quantization error is
$V(X)=\frac{11}{36}$.

The optimal set of two- and three-means for the mixed distribution
$P:=pP_{1}+(1-p)P_{2}$ when $p=\frac{1}{3}$ are given by the following two
propositions.

######  Proposition 2.9.1.

The optimal set of two-means is $\\{\frac{1}{2},\frac{3}{2}\\}$ with
quantization error $V_{2}=\frac{1}{12}$.

###### Proof.

Let $\alpha:=\\{a_{1},a_{2}\\}$ be an optimal set of two-means, where
$0<a_{1}<a_{2}<2$. If $P=\frac{1}{2}P_{1}+\frac{1}{2}P_{2}$, i.e., if $P$ is a
uniform distribution, then by Proposition 2.4, we know that the optimal set of
two-means is given by $\\{\frac{1}{2},\frac{3}{2}\\}$. Notice that here the
mixed distribution is given by $P=\frac{1}{3}P_{1}+\frac{2}{3}P_{2}$, i.e., we
are giving more weight to the right-hand interval $J_{2}$. This leads us to
conclude that the Voronoi region of $a_{1}$ may contain some points from
$J_{2}$ yielding the fact that $\frac{1}{2}\leq a_{1}$, i.e, the boundary
$\frac{1}{2}(a_{1}+a_{2})$ of the two Voronoi regions must satisfy
$1\leq\frac{1}{2}(a_{1}+a_{2})$. Under this condition, we have the distortion
error for the optimal set $\alpha=\\{a_{1},a_{2}\\}$ as

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP=\frac{1}{3}\int_{0}^{1}(x-a_{1}){}^{2}\,dx+\frac{2}{3}\int_{1}^{\frac{1}{2}(a_{1}+a_{2})}(x-a_{1}){}^{2}\,dx+\frac{2}{3}\int_{\frac{1}{2}(a_{1}+a_{2})}^{2}(x-a_{2}){}^{2}\,dx$
$\displaystyle=\frac{1}{6}\Big{(}a_{1}^{2}(a_{2}-2)-a_{1}(a_{2}^{2}-2)+a_{1}^{3}-a_{2}^{3}+8a_{2}^{2}-16a_{2}+10\Big{)},$

the minimum value of which is $\frac{1}{12}$, and it occurs when
$a_{1}=\frac{1}{2}$ and $a_{2}=\frac{3}{2}$. Thus, the proposition is yielded.
∎

######  Remark 2.9.2.

It is interesting to see that the optimal set of two-means for the mixed
distribution $P:=\frac{1}{3}P_{1}+\frac{2}{3}P_{2}$ is same as the optimal set
of two-means for the uniform distribution $P$ given by
$P:=\frac{1}{2}P_{1}+\frac{1}{2}P_{2}$.

######  Proposition 2.9.3.

The set $\\{0.380129,1.14039,1.71346\\}$ forms an optimal set of three-means
with quantization error $V_{3}=0.0343006$.

###### Proof.

Let $\alpha:=\\{a_{1},a_{2},a_{3}\\}$ be an optimal set of three-means with
quantization error $V_{3}$. Without any loss of generality, we can assume that
$0<a_{1}<a_{2}<a_{3}<2$. Consider the set of three points
$\beta:=\\{0.380129,1.14039,1.71346\\}$. The distortion error due to the set
$\beta$ is given by

$\int\min_{a\in\beta}(x-a)^{2}dP=0.0343006.$

Since $V_{3}$ is the quantization error for three-means, we have $V_{3}\leq
0.0343006$. As shown in the proof of Proposition 2.8.2, we can show that
$a_{1}<1$ and $1<a_{3}$. We now consider the following possible cases:

Case 1. $0<a_{1}<\frac{1}{2}(a_{1}+a_{2})<a_{2}<\frac{1}{2}(a_{2}+a_{3})\leq
1<a_{3}<2$.

In this case the distortion error is given by $V1(2,1)=0.0625000.$

Case 2. $0<a_{1}<\frac{1}{2}(a_{1}+a_{2})<a_{2}\leq
1\leq\frac{1}{2}(a_{2}+a_{3})<a_{3}<2$.

In this case the distortion error is given by $V2(2,1)=0.037037$.

Case 3. $0<a_{1}<\frac{1}{2}(a_{1}+a_{2})\leq
1<a_{2}<\frac{1}{2}(a_{2}+a_{3})<a_{3}<2$.

In this case the distortion error is given by $V1(1,2)=0.0343006$.

Case 4.
$0<a_{1}<1\leq\frac{1}{2}\left(a_{1}+a_{2}\right)<a_{2}<\frac{1}{2}\left(a_{2}+a_{3}\right)<a_{3}<2$.

In this case the distortion error is given by $V2(1,2)=0.0416667.$

Thus, considering all the possible cases we see that in Case 3 the distortion
error is smallest, and it occurs when $a_{1}=0.380129$, $a_{2}=1.14039$, and
$a_{3}=1.71346$. Thus, the proof of the proposition is complete. ∎

The following two lemmas, which are analogous to Lemma 2.8.3 and Lemma 2.8.4
are also true here.

######  Lemma 2.9.4.

Let $\alpha_{n}$ be an optimal set of $n$-means for $P$ where $n\geq 2$. Then,
$\alpha_{n}$ must contain a point from $[0,1]$ and a point from $(1,2]$, i.e.,
$\alpha_{n}\cap[0,1]\neq\emptyset$ and $\alpha_{n}\cap(1,2]\neq\emptyset$.

######  Lemma 2.9.5.

For $n\geq 2$, let $\alpha_{n}$ be an optimal set of $n$-means for $P$. Assume
that $\text{card}(\alpha_{n}\cap[0,1])=k$ and
$\text{card}(\alpha_{n}\cap(1,2])=m$ for some positive integers $k$ and $m$.
Then, either $\text{card}(\alpha_{n+1}\cap[0,1])=k+1$ and
$\text{card}(\alpha_{n+1}\cap(1,2])=m$, or
$\text{card}(\alpha_{n+1}\cap[0,1])=k$ and
$\text{card}(\alpha_{n+1}\cap(1,2])=m+1$.

######  Remark 2.9.6.

By Lemma 2.9.4 and Lemma 2.9.5, we see that for $n\geq 2$, if $\alpha_{n}$ is
an optimal set of $n$-means for $P$, then there exist two positive integers
$k:=k(n)$ and $m:=m(n)$ such that $\text{card}(\alpha_{n}\cap[0,1])=k$ and
$\text{card}(\alpha_{n}\cap(1,2])=m$, and $k+m=n$. Let $F(k,m)$ be the
corresponding quantization error for $n$-means. Notice that the function
$F(k,m)$ satisfies

$F(k,m)=\min\\{V1(k,m),V2(k,m)\\}.$

Let us now give the following sequence and the algorithm which are analogous
to the sequence and the algorithm defined in the previous subsection.

######  Definition 2.9.7.

Define the sequence $\\{b(n)\\}$ such that
$b(n)=\lfloor\frac{4(n+1)}{7}\rfloor$ for $n\in\mathbb{N}$, i.e.,

$\displaystyle\\{b(n)\\}_{n=1}^{\infty}=$
$\displaystyle\\{1,1,2,2,3,4,4,5,5,6,6,7,8,8,9,9,10,10,11,12,12,13,13,14,14,$
$\displaystyle\qquad 15,16,16,17,17,18,18,19,20,20,21,21,22,22,23,\cdots\\},$

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

####  2.9.8. Algorithm.

For $n\geq 2$ let $k:=n-m$ and $m$ be the positive integers as defined in
Remark 2.9.6, and let $F(n-m,m)$ be the corresponding distortion error. Let
$\\{b(n)\\}$ be the sequence defined by Definition 2.9.7. Then, the algorithm
runs as follows:

$(i)$ Write $m:=b(n)$ and calculate $F(n-m,m)$.

$(ii)$ If $F(n-m+1,m-1)<F(n-m,m)$ replace $m$ by $m-1$ and return, else step
$(iii)$.

$(iii)$ If $F(n-m-1,m+1)<F(n-m,m)$ replace $m$ by $m+1$ and return, else step
$(iv)$.

$(iv)$ End.

When the algorithm ends, then the value of $m$, obtained, is the exact value
of $m$ that $\alpha_{n}$ contains from the interval $(1,2]$.

####  2.9.9. Optimal sets of $n$-means and the $n$th quantization errors for
all $n\geq 2$ when $p=\frac{1}{3}$.

If $n=21$, then $b(n)=12$, and by the algorithm we also obtain $m=12$; if
$n=100$, then $b(n)=57$, and by the algorithm we obtain $m=56$. If $n=500$,
then $b(n)=286$, and by the algorithm, we obtain $m=279$. Thus, we see that
with the help of the sequence and the algorithm, we can easily determine the
exact values of $k:=n-m$ and $m$ for any positive integer $n\geq 2$. Once $k$
and $m$ are known, by using Proposition 2.5 to Proposition 2.7, we can
calculate the optimal sets $\alpha_{n}$ of $n$-means and the corresponding
quantization errors $V_{n}:=F(k,m)$ for any positive integer $n\geq 2$.

##  3\. Mixed distribution $P$ with support the union of the disconnected line
segments $[0,\frac{1}{3}]$ and $[\frac{2}{3},1]$

###  3.1. Basic preliminaries

Let $P_{1}$ and $P_{2}$ be two uniform distributions, respectively, defined on
the intervals given by

$J_{1}:=[0,\frac{1}{3}],\text{ and }J_{2}:=[\frac{2}{3},1].$

Let $f_{1}$ and $f_{2}$ be their respective density functions. Then,

$f_{1}(x)=\left\\{\begin{array}[]{cc}3&\text{ if }x\in[0,\frac{1}{3}],\\\
0&\text{otherwise};\end{array}\right.\text{ and
}f_{2}(x)=\left\\{\begin{array}[]{cc}3&\text{ if }x\in[\frac{2}{3},1],\\\
0&\text{otherwise}.\end{array}\right.$

The underlying mixed distribution considered is given by
$P:=pP_{1}+(1-p)P_{2}$, where $0<p<1$. By $E(X)$, it is meant the expectation
of a random variable $X$ with probability distribution $P$, and $V(X)$
represents the variance of $X$.

######  Proposition 3.2.

Let $P$ be the mixed distribution defined by $P=pP_{1}+(1-p)P_{2}$. Then,
$E(X)=\frac{1}{6}(5-4p)$, and $V(X)=\frac{1}{108}(-48p^{2}+48p+1)$.

###### Proof.

We have

$E(X)=\int xdP=p\int xd(P_{1}(x))+(1-p)\int
xd(P_{2}(x))=p\int_{J_{1}}3x\,dx+(1-p)\int_{J_{2}}3x\,dx$

yielding $E(X)=\frac{1}{6}(5-4p)$, and

$V(X)=\int(x-E(X))^{2}dP=p\int(x-E(X))^{2}d(P_{1}(x))+(1-p)\int(x-E(X))^{2}d(P_{2}(x)),$

implying $V(X)=\frac{1}{108}(-48p^{2}+48p+1)$, and thus, the proposition is
yielded. ∎

######  Remark 3.3.

The optimal set of one-mean is the set $\\{\frac{1}{6}(5-4p)\\}$, and the
corresponding quantization error is the variance $V:=V(X)$ of a random
variable with distribution $P:=pP_{1}+(1-p)P_{2}$.

In the next two subsections, taking $p=\frac{1}{100}$, $p=\frac{2}{5}$, and
$p=\frac{1}{1000}$, for all positive integers $n$, we calculate the optimal
set of $n$-means and the $n$th quantization errors for the mixed distribution
$P:=pP_{1}+(1-p)P_{2}$ with support the union of two disconnected line
segments $[0,\frac{1}{3}]$ and $[\frac{2}{3},1]$.

###  3.4. Quantization for the mixed distribution $P$ when $p=\frac{1}{100}$

Notice that if $p=\frac{1}{100}$, then $E(X)=\frac{62}{75}$ and
$V(X)=\frac{461}{33750}$, i.e., the optimal set of one-mean for
$p=\frac{1}{5}$ is $\\{\frac{62}{75}\\}$ and the corresponding quantization
error is $V(X)=\frac{461}{33750}$.

######  Proposition 3.4.1.

The optimal set of two-means is $\\{0.731517,0.910506\\}$ with quantization
error $V_{2}=0.005682$.

###### Proof.

Let $\alpha:=\\{a_{1},a_{2}\\}$ be an optimal set of two-means. Since the
points in an optimal set are the conditional expectations in their own Voronoi
regions, without any loss of generality, we can assume that $0<a_{1}<a_{2}<1$.
If $\frac{1}{3}<a_{1}<a_{2}<\frac{2}{3}$, then the quantization error can be
strictly reduced by moving the point $a_{1}$ to $\frac{1}{3}$, and $a_{2}$ to
$\frac{2}{3}$. Thus, $\frac{1}{3}<a_{1}<a_{2}<\frac{2}{3}$ is not possible.
Let us now discuss all the possible cases:

Case 1. $0<a_{1}<a_{2}\leq\frac{1}{3}$.

Since the boundary of the Voronoi region is $\frac{1}{2}(a_{1}+a_{2})$, we
have the distortion error as

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP=\int_{0}^{\frac{a_{1}+a_{2}}{2}}(x-a_{1})^{2}dP+\int_{\frac{a_{1}+a_{2}}{2}}^{\frac{1}{3}}(x-a_{2})^{2}dP+\int_{\frac{2}{3}}^{1}(x-a_{2})^{2}dP$
$\displaystyle=\frac{1}{100}\int_{0}^{\frac{1}{2}\left(a_{1}+a_{2}\right)}3\left(x-a_{1}\right){}^{2}\,dx+\frac{1}{100}\int_{\frac{1}{2}\left(a_{1}+a_{2}\right)}^{\frac{1}{3}}3\left(x-a_{2}\right){}^{2}\,dx+\frac{99}{100}\int_{\frac{2}{3}}^{1}3\left(x-a_{2}\right){}^{2}\,dx$
$\displaystyle=\frac{81a_{1}^{3}+81a_{2}a_{1}^{2}-81a_{2}^{2}a_{1}-81a_{2}^{3}+10800a_{2}^{2}-17856a_{2}+7528}{10800},$

the minimum value of which is $\frac{3119}{12150}$ and it occurs when
$a_{1}=\frac{1}{9},$ and $a_{2}=\frac{1}{3}$.

Case 2. $0<a_{1}\leq\frac{1}{3}<a_{2}<\frac{2}{3}$.

In this case, the boundary $\frac{1}{2}(a_{1}+a_{2})$ of the Voronoi regions
of $a_{1}$ and $a_{2}$ must satisfy
$0<a_{1}<\frac{1}{2}(a_{1}+a_{2})<\frac{1}{3}$, otherwise the quantization
error can be strictly reduced by moving the point $a_{2}$ to $\frac{2}{3}$.
Hence, the distortion error in this case is given by

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP$
$\displaystyle=\frac{1}{100}\int_{0}^{\frac{1}{2}\left(a_{1}+a_{2}\right)}3\left(x-a_{1}\right){}^{2}\,dx+\frac{1}{100}\int_{\frac{1}{2}\left(a_{1}+a_{2}\right)}^{\frac{1}{3}}3\left(x-a_{2}\right){}^{2}\,dx+\frac{99}{100}\int_{\frac{2}{3}}^{1}3\left(x-a_{2}\right){}^{2}\,dx$
$\displaystyle=\frac{81a_{1}^{3}+81a_{2}a_{1}^{2}-81a_{2}^{2}a_{1}-81a_{2}^{3}+10800a_{2}^{2}-17856a_{2}+7528}{10800},$

the minimum value of which is $\frac{89}{2430}$ and it occurs when
$a_{1}=\frac{2}{9},$ and $a_{2}=\frac{2}{3}$.

Case 3. $0<a_{1}\leq\frac{1}{3}<\frac{2}{3}\leq a_{2}$.

In this case, the Voronoi region of $a_{1}$ does not contain any point from
$J_{2}$, if it does, then we must have $\frac{1}{2}(a_{1}+a_{2})>\frac{2}{3}$
implying $a_{2}>\frac{4}{3}-a_{1}\geq\frac{4}{3}-\frac{1}{3}=1$, which is a
contradiction as $a_{2}<1$. Similarly, we can show that the Voronoi region of
$a_{2}$ does not contain any point from $J_{1}$. This yields the fact that

$a_{1}=E(X:X\in J_{1})=\frac{1}{6},\text{ and }a_{2}=E(X:X\in
J_{2})=\frac{5}{6},$

with distortion error

$\int\min_{a\in\alpha}(x-a)^{2}dP=\frac{1}{100}\int_{0}^{\frac{1}{3}}3(x-\frac{1}{6})^{2}\,dx+\frac{99}{100}\int_{\frac{2}{3}}^{1}3(x-\frac{5}{6})^{2}\,dx=\frac{1}{108}.$

Case 4. $\frac{1}{3}<a_{1}\leq\frac{2}{3}<a_{2}$.

In this case, the boundary $\frac{1}{2}(a_{1}+a_{2})$ of the Voronoi regions
of $a_{1}$ and $a_{2}$ must satisfy
$\frac{2}{3}<\frac{1}{2}(a_{1}+a_{2})<a_{2}<1$, otherwise the quantization
error can be strictly reduced by moving the point $a_{1}$ to $\frac{1}{3}$.
Hence, the distortion error in this case is given by

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP$
$\displaystyle=\frac{1}{100}\int_{0}^{\frac{1}{3}}3\left(x-a_{1}\right){}^{2}\,dx+\frac{99}{100}\int_{\frac{2}{3}}^{\frac{1}{2}\left(a_{1}+a_{2}\right)}3\left(x-a_{1}\right){}^{2}\,dx+\frac{99}{100}\int_{\frac{1}{2}\left(a_{1}+a_{2}\right)}^{1}3\left(x-a_{2}\right){}^{2}\,dx$
$\displaystyle=\frac{8019a_{1}^{3}+27\left(297a_{2}-788\right)a_{1}^{2}-9\left(891a_{2}^{2}-1580\right)a_{1}-8019a_{2}^{3}+32076a_{2}^{2}-32076a_{2}+7528}{10800},$

the minimum value of which is $\frac{1}{150}$ and it occurs when
$a_{1}=\frac{2}{3},$ and $a_{2}=\frac{8}{9}$.

Case 5. $\frac{2}{3}<a_{1}<a_{2}<1$.

In this case, the distortion error is given by

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP$
$\displaystyle=\frac{1}{100}\int_{0}^{\frac{1}{3}}3\left(x-a_{1}\right){}^{2}\,dx+\frac{99}{100}\int_{\frac{2}{3}}^{\frac{1}{2}\left(a_{1}+a_{2}\right)}3\left(x-a_{1}\right){}^{2}\,dx+\frac{99}{100}\int_{\frac{1}{2}\left(a_{1}+a_{2}\right)}^{1}3\left(x-a_{2}\right){}^{2}\,dx$
$\displaystyle=\frac{8019a_{1}^{3}+27\left(297a_{2}-788\right)a_{1}^{2}-9\left(891a_{2}^{2}-1580\right)a_{1}-8019a_{2}^{3}+32076a_{2}^{2}-32076a_{2}+7528}{10800},$

the minimum value of which is $0.005682$ and it occurs when $a_{1}=0.731517,$
and $a_{2}=0.910506$.

Comparing the distortion errors obtained in all the above possible cases, we
see that the distortion error in Case 5 is smallest. Thus, the optimal set of
two-means is $\\{0.731517,0.910506\\}$ with quantization error
$V_{2}=0.005682$, which is the proposition. ∎

######  Proposition 3.4.2.

Optimal set of three-means is $\\{\frac{1}{6},\frac{3}{4},\frac{11}{12}\\}$
with quantization error $V_{3}=\frac{103}{43200}$.

###### Proof.

Let $\alpha=\\{a_{1},a_{2},a_{3}\\}$ be an optimal set of three-means.
Proposition 1.1 implies that if $\alpha$ contains a point from the open
interval $(\frac{1}{3},\frac{2}{3})$, it cannot contain more than one point
from the open interval $(\frac{1}{3},\frac{2}{3})$. First, we assume that
$\alpha$ contains one point from $J_{1}$, and two points from $J_{2}$. Then,
$0<a_{1}\leq\frac{1}{3}<\frac{2}{3}\leq a_{2}<a_{3}<1$ yielding the fact that
the Voronoi region of $a_{1}$ does not contain any point from $J_{1}$, and the
Voronoi region of $a_{2}$, and so of $a_{3}$ cannot not contain any point from
$J_{1}$. This yields $a_{1}=\frac{1}{6}$, $a_{2}=\frac{3}{4}$, and
$a_{3}=\frac{11}{12}$ with distortion error

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP=\frac{1}{100}\int_{0}^{\frac{1}{3}}3(x-\frac{1}{6})^{2}\,dx+\frac{99}{100}\int_{\frac{2}{3}}^{\frac{5}{6}}3(x-\frac{3}{4})^{2}\,dx+\frac{99}{100}\int_{\frac{5}{6}}^{1}3(x-\frac{11}{12})^{2}\,dx$

yielding

$\int\min_{a\in\alpha}(x-a)^{2}dP=\frac{103}{43200}.$

Since $V_{3}$ is the quantization error for three-means, we have
$V_{3}\leq\frac{103}{43200}=0.00238426$. If $a_{3}<\frac{2}{3}$, then

$V_{3}\geq\frac{99}{100}\int_{\frac{2}{3}}^{1}3(x-\frac{2}{3})^{2}\,dx=\frac{11}{300}>V_{3},$

which is a contradiction. Hence, we can assume that $\frac{2}{3}<a_{3}$.
Suppose that $a_{2}<\frac{2}{3}$. Then,

$V_{3}\geq\int_{J_{3}}\min_{a\in\alpha}(x-a)^{2}dP=\int_{J_{3}}\min_{a\in\\{a_{2},a_{3}\\}}(x-a)^{2}dP\geq\int_{J_{3}}\min_{a\in\\{\frac{2}{3},a_{3}\\}}(x-a)^{2}dP$

implying

$\displaystyle V_{3}$
$\displaystyle\geq\frac{99}{100}\int_{\frac{2}{3}}^{\frac{1}{2}(a_{3}+\frac{2}{3})}3(x-\frac{2}{3})^{2}\,dx+\frac{99}{100}\int_{\frac{1}{2}\left(a_{3}+\frac{2}{3}\right)}^{1}3(x-a_{3}){}^{2}\,dx$
$\displaystyle=-\frac{11(81a_{3}^{3}-270a_{3}^{2}+288a_{3}-100)}{1200},$

the minimum value of which is $\frac{11}{2700}$ and it occurs when
$a_{3}=\frac{8}{9}$. Notice that $\frac{11}{2700}=0.00407407>V_{3}$, which
leads to a contradiction. Hence, we can assume that $\frac{2}{3}\leq
a_{2}<a_{3}<1$. Notice that for $\frac{2}{3}\leq a_{2}<a_{3}<1$, the Voronoi
region of $a_{2}$ does not contain any point from $J_{1}$. Suppose that
$\frac{2}{3}\leq a_{1}$. Then,

$V_{3}>\frac{1}{100}\int_{0}^{\frac{1}{3}}3(x-\frac{2}{3})^{2}\,dx=\frac{7}{2700}=0.00259259>V_{3},$

which leads to a contradiction. So, we can assume that $a_{1}<\frac{2}{3}$.
Suppose that $\frac{1}{3}<a_{1}<\frac{2}{3}$. Then, we must have
$\frac{2}{3}<\frac{1}{2}(a_{1}+a_{2})<a_{2}<a_{3}<1$ yielding the distortion
error as

$\displaystyle\int\min_{a\in\alpha}(x-a)^{2}dP$
$\displaystyle=\frac{1}{100}\int_{0}^{\frac{1}{3}}3\left(x-a_{1}\right){}^{2}\,dx+\frac{99}{100}\int_{\frac{2}{3}}^{\frac{1}{2}\left(a_{1}+a_{2}\right)}3\left(x-a_{1}\right){}^{2}\,dx+\frac{99}{100}\int_{\frac{1}{2}\left(a_{1}+a_{2}\right)}^{\frac{1}{2}\left(a_{2}+a_{3}\right)}3\left(x-a_{2}\right){}^{2}\,dx$
$\displaystyle\qquad\qquad\qquad\qquad+\frac{99}{100}\int_{\frac{1}{2}\left(a_{2}+a_{3}\right)}^{1}3\left(x-a_{3}\right){}^{2}\,dx$
$\displaystyle=\frac{1}{10800}\Big{(}8019a_{1}^{3}+27(297a_{2}-788)a_{1}^{2}-9(891a_{2}^{2}-1580)a_{1}-8019a_{3}^{3}-8019(a_{2}-4)a_{3}^{2}$
$\displaystyle\qquad\qquad\qquad\qquad+8019(a_{2}^{2}-4)a_{3}+7528\Big{)}$

the minimum value of which is $\frac{137}{33750}$, and it occurs when
$a_{1}=\frac{2}{3}$, $a_{2}=\frac{4}{5}$, and $a_{3}=\frac{14}{15}$. Notice
that $\frac{137}{33750}=0.00405926>V_{3}$, and thus a contradiction arises.
Hence, we can assume that $a_{1}\leq\frac{1}{3}$. Thus, as described before,
we deduce that $a_{1}=\frac{1}{6}$, $a_{2}=\frac{3}{4}$, and
$a_{3}=\frac{11}{12}$, and the quantization error for three-means is
$V_{3}=\frac{103}{43200}$. Thus, the proof of the proposition is complete. ∎

######  Lemma 3.4.3.

For $n\geq 3$ , let $\alpha_{n}$ be an optimal set of $n$-means for $P$. Then,
$\alpha_{n}\cap J_{1}\neq\emptyset$ and $\alpha_{n}\cap J_{2}\neq\emptyset$.

###### Proof.

By Proposition 3.4.2, the lemma is true for $n=3$. Let us now prove the lemma
for $n\geq 4$. The distortion error due to the set
$\beta:=\\{\frac{1}{6},\frac{2}{3}+\frac{1}{18},\frac{2}{3}+\frac{3}{18},\frac{2}{3}+\frac{5}{18}\\}=\\{\frac{1}{6},\frac{13}{18},\frac{15}{18},\frac{17}{18}\\}$
is given by

$\int\min_{a\in\beta}(x-a)^{2}dP=\int_{J_{1}}(x-\frac{1}{6})^{2}dP+\int_{J_{2}}\min_{a\in(\beta\setminus\\{\frac{1}{6}\\})}(x-a)^{2}dP=\frac{1}{900}.$

Since $V_{n}$ is the quantization error for $n$-means, where $n\geq 4$, we
have $V_{n}\leq\frac{1}{900}=0.00111111$. For $n\geq 4$, let
$\alpha_{n}:=\\{a_{1},a_{2},\cdots,a_{n}\\}$ be an optimal set of $n$-means
for $P$ such that $0<a_{1}<a_{2}<\cdots<a_{n}<1$. If $a_{n}<\frac{2}{3}$, then

$V_{n}>\int_{J_{2}}(x-\frac{2}{3})^{2}dP=\frac{11}{300}=0.0366667>V_{n},$

which leads to a contradiction. Hence, we can assume that
$a_{n}\geq\frac{2}{3}$, i.e., $\alpha_{n}\cap J_{2}\neq\emptyset$. We now show
that $\alpha_{n}\cap J_{1}\neq\emptyset$. If $\frac{1}{2}\leq a_{1}$, then

$V_{n}>\int_{J_{1}}(x-\frac{1}{2})^{2}dP=\frac{13}{10800}=0.0012037>V_{n},$

which yields a contradiction. Hence, we can assume that $a_{1}<\frac{1}{2}$.
Assume that $\frac{1}{3}\leq a_{1}<\frac{1}{2}$. Then, by Proposition 1.1, we
must have $\frac{1}{2}(a_{1}+a_{2})>\frac{2}{3}$ yielding
$a_{2}>\frac{4}{3}-a_{1}\geq\frac{4}{3}-\frac{1}{2}=\frac{5}{6}$, and so,

$V_{n}>\int_{J_{1}}(x-\frac{1}{3})^{2}dP+\int_{[\frac{2}{3},\frac{5}{6}]}(x-\frac{1}{2})^{2}dP=\frac{2681}{21600}=0.12412>V_{n},$

which leads to a contradiction. Hence, we can assume that $a_{1}<\frac{1}{3}$,
i.e., $\alpha_{n}\cap J_{1}\neq\emptyset$. Thus, the proof of the lemma is
complete. ∎

######  Lemma 3.4.4.

An optimal set of four-means does not contain any point from the open interval
$(\frac{1}{3},\frac{2}{3})$.

###### Proof.

Let $\alpha:=\\{a_{1},a_{2},a_{3},a_{4}\\}$, where
$0<a_{1}<a_{2}<a_{3}<a_{4}<1$, be an optimal set of four-means. As mentioned
in the proof of Lemma 3.4.3, we have $V_{4}\leq\frac{1}{900}=0.00111111$.
Suppose that $a_{2}\leq\frac{2}{3}$. Then,

$\displaystyle V_{4}$
$\displaystyle>\frac{99}{100}\int_{\frac{2}{3}}^{\frac{1}{2}\left(a_{3}+\frac{2}{3}\right)}3(x-\frac{2}{3})^{2}\,dx+\frac{99}{100}\int_{\frac{1}{2}\left(a_{3}+\frac{2}{3}\right)}^{\frac{1}{2}\left(a_{3}+a_{4}\right)}3\left(x-a_{3}\right){}^{2}\,dx+\frac{99}{100}\int_{\frac{1}{2}\left(a_{3}+a_{4}\right)}^{1}3\left(x-a_{4}\right){}^{2}\,dx$
$\displaystyle=\frac{11\left(-81a_{4}^{3}+324a_{4}^{2}-324a_{4}+27a_{3}^{2}\left(3a_{4}-2\right)-9a_{3}\left(9a_{4}^{2}-4\right)+100\right)}{1200}$

the minimum value of which is $\frac{11}{7500}$, and it occurs when
$a_{3}=\frac{4}{5}$ and $,a_{4}=\frac{14}{15}$ implying

$V_{4}>\frac{11}{7500}=0.00146667>V_{4},$

which leads to a contradiction. Thus, we can assume that $\frac{2}{3}<a_{2}$,
and so $\frac{2}{3}<a_{2}<a_{3}<a_{4}<1$. Again, by Lemma 3.4.3, we see that
$a_{1}<\frac{1}{3}$. Hence, an optimal set of four-means does not contain any
point from the open interval $(\frac{1}{3},\frac{2}{3})$, which is the lemma.
∎

######  Remark 3.4.5.

Proceeding in the similar way as Lemma 3.4.4, we can show that the optimal set
of five-means does not contain any point from the open interval
$(\frac{1}{3},\frac{2}{3})$.

######  Proposition 3.4.6.

For $n\geq 3$ , let $\alpha_{n}$ be an optimal set of $n$-means for $P$. Then,
$\alpha_{n}$ does not contain any point from the open interval
$(\frac{1}{3},\frac{2}{3})$. Moreover, the Voronoi region of any point in
$\alpha_{n}\cap J_{1}$ does not contain any point from $J_{2}$, and the
Voronoi region of any point in $\alpha_{n}\cap J_{2}$ does not contain any
point from $J_{1}$.

###### Proof.

By Proposition 3.4.2, Lemma 3.4.4 and Remark 3.4.5, the proposition is true
for $n=3,4,5$. We now prove the proposition for $n\geq 6$. Let
$\alpha_{n}:=\\{a_{1},a_{2},\cdots,a_{n}\\}$, where $n\geq 6$, be an optimal
set of $n$-means. Without any loss of generality, we can assume that
$0<a_{1}<a_{2}<\cdots<a_{n}<1$. Let us now consider the set of six points
$\beta:=\\{\frac{1}{6},\frac{7}{10},\frac{23}{30},\frac{5}{6},\frac{9}{10},\frac{29}{30}\\}$.
By routine calculation, the distortion error due to the set $\beta$ is given
by

$\int\min_{a\in\beta}(x-a)^{2}dP=\frac{1}{100}\int_{J_{1}}(x-\frac{1}{6})^{2}dP_{1}+\frac{99}{100}\int_{J_{2}}\min_{b\in(\beta\setminus{\\{\frac{1}{6}\\}})}(x-b)^{2}dP_{2}=\frac{31}{67500},$

and so, $V_{6}\leq\frac{31}{67500}=0.000459259$. Since $V_{n}$ is the
quantization error for six-means with $n\geq 6$, we have $V_{n}\leq V_{6}\leq
0.000459259$. By Lemma 3.4.3, we know that $a_{1}<\frac{1}{3}$ and
$a_{n}>\frac{2}{3}$. Let $k$ be the largest positive integer such that
$a_{k}\leq\frac{1}{3}$. For the sake of contradiction, assume that
$\alpha_{n}$ contains a point from the open interval
$(\frac{1}{3},\frac{2}{3})$. Then, by Proposition 1.1, we must have
$a_{k+1}\in(\frac{1}{3},\frac{2}{3})$, and $\frac{2}{3}\leq a_{k+2}$. The
following two cases can arise:

Case 1. $\frac{1}{3}<a_{k+1}\leq\frac{1}{2}$.

Then, the Voronoi region of $a_{k+1}$ must contain points from $J_{2}$, i.e.,
$\frac{1}{2}(a_{k+1}+a_{k+2})\geq\frac{2}{3}$ implying
$a_{k+2}\geq\frac{4}{3}-a_{k+1}\geq\frac{4}{3}-\frac{1}{2}=\frac{5}{6}$,
otherwise the quantization error can be strictly reduced by moving the point
$a_{k+1}$ to $\frac{1}{3}$. Then,

$\displaystyle
V_{n}\geq\int_{[\frac{2}{3},\frac{5}{6}]}(x-\frac{5}{6})^{2}dP=\frac{11}{2400}=0.00458333>V_{n},$

which is a contradiction.

Case 2. $\frac{1}{2}\leq a_{k+1}<\frac{2}{3}$.

Then, we must have $\frac{1}{2}(a_{k}+a_{k+1})\leq\frac{1}{3}$ implying
$a_{k}\leq\frac{2}{3}-a_{k+1}\leq\frac{2}{3}-\frac{1}{2}=\frac{1}{6}$, and so

$\displaystyle
V_{n}\geq\int_{[\frac{1}{6},\frac{1}{3}]}(x-\frac{1}{6})^{2}dP=\frac{1}{21600}=0.0000462963>V_{n},$

which leads to a contradiction.

By Case 1 and Case 2, we deduce that $\alpha_{n}$ does not contain any point
from the open interval $(\frac{1}{3},\frac{2}{3})$. Thus, $\frac{2}{3}\leq
a_{k+1}$. To complete the proof, assume that the Voronoi region of $a_{k}$
contains points from $J_{2}$. Then, $\frac{1}{2}(a_{k}+a_{k+1})>\frac{2}{3}$
implying $a_{k+1}>\frac{4}{3}-a_{k}\geq\frac{4}{3}-\frac{1}{3}=1$, which is a
contradiction. Similarly, we can show that if the Voronoi region of $a_{k+1}$
contains points from $J_{1}$, then a contradiction arises. Thus, the proof of
the proposition is complete. ∎

We are now ready to prove the following theorem.

######  Theorem 3.4.7.

For $n\geq 3$ , let $\alpha_{n}$ be an optimal set of $n$-means for $P$. Let
$\text{card}(\alpha_{n}\cap J_{1})=k$. Then, $\alpha_{n}$ contains $k$
elements from $J_{1}$, and $(n-k)$ elements from $J_{2}$, i.e.,
$\alpha_{n}(P)=\alpha_{k}(P_{1})\cup\alpha_{n-k}(P_{2})$ with quantization
error

$V_{n}(P)=\frac{1}{324}\Big{(}\frac{1}{k^{2}}+\frac{2}{(n-k)^{2}}\Big{)}.$

###### Proof.

By Proposition 3.4.6, we have $\alpha_{n}\cap J_{1}\neq\emptyset$ and
$\alpha_{n}\cap J_{2}\neq\emptyset$. Thus, there exist two positive integers
$n_{1}$ and $n_{2}$ such that $\text{card}(\alpha_{n}\cap J_{1})=n_{1}$, and
$\text{card}(\alpha_{n}\cap J_{2})=n_{2}$. Again, by Proposition 3.4.6,
$\alpha_{n}$ does not contain any point from the open interval
$(\frac{1}{3},\frac{2}{3})$, and so we have $n=n_{1}+n_{2}$. Hence, by taking
$n_{1}=k$, we see that $\alpha_{n}$ contains $k$ elements from $J_{1}$, and
$(n-k)$ elements from $J_{2}$. Again, by Proposition 3.4.6, we know that the
Voronoi region of any point in $\alpha_{n}\cap J_{1}$ does not contain any
point from $J_{2}$, and the Voronoi region of any point from $\alpha_{n}\cap
J_{2}$ does not contain any point from $J_{1}$. This implies the fact that
$\alpha_{n}(P)=\alpha_{k}(P_{1})\cup\alpha_{n-k}(P_{2})$, and the
corresponding quantization error is given by

$V_{n}(P)=\frac{1}{100}V_{k}(P_{1})+\frac{99}{100}V_{n-k}(P_{2})=\frac{1}{10800}\Big{(}\frac{1}{k^{2}}+\frac{99}{(n-k)^{2}}\Big{)}.$

Thus, the proof of the theorem is complete. ∎

######  Remark 3.4.8.

Let $k$ be the positive integer as stated in Theorem 3.4.7. Then,
$\alpha_{k}(P_{1})$ and $\alpha_{n-k}(P_{2})$ are known by Theorem 3.4.7.
Thus, once $k$ is known, we can easily determine the optimal sets of $n$-means
and the $n$th quantization errors for all $n\in\mathbb{N}$ with $n\geq 3$. For
$n\geq 3$, consider the real valued function

$F(n,x)=\frac{1}{10800}\Big{(}\frac{1}{x^{2}}+\frac{99}{(n-x)^{2}}\Big{)}$

defined in the domain $1\leq x\leq n-1$. Notice that $F(n,x)$ is concave
upward, and so $F(n,x)$ attains its minimum at a unique $x$ in the interval
$[1,n-1]$. Thus, we can say that for a given positive integer $n\geq 3$, there
exists a unique positive integer $k$ for which $F(n,k)$ is minimum if $x$
ranges over the positive integers in the interval $[1,n-1]$.

######  Remark 3.4.9.

For any positive integer $n\geq 3$, let us write

$V(j,n-j):=\frac{1}{100}V_{j}(P_{1})+\frac{99}{100}V_{n-j}(P_{2}),$

where $1\leq j\leq n-1$. For a given $n$ let $k:=k(n)$ be the positive integer
as stated in Theorem 3.4.7. Then, we have $V_{n}=V(k,n-k)$. Notice that

$V_{n}=V(k,n-k)=\min\\{V(j,n-j):1\leq j\leq n-1\\}.$

Moreover, if we order the elements of the set $\\{V(j,n-j):1\leq j\leq n-1\\}$
in a sequence as

$\\{V(1,n-1),V(2,n-2),\cdots,V(n-1,1)\\}$

then $V(k,n-k)$ is the $k$th term in the sequence. Using this fact, for a
given $n$ we can easily determine the value of the positive integer $k$ as
follows:

Define the function

(1) $f:\mathbb{N}\to\mathbb{N}\text{ such that }f(n)=k,$

where $k$ is the unique positive integer such that

$V_{n}:=V(k,n-k)=\min\\{V(j,n-j):j\in\mathbb{N},\,1\leq j\leq n-1\\}.$

For a given positive integer $n$, once $k:=k(n)$ is known, using Theorem
3.4.7, we can determine the optimal set of $n$-means and the corresponding
quantization error.

In the following example, we calculate the values of $k$ for different values
of $n$. For such calculations, we have used Mathematica.

######  Example 3.4.10.

Recall the function $f$ defined by (1). Then, we see that

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

In fact,

$\displaystyle\\{f(n)\\}_{n=4985}^{5011}$
$\displaystyle=\\{886,886,886,887,887,887,887,887,887,888,888,888,888,888,889,889,889,$
$\displaystyle 889,889,889,890,890,890,890,890,890,891\\}.$

###  3.5. Quantization for the mixed distribution $P$ when $p=\frac{2}{5}$,
and $p=\frac{1}{1000}$

Let $P_{1}$ and $P_{2}$ be two uniform distributions, respectively, on the
intervals given by

$J_{1}:=[0,\frac{1}{3}],\text{ and }J_{2}:=[\frac{2}{3},1].$

Let $P:=pP_{1}+(1-p)P_{2}$ be the mixed distribution generated by
$(P_{1},P_{2})$ associated with the probability vector $(p,1-p)$, where
$0<p<1$. For $n\in\mathbb{N}$ , let $\alpha_{n}$ be an optimal set of
$n$-means for $P$. Using the similar technique as given in Subsection 3.4, we
can show that if $p=\frac{2}{5}$, then

$\displaystyle\alpha_{1}=\\{\frac{17}{30}\\}\text{ with }V_{1}$
$\displaystyle=\frac{313}{2700},\,\alpha_{2}=\\{\frac{1}{6},\frac{5}{6}\\}\text{
with
}V_{2}=\frac{1}{108},\,\alpha_{3}=\\{\frac{1}{6},\frac{3}{4},\frac{11}{12}\\}\text{
with }V_{3}=\frac{11}{2160},$ $\displaystyle\alpha_{4}$
$\displaystyle=\\{\frac{1}{12},\frac{1}{4},\frac{3}{4},\frac{11}{12}\\}\text{
with }V_{4}=\frac{1}{432},\text{ and so on}.$

On the other hand, if $p=\frac{1}{1000}$, then we see that

$\displaystyle\alpha_{1}$ $\displaystyle=\\{\frac{1249}{1500}\\}\text{ with
}V_{1}=0.00970326,$ $\displaystyle\alpha_{2}$
$\displaystyle=\\{0.74824116,0.91608039\\}\text{ with }V_{2}=0.0026610135,$
$\displaystyle\alpha_{3}$ $\displaystyle=\\{0.719398,0.831639,0.94388\\}\text{
with }V_{3}=0.00134412,$ $\displaystyle\alpha_{4}$
$\displaystyle=\\{0.704407,0.788862,0.873317,0.957772\\}\text{ with
}V_{4}=0.00087869,$ $\displaystyle\alpha_{5}$
$\displaystyle=\\{\frac{1}{6},\frac{17}{24},\frac{19}{24},\frac{7}{8},\frac{23}{24}\\}\text{
with }V_{5}=0.000587384\text{ and so on}.$

The function $f:\mathbb{N}\to\mathbb{N}$, defined in (1), is also true here
under the condition that $V(j,n-j)$ in this section is defined as follows:

$V(j,n-j):=\left\\{\begin{array}[]{ll}\frac{2}{5}V_{j}(P_{1})+\frac{3}{5}V_{n-j}(P_{2})&\text{
if }p=\frac{2}{5},\\\
\frac{1}{1000}V_{j}(P_{1})+\frac{999}{1000}V_{n-j}(P_{2})&\text{ if
}p=\frac{1}{1000}\end{array}\right.$

where $1\leq j\leq n-1$. Now, we give the following two examples which are
analogous to Example 3.4.10 given in the previous section.

######  Example 3.5.1.

For $p=\frac{2}{5}$, we have

$\displaystyle\\{f(n)\\}_{n=2}^{\infty}$
$\displaystyle=\\{1,1,2,2,3,3,4,4,5,5,6,6,7,7,7,8,8,9,9,10,10,11,\
11,12,12,13,13,14,14,\cdots.\\}.$

In fact,

$\displaystyle\\{f(n)\\}_{n=4985}^{5011}$
$\displaystyle=\\{2324,2325,2325,2326,2326,2327,2327,2328,2328,2329,2329,2329,2330,2330,$
$\displaystyle
2331,2331,2332,2332,2333,2333,2334,2334,2335,2335,2336,2336,2336\\}.$

######  Example 3.5.2.

For $p=\frac{1}{1000}$, we have

$\displaystyle\\{f(n)\\}_{n=5}^{\infty}$
$\displaystyle=\\{1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,\
3,3,3,3,3,3,3,3,3,3,3,3,4,4,$ $\displaystyle
4,4,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6\cdots\\}.$

In fact,

$\displaystyle\\{f(n)\\}_{n=4985}^{5011}$
$\displaystyle=\\{453,453,454,454,454,454,454,454,454,454,454,454,454,455,455,$
$\displaystyle 455,455,455,455,455,455,455,455,455,456,456,456\\}.$

######  Remark 3.6.

By the results in Subsection 3.4 and Subsection 3.5, we see that for
$p=\frac{1}{100}$ and $p=\frac{1}{1000}$ the optimal sets of two-means do not
contain any point from $J_{1}$, but for $p=\frac{2}{5}$ it contains a point
from $J_{1}$. Moreover, we see that for $p=\frac{1}{100}$ and $p=\frac{2}{5}$
the optimal sets of three-means contain points from $J_{1}$, but for
$p=\frac{1}{1000}$, the optimal set of three-means, and four-means do not
contain any point from $J_{1}$. Using the similar technique as given in
Subsection 3.4, it can be shown that Lemma 3.4.3 and Proposition 3.4.6, and
Theorem 3.4.7 are also true for $p=\frac{2}{5}$ and $p=\frac{1}{1000}$. The
main difference is that for $p=\frac{1}{100}$, they are true for all $n\geq
3$, but for $p=\frac{2}{5}$, they are true for all $n\geq 2$, on the other
hand, for $p=\frac{1}{1000}$ they are true for all $n\geq 5$.

Let us now give the following conjecture and the open problems.

######  Conjecture 3.7.

Let $P_{1}$ and $P_{2}$ be two uniform distributions defined on any two closed
intervals $[a,b]$ and $[c,d]$, where $a<b<c<d$. Let $P:=pP_{1}+(1-p)P_{2}$ be
a mixed distribution generated by $(P_{1},P_{2})$ associated with any
probability vector $(p,1-p)$, where $0<p<1$. Then, we conjecture that for each
probability vector $(p,1-p)$ there exists a positive integer $N$ such that for
all $n\geq N$, the optimal sets $\alpha_{n}$ contain points from both the
intervals $[a,b]$ and $[c,d]$, and do not contain any point from the open
interval $(b,c)$. This yields the fact that if
$\text{card}(\alpha_{n}\cap[a,b])=k:=k(n)$, then $\alpha_{n}$ contains $k$
elements from $[a,b]$, and $(n-k)$ elements from $[c,d]$, i.e.,
$\alpha_{n}(P)=\alpha_{k}(P_{1})\cup\alpha_{n-k}(P_{2})$ for all $n\geq N$
with quantization error

$V_{n}(P)=\frac{1}{108}\Big{(}\frac{p}{k^{2}}+\frac{1-p}{(n-k)^{2}}\Big{)}.$

######  Open 3.8.

Let $P:=pP_{1}+(1-p)P_{2}$ be the mixed distributions generated by the two
uniform distributions $P_{1}$ and $P_{2}$ defined on the closed intervals
$[0,\frac{1}{3}]$ and $[\frac{2}{3},1]$ associated with the probability
vectors $(p,1-p)$. It is still not known whether there is any probability
vector $(p,1-p)$, or what is the range of $p$, for which an optimal set
$\alpha_{2}$ of two-means for the mixed distributions $P$ will contain a point
from the open interval $(\frac{1}{3},\frac{2}{3})$, and an optimal set
$\alpha_{3}$ of three-means will contain points from $[0,\frac{1}{3}]$ and
$[\frac{2}{3},1]$, and also from the open interval
$(\frac{1}{3},\frac{2}{3})$.

If the answer of the above open problem comes in the negative, then it leads
to investigate the following open problem.

######  Open 3.9.

It is still not known whether there is a set of values $a,b,c,d$, and $p$,
where $a<b<c<d$ and $0<p<1$, such that if $P:=pP_{1}+(1-p)P_{2}$ is the mixed
distribution generated by the two uniform distributions $P_{1}$ and $P_{2}$
defined on the closed intervals $[a,b]$ and $[c,d]$ associated with the
probability vector $(p,1-p)$, then an optimal set $\alpha_{2}$ of two-means
for the mixed distributions $P$ will contain a point from the open interval
$(b,c)$, and an optimal set $\alpha_{3}$ of three-means will contain points
from $[a,b]$ and $[c,d]$, and also from the open interval $(b,c)$.

######  Conjecture 3.10.

We conjecture that the answer of the open problem Open 3.8 will be negative.

###  3.11. Observation

Under Conjecture 3.10, in this subsection, we try to give a partial answer of
the open problem ‘Open 3.9’. Let us choose the two closed intervals as
follows:

$[a,b]=[0,\frac{7}{15}]\text{ and }[c,d]=[\frac{8}{15},1].$

Let $P_{1}$ and $P_{2}$ be the uniform distributions defined on the closed
intervals $[a,b]$ and $[c,d]$. Then, the density functions $f_{1}$ and $f_{2}$
of the uniform distributions $P_{1}$ and $P_{2}$ are, respectively, given by

$f_{1}(x)=\left\\{\begin{array}[]{cc}\frac{15}{7}&\text{ if
}x\in[0,\frac{7}{15}],\\\ 0&\text{otherwise};\end{array}\right.\text{ and
}f_{2}(x)=\left\\{\begin{array}[]{cc}\frac{15}{7}&\text{ if
}x\in[\frac{8}{15},1],\\\ 0&\text{otherwise}.\end{array}\right.$

We now give the following two propositions.

######  Proposition 3.11.1.

Let $p=\frac{51}{500}$, and let $P:=pP_{1}+(1-p)P_{2}$ be the mixed
distribution generated by the two uniform distributions $P_{1}$ and $P_{2}$
defined on the closed intervals $[a,b]$ and $[c,d]$. Then, an optimal set
$\alpha_{2}$ of two-means for the mixed distribution $P$ contains a point from
the open interval $(b,c)$.

###### Proof.

Proceeding in the similar way as Proposition 3.4.1, we see that an optimal set
of two-means for the mixed distribution $P:=pP_{1}+(1-p)P_{2}$, where
$p=\frac{51}{500}$, is given by $\alpha_{2}:=\\{0.488570,0.829523\\}$ with
quantization error $V_{2}=0.0179722$. Notice that

$b<0.488570<c<0.829523<d,$

and so the assertion of the proposition follows. ∎

######  Proposition 3.11.2.

Let $p=\frac{225}{500}$, and let $P:=pP_{1}+(1-p)P_{2}$ be the mixed
distribution generated by the two uniform distributions $P_{1}$ and $P_{2}$
defined on the closed intervals $[a,b]$ and $[c,d]$. Then, an optimal set
$\alpha_{3}$ of three-means for the mixed distribution $P$ contains points
from $[a,b]$ and $[c,d]$, and also from the open interval $(b,c)$.

###### Proof.

Proceeding in the similar way as Proposition 3.4.2, we see that an optimal set
of three-means for the mixed distribution $P:=pP_{1}+(1-p)P_{2}$, where
$p=\frac{225}{500}$, is given by
$\alpha_{3}:=\\{0.174089,0.522267,0.840756\\}$ with quantization error
$V_{3}=0.00985931$. Since

$a<0.174089<b<0.522267<c<0.840756<d,$

and so the assertion of the proposition follows. ∎

######  Remark 3.11.3.

Notice that the two mixed distributions $P:=pP_{1}+(1-p)P_{2}$ considered in
Proposition 3.11.1 and Proposition 3.11.2, are different. It is worthwhile to
investigate whether the two mixed distributions can be same. In other words,
whether there is a mixed distribution $P:=pP_{1}+(1-p)P_{2}$, where $P_{1}$
and $P_{2}$ are two uniform distributions on two different closed intervals
$[a,b]$ and $[c,d]$ associated with a probability vector $(p,1-p)$ with
$a<b<c<d$ and $0<p<1$, for which the open problem ‘Open 3.9’ is true.

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