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Singularity in signal theoryB. Picinbono aa Laboratoire des Signaux et Systèmes Supélec, Plateau de Moulon, 91192 Gif surYvette, France

Online Publication Date: 01 March 2008To cite this Article: Picinbono, B. (2008) 'Singularity in signal theory', InternationalJournal of Control, 81:3, 495 - 506To link to this article: DOI: 10.1080/00207170701556906URL: http://dx.doi.org/10.1080/00207170701556906

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International Journal of Control

Vol. 81, No. 3, March 2008, 495–506

Singularity in signal theory

B. PICINBONO*

Laboratoire des Signaux et Systemes Supelec, Plateau de Moulon, 91192 Gif sur Yvette, France

(Received 5 July 2007; in final form 6 July 2007)

A discrete-time random signal is singular if its values are singular random variables defined by

a distribution function continuous but with a derivative equal to zero almost everywhere.

Singular random signals can be obtained at the output of some linear filters when the input is a

discrete-valued white noise. Sufficient conditions for singularity are established. In particular

it is shown that if the poles of the filter are inside a circle called the circle of singularity and if

the input is white and discrete-valued the output is singular. Computer experiments using

histograms at different scales exhibit the structure of singular signals. The influence of input

correlation is also analysed. It is shown that when the input is not white, but has a specific

Markovian structure, the output can be singular. This is also verified by computer

experiments. Finally, mixtures of singular and discrete-valued random signals are analysed.

1. Introduction

Singularity is a concept introduced in probability theoryand related to properties of distribution functions ofrandom variables. Usually only two kinds of randomvariables are considered: those that are continuous andhaving a distribution function (DF) with a derivative(probability density function) and those that are discretewith a DF varying by steps defining possible values andtheir corresponding probabilities. However there is athird kind of random variables: those that are singular.A random variable (RV) is said to be singular if its

DF is continuous but with a derivative equal to zeroalmost everywhere. Then such a RV is neither contin-uous (no probability density function) nor discretebecause its DF is continuous and not a stepwisefunction.Singular RVs are often considered as mathematical

curiosities without interest in signal theory and engi-neering sciences. They are rarely introduced in standardtextbooks (Papoulis 1984, Ochi 1990, Pfeifer 1990,Helstrom 1991, Picinbono 1993) and appear only inmathematically oriented books (Loeve 1977, Wong andHajek 1985). Very simple RVs however can be singular.The best example comes from an old result known formore than sixty years. It says that if wk is a set of

independent and identically distributed (IID) RVs

taking the values �1 with the same probability, then

the sum of the seriesP1

k¼0 akwk is singular as soon as

jaj51/2. So one of the simplest RV that can be

considered is singular. This result is often omitted

because its usual proof requires very abstract reasoning.

One of the first tasks of this paper is to show its origin

and to introduce an elementary proof that can after-

wards be used in signal theory. Indeed the series

considered above is similar to the output of an

exponential discrete-time filter whose input is a discrete

valued white noise. This is the case of autoregressive

signals widely used in signal processing. Then the

question of knowing whether or not this can be extended

to other class of signals and systems appears

immediately.The main result of this paper is that singularity of the

output of a linear filter depends on two facts: the

discrete character of the input, which is a common

situation in communication theory, and some specific

properties of the filter such as the location of its poles in

a circle called the circle of singularity. In order to

provide a better understanding of a problem usually

widely ignored, and to visualize how singularity can

appear for rather simple signals, number of computers

experiments are presented.In the last part of the paper it is also shown that the

assumption of whiteness, fundamental in the proof of*Email: bernard.picinbono@lss.supelec.fr

International Journal of ControlISSN 0020–7179 print/ISSN 1366–5820 online � 2008 Taylor & Francis

http://www.tandf.co.uk/journalsDOI: 10.1080/00207170701556906

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the basic result, can be partially deleted and we presentsome examples of coloured input signals leading also tosingular outputs. This is especially the case of someMarkov processes. Finally mixtures of discrete and andsingular random signals can be obtained depending onthe properties of the correlation function of the input,and theoretical and experimental examples of suchsignals are presented.

2. Statement of the problem and review of

known results

Let X(!) be a RV defined on some probability space. LetF(x) be its distribution function (DF). In all of whatfollows we do not make any difference between two RVsdistinct but equal with probability 1. This means thatX(!) is entirely defined by its DF F(x). The RV X(!) issaid to be singular if its DF F(x) is continuous but with aderivative equal to zero almost everywhere.At the beginning let us remind some fundamental

results of probability theory used in the discussion thatfollows and especially the Lebesgue decompositiontheorem (Lukacs 1970). For this let us recall that aRV is said to be continuous if its DF is continuous andhas a derivative which is its probability density function(PDF). On the other hand a RV is said to be discrete ifits DF varies only by steps of amplitudes pi at points xi.This means that the RV takes only the values xi with theprobabilities pi. The Lebesgue theorem says that any DFF(x) can be decomposed in an unique way as a sum ofthree terms, or that

FðxÞ ¼ a1FcðxÞ þ a2FdðxÞ þ a3FsðxÞ;

a1 þ a2 þ a3 ¼ 1; ai � 0: ð1Þ

In this equation the three functions Fi(x) are DFs andFc(x), Fd(x), and Fs(x) are the continuous, discrete andsingular components of F(x) respectively. If a1¼ 1, theRV is continuous and its PDF is the derivative of Fc(x).If a2¼ 1, X is a discrete RV. If, finally, a3¼ 1, the RV Xis singular. If one of the coefficients ai is equal to 1, theDF is said to be pure. The spectrum SF of a RV is the setof the points of increase of its DF F(x) and the spectralmeasure (SM) is the Lebesgue measure L(SF) of this set.It is clear that if the SM is zero the continuous part in (1)is zero, or a1¼ 0. Then the RV can be discrete (a3¼ 0),singular (a2¼ 0 ), or a mixture of a discrete and asingular parts. Then to show that a RV is singular itsuffices to show that its SM is zero and that there is nodiscrete part in the decomposition of its DF.The fundamental theorem opening all this discussion

is the following. Consider a set of RVs wk independentand identically distributed (IID) with a symmetricBernoulli distribution. This means that the wks take

only two values with the same probabilities 1/2. Whenthis is not otherwise explicitly indicated the two possiblevalues of each RV wk are �1. Consider the RV X definedby the series

X ¼X1

k¼0

akwk; ð2Þ

which is convergent provided that jaj51. Note that thepossible values of wk imply that X is symmetric, whichmeans that X and �X have the same DF. Furthermore,

for the same reasons, changing a in �a does not changethe DF of X. Then we can assume that a� 0. Thefollowing results holds:

(1) if a51/2, X is singular;(2) if a¼ 1/2, X is uniformly distributed in [�2, þ2];(3) if 1/25a51, X is in general continuous, but can

also be singular for values of a belonging to a set ofzero measure.

Point 2 can be shown directly by a simple calculation.Point 3 is without interest for this paper and thecomplete set of points of singularity is still a subject ofresearch. Point 1 was shown long time ago in theframework of infinite products of convolutions(Kershner and Wintner 1935) and discussed morerecently (Peres and Solomyak 1998, Solomyak et al.2000). Because of its importance for all what follows, weshall now present a direct proof.

For this purpose consider the partial sum XN and therest RN defined by

XN ¼XN

k¼0

akwk; RN ¼X1

k¼Nþ1

akwk ¼ aNþ1X1

k¼0

akwNþkþ1:

ð3Þ

The RV XN takes 2Nþ 1 distinct values vNi with the sameprobability 1/2Nþ 1. The rest RN satisfies jRNj� aNþ 1/(1� a), this limit being obtained if all the wNþ kþ 1 areequal to þ1 or �1. As a consequence the SM of RN issmaller than aNþ 1 [2/(1� a)]. Since there are 2Nþ 1

distinct values vNi of XN, the SM of X is smaller than(2a)Nþ 1 [2/(1� a)]. As this is valid for all N, this SM iszero whenever a51/2.

Note that this property of the SM is due to the valueof a and to the fact that the input wk has only twopossible values. On the other hand the probabilities ofthese outcomes do not play any role. We shall see laterthat this is general.

This result means that there is no continuous part inthe decomposition of the DF of X, or that a1¼ 0 in (1).Let us see now that there is also no discrete component,or a2¼ 0. Indeed suppose that this is not the case. Thiswould mean that there is a value x0 such that theprobability that X¼ x0 is positive. But as x0 is a value of

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X, there exists a set of numbers �k taking only the values

�1 such that x0 ¼P1

k¼0 ak�k. Furthermore, as the

values vNi are distinct, this set is unique. Since the RVs

wk are IID and P[wk¼ �k]¼ 1/2, we deduce

P½X ¼ x0� ¼Y1

k¼0

P½wk ¼ �k� ¼ 0: ð4Þ

As a consequence the RV X is singular, or a3¼ 1. On the

contrary to the property of the SM, it is clear that this

proof uses the fact that the wks are independent and that

their two values have the same probability 1/2. It is

simple to see however that the result remains valid if one

of the values has the probability p, except when p¼ 0 or

p¼ 1, which corresponds to a situation where the input

signal wk is no longer random.For the discussion that follows it is important to

understand that the singularity comes from two

completely distinct properties. The first one depends

only on a and on the fact that wk takes only two values

�1. But it is insufficient to imply singularity and we shall

see later that for some particular probability distribu-

tions of the wk the sum X can be discrete or a mixture of

discrete and singular parts. The second properties arises

from the whiteness of wk. It is the combination of these

two properties which ensures that X is singular.Let now present an interpretation of XN by its tree of

construction presented in figure 1. For each value vNi of

XN we can associate the value �vNi . It is obtained simply

by changing the signs of the wks appearing in (3). This

means, as noted above, that the the RV X is symmetric.

As a consequence we can consider only positive vNi s. For

the same reason it is always possible to assume that

a40. With these assumptions we have v00 ¼ 1. The two

positive values of v1i are 1� a and 1þ a. The construc-

tion of the 8 positive values of X3 appears in the tree of

figure 1. Let us now see that the assumption a51/2

means that there is no crossing of the branches of the

tree. Indeed consider the two branches of the tree

starting from a point vNi . There is no crossing between

all the branches starting from this point if

vNi � aNþ1 þ aNþ2=ð1� aÞ < vNi þ aNþ1 � aNþ2=ð1� aÞ.

This yields a51/2. It is clear that the construction of this

tree is similar to the one of Cantor sets. This is why it is

sometimes said that the RV X has a Cantor-type

distribution (Wittke et al. 1988).Singularity is not limited to random geometric series

like (2) but can appear with RVs such as

X ¼X1

k¼0

hkwk; ð5Þ

where hk40 and the wks have the same properties as in

(2). In this case XN and RN of (3) are written simply by

replacing ak by hk. The last equality of (3) does not hold.It is shown without complete proof in p. 66 of

(Lukacs 1970) that if

�n ¼X1

k¼nþ1

hk < hn; 8n; ð6Þ

then X is singular. In the case where hk¼ ak, this yields

a51/2. It is possible to construct a tree as in figure 1

with the hks instead of the aks. One can then see that the

condition (6) implies that there is no crossing between

the branches of the tree, which introduces again a

Cantor structure (Picinbono and Tourneret 2005).Before leaving this section let us present a short review

of some papers from the engineering literature where the

problem of singularity is discussed. The treatment of

sequences of Bernoulli RVs appear frequently in the

context of digital communications. The first discussion

concerning consequences of singularity was presented in

(Hill and Blanco 1973). The discussion was limited to

geometric series like (2) and the purpose was to obtain

an approximation of the DF for the calculation of the

error probability or the performance of communication

systems. Extensions of the same problem to Cantor-type

distribution was presented in (Wittke et al. 1988). In this

paper the condition (6) is explicitly used and various

examples of filters satisfying this condition are intro-

duced. However, as indicated by the authors, the

condition hk40 used in (5) and introducing the Cantor

structure is very restrictive and it is not satisfied by a

large class of filters containing for example terms like

ak cos(!k). The principle of calculation of the distribu-

tion function and of some expectations is then presented

and used for the evaluation of the error probability. A

rather more theoretical approach of the same problem is

presented in (Smith et al. 1993). Finally other calcula-

tions of expectations of singular RVs are discussed in

(Campbell et al. 1995). In this paper the singularity is

introduced from some properties of the entropy of the

RVs by using an approach introduced in (Garsia 1962).

Similar discussions appear in (Naraghi-Pour et al. 1990,

Tourneret et al. 1994).The first purpose of the present paper is to show that

singularity can be introduced from considerations ofFigure 1. Tree of construction of the successive values vNi .

Singularity in signal theory 497

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properties of the poles of a linear filter. Furthermore inall these papers it is assumed that the RVs wk areindependent, and this assumption is a corner stone forthe introduction of singularity. Then it arises immedi-ately the question of knowing whether or not it can berelaxed without suppressing singularity. This question isdiscussed in the second part of the paper.

3. Singularity of autoregressive of order 1 signals

The previous discussion has an immediate application inthe case of autoregressive signals. A signal xk is said tobe autorregressive of order 1 [AR(1)] if it is deducedfrom a white noise wk by the first order recursionxk¼ axk�1þwk. It is then defined by the parameter acalled the regression coefficient and by the DF of wk.This recursion introduces a linear filter and its input-output relationship is

xn ¼X1

k¼0

akwn�k: ð7Þ

It is then a discrete time causal filter with the impulseresponse ak. If the input signal is a white symmetricBernoulli signal, it results from the previous discussionthat the RVs xk are singular as soon that jaj51/2, andwe say that the signal xk is singular. Indeed in order tocome at (2) it suffices to introduce wk ¼ wn�k and it isclear that the RVs wk are still IID Bernoulli. Thesimplicity of this signal explains why it was saidpreviously that singularity is a common phenomenon.Five trajectories of the signal xk are represented in

figure 2. The corresponding values of a are : 0: (1); 0.2:(2); 1/3; (3); 0.4; (4); 0.5; (5). Trajectory 1 corresponds tothe white noise wk. The only specific property apparenton these trajectories is the damping phenomenoncharacterized by the fact that the speed of variation ofxk decreases when a increases. There is however noelement which can illustrate the singularity.For this purpose it is necessary to study the DF of the

signals and the appropriate mean for this purpose is torealize histograms at various scales. A histogram isrelated to the DF according to the following property.Let n(x, �x) be the number of samples of the signal xkrecorded in an interval ]x, xþ�x]. Let S be the totalnumber of samples of the signal analysed in thehistogram. According to the law of large numbers, theratio n(x,�x)/S tends to the increment F(xþ�x)�F(x)of the DF when S!1. If the RV xk is continuous(a1¼ 1) the histogram yields an estimation of the PDF,and this is the usual way for this purpose. On theother hand if xk is singular the PDF is zero and this iswhy an analysis at different scales is necessary. For thislet us remind that a histogram is characterized by

three parameters. The first is the total number S ofsamples analysed. The second is the interval of analysis[�,�]. This means that only the samples xk satisfying�5xk5� are taking into account in the histogram. Thethird is the number B of adjacent bins in this interval.The bin i of this histogram corresponds to the interval[�þ i(���)/B, �þ (iþ 1) (���)/B], 0� i�B� 1.

In order to analyse singularity by histograms weconsider the successive intervals centered at somenodes vNi of the tree of figure 1 and defined by½vNi � �N; v

Ni þ �N�, where �N is defined by (6). For

hk¼ ak, it satisfies �N¼ aNþ1/(1� a).In the computer experiments presented in what

follows S is of the order of 107 and B¼ 400. Infigure 3 four histograms corresponding to variousvalues of the regression coefficient a are presented. Inthese figures the centres of the histograms are the originand we verify that X is symmetric. We observe clearly anevolution from the case where there are only two lines(a¼ 0), which is not represented, to a uniform distribu-tion between [�2, þ2] obtained for a¼ 1/2, as predictedby the theory.

A more detailed analysis of the histogram 2 of thisfigure corresponding to a¼ 1/3 is presented in figure 4.In this case the extreme values of X are �3/2.

2000 2050 2100 2150 2200−2

0

2

1

2000 2050 2100 2150 2200−2

0

2

2

2000 2050 2100 2150 2200−2

0

2

3

2000 2050 2100 2150 2200−2

0

2

4

2000 2050 2100 2150 2200−2

0

2

5

Figure 2. Trajectories of xk for various values of the

regression coefficient a: 1. a¼ 0; 2. a¼ 0.2; 3. a¼ 1/3;

4. a¼ 0.4; 5. a¼ 0.5.

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The number of samples of the signal in the experiment isS¼ 2.106. The centres vNi of the histograms are obtainedby taking i¼ 2N� 1. These successive centres of theseven histograms are 0, 1, 1.133, 1.444, 1.4815, 1.4938,1.4979. The ratio of the two extreme window widths is729, and even with this very great diminution of theinterval of analysis the structure of the histogramsremains the same. Of course it would be possible tocontinue the experiment, but this would require a muchgreater value of S. This figure exhibits a fractalproperty explaining why the DF has no derivative(Falconer 1990).In order to have a more precise idea of the

phenomenon we present the result of the same experi-ment realized with a¼ 1/2 where the theory predicts auniform distribution in [�2,þ2]. This appears in figure 5which indicates clearly this uniform distribution. Thefluctuations appearing in the last histogram are simplydue to too small number of samples appearing in theinterval of analysis.

4. Extensions to other filters

It is intuitive that the previous results are not restricted toAR(1) signals with symmetric Bernoulli inputs. In this

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

0

5

10

15× 105

× 105

× 105

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

× 105

1

0

2 2

0

1

2

3

0

5

4

Figure 3. Histograms of xk for various values of the

regression coefficient a: 1. a¼ 0.2; 2. a¼ 1/3; 3. a¼ 0.4;

4. a¼ 0.5.

1.5 1 0.5 0 0.5 1 1.50

5× 104

× 104

1

0.5 1 1.50123

2

1.15 1.2 1.25 1.3 1.35 1.4 1.45 1.50

50001000015000

3

1.4 1.42 1.44 1.46 1.48 1.50

5000

4

1.465 1.47 1.475 1.48 1.485 1.49 1.495 1.50

2000

5

1.488 1.49 1.492 1.494 1.496 1.498 1.50

1000

6

1.496 1.4965 1.497 1.4975 1.498 1.4985 1.499 1.4995 1.50

500

7

Figure 4. Histograms with scaling effect for an AR (1) signal

with a¼ 1/3.

−1.5−2 −1 0.5 0 0.5 1 1.5 20

5000

10000

1

0 0.5 1 1.5 20

5000

2

1 1.2 1.4 1.6 1.8 20

100020003000

3

1.5 1.6 1.7 1.8 1.9 20

50010001500

4

1.75 1.8 1.85 1.9 1.95 20

500

5

1.88 1.9 1.92 1.94 1.96 1.98 20

200

6

1.94 1.95 1.96 1.97 1.98 1.99 20

100

200

7

Figure 5. Histograms with scaling effect for an AR (1) signal

with a¼ 1/2.

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section we shall effectively show that they can beextended to a large number of other signals. This canbe done either by changing the filter that yields xk fromwk

or by changing the statistical properties of the input wk.Let us first present some preliminary considerations.

We restrict our discussion to dynamical discrete timefilters. These filters are characterized by the facts thatthey are causal and that their transfer function H(z) is arational function of z. Such filters are defined by theirpoles and their zeros and almost all the filters used insignal processing are dynamical. We consider furtheronly IIF filters, because RIF filters cannot introducesingularity. Indeed this property, as seen previously, isdue to a series and the input-output relationships in RIFfilters is a simple sum. This means that we exclude fromour analysis transfer functions with only one pole atthe origin.We consider also white discrete-valued inputs signals

wk, or sequences of IID random variables taking only afinite number q of possible values.Let finally introduce a circle called circle of

singularity with center O and with the radius equalto 1/q. This allows to introduce the followingtheorem.

Theorem 1: Let xk be the output of a dynamical non-RIFfilter generated by the input wk. If wk is a white signaltaking only q values and if the poles of the filter lie insidethe circle of singularity, then the output xk is singular.

Proof: As in the previous section it is made in twosteps: 1. Proof that the SM is zero, 2. Proof that there isno discrete component in the DF.

Let F be a dynamical filter defined by its transferfunction H(z) or its impulse response hk. Let F

0 be thefilter with the impulse response gk¼ qkhk. It is obviousthat its transfer function is G(z)¼H(z/q). This impliesthat if the poles of F are zi, those of F0 are qzi. Theassumption that the poles of F are inside the circle ofsingularity implies that the poles of F0 are inside the unitcircle, or that F0 is a dynamical filter. As such a filter isstable we deduce that

P1k¼0 jgkj <1. The signal xp is

defined by xp ¼P1

k¼0 hkwp�k and by introducing xp¼Xand wk ¼ wp�k we have X ¼

P1k¼0 hkwk, where the wks

have the same properties as the wks or are IID and takeonly q values. As indicated above the finite sum XN andthe rest RN are defined in (3) where ak is replaced by hkand wk by wk. Let A be the greatest possible value of wk.We have then jRNj � A�N with �N ¼

P1k¼Nþ1 jhkj. As

XN can take only qN values, the SM S of X satisfiesS � 2AqN�N. This is valid for all N. ThenS� 2A limN!1(q

N�N), but we have

qN�N ¼ qNX1

k¼Nþ1

jhkj <X1

k¼Nþ1

qkjhkj ¼X1

k¼Nþ1

jgkj; ð8Þ

and the limit is 0 because the filter F0 is stable orP1k¼0 jgkj < þ1. This implies that S ¼ 0, or a1¼ 0. The

proof that a2¼ 0 is exactly the same as previously and

comes only from the whiteness of the wks or from their

independence. Another proof of the fact that there is

non discrete component is given by Lukacs (1970, p. 65).It is important to note that this theorem yields only a

sufficient condition of singularity. However the condi-

tion that the poles lie in the singularity circle is not at all

necessary. Let us discuss this point on a particular

example.Consider the filter defined by the transfer function

H(z)¼ z4/(z4� a4). It has four poles located on the circle

of radius a. It impulse response is such that h4k¼ a4k and

the other hk are zero. It introduces the RV

X ¼P1

k¼0 a4kw4k. It results from the previous result

that X is singular if a451/2, or a52�1/4¼ 0.8409 and

uniformly distributed if a¼ 2�1/4. Then singularity

appears for poles outside the singularity circle. This is

presented in figure 6 which is very similar to figure 4.

The histograms corresponding to the values of a equal to

0.5, 0.75, and 0.8 exhibit singular behaviour.Let us now present some comments on this theorem.

The first point to note is that it implies the result

analysed in the previous section. Indeed the causal filter

0

1

2

3× 106

1

0

1

2

3

4× 105

2

0

0.5

1

1.5

2× 105

3

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

0

2

4

6× 104

4

Figure 6. Four poles zi¼ a1/4: 1. a¼ 0.5, 2. a¼ 0.75;

3. a¼ 0.8; 4. a¼ 2�1/4¼ 0. 8409.

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with impulse response hk¼ ak is obviously dynamical

because its transfer function is simply H(z)¼ z / (z� a).

Then if q¼ 2, or if the input signal wk takes only two

distinct values, the radius of the circle of singularity is

1/2 and we find again the condition a51/2.The second point to note is that the possible values of

wk do not play any role in the result. However for a

given set of values it is sometimes possible to extend the

domain of singularity by more abstract methods

(Peres and Solomyak 1998). But if we impose the

singularity for all the possible sets of values, we come

back to the conditions of Theorem 1.The final point is that it is in general no longer

possible to interpret the result by a tree of construction.

This is especially the case when the dynamical filter has

complex poles as for example when the impulse response

is of the kind ak cos(k�).In order to illustrate the results of this section let us

consider the filter defined by H(z)¼ z3 / (z� a)3,

05a51. Its impulse response is hk¼ (1/2)k(kþ 1)ak. It

has a triple pole at z¼ a. Suppose that the input is a

Bernoulli white noise as in the previous sections. In this

case if a51/2 the pole is inside the circle of singularity

and the output is singular. This is illustrated in figure 7

calculated for a¼ 1/3 as in figure 4. The results are quite

different and the fractal structure appears only in the

last three histograms. It is interesting to note that the

condition (6) requires that a50.2063, which is notsatisfied in this example. This means that the DF is not aCantor-type distribution, and this appears in the figure.This is a good example among many others that thecondition (6) is not required to introduce singularity.

The results of various other computer experiments arepresented in (Picinbono and Tourneret 2005) and thefractal behaviour of the histograms appears clearly, butin a less regular form than in figure 4 valid in the case ofthe exponential filter. All these results show the greatvariety of possible singular signals.

5. The influence of the correlation

The assumption of independence, or of whiteness of theinput, plays a fundamental role in the previous results. Itis introduced in all the papers indicated in the list ofreferences. This assumption allows us to show that, evenif the SM is zero, there is no discrete component in theDF, of that a2¼ 0. Thus appears immediately thequestion of knowing whether it is still possible to meetsingularity in the case of coloured inputs.

Consider a filter F defined by its impulse response hkand satisfying the conditions of Theorem 1. Theproblem of singularity of the output depends on theproperties of the RV X given by (5) where the conditionhk40 is relaxed. We assume that the wks are symmetricBernoulli, but not necessarily independent. The max-imum value of X is Xm ¼

P1k¼0 jhkj which is finite

because F is stable. Let XN be the partial sum analogueto (3) and defined by

XN ¼XN

k¼0

hkwk: ð9Þ

It takes at the maximum 2Nþ 1 distinct values vNi and weassume that this maximum is reached. This assumptionof distinct values is obviously satisfied when there is nocrossing of the branches of the tree constructed withthe hk. This appears with filters satisfying (6), or for theCantor-type structure. It is clear that this assumptiondepends only of the impulse response hk of the filter. It ishowever satisfied by a large class of filters which is notdiscussed here. This means that for any N there is nopair (i, j), i 6¼ j, such that vNi ¼ vNj . Because of thesymmetry of the wks, the RVs X and XN are alsosymmetric and this implies that vNi 6¼ 0. Indeedthe symmetry and the existence of a zero value wouldimply that the number of distinct values is odd, which isnot the case.

To each value vNi we associate a node VNi in the tree of

construction similar to the one appearing in figure 1.The assumption of distinct values vNi means that thenodes of the tree are single, which means that each node

3 2 1 0 1 2 30

500010000

1 0 1 2 30

5000

0.5 1 1.5 2 2.5 3 3.50

20004000

2 2.2 2.4 2.6 2.8 3 3.2 3.40

1000

2.7 2.8 2.9 3 3.1 3.2 3.3 3.40

500

3.1 3.15 3.2 3.25 3.3 3.350

200

3.24 3.26 3.28 3.3 3.32 3.34 3.36 3.380

100

Figure 7. Histograms with scaling effect for an AR (3) signal

with a¼ 1/3.

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VNi is reached by only one path coming from only one

node VN�1j at the step N� 1. This is obviously satisfied

when there is no crossing of the branches of the tree

constructed with the hns. This especially appears if

condition (6) is satisfied. However, since hk is not

necessarily positive as in figure 1, it is not possible to

restrict this tree to the nodes VNi corresponding to

positive values vNi . Then we assume that there are 2Nþ 1

distinct nodes satisfying vN0 < vN1 < � � � < vN2Nþ1�1

. Finally

we assume that the nodes V00 and V0

1 defined by

�v00 ¼ v01 ¼ jh0j are generated from an origin node V

which does not correspond to a value of XN.The fundamental consequence of the assumption of

distinct values vNi is that for any N and i there is a unique

path going from V to VNi . Let i

Nk ðiÞ, 0� k�N� 1, be the

indices j characterizing the nodes Vkj of this path. These

nodes can then be written VkiNkðiÞ.

The problem is to calculate the probabilities

pNðiÞ ¼4P½XN ¼ vNi � ; 0 � i � 2Nþ1 � 1: ð10Þ

When the wks are IID this probability is 1/2Nþ 1. When

they are no longer independent, its calculation is much

more complicated.For this we introduce the conditional probability

pNði; jÞ ¼4pN½XN ¼ vNi jXN�1 ¼ vN�1j �; ð11Þ

called also transition probability. It has two funda-

mental properties for the discussion that follows.The first comes from the fact that, as any probability,

it is normalized or satisfies for all j the relationP2Nþ1�1i¼0 pði; jÞ ¼ 1. However a node VN�1

j of the tree of

construction generates only two nodes VNi characterized

by the indices i+( j) and i�( j) and, according to (9),

corresponding to the values vNi�ð jÞ ¼ vN�1j � hNwN. As a

consequence for a given j there is only two terms in the

previous sum and we have

pN½iþð jÞ; j� þ pN½i�ð jÞ; jÞ� ¼ 1: ð12Þ

The second starts from the fact that there is only one

VN�1j at the step N� 1 of the tree generating VN

i and

called j(i). Thus pN(i, j) is zero except when j¼ j(i), and

the only non-zero values of pN(i, j) are

qNðiÞ ¼ pN½i; jðiÞ� ð13Þ

for N40 and q0(i)¼ 1/2.It results from (11) and from the unicity of the path

between V and VNi that

pN½ðXN¼ vNi Þ\ ðXN�1¼ vN�1j Þ� ¼ pN�1ð jÞ : pNði; jÞ �½j� jðiÞ�;

ð14Þ

where �[�] is the Kronecker delta symbol. By a

summation on j, which contains only one term,

we obtain

pNðiÞ ¼ pN�1½jðiÞ� qNðiÞ: ð15Þ

By repeating this at all the nodes of the unique path

between V and VNi characterized by the indices iNk ðiÞ we

obtain

pNðiÞ ¼YN

k¼0

qk½iNk ðiÞ�: ð16Þ

When the RVs wk are IID we have of course

qN½iNk ðiÞ� ¼ 1=2, and we find again that the values vNi

have equal probabilities 1/2Nþ 1.The probabilities pN(i) of (16) are normalized, or

�ipN(i)¼ 1, where the sum is extended to all the indices i

from 0 to 2N� 1. This property is valid for N¼ 0

because q0(i)¼ 1/2. Suppose that it is valid at the step

N� 1. Since each node VN�1j generates only two nodes

VNi�ð jÞ

and VNiþð jÞ

the result comes from (12).

The relation (16) is the basis for the discussion of the

singularity. Indeed if all the pN(i) tend to 0 when

N!1, there is no value v1i with a finite probability,

and this means that the RV cannot have a discrete

component and then is singular. This can be specified by

the following theorem.

Theorem 2: Let X be the RVP1

k¼0 hkwk, where hk is the

impulse response of a dynamical non-RIF filter F and wk

a sequence of Bernoulli RVs. If the poles of F are inside

the circle of singularity, if the possible values vNi of the

partial sums XN are distinct and if the transition

probabilities qN(i) defined by (13) satisfy

0 < qNðiÞ < B < 1; ð17Þ

then the RV X is singular, or a1¼ a2¼ 0.

Proof: If the poles are inside the circle of singularity,

Theorem 1 shows that the SM is zero or a1¼ 0. It

remains to show that a2¼ 0. This is a direct consequence

of (16) and (17) because pN(i)5(1/2)BN, which tends to

zero when N!1.

Comments: It is clear that this situation appears for

white input because in this case qN(i)¼ 1/2. The question

that remains concerns the conditions of the theorem on

the filter. It is clear from the previous discussion and

according to figure 1 that if hk¼ ak with jaj51/2, these

conditions are satisfied. There is a large class of filters

satisfying also these conditions. However the question of

characterizing all the dynamical filters with poles inside

the circle of singularity and introducing distinct values

vNi remains open. As a matter of fact it is possible to

extend this theorem to the case where these values are

not distinct, but this introduces other conditions that

cannot be presented in this paper.

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When the SM is zero the only consequence is that

there is no continuous component, or a1¼ 0. It is now

interesting to discuss conditions ensuring that there is

only a discrete component, or that a1¼ a3¼ 0. This

is partially solved by the following theorem.

Theorem 3: Let X be the RVP1

k¼0 hkwk, where hk is the

impulse response of a dynamical non-RIF filter F and wk

a sequence of Bernoulli RVs. If the poles of F are inside

the circle of singularity and if there exists an N0 such that

for N4N0

pN½i; j� ¼ 0 or 1; 8 i; j; ð18Þ

then the RV X is discrete, or a1¼ a3¼ 0.

Proof: Let s be the number of values of j for which

pN( j) 6¼ 0 at the step N0. Let vN0

j be one of these values

and VN0

j be the corresponding node of the tree. It

generates at the step N0þ 1 two nodes VN0þ1i character-

ized by the indices iþ( j) and i�( j). It results from (12)

and (18) that one of the two pN0þ1ði�Þ is pN0ð jÞ and

the other 0. Then at the step N0þ 1 there are also

only s non-zero probabilities pN0þ1ðiÞ equal to those of

the step N0.By repeating this till infinity, we deduce that the RV X

takes only s values with the same probabilities as at the

step N0, which means that it is discrete.

Let us present a very simple example of this situation.

Consider the signal wk¼w0(�1)k, where w0 is a

symmetric Bernoulli RV, and the filter hk¼ ak. It is

clear that X¼w0[1/(1þ a)]. Then X is a discrete RV

taking only the values �1/(1þ a) with equal probabil-

ities. It is obvious that (18) is satisfied. In this case

N0¼ 1 and s¼ 1 at step 1. The unique paths going from

V to infinity are characterized at step N by

vN� ¼ 1� ½�aþ a2 þ � � � þ ð�aÞN�.Finally let us now present a relation between these

considerations and the prediction of signals. This comes

from the fact that the transition probability (11) can be

expressed in terms of the RVs wk. Indeed for a given

vN�1j , or a given node VN�1j , there is only one path

coming from V to VN�1j , or one sequence of values of wk,

05k�N� 1. Let us note SN� 1 this sequence. The

transition probability pN(i, j) is simply related to

the conditional probabilities of the wk by

pN½i�ð jÞ; j� ¼ P½wN ¼ �1jSN�1�;

pN½iþð jÞ; j� ¼ P½wN ¼ 1jSN�1�; ð19Þ

where i�( j) and iþ( j) are the two indices i defining the

two nodes VNi generated by the node VN�1

i .This is related with the time prediction of the signal

wk. Indeed the prediction of wk in terms of its past values

is entirely described by the conditional probabilities

appearing in (19). In particular if the signal is

predictable with a finite past P we have for N4Peither pN[i�( j)]¼ 1 or pN[iþ( j)]¼ 1, which impliespN[iþ( j)]¼ 0 or pN[i�( j)]¼ 0 respectively.

Theorem 4: Let X be the RVP1

k¼0 hkwk, where hk is theimpulse response of a dynamical non-RIF filter F and Xk

a sequence of Bernoulli RVs. If the poles of F are insidethe circle of singularity and if wk is predictable with afinite past P, then X is discrete, or a1¼ a3¼ 0.

Proof: The assumption of predictability means that,for all k4P, wk satisfies

wk ¼ fðw1;w2, . . . ,wPÞ: ð20Þ

Since the RVs wi take only 2 values, X of (5) takes at themaximum 2P distinct positive values which means that itis a discrete RV.

6. Singularity and Markovian inputs

A symmetric Bernoulli Markovian signal of order P is asignal taking the values �1 with equal probabilities andwhich can be expressed as

wk ¼ fðwk�1;wk�2, . . . ,wk�P; bkÞ; ð21Þ

where bk is a vectorial white noise. Its value at time kconditionally to the whole past depends only on wk�1,wk�2 , . . . ,wk�P. The function f(�) and the noise bk mustsatisfy conditions ensuring that if the wks are signalstaking the values �1 with equal probabilities, wk has thesame property.

The transition probabilities pN(i, j) are given by (19).The Markov assumption means that the sequence SN� 1

can be replaced by a sequence using only the past oforder P. This can be written

pN½i�ð jÞ; j� ¼ P½wN ¼ �jwN�1, . . . ,wN�P�: ð22Þ

Since the RVs wk take only the values �1, there are onlyat most 2P distinct values of pN[i�( j), j]. If we assumethat the signal is not predictable there is no value of pNequal to 1. Then the bound of (17) is simply themaximum value of the finite number of transitionprobabilities pN(i, j). Then Theorem 1 can be appliedand the RV X is singular.

In order to simplify the discussion and the experi-ments we shall consider only linear Markovian signals oforder 1 defined as follows. Let uk be a strictly whitenoise taking only the values 0 or 1 with probabilities1� p and p respectively. Similarly let vk be a strictlywhite noise taking the values �1 with the sameprobabilities 1/2 and independent of uk. Consider nowthe signal

wk ¼ ukwk�1 þ �ukvk; ð23Þ

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where �uk ¼ 1� uk. It has the form (21) where P¼ 1 andbk¼ [uk, vk].It is obvious that if wk� 1 takes only the values �1

with probabilities 1/2, wk has also the same property. Itcan be shown that, whatever the values of w1, wk tendsto have this property. It can also be shown that thecovariance function �k of wk is pjkj, so it is anexponential covariance function. For p¼ 0, uk¼ 0, andwk is the white noise vk with a zero covariance function.On the other hand if p¼ 1, uk¼ 1 and wk¼w1, whichintroduces constant covariance function or long rangememory.Let us now consider the signal xk defined by the

recursion xk¼ axk� 1þwk, where wk is given by (23).Some of its main properties appear in figures 8 and 9.The results of figure 8 are obtained by the sameprocedure as for figure 4 with the same value of theregression coefficient a and the same value ofthe number N of cells in each histogram. Note thathistograms 1 and 2 of figure 8 are the same except adifferent scale of the y axis.The first point to note is that there is no discrete

component, at least within the precision of ourmeasurements. This means that, as expected, xk issingular.The structures of histograms at various scales are the

same as when wk is white and they indicate a fractal

property which is the origin of the singularity of xk. Butthere is a clear difference between these two series ofhistograms. In the case where the input signal wk iswhite, the amplitudes, or numbers of samples recordedin each cell of the histograms, are constant. This is nolonger true when wk is a Markovian signal. Finally notethat these amplitudes are maximum at the extremevalues of the histograms. Even if the calculation ofthe DF of xk is very complicated, the origin of thismaximum can be explained (Picinbono 2006).

Note that when the input is white the histograms aresymmetric with respect to their centres. This is no longertrue when Markovian inputs are used, except for thehistogram centered at 0 because X is symmetric and itshistograms are symmetric with respect to 0. Thisexplains that only histograms of positive values of xkare represented.

The influence of the parameter p of the model isshown in figure 9 in the case where a¼ 1/2. When p¼ 0the input wk is white and we find again the results ofhistogram 4 of figure 3. The uniform distribution isstrongly modified for non-zero values of p and this alsocan be explained (Picinbono 2006).

All the previous results can be extended to higherorder linear Markovian inputs. For this purpose itsuffices to replace ukwk� 1 of (23) by a termdepending on wk� 1, wk� 2 , . . . ,wk�m. It is easy to

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

1000

2000

1

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

2000

4000

6000

2

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

5000

10000

3

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

2

4

6

× 104

4

Figure 9. Influence of p for a¼ 1/2: 1. p¼ 0; 2. p¼ 0.1;

3. p¼ 0.2; 4. p¼ 0.4; S¼ 107.

−1.5 −1 −0.5 0 0.5 1 1.50

5× 104

× 104

× 104

1

−1.5 −1 −0.5 0 0.5 1 1.50

2

2

0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.50

1

2

3

1.15 1.2 1.25 1.3 1.35 1.4 1.45 1.50

500010000

4

1.4 1.42 1.44 1.46 1.48 1.50

5000

10000

5

1.465 1.47 1.475 1.48 1.485 1.49 1.495 1.50

5000

6

1.488 1.49 1.492 1.494 1.496 1.498 1.50

5000

7

Figure 8. Histograms of xk at different scales: a¼ 1/3,

p¼ 1/2, S¼ 107.

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construct models describing this situation. The calcula-tions however are much more complicated and cannotbe presented here.

7. Mixture of discrete and singular components

Consider the signal uk¼ u0(�1)k introduced above and

the white signal vk taking the values �1 with equalprobabilities. Suppose now that wk is equal either to ukor to vk with the probabilities � and 1� � respectively.This implies that its covariance function is�k¼ �(�1)

nþ (1��)�[k]. The DF of the RVs wk is

F(x)¼ �Fd(x)þ (1� �)Fs(x). Applying this signal againas the input to the exponential causal filter with impulseresponse hk¼ ak with a51/2 yields an output signaltaking the values �1/(1þ a) with probability � andbeing singular with probability 1� �.Results of computer experiments on this signal are

presented in figures 10 and 11. In these experimentsa¼ 1/3, as in figure 4, and �¼ 0.5. This implies thatxk takes the values �0.75 with probability 0.25,which means that its DF is discontinuous forx¼�0.75 and the amplitude of the discontinuity is0.25. The first histogram of figure 10 exhibits thesetwo discontinuities characterized by two lines at thepoints �0.75. One of these lines disappears in thesecond histogram of this figure and there is no longera line in the last two histograms. The fractalstructure introducing a singular component appearsclearly in these histograms. In order to verify whether

or not the discontinuity of the DF is an experimentalartefact, we present in figure 11 four histogramsisolating the point 0.75 and with decreasing widths ofthe cells. We observe still a residual contribution ofthe singular part of the DF in the first histogram.There is however a line at 0.75 and the fact that itsamplitude is constant indicates clearly that there iseffectively a discrete component. As the experimentuses 106 samples, the discontinuity corresponding tothe probability 0.25 must be equal to 2.5.105, whichclearly appears in the four histograms of figure 11.This shows that the signal xk is effectively a mixtureof a discrete-valued signal and a singular signal.

8. Conclusion

When the input of a causal discrete-time exponentiallinear filter with impulse response ak is a Bernoulliwhite noise, the output is singular for a51/2. Asimple proof of this result was presented and itexhibits two steps. The condition on the parameter aand the fact that the input is a discrete-valued signalimplies that the SM of the output is zero, which

0.5 0.6 0.7 0.8 0.9 10

1

2

3× 105

× 105

× 105

× 105

0.65 0.7 0.75 0.8 0.850

1

2

3

0.72 0.73 0.74 0.75 0.76 0.77 0.780

1

2

3

0.74 0.745 0.75 0.755 0.760

1

2

3

Figure 11. Mixture discrete-singular, �¼ 0.5. Analysis of the

neighbourhood of 0.75, S¼ 106.

−1.5 −1 −0.5 0 0.5 1 1.50

5

1

0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.50

2

x 104

x 104

2

1.15 1.2 1.25 1.3 1.35 1.4 1.45 1.50

10000

3

1.4 1.42 1.44 1.46 1.48 1.50

5000

4

Figure 10. Mixture discrete-singular, �¼ 0.5. Fractal part of

the histogram, S¼ 107.

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means that there is no continuous component in itsDF. The fact that this DF does not contain a discretecomponent arises from the whiteness of the input.This result is not specific to exponential filters. Wehave shown that singularity can appear in many othersituations and we have established a sufficient condi-tion for singularity by using the positions of the polesof the filter with respect to the circle of singularityand the whiteness of the input.This assumption of whiteness can however be

partially relaxed without changing the singularity ofthe output. Some sufficient conditions ensuring thesingularity with coloured inputs have been established.These conditions are obviously satisfied not only bywhite noise but also by a large class of correlated signals.It is especially the case of some Markovian signals offinite order. The theoretical analysis also shows that theoutput generated by coloured inputs can be a mixture ofa discrete and a singular distribution. Computerexperiments in order to verify the theoretical resultshave been realized and discussed. For this purpose aspecific model of linear Markovian signal of order onewas introduced and the experimental results are incomplete agreement with the theory. Finally a model ofBernoulli input ensuring that the output contains adiscrete and a singular part was introduced and here alsothe experimental results are in perfect agreement withthe theory.

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506 B. Picinbono