MIXED SPECTRA FOR STABLE SIGNALS FROM DISCRETE OBSERVATIONS

Signal & Image Processing: An International Journal (SIPIJ) Vol.12, No.5, October 2021
DOI: 10.5121/sipij.2021.12502 21
MIXED SPECTRA FOR STABLE SIGNALS
FROM DISCRETE OBSERVATIONS
Rachid Sabre
Biogeosciences (UMR CNRS/uB 6282),
University of Burgundy, 26, Bd Docteur Petitjean, Dijon, France
ABSTRACT
This paper concerns the continuous-time stable alpha symmetric processes which are inivitable in the
modeling of certain signals with indefinitely increasing variance. Particularly the case where the spectral
measurement is mixed: sum of a continuous measurement and a discrete measurement. Our goal is to
estimate the spectral density of the continuous part by observing the signal in a discrete way. For that, we
propose a method which consists in sampling the signal at periodic instants. We use Jackson's polynomial
kernel to build a periodogram which we then smooth by two spectral windows taking into account the
width of the interval where the spectral density is non-zero. Thus, we bypass the phenomenon of aliasing
often encountered in the case of estimation from discrete observations of a continuous time process.
KEYWORDS
Spectral density, stable processes, periodogram, smoothing estimate, aliasing.
1. INTRODUCTION
The multiple applications where the random signals whose variance increases indefinitely impose
the interest of using stable alpha processes by several research authors in various fields. In
particular, Stable symmetric harmonizable processes and their properties have been widely
studied by many authors such as [1] - [10] to name a few.
Concrete applications of stable symmetric processes cover a wide spectrum of fields such as:
physics, biology, electronics and electricity, hydrology, economics, communications and radar
applications., ...ect. See: [11]-[22]. In this paper we consider a symmetric alpha stable
harmonizable process = { ( ): }
X X t t R
 . Alternatively X has the integral representation:
 
( ) = exp ( ) ( )
X t i t d
  
 (1)
where 1< < 2
 and  is a complex valued symmetric  -stable random measure on R with
independent and isotropic increments. The control measure is defined by ( ) =| ( ) |
m A A 

 (see
[4]) is called spectral measure. The estimation of the spectral density function was already
studied in different cases: by E.Masry and S.Combanis [4] when the time of the process is
continuous, by Sabre [23] when the time of the process is discrete and by R.Sabre [24]-[25]when
the time of the process is p-adic.
In this paper we consider a general case where the spectral measure is the sum of an absolutely
continuous measure with respect to Lebesgue measure and a discrete measure:

22
=1
( ) = ( )
q
i wi
i
d x dx c
   
+ 
where  is a Dirac measure, the specral density  is nonnegative integrable and bounded
function. i
c is an unknown positive real number and i
w is an unknown real number. Assume
that 0
i
w  . The discrete measurement comes from the repeated random energy jumps during the
experimental measurements. Spectral density represents the distribution of the energy carried by
the signal.
Our objective is to propose a nonparametric estimator of the spectral density after discrete
sampling of the process ( )
X t . This work is motivated by the fact that in practice it is impossible
to observe the process over a continuous time interval. However, we sampled the process at
equidistant times, i.e., =
n
t n , > 0
 . It is known that aliasing of  occurs. For more details
about aliasing phenomenon, see [26]. To avoid this difficulty, we suppose that the spectral
density  is vanishing for | |>
  where  is a nonnegative real number. We create an
estimate of the spectral density based on smoothing methods. We show that it is asymptotically
unbiased and consistent.
Briefly, the organization of this paper will be as follows: in the second section two technical
lemmas will be presented as well as a preiodogram and we will show that this periodogram is an
asymptotically unbiased but inconsistent estimator. In the third section, the periodogram will be
smoothed by two well chosen spectral windows to estimate the spectral density at the jump
points. We show that the smoothing periodogram is a consistent estimator. The fourth section
gives conclusions and working perspectives.
2. THE PERIODOGRAM AND ITS PROPRIETIES
In this section we give some basic notations and properties of the Jackson's polynomial kernel.
Let N is the size of sample of X . Let k and n are the numbers satisfying:
1
1= 2 ( 1)
2
N k n with n N k N
 
− −     
 
if
1
=
2
k then 1 1
= 2 1,
n n n N
−  .
The Jackson's polynomial kernel is defined by: ( )
( ) = ( )
N
N N
H A H


  where
2 2
( )
,
,
sin sin
1 1
2 2
( ) = = .
2
sin sin
2 2
k k
N
k n
k n
n n
H with q d
q


 
 
 
 −
   
   
   
   
   

where
1
,
= ( )
N N
A B 

−
with ( )
, = ( ) .
N
N
B H d
 
 
 
−

We cite two lemmas which are used in this paper. Their proof are given in [23].
Lemma 2.1 There is a non negative function k
h such as:

23
( 1)
( )
= ( 1)
( ) = cos( )
k n
N
k
m k n
m
H h m
n
 
−
− −
 
 
 

Lemma 2.2 Let
2
, ,
sin
2
= = | | | ( ) | ,
sin
2
k
N N N
n
B d and J u H u du

 
 
 
 



− −
   where
]0,2],
  then
2
2 1
,
2 1
2
2 0 < < 2
4 1
< < 2
2 1 2
k
k
N
k
n if
B
k
n if
k k




 

 


−
−
  

  
  
 

 −

and
2
2 2 1
, 2
2
1 1 1
< <
2 ( 2 1) 2 2
2 1 1
< < 2
2 ( 1)(2 1) 2
k
k k
N k
k
if
k n k k
J
k
if
k n k
 
 
  
 
 

 
 

  
+
−
+
 +
 − +

 
+

 + − −

This paragraph gives a periodogram and develops its proprieties. Consider the process ( ),
X t
defined in (1), observed at instants =
j
t j , =1,2,...
j N and
2
=



, where  is a real
number strictly greater than 2 . Define the periodogram ˆ
N
I on ] , [
−   as follows:
( )
,
ˆ ( ) = | | , 0 < <
2
p
N p N
I C I p


 
Where
1 = ( 1)
= ( 1)
( ) = [ ] e[ exp{ ( )} ( ( 1) )],
' '
n k n
' '
N N k
'
n k n
n
I A R h i n X n k n
n

    
−
− −
 
− + −
 
 

The normalisation constant ,
p
C  is given by , /
,
=
[ ]
p
p p
p
D
C
F C
 
 
, with
| |
,
1 1
1 cos( ) 1
= a =
| | | |
a
u
p p
p p
u e
D du nd F du
u u

−
 
+ +
− −
− −
  .
Lemma 2.3

24
The characteristic function of ( )
N
I  ,  
exp ( )
N
E irI  , converges to
,1 ,2
exp[ | | ( ( ) ( )].
N N
C r 
    
− + where
( )
,1( ) = | |
N N
y
H y dy



   

−
 
−  
 
 and ,2
=1
( ) = ( )
q
N i N i
i
c H w

  
−

Proof
From (1) and the expression of N
I , we have:
   
1 = ( 1)
= ( 1)
( ) = [ ] e exp [ ( )] exp [ ( 1)] ( ).
' '
n k n
'
N N k
R
'
n k n
n
I A R h i n u i uk n d u
n

     
−
− −
 
− −
 
 


Using [1] and the definition of the Jackson polynomial kernel we obtain that the characteristic
function is the form:
 
exp ( ) = exp | | ( ) .
a
N a N
E irI C r
  
 
−
  (2)
where ,1 ,2
( ) = ( ) ( )
N N N
     
+ with
( )
,1( ) = | |a
N N
R
v
H v dv
   

 
−  
 
 and ,2
=1
( ) = ( )
q
N i N i
i
c H w

  
−

( )
,1( ) = | |a
N N
R
v
H v dv
   

 
−  
 
 ( )
(2 1)
(2 1)
= | | .
j
a
N
j
j Z
v
H v dv


 

+
−

 
−  
 

Putting = 2
v y j

− and using the fact that N
H is 2 -periodic, we get
( )
( ) = | | ( ) ,
a
N N j
j Z
H y y dy


   
 −
−
  where
2
( ) = .
j
y
y j

 
 
 
−
 
 
Let j be an integer
such that
2
< <
y j


−
−  . Since <
 
 and | |<
y  , we get
1
| |< <1
2 2
j



+ and then
= 0
j . Therefore
( )
,1( ) = | | .
N N
y
H y dy



   

−
 
−  
 
 (3)
Theorem 2.4 Let < <

−  then ( )
ˆ ( ) = ,
p
N N
E I 
  
   
 
 
Proof
From the following equality used in [4], for all real x and 0 < < 2
p ,
1 1
1 1
1 cos( ) 1
| | = = e ,
| | | |
ixu
p
p p
p p
xu e
x D du D R du
u u
 
− −
+ +
− −
− −
  (4)

25
and replacing x by N
I , we obtain
/ 1
,
1 exp{ ( )}
1
ˆ ( ) = e ,
[ ] | |
N
N p p
p a
iuI
I R du
F C u





+
−
−
 (5)
The equation (5) and the definition of the , ,
p
F  give
 
 
/ 1
,
/
1 exp | | ( )
1
ˆ ( ) = .
[ ] | |
= ( ) .
a N
N p p
R
p a
p
N
C u
EI du
F C u




 

 
+
− −

3. SMOOTHING PERIODOGRAM
In order to obtain a consistent estimate of [ ( )]
p

  , we smooth the periodogram via spectral
windows depending on whether  is a jump point or not( )
i
w
  .
(1)
1 2
(2) (1)
( ) { , , , }
( ) = ( ) ( )
1
N q
N N N
f if w w w
f f cf
else
c
 
  
 

 −

−

where (1) (1) ˆ
( ) = ( ) ( )
N N N
f W u I u du


 
−
−
 and (2) (2) ˆ
( ) = ( ) ( ) .
N N N
f W u I u du


 
−
−

The spectral windows (1)
N
W and (2)
N
W are defined by: (1) (1) (1)
( ) = ( )
N N N
W x M W M x and
(2) (2)) (2)
( ) = ( )
N N N
W x M W M x with W is an even nonnegative, continuous function , vanishing for
| |>1
 such that
1
1
( ) =1
W u du
−
 . The bandwidths (1)
)
N
M and (2)
)
N
M satisfying:
(2)
(1)
N
N
M
c
M
=
( )
lim
i
N N
M
→ = +,
( )
= 0
lim
i
N
N
M
N
→ for 1,2
i = ,
(2)
(1)
= 0
lim
N
N
N
M
M
→+ and such that
(2) (1)
(1) (1)
1 1
( ) = ( ) ,
N N
N N
W M W M
M M
  
 
  −
 
 
.
Th following theorem shows that ( )
N
f  is an asymptotically unbiased estimator of [ ( )]
p

  for
< <

−  and 1 2
{ , , , }
q
w w w
  .
Theorem 3.1
Let < <

− , such that 1 2
{ , , , }
q
w w w
  . Then,     ( )
( ) ( ) = o 1 .
p
N
E f 
  
−

26
If  satisfies the hypothesis ( ) ( ) ( ) ,
x y cste x y

 
−
−  − with < 2 1
k
  − , then,
 
(2 1) (1)
(1) (2 1) (1) 2 1
1 1
= 0
( ) [ ( )] = .
1 1 1
= 0
k
p
N
N
k k
N N
O if
n M
E f
O if
M n M n
 

  

  

−
− −
  
+
  
  
− 
 
 + +
 

 

Proof
It is easy to see that:
  (1) (1) ˆ
( ) = ( ) ( ) .
N N N N
R
E f M W M u E I u du
   
 
−
   

Let (1)
( ) =
N
M u v
 − , we obtain:
1
(1)
1
[ ( )] = ( ) .
p
N N
N
v
E f W v dv
M

  
−
 
 
−
 
 
 
 
 (7)
Since
1
1
( ) =1
W u du
−
 and the inequality (3), we get:
   
1
(1)
1
( ) ( ) ( ) ( ) .
p
p
N N
N
v
E f W v dv
M


      
−
 
−  − −
 
 

As < 1
p

, we obtain
,1 ,2
(1) (1) (1)
( ) ( ) |
p p p
N N N
N N N
v v v
M M M
  
         
     
− −  − − + −
     
     
We now examine the limit of ,1 (1)
N
N
v
M
 
 
−
 
 
. From (3) we get:
,1 (1) (1)
= .
N N
N N
v v u
H u du
M M

    


−
 
     
− − −
 
     
   
   
 

Let (1)
=
N
v
u y
M
 
 
− −
 
 
, we obtain:

27
,1 (1) (1)
(2 1)
(1)
(2 1)
= ( )
= ( ) .
N N
R
N N
j
N
j
j Z N
v v y
H y dy
M M
v y
H y dy
M

 

   

 

+
−

   
− − +
   
   
 
− +
 
 

 (8)
Let 2 =
y j s

− . Since | (.) |
N
H 
is 2 −periodic function, we get
,1 (1) (1)
2
= ( ) .
N N
j Z
N N
v v s
H s j ds
M M
 


   
 
+
−

   
− − + +
   
   

The function  is uniformly continuous on [ , ]
−  and since | |
N
H 
is a kernel, the right hand
side of the last equality converges to
2
.
j Z
j

 


 
+
 
 
 Let j be an integer such that
2
< <
j
 

+
− . The definition of  implies that | |<| |<
  
 . It is easy to see that
| |< 1
j and then = 0
j . Since (1)
N
H is a kernel, we obtain that ,1 (1)
N
N
v
M
 
 
−
 
 
converges to
( )
  . On the other hand, (1)
,2 (1) (1)
=1
=
q
N N i
i
N N
v v
H w
M M

   
 
   
− − −
 
   
 
   
 

Since i
w is different from  and from the lemma 2.2, we get
1
2 (1)
2 1
,2 (1)
=1
2 1
2 , where cte =inf sin .
2
k i
q
N
k
N i
i
N
v
w
M
v
n c
M cte



 
  

−
−
 
 
− −
 
 
 
     
 
−   
     
 
 
     
 
 

Therefore, ,2 (1) 2 1
1
=
N k
N
v
O
M n 
  −
   
−
   
 
 
. Thus, we have    
( ) ( ) = (1).
p
N
E f o

  
−
The rate of convergence:
Assume that the spectral density  satisfies the hypothesis H . Let
/
=| ( ( )) |=| [ ( )] [ ( )] |
p
N N
F Bias f E f 
   
− . It follows that
1
1
1
(1) (1)
1
( ) [ ( )] ( ) .
2
p
p
N N
N N
p v v
F W v dv
M M


       

−
−
−
 
 
   
 
 − + − −
 
   
 
   
 
 
 


28
Since N
N
v
M
 
 
−
 
 
converges to ( )
  , in order to study the rate of the convergence for F
we examine the rate of convergence of
1
(1)
1
( ) ( )
N
N
v
W v dv
M
   
−
 
− −
 
 
 . Indeed, from (3), we
obtain (1) (1)
= | | .
N N
N N
v v y
H y dy
M M




   

−
     
− − +
     
 
   

Let ( ) (1)
, = ( )
N N
N
v
M
     
 
 − −
 
 
. Putting (1)
=
N
v
t y
M


 
− − +
 
 
and using the condition
H , we get ( ) ( )
(1)
1 (1)
(1)
, .
v
MN
v
N N
N
MN
v t
C H t dt
M
 
 


 
 

− +
− −
  +

We can maximize as follows:
( )
1 1
1 (1)
1 1
1
(1)
1
1
1
1 (1)
1
( ) ( , ) 2 ( ) | |
2 ( )
N
N
v
MN
v N
MN
W v dv C W v v dv
M
C
W v H t t dtdv

 

 
 


   
 

− −
− +
− − −
 
+
 
 
The second integral of the right hand side is bounded as follows:
( ) ( )
( )
( )
(1)
| | | |
(1)
(1)
| | | |
(1)
.
v
MN
v
v N N
M
M N
N
N
v
MN
N
H t t dt H t t dt
H t t dt
H t t dt

 

 
 


 
 
  


 
 

− +
−
− − −
− −
−
+ +

+
+
 


(9)
Since | (.) |
N
H is even, the first and the last integrals in the right hand side of (9) are equal. As
N
v
M

converges to zero and < <
  
 , for a large N we have:

29
( ) ( )
| | | | | | | |
(1) (1)
(1)
2
,
(1)
(2 )
| |
(2 )
sin | |
v v
M M
N N
N N
N
k
'
N
N
H t t dt H t dt
M
B
M
 
   
 
 
 








 
+ + + +

+

 
+ +
 
 
 
The lemma 2.1 gives:
( )
2 1
| | | |
(1)
(1) 2 1
1
O 0,
=
1
O = 0,
k
v
MN
N
k
N
if
n
H t t dt
if
M n


 
 




−
+ +
−
  

  
 




 
  
  


Thus, we obtain the result of the theorem.
Theorem 3.2. Let  a real nuber belonging to ] , [
−   , and = i
w
 . Choose k such that
( )
2
(1)
2 1
= 0
lim
k
N
N k
M
n


→ −
. Then,
i) [ ( )] [ ( )] = (1)
p
N
E f o

  
−
ii) If  satisfiies the hypothesis ( ) ( ) ( ) ,
x y cste x y

 
−
−  − with
1
< < 2
2k


+
, then
( )
( )
( )
2
(1)
2 1
(2)
2
(1)
(2) 2 1
1
0 < 1
[ ( )] [ ( )] =
1
1< 2
k
N
k
p
N
N k
N
k
N
M
O if
n
M
E f
M
O if
M n

 




  

−
−
  
  
+ 
  
  
− 
 

 
 + 
 

 

Proof :
The form of estimator gives:

30
(2)
(2) (1)
(1)
(2)
(1)
( ) ( )
[ ( )] = [ ( )]
1
N
N N p
N
N N
N
N
M
W u W u
M
E f u du
M
M



 
 
−
− − −
−

1 1
(1) (1)
1
1 1 2 3
(1)
(1)
[ ( )] =
M M
N N
N
M
M N
N
E f E E E
 

 


− +
− +
−
+ + = + +
  
Put
1
(2) (2) (1)
(1)
1
2 (2)
(1)
(1)
[ ] [ ]
= , = [ ( )] .
1
p
N n N
MN
N
N
MN
N
M W M v W M v
u v E v dv
M
M

  
−
 
−
 
− −
−
 Therefore 2 0
E =
for a large N.
1
(2) (1)
(2)
1 (2)
(1)
= [ ( )][ ( )]
1
p
M
N N
N N
N
N
M
E W M u u du
M
M



 
−
−
− −
−

1
(2) (1)
(1)
(2)
(1)
[ ( )][ ( )]
1
p
M
N N
N N
N
N
M
W M u u du
M
M



 
−
−
−
−

Put (2)
( ) =
N
M u v
 − in the first integral and put (1)
( ) = ,
N
M u w
 − in the second integral and for
a large N, we have (1) (2)
( ) >1 ( ) >1.
N N
M et M
   
+ + As W is null outside of [ 1,1]
− , for
large N, the second integral of 1
E is zero.Therefore,
1
(2)
1 (2) (2)
(1)
(1)
1
= ( ) .
1
p
N
MN
N N
MN
N
v
E W v dv
M M
M

 
 
 
−
 
 
 
 
−
 (10)
(2) (2)
(2) (1)
1 1
3 (2) (2)
(1) (1)
(1) (1)
= [ ( )][ ( )] [ ( )][ ( )]
1 1
p p
N N
N N N N
N N
M M
N N
N N
M M
E W M u u du W M u u du
M M
M M
 
 
 
   
+ +
− − −
− −
 
Putting (2)
( ) =
N
M u v
 − in the first integral and (1)
( ) =
N
M u w
 − in the second integral, we
obtain
(2)
(2)
(1)
(1) 1
(2) (1)
3 (2) (2)
(2) (1)
( ) ( )
(1) (1)
1
= ( ) ( ) .
1 1
p p
N
MN
M N
N
N N
M M
N N
N N
N N
N N
M
M
v w
E W v dv W w dw
M M
M M
M M
 
   
   
−
−
− −
   
   
− − −
   
   
   
   
− −
 

31
(2) (1)
For large we have, ( ) < 1 ( ) < 1.
N N
N M and M
   
− − − −
1
(2)
3 (2) (2)
(1)
(1)
1
= ( ) .
1
p
N
MN
N N
MN
N
v
E W v dv
M M
M

 
 
 
+
 
 
 
 
−
 (11)
It is easy to show that for a large N
1
(2)
(2)
(1)
(1)
1 1
( ) = .
2
1
MN
N
MN
N
W v dv
M
M
−
 (12)
1 3
[ ( )] [ ( )] = [ ( )]
p p
N
E f E E
 
    
− + − 1 3
1 1
[ ( )] [ ( )]
2 2
p p
E E
 
   
 − + −
From (10) and (12), for a large N, we have
1
(2)
1 ,1
(2) (2) (2)
2
(1)
(1)
1 1
[ ( )] ( ) ( )
2
1
p
p
N N
MN
N N N
MN
N
v v
E W v dv
M M M
M


       
   
−  − + − −
   
   
−

As 1
p

 , we obtain
1
(2)
1 ,1
(2) (2)
(1)
(1)
1 1
[ ( )] ( ) ( )
2
1
p
p
N
MN
N N
MN
N
v
E W v dv
M M
M


     
 
−  − −
 
 
−

+
1
(2) ,2
(2) (2)
(1)
(1)
1
( ) .
1
p
N
MN
N N
MN
N
v
W v dv
M M
M

 
 
−
 
 
−

On the other hand,
1 1
(2) ,1 ,1
(2) (2)
0
(1)
( ) ( ) ( ) ( )
p p
N N
MN N N
MN
v v
W v dv W v dv
M M
 
       
   
 −   −
   
   
 
For all  belonging to ] , [
 
− ,1 (2)
N
N
v
M
 
 

 
 
converges to ( )
  , uniformly in
[ 1,1].
v − Therefore,
1
(2) ,1
(2) (2)
(1)
(1)
1
( ) ( )
1
p
N
MN
N N
MN
N
v
W v dv
M M
M

   
 
 −
 
 
−
 converge to zero.

32
Since = ,
i
w ,2 2
(2)
=1 ,
=
(2)
1
1
sin
2
m
N i k
m
N N
m i i m
N
q
a
v
M B v
w w
M


 

 
 
 

   
 
 −
 
 
 
 

2
,
(2)
1
.
1
sin
2
i
k
N
N
a
B v
M

 
+
  
 

 
 
 
 
For all =
m i ,
( )
2 2
[ 1,1]
(2)
1 1
= .
sup
lim
1
1 sin
sin
2
2
k k
N v
i m
i m
N
v w w
w w
M
 

→+  −    
  −
 −
   
   
 
 
Thus, for large N, we get
( )
2 2
=1 =1
, ,
= =
(2) {1,2, , } { }
1 1
.
1
1 sin
inf
sin
2
2
m m
k k
m m
N N
m i m i
i m
i m m q i
N
q q
a a
B B
v w w
w w
M
 
 


 −
 
 
 
 +
 
 
   
   
−
 −  
   
   
 
 
 
 
The lemma 2.1 gives 2 2 1
=1 ,
=
(2)
1 1
= .
1
sin
2
m
k k
m N
m i i m
N
q
a
O
B n
v
w w
M
 
 
−
 
 
  
 
 
 −
 
 
 
 
 (13)
For large N, we have ]0,1] (2) (2)
= <
supv
N N
v
M M
 


. Consequently
2
2 2
, ,
(2)
(2)
1
.
1
sin
2
k
i i
k k
N N
N
N
a a
B B v
v
M
M

 
 




 
 
 

 
 
 
 
As (1) (2) (2)
N N N
v
M M M
  
  , we obtain 2
,
(2)
1
1
sin
2
i
k
N
N
a
B v
M

 
  
 

 
 
 
 
2
2
,
(1)
.
k
i
k
N
N
a
B
M







Frome the lemma 2.2 , we obtain

33
( )
2
(1)
2 2 1
,
(2)
1
= .
1
sin
2
k
N
i
k k
N
N
M
a
O
B n
v
M

 
 
−
 
 
 
  
   
+ −
 
 
 
 
(14)
Therefore, we get
( )
2
(1)
1
(2) ,2
(2) (2) 2 1
(2 1)
(1)
(1)
2 1
( ) =
1
p
p
p k
N
N p k
M k
N
N N
MN
N
M
v
W v dv O
M M n
n
M
 






  −
−
 
 
 
   
 +
 
   
   
−  
 
 
 (15)
Choosing (1)
N
M such that
( )
2
(1)
2 1
k
N
k
M
n

−
converges to 0 . For example (1)
= b
N
M n with
1
0 < <1
2
b
k
− . Thus, [ ( )] [ ( )] = 0.
lim
p
N
N
E f 
  
→+
−
Theorem 4.2 Let < <

−  such that ( ) > 0
  . Then,  
( ))
N
var f  converges to zero.
If (1)
= c
N
M n with 2 2
1 1
< <
2 2
c
k 
, then   (1 2 )
1
( )) = O .
N c
var f
n
 −
 
 
 
Proof
Fisrt suppose that 1 2
{ , , , }
q
w w w
  . It is clear that the variance of N
f can be written as
follows:
(1) (1)
1
2
ˆ ˆ
[ ( )] = ( ) ( )c ( ), ( ) .
' ' '
N N N N N
R
var f W u W u ov I u I u dudu
    
− −  

Since W is zero for | |>1
 , for large N , we have
1
1 1
1 1 1 1
1
ˆ ˆ
[ ( )] = , ( ) ( ) .
'
' '
N N N
N N
x x
var f cov I I W x W x dx dx
M M
  
−
 
   
− −
 
   
   
 

Define two subsets of the 2
[ 1,1]
− by:
•  
2
1 1 1 1 1
= ( , ) [ 1,1] ; | |> ,
' '
N
L x x x x 
 − −
•  
2
2 1 1 1 1
= ( , ) [ 1,1] ; | | ,
' '
N
L x x x x 
 − − 

34
where N
 is a nonnegative real, converging to 0 . We split the integral into an integral over the
subregion 2
L and an integral over 1
L : 1 2
2 1
var[ ( )] = = .
N L L
f J J


+ +
 
Cauchy-Schwartz inequality and theorem 3.1, give
1 1 1 1 1
| |
1 1
( ) ( ) .
' '
'
x x N
J C W x W x dx dx

− 
  where C is constant. Thus, we obtain
1 = ( )
N
J O  (16)
It remains to show that 2
J converges to zero. For simplicity, we define
1 1
1 2
(1) (1)
= ; = ,
'
N N
x x
M M
   
− − and 1 1
(1) (1)
ˆ ˆ
( ) = c , .
'
N N
N N
x x
C ov I I
M M
  
 
   
− −
 
   
   
 
We first show that ( )
C  converges to zero uniformly in 1 1
, [ 1,1]
'
x x  − . Indeed, from lemma
2.3, we have
( )
( ) | | ( )
1 /
, 1
e
ˆ ˆ
( ) ( ) = [ ] .
| |
iuI v C u v
N N
p
N N p p
R e e
EI v I v F C du
u



 
−

− −
+
−
−
− 
Thus, the expression of the covariance becomes
( )
 
2 2
2
, 2
=1
2 1 2
1
=1
1 2
( ) = cos ( )
exp | | ( )
| |
p
p k N k
R
k
k N k p
k
C F C E u I
du du
C u
u u

 


 
 
−
−
+
 
 
 
− −



Using the following equality: 2cos cos = cos( ) cos( )
x y x y x y
+ + − , we have
( )
1
2 2
=1
=1
1
2 1
=1
1
cos ( ) = exp ( ) ( ) ( )
2
1
exp ( ) ( 1) ( ) ( ) .
2
k N k k N k
k
k
k
k N k
k
E u I C u H v d v
C u H v d v






    
   
+
 
   
− −
   
 
 
 
+ − − −
 
 

 


By substituting the expression for ( )
C  and changing the variable 2
u to 2
( )
u
− in the second
term, we obtain
( )
2
2 1 2
, 2 1
1 2
( ) = ,
| |
p
'
K K
p p
R
du du
C F C e e
u u

 

−
− − −
+
−
 (17)

35
where
1
2
=1
= ( ) ( ) ( )
k N k
k
R
K C u H v d v


    
−

 and
2
=1
= | | ( ) ( )
'
k N k
k R
K C u H v v dv


   
−
 
Since , > 0
'
K K , exp{ }
'
K K ' ' '
e e K K K K K
− −
−  − − − , we obtain:
2
1 2
2 | | ( ; ),
'
N
K K C u Q

  
−  where
2 2
1 2 1 2
( ; ) = | ( ) | | ( ) | ( )
N N N
Q H u H u d u
 
      

−
− −

Now, let us show that 1 2
( ; )
N
Q   converges to zero. Indeed, since  is bounded on [ , ]
−  ,
we have
2
1 2 1 2
2
1 2
=1
( ; ) sup( ) | ( ) ( ) |
( ) ( )
N N N
q
i N i N i
i
Q H u H u du
c H w H w


      
   

−
 − −
+ − −


(18)
From the definition of N
H , we can write
( ) ( )
( )
( )
( )
( )
1 2
2
1 2
,
1 2
sin sin
1 2 2
| | = .
1 1
sin sin
2 2
k k
N N '
N
n n
v v
H v H v dv dv
B
v v
 


   
   
   
 
− −
   
− −
   
   
− −
   
− −
   
   
 
a) First step:
We show that the denominators of the first and second terms under the last integral do not vanish
for the same v, so we suppose thatv exists, belonging to [ , ]
−  and , '
z z Z
 such as:
1 2
= 2 = 2 '
v z and v z
     
− − . Since 1 2
=
  , then z and z are different.
Therefore, ( )
1 2
=
2
'
z z

 

− − . Hence, 1 2
1
| |= .
'
z z
w
 
− − As 1 2 =0
limN  
→ − ,
consequently, for a large N we get:
1
<
2
'
z z
− .
Thus, we obtain a contradiction with the fact that z and '
z are different integers.
b) Second step:
Asuume there exist q points, 1 2
, , , [ , ]
q
V V V  −  such as:

36
for 1
=1,2, , 2
j
j q V Z
  
−  then 1 j
V
Z
w w

−  , and we assume that there exist '
q points
1 2
, , , [ , ]
' ' '
'
q
V V V  −  such that for =1,2, , '
i q 2
'
i
V
Z
w w

−  . Showing that, | |=
j
V  ,
| |=
i
V  for 1 j q
  and 1 '
i q
  . Indeed, 1< < 0
w
 − 
− and 0 < <1
w
 + 
because
> 2
w  . Hence Z
w
 − 
 and Z
w
 + 
 . For a large N , we get that
1
< <1 ,
E E
w w w w

 
−  −
   
− +
   
   
where  E
x is the integer part of x . Hence,
1
Z
w w
 
−  . In the same manner, we show that 1
Z
w w
 
+  . Similarly, it can be shown that
2
Z
w
 + 
 . Thus, | |=
j
V  and | |=
i
V  .
c) Third step:
We classify j
V and '
i
V by increasing order:
1 2
< < < < <
j j j '
q q
V V V
+
−  , and we write
the integral in the following manner:
( )
( )
( )
( )
1 2 1
1 2, 3, 4
=1 =1
1 2 1
sin sin
2 2
=
1 1
sin sin
2 2
k k
' '
q q q q
i i
i i
n n
v v
dv I I I I
v v
 
   
   
+ + −

−
   
− −
   
    + + +
   
− −
   
   
 

where

37
( )
( )
( )
( )
( )
( )
( )
( )
1 2
( )
1
1
1
1 2
1 2
( )
2, ( )
1 2
3, ( )
sin sin
2 2
=
1 1
sin sin
2 2
sin sin
2 2
=
1 1
sin sin
2 2
=
k k
V N
j
k k
V N
ji
i V N
ji
Vj
i V N
ji
n n
v v
I dv
v v
n n
v v
I dv
v v
I
 

 



   
   
   
   
−
−
+
−
+
   
− −
   
   
   
− −
   
   
   
− −
   
   
   
− −
   
   


( )
( )
( )
( )
( )
( )
( )
( )
1 2
( )
1
1 2
1 2
4 ( )
1 1 2
sin sin
2 2
1 1
sin sin
2 2
sin sin
2 2
=
1 1
sin sin
2 2
k k
N
i
k k
V N
j '
q q
n n
v v
dv
v v
n n
v v
I dv
v v
 

 

   
   
   
    
−
+

+
+
   
− −
   
   
   
− −
   
   
   
− −
   
   
   
− −
   
   


where ( )
N
 is a nonnegative real number converging to zero and satisfying:
1 1 2 2
< ( ) < ( ) < ( ) < ( ) < < ( ) < ( ) < ,
j j j j j j
' '
q q q q
V N V N V N V N V N V N
     
+ +
− − + − + − +  a
nd 1 2
( ) < .
2
N
 

−
Showing that the first integral converges to zero. For a large N , it easy to see that 1 <
 .
Without loss of generality, we assume that 1 = 2
j i
i
V k
  
− for all i , with i
k Z
 . The fact that
there is no v between − and
1
( )
j
V N

− on which the denominators are vanishing, gives
1
1
1 2 1 2
( ) 1
.
( ) ( ( ))
( ) ( )
inf sin , sin inf sin , sin
2 2 2 2
j
k k
k k
j
V N
I
V N
N
 
 

    
  
− + 

   
+  − + + 
   
   
 
 
Substituting
1
j
V in the last inequality, we get
2 1 2 1
( ( )) | ( ) |
sin = sin
2 2
k k
j
V N N
 
      
− + − +
. For a large N ,we obtain

38
2 1
| ( ) | ( ) ( )
<
2 2 2
N N N
     
 
− +
  + − .
On the other hand, two cases are possible:
1. 2 1 > 0
 
− , then we have 2 1 2 1
| ( ) |= ( ) > ( )
N N N
      
− + − +
2. 2 1 < 0
 
− , since 2 1
| |> 2 ( )
N
  
− , we have
2 1 1 2
| ( ) |= ( ) > ( )
N N N
      
− + − − .
Therefore, 2 1
| ( ) |
( ) ( )
< <
2 2 2
N
N N
   
 

− +
− . For a large N , we have
1
( ) < 2 ( )
N
   
− +  and 2
( ) < 2 ( )
N
   
− +  . Then,
1
( )
( ) ( )
< <
2 2 2
N N
 
 

+ 
− and 2
( )
( ) ( )
< < .
2 2 2
N N
 
 

+ 
−
Therefore,
1
1 2
( )
.
( )
sin
2
j
k
V N
I
N



− + 

For the integral 2,i
I , we bound the first fraction under integral by k
n 
:
( )
( )
2, ( )
2
1
1
sin
2
V N
j
k i
i k
V N
ji
I n dv
v




 
+
−

 
−
 
 
 . By substituting for ji
V in the last
inequality and putting
2
=
k
v u


− , we get
( )
( )
1
2, ( )
1
2
1
1
sin
2
N
k
i k
N
I n du
u
 


 
 
+
−

 
−
 
 
 . Since 1
| |< ( )
u N
 
− , it is easy to see that
:
2 1
2 2 1 1 2 1
| |
| | | | | | | | ( ) >
2
u u N
 
      
−
−  − − −  − −
Since ( )
N
 converges to zero, for a large N we have 2 1
2
( ) < | |
2
N

   

 
− −
 
 
, therefore
2 1 2 2 1
| | | | | | ( )
0 < < < < .
4 2 2
u N
     
   
− − − +
Consequently:

39
2,
1 2 1 2
2 ( )
,
| | | | ( )
inf sin ; sin
4 2
k
i
N n
I
N


       

− − +
 
 
 
where 2 1
2 1
| |=
N
x x
M
 
−
− . Then, for a large N , we have 2 1
3
< 2 .
2

  
− Therefore,
2 1 2 1 2 1
| | | | ( ) | |
< < .
4 2 4
N
       
  
− − + −
− Thus, we bound the integral as follows:
2,
2 1
2 ( )
| |)
sin
4
k
i k
N n
I



  

−
Since there is no v between ( )
ji
V N

+ and
1
( )
ji
V N

+
− on which the denominators are
vanishing, we get:
1
3,
2 ( )
,
j j
i i
i
V V N
I
A B

+
− −


where
1 1 1
2 2 1
( ) ( )
= inf sin , sin
2 2
( ) ( )
= inf sin , sin
2 2
k k
j j
i i
k k
j j
i i
V N V N
A
V N V N
B
 
 
     
     
+
+
 
− − − +
 
 
 
 
− − − +
 
 
 
It follows from the hypothesis on ( )
N
 that
1 2 1 2 1 2
| | | ( ) | | |
( ) | ( ) | | ( ) | ( )
< < < < .
2 2 2 2 2 2 2
N
N N N N
         
     

− − − −
− + −
The definition of ji
V gives

40
2
1 1
< .
( ) ( )
sin
sin 2
2
k k
ji
V N N
 
   
− −
Similarly we bound the other terms:
1
3, 2
2 ( )
.
( )
sin
2
j j
i i
i k
V V N
I
N



+
− −

We can show by same majoration that :
1
4
2 ( )
,
j j
i i
V V N
I
E F

+
− −


where
1
1
2
2
( )
= inf sin , sin
2 2
( )
= inf sin , sin
2 2
k
k
jq q
k
k
jq q
V N
E
V N
F




  
 
  
 

+

+
 
− −
− 
 
 
 
 
− −
− 
 
 
 
Since ( )
N
 converges to zero, for a large N , we have 1
2
( ) < | |
2
N

  

 
− − 
 
 
, and
1 2
2
( ) < | |
2
N

   

 
− −
 
 
. It follows that: 4 2
( )
.
( )
sin
2
jq q
k
V N
I
N




+
 − −
 We recapitulate, from the
previous increases, we obtain
( )
( )
( )
( )
1 2
1
2
=1 2 1
1 2
sin sin ( ) 2 ( )
2 2
1 1 ( ) | |
sin sin sin sin
2 2 2 4
k k
k
q q
j
k k
i
n n
v v V N n N
dv
N
v v
 

 
     
   
   

+

−
   
− −
     + −
     +
    −
− −
   
   


1
1
2 2
=1
( )
( )
( ) ( )
sin sin
2 2
q q
j
j j q q
i i
k k
i
V N
V V N
N N
 


 

+ −

+
+
 − −
− −
+ +

After simplification, we have

41
( )
( )
( )
( )
1 2
2
2 1
1 2
sin sin
2 2( 1) ( ) 2 ( )( )
2 2
.
1 1 ( ) | |
sin sin sin sin
2 2 2 4
k k
k
k k
n n
v v
q q N n N q q
dv
N
v v
 

 
   
 
   
   

−
   
− −
     
− + + +
     +
    −
− −
   
   

Using the following inequality
| |
sin
2
x x

 , we get
( )
( )
( )
( ) ( )
2
1 2
2
2 1
1 2
sin sin
2 2 ( )( )(2 )
2 2
.
1 1 ( ) | |
sin sin
2 2
k k
k k k
k k
N
n n
v v
n N q q
dv
N x x
v v
M
 
  
 
   
  
 
   

−
   
− −
    
 +
     +
     
−
− −
     
   
 

The lemma 2.1 gives
( )
( )
( )
( )
( )
1 2
,
1 2
2 2
2
2 1
1
sin sin
1 2 2
1 1
sin sin
2 2
1 2 2 ( )( )(2 )
2 ( )
k k
N
k k k
k k
k
k N
N
n n
v v
dv
B
v v
N q q
n N
n
M
 

  
 


   
   
   
  

−
−
−
   
− −
   
    
    
− −
   
   
 
 

 +
 
 
+
   
   

 

 
 

.


(19)
In order to obtain the convergence of the last expression to zero, we choose
( ) = , > 0
N n 
 
−
, such as
2
2 1
1
1
= 0 = 0.
lim lim
k
k
k
n n
k N
N
n
and
n
n
M



  
−
→ →
+ −  
 
 
(20)
Thus, from (18) N
Q converges to zero. On the other hand,
2
2 | | 1 2
, 1
1 2
( ) | | ,
| |
p
K K K
p p
du du
C F C K K e
u u

 

−    
− − −
+
− −

 −
 
where ( )
2
1 2
=1
| | | | ( ; )
k N k N
k
K K K C u Q

     
 

 
− −  − −
 .
We denote by: ( )
( , ) 1 2
= ( ; )
N k N k N
Q
     
  − . It follows from (18) and (20) that ( , )
N k
 
converges to ( )
  . Hence,

42
2 2
2
, 1 2 ( , )
0 1
=1 2
( ) 2 ( ; )4 exp ( ) | | .
( )
p
k
p a N k N k
p
k
k
du
C F C C Q C u
u


   
    
− 
− 
 
+ −


 
 −
 

Putting ( )
1
, = ,
k N k
u v


  we obtain
2
| |
2
2 1 2
, 1
1
2 2
( ,1) ( ,2)
( ; )
( ) 2 .
| |
C v
p
N
p p
p
N N
Q e
C F C C dv
v



   

 
 
 
−
− 
−
−
− + −
 
 

 
 
   
 
 (20)
Since ( ) > 0
  , ( )
C  converges uniformly in 1 1
, [ 1,1]
x x − to zero. From (20), we obtain
2 2 (1 ) 1
1
1 1
= .
k
k
k N
N
J O
n
n
M

 
  
− −
+ −
 
 
 
+
 
 
 
 
 
 
 
Thus, [ ( )]
N
var f  converges to zero and then, ( )
N
f 
is an asymptotically unbiased and consistent estimator.
4. CONCLUSIONS
This paper gives an estimate of the spectral density of a mixed continuous-time stable process
from observations at discrete instants. To avoid the phenomenon of aliasing we have assumed
that the spectral density is a compact support. The applications of these processes are found in
various fields. For example:
- The study of soil cracking where the observed signal is the resistance of the soil. This signal
encounters random jumps due to the encounter of certain stones in the ground. The spectral
measurement will therefore be composed of two parts, one continuous and the other discrete.
This last corresponds to the resistance jumps encountered during the measurement.
- The growth of fruits on a tree can be seen as a continuous distribution, and when there is a fall
of a fruit, the other fruits remaining on the tree absorb more energy and their growth will have a
jump in value.
As perspective of this work is to find the optimum smoothing parameters to have a better rate of
convergence. For this purpose, the cross-validation method will be the most appropriate tool.
We intend to complete this work by studying the case where the process is observed with random
errors. For this, we will use the deconvolution methods known for their efficiency in the presence
of random errors.
We think to give an estimator of the mode of the spectral density representing the frequency
where the spectral density reaches the maximum of energy.

43
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44
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AUTHORS
Rachid Sabre received the PhD degree in statistics from the University of Rouen, France, in 1993 and
Habilitation to direct research (HdR) from the University of Burgundy, Dijon, France, in 2003. He joined
Agrosup Dijon, France, in 1995, where he is an Associate Professor. From 1998 through 2010, he served as
a member of Institut de Mathématiques de Bourgogne, France. He was a member of the Scientific Council
AgroSup Dijon from 2009 to 2013. From 2012 to 2019, he has been a member of Laboratoire Electronic,
Informatique, and Image (LE2I), France. Since 2019 he has been a member of Laboratory Biogeosciences
UMR CNRS University Burgundy. He is author/co-author of numerous papers in scientific and technical
journals and conference proceedings. His research interests lie in areas of statistical process and spectral
analysis for signal and image processing.

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